Topics on Continua (Chapman & Hall CRC Pure and Applied Mathematics)

Topics on Continua Sergio Macías Instituto de Matemáticas Universidad Nacional Autónoma de México Ciudad Universitaria,...

Author: Sergio Macias

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Topics on

Continua Sergio Macías Instituto de Matemáticas Universidad Nacional Autónoma de México Ciudad Universitaria, México

Boca Raton London New York Singapore

Copyright © 2005 Taylor & Francis Group, LLC

DK6026_Discl.fm Page 1 Thursday, April 21, 2005 10:17 AM

Published in 2005 by Chapman & Hall/CRC Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2005 by Taylor & Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-10: 0-8493-3738-0 (Hardcover) International Standard Book Number-13: 978-0-8493-3738-3 (Hardcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

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and the CRC Press Web site at http://www.crcpress.com

To Elsa


León Felipe escribi´ o un tributo, no al héroe de la historia, sino a su fiel caballo Rocinante, quien lo llev´ o en su lomo por las tierras de Espa˜ na. El héroe es, por supuesto, Don Quijote de la Mancha: “El Caballero de la Triste Figura” ¡Yo quer´ıa ese nombre! pero me lo ganaron, llegué a este mundo casi trescientos cincuenta a˜ nos tarde... Ya s´ olo me queda ser: “El Caballero de la Triste Locura...” S. M.


Preface

My aim is to present four of my favorite topics in continuum theory: inverse limits, Professor Jones’s set function T , homogeneous continua and n–fold hyperspaces. Most topics treated in this book are not covered in Professor Sam B. Nadler Jr.’s book: Continuum Theory: An Introduction, Monographs and Textbooks in Pure and Applied Math., Vol. 158, Marcel Dekker, New York, Basel, Hong Kong, 1992. The reader is assumed to have taken a one year course on general topology. The book has seven chapters. In Chapter 1, we include the basic background to be used in the rest of the book. The experienced readers may prefer to skip this chapter and jump right to the study of their favorite subject. This can be done without any problem. The topics of Chapter 1 are essentially independent of one another and can be read at any time. Chapter 2 is for most part about inverse limits of continua. We present the basic results on inverse limits. Some theorems are stated without proof in Professor W. Tom Ingram’s book: Inverse Limits, Aportaciones Matemáticas, Textos # 18, Sociedad Matemática Mexicana, 2000. We show that the operation of taking inverse limits commutes with the operations of taking finite products, cones and hyperspaces. We also include some applications of inverse limits. In Chapter 3 we discuss Professor F. Burton Jones’s set function T . After giving the basic properties of this function, we present properties of continua in terms of T , such as connectedness im kleinen, local connectedness and semi–local connectedness. We also study continua for which the set function T is continuous. In the last section we present some applications of T . In Chapter 4 we start our study of homogeneous continua. We present a topological proof of a Theorem of E. G. Effros given by Copyright © 2005 Taylor & Francis Group, LLC

F. D. Ancel. We include a brief introduction to topological groups and group actions. Chapter 5 contains our main study of homogeneous continua. We present two Decomposition Theorems of such continua, whose proofs are applications of Jones’s set function T and Effros’s Theorem. These theorems have narrowed the study of homogeneous continua in such a way that they may hopefully be eventually classified. We also give examples of nontrivial homogeneous continua and their covering spaces. In Chapter 6 we present most of what is known about n–fold hyperspaces. This chapter is slightly different from the other chapters because the proofs of many of the theorems are based on results in the literature that we do not prove; however, we give references to the appropriate places where proofs can be found. This Chapter is a complement of the two existing books–Sam B. Nadler, Jr., Hyperspaces of Sets: A Text with Research Questions, Monographs and Textbooks in Pure and Applied Math., Vol. 49, Marcel Dekker, New York, Basel, 1978 and Alejandro Illanes and Sam B. Nadler, Jr., Hyperspaces: Fundamentals and Recent Advances, Monographs and Textbooks in Pure and Applied Math., Vol. 216, Marcel Dekker, New York, Basel, 1999, in which a thorough study of hyperspaces is done. In Chapter 6, we also prove general properties of n–fold hyperspaces. In particular, we show that n–fold hyperspaces are unicoherent and finitely aposyndetic. We study the arcwise accessibility of points of the n–fold symmetric products from their complement in n–fold hyperspaces. We give a treatment of the points that arcwise disconnect n–fold hyperspaces of indecomposable continua. Then we study continua for which the operation of taking n–fold hyperspaces is continuous (Cn∗ –smoothness). We also investigate continua for which there are retractions between their various hyperspaces. Next, we present some results about the n–fold hyperspaces of graphs. We end Chapter 6 by studying the relation between n–fold hyperspaces and cones over continua. We end the book with a chapter containing open questions on each of the subjects presented in the book (Chapter 7). Copyright © 2005 Taylor & Francis Group, LLC

We include figures to illustrate definitions and aspects of proofs.

The book originates from two sources – class notes I took from the course on continuum theory given by Professor James T. Rogers, Jr. at Tulane University, in the Fall Semester of 1988 and the one– year courses on continuum theory I have taught in the graduate program of mathematics at the Facultad de Ciencias of the Universidad Nacional Autónoma de México, since the Spring of 1993. I thank all the students who have taken such courses. I thank Mar´ıa Antonieta Molina and Juan Carlos Mac´ıas for letting me include part of their thesis in the book. Ms. Molina’s thesis was based on two talks on the set function T given by Professor David P. Bellamy in the IV Research Workshop on Topology, celebrated in Oaxaca City, Oaxaca, Mexico, November 14 through 16, 1996. I thank Professors Sam B. Nadler, Jr. and James T. Rogers, Jr. for reading parts of the manuscript and making valuable suggestions. I also thank Ms. Gabriela Sanginés and Mr. Leonardo Espinosa for answering my questions about LATEX, while I was typing this book. I thank Professor Charles Hagopian and Marvi Hagopian for letting me use their living room to work on the book during my visit to California State University, Sacramento. I thank the Instituto de Matemáticas of the Universidad Nacional Autónoma de México and the Mathematics Department of West Virginia University, for the use of resources during the preparation of the book. Finally, I thank the people at Marcel Dekker, Inc., especially Ms. Maria Allegra and Mr. Kevin Sequeira, who were always patient and helpful.

Sergio Mac´ıas


Contents 1 PRELIMINARIES 1.1 Product Topology . . . . . . . . . . . 1.2 Continuous Decompositions . . . . . 1.3 Homotopy and Fundamental Group . 1.4 Geometric Complexes and Polyhedra 1.5 Complete Metric Spaces . . . . . . . 1.6 Compacta . . . . . . . . . . . . . . . 1.7 Continua . . . . . . . . . . . . . . . . 1.8 Hyperspaces . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . .

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2 INVERSE LIMITS AND RELATED TOPICS 2.1 Inverse Limits . . . . . . . . . . . . . . . . . . . 2.2 Inverse Limits and the Cantor Set . . . . . . . . 2.3 Inverse Limits and Other Operations . . . . . . . 2.4 Chainable Continua . . . . . . . . . . . . . . . . 2.5 Circularly Chainable and P–like Continua . . . 2.6 Universal and A–H Essential Maps . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . 3 JONES’S SET FUNCTION 3.1 The Set Function T . . . . 3.2 Continuity of T . . . . . . 3.3 Applications . . . . . . . . References . . . . . . . . . . . .

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1 1 8 21 35 38 42 45 58 66

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69 69 92 100 105 117 126 140

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143 . 143 . 174 . 188 . 199

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4 A THEOREM OF E. G. EFFROS 203 4.1 Topological Groups . . . . . . . . . . . . . . . . . . . 203 4.2 Group Actions and a Theorem of Effros . . . . . . . . 209 Copyright © 2005 Taylor & Francis Group, LLC

References . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 5 DECOMPOSITION THEOREMS 5.1 Jones’s Theorem . . . . . . . . . . . . 5.2 Detour to Covering Spaces . . . . . . . 5.3 Rogers’s Theorem . . . . . . . . . . . . 5.4 Case and Minc–Rogers Continua . . . . 5.5 Covering Spaces of Some Homogeneous References . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . Continua . . . . . .

. . . . . .

225 . 225 . 242 . 249 . 263 . 270 . 283

6 n–FOLD HYPERSPACES 6.1 General Properties . . . . . . . 6.2 Unicoherence . . . . . . . . . . 6.3 Aposyndesis . . . . . . . . . . . 6.4 Arcwise Accessibility . . . . . . 6.5 Points that Arcwise Disconnect 6.6 Cn∗ –smoothness . . . . . . . . . 6.7 Retractions . . . . . . . . . . . 6.8 Graphs . . . . . . . . . . . . . . 6.9 Cones . . . . . . . . . . . . . . References . . . . . . . . . . . . . . .

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287 . 287 . 295 . 296 . 298 . 302 . 312 . 320 . 328 . 333 . 342

7 QUESTIONS 7.1 Inverse Limits . . . . . 7.2 The Set Function T . . 7.3 Homogeneous Continua 7.4 n–fold Hyperspaces . . References . . . . . . . . . .

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347 347 349 352 354 355

Chapter 1 PRELIMINARIES

We gather some of the results of topology of metric spaces which will be useful for the rest of the book. We assume the reader is familiar with the notion of metric space and its elementary properties. We present the proofs of most of the results; we give an appropriate reference otherwise. The topics reviewed in this chapter are: product topology, continuous decompositions, homotopy, fundamental group, geometric complexes, polyhedra, complete metric spaces, compacta, continua and hyperspaces.

1.1

Product Topology

The symbols IN, ZZ, Q, I IR and C I denote the positive integers, integers, rational numbers, real numbers and complex numbers, respectively. The word map means a continuous function. A compactum is a compact metric space.


1

2

CHAPTER 1. PRELIMINARIES

1.1.1. Definition. Given a sequence, {Xn }∞ n=1 , of nonempty sets, ∞ we define its Cartesian product, denoted by Xn , as the set: n=1 ∞

Xn = {(xn )∞ n=1 | xn ∈ Xn for each n ∈ IN}.

n=1

For each m ∈ IN, there is a function πm :

∞

Xn → → Xm

n=1

defined by πm ((xn )∞ n=1 ) = xm . This function πm is called the mth– projection map.

1.1.2. Remark. Given a metric space (X, d ), there is a metric, d, which generates the same topology as d , with the property that d(x, x ) ≤ 1 for each pair of points x and x of X. This metric d is called bounded metric. An example of such metric is given by d(x, x ) = min{1, d (x, x )}.

1.1.3. Notation. Given a metric space (X, d) and a subset A of X, ClX (A), IntX (A) and BdX (A) denote the closure, interior and boundary of A, respectively. We omit the subindex if there is no confusion. If ε is a positive real number, then the symbol Vεd (A) denotes the open ball of radius ε about A. If A = {x}, for some x ∈ X, we write Vεd (x) instead of Vεd ({x}).

1.1.4. Definition. If {(Xn , dn )}∞ n=1 is a sequence of metric spaces, with bounded metrics, we define a metric ρ, for its Cartesian product as follows: ∞ ρ((xn )∞ n=1 , (xn )n=1 )


∞ 1 = d (x , xn ). n n n 2 n=1

1.1. PRODUCT TOPOLOGY

3

1.1.5. Remark. Since the metrics, dn , in Definition 1.1.4 are bounded, ρ is well defined.

1.1.6. Lemma. If {(Xn , dn )}∞ n=1 is a sequence of metric spaces, with bounded metrics, then ρ (Definition 1.1.4) is a metric and for each m ∈ IN, πm is a continuous function. Proof. The proof of the fact that ρ is, in fact, a metric is left to the reader. Let m ∈ IN be given. We show that πm is continuous. Let ε > 0 ∞ 1 ∞ and let δ = m ε. If (xn )∞ and (x ) are two points of Xn n=1 n n=1 2 n=1 1 ∞ ∞ such that ρ((xn )n=1 , (xn )n=1 ) < δ, then, since m dm (xm , xm ) ≤ 2 ∞ 1 1 d (x , xn ), we have that m dm (xm , xm ) < δ. Hence, n n n 2 2 n=1 dm (xm , xm ) < 2m δ = ε. Therefore, πm is continuous. Q.E.D. 1.1.7. Lemma. If {(Xn , dn )}∞ n=1 is a sequence of metric spaces, with bounded metrics, then given ε > 0 and a point (xn )∞ n=1 ∈ ∞ Xn , there exist N ∈ IN and N positive real numbers, ε1 , . . . , εN , n=1

such that

N

πj−1 (Vεdjj (xj )) ⊂ Vερ ((xn )∞ n=1 ).

j=1

ε 1 < . For each j ∈ n 2 2 n=N +1 N We assert that πj−1 Vεdjj (xj ) ⊂

Proof. Let N ∈ IN be such that {1, . . . , N }, let εj =

ε . 2N

∞

∞ Vερ ((xn )∞ n=1 ). To see this, let (yn )n=1 ∈

j=1

N j=1


πj−1 Vεdjj (xj ) . We want

4


∞ to see that ρ((xn )∞ n=1 , (yn )n=1 ) < ε. Note that ∞ 1 d (x , y ) = n n n n 2 n=1

∞ ρ((xn )∞ n=1 , (yn )n=1 ) = N 1 d (x , y ) + n n n n 2 n=1

∞

1 d (x , y ) < n n n n 2 n=N +1 N 1 1 1 1 1 1 1 1 ε+ ε = ε + ε ≤ ε + ε = ε. 1− N n N N 2 2 2 2 2 2 2 2 n=1 Q.E.D. 1.1.8. Lemma. If {(Xn , dn )}∞ n=1 is a sequence of metric spaces, with bounded metrics, then given a finite number of positive real ∞ ∞ numbers ε1 , . . . , εk and a point (xn )n=1 ∈ Xn , there exists ε > 0 such that Vερ ((xn )∞ n=1 ) ⊂

k

n=1

πj−1 (Vεdjj (xj )).

j=1

Proof. Let

(xn )∞ n=1

∈

∞

Xn , and let U =

n=1

k

πj−1 Vεdjj (xj ) . Take

j=1

ε = min

1 1 ε1 , . . . , k ε k . 2 2

∞ ρ ∞ We show Vερ ((xn )∞ n=1 ) ⊂ U . Let (yn )n=1 ∈ Vε ((xn )n=1 ). Then ∞ ρ((xn )∞ n=1 , (yn )n=1 )

∞ 1 < ε, i.e., dn (xn , yn ) < ε. 2n n=1

1 1 dj (xj , yj ) < ε ≤ j εj for each j ∈ {1, . . . , k}. Thus, if j 2 2 j ∈ {1, . . . , k}, then dj (xj , yj ) < εj . Therefore, Vερ ((xn )∞ n=1 ) ⊂ U . Q.E.D.

Hence,


1.1. PRODUCT TOPOLOGY

5

1.1.9. Theorem. Let Z be a metric space. If {(Xn , dn )}∞ n=1 is ∞ a sequence of metric spaces, then a function f : Z → Xn is n=1

continuous if and only if πn ◦ f is continuous for each n ∈ IN. Proof. Clearly, if f is continuous, then πn ◦ f is continuous for each n ∈ IN. k Suppose πn ◦ f is continuous for each n ∈ IN. Let πj−1 (Uj ) be a basic open subset of

∞

j=1

Xn . Since

n=1

f

−1

k

πj−1 (Uj )

=

j=1

k

f −1 (πj−1 (Uj ))

j=1

=

k

(πj ◦ f )−1 (Uj ),

j=1

we have that f −1

k

πj−1 (Uj )

is open in Z. Hence, f is continu-

j=1

ous.

Q.E.D.

∞ 1.1.10. Theorem. Let {Xn }∞ n=1 and {Yn }n=1 be two countable collections of metric spaces. Suppose that for each n ∈ IN, there ∞ ∞ exists a map fn : Xn → Yn . Then the function fn : Xn → ∞ n=1

Yn given by

∞

n=1

n=1

∞ fn ((xn )∞ n=1 ) = (fn (xn ))n=1 is continuous.

n=1

Proof. For each m ∈ IN, let πm :

∞ n=1

be the projection maps. Copyright © 2005 Taylor & Francis Group, LLC

Xn → → Xm and

πm :

∞ n=1

Yn → → Ym

6

CHAPTER 1. PRELIMINARIES Let (xn )∞ n=1 ∈

∞

Xn , and let m ∈ IN. Then πm ◦

n=1

∞

fn ((xn )∞ n=1 )

n=1

∞ = πm ((fn (xn ))∞ n=1 ) = fm (xm ) = fm ◦ πm ((xn )n=1 ). Hence, by The∞ orem 1.1.9, fn is continuous. n=1

Q.E.D. The following result is a particular case of Tychonoff ’s Theorem, which says that the Cartesian product of any family of compact topological spaces is compact. The proof of this theorem uses the Axiom of Choice. However, the case we show only uses the fact that compactness and sequential compactness are equivalent in metric spaces (Remark 3 (p. 3) of [13]). 1.1.11. Theorem. If {(Xn , dn )}∞ n=1 is a sequence of compacta, then ∞ Xn is compact. n=1

Proof. By Lemma 1.1.6, any sequence of points of

∞

Xn is a metric space. We show that

n=1 ∞

Xn has a convergent subsequence.

n=1

Let {pk }∞ k=1 be a sequence of points of

∞ n=1

Xn , where pk =

k ∞ (pkn )∞ n=1 for each k ∈ IN (in this way, if we keep n fixed, {pn }k=1 is a sequence of points of Xn ). Since (X1 , d1 ) is sequentially comkj ∞ pact, {pk1 }∞ k=1 has a convergent subsequence {p1 }j=1 converging to a point q1 of X1 . Let us note that, implicitly, we have defined a k ∞ subsequence {pkj }∞ j=1 of {p }k=1 . Now, suppose, inductively, that for some m ∈ IN, we have dek ∞ ki ∞ fined a subsequence {pki }∞ i=1 of {p }k=1 such that {pm }i=1 converges to a point qm of Xm . Since (Xm+1 , dm+1 ) is sequentially compact, kij ∞ i }∞ {pkm+1 i=1 has a convergent subsequence {pm+1 }j=1 such that it converges to a point qm+1 of Xm+1 . Hence, by the Induction Principle, we have defined a sequence of subsequences of {pk }∞ k=1 in such a way that each subsequence is a subsequence of the preceding one.


1.1. PRODUCT TOPOLOGY kji

, . . . }. Clearly, Σ is a subsequence ∞ which converges to the point (qn )∞ . Therefore, Xn n=1

Now, let Σ = {p1 , pk2 , pkj3 , p of {pk }∞ k=1

7

4

n=1

is compact. Q.E.D. 1.1.12. Definition. Let Q =

∞

[0, 1]n , where [0, 1]n = [0, 1], for

n=1

each n ∈ IN. Then Q is called the Hilbert cube. 1.1.13. Theorem. The Hilbert cube is a connected compactum. Proof. By Lemma 1.1.6, Q is a metric space. By Theorem 1.1.11, Q is compact. By Theorem 11 (p. 137) of [13], Q is connected. Q.E.D. 1.1.14. Definition. Let f : X → Y be a map between metric spaces. We say that f is an embedding if f is a homeomorphism onto f (X). 1.1.15. Definition. A map f : X → Y between metric spaces is said to be closed provided that for each closed subset K of X, f (K) is closed in Y . The next Theorem says that there is a “copy” of every compactum inside the Hilbert cube. 1.1.16. Theorem. If X is a compactum, then X can be embedded in the Hilbert cube Q. Proof. Let d be the metric of X. Without loss of generality, we assume that diam(X) ≤ 1. Since X is a compactum, it contains a countable dense subset, {xn }∞ n=1 . Let h : X → Q be given by ∞ h(x) = (d(x, xn ))n=1 . By Theorem 1.1.9, h is continuous. Clearly, h is one–to–one. Since X is compact and Q is metric, h is a closed map. Therefore, h is an embedding. Q.E.D.


8

1.2


Continuous Decompositions

We present a method to construct “new” spaces from “old” ones by “shrinking” certain subsets to points. 1.2.1. Definition. A decomposition of a set X is a collection of nonempty, pairwise disjoint sets whose union is X. The decomposition is said to be closed if each of its element is a closed subset of X.

1.2.2. Definition. Let G be a decomposition of a metric space X. We define X/G as the set whose elements are the elements of the decomposition G. X/G is called the quotient space. The function q: X → → X/G, which sends each point x of X to the unique element G of G such that x ∈ G, is called the quotient map.

1.2.3. Remark. Given a decomposition of a metric space X, note that q(x) = q(y) if and only if x and y belong to the same element of G. We give a topology to X/G in such a way that the function q is continuous and it is the biggest with this property.

1.2.4. Definition. Let X be a metric space, let G be a decomposition of X and let q : X → → X/G be the quotient map. Then the topology U = {U ⊂ X/G | q −1 (U ) is open in X} is called the quotient topology for X/G.

1.2.5. Remark. Let G be a decomposition of a metric space X, and let q : X → → X/G be the quotient map. Then a subset U of X/G is open (closed, respectively) if and only if q −1 (U ) is an open (closed, respectively) subset of X. Copyright © 2005 Taylor & Francis Group, LLC

1.2. CONTINUOUS DECOMPOSITIONS

9

1.2.6. Definition. Let f : X → → Y be a surjective map between metric spaces. Since f is a function, Gf = {f −1 (y) | y ∈ Y } is a decomposition of X. The function ϕf : X/Gf → Y given by ϕf (q(x)) = f (x) is of special interest. Note that ϕf is well defined; in fact, it is a bijection and the following diagram: f

−→

X q

X/Gf

Y ϕf

is commutative. The next Lemma is a special case of the Transgression Lemma (3.22 of [23]). 1.2.7. Lemma. Let f : X → → Y be a surjective map between metric spaces. If X/Gf has the quotient topology, then the function ϕf is continuous. −1 Proof. If U is an open subset of Y , then ϕ−1 (U ). Since f (U ) = qf −1 −1 −1 −1 −1 −1 q ϕf (U ) = q qf (U ) = f (U ) and f (U ) is an open subset of X, we have, by the definition of quotient topology, that ϕ−1 f (U ) is an open subset of X/Gf . Therefore, ϕf is continuous. Q.E.D.

1.2.8. Example. Let X = [0, 2π) and let f : X → → S 1 , where S 1 is it the unit circle, be given by f (t) = exp(t) = e . Then f is a continuous bijection. Since Gf is, “essentially,” X, it follows that X/Gf is homeomorphic to X. On the other hand, X is not homeomorphic to S 1 , since X is not compact and S 1 is. Therefore, ϕf is not a homeomorphism.

1.2.9. Definition. A map f : X → Y between metric spaces is said to be open provided that for each open subset K of X, f (K) is open in Y . Copyright © 2005 Taylor & Francis Group, LLC

10


The following Theorem gives sufficient conditions to ensure that ϕf is a homeomorphism: 1.2.10. Theorem. Let f : X → → Y be a surjective map between metric spaces. If f is open or closed, then ϕf : X/Gf → → Y is a homeomorphism. Proof. Suppose f is an open map. Since ϕ is a bijective map, it is enough to show that ϕ is open. Let A be an open subset of X/G. Since ϕf (A) = f q −1 (A), ϕf (A) is an open subset of Y . Therefore, ϕf is an open map. The proof of the case when f is closed is similar. Q.E.D. Decompositions are also used to construct the cone and suspension over a given space. 1.2.11. Definition. Let X be a metric space and let G = {{(x, t)} | x ∈ X and t ∈ [0, 1)} ∪ {(X × {1})}. Then G is a decomposition of X × [0, 1]. The cone over X, denoted by K(X), is the quotient space (X × [0, 1])/G. The element {X × {1}} of (X × [0, 1])/G is called the vertex of the cone and it is denoted by νX .

q

A proof of the following Proposition may be found in 5.2 (p. 127) of [9].



11

1.2.12. Proposition. Let f : X → Y be a map between metric spaces. Then f induces a map K(f ) : K(X) → K(Y ) by if ω = νX ∈ K(X); νY K(f )(ω) = (f (x), t) if ω = (x, t) ∈ K(X) \ {νX }.

1.2.13. Definition. Let X be a metric space and let G = {{(x, t)} | x ∈ X and t ∈ (0, 1)} ∪ {(X × {0}), (X × {1})}. Then G is a decomposition of X × [0, 1]. The suspension over X, denoted by Σ(X), is the quotient space (X × [0, 1])/G.

q

1.2.14. Definition. Let X be a metric space and let G be a decomposition of X. We say that G is upper semicontinuous if for each G ∈ G and each open subset U of X such that G ⊂ U , there exists an open subset V of X such that G ⊂ V and such that if G ∈ G and G ∩ V = ∅, then G ⊂ U . We say that G is lower semicontinuous provided that for each G ∈ G any two points x and y of G and each open set U of X such that x ∈ U , there exists an open set V of X such that y ∈ V and such that if G ∈ G and G ∩ V = ∅, then G ∩ U = ∅. Finally, we say that G is continuous if G is both upper and lower semicontinuous.


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1.2.15. Example. Let X = ([−1, 1] × [0, 1]) ∪ ({0} × [0, 2]). For each t ∈ [−1, 1] \ {0}, let Gt = {t} × [0, 1], and for t = 0, let G0 = {0} × [0, 2]. Let G = {Gt | t ∈ [0, 1]}. Then G is an upper semicontinuous decomposition of X.

X

Upper semicontinuous decompostion

1.2.16. Example. Let

1 2 X = ([0, 1] × [0, 1)) ∪ {1} × 0, ∪ {1} × , 1 . 3 3

1 and For each t ∈ [0, 1), let Gt = {t} × [0, 1], let G1 = {1} × 0, 3

2 let G1 = {1} × , 1 . Let G = {Gt | t ∈ [0, 1]} ∪ {G1 }. Then G is 3 a lower semicontinuous decomposition of X. X

Lower semicontinuous decomposition



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The following Theorem gives us a way to obtain upper semicontinuous decompositions of compacta. 1.2.17. Theorem. Let f : X → → Y be a surjective map between compacta. If Gf = {f −1 (y) | y ∈ Y }, then Gf is an upper semicontinuous decomposition of X. Proof. Let U be an open subset of X such that f −1 (y) ⊂ U . Note that X \ U is a closed subset, hence compact, of X. Then f (X \ U ) is a compact subset, hence closed, of Y such that y ∈ f (X \ U ). Thus, Y \ f (X \ U ) is an open subset of Y containing y. If V = f −1 (Y \ f (X \ U )), then V is an open subset of X such −1 that f (y) ⊂ V ⊂ U . Since V = {f −1 (y) | y ∈ Y \ f (X \ U )}, clearly V satisfies the required property of the definition of upper semicontinuous decomposition. Q.E.D. 1.2.18. Remark. Let us note that Theorem 1.2.17 is not true without the compactness of X. Let X be the Euclidean plane IR2 and let π : IR2 → → IR be given by π((x, y)) = x. Then Gπ is a decomposition To see this, of X which is not upper semicontinuous. 1 let U = (x, y) ∈ X x = 0 and y < ∪ {0} × IR. Then U is an x open set of X such that π −1 (0) ⊂ U , whose boundary is asymptotic to π −1 (0). Hence, for each t ∈ IR \ {0}, π −1 (t) ∩ (X \ U ) = ∅. The next Theorem gives three other ways to think about upper semicontinuous decompositions. 1.2.19. Theorem. If X is a metric space and G is a decomposition of X, then the following conditions are equivalent: (a) G is an upper semicontinuous decomposition; (b) the quotient map q : X → → X/G is closed; (c) if U is an open subset of X, then WU = {G ∈ G | G ⊂ U } is an open subset of X; Copyright © 2005 Taylor & Francis Group, LLC

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(d) if D is a closed subset of X, then KD = is a closed subset of X.

{G ∈ G | G ∩ D = ∅}

Proof. Suppose G is an upper semicontinuous decomposition. Let D be a closed subset of X. By Remark 1.2.5, we have that q(D) is closed in X/G if and only if q −1 (q(D)) is closed in X. We show that X \ q −1 (q(D)) is open in X. Let x ∈ X \ q −1 (q(D)). Then q(x) ∈ X/G \ q(D) and hence q −1 (q(x)) ⊂ X \ D. Therefore, since X \ D is open, by Definition 1.2.14, there exists an open set V of X such that q −1 (q(x)) ⊂ V and for each y ∈ V , q −1 (q(y)) ⊂ X \ D. Clearly, x ∈ V and q(V ) ⊂ X/G \ q(D). Thus, V ⊂ X \ q −1 (q(D)). Therefore, X \ q −1 (q(D)) is open, since x ∈ V ⊂ X \ q −1 (q(D)). Now, suppose q is a closed map. Let U be an open subset of X. Since q is a closed map, we have that q −1 (X/G \ q(X \ U )) is an open subset of X such that q −1 (X/G \ q(X \ U )) = WU . (If x ∈ q −1 (X/G \ q(X \ U )), then q(x) ∈ X/G \ q(X \ U ). Hence, q −1 (q(x)) ⊂ X \ q −1 (q(X \ U )) ⊂ X \ (X \ U )) = U . Thus, x ∈ WU . The other inclusion is obvious.) Next, suppose WU is open for each open subset U of X. Let D be a closed subset of X. Then X \ D is open in X. Hence, WX\D is open in X. Since, clearly, KD = X \ WX\D , we have that KD is closed. Finally, suppose KD is closed for each closed subset D of X. To see G is upper semicontinuous, let G ∈ G and let U be an open subset of X such that G ⊂ U . Note that X \ U is a closed subset of X. Hence, KX\U is a closed subset of X. Let V = X \ KX\U . Then V is open, G ⊂ V ⊂ U and if G ∈ G and G ∩ V = ∅, then G ⊂ V . Therefore, G is upper semicontinuous. Q.E.D. 1.2.20. Corollary. Let X be a metric space. If G is an upper semicontinuous decomposition of X, then the elements of G are closed. Proof. Let G ∈ G. Take x ∈ G and let q : X → → X/G be the quotient map. Since X is a metric space, {x} is closed in X. By Theorem 1.2.19, q({x}) is closed in X/G. Since q is continuous and q −1 (q({x})) = G, G is a closed subset of X. Q.E.D.



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1.2.21. Theorem. If X is a compactum and G is an upper semicontinuous decomposition of X, then X/G has a countable basis. Proof. Let q : X → → X/G be the quotient map. Since X is a compactum, it has a countable basis U. Let n B= Uj U1 , . . . , Un ∈ U and n ∈ IN . j=1

Note that B is a countable family of open subsets of X. Let B = {X/G \ q(X \ U ) | U ∈ B}. We see that B is a countable basis for X/G. Clearly, B is a countable family of open subsets of X/G. Let U be an open subset of X/G and let x ∈ U . Then q −1 (U ) is an open subset of X and q −1 (x) ⊂ q −1 (U ). Since q −1 (x) is compact, there exist U1 , . . . , Uk ∈ U such k k −1 −1 that q (x) ⊂ Uj ⊂ q (U ). Let U = Uj . Then U ∈ B. j=1

j=1

Hence, X/G \ q(X \ U ) ∈ B. Also, x ∈ X/G \ q(X \ U ) ⊂ U . Therefore, B is a countable basis for X/G. Q.E.D. 1.2.22. Corollary. If X is a compactum and G is an upper semicontinuous decomposition of X, then X/G is metrizable. Proof. By Theorem 1.2.21, we have that X/G has a countable basis. By Theorem 1 (p. 241) of [12], it suffices to show that X/G is a Hausdorff space. Let x and y be two distinct points of X/G. Then q −1 (x) and q −1 (y) are two disjoint closed subsets of X. Since X is a metric space, there exist two disjoint open subsets, U1 and U2 , of X such that q −1 (x) ⊂ U1 and q −1 (y) ⊂ U2 . Note that, by Theorem 1.2.19 (c), WU1 and WU2 are open subsets of X such that q −1 (x) ⊂ WU1 ⊂ U1 , q −1 (y) ⊂ WU2 ⊂ U2 , and q(WU1 ) and q(WU2 ) are open subsets of X/G. Since U1 ∩ U2 = ∅, q(WU1 ) ∩ q(WU2 ) = ∅. Therefore, X/G is a Hausdorff space. Q.E.D. The next Theorem gives a characterization of lower semicontinuous decompositions.


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1.2.23. Theorem. Let X be a metric space and let G be a decomposition of X. Then G is lower semicontinuous if and only if the quotient map q : X → → X/G is open. Proof. Suppose G is lower semicontinuous. Let U be an open subset of X. We show q(U ) is an open subset of X/G. To this end, by Remark 1.2.5, we only need to show that q −1 (q(U )) is an open subset of X. Let y ∈ q −1 (q(U )). Then q(y) ∈ q(U ), and there exists a point x in U such that q(x) = q(y). Since G is a lower semicontinuous decomposition, there exists an open subset V of X containing y such that if G ∈ G y G ∩ V = ∅, then G ∩ U = ∅. Hence, V ⊂ q −1 (q(U )). Therefore, q is open. Now, suppose q is open. Let G ∈ G. Take x, y ∈ G and let U be an open subset of X such that x ∈ U . Since q is open, V = q −1 (q(U )) is an open subset of X such that G ⊂ V . In particular, y ∈ V . Let G ∈ G such that G ∩ V = ∅. Then G ⊂ V . Thus, q(G ) ∈ q(U ). Hence, there exists u ∈ U such that q(u) = q(G ). Since q −1 (q(G )) = G , u ∈ G . Thus, G ∩ U = ∅. Therefore, G is lower semicontinuous. Q.E.D. The following Corollary is a consequence of Theorems 1.2.19 and 1.2.23: 1.2.24. Corollary. Let X be a metric space and let G be a decomposition of X. Then G is continuous if and only if the quotient map is both open and closed. The following Theorem gives us a necessary and sufficient condition on a map f : X → → Y between compacta, to have that Gf = −1 {f (y) | y ∈ Y } is a continuous decomposition. 1.2.25. Theorem. Let X and Y be compacta and let f : X → →Y −1 be a surjective map. Then Gf = {f (y) | y ∈ Y } is continuous if and only if f is open. Proof. If Gf is a continuous decomposition of X, by Theorem 1.2.23, the quotient map q : X → → X/Gf is open. By Theorem 1.2.10, Copyright © 2005 Taylor & Francis Group, LLC


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→ Y is a homeomorphism. Hence, f = ϕf ◦ q is an ϕf : X/Gf → open map. Now, suppose f is open. By Theorem 1.2.17, Gf is upper semicontinuous. Since q = ϕ−1 f ◦ f and f is open, q is open. By Theorem 1.2.23, Gf is a lower semicontinuous decomposition. Therefore, Gf is continuous. Q.E.D. In the following Definition a notion of convergence of sets is introduced. 1.2.26. Definition. Let {Xn }∞ n=1 be a sequence of subsets of the metric space X. Then: (1) the limit inferior of the sequence {Xn }∞ n=1 is defined as follows: lim inf Xn = {x ∈ X | for each open subset U of X such that x ∈ U , U ∩ Xn = ∅ for each n ∈ IN, save, possibly, finitely many}. (2) the limit superior of the sequence {Xn }∞ n=1 is defined as follows: lim sup Xn = {x ∈ X | for each open subset U of X such that x ∈ U , U ∩ Xn = ∅ for infinitely many indices n ∈ IN}. Clearly, lim inf Xn ⊂ lim sup Xn . If lim inf Xn = lim sup Xn = L, then we say that the sequence {Xn }∞ n=1 is a convergent sequence with limit L = lim Xn . n→∞

lim inf X n


lim sup X n

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1.2.27. Lemma. Let {Xn }∞ n=1 be a sequence of subsets of the metric space X. Then lim inf Xn and lim sup Xn are both closed subsets of X. Proof. Let x ∈ Cl(lim inf Xn ). Let U be an open subset of X such that x ∈ U . Since x ∈ Cl(lim inf Xn ) ∩ U , we have that lim inf Xn ∩ U = ∅. Hence, U ∩ Xn = ∅ for each n ∈ IN, save, possibly, finitely many. Therefore, x ∈ lim inf Xn . The proof for lim sup is similar. Q.E.D. The next Theorem tells us that separable metric spaces behave like sequentially compact spaces using the notion of convergence just introduced. 1.2.28. Theorem. Each sequence {Xn }∞ n=1 of closed subsets of a separable metric space X has a convergent subsequence. 1 ∞ Proof. Let {Um }∞ m=1 be a countable basis for X. Let {Xn }n=1 = {Xn }∞ n=1 . Suppose, inductively, that we have defined the sequence m ∞ {Xn }n=1 . We define the sequence {Xnm+1 }∞ n=1 as follows: m ∞ m (1) If {Xnm }∞ n=1 has a subsequence {Xnk }k=1 such that lim sup Xnk ∩ m ∞ Um = ∅, then let {Xnm+1 }∞ n=1 be such subsequence of {Xn }n=1 . m ∞ (2) If for each subsequence {Xnmk }∞ k=1 of {Xn }n=1 , we have that m m+1 ∞ lim sup Xnk ∩ Um = ∅, we define {Xn }n=1 as {Xnm }∞ n=1 .

Since we have the subsequences {Xnm }∞ n=1 , let us consider the “dim ∞ m ∞ agonal subsequence” {Xm }m=1 . By construction, {Xm }m=1 is a ∞ m ∞ subsequence of {Xn }n=1 . We see that {Xm }m=1 converges. m ∞ Let us assume that {Xm }m=1 does not converge. Hence, there m m exists p ∈ lim sup Xm \ lim inf Xm . Let Uk be a basic open set such m m ∞ that p ∈ Uk and Uk ∩ Xm = ∅ for some subsequence {Xm } =1 m ∞ of {Xm }m=1 (lim inf Xn is a closed subset of X by Lemma 1.2.27). m ∞ k ∞ Clearly, {Xm } is a subsequence of {Xnk }∞ n=1 . Thus, {Xn }n=1 sat =1 isfies condition (1), with k in place of m. Hence, lim sup Xnk+1 ∩Uk = m ∞ m ∅. Since {Xm }m=1 is a subsequence of {Xnk+1 }∞ n=1 and lim sup Xm ⊂ k+1 m lim sup Xn , it follows that lim sup Xm ∩ Uk = ∅. Now, recall that m p ∈ lim sup Xm ∩ Uk . Thus, we obtain a contradiction. Therefore, m ∞ {Xm }m=1 converges. Q.E.D.



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1.2.29. Theorem. Let X be a compactum. If {Xn }∞ n=1 is a sequence of connected subsets of X and lim inf Xn = ∅, then lim sup Xn is connected. Proof. Suppose, to the contrary, that lim sup Xn is not connected. Since lim sup Xn is closed, by Lemma 1.2.27, we assume, without loss of generality, that there are two disjoint closed subsets A and B of X such that lim sup Xn = A ∪ B. Since X is a metric space, there exist two disjoint open subsets U and V of X such that A ⊂ U and B ⊂ V . Then there exists N ∈ IN such that if n ≥ N , then Xn ⊂ U ∪ V . To show this, suppose it is not true. Then for each n ∈ IN, there is mn > n such that Xmn \ (U ∪ V ) = ∅. Let xmn ∈ Xmn \ (U ∪ V ) for each n ∈ IN. Since X is compact, without loss of generality, we assume that the sequence {xmn }∞ n=1 converges to a point x of X. Note that x ∈ X \ (U ∪ V ) and, by construction, x ∈ lim sup Xn , a contradiction. Therefore, there exists N ∈ IN such that if n ≥ N , then Xn ⊂ U ∪ V . Since lim inf Xn = ∅ and lim inf Xn ⊂ lim sup Xn , we assume, without loss of generality, that lim inf Xn ∩U = ∅. Then there exists N ∈ IN such that if n ≥ N , U ∩ Xn = ∅. Let N = max{N , N }. Hence, if n ≥ N , then Xn ⊂ U ∪ V and U ∩ Xn = ∅. Since Xn is connected for every n ∈ IN, Xn ∩ V = ∅ for each n ≥ N , a contradiction. Therefore, lim sup Xn is connected. Q.E.D. The following Theorem gives us a characterization of an upper semicontinuous decomposition of a compactum in terms of limits inferior and superior. 1.2.30. Theorem. Let X be a compactum, with metric d. Then a decomposition G of X is upper semicontinuous if and only if G is a closed decomposition and for each sequence {Xn }∞ n=1 of elements of G and each element Y of G such that lim inf Xn ∩ Y = ∅, then lim sup Xn ⊂ Y . Proof. Suppose G is an upper semicontinuous decomposition of X. By Corollary 1.2.20, G is a closed decomposition. Let {Xn }∞ n=1 be a sequence of elements of G and let Y be an element of G such that lim inf Xn ∩ Y = ∅. Copyright © 2005 Taylor & Francis Group, LLC

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Suppose there exists p ∈ lim sup Xn \ Y . Since Y is closed and p is not an element of Y , there exists an open set W of X such that p ∈ W and Cl(W ) ∩ Y = ∅. Let U = X \ Cl(W ). Since G is upper semicontinuous, there is an open set V of X such that Y ⊂ V and if G ∈ G such that G ∩ V = ∅, G ⊂ U . Let q ∈ lim inf Xn ∩Y . Then q ∈ lim inf Xn ∩V . Hence, V ∩Xn = ∅ for each n ∈ IN, save, possibly, finitely many. Thus, W ∩ Xn = ∅ for each n ∈ IN, save, possibly, finitely many. This contradicts the fact that p ∈ W ∩ lim sup Xn . Therefore, lim sup Xn ⊂ Y . Now, suppose G is a closed decomposition and let Y be an element of G. Suppose that if {Xn }∞ n=1 is a sequence of elements of G such that lim inf Xn ∩ Y = ∅, then lim sup Xn ⊂ Y . To see G is upper semicontinuous, let U be an open subset of X such that Y ⊂ U . For each n ∈ IN, let Vn = V d1 (Y ). Suppose that n for each n ∈ IN, there is an element Xn of G such that Xn ∩ Vn = ∅ and Xn ⊂ U . For each n ∈ IN, let pn ∈ Xn ∩Vn . Since X is compact, ∞ {pn }∞ n=1 has a convergent subsequence {pnk }k=1 . Let p be the point of convergence of {pnk }∞ k=1 . Note that p ∈ lim inf Xnk ∩ Y . Hence, lim sup Xnk ⊂ Y . For each k ∈ IN, let qk ∈ Xnk \ U . Since X is compact, the ∞ sequence {qnk }∞ k=1 has a convergent subsequence {qnk }=1 . Let q be the point of convergence of {qnk }∞ =1 . Note that q ∈ Y and q ∈ lim sup Xnk ⊂ lim sup Xnk , a contradiction. Therefore, G is upper semicontinuous. Q.E.D. The following Theorem gives us a characterization of a continuous decomposition of a compactum in terms of limits inferior and superior. 1.2.31. Theorem. Let X be a compactum, with metric d. Then a decomposition G of X is continuous if and only if G is a closed decomposition and for each sequence {Xn }∞ n=1 of elements of G and each element Y of G such that lim inf Xn ∩Y = ∅, then lim sup Xn = Y. Proof. Suppose G is a continuous decomposition of X. By Corollary 1.2.20, G is a closed decomposition. Let {Xn }∞ n=1 be a sequence of elements of G and let Y be an element of G such that Copyright © 2005 Taylor & Francis Group, LLC

1.3. HOMOTOPY AND FUNDAMENTAL GROUP

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lim inf Xn ∩ Y = ∅. By Theorem 1.2.30, lim sup Xn ⊂ Y . Suppose there exists p ∈ Y \ lim sup Xn . Let U be an open subset of X such that p ∈ U and U ∩ lim sup Xn = ∅ (by Lemma 1.2.27 lim sup Xn is closed). Let q ∈ lim sup Xn ⊂ Y . Since G is a lower semicontinuous decomposition, there is an open set V of X such that q ∈ V and if G ∈ G and G ∩ V = ∅, then G ∩ U = ∅. Since q ∈ lim sup Xn and V is an open set containing q, V ∩ Xn = ∅ for infinitely many indices n ∈ IN. Hence, U ∩ Xn = ∅ for infinitely many indices n ∈ IN. Then U ∩ lim sup Xn = ∅, a contradiction. Therefore, lim sup Xn = Y . Now, suppose that G is a closed decomposition of X such that for each sequence {Xn }∞ n=1 of elements of G and each element Y of G such that lim inf Xn ∩ Y = ∅, then lim sup Xn = Y . By Theorem 1.2.30, G is upper semicontinuous. Let Y be an element of G. Let p and q be two points of Y , and let U be an open subset of X such that p ∈ U . For each n ∈ IN, let Vn = V d1 (q). Suppose that for each n ∈ IN, there is an n element Xn of G such that Xn ∩ Vn = ∅ and Xn ∩ U = ∅. Hence, lim sup Xn ∩ U = ∅. For each n ∈ IN, let qn ∈ Xn ∩ Vn . Clearly, {qn }∞ n=1 converges to q. Thus, q ∈ lim inf Xn ∩ Y . By hypothesis, Y = lim sup Xn . Hence, Y ∩ U = ∅, a contradiction. Therefore, G is a continuous decomposition. Q.E.D.

1.3

Homotopy and Fundamental Group

We introduce the fundamental group of a metric space. We show that the fundamental group of the unit circle S 1 is isomorphic to the group of integers ZZ. We assume that the reader is familiar with the elementary concepts of group theory. The reader may find more than enough information about groups in [25]. 1.3.1. Definition. Given a metric space X and two points x and y of X, a path joining x and y is a map α : [0, 1] → X such that Copyright © 2005 Taylor & Francis Group, LLC

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α(0) = x and α(1) = y. In this case, x and y are the end points of α. When α is one–to–one, α is called an arc. Some times, we identify a path or an arc α with its image α([0, 1]).

1.3.2. Definition. Let X and Y be metric spaces. Let g, f : X → Y be two maps. We say that f is homotopic to g provided that there is a map G : X × [0, 1] → Y such that G((x, 0)) = f (x) and G((x, 1)) = g(x) for each x ∈ X. The map G is called a homotopy between f and g. If for each t ∈ [0, 1], G(·, t) : X → → Y is a homeomorphism, G is called an isotopy between f and g.

1.3.3. Example. Let g, f : X → Y be two constant maps between metric spaces, say g(x) = q and f (x) = p for each x ∈ X. Then f and g are homotopic if and only if p and q both belong to the same path component of Y . To show this, suppose first that p and q belong to the same path component of Y . Then there is a map α : [0, 1] → Y such that α(0) = p and α(1) = q. Hence, G : X × [0, 1] → Y given by G((x, t)) = α(t) is a homotopy between f and g. Now, suppose f and g are homotopic. Then there is a map G : X × [0, 1] → Y such that G ((x, 0)) = f (x) = p and G ((x, 1)) = g(x) = q for each x ∈ X. Let x0 ∈ X. Then the map β : [0, 1] → Y given by β(t) = G ((x0 , t)) is a path such that β(0) = p and β(1) = q.

1.3.4. Example. Let X be a metric space. If g, f : X → IRn are two maps, for some n ∈ IN. Then f and g are homotopic. To see this, let G : X × [0, 1] → IRn be given by G((x, t)) = (1 − t)f (x) + tg(x). Then G is a homotopy between f and g.

The next Theorem is known as Borsuk’s homotopy extension theorem. A proof of this result may be found in Theorem 4–4 of [10].



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1.3.5. Theorem. Let A be a closed subset of a separable metric space X, and let g , f : A → S n be homotopic maps of A into the n–sphere S n . If there exists a map f : M → S n such that f |A = f (i.e., f extends f ), then there also exists a map g : M → S n such that g|A = g , and f and g may be chosen to be homotopic also.

1.3.6. Theorem. Let Y be a closed subset of a compactum X. If f : X → S 1 is a map such that f |Y : Y → S 1 is homotopic to a constant map, then there exists an open set U of X such that Y ⊂ U and f |U : U → S 1 is homotopic to a constant map. Proof. Let f : X → S 1 be a map such that f |Y : Y → S 1 is homotopic to a constant map g : Y → S 1 given by g (y) = b for every y ∈ Y . Since f is a map which extends f |Y to X, by Theorem 1.3.5, there exists a map g : X → S 1 such that g|Y = g and f and g are homotopic. Let V be a proper open subset of S 1 such that b ∈ V . By continuity of g, there exists an open set U of X such that Y ⊂ U and g(U ) ⊂ V . Since V is an arc, it is easy to see that g|U is homotopic to a constant map. Since f and g are homotopic, f |U is also homotopic to a constant map. Q.E.D. 1.3.7. Theorem. Let X be a compactum. If f : X → → S 1 is a map not homotopic to a constant map, then there exists a closed connected subset C of X such that f |C : C → → S 1 is not homotopic to a constant map and for each proper closed subset C of C, f |C : C → S 1 is homotopic to a constant map. Proof. Let C be the collection of all closed subsets, K, of X such that f |K : K → → S 1 is not homotopic to a constant map. Since X ∈ C, C = ∅. Define the following partial order on C. If K1 , K2 ∈ C, then K1 > K2 if K1 ⊂ K2 . Let K be a (set theoretic) chain of elements of C. We show that K has an upper bound. Let Y = K. We assert that Y ∈ C. If Y does not belong to C, then K∈K


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f |Y : Y → S 1 is homotopic to a constant map. Hence, by Theorem 1.3.6 there exists an open subset U of X such that f |U : U → S 1 is homotopic to a constant map. Note that {X \ K | K ∈ K} is an open cover of X \ U . Then there exist K1 , . . . , Km ∈ K such m m that X \ U ⊂ X \ Kj . Hence, K = Kj ⊂ U . Since K is j=1

j=1

a chain, K ∈ K. Since K ⊂ U , f |K : K → S 1 is homotopic to a constant map, a contradiction to the fact that K ∈ C. Therefore, Y ∈ C and Y is an upper bound for K. Thus, by Kuratowski–Zorn Lemma, there exists an element C ∈ C such that f |C : C → → S 1 is not homotopic to a constant map and for each proper closed subset C of C, f |C : C → S 1 is homotopic to a constant map. It remains to show that C is connected. Suppose C is not connected. Thus, there exist two closed subsets C1 and C2 of X such that C = C1 ∪C2 . Since C is a maximal element of C, f |Cj : Cj → S 1 is homotopic to a constant map gj : Cj → S 1 by a homotopy Hj such that Hj ((c, 0)) = (f |Cj )(c) and Hj ((c, 1)) = gj (c) for each c ∈ Cj and j ∈ {1, 2}. Since S 1 is arcwise connected, we assume, without loss of generality, that g1 and g2 have the same image {b} ⊂ S 1 . Let g : C → S 1 be given by g(c) = b for each c ∈ C. Let H : C × [0, 1] → S 1 be given by H((c, t)) = Hj ((c, t)) if c ∈ Cj , j ∈ {1, 2}. Then H is a homotopy such that H((c, 0)) = (f |Cj )(c) = (f |C )(c) and H((c, 1)) = gj (c) = b for every c ∈ C. Thus, f |C is homotopic to a constant map, a contradiction. Therefore, C is connected. Q.E.D. 1.3.8. Theorem. Let X and Y be metric spaces. The relation of homotopy is an equivalence relation in the set of maps between X and Y . The equivalence classes of this equivalence relation are called homotopy classes. Proof. Let f : X → Y be a map. Then G : X × [0, 1] → Y , given by G((x, t)) = f (x), is a homotopy between f and f . Hence, the relation is reflexive. Now, let g, f : X → Y be two maps and suppose f is homotopic to g. Then there is a homotopy H : X × [0, 1] → Y such that H((x, 0)) = f (x) and H((x, 1)) = g(x) for each x ∈ X. Hence, Copyright © 2005 Taylor & Francis Group, LLC


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the map K : X × [0, 1] → Y given by K((x, t)) = H((x, 1 − t)) is a homotopy between g and f , since K((x, 0)) = g(x) and K((x, 1)) = f (x) for each x ∈ X. Thus, the relation is symmetric. Finally, let h, g, f : X → Y be three maps and suppose that f is homotopic to g and g is homotopic to h. Then there exist two homotopies J, L : X × [0, 1] → Y such that J((x, 0)) = f (x), J((x, 1)) = g(x), L((x, 0)) = g(x) and L((x, 1)) = h(x) for each x ∈ X. Thus, the map R : X × [0, 1] → Y given by

⎧ 1 ⎪ ⎪ if t ∈ 0, ; ⎨J((x, 2t)) 2

R((x, t)) = 1 ⎪ ⎪ ⎩L((x, 2t − 1)) if t ∈ , 1 2 is a homotopy between f and h, since for each x ∈ X, R((x, 0)) = f (x) and R((x, 1)) = h(x). Hence, the relation is transitive. Q.E.D. 1.3.9. Notation. If f : X → Y is a map, then the homotopy class to which f belongs is denoted by [f ].

1.3.10. Definition. A metric space X is said to be contractible provided that the identity map, 1X , of X is homotopic to a constant map g. We say that X is locally contractible at p if for each neighborhood U of p in X, there exist a neighborhood V of p in X and a homotopy G : V × [0, 1] → U such that G((x, 0)) = x and G((x, 1)) = x0 for each x ∈ V and some x0 ∈ U . The metric space X is locally contractible if it is locally contractible at each of its points. The following two Theorems present some consequences of the contractibility of a space. 1.3.11. Theorem. If X is a contractible metric space, then X is path connected. Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Since X is contractible, there exists a map G : X × [0, 1] → X such that G((x, 0)) = x and G((x, 1)) = p for each x ∈ X and some point p of X. Let x and y be two points of X. Then the map α : [0, 1] → X given by

⎧ 1 ⎪ ⎪ if t ∈ 0, ; ⎨G((x, 2t)) 2

α(t) = 1 ⎪ ⎪ ⎩G((y, 2 − 2t)) if t ∈ , 1 2 is a path such that α(0) = x and α(1) = y. Therefore, X is path connected. Q.E.D. 1.3.12. Theorem. Let X and Y be metric spaces, where Y is arcwise connected. If either X or Y is contractible and if g, f : X → Y are two maps, then f and g are homotopic. Proof. Suppose Y is contractible. Then there exists a map G : Y × [0, 1] → Y such that G((y, 0)) = y and G((y, 1)) = q for each y ∈ Y and some point q of Y . Then the map K : X × [0, 1] → Y given by

⎧ 1 ⎪ ⎪ if t ∈ 0, ; ⎨G((f (x), 2t)) 2

K((x, t)) = 1 ⎪ ⎪ ⎩G((g(x), 2 − 2t)) if t ∈ , 1 2 is a homotopy between f and g. The proof when X is contractible is similar. Q.E.D. Now, we consider a particular case of homotopy; namely, we study the homotopies between paths. 1.3.13. Definition. Let X be a metric space. We say that two paths β, α : [0, 1] → X are homotopic relative to {0, 1} provided that there is a homotopy G : [0, 1]×[0, 1] → X such that G((s, 0)) = α(s), G((s, 1)) = β(s), G((0, t)) = α(0) = β(0) and G((1, t)) = α(1) = β(1) for each s, t ∈ [0, 1]. Copyright © 2005 Taylor & Francis Group, LLC


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In order to define the fundamental group we need the following definitions: 1.3.14. Definition. Let X be a metric space and let x0 be a point in X. The pair (X, x0 ) is called pointed space.

1.3.15. Definition. Let X be a metric space. We say that a path α : [0, 1] → X is closed provided that α(0) = α(1).

1.3.16. Notation. If (X, x0 ) is a pointed space and α : [0, 1] → X is a closed path, we assume that α(0) = α(1) = x0 . The point x0 is called the base of the closed path.

1.3.17. Definition. Let X be a metric space. Given two closed paths β, α : [0, 1] → X such that α(0) = β(0), we define their product, denoted by α ∗ β, as the closed path given by:

⎧ 1 ⎪ ⎪ if t ∈ 0, ; ⎨α(2t) 2

(α ∗ β)(t) = 1 ⎪ ⎪ ⎩β(2t − 1) if t ∈ , 1 . 2

1.3.18. Definition. Let X be a metric space. Given a closed path α : [0, 1] → X, we define its inverse, denoted by α−1 , as the closed path α−1 : [0, 1] → X given by α−1 (t) = α(1 − t).

1.3.19. Theorem. Let X be a metric space. If β , β, α , α : [0, 1] → X are closed paths such that α(0) = α (0) = β(0) = β (0) and such that α is homotopic to α relative to {0, 1} and β is homotopic to β relative to {0, 1}, then α ∗ β is homotopic to α ∗ β relative to {0, 1} and α−1 is homotopic to (α )−1 relative to {0, 1}. Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Since α is homotopic to α relative to {0, 1} and β is homotopic to β relative to {0, 1}, there exist two homotopies K, G : [0, 1]× [0, 1] → X such that G((s, 0)) = α(s), G((s, 1)) = α (s), G((0, t)) = G((1, t)) = α(0) = α (0), K((s, 0)) = β(s), K((s, 1)) = β (s), and K((0, t)) = K((1, t)) = β(0) = β (0). Let L : [0, 1] × [0, 1] → X be given by

⎧ 1 ⎪ ⎪ if s ∈ 0, ; ⎨G((2s, t)) 2

L((s, t)) = 1 ⎪ ⎪ ⎩K((2s − 1, t)) if s ∈ , 1 . 2 Then L is the required homotopy between α ∗ β and α ∗ β , relative to {0, 1}. Next, let R : [0, 1] × [0, 1] → X be given by R((s, t)) = G((1 − s, t)). Then R is the required homotopy between α−1 and (α )−1 , relative to {0, 1}. Q.E.D. We are ready to define the fundamental group of a metric space.

1.3.20. Definition. Let (X, x0 ) be a pointed space. The fundamental group of (X, x0 ), denoted by π1 (X, x0 ), is the family of all homotopy classes of closed paths whose base is x0 . The group operation is given by [α] ∗ [β] = [α ∗ β].

1.3.21. Remark. If (X, x0 ) is a pointed space, then, by Theorem 1.3.19, the operation defined on π1 (X, x0 ) is well defined. Clearly, the identity element of π1 (X, x0 ) is the homotopy class of the constant path “x0 .”

1.3.22. Notation. Let (X, x0 ) and (Y, y0 ) be two pointed spaces. By a map, f : (X, x0 ) → (Y, y0 ), between the pointed spaces (X, x0 ) and (Y, y0 ), we mean a map f : X → Y such that f (x0 ) = y0 .



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1.3.23. Lemma. Let f : (X, x0 ) → (Y, y0 ) be a map between pointed spaces. If α and β are two closed paths whose base is x0 which are homotopic relative to {0, 1}, then f ◦ α and f ◦ β are two closed paths whose base is y0 which are homotopic relative to {0, 1}. Proof. Clearly, f ◦ α and f ◦ β are two closed paths whose base is y0 . Since α and β are homotopic relative to {0, 1}, there exists a homotopy G : [0, 1]×[0, 1] → X such that G((s, 0)) = α(s), G((s, 1)) = β(s) and G((0, t)) = G((1, t)) = x0 for each s, t ∈ [0, 1]. Then the map K : [0, 1] × [0, 1] → Y given by K((s, t)) = f (G((s, t))) is a homotopy between f ◦ α and f ◦ β relative to {0, 1}. Q.E.D. 1.3.24. Definition. If f : (X, x0 ) → (Y, y0 ) is a map between pointed spaces, then f induces a homomorphism π1 (f ) : π1 (X, x0 ) → π1 (Y, y0 ) given by π1 (f )([α]) = [f ◦ α]. 1.3.25. Remark. Let f : (X, x0 ) → (Y, y0 ) be a map between pointed spaces. If α and β are two closed paths whose base is x0 , then, clearly, f ◦ (α ∗ β) = (f ◦ α) ∗ (f ◦ β). Hence, the induced map defined in Definition 1.3.24 is a well defined group homomorphism. The next Lemma says that the induced map of a composition is the composition of the induced maps. 1.3.26. Lemma. If f : (X, x0 ) → (Y, y0 ) and g : (Y, y0 ) → (Z, z0 ) are maps between pointed spaces, then π1 (g ◦ f ) = π1 (g) ◦ π1 (f ). Proof. Let [α] ∈ π1 (X, x0 ). Then π1 (g ◦ f )([α]) = [(g ◦ f ) ◦ α] = [g ◦ (f ◦ α)] = π1 (g)([f ◦ α]) = π1 (g) (π1 (f )([α])) = (π1 (g) ◦ π1 (f )) ([α]). Q.E.D. In order to show that the fundamental group of the unit circle S is isomorphic to ZZ, we associate to each closed path α in S 1 a number η(α), which is called the degree of α, in such a way that two closed paths are homotopic if and only if they have the same degree. We use the exponential map exp : IR → → S 1 given by exp(t) = eit . 1


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1.3.27. Remark. Recall that given z ∈ S 1 , exp−1 (z) = {t + 2πn | n ∈ ZZ}, where t is any real number such that exp(t) = z. 1.3.28. Notation. Let A be a nonempty subset of IR and let t ∈ IR. Then A + t = {a + t | a ∈ A}. 1.3.29. Lemma. The exponential map is open. Proof. Let U be an open subset of IR, and let F = S 1 \ exp(U ). We show that F is closed in S 1 . Note that exp−1 (exp(U )) = {U + 2πn | n ∈ ZZ}, which is an open subset of IR. Hence, its complement, exp−1 (F ), is closed in IR. Since for each t ∈ exp−1 (F ), there exists t ∈ [0, 2π] such that exp(t ) = exp(t), F = exp(exp−1 (F )) = exp(exp−1 (F ) ∩ [0, 2π]). Since exp−1 (F ) ∩ [0, 2π] is compact, F is compact. Thus, F is closed in S 1 . Q.E.D. 1.3.30. Corollary. If t ∈ IR, then the restriction, → S 1 \ {exp(t)}, exp |(t,t+2π) : (t, t + 2π) → of the exponential map to (t, t + 2π) is a homeomorphism onto S 1 \ {exp(t)}. 1.3.31. Theorem. Let α : [0, 1] → S 1 be a closed path whose base is (1, 0). Then there exists a unique map α : [0, 1] → IR such that α (0) = 0 and α(t) = exp(α (t)). The map α is called the lifting of α beginning at 0. Proof. First, suppose that α([0, 1]) = S 1 . Let A be the component of exp−1 (α([0, 1])) containing 0. Then exp |A : A → → α([0, 1]) is a homeomorphism (Corollary 1.3.30). Hence, the map α : [0, 1] → IR given by α (t) = (exp |A )−1 (α(t)) is the required map. Next, suppose α([0, 1]) = S 1 . Let t0 = 0 < t1 < · · · < tn−1 < tn = 1 be a subdivision of [0, 1] such that α([tj−1 , tj ]) = S 1 for each j ∈ {1, . . . , n}. Copyright © 2005 Taylor & Francis Group, LLC


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Let A0 be the component of exp−1 (α([0, t1 ])) containing 0. Then, as before, exp |A0 : A0 → → α([0, t1 ]) is a homeomorphism. Let α0 : [0, t1 ] → IR be given by α0 (t) = (exp |A0 )−1 (α(t)). Let A1 be the component of exp−1 (α([t1 , t2 ])) containing α0 (t1 ), and let α1 : [t1 , t2 ] → IR be given by α1 (t) = (exp |A1 )−1 (α(t)). Repeating this process, for each j ∈ {0, . . . , n − 1}, we define maps αj : [tj , tj+1 ] → IR given −1 (α(t)). by αj (t) = exp |Aj Let α : [0, 1] → IR be given by α (t) = αj (t) if t ∈ [tj , tj+1 ]. Then α (0) = 0 and α(t) = exp(α (t)). To see α is unique, suppose β : [0, 1] → IR is another map such that β(0) = 0 and α(t) = exp(β(t)). Hence, exp(α (t)) = exp(β(t)) for each t ∈ [0, 1]. Consider the map γ : [0, 1] → IR given by γ(t) = α (t) − β(t) . Then γ([0, 1]) ⊂ ZZ. Since α (0) = β(0), γ([0, 1]) = 2π {0}. Therefore, α (t) = β(t) for each t ∈ [0, 1]. Q.E.D. 1.3.32. Remark. If in Theorem 1.3.31 we do not require that the map α satisfies that α (0) = 0, we may have many liftings of the map α. However, any other lifting α of α satisfies that α (t) = α (t) + 2πk for some k ∈ ZZ and each t ∈ [0, 1].

1.3.33. Definition. Let α : [0, 1] → S 1 be a closed path, and let α be a lifting of α. Then η(α) =

α (1) − α (0) 2π

is an integer, and it is called the degree of α.

1.3.34. Remark. Observe that for each closed path α, the definition of η(α) does not depend on the lifting of α. Since for any two liftings α and α of α, we have that α (1) − α (0) = α (1) − α (0) by Remark 1.3.32. Intuitively, the degree of a closed path “counts” the number of times that the closed path wraps [0, 1] around S 1 .


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1.3.35. Theorem. Let β, α : [0, 1] → S 1 be two closed paths whose base is (1, 0). Then: (1) η(α ∗ β) = η(α) + η(β); (2) If α and β are homotopic relative to {0, 1} if and only if η(α) = η(β); (3) Given k ∈ ZZ, there exists a closed path γ, whose base is (1, 0), such that η(γ) = k. Proof. Let α : [0, 1] → IR be a lifting of α such that α (0) = 0 (Theorem 1.3.31), and let β : [0, 1] → IR be a lifting of β such that β (0) = α (1). Define α ∗ β : [0, 1] → IR by

⎧ 1 ⎪ ⎪ if s ∈ 0, ; ⎨α (2s) 2

(α ∗ β )(s) = 1 ⎪ ⎪ ⎩β (2s − 1) if s ∈ , 1 . 2 Then it is easy to see that α ∗ β is a lifting of α ∗ β. Since 2πη(α ∗ β) = (α ∗ β )(1) − (α ∗ β )(0) = β (1) − α (0) = (β (1) − β (0)) + (α (1) − α (0)) = 2π(η(β) + η(α)), we have that η(α ∗ β) = η(α) + η(β). Suppose η(α) = η(β). Let α and β be liftings of α and β, respectively, such that α (0) = β (0) = 0. Since η(α) = η(β), α (1) − α (0) = β (1) − β (0). In particular, α (1) = β (1). By Example 1.3.4, the map G : [0, 1] × [0, 1] → IR given by G((s, t)) = (1 − t)α (s) + tβ (s) is a homotopy between α and β . Note that for each t ∈ [0, 1], G((1, t)) − G((0, t)) = (1 − t) [α (1) − α (0)] + t [β (1) − β (0)] = (1 − t)2πη(α) + t2πη(β) = 2πη(α). Hence, the map K = exp ◦G is a homotopy between α and β such that K((0, t)) = K((1, t)) = (1, 0). Thus, K is a homotopy between α and β relative to {0, 1}. Now, assume α and β are homotopic relative to {0, 1}. First, suppose that ||α(s)−β(s)|| < 2 for each s ∈ [0, 1], i.e., α(s) and β(s) are never antipodal points. Let α and β be liftings of α and β, respectively, such that α (0) = β (0) = 0. Since ||α(s) − β(s)|| < 2, |α (s) − β (s)| < π for each s ∈ [0, 1]. Hence, 2π|η(α) − η(β)| = |α (1) − α (0) − β (1) + β (0)| ≤ |α (1) − β (1)| + |α (0) − β (0)| < π + π = 2π. Thus, |η(α) − η(β)| = 0, i.e., η(α) = η(β). Next, assume there is an s ∈ [0, 1] such that ||α(s) − β(s)|| = 2. Let Copyright © 2005 Taylor & Francis Group, LLC


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H : [0, 1] × [0, 1] → S 1 be a homotopy between α and β relative to {0, 1}. Since H is uniformly continuous, there exists δ > 0 such that if |t − t | < δ, then ||H((s, t)) − H((s, t ))|| < 2 for each s ∈ [0, 1]. Let t0 = 0 < t1 < · · · < tk−1 < tk = 1 be a subdivision of [0, 1] such that tj − tj−1 < δ for each j ∈ {1, . . . , k}. Let αj : [0, 1] → S 1 be given by αj (s) = H((s, tj )) for each j ∈ {0, . . . , k}. Note that α0 = α and αk = β. By construction, ||αj−1 (s) − αj (s)|| < 2 for each j ∈ {1, . . . , k}. Then, applying the above argument, we obtain that η(α) = η(α0 ) = η(α1 ) = . . . = η(αk ) = η(β). Finally, let k ∈ ZZ. Define γ : [0, 1] → S 1 by γ(s) = exp(2πks). Then γ is a closed path whose base is (1, 0). Note that the map γ : [0, 1] → IR given by γ (s) = 2πks is a lifting of γ such that γ (1) − γ (0) γ (0) = 0. Hence, η(γ) = = k. 2π Q.E.D. Now, we are ready to show that the fundamental group of the unit circle S 1 is isomorphic to ZZ. 1.3.36. Theorem. The fundamental group of the unit circle S 1 is isomorphic to ZZ. Proof. Let Σ : π1 (S 1 ) → → ZZ be given by Σ([α]) = η(α). Note that, by Theorem 1.3.35 (2), Σ is well defined and by (3) of the same Theorem, Σ is, indeed, a surjection. By Theorem 1.3.35 (1), Σ is a homomorphism. Finally, by Theorem 1.3.35 (2), Σ is one–to–one. Therefore, Σ is an isomorphism. Q.E.D. Next, we define the degree of a map between simple closed curves. 1.3.37. Definition. Let f : S 1 → S 1 be a map. The degree of f , denoted by deg(f ), is defined as follows. Consider the induced map π1 (f ) : π1 (S 1 ) → π1 (S 1 ). Then deg(f ) = (Σ ◦ π1 (f )) (Σ−1 (1)), where Σ is defined in Theorem 1.3.36. 1.3.38. Remark. Let f : S 1 → S 1 be a map. If deg(f ) = 0, then f is homotopic to a constant map (see Theorem 7.4 (p. 352) of [9]). Copyright © 2005 Taylor & Francis Group, LLC

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1.3.39. Lemma. If g, f : S 1 → S 1 are two maps, then deg(g ◦f ) = deg(g) · deg(f ). Proof. Recall that, by Lemma 1.3.26, π1 (g ◦ f ) = π1 (g) ◦ π1 (f ). By definition, deg(g ◦ f ) = [Σ ◦ π1 (g ◦ f )] (Σ−1 (1)) = [Σ ◦ (π1 (g) ◦ π1 (f ))] (Σ−1 (1)) = [Σ ◦ π1 (g)] π1 (f )(Σ−1 (1)) = [Σ ◦ π1 (g)] π1 (f )(Σ−1 (1)) ∗ Σ−1 (1) = Σ ◦ π1 (f )(Σ−1 (1)) ∗ Σ ◦ π1 (g)(Σ−1 (1)) = deg(f ) · deg(g). Q.E.D. We end this section with the following Theorem. 1.3.40. Theorem. A map f : S 1 → S 1 is homotopic to a constant map if and only if there exists a map f˜: S 1 → IR such that f = exp ◦f˜. Proof. If f˜: S 1 → IR exists such that f = exp ◦f˜, then f˜ is homotopic to a constant map since IR is contractible (Theorem 1.3.12). Hence, f = exp ◦f˜ is also homotopic to a constant map. Now, suppose f is homotopic to a constant map. Without loss of generality, we assume that f ((1, 0)) = (1, 0). Since f is homotopic to a constant map, deg(f ) = 0. Let g : [0, 1] → → S 1 be given by g(t) = exp(2πt). Since f is homotopic to a constant map, f ◦ g is homotopic to a constant map. By Theorem 1.3.35 (2), η(f ◦ g) = 0. Let ξ : [0, 1] → IR be a map such that exp ◦ ξ = f ◦ g, i.e., ξ is a lifting of f ◦ g (Theorem 1.3.31). Since η(f ◦ g) = 0, ξ(0) = ξ(1). Hence, the function f˜: S 1 → IR given by f˜(z) = ξ(g −1 (z)) is well defined, continuous, and satisfies that (exp ◦f˜)(z) = (exp ◦ ξ)(g −1 (z)) = (f ◦ g)(g −1 (z)) = f (z) for each z ∈ S 1. Q.E.D.


1.4. GEOMETRIC COMPLEXES AND POLYHEDRA

1.4

35

Geometric Complexes and Polyhedra

This is a very small section. We present the definitions of polyhedra and the nerve of a finite collection of sets, which are used in Chapter 2. 1.4.1. Definition. We say that r + 1 points, {x0 , . . . , xr }, of IRn are affinely independent provided that the set {x1 − x0 , . . . , xr − x0 } is linearly independent. The following Theorem gives us an alternative way, in terms of linear algebra, to see affinely independent subsets of IRn . 1.4.2. Theorem. The set of points {x0 , . . . , xr } of IRn is affinely independent if and only if each time r

ξj xj = 0 and

j=0

r

ξj = 0,

j=0

where ξj ∈ IRn , we have that ξj = 0 for each j ∈ {0, . . . , r}. Proof. Suppose {x0 , . . . , xr } is affinely independent. Let ξ0 , . . . , ξr r r be real numbers such that ξj xj = 0 and ξj = 0. Since

r

j=0

ξj = 0, ξ0 = −

j=0 r

ξj xj =

j=1

−

r j=1

ξj

x0 +

r

j=0

ξj . Then: 0 =

j=1 r

r

j=1

j=1

ξj xj =

r

ξj xj = ξ0 x0 +

j=0

ξj (xj − x0 ). Since {x1 −

x0 , . . . , xr − x0 } is linearly independent, ξ1 = . . . = ξr = 0. Since r ξ0 = − ξj , ξ0 = 0. j=1

Next, suppose {x0 , . . . , xr } satisfies the hypothesis of the Theor rem. Let ξ1 , . . . , ξr be real numbers such that ξj (xj − x0 ) = 0. j=1


36


Note that

r

ξj (xj − x0 ) =

r

r ξj xj − ξj x0 . Let ξ0 =

j=1

r

j=1 r r r r − ξj . Then ξj xj − ξj x0 = ξj xj and ξj = 0. j=1

j=1

j=1

j=1

j=0

j=0

Then, by hypothesis, ξ0 = . . . = ξr = 0. In particular, ξ1 = . . . = ξr = 0. Thus, {x1 − x0 , . . . , xr − x0 } is linearly independent. Therefore, {x0 , . . . , xr } is affinely independent. Q.E.D. 1.4.3. Definition. Let {x0 , . . . , xr } be an affinely independent subset of IRn . We define the (geometric) r–simplex generated by {x0 , . . . , xr }, denoted by [x0 , . . . , xr ], as the following subset of IRn : r ξj xj for each j ∈ {0, . . . , r}, [x0 , . . . , xr ] = j=0

ξj ∈ [0, 1] and

r

ξj = 1 .

j=0

The set {x0 , . . . , xr } is called the set of vertexes of the r–simplex. r 1 The point x = xj is called the barycenter of the r–simplex. r+1 j=0 Each s–simplex generated by s + 1 points taken from {x0 , . . . , xr } is called an s–face of [x0 , . . . , xr ].

1.4.4. Remark. It is easy to see that a 0–simplex is a point, a 1– simplex is a line segment, a 2–simplex is a triangle and a 3–simplex is a tetrahedron.

simplexes of dimension one, two and three


1.4. GEOMETRIC COMPLEXES AND POLYHEDRA

37

1.4.5. Definition. A (geometric) complex, K, is a finite collection of simplexes in IRn such that: (1) each face of a simplex in K also belongs to K and (2) the intersection of any two simplexes is either empty or it is a face of both simplexes.

n ot a comp lex

complex

1.4.6. Definition. A polyhedron in IRn is the union of the simplexes of a geometric complex. 1.4.7. Definition. The barycentric subdivision of an r–simplex R is a (geometric) complex K obtained as follows: (a) If r = 0, then K is just R. (b) Suppose r > 0. If S0 , . . . , Sr are the (r − 1)–faces of R and if xb is the barycenter of R, then K consists of all r–simplexes generated by xb and the vertexes of all (r − 1)–simplexes of the barycentric subdivision of Sj for each j ∈ {0, . . . , r}. The barycentric subdivision of a complex K0 is a (geometric) complex K1 consisting of all the simplexes obtained by the barycentric subdivision of each simplex of K0 . 1.4.8. Remark. It is well known that a complex K1 obtained from a complex K0 by barycentric subdivision is, in fact, a complex (15.2 of [20]). Some times the barycentric subdivision is performed several times, i.e., one can apply the barycentric subdivision to K1 to obtain a complex K2 , and then apply the barycentric subdivision to K2 to obtain a complex K3 , etc. It is also possible to show that the diameters of the simplexes of the complex Km , obtained after m barycentric subdivisions, tend to zero as m tends to infinity (15.4 of [20]). Copyright © 2005 Taylor & Francis Group, LLC

38


Barycentric subdivisions

1.4.9. Definition. Let X be a compactum and let U = {U1 , . . . , Un } be a finite collection of subsets of X. Then the nerve of U, denoted by N (U), is the complex defined as follows. To each Uj ∈ U associate the point ej of IRn , where ej = (0, . . . , 0, 1, 0 . . . , 0) (the number “1” appears in the jth coordinate). Hence, the vertexes of the complex are {e1 , . . . , en }. For each subfamily {Uj1 , . . . , Ujk } of k U such that Uj = ∅, we consider the simplex whose vertexes =1

are the points {ej1 , . . . , ejk }. Then N (U) consists of all such possible simplexes. We denote by N (U) the polyhedron associated to N (U).

1.5

Complete Metric Spaces

We present results about complete metric spaces. 1.5.1. Definition. Let X be a metric space, with metric d. A sequence, {xn }∞ n=1 , of elements of X is said to be a Cauchy sequence provided that for each ε > 0, there exists N ∈ IN such that if n, m ≥ N , then d(xn , xm ) < ε.

1.5.2. Definition. A metric space X, with metric d, is said to be complete provided that every Cauchy sequence of elements of X converges to a point of X. We say that X is topologically complete if there exists an equivalent metric d1 for X such that (X, d1 ) is a complete space. Copyright © 2005 Taylor & Francis Group, LLC

1.5. COMPLETE METRIC SPACES

39

The following results give some basic properties of complete metric spaces. 1.5.3. Lemma. Let X be a complete metric space. If F is a closed subset of X, then F is complete. Proof. Let {xn }∞ n=1 be a Cauchy sequence of elements of F . Since X is complete, there exists a point x ∈ X such that {xn }∞ n=1 converges to x. Since F is a closed subset of X, x ∈ F . Therefore, F is complete. Q.E.D. 1.5.4. Proposition. Let X be a metric space. If Y is a subset of X and Y is complete, then Y is closed in X. Proof. Let x ∈ Cl(Y ). Then there exists a sequence {yn }∞ n=1 of elements of Y converging to x. Since a convergent sequence is a Cauchy sequence, and Y is complete, we have that x ∈ Y . Therefore, Y is closed in X. Q.E.D. 1.5.5. Proposition. Let X be a complete metric space, with metric d. If {Fn }∞ n=1 is a sequence of closed subsets of X such that Fn+1 ⊂ Fn and lim diam(Fn ) = 0, then there exists a point x ∈ X n→∞ ∞ such that Fn = {x}. n=1

Proof. For each n ∈ IN, let xn ∈ Fn . Then {xn }∞ n=1 is a sequence of points of X such that for each N ∈ IN, if n, m ≥ N , then xn , xm ∈ FN . Thus, since lim diam(Fn ) = 0, {xn }∞ n=1 is a Cauchy sequence. n→∞ Hence, there exists x ∈ X such that lim xn = x. Note that, since n→∞ ∞ ∞ for every N ∈ IN, {xn }n=N ⊂ FN , x ∈ Fn . Now, suppose there exists y ∈

∞

n=1

Fn \ {x}. Then for each n ∈ IN, 0 < d(x, y) ≤

n=1


40


diam {x}.

∞

Fn

≤ diam(Fn ), a contradiction. Therefore,

n=1

∞

Fn =

n=1

Q.E.D. 1.5.6. Definition. A subset of a metric space X is said to be a Gδ set provided that it is a countable intersection of open sets.

1.5.7. Lemma. Let X be a metric space, with metric d. If F is a closed subset of X, then F is a Gδ subset of X. Proof. It follows from the fact that F =

∞ n=1

V d1 (F ). n

Q.E.D. The following Theorem is due to Mazurkiewicz, a proof of which may be found in Theorem 8.3 (p. 308) of [9]: 1.5.8. Theorem. Let Y be a complete metric space. Then a nonempty subset A of Y is topologically complete if and only if A is a Gδ subset of Y .

1.5.9. Lemma. If X is a complete metric space, with metric d, ∞ ∞ and {Dn }n=1 is sequence of dense open subsets of X, then Dn n=1

is a dense subset of X. Proof. Let U be an open subset of X. We show that U ∩

∞

n=1

Dn

= ∅. Since D1 is a dense subset of X, D1 ∩ U = ∅. In fact, D1 ∩ U is an open subset of X. Hence, there exist x1 ∈ D1 ∩ U and ε1 > 0 such that ε1 < 1 and Cl(Vεd1 (x1 )) ⊂ D1 ∩ U . Since D2 is a dense open subset of X and Vεd1 (x1 ) is open, D2 ∩ Vεd1 (x1 ) = ∅. So, 1 and there exist x2 ∈ D2 ∩ Vεd1 (x1 ) and ε2 > 0 such that ε2 < 2 Copyright © 2005 Taylor & Francis Group, LLC

1.5. COMPLETE METRIC SPACES

41

Cl(Vεd2 (x2 )) ⊂ D2 ∩ Vεd1 (x1 ). If we continue with this process, we construct a decreasing family of closed subset of X, whose diameters tend to zero. Hence, by Proposition

1.5.5, there exists a point x ∈ X ∞ ∞ such that {x} = Cl(Vεdn (xn )) ⊂ Dn ∩ U . n=1 n=1

∞ ∞ Cl(Vεdn (xn )) ⊂ Dn ∩ U , we obtain that U ∩ Since n=1 n=1

∞ Dn = ∅. n=1

Q.E.D. 1.5.10. Definition. Let X be a metric space. A subset A of X is said to be nowhere dense provided that Int(Cl(A)) = ∅. 1.5.11. Definition. Let X be a metric space. A subset A of X is said to be of the first category if it is the countable union of nowhere dense subsets of X. A subset of X that is not of the first category is said to be of the second category. The following result is known as the Baire Category Theorem: 1.5.12. Theorem. If X is a complete metric space, then X is of the second category. Proof. Suppose X is of the first category. Then there exists a sequence {An }∞ n=1 of nowhere dense subsets of X such that X = ∞ An . Since, for each n ∈ IN, An is nowhere dense, each set n=1

X \ Cl(An ) is an open dense subset of X. By Lemma 1.5.9, Cl(An )) = ∅. Since X =

∞

(X \

n=1 ∞

An , we have that

n=1 ∞ n=1

(X \ Cl(An )) = X \

∞

Cl(An ) ⊂ X \

n=1


∞ n=1

An = X \ X = ∅,

42


a contradiction. Therefore, X is of the second category. Q.E.D. The following Theorem is due to Hausdorff, a proof of which may be found in the Appendix of [2]: 1.5.13. Theorem. Let X and Y be metric spaces. If X is complete and f : X → → Y is a surjective open map, then Y is topologically complete.

1.6

Compacta

We give basic properties of compacta. We construct the Cantor set and present some of its properties. A proof of the following Theorem may be found in Corolário 2 (p. 212) of [14]. 1.6.1. Theorem. If X is a compactum, then X is complete. 1.6.2. Lemma. Let X be a compactum, and let A be a closed subset of X with a finite number of components. If x ∈ Int(A) and C is the component of A such that x ∈ C, then x ∈ Int(C). Proof. Let C, C1 , . . . , Cn be the components of A. Since x ∈ Int(A), there exists an open subset U of X such that x ∈ U ⊂ A. Since A is closed in X, each of C, C1 , . . . , Cn is closed in X. Let n V =U∩ X\ Cj . Then V is an open subset of X such that j=1

x ∈ V ⊂ A and V ∩ x ∈ Int(C).

n

Cj

= ∅. Thus, V ⊂ C. Therefore,

j=1

Q.E.D.


1.6. COMPACTA

43

1.6.3. Definition. Let X be a metric space. A subset Y of X is said to be perfect if Y is closed and every point of Y is a limit point of Y .

1.6.4. Example. We construct the Cantorset and prove some of 1 2 , , and let C1 = its properties. Let C0 = [0, 1]. Remove 3 3

1 2 0, ∪ , 1 . Remove the middle thirds of these intervals, and 3

3 1 2 1 2 7 8 let C2 = 0, ∪ , ∪ , ∪ , 1 . Continuing in this 9 9 3 3 9 9 way, we obtain a sequence, {Cn }∞ n=0 , of compact sets such that for each n ∈ IN, Cn+1 ⊂ Cn and Cn is the union of 2n intervals, In,0 , . . . , In,2n−1 , of length 3−n . The set C=

∞

Cn

n=0

is called the Cantor set. C is clearly compact and nonempty. No interval of the form 3k + 1 3k + 2 , , (∗) 3m 3m where k, m ∈ IN, has a point in common with C. Since every interval y−x , C does (x, y) contains an interval of the form (∗), if 3−m < 6 not contain a nondegenerate interval. We show that C is perfect. Let x ∈ C, and let A be any interval containing x. Let In be the interval in Cn that contains x. Let n large enough, so that In ⊂ A. Let xn be an end point of In such that xn = x. It follows from the construction that xn ∈ C. Hence, x is a limit point of C. Therefore, C is perfect. It follows, from the construction and the fact that C does not contain any nondegenerate interval, that C is totally disconnected. In Chapter 2, we present a characterization of the Cantor set as the only totally disconnected and perfect compactum.


44


1.6.5. Definition. Let X be a metric space, and let U be an open cover of X. We say that a number λ > 0 is a Lebesgue number for the open cover U provided that if A is a nonempty subset of X such that diam(A) < λ, then there exists U ∈ U such that A ⊂ U .

1.6.6. Theorem. If X is a compactum, with metric d, and U is an open cover of X, then there exists a Lebesgue number for U. Proof. Suppose that no such number exists. Then for each n ∈ IN, 1 there exists a nonempty subset An of X such that diam(An ) < n and An is not contained in any element of U. Let xn ∈ An . Since X is compact, without loss of generality, we assume that the sequence {xn }∞ n=1 converges to a point x ∈ X. Since U covers X, there exists U ∈ U such that x ∈ U . Hence, there exists ε > 0 such that Vεd (x) ⊂ U (U is open in X). Let n ∈ IN 1 ε ε such that < and d(xn , x) < . Then for each y ∈ An , n 2 2 d(y, x) ≤ d(y, xn ) + d(xn , x)
0 | A ⊂ Vεd (B) and B ⊂ Vεd (A)} is a metric for 2X . It is called the Hausdorff metric. Proof. We show the triangle inequality. The other two properties are obvious. Let A, B and C be three elements of 2X . We show that H(A, C) ≤ H(A, B) + H(B, C). To this end, let α be a positive real number. α α Let βA = H(A, B) + and βB = H(B, C) + . Let us observe 2 2 that: A ⊂ VβdA (B) and B ⊂ VβdB (C). Then, given a ∈ A, there α exists b ∈ B such that d(a, b) < H(A, B) + . For this b, there 2 α exists c ∈ C such that d(b, c) < H(B, C) + . Hence, d(a, c) < 2 H(A, B) + H(B, C) + α. Therefore, if β = H(A, B) + H(B, C) + α, then A ⊂ Vβd (C). A similar argument shows that C ⊂ Vβd (A). Since α was an arbitrary positive number, it follows, from the definition of H, that H(A, C) ≤ H(A, B) + H(B, C). Q.E.D. Copyright © 2005 Taylor & Francis Group, LLC

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Let X be a continuum. Note that given a sequence, {Yn }∞ n=1 , of X elements of 2 , we have two types of convergence of this sequence, namely, the one given in Definition 1.2.26 (the limit belongs to 2X by Lemma 1.2.27), and the other one given by the Hausdorff metric. It is known that both types of convergence coincide, i.e., we have the following theorem, a proof of which may be found in (0.7) of [22]: 1.8.4. Theorem. Let X be a continuum, and let {Yn }∞ n=1 be a sequence of elements of 2X . Then {Yn }∞ converges to Y in the n=1 ∞ sense of Definition 1.2.26 if and only if {Yn }n=1 converges to Y in the Hausdorff metric.

1.8.5. Theorem. If X is a compactum, then 2X and C(X) are compact. Proof. First, we show that 2X is compact. By Theorem 1.8.4, it is enough to prove that every sequence of elements in 2X has a convergent subsequence in the sense of Definition 1.2.26. This was already done in Theorem 1.2.28. We just need to mention that the limit of the subsequence constructed in Theorem 1.2.28 is closed by Lemma 1.2.27, and, since X is compact, such limit is nonempty. To see that C(X) is compact, it suffices to show that C(X) is closed in 2X . This follows from Theorem 1.2.29. Q.E.D. Regarding n–fold symmetric products, we have the following results: 1.8.6. Lemma. Let X be a continuum with metric d, and let n ∈ IN. If Dn denotes the metric on X n given by Dn ((x1 , . . . , xn ), (x1 , . . . , xn )) = max{d(x1 , x1 ), . . . , d(xn , xn )}, then the function fn : X n → Fn (X) given by fn ((x1 , . . . , xn )) = {x1 , . . . , xn } Copyright © 2005 Taylor & Francis Group, LLC

1.8. HYPERSPACES

61

is surjective and satisfies the following inequality: H(fn ((x1 , . . . , xn )), fn ((x1 , . . . , xn ))) ≤ Dn ((x1 , . . . , xn ), (x1 , . . . , xn )), for each (x1 , . . . , xn ) and (x1 , . . . , xn ) in X n . Proof. Clearly the map fn is surjective. Let (x1 , . . . , xn ) and (x1 , . . . , xn ) be two points of X n . Assume that Dn ((x1 , . . . , xn ), (x1 , . . . , xn )) = r and let ε > 0 be given. Hence, Dn ((x1 , . . . , xn ), (x1 , . . . , xn )) < r + ε. This implies that for each j ∈ {1, . . . , n}, d(xj , xj ) < r+ε. Thus, for each xj ∈ fn ((x1 , . . . , xn )), we have that xj ∈ fn ((x1 , . . . , xn )) and d(xj , xj ) < r + ε. This shows that d fn ((x1 , . . . , xn )) ⊂ Vr+ε (fn ((x1 , . . . , xn ))). d (fn ((x1 , . . . , xn ))). Thus, Similarly, fn ((x1 , . . . , xn )) ⊂ Vr+ε

H(fn ((x1 , . . . , xn )), fn ((x1 , . . . , xn ))) ≤ r + ε. Since the ε was arbitrary, we obtain that H(fn ((x1 , . . . , xn )), fn ((x1 , . . . , xn ))) ≤ r. Q.E.D. As an immediate consequence of Lemma 1.8.6, we have the following Corollary: 1.8.7. Corollary. Let X be a continuum and let n ∈ IN. Then the function fn : X n → → Fn (X) given by: fn ((x1 , . . . , xn )) = {x1 , . . . , xn } is continuous.


62


1.8.8. Corollary. If X is a continuum, then Fn (X) is a continuum for each n ∈ IN. 1.8.9. Corollary. If X is a continuum, then 2X is a continuum. Proof. By Corollary 1.8.8, Fn (X) is a continuum for each n ∈ IN. ∞ Then F(X) = Fn (X) is a connected subset of 2X . We show that n=1

F(X) is dense in 2X . To this end, let ε be a positive real number. Since X is compact, there exist x1 , . . . , xm ∈ X, such that X ⊂ m Vεd (xn ). Then {x1 , . . . , xm } ∈ F(X), X ⊂ Vεd ({x1 , . . . , xm }) n=1

and, clearly, {x1 , . . . , xm } ⊂ Vεd (X). Hence, H({x1 , . . . , xm }, X) < ε. Therefore, F(X) is dense in 2X . Thus, 2X is connected. By Theorem 1.8.5, 2X is a compactum. Therefore, 2X is a continuum. Q.E.D. The following Theorem (see (1.13) of [22]) is a better result than Corollary 1.8.9: 1.8.10. Theorem. If X is a continuum, then 2X and C(X) are both arcwise connected continua. Recall that given a compactum X, we defined 2X as the family of all nonempty closed subsets of X with the Hausdorff metric H. X By Theorem 1.8.5, 2X is a compactum. Hence, we define 22 as the family of all nonempty closed subsets of 2X with the Hausdorff metric H2 . The following Lemma, a proof of which may be found X in (1.48) of [22], gives a nice map from 22 onto 2X : 2 X 1.8.11. Lemma. Let X be a compactum. Let σ : 2 → 2 be given by σ(A) = {A | A ∈ A}. Then σ is well defined and satisfies the following inequality: X

H(σ(A1 ), σ(A2 )) ≤ H2 (A1 , A2 ) for each A1 and A2 in 22 . In particular, σ is continuous. X


1.8. HYPERSPACES

63

1.8.12. Corollary. Let X be a continuum. Then Cn (X) is an arcwise connected continuum for each n ∈ IN. Proof. Let n ∈ IN. Since C(X) is an arcwise connected continuum (Theorem 1.8.10), Fn (C(X)) is a continuum (Corollary 1.8.8). It is easy to see that, in fact, Fn (C(X)) is arcwise connected. Let σ be the union map. Then σ (Fn (C(X))) = Cn (X). Hence, Cn (X) is an arcwise connected continuum. Q.E.D. 1.8.13. Notation. Let X be a compactum. Given a finite collection of open sets of X, U1 , U2 , . . . , Um , we define U1 , . . . , Um = m Uk and A ∩ Uk = ∅ for each k ∈ {1, . . . , m} . A ∈ 2X A ⊂ k=1

A proof of the following Theorem may be found in (0.11) of [22]:

1.8.14. Theorem. Let X be a compactum. If B = {U1 , . . . , Un | U1 , . . . , Un are open subsets of X and n ∈ IN}, then B is a basis for a topology of 2X .

1.8.15. Definition. The topology for 2X given by Theorem 1.8.14 is called the Vietoris topology. It is known that the topology induced by the Hausdorff metric and the Vietoris topology coincide (see (0.13) of [22]):


64


1.8.16. Theorem. Let X be a compactum. Then the topology induced by the Hausdorff metric and the Vietoris topology for 2X are the same.

1.8.17. Definition. Let X be a compactum. A Whitney map is a continuous function µ : 2X → → [0, 1] such that: (1) µ(X) = 1, (2) µ({x}) = 0 for every x ∈ X and (3) µ(A) < µ(B) if A, B ∈ 2X and A B. X

2

1

X -1

µ (t) µ

) (X n C

t

C(X)

0 .

1.8.18. Remark. Whitney maps exist; constructions of them may be found in [22] and [11].

1.8.19. Definition. Let X be a compactum, and let A, B ∈ 2X . An order arc from A to B is a one–to–one map α : [0, 1] → 2X such that α(0) = A, α(1) = B and for each s, t ∈ [0, 1] such that s < t, α(s) α(t). Copyright © 2005 Taylor & Francis Group, LLC

1.8. HYPERSPACES

65

The following Theorem tells us when an order arc exists. A proof of it may be found in (1.8) of [22]: 1.8.20. Theorem. Let X be a compactum, and let A, B ∈ 2X . Then there exists an order arc from A to B if and only if A ⊂ B and each component of B intersects A.

1.8.21. Definition. Let f : X → Y be a map between compacta. Then 2f : 2X → 2Y given by 2f (A) = f (A) is called the induced map between 2X and 2Y . For each n ∈ IN, the maps Cn (f ) : Cn (X) → Cn (Y ) and Fn (f ) : Fn (X) → Fn (Y ) given by Cn (f ) = 2f |Cn (X) and Fn (f ) = 2X |Fn (X) are called the induced maps between the hyperspaces Cn (X) and Cn (Y ), and Fn (X) and Fn (Y ), respectively. 1.8.22. Theorem. Let f : X → Y be a map between compacta. Then 2f is continuous. Proof. Let A ∈ 2X , and let ε > 0. Let δ > 0 given by the uniform continuity of f . Then 2f VδHX (A) ⊂ VεHY (2f (A)). Therefore, 2f is continuous. Q.E.D. 1.8.23. Corollary. Let f : X → Y be a map between compacta. Then for each n ∈ IN, Cn (f ) and Fn (f ) are continuous. From the proof of Theorem 20 of [7], we have the following Theorem: 1.8.24. Theorem. Let f : X → Y be an open surjective map between compacta. Then the function (f ) : 2Y → 2X given by ((f ))(B) = f −1 (B) is continuous.


66

REFERENCES REFERENCES

[1] M. Aguilar, S. Gitler and C. Prieto, Algebraic Topology From a Homotopical View, Universitext, Springer–Verlag, New York, Inc., 2002. [2] F. D. Ancel, An Alternative Proof and Applications of a Theorem of E. G. Effros, Michigan J. 34 (1987), 39–55. [3] K. Borsuk and S. Ulam, On Symmetric Products of Topological Spaces, Bull. Amer. Math. Soc., 37 (1931), 875–882. [4] E. Casta˜ neda, A Unicoherent Continuum for Which its Second Symmetric Product is not Unicoherent, Topology Proceedings, 23 (1998), 61–67. [5] E. Casta˜ neda, Productos Simétricos, Tesis Doctoral, Facultad de Ciencias, U. N. A. M., 2003. (Spanish) [6] J. J. Charatonik, On fans, Dissertationes Math. (Rozprawy Mat.), 54 (1967), 1–37. [7] J. J. Charatonik, A. Illanes and S. Mac´ıas, Induced Mappings on the Hyperspaces Cn (X) of a Continuum X, Houston J. Math., 28 (2002), 781–805. [8] H. S. Davis, P. H. Doyle, Invertible Continua, Portugal. Math., 26 (1967), 487–491. [9] J. Dugundji, Topology, Allyn and Bacon, Inc., Boston, 1966. [10] J. G. Hocking and G. S. Young, Topology, Dover Publications, Inc., New York, 1988. [11] A. Illanes and S. B. Nadler, Jr., Hyperspaces: Fundamentals and Recent Advances, Monographs and Textbooks in Pure and Applied Math., Vol. 216, Marcel Dekker, New York, Basel, 1999. [12] K. Kuratowski, Topology, Vol. I, Academic Press, New York, N. Y., 1966. Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

67

[13] K. Kuratowski, Topology, Vol. II, Academic Press, New York, N. Y., 1968. [14] E. L. Lima, Espa¸cos Métricos, terceira edi¸cão, Instituto de Matemática Pura e Aplicada, CNPq, (Projeto Euclides), 1977. (Portuguese) [15] E. L. Lima, Grupo Fundamental e Espa¸cos de Recobrimento, Instituto de Matemática Pura e Aplicada, CNPq, (Projeto Euclides), 1993. (Portuguese) [16] S. Mac´ıas, On Symmetric Products of Continua, Topology Appl., 92 (1999), 173–182. [17] S. Mac´ıas, Aposyndetic Properties of Symmetric Products of Continua, Topology Proc., 22 (1997), 281–296. [18] S. Mac´ıas, On the Hyperspaces Cn (X) of a Continuum X, Topology Appl., 109 (2001), 237–256. [19] S. Mac´ıas, On the Hyperspaces Cn (X) of a Continuum X, II, Topology Proc., 25 (2000), 255–276. [20] J. Munkres, Elements of Algebraic Topology, Addison–Wesley Publishing Company, Inc., Redwood City, California, 1984. [21] J. Munkres, Topology, second edition, Prentice Hall, Upper Saddle River, NJ, 2000. [22] S. B. Nadler, Jr., Hyperspaces of Sets, Monographs and Textbooks in Pure and Applied Math., Vol. 49, Marcel Dekker, New York, Basel, 1978. [23] S. B. Nadler, Jr., Continuum Theory: An Introduction, Monographs and Textbooks in Pure and Applied Math., Vol. 158, Marcel Dekker, New York, Basel, Hong Kong, 1992. [24] S. B. Nadler, Jr., Continuum Theory, class notes, Fall Semester of 1999, Math. 481, West Virginia University. [25] J. J. Rotman, An Introduction to the Theory of Groups, fourth edition, Graduate Texts in Mathematics, Vol. 148, Springer– Verlag, New York, Inc., 1995. Copyright © 2005 Taylor & Francis Group, LLC

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REFERENCES

[26] W. Rudin, Principles of Mathematical Analysis, third edition, International Series in Pure and Applied Mathematics, McGraw–Hill, Inc., New York, 1976.


Chapter 2 INVERSE LIMITS AND RELATED TOPICS

We present basic results about inverse limits and related topics. An excellent treatment of inverse limits, distinct from the one given here, was written by Professor W. Tom Ingram [10]. First, we present some basic results of inverse limits. Then a construction and a characterization of the Cantor set are given using inverse limits. With this technique, it is shown that a compactum is a continuous image of the Cantor set. Next, we show that inverse limits commute with the operation of taking finite products, cones and hyperspaces. We give some properties of chainable continua. We study circularly chainable and P–like continua. We end the chapter presenting several properties of universal maps and AH– essential maps.

2.1

Inverse Limits

We introduce inverse limits of metric spaces. We prove basic properties of inverse limits of compacta and continua.


69

70 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.1.1. Definition. Let {Xn }∞ n=1 be a countable collection of metric n+1 spaces. For each n ∈ IN, let fn : Xn+1 → Xn be a map. Then the sequence {Xn , fnn+1 } of metric spaces and maps is called an inverse sequence. The maps fnn+1 are called bonding maps. n fn−1

f2

fnn+1

1 X1 ←−X 2 ←− · · · ←− Xn−1 ←−Xn ←−Xn+1 ←− · · ·

n m+1 = fm ◦ ··· ◦ 2.1.2. Notation. If m, n ∈ IN and n > m, then fm n n fn−1 and fn = 1Xn , where 1Xn denotes the identity map on Xn .

2.1.3. Definition. If {Xn , fnn+1 } be an inverse sequence of metric spaces, then inverse limit of {Xn , fnn+1 }, denoted by lim{Xn , fnn+1 } ←− ∞ or X∞ , is the subspace of the topological product Xn given by n=1

lim{Xn , fnn+1 } ←−

∞ ∞ = (xn )n=1 ∈ Xn fnn+1 (xn+1 ) = xn n=1

for each n ∈ IN .

2.1.4. Remark. If {Xn , fnn+1 } is an inverse sequence of metric spaces, then, by Lemma 1.1.6, its inverse limit, X∞ , is a metric space.

2.1.5. Definition. Let {Xn , fnn+1 } be an inverse sequence of metric spaces with inverse limit X∞ . Recall that there exist maps ∞ πm : Xn → → Xm called projection maps. For each m ∈ IN, let n=1

fm = πm |X∞ ; then fm is a continuous function, since it is a restriction of a map (Lemma 1.1.6), and it is called a projection map.


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2.1.6. Remark. Let {Xn , fnn+1 } be an inverse sequence of metric spaces with inverse limit X∞ . Even though the maps πm are surjective, the projection maps fm are not, in general, surjective. However, if all the bonding maps fnn+1 are surjective, then the projection maps are surjective, and vice versa. Note that, by definition, for each n ∈ IN, fnn+1 ◦ fn+1 = fn , i.e., the following diagram fn+1 X∞ fn

Xn+1

−→ Xn n+1

fn

is commutative. 2.1.7. Definition. Let {Xn , fnn+1 } be an inverse sequence of metric spaces. For each m ∈ IN, let ∞ Sm = (xn )∞ Xn fkk+1 (xk+1 ) = xk , 1 ≤ k < m . n=1 ∈ n=1

2.1.8. Proposition. Let {Xn , fnn+1 } be an inverse sequence of compacta whose inverse limit is X∞ . Then: ∞ (1) For each m ∈ IN, Sm is homeomorphic to Xn . n=m

(2)

{Sm }∞ m=1

is a decreasing sequence of compacta and X∞ =

∞

Sm .

m=1

In particular, X∞ = ∅. (3) If for each n ∈ IN, Xn is a continuum, then X∞ is a continuum. Proof. We show (1). Let h : Sm →

∞

Xn be given by h((xn )∞ n=1 ) =

n=m

(xn )∞ n=m . Since πn ◦ h is continuous for each n ≥ m, by Theorem 1.1.9, h is continuous. ∞ ∞ Now, let g : Xn → Sm be given by g ((xn )∞ n=m ) = (yn )n=1 , n=m

where yn = xn if n ≥ m and yn = fnm (xm ) if n < m. Since πn ◦ g is continuous for each n ∈ IN, by Theorem 1.1.9, g is continuous. Copyright © 2005 Taylor & Francis Group, LLC

72 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS . Therefore, h is a Note that g ◦ h = 1Sm and h ◦ g = 1 ∞ n=m Xn homeomorphism. We see (2). By definition, Sm+1 ⊂ Sm for each m ∈ IN. By (1), ∞ each Sm is a compactum (Theorem 1.1.11). Clearly, X∞ = Sm . m=1

We prove (3). Note that (3) follows from (2) and Theorem 1.7.2. Q.E.D. 2.1.9. Proposition. Let {Xn , fnn+1 } be an inverse sequence of compacta with inverse limit X∞ . For each n ∈ IN, let Bn = fn−1 (Un ) | Un is an open subset of Xn . If B =

∞

Bn , then B is a basis for the topology of X∞ .

n=1

Proof. Since X∞ is a subspace of a topological product, a basic open subset of X∞ is of the form k

fn−1 (Unj ), j

j=1

where Unj is an open subset of Xnj . With out loss of generality, k −1 fnnjk (Unj ). we assume that nk = max{n1 , . . . , nk }. Let U = j=1

Then U is an open subset of Xnk and

k (U ) = fn−1 (fnnjk )−1 (Unj ) = fn−1 k k k

j=1

fnnjk

◦ fnk

−1

(Unj ) =

j=1 k

fn−1 (Unj ). j

j=1

Q.E.D. The next Corollary says that projection maps and bonding maps share the property of being open.


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2.1.10. Corollary. Let {Xn , fnn+1 } be an inverse sequence of compacta with surjective bonding maps, whose inverse limit is X∞ . Then the projection maps are open if and only if all the bonding maps are open. Proof. Suppose all the bonding maps are open. Let m ∈ IN. Let U be an open subset of X∞ , and let (xn )∞ n=1 ∈ U. By Proposition 2.1.9, there exist N ∈ IN and an open subset UN of XN such that −1 (xn )∞ n=1 ∈ fN (UN ) ⊂ U.

To see that fm is open, we consider two cases. Suppose first that N ≥ m. Then, by Remark 2.1.6, −1 xm = fm ((xn )∞ n=1 ) ∈ fm fN (UN ) = N N fN (fN−1 (UN )) = fm (UN ) ⊂ fm (U). fm

Hence, since the bonding maps are open, xm = fm ((xn )∞ n=1 ) is an interior point of fm (U). Therefore, fm (U) is open since (xn )∞ n=1 was an arbitrary point of U. Now suppose that m > N . Then, by Remark 2.1.6, −1 xm = fm ((xn )∞ ) ∈ f (U ) = f m N n=1 N −1 fm fm (fNm )−1 (UN ) = (fNm )−1 (UN ) ⊂ fm (U).

Hence, since the bonding maps are continuous, xm = fm ((xn )∞ n=1 ) is an interior point of fm (U). Therefore, fm (U) is open since (xn )∞ n=1 was an arbitrary point of U. Next, suppose all the projection maps are open. Let n ∈ IN, and let Un+1 be an open subset of Xn+1 . Since fnn+1 (Un+1 ) = −1 fn fn+1 (Un+1 ) (Remark 2.1.6) and the fact that the projection maps are continuous and open, fnn+1 (Un+1 ) is an open subset of Xn . Therefore, fnn+1 is open. Q.E.D. 2.1.11. Definition. Let X and Y be continua. A surjective map f: X → → Y is said to be monotone if f −1 (y) is connected for each y ∈Y. Copyright © 2005 Taylor & Francis Group, LLC

74 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS The following Lemma gives a very useful characterization of monotone maps. 2.1.12. Lemma. Let X and Y be compacta. If f : X → → Y is a surjective map, then f is monotone if and only if f −1 (C) is connected for each connected subset C of Y . Proof. Suppose f is monotone and let C be a subset of Y such that f −1 (C) is not connected. Then there exist two nonempty subsets A and B of X such that f −1 (C) = A ∪ B, ClX (A) ∩ B = ∅ and A ∩ ClX (B) = ∅. Observe that if y ∈ C and f −1 (y) ∩ A = ∅, then f −1 (y) ⊂ A (f −1 (y) is connected). Let M = {y ∈ C | f −1 (y) ⊂ A}. Hence, A = f −1 (M ). Similarly, if N = {y ∈ C | f −1 (y) ⊂ B}, then B = f −1 (N ). Clearly, C = M ∪ N . Suppose there exists y ∈ ClY (M ) ∩ N . Since y ∈ N , f −1 (y) ⊂ B. Since y ∈ ClY (M ), there exists a sequence {yn }∞ n=1 of points of M converging to y. This implies that f −1 (yn ) ⊂ A for every n ∈ IN. Let xn ∈ f −1 (yn ). Since X is a compactum, without loss of generality, we assume that {xn }∞ n=1 converges to a point x. Note that x ∈ ClX (A) and, by the continuity of f , f (x) = y. Thus, x ∈ ClX (A)∩f −1 (y) ⊂ ClX (A)∩B, a contradiction. Hence, ClY (M ) ∩ N = ∅. Similarly, M ∩ ClY (N ) = ∅. Therefore, C is not connected. The other implication is obvious. Q.E.D. The next result says that projection maps and bonding maps share the property of being monotone. 2.1.13. Proposition. Let {Xn , fnn+1 } be an inverse sequence of compacta with surjective bonding maps and whose inverse limit is X∞ . Then the bonding maps are monotone if and only if the projection maps are monotone. Proof. Suppose the bonding maps are monotone. Let m ∈ IN. We show fm is monotone. To this end, let xm ∈ Xm . For each n ∈ IN, let n −1 (fm ) (xm ) if n ≥ m; Wn = m if n < m. {fn (xm )} Copyright © 2005 Taylor & Francis Group, LLC

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Then {Wn , fnn+1 |Wn+1 } is an inverse sequence of continua. Hence, by Proposition 2.1.8 (3), W = lim{Wn , fnn+1 |Wn+1 } is a subcontinuum ←−

−1 of X∞ . We assert that W = fm (xm ). To see this, let (yn )∞ n=1 ∈ −1 fm (xm ). Then ym = xm , yn = fnm (xm ) for each n < m, and n −1 yn ∈ (fm ) (xm ) = Wn for each n > m. Hence, (yn )∞ n=1 ∈ W . −1 Therefore, fm (xm ) ⊂ W . −1 (xm ). Therefore, W = Since fm (W ) = Wm = {xm }, W ⊂ fm −1 fm (xm ) and fm is monotone. Now, suppose all the projection maps are monotone. Let m ∈ IN. m+1 We prove that fm is monotone. To see this, let xm ∈ Xm . By Rem+1 −1 −1 m+1 −1 mark 2.1.6, (fm ) (xm ) = fm+1 fm (xm ). Hence, (fm ) (xm ) is connected since the projection maps are monotone and continuous. m+1 Therefore, fm is monotone. Q.E.D.

2.1.14. Corollary. Let {Xn , fnn+1 } be an inverse sequence of locally connected continua with surjective bonding maps and whose inverse limit is X∞ . If all the bonding maps are monotone, then X∞ is locally connected. Proof. Let (xn )∞ n=1 ∈ X∞ , and let U be an open subset of X∞ such ∞ that (xn )n=1 ∈ U. By Proposition 2.1.9, there exist N ∈ IN and an −1 open subset UN of XN such that (xn )∞ n=1 ∈ fN (UN ) ⊂ U. Since XN is locally connected, there exists a connected open subset VN of XN −1 −1 such that xN ∈ VN ⊂ UN . Hence, (xn )∞ n=1 ∈ fN (VN ) ⊂ fN (UN ) ⊂ U. Since the bonding maps are monotone, by Proposition 2.1.13, fN is monotone. Hence, fN−1 (VN ) is a connected (Lemma 2.1.12) open subset of X∞ . Therefore, X∞ is locally connected. Q.E.D. The next Proposition says that when the projection maps are surjective, this property is no longer true for the restriction of such maps to proper closed subsets. 2.1.15. Proposition. Let {Xn , fnn+1 } be an inverse sequence with inverse limit X∞ . If Y is a proper closed subset of X∞ , then there exists N ∈ IN, such that fm (Y ) = Xm for each m ≥ N . Copyright © 2005 Taylor & Francis Group, LLC

76 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Proof. Let (xn )∞ n=1 ∈ X∞ \ Y . By Proposition 2.1.9, there exist N ∈ IN and an open subset UN of XN such that −1 (xn )∞ n=1 ∈ fN (UN ) ⊂ X∞ \ Y.

Hence, fN (Y ) ⊂ XN \ UN . Also, if m > N , then fm (Y ) ⊂ Xm \ (fNm )−1 (UN ). To see this, let zm ∈ fm (Y ). Then there exists ∞ m (yn )∞ n=1 ∈ Y such that fm ((yn )n=1 ) = zm . Since Y ⊂ X∞ , fN (zm ) = m −1 m −1 fN ((yn )∞ (fN ((yn )∞ (fN (Y )). n=1 ). Thus, zm ∈ (fN ) n=1 )) ⊂ (fN ) Since fN (Y ) ⊂ XN \ UN , fm (Y ) ⊂ (fNm )−1 (fN (Y )) ⊂ (fNm )−1 (XN \ UN ) ⊂ Xn \ (fNm )−1 (UN ). Therefore, fm (Y ) = Xm . Q.E.D. The following Proposition gives us the image of the inverse limit under a projection map in terms of the factor spaces and the bonding maps. 2.1.16. Proposition. Let {Xn , fnn+1 } be an inverse sequence of ∞ compacta whose inverse limit is X∞ . Then fn (X∞ ) = fnm (Xm ) m=n+1

for each n ∈ IN.

Proof. Let n ∈ IN. Since, for each m > n, fn = fnm ◦ fm , fn (X∞ ) = ∞ m m fn ◦ fm (X∞ ) ⊂ fn (Xm ). Therefore, fn (X∞ ) ⊂ fnm (Xm ). Now, let xn ∈

∞

m=n+1

fnm (Xm ). We show there exists an ele-

m=n+1

ment of X∞ whose nth coordinate is xn . To this end, let K = n−1 ∞ Xk × {xn } × Xk . Recall that, by Proposition 2.1.8 (2), k=1

X∞ =

∞

k=n+1

Sm and {Sm }∞ m=1 is a decreasing sequence. Let m0 ∈ IN.

m=1

Then K ∩ Sm0 = ∅. To see this, let m1 > max{n, m0 }. Since ∞ xn ∈ fnm (Xm ), (fnm1 )−1 (xn ) = ∅. Let xm1 ∈ (fnm1 )−1 (xn ), m=n+1


2.1. INVERSE LIMITS

77

m1 (xm1 ). Now, let (yk )∞ and let xm0 = fm k=1 ∈ 0

∞

Xk such that

k=1

ym1 = xm1 and yk = fkm1 (xm1 ) for each k ∈ {1, . . . , m1 − 1}. Note ∞ that (yk )∞ k=1 ∈ K ∩ Sm0 . Thus, the family {K} ∪ {Sm }m=1 has the finite intersection property. Therefore, K ∩ X∞ = ∅. Hence, there exists an element of X∞ whose nth coordinate is xn . Q.E.D. 2.1.17. Proposition. Let {Xn , fnn+1 } be an inverse sequence of compacta, whose inverse limit is X∞ . For each n ∈ IN, let An be a closed subset of Xn , and suppose that fnn+1 (An+1 ) ⊂ An . Then {An , fnn+1 |An+1 } is an inverse sequence and lim{An , fnn+1 |An+1 } = ←− ∞ fn−1 (An ). n=1

Proof. Clearly, {An , fnn+1 |An+1 } is an inverse sequence of compacta. −1 Observe that, since fnn+1 (An+1 ) ⊂ An , fn+1 (An+1 ) ⊂ fn−1 (An ) −1 ∞ for every n ∈ IN. Hence, {fn (An )}n=1 is a decreasing sequence of compacta. n+1 Let a = (an )∞ |An+1 }. Since fn (a) = an ∈ An , n=1 ∈ lim{An , fn ←− ∞ −1 a ∈ fn (A) for every n ∈ IN. Thus, a ∈ fn−1 (An ). Therefore, lim{An , fnn+1 |An+1 } ⊂ ←−

∞

n=1

fn−1 (An ).

n=1

Next, let b = (bn )∞ n=1 ∈

∞

fn−1 (An ). Then fn (b) = bn ∈ An for

n=1

each n ∈ IN. Hence, b ∈

∞

An . Since b ∈ X∞ , fnn+1 (bn+1 ) = bn

n=1

for every n ∈ IN. This implies that b ∈ lim{An , fnn+1 |An+1 }. Thus, ←− ∞ fn−1 (An ) ⊂ lim{An , fnn+1 |An+1 }. ←−

n=1

Therefore,

∞ n=1

fn−1 (An ) = lim{An , fnn+1 |An+1 }. ←−

Q.E.D.


78 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.1.18. Definition. Let {Xn , fnn+1 } be an inverse sequence of continua. Then {Xn , fnn+1 } is called an indecomposable inverse sequence provided that, for each n ∈ IN, whenever An+1 and Bn+1 are subcontinua of Xn+1 such that Xn+1 = An+1 ∪ Bn+1 , we have that fnn+1 (An+1 ) = Xn or fnn+1 (Bn+1 ) = Xn . The motivation of the name “indecomposable inverse sequence” is given in the following result: 2.1.19. Theorem. Let {Xn , fnn+1 } be an indecomposable inverse sequence whose inverse limit is X∞ . Then X∞ is an indecomposable continuum. Proof. By Proposition 2.1.8 (3), X∞ is a continuum. Now, suppose X∞ is decomposable. Then there exist two proper subcontinua A and B of X∞ such that X∞ = A ∪ B. By Proposition 2.1.15, there exists n ∈ IN such that if m ≥ n, then fm (A) = Xm and fm (B) = Xm . Since, by definition, the bonding maps are surjective, the projection maps are surjective too (Remark 2.1.6). Hence, Xn+2 = fn+2 (X∞ ) = fn+2 (A) ∪ fn+2 (B). This implies that n+2 n+2 n+2 Xn+1 = fn+1 (Xn+2 ) = fn+1 ◦ fn+2 (A) ∪ fn+1 ◦ fn+2 (B). n+2 n+2 By hypothesis fn+1 ◦ fn+2 (A) = Xn+1 or fn+1 ◦ fn+2 (B) = Xn+1 . Thus, fn+1 (A) = Xn+1 or fn+1 (B) = Xn+1 , contradicting the election of n. Therefore, X∞ is indecomposable. Q.E.D.

2.1.20. Proposition. Let {Xn , fnn+1 } be an inverse sequence of metric spaces whose inverse limit is X∞ . If A is a closed subset of X∞ , then the double sequence {fn (A), fnn+1 |fn+1 (A) } is an inverse sequence with surjective bonding maps and !∞ " (∗) lim fn (A), fnn+1 |fn+1 (A) = A = fn (A) ∩ X∞ . ←−


n=1

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79

Proof. By Remark 2.1.6, for each n ∈ IN, fnn+1 ◦ fn+1 = fn . It follows that {fn (A), fnn+1 |fn+1 (A) } is an inverse sequence with surjective bonding maps. Now we show (∗). First observe that " !∞ n+1 lim fn (A), fn |fn+1 (A) = fn (A) ∩ X∞ . ←−

n=1

Also observe that ! A⊂

∞

" fn (A) ∩ X∞ .

n=1

Thus, we need to show !

∞

" fn (A) ∩ X∞ ⊂ A.

n=1

! Let ε > 0, and let y = (yn )∞ n=1 ∈ N ∈ IN such that

∞

∞

" fn (A) ∩ X∞ . Now, let

n=1

1 < ε. n 2 n=N +1

∞ (n) ∈ Since for each n ∈ IN, yn ∈ fn (A), there exists a(n) = am m=1 (n) A such that fn a = yn . Now observe that ∞ 1 (N ) ρ(a , y) = dn an , yn = 2n n=1 ∞ 1 (N ) d an , yn < ε. n n 2 n=N +1 (N )

Since ε was arbitrary, y ∈ Cl(A) = A. Q.E.D. As a consequence of Propositions 2.1.17 and 2.1.20, we have the following Corollary:


80 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.1.21. Corollary. Let {Xn , fnn+1 } be an inverse sequence of compacta, whose inverse limit is X∞ . If A is a closed subset of X∞ , then ∞ A= fn−1 (fn (A)). n=1

2.1.22. Proposition. Let {Xn , fnn+1 } be an inverse sequence of metric spaces whose inverse limit is X∞ . If A and B are two closed subsets of X∞ , C = A ∩ B and Cn = fn (A) ∩ fn (B) for each n ∈ IN, then C = lim{Cn , fnn+1 |Cn+1 }. ←−

n+1 |Cn+1 }. Then, by defiProof. Let x = (xn )∞ n=1 ∈ lim{Cn , fn ←−

nition, xn ∈ Cn = fn (A) ∩ fn (B) for each n ∈ IN. Hence, x ∈ lim{fn (A), fnn+1 |fn+1 (A) } = A and x ∈ lim{fn (B), fnn+1 |fn+1 (B) } = B ←−

←−

(by (∗) of Proposition 2.1.20). Thus, x ∈ C. Now, let y = (yn )∞ n=1 ∈ C = A ∩ B. Then for each n ∈ IN, yn ∈ fn (A) and yn ∈ fn (B). Thus, yn ∈ Cn for each n ∈ IN. Therefore, y ∈ lim{Cn , fnn+1 |Cn+1 }. ←− Q.E.D. 2.1.23. Definition. Let {Xn , fnn+1 } be an inverse sequence of arcs (i.e., for each n ∈ IN, Xn is an arc) with surjective bonding maps. Then the inverse limit, X∞ , of {Xn , fnn+1 } is called an arc–like continuum. Propositions 2.1.20 and 2.1.22 have the following Corollaries: 2.1.24. Corollary. If X is an arc–like continuum, then each nondegenerate subcontinuum of X is arc–like. Proof. Let X be an arc–like continuum, and let A be a nondegenerate subcontinuum of X. Since X is an arc–like continuum, there exists an inverse sequence {Xn , fnn+1 } of arcs such that X = lim{Xn , fnn+1 }. By (∗) of Propositions 2.1.20, we have that ←−

A = lim{fn (h(A)), fnn+1 |fn+1 (h(A)) }. Therefore, A is an arc–like con←− tinuum. Q.E.D.


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2.1.25. Corollary. If {Xn , fnn+1 } is an inverse sequence of unicoherent continua, with surjective bonding maps, whose inverse limit is X∞ , then X∞ is a unicoherent continuum. Proof. Let A and B be two subcontinua of X∞ such that X∞ = A ∪ B. Since the bonding maps are surjective, by Remark 2.1.6, the projection maps are surjective too. Hence, for each n ∈ IN, Xn = fn (A) ∪ fn (B). By the unicoherence of the spaces, fn (A) ∩ fn (B) is connected for every n ∈ IN. Thus, by Propositions 2.1.20 and 2.1.22, {fn (A) ∩ fn (B), fnn+1 |fn+1 (A)∩fn+1 (B) } is an inverse sequence of continua whose inverse limit is a continuum (Proposition 2.1.8 (3)). Since A ∩ B = lim{fn (A) ∩ fn (B), fnn+1 |fn+1 (A)∩fn+1 (B) } (Proposi←−

tion 2.1.22) A ∩ B is connected. Therefore, X∞ is unicoherent. Q.E.D. 2.1.26. Corollary. Let {Xn , fnn+1 } be an inverse sequence of hereditarily unicoherent continua. If X∞ is the inverse limit of {Xn , fnn+1 }, then X∞ is a hereditarily unicoherent continuum. Proof. Let A and B be two subcontinua of X∞ . If A ∩ B = ∅, then A∩B is connected. Thus, suppose that A∩B = ∅. Then for each n ∈ IN, fn (A) and fn (B) are two subcontinua of Xn such that fn (A) ∩ fn (B) = ∅. Since each Xn is hereditarily unicoherent, fn (A)∩fn (B) is connected. Hence, by Propositions 2.1.20 and 2.1.22, {fn (A) ∩ fn (B), fnn+1 |fn+1 (A)∩fn+1 (B) } is an inverse sequence of continua whose inverse limit is a continuum (Proposition 2.1.8 (3)). Since A ∩ B = lim{fn (A) ∩ fn (B), fnn+1 |fn+1 (A)∩fn+1 (B) } (Proposition 2.1.22) A ∩ B ←− is connected. Therefore, X∞ is hereditarily unicoherent. Q.E.D.

2.1.27. Remark. Note that in Corollary 2.1.26 we do not require the bonding maps to be surjective.

2.1.28. Corollary. Each arc–like continuum is hereditarily unicoherent. Copyright © 2005 Taylor & Francis Group, LLC

82 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Proof. The Corollary follows from the easy fact that an arc is hereditarily unicoherent and from Corollary 2.1.26. Q.E.D.

2.1.29. Corollary. If X is an arc–like continuum, then X does not contain a simple closed curve.

2.1.30. Remark. Let {Xn , fnn+1 } be an inverse sequence of arcs, with surjective bonding maps, whose inverse limit is X∞ . Note that, in this case, Sm is homeomorphic to the Hilbert cube Q (be∞ ing a countable product of arcs). Since X∞ = Sm (Proposim=1

tion 2.1.8 (2)), X∞ may be written as a countable intersection of Hilbert cubes.

The next Theorem tells us a way to define a map from a metric space into an inverse limit of compacta. 2.1.31. Theorem. Let Y be a metric space. Let {Xn , fnn+1 } be a sequence of compacta. If for each n ∈ IN, there is a map hn : Y → Xn such that fnn+1 ◦ hn+1 = hn , then there exists a map h∞ : Y → X∞ such that fn ◦ h∞ = hn for each n ∈ IN. The map h∞ is called induced map, and it is denoted, also, by lim{hn }. ←−

Proof. Let h∞ : Y → X∞ be given by h∞ (y) = (hn (y))∞ n=1 . Since fnn+1 ◦ hn+1 = hn for each n ∈ IN, h∞ is well defined. Clearly, fn ◦ h∞ = hn . Hence, by Theorem 1.1.9, h∞ is continuous. Q.E.D. 2.1.32. Theorem. Let Y be a metric space. Let {Xn , fnn+1 } be a sequence of compacta. Suppose that for each n ∈ IN, there is a map hn : Y → Xn such that fnn+1 ◦ hn+1 = hn . If hm is one–to–one for some m ∈ IN, then the induced map h∞ is one–to–one. Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Suppose hm is one–to–one for some m ∈ IN. Let y, y ∈ Y such that y = y . Since hm is one–to–one, hm (y) = hm (y ). Hence, h∞ (y) = h∞ (y ). Therefore, h∞ is one–to–one. Q.E.D. 2.1.33. Theorem. Let Y be a metric space. Let {Xn , fnn+1 } be a sequence of compacta. Suppose that for each n ∈ IN, there is a map hn : Y → Xn such that fnn+1 ◦ hn+1 = hn . If each hn is surjective, then h∞ (Y ) is dense in X∞ . In particular, if Y is compact, then h∞ is surjective. Proof. Let fn−1 (Un ) be a basic open set of X∞ , where Un is an open subset of Xn (Proposition 2.1.9). Since hn is surjective, there exists y ∈ Y such that hn (y) ∈ Un . Hence, h∞ (y) ∈ fn−1 (Un ). Therefore, h∞ (Y ) is dense in X∞ . Q.E.D. 2.1.34. Definition. For each n ∈ IN, let Xn = S 1 , and let fnn+1 : Xn+1 → → Xn be given by fnn+1 (z) = z 2 (complex number multiplication). Let Σ2 = lim{Xn , fnn+1 }. Then Σ2 is called the dyadic ←− solenoid. Note that, by Theorem 2.1.19, Σ2 is an indecomposable continuum. The following Example shows that an induced map may not be surjective. For each n ∈ IN, 2.1.35. Example. Let Σ2 be the dyadic solenoid. 2πt → S 1 be given by hn (t) = exp n−1 . Let n ∈ IN. let hn : IR → 2 2 2πt 2πt n+1 n+1 = exp = exp Then fn ◦ hn+1 (t) = fn n n 2 2 2πt exp n−1 = hn (t). Hence, fnn+1 ◦ hn+1 = hn for each n ∈ IN. By 2 Theorem 2.1.31, there exists the induced map h∞ : IR → Σ2 . Observe that h∞ is not surjective since h∞ (IR) is an arcwise connected subset of Σ2 . But, since Σ2 is indecomposable, it is not arcwise connected (This follows easily from 11.15 of [18]). Copyright © 2005 Taylor & Francis Group, LLC

84 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS The next two Propositions present some expected results. 2.1.36. Proposition. Let {Xn }∞ n=1 be a sequence of compacta such that Xn+1 ⊂ Xn for each n ∈ IN. If fnn+1 : Xn+1 → Xn is the inclu∞ n+1 Xn . sion map, then lim{Xn , fn } is homeomorphic to ←−

n=1

n+1 Proof. Note that if (xn )∞ }, then xn = x1 for n=1 ∈ lim{Xn , fn ←− ∞ every n ∈ IN and x1 ∈ Xn . n=1

For each n ∈ IN, let hn :

∞

Xn → Xn be given by hn (x) = x,

n=1

i.e., hn is the inclusion map. Clearly fnn+1 ◦ hn+1 = hn for every n ∈ IN. By Theorem 2.1.31, there exists the induced map ∞ h∞ : Xn → lim{Xn , fnn+1 }. Since h1 is one–to–one, h∞ is one– n=1

←−

n+1 to–one (Theorem 2.1.32). Now, if (xn )∞ }, then n=1 ∈ lim{Xn , fn ←− ∞ ∞ h∞ (x1 ) = (xn )n=1 . Hence, h∞ is surjective. Since Xn is comn=1

pact, h∞ is a homeomorphism. Q.E.D. 2.1.37. Proposition. Let {Xn , fnn+1 } be an inverse sequence of compacta whose inverse limit is X∞ . If for each n ∈ IN, Xn is homeomorphic to a compactum Y and all the bonding maps are homeomorphisms, then X∞ is homeomorphic to Y . Proof. Let h1 : Y → X1 be any homeomorphism. For n ≥ 2, let hn+1 = (fnn+1 )−1 ◦ hn . Observe that, by definition, fnn+1 ◦ hn+1 = hn for each n ∈ IN. Then, by Theorem 2.1.31, there exists the induced map h∞ : Y → X∞ . Since each hn is a homeomorphism and Y is compact, by Theorems 2.1.32 and 2.1.33, h∞ is a homeomorphism. Q.E.D. The next Theorem says that certain type of subsequences of an inverse sequence converge to the same limit.


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2.1.38. Theorem. Let {Xn , fnn+1 } be an inverse sequence of compacta, with surjective bonding maps, whose inverse limit is X∞ , and let {m(n)}∞ n=1 be an increasing subsequence of IN. If Y∞ = m(n+1) n+1 for each lim{Yn , gn }, where Yn = Xm(n) and gnn+1 = fm(n) ←− n ∈ IN, then Y∞ is homeomorphic to X∞ . Proof. For each k ∈ IN, let hk : X∞ → Yk be given by hk ((xn )∞ n=1 ) = k+1 xm(k) , i.e., hk = fm(k) . Hence, hk is continuous and gk ◦hk+1 = hk . Then, by Theorem 2.1.31, there exists the induced map h∞ : X∞ → Y∞ . Since the bonding maps are surjective, by Remark 2.1.6, the projections are surjective. In particular, each hk is surjective. Thus, h∞ is surjective (Theorem 2.1.33). To see h∞ is one–to–one, it is ∞ enough to observe that if (xn )∞ n=1 and (xn )n=1 are two distinct points of X∞ , then there exists N ∈ IN such that xn = xn for each n ≥ N . Q.E.D. 2.1.39. Notation. We use the notation {Xn , fnn+1 }∞ n=N to denote n+1 the inverse subsequence obtained from {Xn , fn } by removing the first N − 1 factor spaces. Next, we present some consequences of Theorem 2.1.38. First, we need the following definition: 2.1.40. Definition. A continuum X is said to be a triod, provided that there exists a subcontinuum M of X such that X \ M = K1 ∪ K2 ∪ K3 , where each Kj = ∅, j ∈ {1, 2, 3}, and they are mutually separated, i.e., ClX (Kj ) ∩ K = ∅, j, ∈ {1, 2, 3} and j = .

2.1.41. Corollary. If X is an arc–like continuum, then X does not contain a triod. Proof. Since X is an arc–like continuum, there exists an inverse sequence {Xn , fnn+1 } of arcs whose inverse limit is X. Suppose Y is a triod contained in X. Then there exists a subcontinuum M of Y such that Y \ M = K1 ∪ K2 ∪ K3 , where each Kj = ∅, j ∈ {1, 2, 3}, and they are mutually separated. Note that M ∪ Kj is a continuum (Lemma 1.7.18). Copyright © 2005 Taylor & Francis Group, LLC

86 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Let n ∈ IN. Then fn (Y ) = fn (M ∪ K1 ∪ K2 ∪ K3 ) = fn (M ) ∪ fn (K1 ) ∪ fn (K2 ) ∪ fn (K3 ). Since Xn is an arc, there exists jn ∈ {1, 2, 3} such that fn (M ∪ Kjn ) ⊂ fn (M ∪ K1 ) ∪ fn (M ∪ K2 ), 1 , 2 ∈ {1, 2, 3} \ {jn } and 1 = 2 . Since the set of positive integers is infinite and we only have three choices, by Theorem 2.1.38, we assume, without loss of generality, that fn (M ∪ K3 ) ⊂ fn (M ∪ K1 ) ∪ fn (M ∪ K2 ) for each n ∈ IN. Note that, by (∗) of Proposition 2.1.20, M ∪ Kj = lim{fn (M ∪ Kj ), fnn+1 |fn+1 (M ∪Kj ) }. ←−

Since fn (M ∪ K3 ) ⊂ fn (M ∪ K1 ) ∪ fn (M ∪ K2 ) for each n ∈ IN, it follows that M ∪ K3 ⊂ (M ∪ K1 ) ∪ (M ∪ K2 ) = M ∪ K1 ∪ K2 , a contradiction. Therefore, X does not contain a triod. Q.E.D. 2.1.42. Definition. Let {Xn , fnn+1 } be an inverse sequence of simple closed curves with surjective bonding maps. Then the inverse limit, X∞ , of such an inverse sequence is called a circle–like continuum.

2.1.43. Corollary. If X is a circle–like continuum, then each nondegenerate proper subcontinuum of X is arc–like. Proof. Since X is circle–like, there exists an inverse sequence {Xn , fnn+1 } of simple closed curves whose inverse limit is X. Let Z be a nondegenerate proper subcontinuum of X. Note that by (∗) of Proposition 2.1.20, Z = lim{fn (Z), fnn+1 |fn+1 (Z) }. By ←−

Proposition 2.1.15, there exists N ∈ IN such that fn (Z) = Xn for each n ≥ N . Hence, fn (Z) is an arc for every n ≥ N . Applying Theorem 2.1.38, we have that Z is homeomorphic to lim{fn (Z), fnn+1 |fn+1 (Z) }∞ n=N . Therefore, Z is an arc–like continuum. ←− Q.E.D. 2.1.44. Corollary. If X is a circle–like continuum, then X does not contain a triod. Copyright © 2005 Taylor & Francis Group, LLC

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2.1.45. Theorem. Let {Xn , fnn+1 } be an inverse sequence of compacta whose inverse limit is X∞ . If for each n ∈ IN, Xn has at most k components, for some k ∈ IN, then X∞ has at most k components. Proof. Suppose X∞ has more than k components. Let C1 , . . . , Ck+1 be k + 1 distinct components of X∞ . For each n ∈ IN, let us consider the set {fn (C1 ), . . . , fn (Ck+1 )}. Since Xn has at most k components, there exist in , jn ∈ {1, . . . , k + 1} such that fn (Cin ) and fn (Cjn ) are contained in the same component of Xn . Since {1, . . . , k + 1} is a finite set, there exist i0 , j0 ∈ {1, . . . , k + 1} and a subsequence {n }∞ =1 of IN such that for each ∈ IN, fn (Ci0 ) and fn (Cj0 ) are contained in the same component Yn of Xn . Note that, by Theorem 2.1.38 and Proposition 2.1.20, Ci0 is homeomorphic n to Ci0 = lim{fn (Ci0 ), fn+1 |fn+1 (Ci0 ) } and Cj0 is homeomorphic to ←−

Cj 0 = lim{fn (Cj0 ), fn+1 |fn+1 (Cj0 ) }. n

←−

n

Let Y∞ = lim{Yn , fn+1 |Yn+1 }. Then, by Proposition 2.1.8 and ←−

Theorem 2.1.38, Y∞ is a subcontinuum of X∞ = lim{Xn , fn+1 }, n

←−

which is homeomorphic to X∞ (Theorem 2.1.38). Note that Ci0 ∪ Cj 0 ⊂ Y∞ , a contradiction. Therefore, X∞ has at most k components. Q.E.D. The next Theorem gives us a way to define a map between inverse limits. 2.1.46. Theorem. Let {Xn , fnn+1 } and {Yn , gnn+1 } be inverse sequences of compacta, whose inverse limit are X∞ and Y∞ , respectively. If for each n ∈ IN, there is a map kn : Xn → Yn such that kn ◦ fnn+1 = gnn+1 ◦ kn+1 , then there exists a map k∞ : X∞ → Y∞ such that gn ◦ k∞ = kn ◦ fn . The map k∞ is called induced map, and it is denoted, also, by lim{kn }. ←−

Proof. For each n ∈ IN, let hn : X∞ → Yn be given by hn = kn ◦ fn . Note that gnn+1 ◦ hn+1 = hn . Hence, by Theorem 2.1.31, there exists the induced map h∞ : X∞ → Y∞ . Let k∞ = h∞ . Then gn ◦ k∞ = gn ◦ h∞ = hn = kn ◦ fn . Q.E.D.


88 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.1.47. Theorem. Let {Xn , fnn+1 } and {Yn , gnn+1 } be inverse sequences of compacta, whose inverse limit are X∞ and Y∞ , respectively. Suppose that for each n ∈ IN, there is a map kn : Xn → Yn such that kn ◦ fnn+1 = gnn+1 ◦ kn+1 . If all the maps kn are one–to–one, then the induced map k∞ is one–to–one. ∞ Proof. Let (x )∞ =1 and (x )=1 be two distinct points of X∞ . Then there exists m ∈ IN such that xm = xm . Since km is one–to–one, ∞ ∞ km (xm ) = km (xm ). Hence, k∞ ((x )∞ =1 ) = (kn ◦ fn ((x )=1 ))n=1 = ∞ ∞ ∞ (kn (xn ))n=1 = (kn (xn ))n=1 = k∞ ((x )=1 ). Therefore, k∞ is one–to– one. Q.E.D.

2.1.48. Theorem. Let {Xn , fnn+1 } and {Yn , gnn+1 } be inverse sequences of compacta, with surjective bonding maps, whose inverse limit are X∞ and Y∞ , respectively. Suppose that for each n ∈ IN, there is a map kn : Xn → Yn such that kn ◦ fnn+1 = gnn+1 ◦ kn+1 . If all the maps kn are surjective, then the induced map k∞ is surjective. −1 (Vm ) be a basic open set in Y∞ , where Vm is an open Proof. Let gm subset of Ym (Proposition 2.1.9). Since km is surjective, there exists zm ∈ Xm such that km (zm ) ∈ Vm . Since for each n ∈ IN, fnn+1 is surjective, the projection maps, fn , are surjective (Remark 2.1.6). ∞ Hence, there exists (xn )∞ n=1 ∈ X∞ such that fm ((xn )n=1 ) = zm . ∞ Then gm ◦ k∞ ((xn )∞ n=1 ) = km ◦ fm ((xn )n=1 ) = km (zm ) ∈ Vm . Thus, ∞ −1 k∞ ((xn )n=1 ) ∈ gm (Vm ). Therefore, k∞ (X∞ ) is dense in Y∞ . Since X∞ is compact, k∞ (X∞ ) = Y∞ . Q.E.D.

2.1.49. Theorem. Let {Xn , fnn+1 } and {Yn , gnn+1 } be inverse sequences of compacta, whose inverse limit are X∞ and Y∞ , respectively. Suppose that for each n ∈ IN, kn : Yn → Xn and hn+1 : Xn+1 → Yn are surjective maps such that kn ◦ hn+1 = fnn+1 and hn+1 ◦ kn+1 = gnn+1 . Then there exists a map h∞ : X∞ → Y∞ such that −1 h∞ = k∞ , where k∞ = lim{kn }. In particular, k∞ is a homeomor←− phism. Also, X∞ and Y∞ are homeomorphic. ∞ Proof. Let h∞ : X∞ → Y∞ be given by h∞ ((xn )∞ n=1 ) = (hn (xn ))n=2 . By Theorem 1.1.9, h∞ is continuous.


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Now, let (yn )∞ n=1 ∈ Y∞ . Then ∞ ∞ h∞ ◦ k∞ ((yn )∞ n=1 ) = h∞ ((kn (yn ))n=1 ) = (hn kn (yn ))n=2 n ∞ = (gn−1 (yn ))∞ n=2 = (yn )n=1 ,

i.e., h∞ ◦ k∞ = 1Y∞ . Next, let (xn )∞ n=1 ∈ X∞ . Then ∞ ∞ k∞ ◦ h∞ ((xn )∞ n=1 ) = k∞ ((hn (xn ))n=2 ) = (kn−1 hn (xn ))n=2 n ∞ = (fn−1 (xn ))∞ n=2 = (xn )n=1 ,

i.e., k∞ ◦ h∞ = 1X∞ . −1 . Therefore, h∞ = k∞ Q.E.D. Next, we prove a Theorem due to Anderson and Choquet, which tells us when we can embed an inverse limit in a compactum. 2.1.50. Theorem. Let X be a compactum, with metric d. Let {Xn , fnn+1 } be an inverse sequence of closed subsets of X with surjective bonding maps. Assume: (1) For each ε > 0, there exists k ∈ IN such that for all x ∈ Xk , ! " j −1 diam (x) < ε; fk j≥k

(2) For each n ∈ IN and each δ > 0, there exists δ > 0 such that whenever j > n and x, y ∈ Xj , then d(x, y) > δ . !

" ∞ Xm . In Cl Then lim{Xn , fnn+1 } is homeomorphic to ←−

n=1

m≥n

particular, if Xn ⊂ Xn+1 for each n ∈ IN, then lim{Xn , fnn+1 } is ←−

∞ homeomorphic to Cl Xn . n=1

Proof. Let X∞ = lim{Xn , fnn+1 }. If x = (xn )∞ n=1 ∈ X∞ , then, by ←−

∞ (1), we have that (xn )∞ n=1 is a Cauchy sequence in X. Thus, (xn )n=1


90 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS converges to a point in X, which we call h(x). Hence, we have defined a function h : X∞ → X. We see that h is an embedding. First, we show h is continuous. Let x = (xn )∞ n=1 ∈ X∞ , and let ε > 0. Now, choose k ∈ IN as guaranteed by (1). Let = k + 1, and let V = f−1 (Vεd (h(x)) ∩ X ). Since x ∈ X∞ and k satisfies (1), we have from the definition of h that d(x , h(x)) < ε. Hence, x ∈ Vεd (h(x)) ∩ X . Thus, x ∈ V . Next, observe that, since Vεd (h(x)) ∩ X is open in X , V is open in X∞ . Now, let y = (yn )∞ n=1 ∈ V . For the same reasons as for x, d(y , h(y)) < ε. Since y ∈ V , y ∈ Vεd (h(x)). Hence, d(h(y), h(x)) ≤ d(h(y), y ) + d(y , h(x)) < 2ε. Therefore, h is continuous. ∞ To prove h is one–to–one, let x = (xn )∞ n=1 and y = (yn )n=1 be two distinct points of X∞ . Then there exists n ∈ IN, such that d(xn , yn ) xn = yn . Let δ = . Then, by (2), there exists δ > 0 such 2 that whenever j > n, then, since d(fnj (xj ), fnj (yj )) = d(xn , yn ) = 2δ > δ

= h(y). " we have that d(xj , yj ) > δ . Therefore, h(x) !

∞ Xm . For conCl It remains to see that h(X∞ ) = n=1

m≥n

venience, let Z=

∞ n=1

!

Cl

" Xm

.

m≥n

It is clear, from the definition of h, that h(X∞ ) ⊂ Z. To show h(X∞ ) = Z, we prove that h(X∞ ) is dense in Z. For this purpose, let z ∈ Z and let ε > 0. Let k ∈ IN be as guaranteed by (1). Since z ∈ Z, there exists m ≥ k such that d(z, p) < ε for some p ∈ Xm . Since the bonding maps are surjective, by Remark 2.1.6, there exists x = (xn )∞ n=1 ∈ X∞ such that fm (x) = ∞ fm ((xn )n=1 ) = xm = p. Copyright © 2005 Taylor & Francis Group, LLC

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Note that, since m ≥ k, condition (1) holds for m in place of k. Hence, it follows that d(z, h(x)) < 2ε. Thus, h(X∞ ) is dense in Z. Therefore, since X∞ is compact, h(X∞ ) = Z. Q.E.D. We end this section proving that any continuum is homeomorphic to an inverse limit of polyhedra. 2.1.51. Theorem. Each continuum is homeomorphic to an inverse limit of polyhedra. Proof. Let X be a continuum. By Theorem 1.1.16, we assume that X is contained in the Hilbert cube Q. For each n ∈ IN, let dn denote the Euclidean metric on [0, 1]n , and let πn : Q → [0, 1]n and πnn+1 : [0, 1]n+1 → [0, 1]n be the projection maps. By Lemma 1.7.31, there exists a polyhedron P1 in [0, 1] such that π1 (X) ⊂ Int(P1 ) ⊂ P1 ⊂ V1d1 (π1 (X)). Note that π2 (X) ⊂ (π12 )−1 (Int(P1 )). Hence, by Lemma 1.7.31, there exists a polyhedron P2 in [0, 1]2 such that π2 (X) ⊂ Int(P2 ) ⊂ P2 ⊂ V d1 2 (π2 (X)) and π12 (P2 ) ⊂ Int(P1 ). Continue in this way 2

to define a sequence, {Pn }∞ n=1 , of polyhedra such that πn (X) ⊂ Int(Pn ) ⊂ Pn ⊂ V d1n (πn (X)) and πnn+1 (Pn+1 ) ⊂ Int(Pn ). Hence, n

{Pn , πnn+1 |Pn+1 } is an inverse sequence. Let P∞ = lim{Pn , πnn+1 |Pn+1 }. ←− We show that X is homeomorphic to P∞ . For each n ∈ IN, let hn : X → Pn by given by hn = πn |X . Observe that, by construction, if x ∈ X, then πnn+1 (πn+1 (x)) = πn (x) for each n ∈ IN. Thus, πnn+1 (hn+1 (x)) = hn (x) for every n ∈ IN. Hence, the sequence {hn }∞ n=1 induces a map h∞ : X → P∞ (Theorem 2.1.31). To see h∞ is one–to–one, let x and x be two distinct points of X. Then there exists n ∈ IN, such that πn (x) = πn (x ), i.e., hn (x) = hn (x ). Thus, h∞ (x) = h∞ (x ). Therefore, h∞ is one–to–one. Clearly, if z1 and z2 are two points of [0, 1]k+1 , then dk (πkk+1 (z1 ), πkk+1 (z2 )) ≤ dk+1 (z1 , z2 ). Now, let p = (pk )∞ k=1 ∈ P∞ , and let n ∈ IN. Then, by Proposi∞ tion 2.1.16, pn ∈ πnm (Pm ). Since hn (X) = πnn+1 ◦ hn+1 (X), m=n+1


92 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS dn (pn , hn (X)) ≤ dn+1 (pn+1 , hn+1 (X)). Hence, dn (pn , hn (X)) ≤ dm (pm , hm (X)) ≤

1 m

for each m > n. Therefore, pn ∈ πn (X) = hn (X). Finally, suppose h∞ is not surjective. Then there exists a point p = (pk )∞ k=1 ∈ P∞ \h∞ (X). By Proposition 2.1.9, there exist N ∈ IN −1 and an open subset UN of PN such that (pk )∞ k=1 ∈ (πN |P∞ ) (UN ) ⊂ P∞ \ h∞ (X). Hence, pN ∈ UN \ hN (X), a contradiction with the preceding paragraph. Therefore, h∞ is surjective. Q.E.D. Note that in Theorem 2.1.51 the bonding maps are not surjective. The following result tells us that each continuum can be written as an inverse limit of polyhedra with surjective bonding maps; a proof of this may be found in Theorem 2 of [12]. 2.1.52. Theorem. Each continuum is homeomorphic to an inverse limit of polyhedra with surjective bonding maps.

2.2

Inverse Limits and the Cantor Set

The purpose of this section is to characterize the Cantor set C (Example 1.6.4) and to show that every compactum is a continuous image of C. We begin by showing that C is an inverse limit of finite discrete spaces. First, we need the following definition: 2.2.1. Definition. If x ∈ IR, then [x] = max{n ∈ IN | n ≤ x} is called the greatest integer function.


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93

2.2.2. Theorem. For each n ∈ IN, let Xn = {0, 1, . . . , 2n −1} with n+1 n+1 the # x $ discrete topology, and define fn : Xn+1 → Xn by fn (x) = . If X∞ = lim{Xn , fnn+1 }, then X∞ is homeomorphic to C. ←− 2 Proof. For each n ∈ IN, let hn : C → Xn be given by hn (x) = k if x ∈ In,k (Example 1.6.4). Clearly, hn is surjective for each n ∈ IN. To see hn is continuous, it is enough to observe that for each k ∈ Xn , h−1 n (k) = In,k ∩ C, which is an open subset of C. Let n ∈ IN. Observe that, by the construction of C, In+1,2 ∪ I n+1,2+1 ⊂ In, . Hence, if x ∈ C and hn+1 (x) = k, then hn (x) =

k hn+1 (x) . Consequently, if x ∈ C, then fnn+1 ◦ hn+1 (x) = = 2 2 hn (x). Therefore, fnn+1 ◦hn+1 = hn . Thus, by Theorem 2.1.31, there exists the induced map h∞ : C → → X∞ . By Theorem 2.1.33, h∞ is surjective. To see h∞ is one–to–one, let x and x be two distinct points of C. Then there exist n ∈ IN and j, k ∈ {0, . . . , 2n − 1} such that j = k, x ∈ In,j and x ∈ In,k . Hence, hn (x) = hn (x ). Thus, h∞ is one–to–one. Therefore, h∞ is a homeomorphism. Q.E.D. Now, we present some results about totally disconnected compacta. 2.2.3. Proposition. If X is a totally disconnected compactum, then the family of all open and closed subsets of X forms a basis for the topology of X. Proof. Let U be an open subset of X, and let x ∈ U . Since X is totally disconnected, no connected subcontinuum of X intersects both {x} and X \ U . By Theorem 1.6.8, there exist two disjoint open and closed subsets X1 and X2 of X such that X = X1 ∪ X2 , {x} ⊂ X1 and X \ U ⊂ X2 . Hence, x ∈ X1 ⊂ U . Q.E.D. 2.2.4. Definition. Let X be a metric space. If U is a family of subsets of X, then the mesh of U, denoted by mesh(U), is defined by: mesh(U) = max{diam(U ) | U ∈ U}. Copyright © 2005 Taylor & Francis Group, LLC

94 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.2.5. Definition. Let X be a metric space. If U and V are two coverings of X, then V is a refinement of U provided that for each V ∈ V, there exists U ∈ U such that V ⊂ U . If V refines U, we write V ≺ U. 2.2.6. Theorem. If X is a totally disconnected compactum, then there exists a sequence {Un }∞ n=1 of finite coverings of X such that for each n ∈ IN: (i) Un+1 ≺ Un ; 1 (ii) mesh(Un ) < ; n (iii) if U ∈ Un , then U is open and closed in X; and (iv) if U, V ∈ Un and U = V , then U ∩ V = ∅. Proof. We construct U1 first. By Proposition 2.2.3, for each x ∈ X, there exists an open and closed subset Ux of X such that x ∈ Ux and diam(Ux ) < 1. Hence, {Ux | x ∈ X} is an open cover of X. Since k1 X is compact, there exist x11 , . . . , x1k1 ∈ X such that X = Ux1j . j=1

The sets Ux11 , . . . , Ux1k1 are not necessarily pairwise disjoint. But this can be fixed. Let U11 = Ux11 and for j ∈ {2, . . . , k1 }, let j−1 U1 . Then U1 = {U11 , . . . , U1k1 } is a finite family of U1j = Ux1j \ =1

pairwise disjoint open and closed subsets of X covering X. Let λ1 be a Lebesgue number of the open covering U1 (Theo1 rem 1.6.6). Without loss of generality, we assume that λ1 < . By 2 Proposition 2.2.3, for each x ∈ X, there exists an open and closed subset Vx of X such that x ∈ Vx and diam(Vx ) < λ1 . Hence, the family {Vx | x ∈ X} forms an open cover of X. Since X is compact, k2 there exist x21 , . . . , x2k2 ∈ X such that X = Vx2j . Again, the j=1

elements of {Vx21 , . . . , Vx2k2 } may not be pairwise disjoint. Hence, j−1 V2 . let U21 = Vx21 and for each j ∈ {2, . . . , k2 }, let U2j = Vx2j \ =1

Then U2 = {U21 , . . . , U2k2 } is a finite family of pairwise disjoint open Copyright © 2005 Taylor & Francis Group, LLC


95

and closed subsets of X covering X. Note that, by construction, U2 ≺ U1 . Proceeding by induction, we obtain the result. Q.E.D.

2.2.7. Theorem. If X is a totally disconnected compactum, then there exists an inverse sequence {Xn , fnn+1 } of finite discrete spaces such that lim{Xn , fnn+1 } is homeomorphic to X. ←−

Proof. Let {Un }∞ n=1 be a sequence of finite coverings given by Theorem 2.2.6. For each n ∈ IN, let Xn = Un . Give Xn the discrete topology. We define the bonding maps fnn+1 as follows: If Un+1 ∈ Xn+1 , then fnn+1 (Un+1 ) = Un , where Un is the unique element of Un = Xn such that Un+1 ⊂ Un . Hence, {Xn , fnn+1 } is an inverse sequence. Let X∞ = lim{Xn , fnn+1 }. ←−

For each n ∈ IN, let hn : X → Xn be given by hn (x) = U , where U is the unique element of Xn such that x ∈ U . Let n ∈ IN. Note that, by construction, hn is surjective and fnn+1 ◦ hn+1 = hn . Also note that if U ∈ Xn , then h−1 n (U ) = U . Then hn is continuous. By Theorem 2.1.31, there exists the induced map h∞ : X → X∞ . Since X is compact and each hn is surjective, by Theorem 2.1.33, h∞ is surjective. To see h∞ is one–to–one, let x and y be two distinct points of X. Since lim mesh(Un ) = 0, there exist n ∈ IN and U, V ∈ Un such n→∞

that x ∈ U and y ∈ V . Hence, hn (x) = hn (y), since U ∩ V = ∅. Therefore, h∞ is one–to–one. Q.E.D. 2.2.8. Lemma. If X is a totally disconnected and perfect compactum, then for each n ∈ IN, there exist n nonempty pairwise disjoint open and closed subsets of X whose union is X. Proof. The proof is done by induction on n. For n = 1, X itself is open and closed. Let n ≥ 2 and assume that there exist n nonempty pairwise disjoint open and closed subsets, U1 , . . . , Un , of X such n that X = Uj . Let x ∈ Un . Since X is perfect, Un = {x}. By j=1


96 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Proposition 2.2.3, there exists an open and closed subset of X such that x ∈ V ⊂ Un . Hence, X = U1 ∪ . . . Un−1 ∪ V ∪ (Un \ V ). Q.E.D. We are ready to prove the characterization of the Cantor set, C, mentioned above. 2.2.9. Theorem. If X and Y are two totally disconnected and perfect compacta, then X and Y are homeomorphic. In particular, X and Y are homeomorphic to C (Example 1.6.4). Proof. By Proposition 2.2.3 and Lemma 2.2.8, there exist two open covers U1 and U1 of X and Y , respectively, such that the following four conditions hold: (i) The elements of each cover are open and closed. (ii) The elements of each cover are pairwise disjoint. (iii) The mesh of each cover is less than one. (iv) Both covers have the same cardinality. Let h1 : U1 → → U1 be any bijection. Apply Proposition 2.2.3 to obtain refinements U2 and U2 of U1 and U1 , respectively, of mesh 1 less than and satisfying conditions (i) and (ii) above. 2 By Lemma 2.2.8, we assume that if U1 ∈ U1 , then U1 and h1 (U1 ) contain the same number of elements of U2 and U2 , respectively. It follows that there is a bijection h2 : U2 → → U2 such that if U2 ∈ U2 , U1 ∈ U1 and U2 ⊂ U1 , then h2 (U2 ) ⊂ h1 (U1 ). This process can be repeated inductively to obtain two inverse sequences, {Un , fnn+1 } ∞ and {Un , (f )n+1 n }, and a sequence {hn }n=1 of bijections such that we have the following infinite ladder: n fn−1

· · · ←− Un−1 ←− ⏐ ⏐h & n−1 · · · ←− Un−1 ←− n

(f )n−1

fnn+1

U⏐n ←− Un+1 ←− · · · ⏐ ⏐h ⏐h & n & n+1 Un ←− Un+1 ←− · · · n+1 (f )n

:⏐ U∞ ⏐h & ∞ : U∞

where each Un and each Un have the discrete topology, and the functions fnn+1 and (f )n+1 are defined as in the proof of Theon rem 2.2.7. Since each hn is a homeomorphism, by Theorems 2.1.47 Copyright © 2005 Taylor & Francis Group, LLC


97

and 2.1.48, h∞ is a homeomorphism. It follows from the proof of Theorem 2.2.7 that U∞ and U∞ are homeomorphic to X and Y , respectively. Therefore, X and Y are homeomorphic. Q.E.D. 2.2.10. Lemma. If X is a totally disconnected compactum, then X × C is homeomorphic to C. Proof. Let X be a totally disconnected compactum. By Theorem 2.2.9, we only need to show that X × C is perfect. Let (x, t) ∈ X × C, and let U × V be a basic open set of X × C such that (x, t) ∈ U × V . Since C is perfect (Example 1.6.4), there exists t ∈ V \ {t}. Hence, (x, t ) ∈ U × V . Therefore, X × C is perfect. Q.E.D. The characterization of C in Theorem 2.2.9 allows us to show that any compactum is a continuous image of C. 2.2.11. Theorem. If X is a compactum, with metric d, then there exists a surjective map f : C → → X of the Cantor set onto X. Proof. Since X is compact, using Theorem 1.6.6 (Lebesgue numbers), there exists a sequence {Un }∞ n=1 of finite covers of X such that, for each n ∈ IN, the following three conditions are satisfied: (i) each U ∈ Un is the closure of an open subset of X; 1 (ii) mesh(Un ) < n ; and 2 (iii) Un+1 ≺ Un . Let U1 = {U11 , . . . , U1k1 }. The elements of U1 may not be pairwise disjoint. We use the following trick. For each i ∈ {1, . . . , k1 }, let V1i = U1i × {i}. Then we say that a subset W × {i} is open in k1 V1i if W is an open subset of U1i . Let V1 = V1i . We define a metric d1 on V1 as follows:

i=1

d(x, y) if i = j; d1 ((x, i), (y, j)) = 1 if i = j. Copyright © 2005 Taylor & Francis Group, LLC

98 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Now, let U2 = {U21 , . . . , U2k2 }. For each U2j ∈ U2 , let i ∈ {1, . . . , k1 } such that U2j ⊂ U1i . Then let V2ij = U2j × {i} × {j} (whenever U2j ⊂ U1i ). Again, we say that a subset W × {i} × {j} is open in V2ij if W is open in U2j . Let V2 = V2ij . We define a metric i,j

d2 in a similar way as we defined d1 on V1 . Finally, let f12 : V2 → V1 be given by f12 (u, i, j) = (u, i). Clearly, f12 is continuous. Now, let U3 = {U31 , . . . , U3k3 }. For each U3k ∈ U3 , let j ∈ {1, . . . , k1 } and i ∈ {1, . . . , k1 } such that U3k ⊂ U2j ⊂ U1i . Then define V3ijk = U3k × {i} × {j} × {k}. As before, a subset W × {i} × {j}×{k} is open in V3ijk if W is open in U3k . Let V3 = V3ijk . We i,j,k

define a metric on V3 in the same way we defined d1 on V1 . Finally, let f23 : V3 → V2 be given by f23 (u, i, j, k) = (u, i, j). Clearly, f23 is continuous. Although it is notationally complicated, the general inductive step is clear now. Hence, we obtain an inverse sequence {Vn , fnn+1 } of compacta. Let V∞ = lim{Vn , fnn+1 }. Then V∞ is a ←− compactum. V

V

222

224

V

V

221

223

V

212

V 214

V

211

V 213

V

16

V 14

V

V

12

15

V 13

V

11

U

12

U

11

U

13

U

14

U

16

U

15

.

A second inverse sequence, {Xn , gnn+1 }, may be constructed letting for each n ∈ IN, Xn = X and gnn+1 = 1X . It follows, from Copyright © 2005 Taylor & Francis Group, LLC


99

Proposition 2.1.37, that X is homeomorphic to lim{Xn , gnn+1 }. ←−

For each n ∈ IN, let hn : Vn → Xn be given by hn (u, i, j, . . . , p) = u. Thus, the following diagram n fn−1

fnn+1

gn−1

gn

· · · ←− Vn−1 ←− V⏐n ←− Vn+1 ←− · · · ⏐ ⏐ ⏐h ⏐h ⏐h & n−1 & n & n+1 · · · ←− Xn−1 ←− Xn ←− Xn+1 ←− · · · n n+1

:⏐ V∞ ⏐h & ∞ : X∞

is commutative, where X∞ = lim{Xn , gnn+1 } and h∞ = lim{hn }. ←− ←− Since each hn is surjective and each Vn is a compactum, by Theorem 2.1.48, h∞ is surjective. Now, we show that V∞ is totally disconnected. To this end, we prove that for any two distinct points of V∞ , there exists an open and closed subset of V∞ containing one of them and not containing the other. ∞ Let x = (xn )∞ n=1 and y = (yn )n=1 be two distinct points of V∞ . Then there exists n0 ∈ IN such that xn0 = yn0 and n0 is the smallest with this property. First, suppose that n0 = 1, i.e., x1 = y1 . Suppose, also, that x1 = (u, i) and y1 = (u , i ). If u = u , then there exist m ∈ IN and Um , Um ∈ Um such that u ∈ Um , u ∈ Um and Um ∩ Um = ∅. Then the corresponding sets Vmij... and Vmij... are disjoint and −1 both are open and closed subsets of Vm . Hence, fm (Vmij... ) is an open and closed subset of V∞ containing x and not containing y. If u = u , then i = i . Hence, x1 ∈ V1i and y1 ∈ V1i . Note that V1i and V1i are disjoint open and closed subsets of V1 . Then f1−1 (V1i ) is an open and closed subset of V∞ containing x and not containing y. Now, suppose n0 ≥ 2. For simplicity, we assume that n0 = 3. Then x3 = y3 , x3 = (u, i, j, k) and y = (u, i, j, k ), with k = k . Hence, x3 ∈ V3ijk and y3 ∈ V3ijk . Note that V3ijk and V3ijk are disjoint open and closed subsets of V3 . Then f3−1 (Vijk ) is an open and closed subset of V∞ containing x and not containing y. Therefore, V∞ is totally disconnected. Even though V∞ may not be perfect, by Lemma 2.2.10, V∞ × C is perfect. Hence, by Theorem 2.2.9, V∞ × C is homeomorphic to C. Thus, we have the following maps: (a) A homeomorphism ξ : C → V∞ × C; Copyright © 2005 Taylor & Francis Group, LLC

100 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS (b) A surjective map ζ : V∞ × C → V∞ given by ζ(v, c) = v (the projection map); (c) A surjective map h∞ : V∞ → X∞ ; and (d) A homeomorphism ϕ : X∞ → X. Therefore, f = ϕ ◦ h∞ ◦ ζ ◦ ξ is a surjective map from C onto X. Q.E.D.

2.3

InverseLimits andOther Operations

We show that taking inverse limits commute with products, cones and hyperspaces. 2.3.1. Theorem. Let {Xn , fnn+1 } and {Yn , gnn+1 } be inverse sequences of compacta, with surjective bonding maps, whose inverse limits are X∞ and Y∞ , respectively. Then {Xn × Yn , fnn+1 × gnn+1 } is an inverse sequence and lim{Xn ×Yn , fnn+1 ×gnn+1 } is homeomorphic ←− to X∞ × Y∞ . Proof. Let n ∈ IN. By Theorem 1.1.10, the function fn × gn is continuous. For each n ∈ IN, let hn : X∞ × Y∞ → Xn × Yn be given by hn = fn × gn . Then (fnn+1 × gnn+1 )◦ hn+1 = hn . Hence, by Theorem 2.1.31, there exists the induced map h∞ : X∞ × Y∞ → lim{Xn × Yn , fnn+1 × ←−

gnn+1 }. Since all the bonding maps are surjective, hn is surjective for each n ∈ IN. Then h∞ is surjective by Theorem 2.1.33. Clearly, h∞ is one–to–one. Q.E.D. The next Theorem shows that taking inverse limits and taking cones commute. 2.3.2. Theorem. Let {Xn , fnn+1 } be an inverse sequence of compacta whose inverse limit is X∞ . Then {K(Xn ), K(fnn+1 )} is an Copyright © 2005 Taylor & Francis Group, LLC

2.3. INVERSE LIMITS AND OTHER OPERATIONS

101

inverse sequence of cones with induced bonding maps and K(X∞ ) is homeomorphic to lim {K(Xn ), K(fnn+1 )}. ←−

Proof. By Proposition 1.2.12, K(fnn+1 ) is a map from K(Xn+1 ) into K(Xn ) for each n ∈ IN. Hence, {K(Xn ), K(fnn+1 )} is an inverse sequence of continua. Let Y∞ = lim{K(Xn ), K(fnn+1 )}. ←−

Note that for each n ∈ IN, by Proposition 1.2.12, K(fn ) : K(X∞ ) → K(Xn ) is a map. Clearly, K(fnn+1 ) ◦ K(fn+1 ) = K(fn ) for every n ∈ IN. Hence, by Theorem 2.1.31, there exists an induced map K(f )∞ : K(X∞ ) → Y∞ . To see K(f )∞ is a homeomorphism, we show it is a bijection. ∞ Note that K(f )∞ (νX∞ ) = (νXn )∞ n=1 . If ((xn , t))n=1 ∈ Y∞ \ ∞ {(νXn )∞ n=1 }, then ((xn )n=1 , t) is a point of K(X∞ ) satisfying that ∞ K(f )∞ (((xn )n=1 , t)) = ((xn , t))∞ n=1 . Hence, K(f )∞ is surjective. To see K(f )∞ is one–to–one, note that (K(f )∞ )−1 ((νXn )∞ n=1 ) = ∞ ∞ {νX∞ }. Now, if ((xn )n=1 , t) and ((xn )n=1 , t ) are two points of K(X∞ ) such that ∞ K(f )∞ (((xn )∞ n=1 , t)) = K(f )∞ (((xn )n=1 , t )), ∞ then ((xn , t))∞ n=1 = ((xn , t ))n=1 . Hence, xn = xn for each n ∈ IN, ∞ and t = t . Thus, ((xn )∞ n=1 , t) = ((xn )n=1 , t ). Consequently, K(f )∞ is one–to–one. Therefore, K(f )∞ is a homeomorphism. Q.E.D.

The following Lemma gives a base for the hyperspace of subcompacta of an inverse limit in terms of the open subsets of the factor spaces and the projection maps. 2.3.3. Lemma. Let {Xn , fnn+1 } be an inverse sequence of compacta whose inverse limit is X∞ . For each j ∈ IN, let Bj = {fj−1 (U1 ), . . . , fj−1 (Uk ) | U1 , . . . , Uk are open subsets of Xj }. If B =

∞

Bj , then B is a base for the Vietoris topology for 2X∞ .

j=1


102 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Proof. By Proposition 2.1.9 and Theorem 1.8.14, −1 −1 (Un(1) ), . . . , fn(m) (Un(m) ) | Un() is an open B ∗ = {fn(1)

subset of Xn() for each ∈ {1, . . . , m}, m ∈ IN} is a base for the Vietoris topology for 2X∞ . We show that B = B ∗ . Clearly, B ⊂ B ∗ . −1 −1 (Un(1) ), . . . , fn(m) (Un(m) ) ∈ B∗ , and let Let fn(1) k = max{n(1), . . . , n(m)}. k Note that, since fn() ◦ fk = fn() , we have that for every ∈ −1 k )−1 (Un() ). {1, . . . , m}, fn() (Un() ) = fk−1 (fn() k )−1 (Un() ). Then, since For each ∈ {1, . . . , m}, let Vn() = (fn() the bonding maps are continuous, each Vn() is an open subset of Xk . Hence, fk−1 (Vn(1) ), . . . , fk−1 (Vn(m) ) ∈ B. −1 Since fk−1 (Vn() ) = fn() (Un() ) for each ∈ {1, . . . , m}, −1 −1 fk−1 (Vn(1) ), . . . , fk−1 (Vn(m) ) = fn(1) (Un(1) ), . . . , fn(m) (Un(m) ). −1 −1 (Un(1) ), . . . , fn(m) (Un(m) ) ∈ B. Hence, fn(1) ∗ Therefore, B = B .

Q.E.D. 2.3.4. Theorem. Let {Xn , fnn+1 } be an inverse sequence of continua whose inverse limit is X∞ . Then the following hold. n+1

(1) 2X∞ is homeomorphic to lim{2Xn , 2fn }; ←−

(2) Cm (X∞ ) is homeomorphic to lim{Cm (Xn ), Cm (fnn+1 )} for each ←− m ∈ IN; and (3) Fm (X∞ ) is homeomorphic to lim{Fm (Xn ), Fm (fnn+1 )} for each ←− m ∈ IN. Furthermore, there is a homeomorphism n+1

→ 2X∞ h : lim{2Xn , 2fn } → ←−


2.3. INVERSE LIMITS AND OTHER OPERATIONS

103

such that for each m ∈ IN, h lim{Cm (Xn ), Cm (fnn+1 )} = Cm (X∞ ) ←−

and

h lim{Fm (Xn ), Fm (fnn+1 )} = Fm (X∞ ). ←−

Proof. We define the homeomorphism h after making some obsern+1 vations about the points of lim{2Xn , 2fn }. Let

(An )∞ n=1

∈ lim{2

Xn

←−

←− fnn+1

,2

n+1

}. Then, by definition, 2fn (An+1 )

= An . Thus, {An , fnn+1 |An+1 } is an inverse sequence with surjective bonding maps. Since, for each n ∈ IN, An ⊂ Xn , we have that lim{An , fnn+1 |An+1 } ←−

⊂ lim{Xn , fnn+1 } = X∞ . Hence, lim{An , fnn+1 |An+1 } ∈ 2X∞ . ←−

Now, let h : lim{2

Xn

←−

fnn+1

,2

←−

}→ → 2X∞ be given by

n+1 h((An )∞ |An+1 }. n=1 ) = lim{An , fn ←−

By the above considerations, h is well defined. Let K ∈ 2X∞ . Then K is a closed subset of X∞ . Hence, by (∗) of Proposition 2.1.20, K = lim{fn (K), fnn+1 |fn+1 (K) }. Note that n+1

←−

fn (K) ∈ 2Xn and 2fn (fn+1 (K)) = fn (K) for each n ∈ IN. Thus, n+1 Xn (fn (K))∞ , 2fn }, and h((fn (K))∞ n=1 ∈ lim{2 n=1 ) = K. Therefore, ←− h is surjective. n+1 ∞ Xn Next, let (An )∞ , 2fn } n=1 and (Bn )n=1 be two elements of lim{2 ←−

∞ such that h((An )∞ n=1 ) = h((Bn )n=1 ). We prove that Ak = Bk for every k ∈ IN. Let k ∈ IN, and let p ∈ Ak . Since {An , fnn+1 |An+1 } is an inverse sequence with surjective bonding maps, there exists a point ∞ (xn )∞ n=1 ∈ h((An )n=1 ) such that xk = p (Remark 2.1.6). Since ∞ ∞ ∞ h((An )n=1 ) = h((Bn )∞ n=1 ), (xn )n=1 ∈ h((Bn )n=1 ). Hence, xk ∈ Bk , i.e., p ∈ Bk . Therefore, Ak ⊂ Bk . A similar argument shows that Bk ⊂ Ak . Thus, Ak = Bk . Therefore, h is one–to–one. n+1 Now, we see that h is continuous. Let πn : lim{2Xn , 2fn } →

2Xn be the projection map for every n ∈ IN. Copyright © 2005 Taylor & Francis Group, LLC

←−

104 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Let fj−1 (U1 ), . . . , fj−1 (Uk ) ∈ B (Lemma 2.3.3). We show that h−1 (fj−1 (U1 ), . . . , fj−1 (Uk )) n+1

is an open subset of lim{2Xn , 2fn }. To this end, it suffices to prove ←− that h−1 (fj−1 (U1 ), . . . , fj−1 (Uk )) = πj−1 (U1 , . . . , Uk ). n+1

Xn , 2fn }, then fj (h((An )∞ First observe that if (An )∞ n=1 ∈ lim{2 n=1 )) ←− = Aj for each j ∈ IN. Hence,

h−1 (fj−1 (U1 ), . . . , fj−1 (Uk )) = (An )∞ n=1

∈ lim{2

Xn

←−

fnn+1

,2

k ∞ } h((An )n=1 ) ⊂ fj−1 (U ) and =1

−1 h((An )∞ n=1 ) ∩ fj (U ) = ∅, for each ∈ {1, . . . , k}

fnn+1

Xn ,2 (An )∞ n=1 ∈ lim{2 ←−

−1 } h((An )∞ n=1 ) ⊂ fj

k

=

U

=1

and

−1 h((An )∞ n=1 ) ∩ fj (U ) = ∅, for each ∈ {1, . . . , k}

=

k Xn fnn+1 ∞ ∈ lim {2 , 2 } (h((A ) )) ⊂ U and f (An )∞ j n n=1 n=1 ←−

=1

fj (h((An )∞ n=1 )) ∩ U = ∅, for each ∈ {1, . . . , k} (An )∞ n=1

∈ lim{2

Xn

←−

fnn+1

,2

k } Aj ⊂ U and =1

Aj ∩ U = ∅, for each ∈ {1, . . . , k}

πj−1 (U1 , . . . , Uk ). Copyright © 2005 Taylor & Francis Group, LLC

=

=

2.4. CHAINABLE CONTINUA

105

Therefore, h is continuous. Hence, h is a homeomorphism. Observe that by the definition of h and by Theorem 2.1.45, it follows that h lim{Cm (Xn ), Cm (fnn+1 )} = Cm (X∞ ) ←−

and that

h

lim{Fm (Xn ), Fm (fnn+1 )} ←−

= Fm (X∞ )

for every m ∈ IN. Q.E.D.

2.4

Chainable Continua

We discuss an important class of continua, namely, chainable continua. Besides presenting some of its properties, we prove that it coincides with the class of arc–like continua. 2.4.1. Definition. Let X be a compactum. A chain U in X is a finite sequence, U1 , . . . , Un , of subsets of X such that Ui ∩ Uj = ∅ if and only if |i − j| ≤ 1 for each i, j ∈ {1, . . . , n}. Each Uj is called a link of U. If each link of U is open, then U is called an open chain. If ε > 0, U is an open chain and the mesh(U) < ε, then U is called an ε–chain. 2.4.2. Remark. Let us observe that we do not require the links of a chain to be connected. Hence, a chain may look like in the following picture:

U1


U2

U3

106 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.4.3. Definition. A continuum X is said to be chainable provided that for each ε > 0, there exists an ε–chain covering X. If x1 , x2 ∈ X, then X is chainable from x1 to x2 if for each ε > 0 there is an ε–chain U = {U1 , . . . , Un } covering X such that x1 ∈ U1 and x2 ∈ Un .

2.4.4. Example. The unit interval [0, 1] is chainable from 0 to 1.

0

1

2.4.5. Example. Let X = {0} × [−1, 1] ∪

2 1 2 ∈ IR x ∈ 0, . x, sin x π

Then X is called the topologist sine Note that X is a chain curve. 2 ,1 . able continuum from (0, −1) to π

2.4.6. Example. Let X be the topologist sine curve and let X 2 be the reflection of X in IR2 with respect to the line x = . Let π Copyright © 2005 Taylor & Francis Group, LLC


107

Z = X∪ X . Then Z is a chainable continuum from (0, −1) to 4 , −1 . π

2.4.7. Example. The Knaster continuum, K, is defined in the following way. The continuum consists of 2 (a) all semi–circles in IR with nonnegative ordinates, with center 1 at the point , 0 and passing through every point of the Cantor 2 set C. ordinates, which have for (b) all semi–circles in IR2 with nonpositive 5 each n ∈ IN, the center at the point , 0 and passing through 2 · 3n

2 1 each point of the Cantor set C lying in the interval n , n−1 . 3 3 Then K is a chainable continuum.


108 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.4.8. Remark. It is well known that K is an indecomposable continuum (Remark to Theorem 8 (p. 213) of [11]). 2.4.9. Example. Let K be the reflection of K in IR2 with respect to the origin (0, 0). Let M = K ∪ K . Then M is a chainable continuum.

The next Theorem shows that the property of being chainable is hereditary, i.e., each nondegenerate subcontinuum has the property.

2.4.10. Theorem. If X is a chainable continuum and K is a subcontinuum of X, then K is chainable. Proof. Let ε > 0. Since X is chainable, there exists an ε–chain U = {U1 , . . . , Un } covering X. Let i = min{k ∈ {1, . . . , n} | K ∩ Uk = ∅}, and let j = max{k ∈ {1, . . . , n} | K ∩ Uk = ∅}. We show that U = {Ui ∩ K, . . . , Uj ∩ K} is an ε–chain covering K. Clearly, mesh(U ) < ε. Suppose U is not a chain. Then there exists ∈ {i, . . . , j − 1} such that (U ∩ K) ∩ (U+1 ∩ K) = ∅. Hence,

j K⊂ (Um ∩ K) (Um ∩ K) m=1


m=+1

2.4. CHAINABLE CONTINUA and

(Um ∩ K)

109

m=i

j

(Um ∩ K)

= ∅.

m=+1

This contradicts the fact that K is connected since K ∩ Ui = ∅ and K ∩ Uj = ∅. Thus, U is a chain. Therefore, K is chainable. Q.E.D. The following Lemma is known as the Shrinking Lemma. 2.4.11. Lemma. Let X be a compactum. If V = {V1 , . . . , Vm } is a finite open cover of X, then there exists an open cover U = {U1 , . . . , Um } such that Cl(Uj ) ⊂ Vj for each j ∈ {1, . . . , m}. Proof. The construction

m of U is done inductively. Let F1 = X \ Vj . Then F1 ⊂ V1 , and F1 is a closed subset j=2

of X. Since X is a metric space, there exists an open subset U1 of X such that F1 ⊂ U1 ⊂ Cl(U1 ) ⊂ V1 . Suppose Uk−1 has been defined for each k < m. Let Fk = X \

k−1 j=1

Uj ∪

m

Vj

.

j=k+1

Then Fk ⊂ Vk , and Fk is a closed subset of X. Since X is a metric space, there exists an open subset Uk of X such that Fk ⊂ Uk ⊂ Cl(Uk ) ⊂ Vk . Thus, we finish the inductive step. Now, let U = {U1 , . . . , Um }. Then U is a family of m open subsets of X. We show U covers X. To this end, let x ∈ X. Then x belongs to finitely many elements of V, say Vk1 , . . . , Vkn . Let k = max{k1 , . . . , kn }. Now, x ∈ X \ U for every > k, and hence, if x ∈ X \ Uj for every j < k, then x ∈ Fk ⊂ Uk . Thus, in any case, x ∈ Uj for some j ∈ {1, . . . , m}. Therefore, U covers X. Q.E.D.


110 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.4.12. Definition. Let X be a compactum. A cover U of X is said to be essential provided that no proper subfamily of U covers X.

2.4.13. Lemma. If X is a chainable continuum, then there exists a sequence {Un }∞ n=1 of essential covers of X such that for each n ∈ IN, the following conditions hold: 1 (a) Un is a n –chain with the property that disjoint links have 2 disjoint closures; and (b) The closure of the union of any three consecutive links of Un+1 is contained in one link of Un . Proof. Note that if U = {U1 , . . . , Un } is a chain covering X is not essential, by the definition of chain, then U1 ⊂ U2 and/or Un ⊂ Un−1 . Since one can always modify such a chain U so that its end links contain a point not in any other link (by simply removing U1 and/or Un if necessary) we may assume that the chain is essential. The construction of the coverings is done inductively. 1 Since X is chainable, there exists a –chain V1 = {V11 , . . . , V1,m1 } 2 covering X. By Lemma 2.4.11, there exists an open cover U1 = {U11 , . . . , U1m1 } such that Cl(U1j ) ⊂ V1j for each j ∈ {1, . . . , m1 }. 1 Then U1 is a –chain covering X with the property that disjoint 2 links have disjoint closures. Suppose that, for some n ∈ IN, we have constructed open covers U1 , . . . , Un of X satisfying conditions (a) and (b) of the statement above. We construct Un+1 as follows. Let λn+1 be for the cover Un (Theorem 1.6.6). a Lebesgue number 1 1 Let α < min λn+1 , n+1 . Since X is chainable, there is an α– 3 2 chain Vn+1 = {Vn+11 , . . . , Vn+1mn+1 } covering X. By Lemma 2.4.11, there exists an open cover Un+1 = {Un+11 , . . . , Un+1mn+1 } such that 1 Cl(Un+1j ) ⊂ Vn+1j . Then Un+1 is a n+1 –chain covering X with the 2 property that disjoint links have disjoint closures. We need to see that the closure of the union of three consecutive links of Un+1 is contained in a link of Un . Let Un+1j , Un+1j+1 and Copyright © 2005 Taylor & Francis Group, LLC


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Un+1j+2 be three consecutive links of Un+1 . Since diam(Cl(Un+1j ) ∪ Cl(Un+1j+1 ) ∪ Cl(Un+1j+2 )) = diam(Un+1j ∪ Un+1j+1 ∪ Un+1j+2 ) ≤ 3α < λn+1 , there exists a link Unk of Un such that Cl(Un+1j ) ∪ Cl(Un+1j+1 ) ∪ Cl(Un+1j+2 ) ⊂ Unk (since λn+1 is a Lebesgue number for Un ). In this way, we finish the inductive step. Therefore, the existence of the sequence of open coverings is proven. Q.E.D. 2.4.14. Definition. Let X be a chainable continuum. A sequence {Un }∞ n=1 of essential covers satisfying the conditions of Lemma 2.4.13 is called a defining sequence of chains for X.

2.4.15. Definition. Let X be a metric space, and let f : X → X be a map. We say that f has a fixed point if there exists x ∈ X such that f (x) = x.

2.4.16. Definition. Let X be a metric space. We say that X has the fixed point property provided that for each map f : X → X, f has a fixed point. It is a well known fact from calculus that the unit interval [0, 1] has the fixed point property. Hamilton [6] has shown that this property is shared by all chainable continua. To prove this result, we need the following Lemma: 2.4.17. Lemma. Let X be a compactum, with metric d, and let f : X → X be a map. If for each ε > 0, there exists a point xε in X such that d(f (xε ), xε ) < ε, then f has a fixed point. Proof. By hypothesis, for each n ∈ IN, there exists xn ∈ X such 1 that d(xn , f (xn )) < . Since X is compact, without loss of genn erality, we assume that the sequence {xn }∞ n=1 converges to a point x ∈ X. Since f is continuous, the sequence {f (xn )}∞ n=1 converges to f (x). Copyright © 2005 Taylor & Francis Group, LLC

112 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Let ε > 0. Then there exists N ∈ IN such that ε ε 1 ε d(xN , x) < , d(f (xN ), f (x)) < and < . 3 3 N 3 Hence, d(x, f (x)) ≤ d(x, xN ) + d(xN , f (xN )) + d(f (xN ), f (x)) < 1 ε 2 ε ε + + < ε + = ε. 3 N 3 3 3 Since ε was arbitrary, d(x, f (x)) = 0. Thus, f (x) = x. Therefore, f has a fixed point Q.E.D. 2.4.18. Theorem. If X is a chainable continuum, with metric d, then X has the fixed point property. Proof. Let f : X → X be a map. Let {Un }∞ n=1 be a defining sequence of chains for X. For each k ∈ IN, we assume that Uk = {Uk1 , . . . , Uknk }. Let ε > 0. By Lemma 2.4.17, we need to find a point xε ∈ X such that d(xε , f (xε )) < ε. 1 Let k ∈ IN such that k < ε. Consider the chain Uk and define 2 the following subsets of X: A = {x ∈ X | if x ∈ Cl(Ukj ) and f (x) ∈ Cl(Uki ), then j < i}; B = {x ∈ X | x, f (x) ∈ Cl(Uki ) for some i ∈ {1, . . . , nk }}; C = {x ∈ X | if x ∈ Cl(Ukj ) and f (x) ∈ Cl(Uki ), then j > i}. We show that B = ∅. To this end, suppose B = ∅. Let x ∈ X \ A, and let λ be a Lebesgue number of Uk (Theorem 1.6.6). Since X is compact, f is uniformly continuous. Hence, there exists δ > 0 such that δ < λ and such that if y, z ∈ X and d(y, z) < δ, then d(f (y), f (z)) < λ. Let y ∈ X such that d(y, x) < δ. Since δ < λ, there exists Ukm ∈ Uk such that x, y ∈ Ukm ⊂ Cl(Ukm ). Since x ∈ Cl(Ukm ) ∩ (X \ A), then f (x) ∈ Cl(Ukj ), where j < m. Now, since d(f (y), f (x)) < λ, f (y) ∈ Cl(Ukn ), where n ≤ m. Since B = ∅, n < m. Hence, y ∈ X \ A. Consequently, X \ A is an open Copyright © 2005 Taylor & Francis Group, LLC


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subset of X. Therefore, A is closed. Similarly, C is a closed subset of X. Since X = A ∪ C and A ∩ C = ∅, we obtain a contradiction. Therefore, B = ∅. Q.E.D. The next concept is used to define classes of continua. 2.4.19. Definition. Let f : X → → Y be a surjective map between metric spaces, and let ε > 0. We say that f is an ε–map provided that for each y ∈ Y , diam(f −1 (y)) < ε. The following Lemma says that ε–maps between compacta behave in the same way with sets of positive small diameter. 2.4.20. Lemma. Let X and Y be compacta, with metrics d and d , respectively. Let ε > 0. If f : X → → Y is an ε–map, then there −1 exists δ > 0 such that diam(f (U )) < ε for each subset U of Y with diam(U ) < δ. Proof. First note that since diam(U ) = diam(Cl(U )) for any set U , it suffices to prove the lemma for closed sets. Suppose the lemma is not true. Then for each n ∈ IN, there is a 1 closed subset Un of Y such that diam(Un ) < and diam(f −1 (Un )) ≥ n ε. Let xn , xn ∈ f −1 (Un ) such that d(xn , xn ) = diam(f −1 (Un )). Since X is compact, without loss of generality, we assume that there exist two points x, x ∈ X such that lim xn = x and lim xn = x . Note that d(x, x ) ≥ ε. Now, by continuity,

n→∞

n→∞

d (f (x), f (x )) = lim d (f (xn ), f (xn )) ≤ n→∞

1 = 0, n→∞ n→∞ n a contradiction to the fact that f is an ε–map. Therefore, the lemma is true. Q.E.D. lim diam(Un ) ≤ lim


114 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.4.21. Definition. A continuum X is said to be like an arc provided that for each ε > 0, there exists an ε–map f : X → → [0, 1]. The next Theorem gives us the equality of the class of chainable continua and the class of arc–like continua. The proof we present is due to James T. Rogers, Jr. 2.4.22. Theorem. If X be a continuum with metric d, then the following are equivalent: (1) X is chainable. (2) X is arc–like. (3) X is like an arc. Proof. Suppose X is a chainable continuum. We show X is arc– like. Let {Un }∞ n=1 be a defining sequence of chains for X (Lemma 2.4.13). For every n ∈ IN, we use the notation Un = {Un,0 , . . . , Un,k(n) }. For each n ∈ IN, let Xn = [0, 1]. Divide Xn into k(n) equal subintervals with vertexes vn,0 = 0, . . . , vn,k(n) = 1. Note that there is a one–to–one correspondence between the vertexes of the subintervals of Xn and the links of the chain Un . We define the functions fnn+1 : Xn+1 → → Xn as follows: ⎧ vn,j if Un,j is the only link of Un ⎪ ⎪ ⎪ ⎪ ⎨ containing fnn+1 (vn+1,m ) = Un+1,m ; ⎪ ⎪ ⎪ v + v ⎪ n,j n,j+1 ⎩ if Un+1,m ⊂ Un,j ∩ Un,j+1 2 and extend fnn+1 linearly over Xn+1 . Since the chains are essential, the function fnn+1 is well defined for every n ∈ IN. Also, all these functions are continuous and surjective. Let X∞ = lim{Xn , fnn+1 }. Then X∞ is an arc–like continuum. ←− We show X∞ is homeomorphic to X. To this end, let hn : X → C(Xn ) be given by: {vn,j } if Un,j is the only link of Un containing x; hn (x) = [vn,j , vn,j+1 ] if x ∈ Un,j ∩ Un,j+1 . Copyright © 2005 Taylor & Francis Group, LLC


115

Note that for each n ∈ IN, fnn+1 (hn+1 (x)) ⊂ hn (x). To see this, we consider six cases. If x ∈ Un+1,m \(Un+1,m−1 ∪Un+1,m+1 ) and Un+1,m ⊂ Un,j \(Un,j−1 ∪ Un,j+1 ), then fnn+1 (hn+1 (x)) = fnn+1 ({vn+1,m }) = {vn,j } = hn (x). If x ∈ Un+1,m \ (Un+1,m−1 ∪ Un+1,m+1 ) and Un+1,m ⊂ Un,j ∩ vn,j + vn,j+1 Un,j+1 , then fnn+1 (hn+1 (x)) = fnn+1 ({vn+1,m }) = ⊂ 2 [vn,j , vn,j+1 ] = hn (x). If x ∈ Un+1,m \(Un+1,m−1 ∪Un+1,m+1 ), Un+1,m ⊂ Un,j and Un+1,m ∩ Un,j+1 = ∅, then fnn+1 (hn+1 (x)) = fnn+1 ({vn+1,m }) = {vn,j } ⊂ [vn,j , vn,j+1 ] = hn (x). If x ∈ Un+1,m ∩Un+1,m+1 and Un+1,m ∩Un+1,m+1 ⊂ Un,j \(Un,j−1 ∪ Un,j+1 ), then fnn+1 (hn+1 (x)) = fnn+1 ([vn+1,m , vn+1,m+1 ]) = {vn,j } = hn (x). If x ∈ Un+1,m ∩ Un+1,m+1 and Un+1,m ∪ Un+1,m+1 ⊂ Un,j ∩ Un,j+1 , + v v n,j n,j+1 then fnn+1 (hn+1 (x)) = fnn+1 ([vn+1,m , vn+1,m+1 ]) = ⊂ 2 [vn,j , vn,j+1 ] = hn (x). If x ∈ Un+1,m ∩Un+1,m+1 , Un+1,m ∪Un+1,m+1 ⊂ Un,j and Un+1,m+1 ∩ Un,j+1 = ∅, then fnn+1 (hn+1 (x)) = fnn+1 ([vn+1,m , vn+1,m+1 ]) =

vn,j + vn,j+1 vn,j , ⊂ [vn,j , vn,j+1 ] = hn (x). 2 Therefore, fnn+1 (hn+1 (x)) ⊂ hn (x). Note that this implies that −1 {fn (hn (x))}∞ n=1 is a decreasing sequence of closed subsets of X∞ . ∞ Thus, fn−1 (hn (x)) = ∅. n=1

Now, we are ready to define the homeomorphism g : X → → X∞ ∞ as follows: g(x) is the unique point in fn−1 (hn (x)). n=1

1 –maps, we n→∞ 2n have that lim diam(fn−1 (hn (x))) = 0. Hence, g is well defined. Since lim k(n) = ∞ and the projection maps are n→∞

Now, we show that g is one–to–one. Let x, x ∈ X such that x = x . Let Un be an element of the defining sequence chains for X with the property that x and x belong to different links of Un . In this case, hn (x) ∩ h(x ) = ∅. Hence, g(x) = g(x ). Copyright © 2005 Taylor & Francis Group, LLC

116 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS To see g is surjective, let y = (yn )∞ n=1 ∈ X∞ . Let Mn : X∞ → C(Xn ) be given by {vn,j } if yn = vn,m ; Mn (y) = [vn,j , vn,j+1 ] if yn ∈ (vn,m , vn,m+1 ). Let Rn be the union of the links of Un whose vertexes are in Mn (y). Then Rn consists of one link or it is the union of two consec∞ ∞ utive links of Un . Let {x} = Rn = Cl(Rn ) (recall that n=1

n=1

Cl(Rn+1 ) ⊂ Rn for every n ∈ IN). We prove that g(x) = y. Let n ∈ IN. Then either Rn = Un,j or Rn = Un,j ∪ Un,j+1 , and x ∈ Rn . Hence, either hn (x) = {vn,j } or hn (x) = [vn,j , vn,j+1 ]. In either case, hn (x) = Mn (y). Thus, y ∈ fn−1 (Mn (y)) = fn−1 (hn (x)). Hence, ∞ y∈ fn−1 (hn (x)). Therefore, g(x) = y. n=1

Now, we see that g is continuous. To this end, by Theorem 1.1.9, it is enough to prove that for each n ∈ IN, fn ◦ g is continuous. Let n ∈ IN and let ε > 0. Let > n such that for each m ∈ {1, . . . , k() − 2}, diam(fn ([v,m , v,m+2 ])) < ε. Let λ > 0 be a Lebesgue number for the cover U (Theorem 1.6.6). Hence, if x, x ∈ X and d(x, x ) < λ, then there exists j ∈ {1, . . . , k()} such that x, x ∈ U,j . Note that f ◦g(U,j ) ⊂ [v,m , v,m+2 ] for some m ∈ {1, . . . , k()− 2}. Hence, diam(fn ◦ g(U,j )) = diam(fn ◦ f ◦ g(U,j )) < ε. Thus, dn (fn ◦g(x), fn ◦g(x )) < ε, where dn is the metric on Xn . Therefore, fn ◦ g is continuous. Next, suppose X is arc–like. We show that X is like an arc. Since X is arc–like, there exists an inverse sequence {Xn , fnn+1 }, where Xn = [0, 1] and fnn+1 is surjective for each n ∈ IN, and whose ∞ 1 inverse limit is X. Let ε > 0, and let N ∈ IN such that < ε. 2n n=N +1 ∞ To see fN is an ε–map, let z ∈ XN . Let x = (xn )∞ n=1 and y = (yn )n=1 be two points of fN−1 (z). Then ρ(x, y) =

∞ ∞ ∞ 1 1 1 d (x , y ) = d (x , y ) ≤ < ε. n n n n n n n n n 2 2 2 n=1 n=N +1 n=N +1


2.5. CIRCULARLY CHAINABLE CONTINUA

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Therefore, diam(fN−1 (z)) < ε. Finally, suppose X is like an arc. To see X is chainable, let f: X → → [0, 1] be an ε–map. By Lemma 2.4.20, there exists δ > 0 such that if U is a subset of [0, 1] with diam(U ) < δ, then diam(f −1 (U )) < ε. δ 1 Let n ∈ IN such that < . Divide [0, 1] into n equal parts and n 2 let 1 2 1 3 −1 −1 −1 U= f 0, 0, ,f ,f , ,... , n n n n n − 2 n − 1 ,1 , f −1 ,1 . f −1 n n 2 1 3 n−2 n−1 1 , 0, , , ,... , ,1 , ,1 is Since 0, n n n n n n a δ–chain of [0, 1], U is an ε–chain covering X. Since ε was arbitrary, therefore, X is chainable. Q.E.D.

2.5

Circularly Chainable and P–Like Continua

We present basic facts about circularly chainable and P–like continua. 2.5.1. Definition. Let X be a compactum. A circular chain U in X is a finite sequence, U0 , . . . , Un , of subsets of X such that Ui ∩ Uj = ∅ if and only if |i − j| ≤ 1 or i, j ∈ {1, n}. Each Uj is called a link of U. If each link of U is open, then U is called an open circular chain. If ε > 0, U is an open circular chain and the mesh(U) < ε, then U is called a circular ε–chain.


118 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.5.2. Definition. A continuum X is circularly chainable provided that for each ε > 0, there exists a circular ε–chain covering X.

2.5.3. Example. The unit circle S 1 is a circularly chainable continuum.

S

1

2.5.4. Example. Let X be the topologist sine curve (see Exam2 ple 2.4.5). Join the points (0, −1) and , 1 with an arc, Y , in π IR2 . Then W = X ∪ Y is called the Warsaw circle, and it is a circularly chainable continuum.



119

2.5.5. Example. The double Warsaw circle consists of two copies of the topologist sine curve joined by two arcs (see picture below). It is also a circularly chainable continuum.

2.5.6. Example. Some chainable continua are circularly chainable. For instance, let X be either K, the Knaster continuum (Example 2.4.7), or M (Example 2.4.9). Then X is a circularly ε chainable continuum. If ε > 0 and U = {U0 , . . . , Un } is an –chain, 2 then C = {U0 , . . . , Un , U0 ∪ Un } is a circular ε–chain.

2.5.7. Definition. A continuum X is said to be like a circle if for each ε > 0, there exists an ε–map from X onto S 1 .

The following Theorem is analogous to Theorem 2.4.22, for circularly chainable continua, and it is a special case of Theorem 2.5.13:


120 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.5.8. Theorem. Let X be a continuum. Then the following are equivalent: (1) X is a circularly chainable continuum. (2) X is a circle–like continuum. (3) X is a like a circle continuum. The next Theorem says that each circle–like continuum may be written as an inverse limit of circles where the bonding maps have nonnegative degree. 2.5.9. Theorem. If X is a circle–like continuum, then there exists an inverse sequence {Xn , fnn+1 } such that lim{Xn , fnn+1 } is home←−

omorphic to X, where each Xn = S 1 and such that the bonding maps satisfy one of the following three conditions: (a) for each n ∈ IN, deg(fnn+1 ) = 0; (b) for each n ∈ IN, deg(fnn+1 ) = 1; (c) for each n ∈ IN, deg(fnn+1 ) > 1. Proof. Since X is circle–like, there is an inverse sequence {Yn , rnn+1 } of circles with surjective bonding maps such that lim{Yn , rnn+1 } is ←− homeomorphic to X. First, we show that X is homeomorphic to an inverse limit of circles where all the bonding maps have nonnegative degrees. To this end, we construct a new inverse sequence {Yn , gnn+1 } and maps hn : Yn → → Yn such that hn ◦ rnn+1 = gnn+1 ◦ hn+1 and deg(gnn+1 ) ≥ 0 for every n ∈ IN. Let h1 = 1X1 . If deg(r12 ) ≥ 0, then g12 = r12 , and h2 = 1X2 . If deg(r12 ) < 0, then h2 = ℵ, where ℵ is the antipodal map, and g12 = r12 ◦ ℵ. Hence, h1 ◦ r12 = g12 ◦ h2 , and deg(g12 ) ≥ 0. Next, suppose inductively that we have constructed h1 , . . . , hk , k where either hj = 1Xj or hj = ℵ, j ∈ {1, . . . , k}, and g12 , . . . , gk−1 , j+1 j where deg(gj ) ≥ 0, j ∈ {1, . . . , k − 1}, such that hj−1 ◦ rj−1 = j ◦ hj for every j ∈ {2, . . . , k}. To define hk+1 and gkk+1 we gj−1 consider four cases. If hk = 1Xk and deg(rkk+1 ) ≥ 0, then let hk+1 = 1Xk+1 and let gkk+1 = rkk+1 . Copyright © 2005 Taylor & Francis Group, LLC


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If hk = 1Xk and deg(rkk+1 ) < 0, then let hk+1 = ℵ and let gkk+1 = rkk+1 ◦ ℵ. If hk = ℵ and deg(rkk+1 ) ≥ 0, then let hk+1 = ℵ and let gkk+1 = ℵ ◦ rkk+1 ◦ ℵ. If hk = ℵ and deg(rkk+1 ) < 0, then let hk+1 = 1xk+1 and let gkk+1 = ℵ ◦ rkk+1 . In each case, hk ◦ rkk+1 = gkk+1 ◦ hk+1 and deg(gkk+1 ) ≥ 0. In this way we finish with the inductive step. So, we have constructed an inverse sequence {Yn , gnn+1 }, where deg(gnn+1 ) ≥ 0 for all n ∈ IN. By Theorems 2.1.46, 2.1.47 and 2.1.48, X is homeomorphic to lim{Yn , gnn+1 }. ←−

Now, suppose there exists a subsequence {m(n)}∞ n=1 of IN such m(n)+1 that each map gm(n) has degree zero. Then, by Lemma 1.3.39, m(n+1) = 0 for each n ∈ IN. Thus, each bonding map of the deg gm(n) m(n+1) inverse sequence Ym(n) , gm(n) has degree zero and, by Theo m(n+1) rem 2.1.38, lim Ym(n) , gm(n) is homeomorphic to X. Therefore, ←−

m(n+1)

taking Xn = Ym(n) and fnn+1 = gm(n) , {Xn , fnn+1 } is an inverse sequence of circles with surjective bonding maps with degree zero, whose inverse limit is homeomorphic to X. Next, suppose none of the bonding maps gnn+1 has degree zero. Hence, we assume that for each n ∈ IN, deg(gnn+1 ) > 0. If only finitely many bonding maps have degree greater than one, then taking a subsequence, by Theorem 2.1.38, we assume that all the bonding maps have degree one. Thus, X is homeomorphic to the inverse limit of such subsequence. Finally, suppose there exists a subsequence {m(n)}∞ n=1 of IN m(n)+1 such that each map gm(n) has degree greater than one. Then, by m(n+1) > 1 for each n ∈ IN. Thus, each bondLemma 1.3.39, deg gm(n) m(n+1) ing map of the inverse sequence Ym(n) , gm(n) has degree greater m(n+1) than one and, by Theorem 2.1.38, lim Ym(n) , fm(n) is homeo←−

m(n+1)

morphic to X. Therefore, taking Xn = Ym(n) and fnn+1 = gm(n) , {Xn , fnn+1 } is an inverse sequence of circles with surjective bonding maps with degree greater than one, whose inverse limit is homeoCopyright © 2005 Taylor & Francis Group, LLC

122 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS morphic to X. Q.E.D. The following Theorem tells us that each circle–like continuum, expressed as an inverse sequence of circles with bonding maps having degree zero, is arc–like too. 2.5.10. Theorem. Let {Xn , fnn+1 } be an inverse sequence of circles with surjective bonding maps having degree zero. If X∞ = lim{Xn , fnn+1 }, then X∞ is an arc–like continuum. ←−

Proof. For each n ∈ IN, let pn : IR → S 1 be given by pn (t) = exp(2πt). Since deg(fnn+1 ) = 0 for every n ∈ IN, each map fnn+1 is homotopic to a constant map (Remark 1.3.38). Hence, by Theorem 1.3.40, there is a map hn+1 : Xn+1 → IR such that hn+1 ((1, 0)) = 0 and fnn+1 = pn ◦ hn+1 for every n ∈ IN. For each n ∈ IN, let Yn = hn+1 (Xn+1 ). Note that each Yn is an interval of length at least one (the bonding maps fnn+1 are surjective). For each n ∈ IN, let gnn+1 : Yn+1 → → Yn be given by gnn+1 = hn+1 ◦ (pn |Yn+1 ). Note that gnn+1 is surjective since it is the composition of surjective maps. Hence, {Yn , gnn+1 } is an inverse sequence of arcs with surjective bonding maps whose inverse limit, Y∞ , is homeomorphic to X∞ (Theorem 2.1.49). Therefore, X∞ is an arc–like continuum. Q.E.D. 2.5.11. Remark. It is known that circle–like continua expressed as an inverse limit of circles with bonding maps having degree one are planar continua which separate the plane. If the degree of the bonding maps are greater than one, then such continua are not planar [1]. Next, we present a generalization of arc–like and circle–like continua. 2.5.12. Definition. Let P be a collection of continua. If the continuum X is homeomorphic to an inverse limit of elements of P with surjective bonding maps, then X is said to be P–like. If P consists of just one element P , then X is said to be P –like. Copyright © 2005 Taylor & Francis Group, LLC


123

A proof of the following Theorem may be found in [12]. Note that it generalizes Theorems 2.4.22 and 2.5.8: 2.5.13. Theorem. Let P be a collection of polyhedra, and let X be a continuum. Then the following are equivalent: (a) For each ε > 0, there exists a finite open cover U of X such that mesh(U) < ε and the polyhedron, N (U), associated with the nerve of U, is homeomorphic to an element of P; (b) X is a P–like continuum; and (c) For each ε > 0, there exists a ε–map from X onto an element of P.

2.5.14. Corollary. If X is a continuum, then there exists a countable family P of polyhedra such that X is a P–like continuum. Proof. For each n ∈ IN, let Un be a finite open cover of X such that 1 mesh(Un ) < . Let P = {N (Un )}∞ n=1 . Then, by Theorem 2.5.13, n X is a P–like continuum. Q.E.D. 2.5.15. Lemma. Let P and R be two families of continua. If X is a P–like continuum and each element of P is an R–like continuum, then X is an R–like continuum. Proof. Let ε > 0. By Theorem 2.5.13, there exist an element P ∈ P and an ε–map f : X → → P . By Lemma 2.4.20, there exists δ > 0 such that if U is a subset of P with diam(U ) < δ, then diam(f −1 (U )) < ε. Since P is an R–like continuum, there exist an element R ∈ R and a δ–map g : P → → R. Hence, g ◦ f : X → →R is an ε–map from X onto R. Since the ε was arbitrary, by Theorem 2.5.13, X is R–like. Q.E.D. 2.5.16. Theorem. Let T = [−1, 1] × {0} ∪ {0} × [−1, 0] ⊂ IR2 . Then [0, 1] is T –like. Copyright © 2005 Taylor & Francis Group, LLC

124 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Proof. By Theorem 2.5.13, it suffices to show that for each ε > 0, there exists an ε–map from [0, 1] onto T . Let ε > 0. Let f : [0, 1] → → T be given by ⎧ t + ε − 1 ⎪ ⎪ if t ∈ [0, 1 − ε]; ⎪ 0, ⎪ ⎪ 1−ε ⎪ ⎪ ⎨ # ε$ 2 f (t) = if t ∈ 1 − ε, 1 − ; − (t + ε − 1), 0 ⎪ ε 2 ⎪ ⎪ # $ ⎪ ⎪ ε 1 ⎪ ⎪ if t ∈ 1 − , 1 . ⎩ (ε + 4(t − 1)), 0 ε 2 1 (-1 ,0)

(0,0)

1- 3

(1 ,0)

f

0

(0,-1)

Then f is an ε–map. Q.E.D. 2.5.17. Corollary. Let T = [−1, 1] × {0} ∪ {0} × [−1, 0] ⊂ IR2 . If X is an arc–like continuum, then X is T –like. Proof. It follows easily from Theorem 2.5.16 and Lemma 2.5.15. Q.E.D. The following Theorem is a slight modification of the Hahn– Mazurkiewicz Theorem:

2.5.18. Theorem. Let X be a locally connected continuum. If x and y are two points of X, then there exists a surjective map f : [0, 1] → → X such that f (0) = x and f (y) = 1. Copyright © 2005 Taylor & Francis Group, LLC


125

Proof. Since X is locally connected, by Hahn–Mazurkiewicz Theorem (8.14 of [18]), there exists a surjective map g : [0, 1] → → X. Let sx and sy be points of [0, 1] such that g(sx ) = x and g(sy ) = y. Let h : [0, 1] → → [0, 1] be given by ⎧

1 ⎪ ⎪ sx − 3sx t ⎪ if t ∈ 0, ; ⎪ ⎪ 3 ⎪ ⎪

⎨ 1 2 h(t) = 3t − 1 if t ∈ , ; ⎪ 3 3 ⎪

⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎩3(sy − 1)t − 2sy + 3 if t ∈ , 1 . 3 Then h is a surjective map such that h(0) = sx and h(1) = sy . Hence, f = g ◦ h is a surjective map from [0, 1] onto X such that f (0) = x and f (1) = y. Q.E.D. The proof of the following Theorem is due to Ray L. Russo. 2.5.19. Theorem. Let n ∈ IN. If B = {x ∈ IRn | ||x|| ≤ 1}, then each arc–like continuum is B–like. Proof. By Lemma 2.5.15, it suffices to show that [0, 1] is B–like. 1 ε Let ε > 0. Let m ∈ IN such that < . For each k ∈ {1, . . . , m}, m 2 let k−1 k n Ek = x ∈ IR 1 − ≤ ||x|| ≤ 1 − . m m 1 and Ek is an annular m m region, k ∈ {1, . . . , m − 1}. Note, also, that B = Ek . Note that Em is the closed ball of radius

k=1

1 Divide [0, 1] into m equal subintervals of length . By Theo m 1 rem 2.5.18, there exists a surjective map g1 : 0, → → E1 such m 1 that g1 ∈ E1 ∩ E2 . By the same Theorem, there exists a m Copyright © 2005 Taylor & Francis Group, LLC

126 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS

1 2 , m m

1 m

1 m

→ → E2 such that g2 = g1 surjective map g2 : 2 and g2 ∈ E2 ∩ E3 . In general, we have a surjective map m

k−1 k k−1 k−1 gk : , → → Ek such that gk−1 = gk for m m m m each k ∈ {2, . . . , m}.

k−1 k , Let f : [0, 1] → → B be given by f (t) = gk (t) if t ∈ . m m Then f is an ε–map. Hence, by Theorem 2.5.13, [0, 1] is B–like. Q.E.D.

2.6

Universal and A–H Essential Maps

We study universal maps and AH–essential maps. We present some of their properties. In particular, we show that these maps are the same, when the range space is an n–cell. We begin this section with the following Theorem: 2.6.1. Theorem. Let {Xn , fnn+1 } be an inverse of continua with the fixed point property, and surjective bonding maps. For each n ∈ IN, let hn : Xn → Xn be a map such that fnn+1 ◦hn+1 = hn ◦fnn+1 . If X∞ = lim{Xn , fnn+1 }, then h∞ = lim{hn } has a fixed point. ←−

←−

1 Proof. Let ε > 0. Let m ∈ IN such that m < ε. 2 Since each Xm has the fixed point property, there exists xm ∈ Xm such that hm (xm ) = xm . Since all the bonding maps are surjective, by Remark 2.1.6, the projection maps are surjective too. Hence, there exists a point z = (zn )∞ n=1 ∈ X∞ such that fm (z) = zm = xm . Since fnn+1 ◦ hn+1 = hn ◦ fnn+1 for each n ∈ IN, h∞ (z) = (z1 , z2 , . . . , zm−1 , xm , hm+1 (zm+1 ), . . . ). Copyright © 2005 Taylor & Francis Group, LLC

2.6. UNIVERSAL AND A–H ESSENTIAL MAPS

127

Hence, ∞

∞ 1 1 1 d(hn (zn ), zn ) ≤ = m < ε. ρ (h∞ (z), z) = n n 2 2 2 n=m+1 n=m+1

Therefore, by Lemma 2.4.17, h∞ has a fixed point. Q.E.D. 2.6.2. Remark. Let us observe that Theorem 2.6.1 does not imply that the inverse limit of continua with the fixed point property has the fixed point property; compare with Theorem 2.6.14.

2.6.3. Definition. A map f : X → Y between metric spaces is called universal provided that for any other map g : X → Y , there exists a point x ∈ X such that f (x) = g(x). As an easy consequence of the definition of a universal map we have the following two Propositions: 2.6.4. Proposition. Let f : X → Y be a universal map between metric spaces. Then the following hold: (1) f is surjective. (2) Y has the fixed point property. (3) If X0 is a subspace of X and f |X0 : X0 → Y is universal, then f is universal. (4) If g : Y → Z is a map between metric spaces and g◦f is universal, then g is universal.

2.6.5. Proposition. A metric space X has the fixed point property if and only if the identity map, 1X , is universal. 2.6.6. Lemma. Let X and Y be metric spaces, and let f : X → →Y be a universal map. If g : W → → X and h : Y → → Z are homeomorphisms, then f ◦ g and h ◦ f are universal. Copyright © 2005 Taylor & Francis Group, LLC

128 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Proof. We show f ◦ g is universal. The proof for h ◦ f is similar. Let : W → Y be a map. Note that ◦ g −1 : X → Y . Since f is universal, there exists x ∈ X such that ( ◦ g −1 )(x) = f (x). Hence, (g −1 (x)) = (f ◦ g)(g −1 (x)). Therefore, f ◦ g is universal. Q.E.D.

2.6.7. Proposition. For each n ∈ IN, let fn : X → Y be a map between compacta. Suppose the sequence {fn }∞ n=1 converges uniformly to a map f : X → Y . If each fn is universal, then f is universal. Proof. Let g : X → Y be a map. Since for each n ∈ IN, fn is universal, there exists xn ∈ X such that fn (xn ) = g(xn ) for every n ∈ IN. Since X is a compactum, the sequence {xn }∞ n=1 has a convergent subsequence {xnk }∞ . Let x be the limit point of k=1 ∞ {xnk }k=1 . Then f (x) = g(x). Therefore, f is universal. Q.E.D.

2.6.8. Theorem. Let X be a continuum. If f : X → → [0, 1] is a surjective map, then f is universal. Proof. Let g : X → [0, 1] be a map. Note that the sets {x ∈ X | f (x) ≤ g(x)} and {x ∈ X | f (x) ≥ g(x)} are nonempty closed subsets of X whose union is X. Hence, these sets have a point x in common. Thus, f (x) = g(x). Therefore, f is universal. Q.E.D.

2.6.9. Theorem. Let {Yn , gnn+1 } be an inverse sequence of compacta whose inverse limit is Y∞ , and let X be a compactum. For each n ∈ IN, let hn : X → Yn be a map such that gnn+1 ◦ hn+1 = hn . If each hn is universal, then the induced map h∞ = lim{hn } is ←− universal. Copyright © 2005 Taylor & Francis Group, LLC


129

Proof. Let k : X → Y∞ be a map. Since each hn is universal, for each n ∈ IN, there exists a point xn ∈ Xn such that (gn ◦ k)(xn ) = hn (xn ). Since hn (xn ) = (gn ◦ h∞ )(xn ), ρ(h∞ (xn ), k(xn )) = ≤

∞ 1 d ((g ◦ h∞ )(xn ), (g ◦ k)(xn )) 2 =n+1 ∞ 1 1 = 2 2n =n+1

for every n ∈ IN. Since X is a compactum, without loss of generality, we assume that the sequence {xn }∞ n=1 converges to a point x ∈ X. Hence, 1 ρ(h∞ (x), k(x)) = lim ρ(h∞ (xn ), k(xn )) ≤ lim n = 0. Thus, n→∞ n→∞ 2 h∞ (x) = k(x). Therefore, h∞ is universal. Q.E.D. 2.6.10. Corollary. Let {Yn , gnn+1 } be an inverse sequence of arcs, with surjective bonding maps, whose inverse limit is Y∞ . If X is a continuum and f : X → → Y∞ is a surjective map, then f is universal. Proof. Note that f = lim{gn ◦ f }. Since each Yn is an arc, by ←− Theorem 2.6.8, each gn ◦ f is universal. Hence, by Theorem 2.6.9, f is universal. Q.E.D. 2.6.11. Corollary. If X is an arc–like continuum, then X has the fixed point property.

2.6.12. Definition. Let Y be a metric space. We say that Y is an absolute retract (absolute neighborhood retract), provided that for any metric space X, and for any closed subset A of X, every map f : A → Y can be extended over Y (over a neighborhood (depending on f ) of A in X).


130 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.6.13. Remark. The concepts defined in Definition 2.6.12 are absolute extensor and absolute neighborhood extensor. It is known that these concepts are equivalent to the original definition of an absolute retract and an absolute neighborhood retract (1.5.2 of [15]). The following Theorem gives us a sufficient condition to obtain an inverse limit with the fixed point property. 2.6.14. Theorem. Let {Xn , fnn+1 } be an inverse sequence of absolute neighborhood retracts, where all the bonding maps fnm : Xm → → Xn are universal. Then X∞ = lim{Xn , fnn+1 } has the fixed point ←− property. Proof. First, we prove each projection map fn : X∞ → → Xn is ∞ universal. To this end, let n ∈ IN. Let (a )∞ X . For each =1 ∈ m ∈ IN, let im : Xm →

∞

=1

X be given by

=1 m im (x) = (f1m (x), f2m (x), . . . , fm−1 (x), x, am+1 , am+2 , . . . ).

→ im (Xm ) ⊂ Sm (Definition 2.1.7) is a homeoNote that im : Xm → morphism. Also note that given z ∈ Xm , im (z) is not necessarily an element of X∞ . Let k : X∞ → Xn be a map. We show there exists x = (x )∞ =1 ∈ X∞ such that k(x) = fn (x). Since Xn is an absolute neighborhood ∞ retract, there exist an open set G of X containing X∞ and a =1

map kˆ : G → Xn extending k. ∞ Sm (Proposition 2.1.8), by Lemma 1.6.7, there Since X∞ = m=1

exists M ∈ IN such that im (Xm ) ⊂ Sm ⊂ G for each m ≥ M . Without loss of generality, we assume that M ≥ n. Let m ≥ M . Then, by Lemma 2.6.6, fnm ◦ i−1 m is universal. Hence, since im (Xm ) ⊂ G, there exists xm ∈ Xm such that fnm ◦ m ˆ i−1 m (im (xm )) = fn (xm ) = k ◦ im (xm ). Note that this is true for Copyright © 2005 Taylor & Francis Group, LLC

2.6. UNIVERSAL AND A–H ESSENTIAL MAPS each m ≥ M . Since

∞

131

X is compact, without loss of generality,

=1

we assume that the sequence {im (xm )}∞ m=M converges to a point x = (x )∞ ∈ X . ∞ =1 Note that the bonding maps are surjective, since they are universal (Proposition 2.6.4 (1)). Hence, the projection maps are surjective (Remark 2.1.6). Thus, for each m ≥ M , there exists z m = m m (zm )∞ =1 ∈ X∞ such that fm (z ) = xm . Note that lim z = x . In m→∞ consequence: ˆ m (xm ) = lim f m (xm ) = ˆ ) = k( ˆ lim im (xm )) = lim k◦i k(x ) = k(x n m→∞

m→∞

lim fnm fm (z m ) = lim fn (z m ) = fn

m→∞

m→∞

m→∞

lim z m = fn (x ).

m→∞

Therefore, fn is universal. Now, observe that lim{fn } = 1X∞ . Hence, by Theorem 2.6.9, ←− 1X∞ is universal. Therefore, by Proposition 2.6.5, X∞ has the fixed point property. Q.E.D. A proof of the following Theorem may be found in 3.6.11 of [15].

2.6.15. Theorem. Each polyhedron is an absolute neighborhood retract.

2.6.16. Corollary. Let {Xn , fnn+1 } be an inverse sequence of polyhedra, where all the bonding maps fnm : Xm → → Xn are universal. Then X∞ = lim{Xn , fnn+1 } has the fixed point property. ←−

2.6.17. Definition. Let X be a metric space and let A be a subset of X. We say that A is a retract of X if there is a map r : X → →A such that r(a) = a for each a ∈ A. The map r is called a retraction.


132 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.6.18. Corollary. Let {Xn , fnn+1 } be an inverse sequence of absolute neighborhood retracts with the fixed point property. If each bonding map is a retraction, then lim{Xn , fnn+1 } has the fixed point ←− property. Proof. By Theorem 2.6.14, it suffices to show that each bonding map fnm : Xm → Xn is universal. Since each bonding map fnm is a retraction, fnm |Xn : Xn → → Xn is the identity map 1Xn . Since every Xn has the fixed point property, by Proposition 2.6.5, 1Xn is universal. Hence, fnm |Xn = 1Xn is universal. Therefore, by (3) of Proposition 2.6.4, fnm is universal. Q.E.D. From now on, we identify the manifold boundary, ∂ ([0, 1]n ), of the n-cell [0, 1]n with the unit (n − 1)–dimensional sphere S n−1 . 2.6.19. Theorem. Let X be a separable metric space such that every map g : X → S n−1 is homotopic to a constant map. Then a map f : X → → [0, 1]n is universal if and only if the restriction f |f −1 (S n−1 ) : f −1 (S n−1 ) → → S n−1 is not homotopic to a constant map. Proof. Let f : X → → [0, 1]n be a map and suppose that f |f −1 (S n−1 ) is homotopic to a constant map. Hence, by Theorem 1.3.5, there exists a map f : X → → S n−1 such that f |f −1 (S n−1 ) = f |f −1 (S n−1 ) . n−1 Let g : X → → S ⊂ [0, 1]n be given by g(x) = −f (x). Then f (x) = g(x) for any x ∈ X. Therefore, f is not universal. Next, suppose f : X → → [0, 1]n is not universal. Then there exists a map g : X → [0, 1]n such that f (x) = g(x) for any x ∈ X. Let f : X → → S n−1 be the map satisfying the following equation: ||f (x) − f (x)|| + ||f (x) − g(x)|| = ||f (x) − g(x)||.

f'(x)

f(x) g(x)



133

Then f is map such that f |f −1 (S n−1 ) = f |f −1 (S n−1 ) . Note that, by hypothesis, f is homotopic to a constant map. Q.E.D. 2.6.20. Definition. Let n= {nk }∞ k=1 be a sequence of positive integers. For each k ∈ IN, let fkk+1 : S 1 → → S 1 be given by fkk+1 (z) = z nk . The n–solenoid, denoted by Σn , is lim{Xk , fkk+1 }, where Xk = S 1 ←− for every k ∈ IN. Note that, by Theorem 2.1.19, Σn is an indecomposable continuum. 2.6.21. Theorem. Let n= {nk }∞ k=1 be a sequence of positive integers. Then the cone over the n–solenoid, Σn , has the fixed point property. Proof. Let Σn = lim{Xk , fkk+1 }, where every Xk = S 1 . By Theo←−

rem 2.3.2, K(Σn ) is homeomorphic to lim{K(Xk ), K(fkk+1 )}. It is ←−

well known that K(S 1 ) is homeomorphic to {x ∈ IR2 | ||x|| ≤ 1} (which is, in turn, homeomorphic to [0, 1]2 ). Since K(Σn ) is contractible, then every map from K(Σn ) into S 1 is homotopic to a constant map (Theorem 1.3.12). Observe that for each k ∈ IN, K(fk )−1 (S 1 ) = Σn × {0}. Hence, each K(fk )|K(fk )−1 (S 1 ) : K(fk )−1 (S 1 ) → S 1 is not homotopic to a constant map. Thus, by Theorem 2.6.19, every K(fk ) is universal. Since the induced map lim{K(fk )} = 1K(Σn ) , by Theorem 2.6.9, ←−

1K(Σn ) is universal. Therefore, by Proposition 2.6.5, K(Σn ) has the fixed point property. Q.E.D. 2.6.22. Definition. Let X be a metric space. A map f : X → [0, 1]n is said to be AH–inessential provided that there exists a map g : X → S n−1 such that f |f −1 (S n−1 ) = g|f −1 (S n−1 ) . The map f is AH–essential if it is not AH–inessential. 2.6.23. Lemma. Let X be a separable metric space. If f : X → → [0, 1]n is AH-essential, then f |f −1 (S n−1 ) : f −1 (S n−1 ) → → S n−1 is not homotopic to a constant map. Copyright © 2005 Taylor & Francis Group, LLC

134 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS Proof. Suppose f |f −1 (S n−1 ) is homotopic to a constant map g . Then, clearly, g can be extended to a constant map g defined on X. Hence, by Theorem 1.3.5, there exists a map fˆ: X → S n−1 such that fˆ|f −1 (S n−1 ) = f |f −1 (S n−1 ) and fˆ is homotopic to g. Hence, f is AH–inessential. Q.E.D. 2.6.24. Corollary. Let X be a separable metric space. If f : X → → 2 −1 [0, 1] is AH-essential, then there exists a component K of f (S 1 ) such that f |K : K → S 1 is not homotopic to a constant map. Proof. By Lemma 2.6.23, f |f −1 (S 1 ) is not homotopic to a constant map. Hence, by Theorem 1.3.7, there exists a subcontinuum C of f −1 (S 1 ) such that f |C is not homotopic to a constant map. Let K be the component of f −1 (S 1 ) such that C ⊂ K. Then f |K : K → → S1 is not homotopic to a constant map. Q.E.D. 2.6.25. Theorem. Let X be a metric space. Then the map f : X → → [0, 1]n is AH–essential if and only if f is universal. Proof. Suppose f is not universal. Then there exists a map g : X → [0, 1]n such that f (x) = g(x) for any x ∈ X. Let f : X → S n−1 be the map satisfying the equation ||f (x) − f (x)|| + ||f (x) − g(x)|| = ||f (x) − g(x)||. Then f is an extension of f |f −1 (S n−1 ) to X. Therefore, f is AH– inessential. Next, suppose f is AH–inessential. Then there exists a map g : X → S n−1 such that g|f −1 (S n−1 ) = f |f −1 (S n−1 ) . Let h : X → S n−1 be given by h(x) = −g(x). Hence, f (x) = h(x) for any x ∈ X. Therefore, f is not universal. Q.E.D. Now, our goal is to prove Mazurkiewicz’s Theorem about the existence of indecomposable subcontinua in continua of dimension at least two. A proof of the following Theorem may be found in 18.6 of [19].



135

2.6.26. Theorem. Let X be a separable metric space and let n ∈ IN. Then dim(X) ≥ n if and only if there is an AH-essential map of X to [0, 1]n .

2.6.27. Lemma. Let X be a continuum, let f : X → → [0, 1]2 be an AH–essential map, and let J be a simple closed curve contained in [0, 1]2 . If H is the bounded complementary domain (in IR2 ) of J, then f |f −1 (H∪J) : f −1 (H ∪ J) → → H ∪ J is an AH–essential map. Proof. Since H ∪ J is homeomorphic to [0, 1]2 , by Theorem 2.6.25, it suffices to show that f |f −1 (H∪J) : f −1 (H ∪J) → H ∪J is universal. Let g : f −1 (H ∪J) → H ∪J be a map. Since H ∪J is an absolute retract (see 1.5.5 of [15]), there exists a map gˆ : X → H ∪ J such that gˆ|f −1 (H∪J) = g. Since f is a universal map (Theorem 2.6.25), there exists x ∈ X such that f (x) = gˆ(x). Note that f (x) ∈ H ∪ J. Hence, x ∈ f −1 (H ∪ J). Thus, gˆ(x) = g(x). Therefore, f |f −1 (H∪J) is a universal map. Q.E.D. 2.6.28. Corollary. Let X be a continuum, and let f : X → → [0, 1]2 be an AH–essential map. If J is a simple closed curve contained in [0, 1]2 , then there exists a subcontinuum K of X such that f (K) = J. Proof. Note that, by Lemmas 2.6.27 and 2.6.23, f |f −1 (J) : f −1 (J) → → J is not homotopic to a constant map. Hence, by Corollary 2.6.24, there exists a component K of f −1 (J) such that f |K : K → → J is not homotopic to a constant map. In particular, f (K) = J. Q.E.D. A proof of the following Theorem may be found on p. 157 of [13]. 2.6.29. Theorem. If C is a subcontinuum of [0, 1]2 , then there exists a sequence {Jn }∞ n=1 of simple closed curves such that lim Jn = n→∞ C.


136 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.6.30. Definition. A map f : X → Y between continua is said to be weakly confluent provided that for every subcontinuum Z of Y , there exists a subcontinuum W of X such that f (W ) = Z.

2.6.31. Theorem. Let X be a continuum. If f : X → → [0, 1]2 is an AH–essential map, then f is weakly confluent. Proof. Let C be a subcontinuum of [0, 1]2 . Then, by Theorem 2.6.29, there exists a sequence {Jn }∞ n=1 of simple closed curves converging to C. By Corollary 2.6.28, for each n ∈ IN, there exists a subcontinuum Kn of X such that f (Kn ) = Jn . Since C(X) is compact (Theorem 1.8.5), without loss of generality, we assume that the sequence {Kn }∞ n=1 converges to a subcontinuum K of X. Therefore, by Corollary 1.8.23, f (K) = C. Q.E.D. As a consequence of Theorems 2.6.26 and 2.6.31, we have the following: 2.6.32. Corollary. If X is a continuum of dimension greater than one, then there exists a weakly confluent map f : X → → [0, 1]2 . Now, we are ready to show Mazurkiewicz’s Theorem. 2.6.33. Theorem. If X is a continuum of dimension greater than one, then X contains an indecomposable continuum. Proof. Since the dimension of X is greater than one, by Corollary 2.6.32, there exists a weakly confluent map f : X → [0, 1]2 . Let K be an indecomposable subcontinuum (like Knaster’s continuum, Example 2.4.7) of [0, 1]2 . Since f is weakly confluent, there exists a subcontinuum Y of X such that f (Y ) = K. Using Kuratowski–Zorn Lemma, there exists a subcontinuum Y of Y such that f (Y ) = K and for each proper subcontinuum C of Y , f (C) = K. We assert that Y is indecomposable. Suppose this is not true. Then there are two proper subcontinua A and B of Y such that Copyright © 2005 Taylor & Francis Group, LLC


137

Y = A ∪ B. Since A and B are proper subcontinua of Y , f (A) = K and f (B) = K. Note that K = f (Y ) = f (A ∪ B) = f (A) ∪ f (B). Hence, K is decomposable, a contradiction. Therefore, Y is an indecomposable subcontinuum of X. Q.E.D. 2.6.34. Corollary. Each hereditarily decomposable continuum is of dimension one. To finish this section, we describe the Waraszkiewicz spirals [23] and present some applications. Let M be the following subset of IR2 , defined in polar coordinates (e denotes the real exponential map): M = {(1, θ) | θ ≥ 0}∪{(1+e−θ , θ) | θ ≥ 0}∪{(1+e−θ , −θ) | θ ≥ 0}. Hence, M is the union of the unit circle S 1 , a ray R+ spiraling in a counterclockwise direction onto S 1 , and a ray R− spiraling in a clockwise direction onto S 1 . 2.6.35. Definition. The Waraszkiewicz spirals are subcontinua of M. Each spiral is homeomorphic to a compactification of [0, 1) with remainder S 1 . To determine a Waraszkiewicz spiral, begin at the point with Cartesian coordinates (2, 0) and follow R+ or R− . At any time that the positive x–axis is crossed, feel free to change rays. This procedure determines the uncountable collection of continua called the Waraszkiewicz spirals. Waraszkiewicz gave a proof of the following result in [23]. Ray L. Russo has improved the proof [22]: 2.6.36. Theorem. No continuum can be mapped onto all Waraszkiewicz spirals.


138 CHAPTER 2. INVERSE LIMITS AND RELATED TOPICS 2.6.37. Corollary. Given a countable collection of continua, there exists a Waraszkiewicz spiral W such that no member of this collection can be mapped onto W . Proof. Suppose the result is not true. Then there exists a countable family {Xn }∞ n=1 of continua such that for each Waraszkiewicz spiral W , there is n ∈ IN such that Xn can be mapped onto W . Without loss of generality, we assume that the members of the family {Xn }∞ n=1 are pairwise disjoint and lim diam(Xn ) = 0. For n→∞ each n ∈ IN, let xn ∈ Xn . Identify all the points of the sequence {xn }∞ n=1 to a point ω. In this way we obtain an infinite wedge. We call this quotient space X. Note that, by construction, X is a continuum.

Observe that, for each n ∈ IN, there exists a retraction rn : X → → Xn given by ω if x ∈ Xm and n = m; rn (x) = x if x ∈ Xn . Hence, X can be mapped onto all the Waraszkiewicz spirals, a contradiction to Theorem 2.6.36. Q.E.D. 2.6.38. Definition. A continuum X is hereditarily equivalent provided that X is homeomorphic to each of its nondegenerate subcontinuum.

2.6.39. Corollary. Each hereditarily equivalent continuum has dimension one. Copyright © 2005 Taylor & Francis Group, LLC


139

Proof. Let X be a continuum of dimension greater than one. By Corollary 2.6.32, there exists a weakly confluent map f : X → → [0, 1]2 . By Theorem 2.6.36, there exists a Waraszkiewicz spiral W such that X cannot be mapped onto W . We assume that W ⊂ [0, 1]2 . Since f is weakly confluent, there exists a subcontinuum Y of X such that f (Y ) = W . Hence, Y and X are not homeomorphic. Therefore, X is not hereditarily equivalent. Q.E.D. 2.6.40. Corollary. If X is a continuum of dimension greater than one, then X contains uncountably many nonhomeomorphic subcontinua. Proof. Suppose X has countably many nonhomeomorphic subcontinua {Xn }∞ n=1 . By Corollary 2.6.37, there exists a Waraszkiewicz spiral W such that Xn cannot be mapped onto W for any n ∈ IN. We assume that W ⊂ [0, 1]2 . Since X is a continuum of dimension greater than one, by Corollary 2.6.32, there exists a weakly confluent map f : X → → [0, 1]2 . Hence, there exists a subcontinuum Y of X such that f (Y ) = W . This implies that Y is not homeomorphic to Xn for any n ∈ IN, a contradiction. Therefore, X contains uncountably many nonhomeomorphic subcontinua. Q.E.D. An argument for the following result may be found on p. 484 of [21]. 2.6.41. Theorem. For each Waraszkiewicz spiral W , there exists ' such that each nondegenerate subcontinuum a plane continuum W ' can be mapped onto W . of W 2.6.42. Definition. A metric space X is homogeneous provided that for each pair of points x, y ∈ X, there exists a homeomorphism h: X → → X such that h(x) = y. We end this chapter showing that homogeneous hereditarily indecomposable continua are one–dimensional.


140

REFERENCES

2.6.43. Theorem. If X is a hereditarily indecomposable continuum of dimension greater than one, then X is not homogeneous. Proof. Let x be a point of X, and let [[x, X]] be the family of all subcontinua of X containing x, with the topology generated by the Hausdorff metric (Theorem 1.8.3). Note that [[x, X]] is the image of an order arc. Hence, [[x, X]] is homeomorphic to [0, 1]. Let {Xn }∞ n=1 be a countable dense subset of [[x, X]]. By Corollary 2.6.37, there exists a Waraszkiewicz spiral W such that Xn cannot be mapped onto W for every n ∈ IN. By Theo' in [0, 1]2 such that each rem 2.6.41, there exists a continuum W ' is constant for each n ∈ IN. map from Xn into W Suppose X is homogeneous. Since X is of dimension greater than one, by Corollary 2.6.32, there exists a weakly confluent map f: X → → [0, 1]2 . Hence, there exists a subcontinuum Z of X such ' . Since X is homogeneous, we assume that x ∈ Z, that f (Z) = W i.e., Z ∈ [[x, X]]. Since {Xn }∞ n=1 is dense in [[x, X]], there exists ∞ a subsequence {Xnk }∞ of {X n }n=1 such that lim Xnk = Z and k=1 k→∞

Xnk ⊂ Z for every k ∈ IN. Since for each k ∈ IN, x ∈ Xnk , f (Xnk ) = {f (x)}. This implies the contradiction that f (Z) = {f (x)}. Therefore, X is not homogeneous. Q.E.D.

REFERENCES

[1] R. H. Bing, Embedding Circle–like Continua in the Plane, Canad. J. Math., 14 (1962), 113–128. [2] C. E. Capel, Inverse Limit Spaces, Duke Math. J. 21 (1954), 233–245. [3] C. O. Christenson and W. L. Voxman, Aspects of Topology, Monographs and Textbooks in Pure and Applied Math., Vol. 39, Marcel Dekker, New York, Basel, 1977. [4] J. Dugundji, Topology, Allyn and Bacon, Inc., Boston, 1966. Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

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[5] J. Grispolakis and E. D. Tymchatyn, On Confluent Mappings and Essential Mappings–A Survey, Rocky Mountain J. Math., 11 (1981), 131–153. [6] O. H. Hamilton, A Fixed Point Theorem for Pseudo–arcs and Certain Other Metric Continua, Proc. Amer. Math. Soc., 2 (1951), 173–174. [7] J. G. Hocking and G. S. Young, Topology, Dover Publications, Inc., New York, 1988. [8] W. Holszty´ nski, Universal Mappings and Fixed Point Theorems, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys., 15 (1967), 433–438 [9] W. Holszty´ nski, A Remark on the Universal Mappings of 1– dimensional Continua, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys., 15 (1967), 547–549. [10] W. T. Ingram, Inverse Limits, Aportaciones Matemáticas, Textos # 15, Sociedad Matemática Mexicana, 2000. [11] K. Kuratowski, Topology, Vol. II, Academic Press, New York, 1968. [12] S. Mardeˇsić and J. Segal, ε–mappings onto Polyhedra, Trans. Amer. Math. Soc., 109 (1963), 146–164. [13] S. Mazurkiewicz, Sur les Continus Absolutment Indécomposables, Fund. Math., 16 (1930), 151–159. [14] S. Mazurkiewicz, Sur L’existence des Continus Indéscomposables, Fund. Math., 25 (1935), 327–328. [15] J. van Mill, Infinite–Dimensional Topology, North Holland, Amsterdam, 1989. [16] S. B. Nadler, Jr., Multicoherence Techniques Applied to Inverse Limits, Trans. Amer. Math. Soc., 157 (1971), 227–234. [17] S. B. Nadler, Jr., Hyperspaces of Sets, Monographs and Textbooks in Pure and Applied Math., Vol. 49, Marcel Dekker, New York, Basel, 1978. Copyright © 2005 Taylor & Francis Group, LLC

142

REFERENCES

[18] S. B. Nadler, Jr., Continuum Theory: An Introduction, Monographs and Textbooks in Pure and Applied Math., Vol. 158, Marcel Dekker, New York, Basel, Hong Kong, 1992. [19] S. B. Nadler, Jr., Dimension Theory: An Introduction with Exercises, Aportaciones Matemáticas, Textos # 18, Sociedad Matemática Mexicana, 2002. [20] J. T. Rogers, Jr., The Pseudo–circle is not Homogeneous, Trans. Amer. Math. Soc. 148 (1970), 417–428. [21] J. T. Rogers, Jr., Orbits of Higher–Dimensional Hereditarily Indecomposable Continua, Proc. Amer. Math. Soc., 95 (1985), 483–486. [22] R. L. Russo, Universal Continua, Fund. Math., 105 (1979), 41–60. [23] Z. Waraszkiewicz, Sur un Probléme de M. H. Hahn, Fund. Math., 18 (1932), 118–137.


Chapter 3 JONES’S SET FUNCTION T

We prove basic results about the set function T defined by F. Burton Jones [17]. We define this function on compacta and then we concentrate on continua. In particular, we present some of the well known properties (such as connectedness im kleinen, local connectedness, semi–local connectedness, etc.) using the set function T . The notion of aposyndesis was the main motivation of Jones to define this function. We present some properties of a continuum assuming the continuity of the set function T . We also give some applications.

3.1

The Set Function T

3.1.1. Definition. Given a compactum X, the power set of X, denoted by P(X), is: P(X) = {A | A ⊂ X}.


143

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CHAPTER 3. JONES’S SET FUNCTION T

3.1.2. Remark. Let X be a metric space. Let us note that if A ∈ P(X), then its closure satisfies: Cl(A) = {x ∈ X | for each open subset U of X such that x ∈ U, we have that U ∩ A = ∅}. As we see below (Definition 3.1.3), this property is similar to the definition of the function T . 3.1.3. Definition. Let X be a compactum. Define T : P(X) → P(X) by T (A) = {x ∈ X | for each subcontinuum W of X such that x ∈ Int(W ), we have that W ∩ A = ∅}, for each A ∈ P(X). The function T is called Jones’s set function T. 3.1.4. Remark. In general, when working with the set function T , we usually work with complements. Hence, for any compactum X and any A ∈ P(X), we have that: T (A) = X \ {x ∈ X | there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ A}. 3.1.5. Remark. Let X be a compactum. If A ∈ P(X), then A ⊂ T (A), and T (A) is closed in X. Hence, the range of T is 2X (see Definition 1.8.1), i.e.: T : P(X) → 2X . The following proposition gives a relation between aposyndesis and T .


3.1. THE SET FUNCTION T

145

3.1.6. Proposition. Let X be a compactum, and let A ∈ P(X). If x ∈ X \ T (A), then X is aposyndetic at x with respect to each point of A. Proof. Let x ∈ X \ T (A), and let a ∈ A. Since x ∈ X \ T (A), by Definition 3.1.3, there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ A. In particular, W ⊂ X \ {a}. Therefore, X is aposyndetic at x with respect to a. Q.E.D. 3.1.7. Proposition. Let X be a compactum. If A, B ∈ P(X) and A ⊂ B, then T (A) ⊂ T (B). Proof. Let x ∈ X \ T (B). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ B. Since X \ B ⊂ X \ A, x ∈ Int(W ) ⊂ W ⊂ X \ A. Therefore, x ∈ X \ T (A). Q.E.D. 3.1.8. Corollary. Let X be a compactum. If A, B ∈ P(X), then T (A) ∪ T (B) ⊂ T (A ∪ B).

3.1.9. Remark. Let us note that the reverse inclusion of Corollary 3.1.8 is, in general, not true (see Example 3.1.18). It is an open question to characterize continua X such that T (A ∪ B) = T (A) ∪ T (B) for each A, B ∈ P(X). Let us see a couple of examples. 3.1.10. Example. Let X be the Cantor set. Then T (∅) = X. To see this, suppose there is a point x ∈ X \ T (∅). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ ∅ = X. Since X is the Cantor set, X is totally disconnected and perfect. Hence, no subcontinuum of X has interior. Therefore, W cannot exist. Hence, T (∅) = X. Note that, by Proposition 3.1.7, T (A) = X for each A ∈ P(X).


146


∞ 1 3.1.11. Example. Let X = {0} ∪ . Then T (∅) = {0}. n n=1 This follows from the fact that the only subcontinuum of X with empty interior is {0}. Hence, by Remark 3.1.5 and Proposition 3.1.7, T (A) = {0} ∪ A for each A ∈ P(X).

3.1.12. Remark. Note that if X is as in Example 3.1.10 or 3.1.11, then T |2X : 2X → 2X is continuous, where 2X has the topology given by the Hausdorff metric. In Section 3.2, we discuss the continuity of T on continua. The following Theorem gives us a characterization of compacta X for which T (∅) = ∅. 3.1.13. Theorem. Let X be a compactum. Then T (∅) = ∅ if and only if X has finitely many components. Proof. Suppose X has finitely many components. Let x ∈ X, and let C be the component of X such that x ∈ C. Hence, C is a subcontinuum of X and, by Lemma 1.6.2, x ∈ Int(C). Thus, each point of X is contained in the interior of a proper subcontinuum of X. Therefore, T (∅) = ∅. Now, suppose T (∅) = ∅. Then for each point x ∈ X, there exists a subcontinuum Wx of X such that x ∈ Int(Wx ) ⊂ Wx ⊂ X \ ∅. Hence, {Int(Wx ) | x ∈ X} is an open cover of X. Since X is n compact, there exist x1 , . . . , xn ∈ X such that X = Int(Wxj ) ⊂ n

j=1

Wxj ⊂ X. Thus, X is the union of finitely many continua.

j=1

Therefore, X has finitely many components. Q.E.D. 3.1.14. Corollary. If X is a continuum, then T (∅) = ∅. Let us see more examples.



147

3.1.15. Example. Let X be the cone over the Cantor set, with vertex νX and base B. Since the Cantor set is totally disconnected and perfect, it follows that the subcontinua of X having nonempty interior must contain νX . Hence, T ({νX }) = X. If r ∈ X \ {νX }, then T ({r}) is the line segment from r to the base B. Also, T (B) = B. v

X

X r

B

3.1.16. Example. Let Y = X ∪ X , where X is the cone over the Cantor set, with vertex νX and base B, and X is another copy of the cone over the Cantor set, with vertex νX ∈ B. In this case, T ({νX }) = X, T ({νX }) = X and T (T ({νX })) = T 2 ({νX }) = Y . Note that this implies, in general, that the function T is not idempotent (see Definition 3.1.51). v

X

X

Y

v

X'

X'



148

3.1.17. Example. Let X be the topologist sine curve, see Exam2 1 . Since X is ple 2.4.5. Let J = X \ x, sin 0<x≤ x π not locally connected at any point of J, if p ∈ J, then T ({p}) = J. Also, T ({q}) = {q} for each q ∈ X \ J. The following example shows that, in general, T (A ∪ B) = T (A) ∪ T (B) does not hold. 3.1.18. Example. Let X be the suspension over the harmonic ∞ 1 sequence {0} ∪ , with vertices a and b. Note that for n n=1 each x ∈ X, T ({x}) = {x}. In particular, T ({a}) = {a} and T ({b}) = {b}; however T ({a, b}) consists of the limit segment from a to b.

a

b

According to VandenBoss (p. 18 of [24]), the following result is due to H. S. Davis. 3.1.19. Theorem. Let X be a compactum, and let A and B be nonempty closed subsets of X. Then the following are equivalent: (1) T (A) ∩ B = ∅. (2) There exist two closed subsets M and N of X such that A ⊂ Int(M ), B ⊂ Int(N ) and T (M ) ∩ N = ∅. Proof. Assume (1), we show (2). For each b ∈ B, there exists a subcontinuum Wb of X such that b ∈ Int(Wb ) ⊂ Wb ⊂ X \ A. Since B is compact, there exist b1 , . . . , bn ∈ B such that B ⊂ n n Int(Wbj ) ⊂ Wbj ⊂ X \ A. Since X is a metric space, there j=1

j=1



149

exist two open sets U and V of X suchn that A ⊂ U ⊂ Cl(U ) ⊂

n Wbj and B ⊂ V ⊂ Cl(V ) ⊂ Int(Wbj ). X\ j=1

j=1

Let M = Cl(U ) and let N = Cl(V ). Then M and N are closed subsets of X such that A ⊂ Int(M ), B ⊂ Int(N ) and T (M ) ∩ N = ∅. The other implication follows easily from Proposition 3.1.7. Q.E.D. The following Corollary is an easy consequence of proof of Theorem 3.1.19. 3.1.20. Corollary. Let X be a continuum, and let A be a closed subset of X. If x ∈ X \ T (A), then there exists an open subset U of X such that A ⊂ U and x ∈ X \ T (Cl(U )). 3.1.21. Theorem. Let X be a continuum. If W is a subcontinuum of X, then T (W ) is also a subcontinuum of X. Proof. By Remark 3.1.5, T (W ) is a closed subset of X. Suppose T (W ) is not connected. Then there exist two disjoint closed subsets A and B of X such that T (W ) = A ∪ B. Since W is connected, without loss of generality, we assume that W ⊂ A. Since X is a metric space, there exists an open subset U of X such that A ⊂ U and Cl(U ) ∩ B = ∅. U

B

A W

Note that Bd(U )∩T (W ) = ∅. Hence, for each z ∈ Bd(U ), there exists a subcontinuum Kz of X such that z ∈ Int(Kz ) ⊂ Kz ⊂ X \ A. Since Bd(U ) is compact, there exist z1 , . . . , zn ∈ Bd(U ) n n n such that Bd(U ) ⊂ Int(Kzj ) ⊂ Kzj . Let V = U \ Kzj . j=1


j=1

j=1

150


Let Y = X \ V = (X \ U )

n

Kzj . By Theorem 1.7.22, Y has

j=1

a finite number of components. Observe that B ⊂ X \ Cl(U ) ⊂ X \ U ⊂ Y . Hence, B ⊂ Int(Y ). Let b ∈ B, and let C be the component of Y such that b ∈ C. By Lemma 1.6.2, b ∈ Int(C). By construction, C ∩ W = ∅. Thus, b ∈ X \ T (W ), a contradiction. Therefore, T (W ) is connected. Q.E.D. The set function T may be used to define some of the properties of continua like connectedness im kleinen or local connectedness. In the following Theorems we show the equivalence between both types of definitions. The following Theorem gives us a local definition of almost connectedness im kleinen using T . 3.1.22. Theorem. Let X be a continuum, with metric d. If p ∈ X, then X is almost connected im kleinen at p if and only if for each A ∈ P(X) such that p ∈ Int(T (A)), p ∈ Cl(A). Proof. Suppose X is almost connected im kleinen at p. Let A ∈ P(X) such that p ∈ Int(T (A)). Then there exists N ∈ IN such that V d1 (p) ⊂ Int(T (A)) for each n ≥ N . Since X is almost connected n im kleinen at p, for each n ≥ N , there exists a subcontinuum Wn of X such that Int(Wn ) = ∅ and Wn ⊂ V d1 (p) ⊂ Cl(V d1 (p)) ⊂ T (A). n n Hence, Wn ∩ A = ∅ for each n ≥ N . Let xn ∈ Wn ∩ A for each n ≥ N . Note that, by construction, the sequence {xn }∞ n=N converges to p and {xn }∞ ⊂ A. Therefore, p ∈ Cl(A). n=N Now suppose that p ∈ X satisfies that for each A ∈ P(X) such that p ∈ Int(T (A)), p ∈ Cl(A). Let U be an open subset of X such that p ∈ U . Let V an open subset of X such that p ∈ V ⊂ Cl(V ) ⊂ U . If some component of Cl(V ) has nonempty interior, then X is almost connected im kleinen at p. Suppose, then, that all the components of Cl(V ) have empty interior. Let A = Bd(V ). Then A is a closed subset of X and p ∈ X \ A. We show that V ⊂ T (A). To this end, suppose there exists x ∈ V \ T (A). Thus, there exists a subcontinuum W of X such that Copyright © 2005 Taylor & Francis Group, LLC


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x ∈ Int(W ) ⊂ W ⊂ X \ A. Since all the components of Cl(V ) have empty interior, and W is a subcontinuum with nonempty interior, we have that W ∩ (X \ V ) = ∅ and, of course, W ∩ V = ∅. Since W is connected, W ∩ Bd(V ) = W ∩ A = ∅, a contradiction. Hence, V ⊂ T (A). Since p ∈ V ⊂ T (A), by hypothesis, p ∈ Cl(A) = A, a contradiction. Therefore, Cl(V ) has component with nonempty interior, and X is almost connected im kleinen at p. Q.E.D. The next Theorem gives us a global definition of almost connectedness im kleinen using T . 3.1.23. Theorem. A continuum X is almost connected im kleinen if and only if for each closed subset, F , of X, Int(F ) = Int(T (F )). Proof. Suppose X is almost connected im kleinen. Let F be a closed subset of X. Since F ⊂ T (F ) (Remark 3.1.5), Int(F ) ⊂ Int(T (F )). Let x ∈ Int(T (F )). Then, by Theorem 3.1.22, x ∈ Cl(F ) = F . Hence, Int(T (F )) ⊂ F . Therefore, Int(T (F )) ⊂ Int(F ) and Int(T (F )) = Int(F ). Now, suppose that Int(F ) = Int(T (F )) for each closed subset F of X. Let x ∈ X, and let U be an open subset of X such that x ∈ U . Since Int(X \ U ) ∩ U = ∅, by hypothesis, Int(T (X \ U )) ∩ U = ∅. Hence, there exists y ∈ U such that y ∈ X \ T (X \ U ). Thus, there is a subcontinuum K of X such that y ∈ Int(K) ⊂ K ⊂ X \ (X \ U ) = U . Therefore, X is almost connected im kleinen at x. Q.E.D. 3.1.24. Theorem. Let X be continuum. If p ∈ X, then X is connected im kleinen at p if and only if for each A ∈ P(X) such that p ∈ T (A), p ∈ Cl(A). Proof. Let p ∈ X. Assume X is connected im kleinen at p. Let A ∈ P(X) and suppose p ∈ X \ Cl(A). Hence, Cl(A) is a closed subset of X \ {p}. Since X is connected im kleinen at p, there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ A. Therefore, p ∈ X \ T (A). Copyright © 2005 Taylor & Francis Group, LLC

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Now, suppose that p ∈ X satisfies that for each A ∈ P(X) such that p ∈ T (A), p ∈ Cl(A). Let D be a closed subset of X such that D ⊂ X \ {p}. Since p ∈ X \ D, by hypothesis, p ∈ X \ T (D). Then there exists a subcontinuum K of X such that p ∈ Int(K) ⊂ K ⊂ X \ D. Therefore, X is connected im kleinen at p. Q.E.D. 3.1.25. Corollary. Let X be continuum. If p ∈ X, then X is connected im kleinen at p if and only if for each closed subset A of X such that p ∈ T (A), p ∈ A. The following Theorem follows from Definition 1.7.13: 3.1.26. Theorem. Let X be a continuum, and let p, q ∈ X. Then X is semi–aposyndetic at p and q if and only if either p ∈ X \T ({q}) or q ∈ X \ T ({p}). The next Theorem follows from the definition of aposyndesis: 3.1.27. Theorem. Let X be a continuum, and let p, q ∈ X. Then X is aposyndetic at p with respect to q if and only if p ∈ X \T ({q}). The following Theorem is a global version of Theorem 3.1.27. 3.1.28. Theorem. A continuum X is aposyndetic if and only if T ({p}) = {p} for each p ∈ X. Proof. Suppose X is aposyndetic. Let p ∈ X. Then for each q ∈ X \ {p}, X is aposyndetic at q with respect to p. Hence, by Remark 3.1.5 and Theorem 3.1.27, q ∈ X \ T ({p}). Therefore, T ({p}) = {p}. Now, suppose T ({x}) = {x} for each x ∈ X. Let p, q ∈ X such that p = q. Since T ({q}) = {q}, p ∈ X \ T ({q}). Thus, by Theorem 3.1.27, X is aposyndetic at p with respect to q. Since p and q were arbitrary points of X, X is aposyndetic. Q.E.D.



153

3.1.29. Theorem. Let X be a continuum. If p ∈ X, then X is semi–locally connected at p if and only if T ({p}) = {p}. Proof. Suppose X is semi–locally connected at p. By Remark 3.1.5, {p} ⊂ T ({p}). Let q ∈ X \ {p}, and let U be an open subset of X such that p ∈ U and q ∈ X \ Cl(U ). Since X is semi– locally connected at p, there exists an open subset V of X such that p ∈ V ⊂ U and X \ V has finitely many components. By Lemma 1.6.2, q belongs to the interior of the component of X \ V containing q. Hence, q ∈ X \ T ({p}). Therefore, T ({p}) = {p}. Now, suppose T ({p}) = {p}. Let U be an open subset of X such that p ∈ U . Since T ({p}) = {p}, for each q ∈ X \ U , there exists a subcontinuum Wq of X such that q ∈ Int(Wq ) ⊂ Wq ⊂ X \ {p}. Note that {Int(Wq ) | q ∈ X \ U } is an open cover of X \ U . Since this set is compact, there exist . , qn ∈ X \U such that X \U ⊂

n q1 , . . n Int(Wqj ). Let V = X \ Wqj . Then V is an open subset of j=1

j=1

X such that p ∈ V ⊂ U and X \ V has finitely many components. Therefore, X is semi–locally connected at p. Q.E.D. As a consequence of Theorems 3.1.28 and 3.1.29, we have: 3.1.30. Corollary. A continuum X is aposyndetic if and only if it is semi–locally connected. The next Theorem gives us a characterization of local connectedness in terms of the set function T . 3.1.31. Theorem. A continuum X is locally connected if and only if for each closed subset A of X, T (A) = A. Proof. Suppose X is locally connected. Let A be a closed subset of X. Then X \ A is an open subset of X. Let p ∈ X \ A. Then there exists an open subset U of X such that p ∈ U ⊂ Cl(U ) ⊂ X \ A. Since X is locally connected, there exists an open connected subset V of X such that p ∈ V ⊂ U . Hence, Cl(V ) is a subcontinuum Copyright © 2005 Taylor & Francis Group, LLC

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of X such that p ∈ V ⊂ Cl(V ) ⊂ Cl(U ) ⊂ X \ A. Consequently, p ∈ X \ T (A). Thus T (A) ⊂ A. By Remark 3.1.5, A ⊂ T (A). Therefore, T (A) = A. Now, suppose T (A) = A for each closed subset of X. Let p ∈ X, and let U be an open subset of X such that p ∈ U . Then X \ U is closed subset of X such that p ∈ X \ U . Since X \ U = T (X \ U ), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \(X \U ) = U . Thus, by Theorem 1.7.9, X is connected im kleinen at p. Since p was an arbitrary point of X, by Theorem 1.7.12, X is locally connected. Q.E.D. Theorem 3.1.31 may be strengthened as follows:

3.1.32. Theorem. Let X be a continuum. Then X is locally connected if and only if T (Y ) = Y for each subcontinuum Y of X. Proof. If X is locally connected, by Theorem 3.1.31, T (Y ) = Y for each subcontinuum Y of X. Now suppose that for every subcontinuum Y of X, T (Y ) = Y . We show that X is locally connected. To this end, by Lemma 1.7.11, it is enough to see the components of open subsets of X are open. Let V be an open subset of X. Let p ∈ V . Since T ({p}) = {p} and X \V is compact, there exist finitely many subcontinua contained in X\{p} whose interiors cover X\V . Let W = {W1 , . . . , W

m } be such m a collection of smallest possible cardinality. Let U = X \ Wj , j=1

and let P be the component of Cl(U ) such that p ∈ P . We show that p ∈ Int(P ) ⊂ P ⊂ V . By Theorem 1.7.22, each component m of Cl(U ) intersects Wj . By the minimality of m, we assert that j=1

no component of Cl(U ) except P can intersect more than one of the Wj ’s. To see this, let P be a component of Cl(U ), different from P , and suppose that P intersects both Wj and Wk , where j, k ∈ {1, . . . , m} and j = k. Note that, in this case, P ∪ Wj ∪ Wj is a continuum. Let W = {W1 , ..., Wm , P ∪ Wj ∪ Wk } \ {Wj , Wk }. Copyright © 2005 Taylor & Francis Group, LLC


155 ⎛

⎞

⎜ ⎟ Then W has m−1 elements and X \V ⊂ ⎝ Wl ⎠ ∪(P ∪Wj ∪Wk ), l=j l=k

a contradiction to the minimality of m. For each j ∈ {1, . . . , m}, let Aj be the union of all the components of Cl(U ) which intersect Wj . Then Aj ∩ Ak ⊂ P , for each j, k ∈ {1, . . . , m} such that j = k. Now, since p ∈ T (Wj ) = Wj , for each j ∈ {1, . . . , m}, there exists a subcontinuum Kj of X such that p ∈ Int(Kj ) ⊂ Kj ⊂ X \ Wj . Then Kj ∩ (Aj \ P ) = ∅. To see this, suppose there exists x ∈ Kj ∩ (Aj \ P ). Let L be the component of Kj ∩ Cl(U ) such that x ∈ L. By Theorem 1.7.22, L intersects the boundary m of Kj ∩ Cl(U ) in Kj . This boundary is contained in W . (If =1

y ∈ BdKj (Kj ∩ Cl(U )), then y ∈ Cl(U ). If y ∈ U , then y ∈ Kj ∩ U ⊂ Kj ∩ Cl(U ) and (Kj ∩ U ) ∩ (Kj \ (Kj ∩ Cl(U ))) = ∅, a m W .) Hence, there contradiction. Therefore, y ∈ Cl(U ) \ U ⊂ =1

exists k ∈ {1, . . . , m} \ {j} such that L ∩ Wk = ∅. But L ⊂ M , for some component M of Cl(U ), M = P and M ⊂ Aj , so that M ∩ Wk = ∅, a contradiction. m m m Thus, Kj ∩ (Aj \ P ) = ∅, and it follows that Kj ⊂ j=1

P . Since p ∈ Int

j=1 m

j=1

Kj , p ∈ Int(P ). Therefore, P is a con-

j=1

nected neighborhood of p in X contained in V . Since p was an arbitrary point of V , every component of V is open. Q.E.D. 3.1.33. Remark. Note that Theorem 3.1.32 gives a partial answer to Question 7.2.11.

3.1.34. Theorem. Let X be a continuum. Then X is indecomposable if and only if T ({p}) = X for each p ∈ X. Hence, T (A) = X for each closed subset A of X. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Suppose X is indecomposable. By Corollary 1.7.21, each proper subcontinuum of X has empty interior. Therefore, T ({p}) = X for each p ∈ X. Now, suppose X is decomposable. Then there exist two proper subcontinua A and B of X such that X = A ∪ B. Let a ∈ A \ B, and let b ∈ B \ A. Then b ∈ Int(B) ⊂ B ⊂ X \ {a}. Hence, b ∈ X \ T ({a}). Therefore, T ({a}) = X. Q.E.D. 3.1.35. Definition. A continuum X is T –symmetric if for each pair, A and B, of closed subsets of X, A ∩ T (B) = ∅ if and only if B ∩ T (A) = ∅. We say that X is point T –symmetric if for each pair, p and q, of points of X, p ∈ T ({q}) if and only if q ∈ T ({p}).

3.1.36. Theorem. If X is a weakly irreducible continuum, then X is T –symmetric. Proof. Let A and B be two closed subsets of X. Suppose A ∩ T (B) = ∅. Hence, for each a ∈ A, there exists a subcontinuum Wa of X such that a ∈ Int(Wa ) ⊂ Wa ⊂ X \ B. Since A is n compact, there exist a1 , . . . , an ∈ A such that A ⊂ Int(Waj ) ⊂ n

Waj ⊂ X \ B. This implies that B ⊂ X \

j=1

n

Int(Waj )

j=1 n

W aj

⊂ X\

j=1

⊂ X \ A.

j=1

Let b ∈ B. Since X is weakly irreducible, X \

n

W aj

has

j=1

a finite number of components are open subsets of X. Let

which n Waj such that b ∈ C. Then C be the component of X \

b ∈ C ⊂ Cl(C) ⊂ X \

j=1 n

Int(Waj )

j=1

Therefore, B ∩ T (A) = ∅.


⊂ X \A. Thus, b ∈ X \T (A).


157

Similarly, if B ∩ T (A) = ∅, then A ∩ T (B) = ∅. Therefore, X is T –symmetric. Q.E.D. 3.1.37. Corollary. If X is an irreducible continuum, then X is T –symmetric. Proof. Let X be an irreducible continuum. By Theorem 1.7.29, X is weakly irreducible. Hence, by Theorem 3.1.36, X is T –symmetric. Q.E.D. 3.1.38. Theorem. Let X be a T –symmetric continuum, and let p be a point of X. Then X is connected im kleinen at p if and only if X is semi–locally connected at p. Proof. Suppose X is connected im kleinen at p. Let q ∈ T ({p}). Since X is T –symmetric, p ∈ T ({q}). Thus, p ∈ {q} (Corollary 3.1.25). Hence, p = q and T ({p}) = {p}. Therefore, by Theorem 3.1.29, X is semi–locally connected at p. Now, suppose X is semi–locally connected at p. Then T ({p}) = {p} (Theorem 3.1.29). Let A be a closed subset of X such that p ∈ T (A). Hence, since X is T –symmetric, A ∩ T ({p}) = ∅. Thus, p ∈ A. Therefore, by Corollary 3.1.25, X is connected im kleinen at p. Q.E.D. 3.1.39. Definition. A continuum X is T –additive if for each pair, A and B, of closed subsets of X, T (A ∪ B) = T (A) ∪ T (B). The following Theorem gives us a sufficient condition for a continuum X to be T –additive. 3.1.40. Theorem. Let X be a continuum. If for each point x ∈ X and any two subcontinua W1 and W2 of X such that x ∈ Int(Wj ), j ∈ {1, 2}, there exists a subcontinuum W3 of X such that x ∈ Int(W3 ) and W3 ⊂ W1 ∩ W2 , then X is T –additive. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Let A1 and A2 be closed subsets of X, and let x ∈ X \ T (A1 ) ∪ T (A2 ). Hence, there exist two subcontinua W1 and W2 of X such that x ∈ Int(Wj ) ⊂ Wj ⊂ X \Aj , j ∈ {1, 2}. By hypothesis, there exists a subcontinuum W3 of X such that x ∈ Int(W3 ) and W3 ⊂ W1 ∩ W2 . Thus, x ∈ Int(W3 ) ⊂ W3 ⊂ X \ (A1 ∪ A2 ). Hence, x ∈ X \ T (A1 ∪ A2 ). Therefore, by Corollary 3.1.8, T (A1 ∪ A2 ) = T (A1 ) ∪ T (A2 ). Q.E.D. We need the following definition to give a characterization of T –additive continua. 3.1.41. Definition. Let X be a continuum. A filterbase ℵ in X is a family ℵ = {Aω }ω∈Ω of subsets of X having two properties: (a) For each ω ∈ Ω, Aω = ∅, and (b) For each ω1 , ω2 ∈ Ω, there exists ω3 ∈ Ω such that Aω3 ⊂ Aω1 ∩ Aω2 .

3.1.42. Lemma. Let X be a compactum. If Γ is a filterbase of closed subsets of X, then T {G | G ∈ Γ} = {T (G) | G ∈ Γ}. Proof. By Proposition 3.1.7, T {G | G ∈ Γ} ⊂ {T (G) | G ∈ Γ}. Let p ∈ X \ T {G | G ∈ Γ} . Then there exists a subcontin uum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ {G | G ∈ Γ} . Since W is compact, there exist G1 , . . . , Gn ∈ Γ such that W ∩

n Gj = ∅. Since Γ is a filterbase, there exists G ∈ Γ such that j=1

G⊂

n

Gj . Note that W ∩ G = ∅. Thus, p ∈ X \ T (G). Hence,

p ∈ X \ {T (G) | G ∈ Γ}. Therefore, T {G | G ∈ Γ} = {T (G) | G ∈ Γ}. Q.E.D. j=1



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3.1.43. Theorem. Let X be a continuum. Then X is T –additive if and only if for each familyΛ ofclosed subsets of X whose union is closed, T {L | L ∈ Λ} = {T (L) | L ∈ Λ}. Proof. Suppose X is T –additive. By Proposition 3.1.7, {T (L) | L ∈ Λ} ⊂ T {L | L ∈ Λ} . Now, suppose x ∈ X \ {T (L) | L ∈ Λ}. Then for each L ∈ Λ, let F (L) = {A ⊂ X | A is closed and L ⊂ Int(A)}.If L = ∅, then T (L) = {T (A) | A ∈ F (L)}. (Clearly T (L) ⊂ {T (A) | A ∈ F (L)}. Let z ∈ X\T (L). Then there exists a subcontinuum W of X such that z ∈ Int(W ). Let A be any proper closed subset of X \ W . Then A ∈ F (L) and z ∈ X \ T (A). Hence, z ∈ X \ {T (A) | A ∈ F (L)}.) If L = ∅, then F (L) is a filterbase of closed subsets of X. Since {A | A ∈ F (L)} = L, T (L) = {T (A) | A ∈ F (L)}, by Lemma 3.1.42. Hence, for each L ∈ Λ, x ∈ X \ {T (A) | A ∈ F (L)}, and thus, there exists, for each L ∈ Λ, AL ∈ F (L) such that x ∈ X \ T (AL ). Note that {Int(AL ) | L ∈ Λ} is an open cover of {L | L ∈ Λ}. Since this set is compact, there exist L1 , . . . , Lm ∈ Λ such that

{L | L ∈ Λ} ⊂

m

Int(ALj ).

j=1

Since, by hypothesis and mathematical induction, T m

T (ALj ), T

{L | L ∈ Λ} ⊂

j=1

m j=1

m

ALj

=

j=1

T (ALj ). Now, since for ev-

ery j ∈ {1, . . . , m}, x ∈ X \ T (ALj ), it follows that x ∈ X \ T {L | L ∈ Λ} . Thus, we have that T {L | L ∈ Λ} ⊂ {T (L) | L ∈ Λ}. The other implication is obvious. Q.E.D.



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3.1.44. Theorem. Each T –symmetric continuum is T –additive. Proof. Let X be a T –symmetric continuum, and let A and B be two closed subsets of X. By Corollary 3.1.8, T (A) ∪ T (B) ⊂ T (A ∪ B). Let x ∈ T (A ∪ B). Then {x} ∩ T (A ∪ B) = ∅. Since X is T –symmetric, T ({x}) ∩ (A ∪ B) = ∅. Hence, either T ({x}) ∩ A = ∅ or T ({x}) ∩ B = ∅. Thus, since X is T –symmetric, {x} ∩ T (A) = ∅ or {x} ∩ T (B) = ∅, i.e., x ∈ T (A) or x ∈ T (B). Then x ∈ T (A) ∪ T (B). Therefore, X is T –additive. Q.E.D. 3.1.45. Theorem. If X is a hereditarily unicoherent continuum, then X is T –additive. Proof. Let X be a hereditarily unicoherent continuum, and let A and B be two closed subsets of X. By Corollary 3.1.8, T (A) ∪ T (B) ⊂ T (A ∪ B). Let x ∈ X \ (T (A) ∪ T (B)). Then x ∈ T (A) and x ∈ T (B). Hence, there exist subcontinua WA and WB of X such that x ∈ Int(WA ) ∩ Int(WB ) ⊂ WA ∩ WB ⊂ X \ (A ∪ B). Since X is hereditarily unicoherent, WA ∩ WB is a subcontinuum of X. Therefore, x ∈ X \ T (A ∪ B). Q.E.D. The following Corollary is a consequence of Theorem 3.1.43; however, we present a different proof based on Corollary 3.1.20.

3.1.46. Corollary. If X is a T –additive continuum and if A is a closed subset of X, then T (A) = T ({a}). a∈A

Proof. Let A be a closed subset of the T –additive continuum X. If a ∈ A, then {a} ⊂ A and T ({a}) ⊂ T (A), by Remark 3.1.5. Hence, T ({a}) ⊆ T (A). a∈A

Now, let x ∈ X such that x ∈ X \ T ({a}) for each a ∈ A. Then, by Corollary 3.1.20, for each a ∈ A, there exists an open subset Ua Copyright © 2005 Taylor & Francis Group, LLC


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of X such that a ∈ Ua and x ∈ X \ T (Cl(Ua )). Since A is compact and {Ua }a∈A is an open cover of A, there exist a1 , . . . , an ∈ A such n that A ⊂ Uaj . Since for each j ∈ {1, . . . , n}, x ∈ X \T (Cl(Uaj )), x∈X\

j=1 n

T (Cl(Uaj )). Thus, x ∈ X \ T

j=1

is T –additive). Hence, x ∈ X \ T (A).

n

Cl(Uaj )

(since X

j=1

Q.E.D. 3.1.47. Theorem. A continuum X is locally connected if and only if X is aposyndetic and T –additive. Proof. Suppose X is a locally connected continuum. Then, by Theorem 3.1.31, T (A) = A for each closed subset A of X. Hence, T ({p}) = {p} for each point p ∈ X and T (A∪B) = A∪B = T (A)∪ T (B) for each pair of closed subsets A and B of X. Therefore, X is aposyndetic (Theorem 3.1.28) and T –additive. Now, suppose X is aposyndetic and T –additive. Let A be a closed subset of X. Since X is T –additive, by Corollary 3.1.46, T (A) = T ({a}). Since X is aposyndetic, by Theorem 3.1.28, a∈A T ({a}) = T ({a}) = {a} for each a ∈ A. Thus, T (A) = a∈A {a} = A. Therefore, by Theorem 3.1.31, X is locally connected. a∈A

Q.E.D. 3.1.48. Corollary. If X is an aposyndetic hereditarily unicoherent continuum, then X is locally connected. Proof. Let X be an aposyndetic hereditarily unicoherent continuum. Since X is hereditarily unicoherent, by Theorem 3.1.45, X is T –additive. Then, since X is an aposyndetic T –additive continuum, by Theorem 3.1.47, X is locally connected. Q.E.D. 3.1.49. Theorem. A continuum X is T –symmetric if and only if X is point T –symmetric and T –additive. Copyright © 2005 Taylor & Francis Group, LLC

162


Proof. Suppose X is T symmetric. By Theorem 3.1.44, X is T – additive. Clearly, X is point T –symmetric. Now, suppose X is point T –symmetric and T –additive. Let A and B be two closed subsets of X. Suppose that A ∩ T (B) = ∅ and B ∩ T (A) = ∅. Let x ∈ B ∩ T (A). Then x ∈ B and x ∈ T (A). Note that x ∈ X \ A (if x ∈ A, then x ∈ A ∩ B ⊂ A ∩ T (B), a contradiction). Since X is T –additive, T (A) = T ({a}). Thus, a∈A

there exists a ∈ A such that x ∈ T ({a}). This implies that a ∈ T ({x}) (since X is point T –symmetric). Consequently, since x ∈ B, a ∈ T ({x}) ⊂ T (B) (Proposition 3.1.7). Hence, a ∈ A ∩ T (B), a contradiction to our assumption. Therefore, B ∩ T (A) = ∅, and X is T –symmetric. Q.E.D.

3.1.50. Theorem. Let X be a semi–aposyndetic irreducible continuum. Then X is homeomorphic to [0, 1]. Proof. Since X is an irreducible continuum, by Corollary 3.1.37, X is T –symmetric. Let x, y ∈ X. Since X is semi–aposyndetic, by Theorem 3.1.26, either x ∈ X \ T ({y}) or y ∈ X \ T ({x}). This implies that x ∈ X \T ({y}) if and only if y ∈ X \T ({x}), by the T – symmetry of X. Hence, T ({x}) = {x} for each x ∈ X. Therefore, X is aposyndetic (Theorem 3.1.28). Since X is T –symmetric, X is T –additive (Theorem 3.1.44). Since X is an aposyndetic T –additive continuum, X is locally connected (Theorem 3.1.47). Thus, X is an irreducible locally connected continuum. Consequently, X is an irreducible arcwise connected continuum (Theorem 3–15 (p. 116) of [14]). Therefore, X is homeomorphic to [0, 1]. Q.E.D. 3.1.51. Definition. Let X be a continuum. We say that T is idempotent on X provided that T 2 (A) = T (A) for each closed subset A of X.



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3.1.52. Proposition. Let X be a continuum for which T is idempotent, and let Z be a nonempty closed subset of X. If T (Z) = A ∪ B, where A and B are nonempty closed subsets of X, then T (A ∪ B) = T (A) ∪ T (B) = A ∪ B. Proof. By the idempotency of T , Remark 3.1.5 and Proposition 3.1.7, we have T (Z) = A ∪ B ⊂ T (A) ∪ T (B) ⊂ T (A ∪ B) = T 2 (Z) = T (Z). Q.E.D. 3.1.53. Proposition. Let X be a continuum for which T is idempotent. If Z is a nonempty closed subset of X, then T (Z) = {T ({w}) | w ∈ T (Z)}. Proof. Let x ∈ {T ({w}) | w ∈ T (Z)}. Then there exists w ∈ T (Z) such that x ∈ T ({w}). Since T is idempotent, T ({w}) ⊂ 2 T (Z) = T (Z). Hence, x ∈ T (Z), and {T ({w}) | w ∈ T (Z)} ⊂ T (Z). The other inclusion is clear. Q.E.D. 3.1.54. Theorem. Let X be a continuum. Then T is idempotent on X if and only if for each subcontinuum W of X and each point x ∈ Int(W ), there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ M ⊂ Int(W ). Proof. Suppose T is idempotent on X. Let W be a subcontinuum of X, and let x ∈ Int(W ). Hence, x ∈ X \ T (X \ W ). Since T is idempotent, x ∈ X \T 2 (X \W ). Thus, there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ M ⊂ X \ T (X \ W ) ⊂ X \ (X \ W ) = W . Therefore, x ∈ Int(M ) ⊂ M ⊂ Int(W ). Now, suppose the condition stated holds. We show T is idempotent. By Remark 3.1.5, for each A ∈ P(X), T (A) ⊂ T 2 (A). Let B ∈ P(X), and let x ∈ X \T (B). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ B. By hypothesis, there exists a subcontinuum M such that x ∈ Int(M ) ⊂ M ⊂ Int(W ). Copyright © 2005 Taylor & Francis Group, LLC

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Since Int(W ) ⊂ X \ T (B), x ∈ Int(M ) ⊂ M ⊂ X \ T (B). Hence, x ∈ X \ T 2 (B). Therefore, T is idempotent. Q.E.D. 3.1.55. Corollary. Let X be a continuum. If T is idempotent on X, W is a subcontinuum of X and K is a component of Int(W ), then K is open. Proof. Let W be a subcontinuum of X, and let K be a component of Int(W ). If x ∈ K, then, by Theorem 3.1.54, there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ Int(W ). Hence, M ⊂ K and x is an interior point of K. Therefore, K is open. Q.E.D. 3.1.56. Corollary. Let X be a continuum. If T is idempotent on X, x ∈ X and W is a subcontinuum of X such that x ∈ Int(W ), then there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ M ⊂ W and M = Cl(Int(M )) (i.e., M is a continuum domain). Proof. Let W be a subcontinuum of X, and let x ∈ Int(W ). By Corollary 3.1.55, the component, K, of W containing x is open. Let M = Cl(K). Then M = Cl(Int(M )) and x ∈ Int(M ) ⊂ M ⊂ W . Q.E.D. 3.1.57. Theorem. Let X be a continuum. If T is idempotent on X and T ({p, q}) is a continuum for all p, q ∈ X, then X is indecomposable. Proof. Suppose X is decomposable. Then, by Corollary 3.1.56, there exists a subcontinuum W of X such that W = Cl(Int(W )). Let p0 and q0 be any two points in X \ Int(W ). Note that, by Theorem 3.1.54, T ({p0 , q0 }) ∩ Int(W ) = ∅. Since T ({p0 , q0 }) is connected and T ({p0 , q0 }) ∩ Int(W ) = ∅, p0 and q0 belong to the same component of X \ Int(W ). Thus, X \ Int(W ) is a continuum. By Theorem 3.1.54, there exists a subcontinuum M , with nonempty interior, such that M ⊂ Int(X \ Int(W )) ⊂ X \W . Then let K = X \(Int(W )∪Int(M )), and let p1 and q1 be any two points of K. By Theorem 3.1.54, T ({p1 , q1 }) ⊂ Copyright © 2005 Taylor & Francis Group, LLC


165

K. Hence, K is a continuum also. Now, let p ∈ Int(M ) and let q ∈ Int(W ). Observe that, by Theorem 3.1.54, T ({p, q}) ∩ Int(K) = ∅. Then T ({p, q}) ⊂ X \ Int(K) ⊂ W ∪ M . Hence, T ({p, q}) is not a continuum, a contradiction. Q.E.D. 3.1.58. Lemma. Let X be a compactum. If A is a closed subset of X, then p ∈ X \ T (A) if and only if there exist a subcontinuum W and an open subset Q of X such that p ∈ Int(W ) ∩ Q, Bd(Q) ∩ T (A) = ∅ and W ∩ A ∩ Q = ∅. Proof. Let p ∈ X \ T (A). Then there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ A. Since X is a metric space, there exists an open subset Q of X such that p ∈ Q ⊂ Cl(Q) ⊂ Int(W ). Hence, Bd(Q) ∩ T (A) = ∅ and W ∩ A ∩ Q = ∅. Now, suppose there exist a continuum W and an open subset Q of X such that p ∈ Int(W )∩Q, Bd(Q)∩T (A) = ∅ and W ∩A∩Q = ∅. Since Bd(Q) is a compact set and Bd(Q) ∩ T (A) = ∅, there exist a finite collection, W1 , . . . , Wn , of subcontinua of X such that n m Bd(Q) ⊂ Int(Wj ) ⊂ Wj ⊂ X \ A. If W ⊂ Q, then, clearly, j=1

j=1

p ∈ X \ T (A). Assume W \ Q = ∅. By Theorem 1.7.22, the closure of each component of W ∩ Q must intersect at least

m oneof the Wj ’s, since m Wj . Hence, H = (W ∩ Q) ∪ Wj has only a finite Bd(Q) ⊂ j=1

j=1

number of components. Let K be the component of H such that p ∈ K. Since p ∈ Int(W ) ∩ Q, by Lemma 1.6.2, p ∈ Int(K), and, of course, K ∩ A ⊂ H ∩ A = ∅. Thus, p ∈ X \ T (A). Q.E.D. 3.1.59. Lemma. Let X be a compactum, and let A be a subset of X. If T (A) = M ∪ N , where M and N are disjoint closed sets, then T (A ∩ M ) = M ∪ T (∅). Proof. Suppose p ∈ T (A ∩ M ) \ (M ∪ T (∅)). Since p ∈ X \ T (∅), there exists a subcontinuum W of X such that p ∈ Int(W ). Since Copyright © 2005 Taylor & Francis Group, LLC

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X is a metric space, there exists an open subset Q of X such that N ⊂ Q and Cl(Q) ∩ M = ∅. Note that p ∈ Int(W ) ∩ Q and Bd(Q) ∩ T (A ∩ M ) ⊂ Bd(Q) ∩ T (A) = ∅, and W ∩ (A ∩ M ) ∩ Q ⊂ Q ∩ M = ∅. Then, by Lemma 3.1.58, p ∈ X \ T (A ∩ M ), thus contradicting the assumption. Now, suppose p ∈ (M ∪T (∅))\T (A∩M ). Since p ∈ X\T (A∩M ) and ∅ ⊂ A ∩ M , p ∈ X \ T (∅). Hence, p ∈ M . Since X is a metric space, there exists an open subset Q of X such that M ⊂ Q and Cl(Q) ∩ N = ∅. Since p ∈ X \ T (A ∩ M ), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ (A ∩ M ). Observe that p ∈ Int(W ) ∩ Q and Bd(Q) ∩ T (A) = ∅. Since Q∩N = ∅, W ∩A∩Q = W ∩(A∩M ) = ∅. Hence, by Lemma 3.1.58, p ∈ X \ T (A) ⊂ X \ M , thus contradicting our assumption. Q.E.D. 3.1.60. Theorem. Let X be a compactum, and let A be a closed subset of X. If K is a component of T (A), then T (A ∩ K) = K ∪ T (∅). Proof. Let K be a component of T (A). Let Γ(K) = {L ⊂ T (A) | K ⊂ L and L is open and closed in T (A)}. Note that the collection of {A ∩ L | L ∈ Γ(K)} fails to be a filterbase if for some L ∈ Γ(K), A ∩ L = ∅. In this case the conclusion of Lemma 3.1.42 holds (an argument similar to the one given in the proof of Theorem 3.1.43 shows this). Note that Lemma 3.1.59 remains true even if A ∩ M = ∅. Hence, by Lemma 3.1.59, for each L ∈ Γ(K), T (A ∩ L) = L ∪ T (∅). Therefore, by Lemma 3.1.42, the following sequence of equalities establishes the theorem: T (A ∩ K) = T {A ∩ L | L ∈ Γ(K)} = {T (A ∩ L) | L ∈ Γ(K)} = {L ∪ T (∅) | L ∈ Γ(K)} = {L | L ∈ Γ(K)} ∪ T (∅) = K ∪ T (∅). (The fact that {L | L ∈ Γ(K)} = K follows from Theorem 2–14 of [14].) Q.E.D. The following Corollary says that for each nonempty closed subset A of a continuum X, the components of T (A) are also in the image of T .



167

3.1.61. Corollary. Let X be a continuum, and let A be a closed subset of X. If K is a component of T (A), then K = T (A ∩ K). Proof. Let A be a closed subset of X and let K be a component of T (A). By Theorem 3.1.60, T (A ∩ K) = K ∪ T (∅). Since T (∅) = ∅, by Corollary 3.1.14, K = T (A ∩ K). Q.E.D. 3.1.62. Corollary. Let X be a continuum, and let W1 and W2 be subcontinua of X. If T (W1 ∪ W2 ) = T (W1 ) ∪ T (W2 ), then T (W1 ∪ W2 ) is a continuum. Proof. Suppose T (W1 ∪W2 ) is not connected. Then there exist two disjoint closed subsets A and B of X such that T (W1 ∪W2 ) = A∪B. By Lemma 3.1.59 and Corollary 3.1.14, T ((W1 ∪ W2 ) ∩ A) = A and T ((W1 ∪ W2 ) ∩ B) = B. Suppose W1 ⊂ A. If W2 ⊂ A, then A = T ((W1 ∪ W2 ) ∩ A) = T (W1 ∪ W2 ), a contradiction. Thus, W2 ⊂ B. Hence T (W1 ) = A and T (W2 ) = B, which implies that T (W1 ∪ W2 ) = T (W1 ) ∪ T (W2 ), a contradiction. Therefore, T (W1 ∪ W2 ) is connected. Q.E.D. 3.1.63. Corollary. Let X be a continuum. If p, q ∈ X are such that T ({p, q}) = T ({p}) ∪ T ({q}), then T ({p, q}) is a continuum. The following Theorem gives relationships between the images of maps and the images of T . We subscript T to differentiate the continua on which T is defined. 3.1.64. Theorem. Let X and Y be continua, and let f : X → →Y be a surjective map. If A ∈ P(X) and B ∈ P(Y ), then the following hold: (a) TY (B) ⊂ f TX f −1 (B). (b) If f is monotone, then f TX (A) ⊂ TY f (A) and TX f −1 (B) ⊂ f −1 TY (B). (c) If f is monotone, then TY (B) = f TX f −1 (B). (d) If f is open, then f −1 TY (B) ⊂ TX f −1 (B). (e) If f is monotone and open, then f −1 TY (B) = TX f −1 (B). Copyright © 2005 Taylor & Francis Group, LLC


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Proof. We show (a). Let y ∈ Y \ f TX f −1 (B). Then f −1 (y) ∩ TX f −1 (B) = ∅. Thus, for each x ∈ f −1 (y), there exists a subcontinuum Wx of X such that x ∈ Int(Wx ) ⊂ Wx ⊂ X \ f −1 (B). Since f −1 (y) is compact, there exist x1 , ...,xn ∈ f −1 (y) such that

n n f −1 (y) ⊂ Int(Wxj ). Note that Wxj ∩ f −1 (B) = ∅, and for j=1

j=1

each j ∈ {1, ..., n}, f −1 (y) ∩ Wxj = ∅. Hence, f

and f

n

n

∩B = ∅

W xj

j=1

is a continuum (y ∈ f (Wxj ) ∩ f (Wxk ) for every

W xj

j=1

j, k ∈ {1, 2, ..., n}).

Observe that Y \ f

X\

and it is contained in f n

Int(Wxj ) ⊂

f

X\

f

X\

j=1 n

f

X\

X\

Int(Wxj )

X\

Int(Wxj )

j=1

X\

n

and, consequently, Y \ Wxj . Since f is a sur-

Also, since f −1 (y) ⊂

n

X\

X\

X\

Int(Wxj )

Int(Wxj ), {y}∩f

j=1 n

j=1

⊂ f

Int(Wxj ). Then

j=1

j=1 n

n

W xj ⊂ X \

Int(Wxj )

j=1

is an open set of Y

j=1 n

Wxj . Therefore, Y \f

∅. Hence, y ∈ Y \ f

n

j=1

n

Wxj . To see this, note that, since

⊂ Y \f

Int(Wxj )

jection, Y \ f n

⊂f

W xj

j=1

j=1

n

n

W xj , X \

j=1

j=1

j=1

n

n

X\

n

Wxj

=

j=1

⊂f

n

j=1 n

W xj .

Int(Wxj )

=

j=1

Int(Wxj ) . Therefore, y ∈ Y \ TY (B).

j=1

We prove (b). First, we see f TX (A) ⊂ TY f (A). Let y ∈ Y \ TY f (A). Then there exists a subcontinuum W of Y such that y ∈ Copyright © 2005 Taylor & Francis Group, LLC


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Int(W ) ⊂ W ⊂ Y \ f (A). It follows that f −1 (y) ⊂ f −1 (Int(W )) ⊂ f −1 (W ) ⊂ f −1 (Y )\f −1 f (A) ⊂ X \A. Hence, f −1 (y) ⊂ X \A. Since f is monotone, f −1 (W ) is a subcontinuum of X (Lemma 2.1.12). Thus, since f −1 (y) ⊂ f −1 (Int(W )) ⊂ f −1 (W ) ⊂ X \ A, f −1 (y) ∩ TX (A) = ∅. Therefore, y ∈ Y \ f TX (A). Now, we show TX f −1 (B) ⊂ f −1 TY (B). Let x ∈ X \ f −1 TY (B). Then f (x) ∈ Y \ TY (B). Hence, there exists a subcontinuum W of Y such that f (x) ∈ Int(W ) ⊂ W ⊂ Y \ B. This implies that x ∈ f −1 (f (x)) ⊂ f −1 (Int(W )) ⊂ f −1 (W ) ⊂ X \ f −1 (B). Since f is monotone, f −1 (W ) is a subcontinuum of X. Therefore, x ∈ X \ TX f −1 (B). We prove (c). By (a), TY (B) ⊂ f TX f −1 (B). Since TX f −1 (B) ⊂ f −1 TY (B), by (b), f TX f −1 (B) ⊂ f f −1 TY (B) = TY (B). Therefore, f TX f −1 (B) = TY (B). We show (d). Let x ∈ X \ TX f −1 (B). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ f −1 (B). Hence, f (x) ∈ f (Int(W )) ⊂ f (W ) ⊂ Y \ B. Since f is open, f (Int(W )) is an open subset of Y . Therefore, f (x) ∈ Y \ TY (B). Note that (e) follows directly from (b) and (d). Q.E.D. The following four Theorems are applications of Theorem 3.1.64.

3.1.65. Theorem. Let X and Y be continua, and let f : X → →Y be a surjective map. If X is locally connected, then Y is locally connected. Proof. By Theorem 3.1.31, it suffices to show that TY (B) = B for each closed subset B of Y . Let B be a closed subset of Y . By Remark 3.1.5, B ⊂ TY (B). By Theorem 3.1.64 (a), TY (B) ⊆ f TX f −1 (B). Since X is locally connected, TX f −1 (B) = f −1 (B). Hence, TY (B) ⊆ f f −1 (B) = B. Therefore, TY (B) = B, and Y is locally connected. Q.E.D. 3.1.66. Theorem. The monotone image of an indecomposable continuum is an indecomposable continuum. Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Let f : X → → Y be a monotone map, where X is an indecomposable continuum. Let y ∈ Y . Then, by Theorem 3.1.64 (b), TX f −1 ({y}) ⊂ f −1 TY ({y}). Since X is indecomposable, by Theorem 3.1.34, TX f −1 ({y}) = X. Hence, Y = f (X) ⊂ f f −1 TY ({y}) ⊂ Y . Thus, TY ({y}) = Y . Therefore, by Theorem 3.1.34, Y is indecomposable. Q.E.D. 3.1.67. Theorem. The monotone image of a T –symmetric continuum is T –symmetric. Proof. Let X be a T –symmetric continuum, and let f : X → →Y be a monotone map. Let A and B be two closed subsets of Y such that A ∩ TY (B) = ∅. Then, by Theorem 3.1.64 (c), A ∩ f TX f −1 (B) = ∅. Thus, f −1 (A) ∩ TX f −1 (B) = ∅. Since X is T –symmetric, TX f −1 (A) ∩ f −1 (B) = ∅. Hence, f (TX f −1 (A) ∩ f −1 (B)) = f TX f −1 (A) ∩ B = TY (A) ∩ B = ∅. Therefore, Y is T –symmetric. Q.E.D. 3.1.68. Theorem. The monotone image of a T –additive continuum is T –additive. Proof. Let X be a T –additive continuum, and let f : X → → Y be a monotone map. Let A and B be two closed subsets of Y . Then, by Theorem 3.1.64 (c), TY (A ∪ B) = f TX f −1 (A ∪ B) = f TX (f −1 (A) ∪ f −1 (B)) = f (TX f −1 (A) ∪ TX f −1 (B)) = f TX f −1 (A) ∪ f TX f −1 (B) = TY (A) ∪ TY (B). Q.E.D. Now, we present some relations between T and inverse limits due to H. S. Davis [10]. 3.1.69. Lemma. Let {Xn , fnn+1 } be an inverse sequence of continua whose inverse limit is X. Let p = (pn )∞ n=1 ∈ X, and let A be a closed subset of X. If there exist N ∈ IN, two open subsets UN and VN of XN and an inverse sequence {Wn , fnn+1 |Wn+1 } of subcontinua such that pN ∈ UN , fN (A) ⊂ VN and for each n ≥ N , (fNn )−1 (UN ) ⊂ Wn ⊂ Xn \ (fNn )−1 (VN ), Copyright © 2005 Taylor & Francis Group, LLC


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then p ∈ X \ T (A). Proof. Let W = lim{Wn , fnn+1 |Wn+1 }. Then W is a subcontinuum ←−

of X, by Proposition 2.1.8. Note that fN−1 (UN ) ⊂ W since, for each n ≥ N , (fNn )−1 (UN ) ⊂ Cl((fNn )−1 (UN )) ⊂ Cl(fn−1 (fNn )−1 (UN )) ⊂ ∞ fn−1 (Wn ) (Corollary 2.1.21). Since p ∈ fN−1 (UN ), Wn and W = n=N

p ∈ Int(W ). Since, for every n ≥ N , fn−1 (Wn ) ⊂ X \ fn−1 (fNn )−1 (VN ) = X \ ∞ −1 −1 fn−1 (Wn ) ⊂ X \ A. Thus, fN (VN ) = fN (XN \ VN ) ⊂ X \ A, n=N

W ∩ A = ∅. Therefore, p ∈ X \ T (A).

Q.E.D. 3.1.70. Lemma. Let {Xn , fnn+1 } be an inverse sequence of continua whose inverse limit is X. Let p = (pn )∞ n=1 ∈ X, and let A be a closed subset of X. If p ∈ X \ T (A), then there exist N ∈ IN, two open subsets UN and VN of XN and an inverse sequence {Wn , fnn+1 |Wn+1 } of subcontinua such that pN ∈ UN , fN (A) ⊂ VN and for every n ≥ N , (fNn )−1 (UN ) ∩ fn (X) ⊂ Wn ⊂ fn (X) \ (fNn )−1 (VN ). Proof. Since p ∈ X \ T (A), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ A. By Proposition 2.1.9, there exist N (p) ∈ IN and an open subset UN (p) of XN (p) such that p ∈ fN−1(p) (UN (p) ) ⊂ Int(W ). Similarly, for each a = (an )∞ n=1 ∈ A, there exist N (a) ∈ IN and an open subset UN (a) of XN (a) such that a ∈ fN−1(a) (UN (a) ) ⊂ X \ W . Since A is compact, there exm fN−1(aj ) (UN (aj ) ). Let N = ist a1 , . . . am ∈ A such that A ⊂ j=1

max{N (p), N (a1 ), . . . , N (am )}. Define UN = (fNN(p) )−1 (UN (p) ) and m (fNN(aj ) )−1 (UN (aj ) ). VN = j=1

For each n ∈ IN, let Wn = fn (W ). Let n ≥ N , and let zn ∈ n −1 (fN ) (UN )∩fn (X). Since zn ∈ fn (X), there exists z ∈ X such that Copyright © 2005 Taylor & Francis Group, LLC


172

fn (z) = zn . Since fNn (p) (zn ) ∈ UN (p) , fN (p) (z) ∈ UN (p) . Therefore, z ∈ fN−1(p) (UN (p) ) ⊂ Int(W ) ⊂ W . Thus, zn = fn (z) ∈ fn (W ) = Wn . Hence, (fNn )−1 (UN ) ∩ fn (X) ⊂ Wn . Now, let wn ∈ Wn = fn (W ). Then there exists w ∈ W such that fn (w) = wn . Clearly, wn ∈ fn (X). Suppose wn ∈ (fNn )−1 (VN ). Then there exists j ∈ {1, . . . , m} such that fNn (aj ) (wn ) ∈ UN (aj ) . Hence, fN (aj ) (w) ∈ UN (aj ) . Therefore, w ∈ X \ W . This contradiction establishes that Wn ⊂ fn (X) \ (fNn )−1 (VN ). Q.E.D. 3.1.71. Theorem. Let {Xn , fnn+1 } be an inverse sequence of continua whose inverse limit is X. Let p = (pn )∞ n=1 ∈ X, and let A be a closed subset of X. Then the following are equivalent: (a) p ∈ X \ T (A), (b) there exist N ∈ IN, two open subsets UN and VN of XN and an inverse sequence {Wn , fnn+1 |Wn+1 } of subcontinua such that pN ∈ UN , fN (A) ⊂ VN and for every n ≥ N , (fNn )−1 (UN ) ∩ fn (X) ⊂ Wn ⊂ fn (X) \ (fNn )−1 (VN ). Proof. Suppose (a). Then Lemma 3.1.70 shows (b). Now, suppose (b). Note that if W = lim{Wn , fnn+1 |Wn+1 }, then ←−

for each n ∈ IN, fn (W ) = fn (X) ∩ Wn , fn (W ) is a continuum and W = lim{fn (W ), fnn+1 |fn+1 (W ) } (Proposition 2.1.20). ←−

The relations in Lemma 3.1.69 imply that (fNn )−1 (UN )∩fn (X) ⊂ Wn ⊂ fn (X) \ (fNn )−1 (VN ). Q.E.D. 3.1.72. Theorem. Let {Xn , fnn+1 } be an inverse sequence of continua, with surjective bonding maps, whose inverse limit is X. Let p = (pn )∞ n=1 ∈ X, and let A be a closed subset of X. Then the following are equivalent: (a) p ∈ X \ T (A), (b) there exist N ∈ IN, two open subsets UN and VN of XN and an inverse sequence {Wn , fnn+1 |Wn+1 } of subcontinua such that pN ∈ UN , fN (A) ⊂ VN and for every n ≥ N , (fNn )−1 (UN ) ⊂ Wn ⊂ Xn \ (fNn )−1 (VN ). Copyright © 2005 Taylor & Francis Group, LLC


173

Proof. Note that, since the bonding maps are surjective, the projection maps are surjective also (Remark 2.1.6). Thus, the Theorem follows directly from Theorem 3.1.71 Q.E.D. 3.1.73. Corollary. Let {Xn , fnn+1 } be an inverse sequence of continua, with surjective bonding maps, whose inverse limit is X. Let A be a closed subset of X, and let N0 ∈ IN. Then ∞

fn−1 (TXn (fn (A))) ⊂ TX (A).

n=N0

Proof. Let p = (pn )∞ n=1 ∈ X \ TX (A). By Theorem 3.1.72, there exist N ∈ IN, two open subsets UN and VN of XN and an inverse sequence {Wn , fnn+1 |Wn+1 } of subcontinua such that pN ∈ UN , fN (A) ⊂ VN and for every n ≥ N , (fNn )−1 (UN ) ⊂ Wn ⊂ Xn \ (fNn )−1 (VN ). Let m ≥ max{N, N0 }. Then pm ∈ (fNm )−1 (UN ) ⊂ Wm ⊂ Xm \ (fNm )−1 (VN ) ⊂ Xm \ fm (A). Thus, pm ∈ Xm \ TXm (fm (A)). Hence, ∞ −1 (TXm (fm (A))). Therefore, p ∈ X \ fn−1 (TXn (fn (A))). p ∈ X \fm n=N0

Q.E.D. 3.1.74. Corollary. Let {Xn , fnn+1 } be an inverse sequence of continua, with surjective bonding maps, whose inverse limit is X. Let A be a closed subset of X, and let N0 ∈ IN. If the bonding maps are monotone, then ∞

fn−1 (TXn (fn (A))) = TX (A).

n=N0

Proof. Let p =

(pn )∞ n=1

∈ X\

∞ n=N0

fn−1 (TXn (fn (A)). Then there

exist N ≥ N0 such that pN ∈ XN \ TXN (fN (A)). Thus, there exist a subcontinuum WN of XN and two open subsets UN and VN of XN such that pN ∈ UN ⊂ WN ⊂ XN \ VN ⊂ XN \ fN (A). Copyright © 2005 Taylor & Francis Group, LLC


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Now, for each n ∈ IN, let (fNn )−1 (WN ) if n ≥ N ; Wn = if n ≤ N . fnN (WN ) Since the bonding maps are monotone, Wn is a continuum for each n ∈ IN (Lemma 2.1.12). By construction, {Wn , fnn+1 |Wn+1 } is an inverse sequence of subcontinua and (fNn )−1 (UN ) ⊂ Wn ⊂ Xn \ (fNn )−1 (VN ). Hence, by Theorem 3.1.72, p ∈ X \ TX (A). The other inclusion is given by Corollary 3.1.73. Q.E.D. The next Corollary is a consequence of Davis’s work: 3.1.75. Corollary. Let {Xn , fnn+1 } be an inverse sequence of aposyndetic continua, with surjective bonding maps, whose inverse limit is X. If the bonding maps are monotone, then X is aposyndetic. Proof. Let p = (pn )∞ n=1 ∈ X. By Corollary 3.1.74, we have that ∞ TX ({p}) = fn−1 (TXn ({pn })). Since each factor space is aposynden=1

tic, by Theorem 3.1.28,

∞

fn−1 (TXn ({pn })) =

n=1

∞

fn−1 ({pn }) = {p}

n=1

(Corollary 2.1.21). Therefore, TX ({p}) = {p}. Since p was arbitrary, by Theorem 3.1.28, X is aposyndetic. Q.E.D.

3.2

Continuity of T

Let X be a continuum. By Remark 3.1.5, the image of any subset of X under T is a closed subset of X. Then we may restrict the domain of T to the hyperspace, 2X , of closed subsets of X. Since 2X has a topology, we may ask if T : 2X → 2X is continuous. The Copyright © 2005 Taylor & Francis Group, LLC

3.2. CONTINUITY OF T

175

answer to this question is negative, as can be easily seen from Example 3.1.15. On the other hand, by Theorems 3.1.31 and 3.1.34, T is continuous for locally connected continua and for indecomposable continua, respectively. In this section we present some results related to the continuity of T . In particular, we show that if a continuum X is almost connected im kleinen at each of its points and T is continuous, then X is locally connected. Most of the results of this section are taken from [1]. We begin with the following Theorem which says that T is always upper semicontinuous.

3.2.1. Theorem. Let X be a continuum. If U is an open subset of X, then U = {A ∈ 2X | T (A) ⊂ U } is open in 2X . Proof. Let U be an open subset of X and let U = {A ∈ 2X | T (A) ⊂ U }. We show U is open in 2X . Let B ∈ Cl2X (2X \U). Then there exX ists a sequence {Bn }∞ n=1 of elements of 2 \U converging to B. Note that for each n ∈ IN, T (Bn )∩(X \U ) = ∅. Let xn ∈ T (Bn )∩(X \U ). Without loss of generality, we assume that {xn }∞ n=1 converges to a point x ∈ X. Note that x ∈ X \ U . We assert that x ∈ T (B). Suppose x ∈ X \ T (B). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ B. Since {Bn }∞ n=1 converges to B and {xn }∞ converges to x, there exists N ∈ IN such that for n=1 each n ≥ N , Bn ⊂ X \ W and xn ∈ Int(W ). Let n ≥ N . Then x ∈ Int(W ) ⊂ W ⊂ X \ Bn . This implies that xn ∈ X \ T (Bn ), a contradiction. Thus, x ∈ T (B) ∩ (X \ U ). Hence, B ∈ 2X \ U. Therefore, 2X \ U is closed in 2X . Q.E.D. The next two Theorems give sufficient conditions for T to be continuous for a continuum.

3.2.2. Theorem. Let X and Z be continua, where Z is locally connected. If f : X → → Z is a surjective, monotone and open map such that for each proper subcontinuum W of X, f (W ) = Z, then TX is continuous for X. Copyright © 2005 Taylor & Francis Group, LLC


176

Proof. Let A be a closed subset of X. Since Z is locally connected, TZ (f (A)) = f (A) (Theorem 3.1.31). Hence, f TX f −1 f (A) = f (A), by Theorem 3.1.64 (c). Thus, f −1 f TX f −1 f (A) = f −1 f (A) and TX f −1 f (A) ⊂ f −1 f (A). Then TX f −1 f (A) = f −1 f (A), by Remark 3.1.5. Since A ⊂ f −1 f (A), TX (A) ⊂ TX f −1 f (A) = f −1 f (A) (the inclusion holds by Proposition 3.1.7). Suppose there exists x ∈ f −1 f (A) \ TX (A). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ A. Since f is open, f (x) ∈ Int(f (W )). For each z ∈ Z \ f (Int(W )), there exists a subcontinuum Mz of Z such that z ∈ Int(Mz ) ⊂ Mz ⊂ Z \ {f (x)} (TZ ({z}) = {z}). Since Z \ f (Int(W )) is compact, there exist z1 , . . . , zn ∈ Z \ f (Int(W )) such that Z \ f (Int(W )) ⊂ n n Int(Mzj ) ⊂ Mzj . By choosing n as small as possible, we j=1

j=1

may assume

that Mzj ∩ f (W ) = ∅ for each , n}. Then

nj ∈ {1, . . . n Mzj , and f −1 f (W ) ∪ f −1 (Mzj ) = X. Z = f (W ) ∪ j=1

j=1

For each j ∈ {1, . . . , n}, let qj ∈ f (W ) ∩ Mzj . Then there exists pj ∈ W such that f (pj ) = qj . Since p j ∈ f −1 (Mzj ),W ∩f −1 (Mzj ) = n f −1 (Mzj ) . Then Y is a ∅, j ∈ {1, . . . , n}. Let Y = W ∪ j=1

subcontinuum of X. We assert that Y = X. To see this, recall that x ∈ f −1 f (A). Thus, there exists A such that f (y) = f (x).

n y ∈ Then y ∈ X \ W and f (y) ∈ Z \ Mzj (recall the construction

of the Mzj ’s). Hence, y ∈ X \

j=1 n

j=1

f −1 (Mzj ) . However, f (Y ) =

Z, contrary to our hypothesis. Therefore, TX (A) = f −1 f (A), i.e., TX = (f ) ◦ 2f . Since f is continuous and open, (f ) and 2f are continuous Copyright © 2005 Taylor & Francis Group, LLC


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(Theorems 1.8.24 and 1.8.22, respectively). Therefore, TX is continuous. Q.E.D. 3.2.3. Remark. As a consequence of Theorem 3.2.2, we have that the circle of pseudo–arcs [7] is a decomposable nonlocally connected continuum for which T is continuous. For other decomposable nonlocally connected continua for which T is continuous see [20]. 3.2.4. Definition. Let f : X → Z be a map. We say that f is TXZ –continuous provided that always f TX (A) ⊂ TZ f (A) for every A ∈ P(X), or equivalently, f −1 TZ (B) ⊃ TX f −1 (B) for every B ∈ P(Z). 3.2.5. Theorem. Let X be a continuum for which TX is continuous, and let Z be a continuum. If f : X → → Z is a TXZ –continuous surjective open map, then TZ is continuous for Z. Proof. Since f is TXZ –continuous, by Theorem 3.1.64 (a), f TX f −1 (B) = TZ (B) for each B ∈ P(Z). Thus, TZ = 2f ◦TX ◦(f ). Since f is continuous and open, (f ) and 2f are continuous (Theorems 1.8.24 and 1.8.22, respectively). Hence, TZ is a composition of three maps. Therefore, TZ is continuous. Q.E.D. The next Lemma says that given a continuum X for which T is continuous, the family {T ({x}) | x ∈ X} resembles an upper semicontinuous decomposition. 3.2.6. Lemma. Let X be a continuum, with metric d, for which T is continuous. If U is an open subset of X and K is a closed subset of X, then: (1) A = {x ∈ X | T ({x}) ⊂ U } is open in X. (2) B = {x ∈ X | T ({x}) ∩ U = ∅} is open in X. (3) C = {x ∈ X | T ({x}) ∩ K = ∅} is closed in X. (4) D = {x ∈ X | T ({x}) ⊂ K} is closed in X. Copyright © 2005 Taylor & Francis Group, LLC


178

Proof. Let x ∈ A. Then T ({x}) ⊂ U , and there exists ε > 0 such that Vεd (T ({x})) ⊂ U . Let δ > 0 be given by the uniform continuity of T for ε. Let w ∈ Vδd (x). Then {w} ∈ VδH ({x}). Thus, H(T ({w}), T ({x})) < ε. Hence, T ({w}) ⊂ Vεd (T ({x})) ⊂ U . This implies that w ∈ A. Therefore, A is open. Let x ∈ B. Then T ({x}) ∩ U = ∅. Let y ∈ T ({x}) ∩ U and let ε > 0 such that Vεd (y) ⊂ U . Let δ > 0 be given by the uniform continuity of T for ε. Let z ∈ Vδd (x). Then {z} ∈ VδH ({x}). Hence, H(T ({z}), T ({x})) < ε. This implies that there exists z ∈ T ({z}) such that d(z , y) < ε. Thus, T ({z}) ∩ U = ∅. Hence, z ∈ B. Therefore, B is open. Now, let Z be either C or D. Let x ∈ Cl(Z). Then there exists a sequence {xn }∞ n=1 of elements of Z converging to x. Since T is continuous, lim T ({xn }) = T ({x}). n→∞

If Z is C, then for each n ∈ IN, there exists kn ∈ T ({xn }) ∩ K. Hence, without loss of generality, we assume that {kn }∞ n=1 converges to a point k. Since lim T ({xn }) = T ({x}) and K is closed, k ∈ n→∞

T ({x}) ∩ K. Thus, x ∈ C. Therefore, C is closed. If Z is D, then for each n ∈ IN, T ({xn }) ⊂ K. Since lim T ({xn }) n→∞

= T ({x}) and K is closed, T ({x}) ⊂ K. Hence, x ∈ D. Therefore, D is closed. Q.E.D.

3.2.7. Notation. Let X be a compactum. If A is any subset of X, then CL(A) = {B ∈ 2X | B ⊂ A}. The following Theorem tells us that the idempotency of T is related to its continuity. 3.2.8. Theorem. If X is a continuum for which T is continuous, then T is idempotent on X. Proof. Let W be a subcontinuum of X, and let x ∈ Int(W ). Since X \ Int(W ) is a closed subset of X, it is easy to see that Copyright © 2005 Taylor & Francis Group, LLC


179

CL(X \ Int(W )) is closed in 2X . Then T −1 (CL(X \ Int(W ))) is a closed subset of 2X by the continuity of T . Note that CL(X \ W ) ⊂ T −1 (CL(X \ Int(W ))) because X \ W ⊂ X \ Int(W ), and Remark 3.1.5. Since, for W = X, X \ Int(W ) is a limit point of CL(X \ W ) (recall that Cl(X \ W ) = X \ Int(W )), it follows that T (X \ Int(W )) ⊂ X \ Int(W ). Then there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ M ⊂ Int(W ). If W = X, then it suffices to choose M = X. Thus, in either case, the Theorem follows from Theorem 3.1.54. Q.E.D. 3.2.9. Theorem. If X is a point T –symmetric continuum for which T is continuous, then T ({p, q}) = T ({p}) ∪ T ({q}) for every p, q ∈ X. Proof. Let p ∈ X. For each x ∈ X, define A(x) = {y ∈ X | T ({x, y}) ∈ C(X)} and B(x) = {y ∈ X | T ({x, y}) = T ({x}) ∪ T ({y})}. We assert that A(x) and B(x) are both closed for every x ∈ X. Let x ∈ X, and let Z(x) ∈ {A(x), B(x)}. If y ∈ Cl(Z(x)), then there exists a sequence {yn }∞ n=1 of elements of Z(x) converging to y. Hence, the sequence {{x, yn }}∞ n=1 converges to {x, y} (note that for each z, w ∈ X, {z, w} = σ({{z}, {w}}). Since σ is continuous (Lemma 1.8.11), lim {x, yn } = lim σ({{x}, {yn }}) = σ({{x}, {y}}) = {x, y}.

n→∞

n→∞

Hence, by the continuity of T , the sequence {T ({x, yn })}∞ n=1 converges to T ({x, y}). If Z(x) is A(x), then, since C(X) is compact (Theorem 1.8.5), T ({x, y}) ∈ C(X). If Z(x) is B(x), then T ({x, y}) = lim T ({x, yn }) n→∞

= lim [T ({x}) ∪ T ({yn })] = T ({x}) ∪ lim T ({yn }) = T ({x}) ∪ n→∞

n→∞

T ({y}) (the third equality follows from the continuity of the union Copyright © 2005 Taylor & Francis Group, LLC

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map, σ). Hence, A(x) and B(x) are both closed subsets of X. Note that, by Corollary 3.1.63, X = A(p) ∪ B(p). Let x ∈ T ({p}). Then T ({x}) ⊂ T 2 ({p}) = T ({p}) (by Proposition 3.1.7 and the fact that T is idempotent, by Theorem 3.2.8). Since X is point T –symmetric and x ∈ T ({p}), p ∈ T ({x}). Hence, T ({p}) ⊂ T ({x}). Therefore, T ({p}) = T ({x}). Also, since {p, x} ⊂ T ({p}), T ({p, x}) ⊂ T ({p}). Thus, T ({p, x}) = T ({p}). Since T ({p}) is a continuum (Theorem 3.1.21), x ∈ A(p) ∩ B(p). Now, let x ∈ A(p) ∩ B(p). Then T ({p, x}) = T ({p}) ∪ T ({x}), and this set is a continuum. Hence, T ({p}) ∩ T ({x}) = ∅. Let y ∈ T ({p}) ∩ T ({x}). Then x ∈ T ({y}) ⊂ T 2 ({p}) = T ({p}). Therefore, A(p) ∩ B(p) = T ({p}). Now, suppose B(p) = T ({p}). Let y ∈ A(p), and let x ∈ B(p) \ T ({p}) be arbitrary points. Then T ({p, x}) = T ({p}) ∪ T ({x}) and T ({p}) ∩ T ({x}) = ∅. Hence, T ({x}) ∩ A(p) = ∅. (Suppose there exists z ∈ T ({x}) ∩ A(p), then, as before, T ({z}) = T ({x}). Since {p, z} ⊂ T ({p}) ∪ T ({z}) = T ({p}) ∪ T ({x}) ⊂ T ({p, x}), T ({p, z}) ⊂ T 2 ({p, x}) = T ({p, x}). Similarly, we have that T ({p, x}) ⊂ T ({p, z}). Hence, T ({p, z}) = T ({p, x}). Since z ∈ A(p), T ({p, z}) is connected. Thus, T ({p, x}) is connected, a contradiction.) Let U be an open set such that T ({x}) ⊂ U and Cl(U ) ∩ A(p) = ∅. Now, let q ∈ Bd(U ). Then q ∈ (X \ A(p)) ∩ (X \ T ({x})). Hence, q ∈ X \ T ({p, x}). Since T is idempotent, q ∈ X \T 2 ({p, x}). Thus, there exists a subcontinuum W of X such that q ∈ Int(W ) ⊂ W ⊂ X \ T ({p, x}). Then W ⊂ B(p), since otherwise (W ∩ A(p)) ∪ (W ∩ B(p)) is a separation of W (recall that A(p) ∩ B(p) = T ({p}) and W ∩ T ({p}) = ∅). Therefore, y ∈ X \ W , and q ∈ X \ T ({x, y}) (if q ∈ T ({x, y}), then {x, y} ∩ W = ∅, a contradiction). Since q was an arbitrary point of Bd(U ), T ({x, y}) = (T ({x, y}) ∩ U ) ∪ (T ({x, y}) ∩ (X \ Cl(U ))), where T ({x, y}) ∩ U and T ({x, y}) ∩ (X \ Cl(U )) are separated. Thus, T ({x, y}) is not connected. Hence, T ({x, y}) = T ({x}) ∪ T ({y}), by Corollary 3.1.63. So, x ∈ B(y). Thus, B(p) \ T ({p}) ⊂ B(y). Note that p ∈ Cl(B(p) \ T ({p})). If not, p ∈ Int(A(p)), and since A(p) is a continuum and T is idempotent (Theorem 3.2.8), there is a continuum M such that p ∈ Int(M ) ⊂ M ⊂ Int(A(p)) Copyright © 2005 Taylor & Francis Group, LLC


181

(Theorem 3.1.54). Hence, M misses some point q ∈ T ({p}), and p ∈ X \ T ({q}), contradicting the point T –symmetry of X. Thus, p ∈ Cl(B(p) \ T ({p})). Since B(y) is closed and p ∈ Cl(B(p) \ T ({p})), it follows that p ∈ B(y), or that y ∈ B(p). But y ∈ A(p), so that y ∈ T ({p}), and A = T ({p}). By contrapositive, if A(p) = T ({p}), then B(p) = T ({p}). So, for each p ∈ X, either A(p) = X or B(p) = X. Suppose there exists p ∈ X such that A(p) = X. Let q ∈ X be arbitrary. If q ∈ T ({p}), then A(q) = A(p) = X. (As we did before, if z ∈ A(p) ∪ A(q), then, since T ({p}) = T ({q}), T ({p, z}) = T ({q, z}).) Suppose q ∈ X \ T ({p}). Then q ∈ A(p), so p ∈ A(q). Since p ∈ X \ T ({q}), A(q) = T ({q}). Hence, A(q) = X. Thus, either A(p) = X for every p ∈ X or B(p) = X for every p ∈ X. If B(p) = X for every p ∈ X, the theorem is proved. So, suppose A(p) = X for all p ∈ X. Then for each p, q ∈ X, T ({p, q}) is a continuum. Hence, by Theorem 3.1.57, X is indecomposable. Thus, by Theorem 3.1.34, B(p) = X for all p ∈ X in this case also. Q.E.D. 3.2.10. Theorem. If X is a point T –symmetric continuum for which T is continuous, then X is T –additive. Proof. Let F(X) =

∞

Fn (X). Then F(X) is the family of all

n=1

finite subsets of X. We show first that for each A ∈ F(X), T (A) = T ({a}). a∈A

Suppose this is not true. Then there exists an M ∈ F(X) such T ({p}). Take M with the smallest cardinality. that T (M ) = p∈M

By Theorem 3.2.9, M has at least three elements. We assert that T (M ) is a continuum. If T (M ) is not connected, then there exist two disjoint closed subsets of A and B of X such that T (M ) = A∪B. Then, by Lemma 3.1.59, Corollary 3.1.14 and the minimality of M : T (M ) = A ∪ B = T (M ∩ A) ∪ T (M ∩ B) = T ({p}) ∪ T ({p}) = T ({p}), p∈M ∩A

p∈M ∩B


p∈M


182

contrary to the choice of M . Further, if p, q ∈ M are distinct points, T ({p}) ∩ T ({q}) = ∅; otherwise, by point T –symmetry and idempotency, T ({p}) = T ({q}) and then, since M ⊂ T (M \ {p}): T (M ) ⊂ T 2 (M \ {p}) = T (M \ {p}) = T ({r}) ⊂ T (M ). r∈M \{p}

Thus, T (M ) =

T ({p}). This also contradicts the choice of M .

p∈M

Now, let p ∈ M be arbitrary, and let N = M \ {p}. Then N has at least two elements. Since N has cardinality smaller than the cardinality of M , T (N ) = T ({r}), i.e., T (N ) is a finite union r∈N

of pairwise disjoint subcontinua. Hence, T (N ) is not a continuum. Now, define L = x ∈ X T (N ∪ {x}) = T ({r}) ∪ T ({x}) r∈N

and K = {x ∈ X | T (N ∪ {x}) ∈ C(X)}. Note that N ⊂ L and p ∈ K. Thus, L = ∅ and K = ∅. We claim that L and K are closed subsets of X. To see this, let Z be either L or K. Let x ∈ Cl(Z). Then there exists a sequence {xn }∞ n=1 of elements of Z converging to x. If Z is L, since T and the union map σ (Lemma 1.8.11) are continuous,

T (N ∪{x}) = lim T (N ∪{xn }) = lim T ({r}) ∪ T ({xn }) = n→∞

r∈N

n→∞

T ({r}) ∪ lim T ({xn }) = n→∞

r∈N

T ({r}) ∪ T ({x}).

r∈N

Hence, x ∈ L. Therefore, L is closed. If Z is K, since T and the union map σ are continuous, the sequence {T (N ∪ {xn })}∞ n=1 is a sequence of continua converging to T (N ∪ {x}) and C(X) is closed in 2X (Theorem 1.8.5), then T (N ∪ {x}) ∈ C(X). Hence, x ∈ K. Therefore, K is closed. Copyright © 2005 Taylor & Francis Group, LLC


183

If y ∈ K ∩ L, then T (N ∪ {y}) =

T ({r}) ∪ T ({y}) and this

r∈N

set is connected. Thus, T ({y}) ∩ T ({r}) = ∅ for each r ∈ N . Then, by point T –symmetry and idempotency, T ({y}) = T ({r}) for every r ∈ N , a contradiction to the fact that for r1 , r2 ∈ N , if r1 = r2 , then T ({r1 }) = T ({r2 }). Thus, K ∩ L = ∅. Let x ∈ X \ L. Then T (N ∪ {x}) is a continuum; otherwise, there exist two disjoint closed subsets A and B of X such that T (N ∪{x}) = A∪B. Hence, by Lemma 3.1.59 and Corollary 3.1.14: T (N ∪ {x}) = A ∪ B = T ((N ∪ {x}) ∩ A) ∪ T ((N ∪ {x}) ∩ B) = T ({r}) ∪ T ({r}) = T ({r}) ∪ T ({x}), r∈(N ∪{x})∩A

r∈(N ∪{x})∩B

r∈N

a contradiction to the fact that x ∈ X \ L. Therefore, T (N ∪ {x}) is a continuum. Thus, x ∈ K. Hence, X = K ∪ L, and K and L are disjoint closed sets, a contradiction to the fact that X is connected. Therefore, T (A) = T ({a}) for each A ∈ F(X). a∈A

Now, let B ∈ 2X , and let ε > 0. Let δ > 0 given by the ε uniform continuity of T for . Since F(X) is dense in 2X (proof of 2 Corollary 1.8.9), there exists A ∈ F(X) such that H(B, A) < δ. Since H(B, A) < δ, H2 (F1 (B), F1 (A)) < δ. Hence, by Theε orem 1.8.22, H2 (2T (F1 (B)), 2T (F1 (A))) < . This implies, by 2 Lemma 1.8.11, that

ε T ({b}), T ({a}) < . H 2 a∈A b∈B Therefore:

H T (B),

H(T (B), T (A)) + H

T ({b})

b∈B

T ({b}),

b∈B


a∈A

≤

T ({a})

tB , since tA = tB is impossible. If tA < tB , applying Lemma 3.3.17 to H|X×[0,tA ] , we obtain an element t < tA such that H((r, t )) ∈ B, a contradiction. Similarly, if tA > tB , applying Lemma 3.3.17 to H|X×[0,tB ] , we obtain an element t < tB such that H((r, t )) ∈ A, a contradiction. Therefore, X is not contractible. Q.E.D. Now, we present some connectivity properties of T . 3.3.19. Lemma. Let X be a continuum. If S is a nonempty totally disconnected closed subset of X such that there is a point p ∈ T (S)\S and such that for each proper subset S of S, p ∈ X \T (S ), then T (S) is connected. Proof. Let S be a totally disconnected closed subset of X satisfying the properties stated. Let S0 be a nonempty subset of S which is both open and closed in S. Since p ∈ X \ T (S \ S0 ), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ (S \ S0 ). ∞ Let {Un }∞ n=1 and {Vn }n=1 be decreasing sequences of open subsets Copyright © 2005 Taylor & Francis Group, LLC

3.3. APPLICATIONS

195

of X such that for each n ∈ IN, S \ S0 ⊂ Un , S0 ⊂ Vn , U1 ∩ Cl(V1 ) = ∞ ∞ U1 ∩ W = Cl(V1 ) ∩ {p} = ∅, S \ S0 = Un and S0 = Vn . n=1

n=1

For each n ∈ IN, let Cn be the component of X \ Un such that W ⊂ Cn . Since p does not belong to the interior of the component of Cn \ Vn in which it lies, there is a sequence {Dnj }∞ j=1 of distinct j components of Cn \ Vn such that p ∈ Dn = lim Dn . Since, by Thej→∞

orem 1.7.22, Dnj ∩ Cl(Vn ) = ∅, for each n ∈ IN, Dn is a continuum, Dn ⊂ Cn \ Vn , p ∈ Dn and Dn ∩ Cl(Vn ) = ∅. Let D = lim Dn . n→∞

Then D is a continuum containing p and D ∩ S0 = ∅. We assert that D ⊂ T (S). Suppose this is not true. Then there exist q ∈ D \ T (S) and a subcontinuum W of X such that q ∈ Int(W ) ⊂ W ⊂ X \S. It follows that there exists N1 ∈ IN such that for n > N1 , W ⊂ X \ (Un ∪ Vn ). Since q ∈ Int(W ) ∩ D, there exists N2 ∈ IN such that for n > N2 , (Int(W )) ∩ Dn = ∅. Let m > max{N1 , N2 }. Since (Int(W )) ∩ Dm = ∅, there exists N3 ∈ IN such j that for j > N3 , (Int(W )) ∩ Dm

= ∅. Let j > max{N1 , N2 , N3 }. j Then Dm ⊂ Cm and W ∩ Cm = ∅. Thus, W ⊂ Cm \ Vm , a j contradiction, since j > max{N1 , N2 , N3 } and (Int(W )) ∩ Dm

= ∅. Therefore, D ⊂ T (S). Now, let s ∈ S. Then there exists a sequence {Sn }∞ n=1 of subsets of S such that for each n ∈ IN, Sn is open and closed in S, and ∞ {s} = Sn . By the previous construction, for each n ∈ IN, there n=1

is a subcontinuum An of T (S) such that p ∈ An and An ∩ Sn = ∅. Since C(X) is compact (Theorem 1.8.5), without loss of generality, we assume that the sequence {An }∞ n=1 converges. Let As = lim An . n→∞

Then As is a subcontinuum of T (S) and {p, s} ⊂ As . Let A = As . Then A is connected (for each s ∈ S, p ∈ As ) s∈S

and S ⊂ A ⊂ T (S). For each x ∈ T (S), let Cx be the component of T (S) such that x ∈ Cx . By Corollary 3.3.4, Cx ∩ S = ∅. Thus, Cx ∩ A = ∅ for each x ∈ T (S), and it follows that ⎛ T (S) = A ∪ ⎝

x∈T (S)


⎞ Cx ⎠

196


is connected. Q.E.D. The following Theorem shows that we may remove the “totally disconnected” hypothesis in Lemma 3.3.19.

3.3.20. Theorem. Let X be a continuum. Suppose A is a nonempty closed subset of X, and there exists a point p ∈ TX (A) \ A such that for each proper closed subset A of A, p ∈ X \ TX (A ). Then TX (A) is connected. Proof. If TX (A) = X, there is nothing to do. Then suppose TX (A) is a proper closed subset of X. Let {Aλ }λ∈Λ be the family of components of A. Then G = {Aλ }λ∈Λ ∪ {{x} | x ∈ X \ A} is an upper semicontinuous decomposition of X and X/G is a continuum (Theorem 1.7.3). Let q : X → → X/G be the quotient map. Note that q is monotone.

Since A is a closed subset of X, q(A) = q Aλ = q(Aλ ) λ∈Λ

λ∈Λ

is a closed totally disconnected subset of X/G. Thus, TX/G (q(A)) is connected, by Lemma 3.3.19. Since q is monotone, by Theorem 3.1.64 (b), we have that TX (A) = TX (q −1 q(A)) ⊂ q −1 TX/G (q(A)). By Theorem 3.1.64 (c), we have q −1 TX/G (q(A)) = q −1 qTX (q −1 (q(A))) = q −1 qTX (A). Since A ⊂ T (A), q −1 qTX (A) = TX (A). Hence, q −1 TX/G (q(A)) = TX (A). Therefore, TX (A) is connected. Q.E.D.

3.3.21. Theorem. Let X be a continuum. Suppose A is a nonempty closed subset of X and there exists x ∈ T (A) \ A. Then there exists a closed subset D of X such that: (a) D ⊂ A, (b) x ∈ T (D), (c) if E is a nonempty closed subset of X and E D, then x ∈ X \ T (E), and (d) T (D) is a continuum. Copyright © 2005 Taylor & Francis Group, LLC

3.3. APPLICATIONS

197

Proof. Let {Bn }∞ n=1 be a decreasing sequence of nonempty closed ∞ subsets of X such that for each n ∈ IN, x ∈ T (Bn ). Let B = Bn . n=1

Then B is a nonempty closed subset of X. If x ∈ X \ T (B), then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ B. Since W is compact, there exists N ∈ IN such that W ⊂ N (X \ Bn ), a contradiction to the fact that x ∈ T (BN ). Hence, n=1

x ∈ T (B). By the Brouwer Reduction Theorem (see (11.1) (p. 17) of [26]), there exists a minimal element D satisfying conditions (a), (b) and (c). Condition (d) follows from Theorem 3.3.20. Q.E.D. In 5.12 of [23] it is shown that if X is a continuum and x is a point at which X is not connected im kleinen, then there is a subcontinuum K of X such that x ∈ K and X is not connected im kleinen at any of the points of K. In the following theorem, we show that something similar happens with the images of T . 3.3.22. Theorem. Let X be a continuum. Suppose A is a nonempty closed subset of X such that there exists x ∈ T (A) \ A. Then there exists a nondegenerate subcontinuum K of X such that x ∈ K ⊂ T (A). Proof. Let A be a nonempty closed subset of X and let x ∈ T (A)\A. By Theorem 3.3.21, there exists a nonempty closed subset D of X satisfying conditions (a), (b), (c) and (d). Let V be an open subset of X such that x ∈ V ⊂ Cl(V ) ⊂ X \ A. Let H be the component of V ∩ T (D) containing x, and let K = Cl(H). Since V ∩ T (D) is open in T (D), by Theorem 1.7.22, K intersects Bd(V ). Thus, K is a nondegenerate subcontinuum of X contained in T (A). Q.E.D. The following Theorem characterizes local connectedness in unicoherent continua. 3.3.23. Theorem. Let X be a unicoherent continuum. Then X is locally connected if and only if for each nonempty closed subset C of X which separates X between two points x and y, there exists a component E of C which separates X between x and y. Copyright © 2005 Taylor & Francis Group, LLC


198

Proof. First, suppose that X is unicoherent and locally connected. Let C be a nonempty closed subset of X which separates X between x and y. Let A be the component of X \ C containing x, and let B be the component of X \ Cl(A) containing y. Since X is locally connected, by Lemma 1.7.11, A and B are open subsets of X. Also, X \ (Cl(A) ∩ Cl(B)) ⊂ (X \ Cl(B)) ∪ B, x ∈ X \ Cl(B), and y ∈ B. Thus, Cl(A) ∩ Cl(B) is a closed subset of X which separates X between x and y. Let {Sλ }λ∈Λ be the family of components of X \ Cl(A) such that Sλ ∩ B = ∅ for each λ ∈ Λ. Then, by Theorem 1.7.22, Cl(A) ∪ Cl(Sλ ) is connected. Note that X = Cl(A) ∪ λ∈Λ

Cl(Sλ ) ∪Cl(B). Since X is unicoherent, Cl(A)∩Cl(B) = Cl λ∈Λ

Cl(Sλ ) ∩ Cl(B) is a continuum. Let E be the Cl(A) ∪ Cl λ∈Λ

component of C that contains Cl(A) ∩ Cl(B). If x and y belong to the same component of X \ E ⊂ X \ (Cl(A) ∩ Cl(B)), then x and y are in the same component of X \ (Cl(A) ∩ Cl(B)). Since this is not true, E separates X between x and y. Now, suppose X is a unicoherent continuum satisfying the conditions stated, and suppose X is not locally connected, hence not connected im kleinen at some point p ∈ X. Then, by Theorem 3.1.24, there exists a nonempty closed subset A of X such that p ∈ T (A)\A. By Theorem 3.3.21, there exists a nonempty closed subset B of X satisfying conditions (a), (b), (c) and (d). Let x ∈ B and let U0 be an open subset of X such that p ∈ U0 ⊂ Cl(U0 ) ⊂ X \ B. Then Bd(U0 ) is a closed set which separates X between x and p. Hence, there exists a component N of Bd(U0 ) which separates X between x and p. Let U and V be disjoint open sets such that X \ N = U ∪ V , p ∈ U and x ∈ V . Then, by Lemma 1.7.18, H = N ∪ U and K = N ∪ V are continua, and X = H ∪ K. Since x ∈ B ∩ K, then B ∩ K = ∅. If B ∩ H = ∅, then p ∈ X \ T (B), which is contrary to (b) of Theorem 3.3.21. Thus, B ∩ H = ∅. Since B ∩ K and B ∩ H are two nonempty closed proper subsets of B, it follows from (c) of Theorem 3.3.21 that there exist two subcontinua L1 Copyright © 2005 Taylor & Francis Group, LLC

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and L2 of X such that p ∈ Int(L1 ) ⊂ L1 ⊂ X \ (B ∩ H) and p ∈ Int(L2 ) ⊂ L2 ⊂ X \ (B ∩ K). Again, if L1 ∩ (B ∩ K) = ∅ or L2 ∩ (B ∩ H) = ∅, it follows that p ∈ X \ T (B). Thus, assume that L1 ∩ (B ∩ K) = ∅ and L2 ∩ (B ∩ H) = ∅. Let X1 = L1 ∪ K and X2 = L2 ∪ H. Then X1 and X2 are continua and X = X1 ∪ X2 . Now, since X is unicoherent, X1 ∩ X2 is a continuum. Note that p ∈ Int(X1 ∩ X2 ) ⊂ (X1 ∩ X2 ) ⊂ X \ B. Hence, p ∈ X \ T (B), which is contrary to condition (b) of Theorem 3.3.21. Therefore, X is locally connected. Q.E.D.

REFERENCES

[1] D. P. Bellamy, Continua for Which the Set Function T is Continuous, Trans. Amer. Math. Soc., 1511 (1970), 581–587. [2] D. P. Bellamy, Set Functions and Continuous Maps, in General Topology and Modern Analysis, (L. F. McAuley and M. M. Rao, eds.), Academic Press, (1981), 31–38. [3] D. P. Bellamy, Some Topics in Modern Continua Theory, in Continua Decompositions Manifolds, (R. H. Bing, W. T. Eaton and M. P. Starbird, eds.), University of Texas Press, (1983), 1–26. [4] D. P. Bellamy and J. J. Charatonik, The Set Function T and Contractibility of Continua, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys., 25 (1977), 47–49. [5] D. P. Bellamy and H. S. Davis, Continuum Neighborhoods and Filterbases, Proc. Amer. Math. Soc., 27 (1971), 371–374. [6] D. E. Bennet, A Characterization of Locally Connectedness By Means of the Set Function T , Fund. Math. 86 (1974), 137–141. [7] R. H. Bing and F. B. Jones, Another Homogeneous Plane Continuum, Trans. Amer. Math. Soc., 90 (1959), 171–192. Copyright © 2005 Taylor & Francis Group, LLC

200

REFERENCES

[8] K. Borsuk and S. Ulam, On Symmetric Products of Topological Spaces, Bull. Amer. Math. Soc., 37 (1931), 875-882. [9] H. S. Davis, A Note on Connectedness im Kleinen, Proc. Amer. Math. Soc., 19 (1968), 1237–1241. [10] H. S. Davis, Relationships Between Continuum Neighborhoods in Inverse Limit Spaces and Separations in Inverse Limit Sequences, Proc. Amer. Math. Soc., 64 (1977), 149– 153. [11] H. S. Davis and P. H. Doyle, Invertible Continua, Portugal. Math., 26 (1967), 487–491. [12] R. W. FitzGerald, The Cartesian Product of Non–degenerate Compact Continua is n–point Aposyndetic, Topology Conference (Arizona State Univ., Tempe, Ariz., 1967), Arizona State University, Tempe, Ariz., (1968), pp. 324–326. [13] J. T. Goodykoontz, Jr., Aposyndetic Properties of Hyperspaces, Pacific J. Math., 47 (1973), 91–98. [14] J. G. Hocking and G. S. Young, Topology, Dover Publications, Inc., New York, 1988. [15] F. B. Jones, Concerning the Boundary of a Complementary Domain of a Continuous Curve, Bull. Amer. Math. Soc., 45 (1939), 428–435. [16] F. B. Jones, Aposyndetic Continua and Certain Boundary Problems, Amer. J. Math., 53 (1941), 545–553. [17] F. B. Jones, Concerning Nonaposyndetic Continua, Amer. Math., 70 (1948), 403–413. [18] K. Kuratowski, Topology, Vol. II, Academic Press, New York, N. Y., 1968. [19] S. Mac´ıas, Aposyndetic Properties of Symmetric Products of Continua, Topology Proc., 22 (1997), 281–296. Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

201

[20] S. Mac´ıas, A Class of One–dimensional Nonlocally Connected Continua for Which the Set Function T is Continuous, to appear in Houston Journal of Mathematics. [21] M. A. Molina, Algunos Aspectos Sobre la Función T de Jones, Tesis de Licenciatura, Facultad de Ciencias, U. N. A. M., 1998. (Spanish) [22] S. B. Nadler, Jr., Hyperspaces of Sets, Monographs and Textbooks in Pure and Applied Math., Vol. 49, Marcel Dekker, New York, Basel, 1978. [23] S. B. Nadler, Jr., Continuum Theory: An Introduction, Monographs and Textbooks in Pure and Applied Math., Vol. 158, Marcel Dekker, New York, Basel, Hong Kong, 1992. [24] E. L. VandenBoss, Set Functions and Local Connectivity, Ph. D. Dissertation, Michigan State University (1970). University Microfilms # 71–11997. [25] G. T. Whyburn, Semi–locally–connected Sets, Amer. J. Math., 61 (1939), 733–749. [26] G. T. Whyburn, Analytic Topology, Amer. Math. Soc. Colloq. Publ., vol. 28, Amer. Math. Soc., Providence, R. I., 1942. [27] S. Willard, General Topology, Addison–Wesley Publishing Co., 1970.


Chapter 4 A THEOREM OF E. G. EFFROS

We present a topological proof of a Theorem by E. G. Effros [4] and a consequence of it, due to C. L. Hagopian [5], which has been very useful in the theory of homogeneous continua. We present the proof of Effros’s result given by Fredric G. Ancel [1]. Before showing Effros’s Theorem, we present some background on topological groups and actions of topological groups on metric spaces.

4.1

Topological Groups

We introduce topological groups and give some of its elementary properties. 4.1.1. Definition. We say that a group G, which has a topology τ , is a topological group provided that the group operations are continuous, i.e., the functions given by: π: G × G → G Copyright © 2005 Taylor & Francis Group, LLC

π((g1 , g2 )) = g1 · g2 203

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and ξ: G → G

ξ(g) = g −1

are continuous. 4.1.2. Proposition. Let G be a group with a topology τ . Then G is a topological group if and only if the function ω : G × G → G given by ω((g1 , g2 )) = g1 · g2−1 is continuous. Proof. Suppose G is a topological group. Observe that ω = π ◦ (1G × ξ), where 1G denotes the identity map of G. Hence, ω is continuous. Now, suppose that ω is continuous. Note that ξ(g) = ω((eG , g)), where eG is the identity element of G. Thus, ξ is continuous. Since π = ω ◦ (1G × ξ), π is continuous. Q.E.D. 4.1.3. Notation. Let G be a group. If H and K are nonempty subsets of G, then define H · K = {h · k | h ∈ H and k ∈ K} and

H −1 = {h−1 | h ∈ H}.

If H = {h}, then we write h · K instead of {h} · K. Similarly, we write H ·k instead of H ·{k}, when K = {k}. We define, inductively, H n+1 = H · H n for each n ∈ IN. 4.1.4. Notation. If G is a topological group and g ∈ G, then N (g) denotes the family of all neighborhoods of g in G. Also, eG denotes the identity element of G. 4.1.5. Remark. If G is a topological group, then we may describe the continuity of the group operations as follows: Let g1 and g2 be two elements of G. For the continuity of π: for each U ∈ N (g1 · g2 ), there exist V ∈ N (g1 ) and W ∈ N (g2 ) such that V · W ⊂ U . For the continuity of ξ: for each U ∈ N (g1−1 ), there exists V ∈ N (g1 ) such that V −1 ⊂ U . Copyright © 2005 Taylor & Francis Group, LLC

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I, +) and (C I \ 4.1.6. Example. It is easy to see that (IRn , +), (C {0}, ·) are topological groups with the usual operations and the usual topologies.

4.1.7. Proposition. Let G be a topological group. If g0 is a fixed element of G, then the functions ψg0 , ϕg0 : G → G given by ψg0 (g) = g0 · g and ϕg0 (g) = g · g0 are homeomorphisms. Proof. Note that ψg0 and ϕg0 are restrictions of the operation π to {g0 } × G and G × {g0 }, respectively. Hence, both functions are continuous. Also note that ψg0 ◦ψg0−1 = ψg0−1 ◦ψg0 = 1G and that ϕg0 ◦ ϕg0−1 = ϕg0−1 ◦ ϕg0 = 1G . Therefore, ψg0 y ϕg0 are homeomorphisms, since, clearly, ψg0−1 and ϕg0−1 are continuous functions. Q.E.D. 4.1.8. Definition. Let G be a topological group. The maps ψg0 and ϕg0 of Proposition 4.1.7 are called left translation by g0 and right translation by g0 , respectively. 4.1.9. Corollary. If G is a topological group and ψg0 and ϕg0 are the left and right translation by g0 , respectively, then ψg−1 = ψg0−1 0 −1 and ϕg0 = ϕg0−1 . 4.1.10. Proposition. Let G be a topological group. Then the map ξ : G → G given by ξ(g) = g −1 is a homeomorphism. Proof. By Definition 4.1.1, ξ is continuous. Since ξ ◦ ξ = 1G , ξ is a homeomorphism. Q.E.D. 4.1.11. Corollary. Each topological group is a homogeneous space. Proof. Let g1 and g2 be two elements of G. Let ζ : G → G be given by ζ(g) = g · g1−1 · g2 . Note that ζ(g1 ) = g2 and that ζ = ϕg1−1 ·g2 . Therefore, ζ is a homeomorphism. Q.E.D.


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4.1.12. Proposition. Let G be a topological group, and let N ◦ (eG ) be a local basis of open sets of G containing eG . Let Lg = {g·U | U ∈ N ◦ (eG )} and let Rg ={U · g | U ∈ N ◦ (eG )}, for each g ∈ G. If L= Lg and R = Rg , then L and R both form a basis for g∈G

g∈G

the topology of G. Proof. Let W be an open subset of G and let g0 be an element of W . Since ψg0−1 is a homeomorphism (Proposition 4.1.7), ψg0−1 (W ) = g0−1 · W is an open subset of G such that eG ∈ g0−1 · W . Since N ◦ (eG ) is a local basis of eG , there exists U ∈ N ◦ (eG ) such that U ⊂ g0−1 · W . Hence, ψg0 (eG ) ∈ ψg0 (U ) ⊂ ψg0 (g0−1 · W ), i.e., g0 ∈ g0 · U ⊂ (g0 · g0−1 ) · W = W . Therefore, L is a basis for the topology of G. The proof of the fact that R forms a basis for the topology of G is similar. Q.E.D. 4.1.13. Definition. Let G be a topological group and let S be a nonempty subset of G. We say that S is symmetric if S = S −1 . The family of symmetric neighborhoods of eG is denoted by N ∗ (eG ), i.e., N ∗ (eG ) = {V ∈ N (eG ) | V = V −1 }. The next three Propositions say that each of the families of symmetric neighborhoods, powers of neighborhoods and closure of neighborhoods of the identity element of a topological group forms a local basis. 4.1.14. Proposition. If G is a topological group and if U ∈ N (eG ), then there exists V ∈ N ∗ (eG ) such that V ⊂ U . Proof. Let U ∈ N (eG ). Since ξ is a homeomorphism (Proposition 4.1.10), ξ(U ) ∈ N (eG ). Note that ξ(U ) = U −1 . Let V = U ∩ U −1 . Then, clearly, V ∈ N ∗ (eG ) and V ⊂ U . Q.E.D. 4.1.15. Proposition. If G is a topological group and U ∈ N (eG ), then for each n ∈ IN, there exists V ∈ N (eG ) such that V n ⊂ U . Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Let U ∈ N (eG ). We do the proof by induction over n. For n = 1, let V = U . Let n ≥ 2 and suppose there exists W ∈ N (eG ) such that W n ⊂ U . We show there exists V ∈ N (eG ) such that V n+1 ⊂ U . Since the operation π is continuous and eG · eG = eG , there exist V1 , V2 ∈ N (eG ) such that V1 · V2 ⊂ W . Let V = V1 ∩ V2 . Then V ∈ N (eG ) and V 2 ⊂ W . Thus, V n+1 = V 2 · V n−1 ⊂ W · W n−1 = W n ⊂ U. Q.E.D. 4.1.16. Proposition. If G is a topological group and U ∈ N (eg ), then there exists V ∈ N (eG ) such that Cl(V ) ⊂ U . Proof. Let U ∈ N (eG ). By Proposition 4.1.15, there exists W ∈ N (eG ) such that W 2 ⊂ U . By Proposition 4.1.14, there exists V ∈ N ∗ (eG ) such that V ⊂ W . Hence, V 2 ⊂ U . Let g ∈ Cl(V ). Then g · V ∈ N (g) (Proposition 4.1.12). Thus, g · V ∩ V = ∅. This implies that there exist g1 , g2 ∈ V such that g · g1 = g2 . Hence, g = g2 · g1−1 ∈ V · V −1 = V 2 ⊂ U . Therefore, Cl(V ) ⊂ U . Q.E.D.

4.1.17. Proposition. Let G be a topological group. Let S, T , U and R be nonempty subsets of G, and let g0 ∈ G. Then the following statements are true: (1) If U is an open subset of G, then the sets: g0 · U , U · g0 , U −1 , R · U and U · R are open subsets of G. (2) If S is a closed subset of G, then the sets: g0 · S, S · g0 and S −1 are closed subsets of G. (3) If S and T are compact subsets of G, then the sets: S · T and S −1 are compact subsets of G. (4) If S is a compact subset of G and T is a closed subset of G, then S · T and T · S are closed subsets of G. S · W and Cl(S) = W · S. (5) Cl(S) = W ∈N (eG )


W ∈N (eG )

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Proof. We show (1) is true. Note that, since the maps ψg0 , ϕg0 (Proposition 4.1.7) and ξ (Proposition 4.1.10) are homeomorphisms, −1 g0 · U , U · g0 and U of G. are open subsets r · U and U · R = U · r, R · U and U · R are Since R · U = r∈R

r∈R

open subsets of G. The proof of (2) is similar to the one given in (1). Note that (3) follows from the continuity of the operations π and ξ and the fact that S and T are compact sets. We show (4) is true. To this end, we see that T · S is closed. The proof of the fact that S · T is closed is similar. To see T · S is closed, we show G \ (T · S) is an open subset of G. Let g ∈ G \ (T · S). Then for each s ∈ S, T · s is a closed subset of G (part (2)). Thus, there exists Us ∈ N ∗ (eG ) such that g·Us ∩T ·s = ∅. By Proposition 4.1.15, there exists Ws ∈ N (eG ) such that Ws2 ⊂ Us . By Proposition 4.1.14, there exists Vs ∈ N ∗ (eG ) such that Vs ⊂ Ws . Hence, for each s ∈ S, Vs2 ⊂ Us . Note that g · Vs ∩ (T · s) · Vs = ∅ for every s ∈ S. Now, observe that the family of neighborhoods {s · Vs | s ∈ S} covers S. Since S is compact, there exist s1 , . . . , sn ∈ S such that n n S⊂ sj ·Vsj . Let W = Vsj . Then W ∈ N ∗ (eG ), g ·W ∈ N (g) j=1

j=1

and g · W ∩ (T · sj ) · Vsj = ∅, for each j ∈ {1, . . . , n}. Thus, g · W ∩ T · S = ∅. Therefore, G \ (T · S) is open. We prove (5) is satisfied. Let W ∈ N (eG ). Then there exists V ∈ N ∗ (eG ) such that V ⊂ W . Note that S ⊂ S · V ⊂ S · W . Let g ∈ Cl(S). Since V ∈ N (eG ), g · V ∩ S = ∅. Thus, there exist v ∈ V and s ∈ S such that g · v = s. Hence, g = s · v −1 ∈ S · V −1 = S · V ⊂ S · W . Therefore, Cl(S) ⊂ S · W. Next, let g ∈

W ∈N (eG )

S · W , and let V ∈ N (g). We show that

W ∈N (eG )

V ∩ S = ∅, which implies that g ∈ Cl(S). To this end, note that V −1 · g ∈ N (eG ). Then g ∈ S · (V −1 · g). Thus, there exist s ∈ S and v ∈ V such that g = s · v −1 · g. Hence, s = v. Therefore, S ∩ V = ∅.


4.2. GROUP ACTIONS AND A THEOREM OF EFFROS

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The proof of the other equality is similar. Q.E.D.

4.2

Group Actions and a Theorem of Effros

We present elementary properties of action of a topological group on a topological space. We also give a topological proof of a Theorem of E. G. Effros (Theorem 4.2.25). 4.2.1. Definition. An action of a topological group G on a metric space X is a map θ : G × X → X such that: (1) θ(g, θ(g , x)) = θ(g · g , x) for each g , g ∈ G and each x ∈ X; and (2) θ(eG , x) = x for each x ∈ X. 4.2.2. Notation. Instead of θ(g, x), we write g · x. In this way, the properties (1) and (2) of Definition 4.2.1 become: (1) g · (g · x) = (g · g ) · x; and (2) eG · x = x. 4.2.3. Remark. Let G be a topological group acting on a metric space X. Note that for each g ∈ G, the function ηg : X → → X, given by ηg (x) = g · x, is a homeomorphism, whose inverse is the map ηg−1 .

4.2.4. Example. (1) If G is topological group, then G acts on Gn , for each n ∈ IN. I n, (2) If G is IR \ {0} (or G is CI \ {0}), then G acts on IRn (or on C respectively). Copyright © 2005 Taylor & Francis Group, LLC

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(3) If G is S 1 , then G acts on CI.

4.2.5. Notation. Let G be a topological group acting on a metric space X. If K is a nonempty subset of G and x ∈ X, then K · x = {g · x | g ∈ K}.

4.2.6. Definition. Let G be a topological group acting on a metric space X. We say that G acts transitively on X, if G · x = X for each x ∈ X. We say that G acts micro–transitively on X if for each x ∈ X and each U ∈ N (eG ), U · x is a neighborhood of x in X.

4.2.7. Remark. From now on, G denotes a topological group whose topology is given by a complete metric acting on a metric space X. We call such a group a complete metric group.

4.2.8. Definition. Let G be a complete metric group acting on a metric space X. For each x ∈ X, define the map γx : G → → X by γx (g) = g · x for every g ∈ G.

4.2.9. Lemma. If G is a complete metric group acting transitively on a metric space X, then the following are equivalent: (a) G acts micro–transitively on X; (b) γx is an open map for each x ∈ X; (c) γx is an open map for some x ∈ X. Proof. First assume (a). We show (b). Let x be a point of X and let U be an open subset of G. Take g ∈ U . Note that g −1 · U is an open subset of G such that eG ∈ g −1 · U (Proposition 4.1.7). Since G acts micro–transitively on X, (g −1 · U ) · x is a neighborhood of x in X. Hence, U · x = γx (U ) is a neighborhood of g · x = γx (g). Therefore, γx is an open map. Copyright © 2005 Taylor & Francis Group, LLC


211

Obviously, if (b) is true, then (c) is true. Assume (c). We prove (b). Suppose γx is an open map for some x ∈ X. Let z ∈ X. We see that γz is also an open map. Since G acts transitively on X, there exists g0 ∈ G such that z = g0 · x. Since ϕg0 is a homeomorphism (Proposition 4.1.7) and γz = γx ◦ ϕg0 , γz is an open map. Finally, assume (b). We see (a). Let x ∈ X and let U ∈ N (eg ). Since γx is an open map, γx (IntG (U )) = (IntG (U )) · x is an open subset of X such that x ∈ (IntG (U )) · x. Hence, U · x is a neighborhood of x in X. Therefore, G acts micro–transitively on X. Q.E.D. 4.2.10. Definition. Let G be a complete metric group acting on a metric space X. We say that X is G–countably covered if for each x ∈ X and each U ∈ N (eG ), there is a sequence {hn }∞ n=1 of homeomorphisms of X such that the family {hn (U · x)}∞ covers n=1 X. 4.2.11. Definition. Let G be a complete metric group acting on a metric space X. We say that G acts weakly micro–transitively on X if for each x ∈ X and each U ∈ N (eG ), Cl(U · x) is a neighborhood of x in X. The following Lemma provides examples of metric spaces which are G–countably covered. 4.2.12. Lemma. Let G be a complete metric group acting transitively on a metric space X. If G is separable, then X is G–countably covered. Proof. Suppose G is separable. Let x ∈ X and let U ∈ N (eG ). Then the collection {g · U | g ∈ G} covers G. Since G is a separable metric space, there is a sequence, {gn }∞ n=1 , of elements of G such ∞ that the sequence of open sets {gn · U }n=1 (Proposition 4.1.7) covers G. Because G acts transitively on X, {(gn · U ) · x}∞ n=1 covers X. Note that for each n ∈ IN, (gn · U ) · x = ηgn (U · x) and each ηgn is a homeomorphism (Remark 4.2.3). Therefore, X is G–countably covered. Q.E.D.


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4.2.13. Lemma. Let G be a complete metric group acting on a metric space X. If X is of the second category and is G–countably covered, then G acts weakly micro–transitively on X. Proof. Assume X is of the second category and is G–countably covered. Let x ∈ X, and let U ∈ N (eG ). We show that Cl(U · x) is a neighborhood of x in X. Since G is a topological group, there exists V ∈ N (eG ) such that V · V −1 ⊂ U (Proposition 4.1.2). By hypothesis, there is a sequence, {hn }∞ n=1 , of homeomorphisms of X such that the family {hn (V · x)}∞ n=1 covers X. Since X is of the second category, there exists n ∈ IN such that Cl(hn (V · x)) has nonempty interior. Hence, Cl(V · x) has nonempty interior. Since any open nonempty subset of Cl(V · x) must intersect V · x, it follows that there exists g ∈ V such that g · x ∈ Int(Cl(V · x)). Thus, x ∈ g −1 · (Int (Cl(V · x))) = Int (Cl((g −1 · V ) · x)) ⊂ Int (Cl((V −1 · V ) · x)) ⊂ Int (Cl(U · x)). Therefore, Cl(U · x) is a neighborhood of x in X. Q.E.D. Let G be a complete metric group acting transitively on a metric space X. If G also acts micro–transitively on X, then G acts weakly micro–transitively on X. The next Lemma shows that the converse is also true. 4.2.14. Lemma. Let G be a complete metric group acting transitively on a metric space X. If G acts weakly micro–transitively on X, then G acts micro–transitively on X. Proof. Suppose G acts weakly micro–transitively on X. Let ρX be a metric on X and let ρG be a complete metric on G. Let x0 ∈ X and let U ∈ N (eG ). We show that U · x0 is a neighborhood of x0 in X. To this end, let U0 ∈ N (eG ) such that (Cl(U0 ))−1 · (Cl(U0 )) ⊂ U (Propositions 4.1.2 and 4.1.16). Since G acts weakly micro–transitively on X, there exists an open subset M0 of X such that x0 ∈ M0 ⊂ Cl(U0 · x0 ). We prove that M0 ⊂ U · x0 . Let y0 ∈ M0 ; we must find g ∈ U such that g · x0 = y0 . At this point we follow a technique introduced by Homma [7]. What we do is to move from x0 to y0 and from y0 to x0 , planning ahead each movement so that the next movement is possible and is close to eG . Copyright © 2005 Taylor & Francis Group, LLC


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First, let V0 = U0 . Invoke the weak micro–transitivity action of G on X to obtain an open subset N0 such that y0 ∈ N0 ⊂ Cl(V0 ·y0 ). We construct eight sequences, namely: two two two G and two

∞ sequences, {gn }∞ n=1 and {hn }n=1 , of elements of G, ∞ sequences, {xn }∞ n=0 and {yn }n=0 , of elements of X, ∞ sequences, {Un }∞ n=0 and {Vn }n=0 , of neighborhoods of eG in ∞ sequences, {Mn }∞ n=0 and {Nn }n=0 , of open subsets of X.

These eight sequences are constructed to satisfy the following thirteen properties: (1n ) gn ∈ Un−1 . (2n ) hn ∈ Vn−1 . (3n ) xn = gn · xn−1 . (4n ) yn = hn · yn−1 . (5n ) xn ∈ Nn−1 . (6n ) yn ∈ Mn . (7n ) Un · gn · . . . · g1 ⊂ U0 . (8n ) Vn · hn · . . . · h1 ⊂ V0 . 1 1 (9n ) diam(Un ·gn ·. . .·g1 ) < n . (10n ) diam(Vn ·hn ·. . .·h1 ) < n . 2 2 (11n ) xn ∈ Mn ⊂ Cl(Un · xn ). (12n ) yn ∈ Nn ⊂ Cl(Vn · yn ). 1 (13n ) diam(Mn ) < . n The construction of these eight sequences is done by induction. We already have chosen U0 , V0 , x0 , y0 , M0 and N0 . Let n ≥ 1 and inductively assume that for k ∈ {1, . . . , n − 1}, we have chosen gk , hk , xk , yk , Uk , Vk , Mk and Nk in such a way that (1k ), . . . , (13k ) are satisfied. Note that from (6n−1 ), (12n−1 ) and (11n−1 ), we have that yn−1 ∈ Mn−1 ∩ Nn−1 ⊂ Cl(Un−1 · xn−1 ). Hence, there exists gn ∈ Un−1 such that gn · xn−1 ∈ Nn−1 ; so (1n ) holds. Let xn = gn · xn−1 . Then (3n ) and (5n ) also hold. From (7n−1 ) and (1n ) follows that gn · gn−1 · . . . · g1 ∈ U0 . Thus, we may choose Un ∈ N (eG ) in such a way that (7n ) and (9n ) are satisfied. Since G acts weakly micro–transitively on X, there exists an open subset Mn of X such that satisfies (11n ) and (13n ). So far, we have chosen gn , xn , Un and Mn in such a way that (1n ), (3n ), (5n ), (7n ), (9n ), (11n ) and (13n ) hold. From (5n ), (11n ) and (12n−1 ), we conclude that xn ∈ Nn−1 ∩ Mn ⊂ Cl(Vn−1 · yn−1 ). Copyright © 2005 Taylor & Francis Group, LLC

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Hence, there exists hn ∈ Vn−1 such that hn · yn−1 ∈ Mn , so (2n ) holds. Let yn = hn · yn−1 . Then (4n ) and (6n ) are also satisfied. From (8n−1 ) and (2n ), we obtain that hn · hn−1 · . . . · h1 ∈ V0 . Thus, we may choose Vn ∈ N (eG ) such that (8n ) and (10n ) hold. Since G acts weakly micro–transitively on X, we may find an open subset Nn of X satisfying (12n ). So, we have completed the inductive construction of the eight sequences. ˜ n = hn ·. . .·h1 . From For each n ≥ 1, let gñ = gn ·. . .·g1 and let h ˜ ∞ (1n ), (2n ), (9n ) and (10n ), we conclude that {˜ gn }∞ n=1 and {hn }n=1 are Cauchy sequences with respect to the metric ρG of G. Since ρG ˜ ∞ is a complete metric, it follows that {˜ gn }∞ n=1 and {hn }n=1 converge to elements g and h of G, respectively. It follows from (7n ) and ˜ ∞ (8n ) that {˜ gn }∞ n=1 ⊂ U0 and {hn }n=1 ⊂ V0 . Therefore, g ∈ Cl(U0 ) and h ∈ Cl(V0 ) = Cl(U0 ). By the election of U0 , we have that h−1 · g ∈ U . It follows from (3n ) and (4n ) that for each n ≥ 1, xn = gñ · x0 ˜ n · y0 . Hence, the sequence {xn }∞ converges to g · x0 , and yn = h n=1 and the sequence {yn }∞ converges to h · y . Since for each n ∈ IN, 0 n=1 it follows from (6n ) and (11n ) that both xn and yn belong to Mn , 1 then, from (13n ), we have that ρX (xn , yn ) < . Thus, g · x0 = h · y0 . n Therefore, h−1 · g ∈ U and (h−1 · g) · x0 = y0 . Q.E.D. 4.2.15. Theorem. Let G be a complete metric group acting transitively on a metric space X. If G is separable, then the following are equivalent: (a) G acts micro–transitively on X. (b) X is topologically complete. (c) X is of the second category. Proof. First, suppose G acts micro–transitively on X. To see X has a complete metric, let x ∈ X and consider the map γx : G → →X (Definition 4.2.8). Since G acts transitively on X, γx is surjective. Since G acts micro–transitively on X, by Lemma 4.2.9, γx is an open map. Hence, by Theorem 1.5.13, X has a complete metric. By the Baire Category Theorem (Theorem 1.5.12), if X has a complete metric, then X is of the second category. Copyright © 2005 Taylor & Francis Group, LLC


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Finally, suppose X is of the second category. Since G is separable, by Lemma 4.2.12, X is G–countably covered. Hence, since X is of the second category and is G–countably covered, by Lemma 4.2.13, G acts weakly micro–transitively on X. Therefore, by Lemma 4.2.14, G acts micro–transitively on X. Q.E.D. Now, we present some definitions needed for Effros’s Theorem. 4.2.16. Definition. Let G be a complete metric group acting on a metric space X. Then G · x is called the orbit of x under the action of G on X. Note that distinct orbits are disjoint. The set {G · x | x ∈ X} of orbits is called the orbit space determined by the action of G on X, and it is denoted by X/G. We give X/G the quotient topology. The function q : X → → X/G given by q(x) = G · x is the quotient map.

4.2.17. Lemma. If G is a complete metric group acting on a metric space X, then quotient map q : X → → X/G is open. Proof. It is enough to observe that if U is an open subset of X, then −1 q (q(U )) = G · U . Since G · U = {g · U | g ∈ G} = {ηg (U ) | g ∈ G} and each ηg is a homeomorphism (Remark 4.2.3), q −1 (q(U )) is open in X. Q.E.D. 4.2.18. Definition. Let G be a complete metric group acting on a metric space X. If x ∈ X, then Gx = {g ∈ G | g · x = x} is a subgroup of G called the stabilizer subgroup of x.


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4.2.19. Remark. Note that Gx acts on G by multiplication on the right. The orbit space of this action, G/Gx , is the set G/Gx = {g · Gx | g ∈ G} and consists of all the left cosets of Gx . We give G/Gx the quotient topology. The function qx : G → G/Gx given by qx (g) = g · Gx is the quotient map.

4.2.20. Lemma. Let G be a complete metric group acting on a metric space X. If x ∈ X, then quotient map qx : G → → G/Gx is open. Proof. It is enough to observe that if U is an open subset of G, then qx−1 (qx (U )) = U · Gx . By Proposition 4.1.17 (1), U · Gx is open in G. Q.E.D.

4.2.21. Definition. Let G be a complete metric group acting on a metric space X. If x ∈ X, define the function ψx : G/Gx → G · x by ψx (g · Gx ) = g · x. In the following Lemma we show that ψx is a well defined function which is a continuous bijection.

4.2.22. Lemma. Let G be a complete metric group acting on a metric space X. If x ∈ X, then the function ψx : G/Gx → G · x of Definition 4.2.21 is a well defined continuous bijection. Moreover, the following diagram qx

G

γx

G/Gx −→(G · x) ψx

is commutative. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Let x ∈ X. For g , g ∈ G, g · Gx = g · Gx if and only if g · x = g · x. Hence, ψx is well defined and one–to–one. To see ψx is surjective, let g · x ∈ G · x. Then ψx (g · Gx ) = g · x. Therefore, ψx is a bijection. Next, let g ∈ G. Then ψx ◦ qx (g) = ψx (qx (g)) = ψx (g · Gx ) = g · x = γx (g). Therefore, the diagram is commutative. Finally, since ψx ◦ qx = γx , qx is open (Lemma 4.2.20) and γx is continuous, we have that ψx is continuous. Q.E.D. 4.2.23. Lemma. Let G be a complete metric group acting on a metric space X. If x ∈ X, then G acts micro–transitively on the orbit G·x if and only if the map ψx : G/Gx → G·x of Definition 4.2.21 is a homeomorphism. Proof. By Lemma 4.2.22, ψx is continuous. Since ψx ◦ qx = γx and qx is continuous and open (Lemma 4.2.20), ψx is open if and only if γx is open. By Lemma 4.2.9, γx is open if and only if G acts micro–transitively on the orbit G · x. Q.E.D. 4.2.24. Lemma. If G is a complete metric group acting on a separable complete metric space X, then each orbit is a Gδ subset of X if and only if X/G is a T0 space. Proof. First, assume that each orbit is a Gδ subset of X. We show that X/G is a T0 space. Let x, y ∈ X. Suppose that q(x) ∈ ClX/G ({q(y)}) and q(y) ∈ ClX/G ({q(x)}). We see that q(x) = q(y). We assert that G · x ⊂ ClX (G · y). Indeed, suppose there exists g ∈ G such that g · x ∈ ClX (G · y). Then there exists an open subset U of X such that g · x ∈ U and U ∩ G · y = ∅. Since the map q : X → X/G is open (Lemma 4.2.17), q(U ) is an open subset of X/G such that q(g · x) = q(x) ∈ q(U ). Note that q(G · y) = {q(y)} and q(y) ∈ q(U ). A contradiction to the fact that q(x) ∈ ClX/G ({q(y)}). Therefore, G · x ⊂ ClX (G · y). Similarly, G · y ⊂ ClX (G · x). We conclude that G · x ∪ G · y ⊂ ClX (G · x) ∩ ClX (G · y). Since G · x is dense in ClX (G · x) and G · y is dense in ClX (G · y), G·x and G·y both are dense in ClX (G·x)∩ClX (G·y). By hypothesis, Copyright © 2005 Taylor & Francis Group, LLC

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G · x and G · y are both Gδ subsets of ClX (G · x) ∩ ClX (G · y). Since ClX (G · x) ∩ ClX (G · y) is a closed subset of X and X is a complete metric space, ClX (G · x) ∩ ClX (G · y) has a complete metric (Lemma 1.5.3). By Lemma 1.5.9, G · x ∩ G · y is dense in ClX (G · x) ∩ ClX (G · y). In particular, G · x ∩ G · y = ∅. Hence, G · x = G · y and q(x) = q(y). Therefore, X/G is a T0 space. Next, suppose X/G is a T0 space. Since X is a separable metric ∞ space, X has a countable basis {Un }∞ n=1 . We assert that {q(Un )}n=1 is a countable basis for X/G. Clearly, this family is countable. Since the map q : X → X/G is open (Lemma 4.2.17), q(Un ) is an open subset of X/G for each n ∈ IN. Furthermore, if x ∈ X and V is an open subset of X/G such that q(x) ∈ V , then there exists n ∈ IN such that q(x) ∈ q(Un ) ⊂ V (by the continuity of q and the fact ∞ that {Un }∞ n=1 is a basis for X). Thus, {q(Un )}n=1 is a countable basis for X/G. Now, take x ∈ X. For each n ∈ IN, let q(Un ) if q(x) ∈ q(Un ), Vn = X/G \ q(Un ) if q(x) ∈ q(Un ). Since X/G is a T0 space,

∞

Vn = {q(x)}. Hence,

n=1

∞

q −1 (Vn ) =

n=1

q −1 (q(x)) = G · x. Since each Vn is either open or closed in X/G, q −1 (Vn ) is either open or closed in X. In either case, q −1 (Vn ) is a Gδ subset of X (see Lemma 1.5.7 when q −1 (Vn ) is closed). Therefore, G · x is a Gδ subset of X. Q.E.D. We are ready to prove Effros’s Theorem [4]. 4.2.25. Theorem. If G is a separable complete metric group acting on a separable complete metric space X, then the following are equivalent: (a) For each x ∈ X, the map ψx : G/Gx → G · x of Definition 4.2.21 is a homeomorphism. (b) G acts micro–transitively on each orbit.



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(c) Each orbit is of the second category (in itself). (d) Each orbit is a Gδ subset of X. (e) X/G is a T0 space. Proof. By Lemma 4.2.23, (a) and (b) are equivalent. By Theorem 4.2.15, (b) and (c) are equivalent. By Theorem 1.5.8, (d) is equivalent to the fact that each orbit is topologically complete. This latter statement is equivalent to (b) by Theorem 4.2.15. Hence, (d) is equivalent to (b). Finally, by Lemma 4.2.24, (d) is equivalent to (e). Q.E.D. 4.2.26. Definition. Let X and Y be continua. Define C(X, Y ) = {f : X → Y | f is a map}. We topologize C(X, Y ) with the sup metric, ρ, given by ρ((f, g)) = sup{d(f (x), g(x)) | x ∈ X}, for every g, f ∈ C(X, Y ). 4.2.27. Definition. If X is a continuum, then H(X) denotes the group of homeomorphisms of X. H(X) acts on X as follows: If h ∈ H(X) and x ∈ X, then θ((h, x)) = h(x). 4.2.28. Theorem. If X is a continuum, then C(X, X) is a separable complete metric space and H(X) is a Gδ subset of C(X, X). Proof. By Theorem 1 (p. 244) of [8], C(X, X) is a separable space. By Theorem 3 (p. 90) of [9], C(X, X) is a complete metric space. By Theorem 1 (p. 91) of [9], H(X) is a Gδ subset of C(X, X). Q.E.D. 4.2.29. Corollary. If X is a continuum, then H(X) has a separable complete metric, ρ . Copyright © 2005 Taylor & Francis Group, LLC

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Proof. By Theorem 4.2.28, H(X) is a Gδ subset of C(X, X). Hence, by Theorem 1.5.8, H(X) has a complete metric. In fact, it is known that ρ is given by ρ (f, g) = ρ(f, g) + ρ(f −1 , g −1 ) for every f, g ∈ H(X). Q.E.D. 4.2.30. Definition. A metric space X has the Property of Effros provided that for each ε > 0, there exists δ > 0 such that if x, y ∈ X and d(x, y) < δ, then there exists h ∈ H(X), such that h(x) = y and d(z, h(z)) < ε for every z ∈ X. The number δ is called an Effros number for the given ε. A homeomorphism h ∈ H(X) satisfying that d(z, h(z)) < ε for each z ∈ X is called an ε–homeomorphism. The following Theorem is known as Effros’s Theorem in the theory of homogeneous continua. A slightly different version of this result was first shown by C. L. Hagopian [5]. 4.2.31. Theorem. If X is a homogeneous continuum, with metric d, then X has the property of Effros. Proof. By Corollary 4.2.29, H(X) is a separable complete metric space. Since X is a homogeneous continuum, H(X) acts transitively on X. Let x ∈ X, by Theorem 4.2.25, the map → H(X) · x ψx : H(X)/H(X)x → is a homeomorphism. By Lemma 4.2.23, H(X) acts micro–transitively on X. Hence, by Lemma 4.2.9, the map γx is open. Let ε > 0 be given and let U = V ρε (1X ), where 1X is the identity 2 map of X. Then for each h ∈ U , h is an 2ε –homeomorphism of X. Since γx is open, γx (U ) is an open subset of X such that x ∈ γx (U ). Let δx > 0 such that Vδdx (x) ⊂ γx (U ). Thus, if y ∈ Vδdx (x), then there exists h ∈ U such that γx (h) = y, i.e., h is an 2ε –homeomorphism such that h(x) = y. Note that {Vδdx (x) | x ∈ X} is an open cover of X. Let δ be a Lebesgue number for this cover (Theorem 1.6.6). Let x, y ∈ X such that d(x, y) < δ. Then there exists z ∈ X such that x, y ∈ Vδdz (z). By the previous paragraph, there exist h1 , h2 ∈ γz (U ) such that Copyright © 2005 Taylor & Francis Group, LLC


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→ X h1 (z) = x and h2 (z) = y. Let h = h2 ◦ h−1 1 . Then h : X → −1 is a homeomorphism such that h(x) = y (h(x) = h2 ◦ h1 (x) = −1 h2 h1 (x) = h2 (z) = y). Let w ∈ X. Since d(w, h(w) = d w, h2 h−1 1 (w) −1 −1 ≤ d w, h−1 (w) + d h (w), h (w) h 2 1 −1 1 −11 = d h1 h−1 (w) , h (w) + d h1 (w), h2 h−1 1 1 1 (w) ε ε < + = ε, 2 2 h is an ε–homeomorphism. Q.E.D. As our first application of Effros’s Theorem, we show that Jones’s set function T is idempotent on homogeneous continua. 4.2.32. Theorem. If X is a homogeneous continuum, with metric d, then for each closed subset A of X, T 2 (A) = T (A). Proof. Let A be a closed subset of X. Note that, by Remark 3.1.5, T (A) ⊂ T 2 (A). Next, let x ∈ X \ T (A). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ A. Let ε > 0 such that ε < d(W, A) and Vεd (x) ⊂ Int(W ). Let δ > 0 be an Effros number for this ε. We assume that δ < ε. Since W is compact, there exist m w1 , . . . , wm ∈ W such that W ⊂ Vδd (wj ). j=1

Let j ∈ {1, . . . , m}. Then for each y ∈ Vδd (wj ), there exists an ε–homeomorphism hy : X → → X such that hy (wj ) = y. When y = wj , we let hy = 1X , the identity map of X. Let ⎛ ⎞ hy (W )⎠ . Mj = Cl ⎝ y∈Vδd (wj )

Then Mj is a subcontinuum of X such that wj ∈ Vδd (wj ) ⊂ Mj ⊂ m X \ A. Let M = Mj . Then M is a subcontinuum of X and W ⊂ j=1


222 m

REFERENCES

Vδd (wj ) ⊂ Int(M ) ⊂ M ⊂ X \ A. Since Int(M ) ⊂ X \ T (A),

j=1

x ∈ Int(W ) ⊂ W ⊂ X \ T (A). Hence, x ∈ X \ T 2 (A). Therefore, T is idempotent. Q.E.D. We finish this chapter showing that any connected metric space with the property of Effros is homogeneous. 4.2.33. Theorem. If X is a connected metric space, with metric d, satisfying the property of Effros, then X is homogeneous. Proof. Let x, y ∈ X, and let ε > 0. Since X has the property of Effros, there exists an Effros number δ > 0 for this ε. Since X is connected, there exist finitely many points z1 = x, z2 , . . . , zn−1 , zn = y in X such that d(zj−1 , zj ) < δ for each j ∈ {2, . . . , n}. Hence, there exists a homeomorphism hj : X → → X such that hj (zj−1 ) = zj , j ∈ {2, . . . , n}. Then h = hn ◦ . . . ◦ h2 is a homeomorphism of X onto itself such that h(x) = y. Therefore, X is homogeneous. Q.E.D.

REFERENCES

[1] F. D. Ancel, An Alternative Proof and Applications of a Theorem of E. G. Effros, Michigan Math. J., 34 (1987), 39–55. [2] G. E. Bredon, Introduction to Compact Transformation Groups, Monographs and Textbooks in Pure and Applied Mathematics, Vol. 46, Academic Press, Inc., New York, 1972. [3] G. E. Bredon, Topology and Geometry, Graduate Texts in Mathematics, Vol. 139, Springer–Verlag, New York, Inc., 1993. [4] E. G. Effros, Transformation Groups and C ∗ –Algebras, Ann. of Math., (2) 81 (1965), 38–55. Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

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[5] C. L. Hagopian, Homogeneous Plane Continua, Houston J. Math., 1 (1975), 35–41. [6] C. Hernández, O. J. Rendón, M. Tkaˇcenko and L. M. Villegas, Grupos Topol´ ogicos, Libros de Texto, Manuales de Prácticas y Antolog´ıas, Universidad Autónoma Metropolitana, Unidad Iztapalpa, 1997. (Spanish) [7] T. Homma, On the Embedding of Polyhedra in Manifolds, Yokohama Math. J., 10 (1962), 5–10. [8] K. Kuratowski, Topology, Vol. I, Academic Press, New York, N. Y., 1966. [9] K. Kuratowski, Topology, Vol. II, Academic Press, New York, N. Y., 1968. [10] G. McCarty, Topology: An Introduction with Applications to Topological Groups, Dover Publications, Inc., New York, 1988.


Chapter 5 DECOMPOSITION THEOREMS

We present a proof of Jones’s Aposyndetic Decomposition Theorem and Rogers’s Terminal Decomposition Theorem. These theorems are proven using Jones’s set function T (Chapter 3) and Effros’s Theorem (Chapter 4). We also give a construction of the Case continuum and present a sketch of the construction of the Minc– Rogers continua. Finally, we study covering spaces of any solenoid, the Menger curve, the Case continuum and one of the Minc–Rogers examples.

5.1

Jones’s Theorem

We give a proof of Jones’s Aposyndetic Decomposition Theorem (Theorem 5.1.19). We begin this section proving that the set function T commutes with homeomorphisms.


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5.1.1. Lemma. Let X be a compactum. If h : X → → X is a homeomorphism, then h(T (A)) = T (h(A)) for each closed subset A of X. Proof. Let A be a closed subset of X. Let x ∈ X \ h(T (A)). Then h−1 (x) ∈ X \ T (A). Hence, there exists a subcontinuum W of X such that h−1 (x) ∈ Int(W ) ⊂ W ⊂ X \ A. This implies that x ∈ Int(h(W )) ⊂ h(W ) ⊂ X \ h(A). Thus, x ∈ X \ T (h(A)). Now, let x ∈ X \ T (h(A)). Then there exists a subcontinuum K of X such that x ∈ Int(K) ⊂ K ⊂ X \ h(A). This implies that h−1 (x) ∈ Int(h−1 (K)) ⊂ h−1 (K) ⊂ X \ A. Thus, h−1 (x) ∈ X \ T (A). Hence, x ∈ X \ h(T (A)). Therefore, h(T (A)) = T (h(A)). Q.E.D. The following Theorem says that, for a homogeneous continuum, the images of all singletons under T form a decomposition of the continuum. 5.1.2. Theorem. Let X be a homogeneous continuum with metric d. If G = {T ({x}) | x ∈ X}, then G is a decomposition of X. Proof. Let x be a point of X. We show that if y ∈ T ({x}), then T ({x}) = T ({y}). Let y ∈ T ({x}). By Proposition 3.1.7 and the fact that T is idempotent (Theorem 4.2.32), T ({y}) ⊂ T 2 ({x}) = T ({x}). To show T ({x}) ⊂ T ({y}) we use an argument of Bellamy and Lum in Lemma 5 of [6]. We find a point x0 of X such that T ({x0 }) ⊂ T ({z}) for each z ∈ T ({x0 }). Then, by Lemma 5.1.1, the homogeneity of X ensures that the same is true for every point of X. We give a partial order to G as follows. T ({z}) < T ({w}) if T ({w}) ⊂ T ({z}). Let K = {T ({xλ })}λ∈Λ be a (set theoretic) chain of elements of G. We show that K has an upper bound. Since each (Theorem 3.1.21), and K is a chain, T ({xλ }) is a continuum T ({xλ }) = ∅. Let x ∈ T ({xλ }). Then, by the previous ∈Λ

λ∈Λ

argument, for each λ ∈ Λ, T ({x }) ⊂ T ({xλ }). Hence, T ({x }) ⊂ T ({xλ }). Thus, K has an upper bound. By Kuratowski–Zorn λ∈Λ


5.1. JONES’S THEOREM

227

Lemma, G has a maximal element, i.e., there exists x0 ∈ X such that for each z ∈ T ({x0 }), T ({x0 }) ⊂ T ({z}). Since X is homogeneous, there is a homeomorphism h : X → →X such that h(x0 ) = x. Since y ∈ T ({x}), by Lemma 5.1.1, h−1 (y) ∈ T ({x0 }). Hence, by the previous paragraph, we have that T ({x0 }) ⊂ T ({h−1 (y)}). Thus, T ({x}) = h(T ({x0 })) ⊂ h(T ({h−1 (y)})) = T ({y}) (Lemma 5.1.1). Consequently, T ({x}) ⊂ T ({y}). Therefore, G is a decomposition of X. Q.E.D. 5.1.3. Definition. Let G be a decomposition of the compactum X. Let H be a family of homeomorphisms of X. We say H respects G provided that for each pair G1 , G2 ∈ G and each h ∈ H, either h(G1 ) = G2 or h(G1 ) ∩ G2 = ∅.

5.1.4. Theorem. Let X be a homogeneous continuum, with metric d, and let G be a decomposition of X such that the elements of G are continua. If the homeomorphism group, H(X), of X respects G, then the following hold: (1) G is a continuous decomposition of X. (2) The elements of G are homogeneous mutually homeomorphic continua. (3) The quotient space X/G is a homogeneous continuum. Proof. First, we show that G is continuous. To this end, we prove first that G is upper semicontinuous. Let G ∈ G, and let U be an open subset of X such that G ⊂ U . Let ε = d(G, X \U ). Since G and X \U are disjoint compacta, ε > 0. Let δ > 0 be an Effros number for this ε (Theorem 4.2.31); without loss of generality, we assume that δ < ε. Let V = Vδd (G). Then V is an open subset of X contained in U . Let G ∈ G such that G ∩ V = ∅. Let y ∈ G ∩ V , and let x ∈ G such that d(x, y) < δ. Since δ is an Effros number, there exists an ε–homeomorphism h : X → →X such that h(x) = y. Since H(X) respects G and h(G) ∩ G = ∅, h(G) = G . Hence, G ⊂ Vεd (G) (h is an ε–homeomorphism), i.e., G ⊂ U . Thus, G is upper semicontinuous. Copyright © 2005 Taylor & Francis Group, LLC

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Next, we show that G is lower semicontinuous. Let G ∈ G, let p, q ∈ G and let U be an open subset of X such that p ∈ U . Let ε > 0 such that Vεd (p) ⊂ U . Let δ be an Effros number for this ε, and let V = Vδd (q). Take G ∈ G such that G ∩ V = ∅, and take z ∈ G ∩ V . Hence, d(q, z) < δ. Since δ is an Effros number, there exists an ε–homeomorphism k : X → → X such that k(q) = z. In particular, d(p, k(p)) < ε. Thus, k(p) ∈ U . Since H(X) respects G, and k(q) ∈ G , k(G) = G . Hence, G ∩ U = ∅. Thus, G is lower semicontinuous. Therefore, G is continuous. Observe that since H(X) respects G, all the elements of G are homeomorphic. Let G ∈ G, and let x, y ∈ G. Since X is homogeneous, there exists a homeomorphism : X → → X such that (x) = y. Note that, since H(X) respects G, (G) = G. Hence, |G : G → →G is a homeomorphism sending x to y. Therefore, G is homogeneous. Finally, we prove that the quotient space X/G is a homogeneous continuum. By Theorem 1.7.3, X/G is a continuum. Let q: X → → X/G be the quotient map. Let χ1 , χ2 ∈ X/G be two points. Let x1 , x2 ∈ X be such that q(x1 ) = χ1 and q(x2 ) = χ2 . Since X is homogeneous, there exists a homeomorphism h : X → → X such that h(x1 ) = x2 . Since H(X) respects G, x1 ∈ q −1 (χ1 ), x2 ∈ q −1 (χ2 ), and h(x1 ) = x2 , we have that h(q −1 (χ1 )) = q −1 (χ2 ). Define f : X/G → → X/G by f (χ) = q ◦ h q −1 (χ) . Then f is well defined (because H(X) respects G) and f (χ1 ) = χ2 . Since G is a continuous decomposition, q is an open map (Theorem 1.2.23). Hence, since q is open and h is continuous, f is continuous. Note that f −1 : X/G → → X/G is given by f −1 (χ) = −1 −1 q ◦ h (q (χ)) and it is continuous also. Thus, f is a homeomorphism. Therefore, X/G is homogeneous. Q.E.D. 5.1.5. Definition. A surjective map f : X → → Y between continua is completely regular if for each ε > 0 and each point y ∈ Y , there exists an open set V in Y containing y such that if y ∈ V , then there exists a homeomorphism h : f −1 (y) → → f −1 (y ) such that for each x ∈ f −1 (y), d(x, h(x)) < ε. Copyright © 2005 Taylor & Francis Group, LLC


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5.1.6. Remark. Note that the fibres of completely regular maps are all homeomorphic.

5.1.7. Proposition. Let X and Y be continua. If g : X → → Y is a completely regular map, then g is open. Proof. Let U be an open subset of X. Let y ∈ g(U ). Then there exists x ∈ U such that g(x) = y. Let ε > 0 such that Vεd (x) ⊂ U . Since g is completely regular, there is an open subset V of Y such that y ∈ V and if y ∈ V , then there exists a homeomorphism h : g −1 (y) → → g −1 (y ) such that d(z, h(z)) < ε for all z ∈ g −1 (y). In particular, d(x, h(x)) < ε. Thus, h(x) ∈ U . Thus, V ⊂ g(U ). Since y was an arbitrary point of g(U ), g(U ) is open. Therefore, g is an open map. Q.E.D. 5.1.8. Theorem. Let X be a homogeneous continuum, with metric d, and let G be a decomposition of X whose elements are proper nondegenerate subcontinua of X. If the homeomorphism group, H(X), of X respects G, then the elements of G are nowhere dense, and the quotient map q : X → → X/G is completely regular. Proof. Let G ∈ G and suppose Int(G) = ∅. Let g ∈ Int(G), and let ε > 0 such that Vεd (g ) ⊂ G. Let δ > 0 be an Effros number for this ε (Theorem 4.2.31). We assume that δ < ε. Let x ∈ X \ G such that d(x, G) < δ. Since G is compact, there exists g ∈ G such that d(x, g) = d(x, G). Now, since d(x, g) < δ, there exists an ε–homeomorphism h : X → → X such that h(g) = x. Hence, since h(g ) ∈ Vεd (g ) ⊂ G and H(X) respects G, h(G) = G. Thus, x ∈ h(G) \ G, a contradiction. Therefore, Int(G) = ∅. To show the quotient map q : X → → X/G is completely regular, let ε > 0, and let δ > 0 be an Effros number for this ε (Theorem 4.2.31). Since H(X) respects G, by Theorem 5.1.4 (1), G is a continuous decomposition. Thus, by (Theorem d 1.2.23), q is open. Let χ ∈ X/G, and let V = q Vδ (x) , where x ∈ q −1 (χ). Then V is an open subset of X/G containing χ. Let χ ∈ V , and let x ∈ q −1 (χ ) ∩ Vδd (x). Then since d(x, x ) < δ, there exists an ε–homeomorphism h : X → → X such that h(x) = x . Since Copyright © 2005 Taylor & Francis Group, LLC

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H(X) respects G and h(q −1 (χ)) ∩ q −1 (χ ) = ∅, h(q −1 (χ)) = q −1 (χ ). Thus, h|q−1 (χ) : q −1 (χ) → → q −1 (χ ) is a homeomorphism such that d(z, h(z)) < ε for each z ∈ q −1 (χ). Therefore, q is completely regular. Q.E.D. 5.1.9. Definition. Let X be a continuum. A subcontinuum Z of X is said to be terminal if each subcontinuum Y of X that intersects Z satisfies either Y ⊂ Z or Z ⊂ Y . A decomposition G of X such that the elements of G are continua is said to be terminal if each element of G is a terminal subcontinuum of X.

5.1.10. Remark. It is easy to see that in the topologist sine curve X (Example 2.4.5), {0} × [−1, 1] is a terminal subcontinuum of X.

5.1.11. Lemma. Let X and Y be continua. If f : X → → Y is a monotone surjective map and W is a terminal subcontinuum of X, then f (W ) is a terminal subcontinuum of Y . Proof. Let W be a terminal subcontinuum of X. Let K be a subcontinuum of Y such that K ∩ f (W ) = ∅. Since f is a monotone map, f −1 (K) is a subcontinuum of X (Lemma 2.1.12) such that f −1 (K) ∩ W = ∅. Hence, since W is a terminal subcontinuum of X, either W ⊂ f −1 (K) or f −1 (K) ⊂ W . This implies that either f (W ) ⊂ K or K ⊂ f (W ) (f is surjective). Therefore, f (W ) is a terminal subcontinuum of Y . Q.E.D. 5.1.12. Corollary. Let X be a continuum. If Z is a terminal subcontinuum of X and h : X → → X is a homeomorphism, then h(Z) is a terminal subcontinuum of X.

5.1.13. Lemma. Let X be a continuum. If X is aposyndetic, then X does not contain nondegenerate proper terminal subcontinua. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Suppose Y is a nondegenerate proper terminal subcontinuum of X. Let y ∈ Y . Let y ∈ Y \ {y}. Then since X is aposyndetic, there exists a subcontinuum W of X such that y ∈ Int(W ) ⊂ W ⊂ X \ {y }. Since Y is terminal and Y \ W = ∅, W ⊂ Y . Thus, y is an interior point of Y . Since y was an arbitrary point of Y , all the points of Y are interior points. Hence, Y is a nonempty open and closed proper subset of X. This contradicts the fact that X is connected. Therefore, X does not contain nondegenerate proper terminal subcontinua. Q.E.D. 5.1.14. Theorem. Let X be a continuum. If A is a terminal subcontinuum of X, if B is a subcontinuum of X disjoint from A, and if f : A → Y is a map from A into the absolute neighborhood retract Y , then there exists a map F : X → Y such that F |A = f and F |B is homotopic to a constant map. Proof. Since Y is an absolute neighborhood retract, there exist an open subset U of X containing A and a map g : U → Y such that g|A = f . Without loss of generality, we assume that U ∩ B = ∅. By Theorem 7.1 (p. 96) of [13], Y is locally contractible. Hence, there exists an open neighborhood V of a point a of A such that V ⊂ U and g|V is homotopic to a constant map. Note that A \ V and X \ U are two closed subsets of X \ V such that no connected subset of X \ V intersects both A \ V and X \ U . (If K is a connected subset of X \ V such that K ∩ (A \ V ) = ∅ and K ∩ (X \ U ) = ∅, then Cl(K) is a continuum in X \ V intersecting A and X \ A. Since A is terminal, A ⊂ Cl(K). This implies that V ∩ Cl(K) = ∅, a contradiction.) By Theorem 1.6.8, there exist two disjoint closed subsets X1 and X2 of X such that X \ V = X1 ∪ X2 , A \ V ⊂ X1 and X \ U ⊂ X2 . Note that X1 ∪ A and X2 are two disjoint closed subsets of X. Then, by Urysohn’s Lemma, there exists a map h : X → → [0, 1] such 1 −1 that h(X1 ∪ A) = {0} and h(X2 ) = {1}. Let M = h 0, 2 1 , 1 . Then X = M ∪ N , A ⊂ M , X \ U ⊂ N and let N = h−1 2 and M ∩ N ⊂ V \ A. Since g|M ∩N is homotopic to a constant map (because M ∩ N ⊂ V and g|V is homotopic to a constant map), Copyright © 2005 Taylor & Francis Group, LLC

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by (15.A.1) of [8], there exists a map k : N → Y such that k is homotopic to a constant map and k|M ∩N = g|M ∩N . Let F : X → Y be given by g(x) if x ∈ M , F (x) = k(x) if x ∈ N . Then F is well defined and continuous. Note that F |A = g|A = f and F |B = k|B . Since k|B is homotopic to a constant map (because B ⊂ X \ U ⊂ N and k is homotopic to a constant map), F is the desired extension of f . Q.E.D. 5.1.15. Definition. A continuum X is cell–like if each map of X into a compact absolute neighborhood retract is homotopic to a constant map.

5.1.16. Definition. Let X and Y be a continua. We say that a surjective map f : X → → Y is cell–like if for each y ∈ Y , f −1 (y) is a cell–like continuum.

5.1.17. Theorem. Let X and Z be nondegenerate continua. Suppose that g : X → → Z is a monotone and completely regular map. If z1 is a point of Z such that g −1 (z1 ) is terminal subcontinuum of X, then g is a cell–like map. Proof. Let Y be a compact absolute neighborhood retract, and let f : g −1 (z1 ) → Y be a map. Let z2 ∈ Z \ {z1 }, and let F : X → Y be an extension of f such that F |g−1 (z2 ) is homotopic to a constant map (Theorem 5.1.14). Since Y is a compact absolute neighborhood retract, by Theorem 1.1 (p. 111) of [13], there exists ε > 0 such that for any metric space W and for any two maps k, k : W → Y such that d(k(w), k (w)) < ε for every w ∈ W , k and k are homotopic. Copyright © 2005 Taylor & Francis Group, LLC


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Since F is uniformly continuous, for this ε, there exists δ > 0 such that if d(x, x ) < δ, then d(F (x), F (x )) < ε. Let Z = {z ∈ Z | F |g−1 (z) is homotopic to a constant map}. We show that Z is open in Z. Let z ∈ Z . Since g is completely regular, there exists an open subset V of Z such that z ∈ V and if z ∈ V , then there is a homeomorphism h : g −1 (z) → → g −1 (z ) with −1 d(x, h(x)) < δ for every x ∈ g (z). Let z ∈ V , and let h be the homeomorphism guaranteed by the complete regularity of g. Hence, d(F |g−1 (z ) (x), (F |g−1 (z) ) ◦ h−1 (x)) < ε for every x ∈ g −1 (z ). Thus, F |g−1 (z ) is homotopic to (F |g−1 (z) )◦h−1 . Since z ∈ Z , (F |g−1 (z) )◦h−1 is homotopic to a constant map. Thus, F |g−1 (z ) is homotopic to a constant map. Therefore, z ∈ Z . A similar argument shows that Z is closed in Z. Since Z is a nonempty open and closed subset of Z and Z is connected, Z = Z. This implies that F |g−1 (z1 ) = f is homotopic to a constant map. Thus, each map from g −1 (z1 ) into a compact absolute neighborhood retract is homotopic to a constant map. Hence, g −1 (z1 ) is cell–like. Since fibres of completely regular maps defined on continua are homeomorphic, we have that g is a cell–like map. Q.E.D. A proof of the following result may be found in 2.1 of [15]. The Theorem was originally proved by E. Dyer.

5.1.18. Theorem. Let X and Y be nondegenerate continua. If f: X → → Y is a surjective, monotone and open map, then there exists a dense Gδ subset W of Y having the following property: for each y ∈ W , for each subcontinuum B of f −1 (y), for each x ∈ Intf −1 (y) (B) and for each neighborhood U of B in X, there exist a subcontinuum Z of X containing B and a neighborhood V of y in Y such that x ∈ IntX (Z), (f |Z )−1 (V ) ⊂ U and f |Z : Z → Y is a monotone surjective map.

Now, we are ready to prove Jones’s Aposyndetic Decomposition Theorem.


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5.1.19. Theorem. Let X be a decomposable homogeneous continuum, with metric d, which is not aposyndetic. If G = {T ({x}) | x ∈ X}, then the following hold: (1) G is a continuous, monotone and terminal decomposition of X. (2) The elements of G are indecomposable, cell–like, homogeneous and mutually homeomorphic continua of the same dimension as X. (3) The quotient map q : X → → X/G is completely regular. (4) The quotient space X/G is a one–dimensional aposyndetic homogeneous continuum, which does not contain nondegenerate proper terminal subcontinua. Proof. Since X is decomposable, there exist two points of X such that X is aposyndetic at one of them with respect to the other. Hence, the elements of G are nondegenerate proper subcontinua (Theorem 3.1.21) of X. By Theorem 5.1.2, G is a decomposition of X. By Lemma 5.1.1, the homeomorphism group of X respects G. Hence, by Theorem 5.1.4, G is a continuous decomposition, the elements of G are mutually homeomorphic homogeneous continua and the quotient space X/G is a homogeneous continuum. A proof of the fact that the elements of G have the same dimension as X may be found in Corollary 9 of [32]. Now, we show that all the elements of G are terminal subcontinua. To this end, suppose there exists a point x ∈ X such that T ({x}) is not terminal. Hence, there exists a subcontinuum Y of X such that Y ∩ T ({x}) = ∅, Y \ T ({x}) = ∅ and T ({x}) \ Y = ∅. Without loss of generality, we assume that x ∈ T ({x}) \ Y . Let p ∈ Y \ T ({x}), and let y ∈ Y ∩ T ({x}). Since p ∈ Y \ T ({x}), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ {x}. Let K = Y ∪ W . Then K is a subcontinuum of X, p ∈ Int(K) and K ⊂ X \ {x}. Let ε > 0 d such that Vεd (x) ⊂ K, V2ε (K) ⊂ X \ {x} and ε < d(x, y). Let δ > 0 be an Effros number for this ε (Theorem 4.2.31). Hence, for each y ∈ Vδd (y), there exists an ε–homeomorphism hy : X → → X such Copyright © 2005 Taylor & Francis Group, LLC


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that hy (y) = y (for y = y, we take hy = 1X ). Then ⎞ ⎛ hy (K)⎠ M = Cl ⎝ y ∈Vδ (y)

is a subcontinuum of X such that Vδd (y) ⊂ M ⊂ X \ {x}. This contradicts the choice of y. Therefore, all the elements of G are terminal. Note that, this implies that all the elements of G are cell– like (Theorem 5.1.17) since the quotient map is completely regular (Theorem 5.1.8). Next, we prove that all the elements of G are indecomposable. Suppose this is not true. Then there exists a point x ∈ X such that T ({x }) is decomposable. Hence, since X is homogeneous, T ({x}) is decomposable for each x ∈ X (Lemma 5.1.1). Let Υ be the Gδ dense subset of X/G guaranteed by Theorem 5.1.18. Let x0 ∈ X such that q(x0 ) ∈ Υ . Let B be a subcontinuum of T ({x0 }) such that IntT ({x0 }) (B) = ∅. Let x ∈ IntT ({x0 }) (B), and let n ∈ IN such that T ({x0 }) \ V d1 (x) = ∅. By Theorem 5.1.18, there exist a subn continuum Z of X and a neighborhood Ω of q(x0 ) in X/G such that x ∈ IntX (Z), B ⊂ Z, (q|Z )−1 (Ω) ⊂ V d1 (x) and q|Z : Z → → X/G n

is a monotone surjective map. Since T ({x0 }) = q −1 (q(x0 )) and T ({x0 }) \ V d1 (x) = ∅, we obtain a contradiction. Hence, all proper n subcontinua of T ({x0 }) have empty interior. Thus, T ({x0 }) is indecomposable (Corollary 1.7.21). Therefore, T ({x}) is an indecomposable continuum for each x ∈ X. To finish the proof, we show that X/G is aposyndetic. Let χ1 , χ2 ∈ X/G. We see that X/G is aposyndetic at χ1 with respect to χ2 . Let x1 ∈ q −1 (χ1 ) and let x2 ∈ q −1 (χ2 ). Note that X is aposyndetic at x1 with respect to x2 . Then there exists a subcontinuum W of X such that x1 ⊂ Int(W ) ⊂ W ⊂ X \ {x2 }. Since G is a terminal decomposition, T ({x1 }) ⊂ W and W ∩ T ({x2 }) = ∅ (T ({x}) is nowhere dense for every x ∈ X (Theorem 5.1.8)). Since G is a continuous decomposition, by Theorem 1.2.23, q is an open map. Then χ1 = q(x1 ) ∈ Int(q(W )) ⊂ q(W ) ⊂ X/G \ {χ2 }. Therefore, X/G is aposyndetic. The fact that X/G is one–dimensional may be found in Theorem 3 of [35]. Copyright © 2005 Taylor & Francis Group, LLC

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The fact that X/G does not contain nondegenerate proper terminal subcontinua follows from Lemma 5.1.13. Q.E.D. As a consequence of Jones’s Theorem we have the following three Corollaries: 5.1.20. Corollary. Let X be a decomposable homogeneous continuum. If x ∈ X, then T ({x}) is the maximal terminal proper subcontinuum of X containing x. Proof. If X is aposyndetic, then T ({x}) = {x} (Theorem 3.1.28). Since aposyndetic continua do not contain nondegenerate proper terminal subcontinua (Lemma 5.1.13), T ({x}) is the maximal terminal proper subcontinuum of X containing x. Suppose X is not aposyndetic, and let x ∈ X. Then, by Theorem 5.1.19 (1), T ({x}) is a terminal subcontinuum of X. Suppose K is a terminal proper subcontinuum of X such that T ({x}) K. By Lemma 5.1.11, q(K) is a terminal subcontinuum of X/G. Since X/G does not contain proper nondegenerate terminal subcontinua (Theorem 5.1.19 (4)), q(K) = {q(x)}. Hence, K = q −1 (q(x)) = T ({x}), a contradiction. Therefore, T ({x}) is the maximal terminal proper subcontinuum of X containing x. Q.E.D. 5.1.21. Corollary. If X is a hereditarily decomposable and homogeneous continuum, then X is aposyndetic.

5.1.22. Corollary. If X is an arcwise connected homogeneous continuum, then X is aposyndetic.

5.1.23. Theorem. Let X be a decomposable nonaposyndetic homogeneous continuum. If G = {TX ({x}) | x ∈ X}, then TX (Z) = q −1 TX/G q(Z) for any nonempty closed subset Z of X, where q : X → → X/G is the quotient map. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Let Z be a nonempty closed subset of X. We divide the proof in six steps. Step 1. q −1 q(Z) ⊂ TX (Z). Let x ∈ q −1 q(Z). Then q(x) ∈ q(Z). Thus, there exists z ∈ Z such that q(z) = q(x). This implies that TX ({z}) = TX ({x}). Hence, since TX is idempotent (Theorem 4.2.32), TX ({x}) = TX ({z}) ⊂ TX (Z). Therefore, x ∈ TX (Z), and q −1 q(Z) ⊂ TX (Z). Step 2. TX (Z) = q −1 qTX (Z). Clearly, TX (Z) ⊂ q −1 qTX (Z). Let x ∈ q −1 qTX (Z). Then q(x) ∈ qTX (Z). Hence, there exists y ∈ TX (Z) such that q(y) = q(x). This implies that TX ({y}) = TX ({x}). Thus, since TX is idempotent (Theorem 4.2.32), we have that TX ({y}) ⊂ TX (Z). Hence, x ∈ TX (Z), and q −1 qTX (Z) ⊂ TX (Z). Therefore, TX (Z) = q −1 qTX (Z). Step 3. If Z is connected and IntX (Z) = ∅, then Z = q −1 q(Z). Clearly, Z ⊂ q −1 q(Z). Let x ∈ q −1 q(Z). Then q(x) ∈ q(Z). Thus, there exists z ∈ Z such that q(z) = q(x). This implies that TX ({z}) = TX ({x}). Since TX ({x}) is a nowhere dense terminal subcontinuum of X (Lemma 5.1.1, Theorem 5.1.8 and Theorem 3.1.21), TX ({x}) ∩ Z = ∅ and IntX (Z) = ∅, we have that TX ({x}) ⊂ Z. In particular, x ∈ Z. Therefore, Z = q −1 q(Z). Step 4. qTX (Z) ⊂ TX/G q(Z). Let χ ∈ X/G \ TX/G q(Z). Then there exists a subcontinuum W of X/G such that χ ∈ IntX/G (W) ⊂ W ⊂ X/G \ q(Z). From these inclusions we obtain that q −1 (χ) ⊂ IntX (q −1 (W)) ⊂ q −1 (W) ⊂ X \ q −1 q(Z) ⊂ X \ Z. Hence, since q is monotone (Theorem 5.1.19), q −1 (χ) ∩ TX (Z) = ∅. Thus, qq −1 (χ) ∩ qTX (Z) = ∅. Therefore, χ ∈ X/G \ qTX (Z), and qTX (Z) ⊂ TX/G q(Z). Step 5. TX/G q(Z) ⊂ qTX (Z). Let χ ∈ X/G \ qTX (Z). Then {χ} ∩ qTX (Z) = ∅. This implies that q −1 (χ)∩q −1 qTX (Z) = ∅. Hence, by Step 2, q −1 (χ)∩TX (Z) = ∅. Since TX is idempotent (Theorem 4.2.32), q −1 (χ) ∩ TX2 (Z) = ∅. Thus, there exists a subcontinuum W of X such that q −1 (χ) ⊂ IntX (W ) ⊂ W ⊂ X \ TX (Z) ⊂ X \ Z. From these inclusions, since q is an open map (Theorem 5.1.19 and Proposition 5.1.7) we obtain that {χ} = qq −1 (χ) ⊂ IntX/G (q(W )) ⊂ q(W ) ⊂ q(X \ Z). To finish, we need to show that q(W ) ∩ q(Z) = ∅. Suppose there exists χ ∈ q(W ) ∩ q(Z). Then, by Steps 3 and 1, q −1 (χ ) ⊂ q −1 q(W ) ∩ Copyright © 2005 Taylor & Francis Group, LLC

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q −1 q(Z) = W ∩ q −1 q(Z) ⊂ W ∩ TX (Z), a contradiction to the election of W . Hence, q(W ) ∩ q(Z) = ∅, and χ ∈ X/G \ TX/G q(Z). Therefore, TX/G q(Z) ⊂ qTX (Z). Step 6. qTX (Z) = TX/G q(Z). The equality follows from Steps 4 and 5. From Steps 2 and 6, we have that TX (Z) = q −1 qTX (Z) = q −1 TX/G q(Z). Therefore, TX (Z) = q −1 TX/G q(Z). Q.E.D. 5.1.24. Corollary. Let X be a decomposable nonaposyndetic homogeneous continuum and let G = {TX ({x}) | x ∈ X}. Then TX is continuous if and only if X/G is locally connected. Proof. Let q : X → → X/G be the quotient map. Suppose TX is continuous for X. Note that, by Step 4 of Theorem 5.1.23, q is TXX/G – continuous (Definition 3.2.4). Thus, TX/G is continuous for X/G (Theorem 3.2.5). Since X/G is an aposyndetic continuum (Theorem 5.1.19) for which TX/G is continuous, X/G is locally connected (Corollary 3.2.16). Next, suppose X/G is locally connected. Recall that q : X → → X/G is a monotone open map (Theorem 5.1.19). Let W be a proper subcontinuum of X, and let x ∈ X \ W . Since TX ({x}) is a terminal subcontinuum of X (Theorem 5.1.19), either W ⊂ TX ({x}) or TX ({x}) ∩ W = ∅. In either case, q(W ) is a proper subcontinuum of X/G. Therefore, by Theorem 3.2.2, TX is continuous. Q.E.D. 5.1.25. Remark. Note that Corollary 5.1.24 gives a partial answer to Question 7.2.2. As a consequence of Theorem 3.1.34, Corollary 3.2.16 and Corollary 5.1.24, we have the following characterization of homogeneous continua for which the set function T is continuous:



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5.1.26. Theorem. Let X be a homogeneous continuum. Then T is continuous for X if and only if one of the following conditions holds: (1) X is indecomposable. (2) X is not aposyndetic and X/G is locally connected, where G = {T ({x}) | x ∈ X}. (3) X is locally connected.

Next, we prove an alternate version of Theorem 5.1.19. The awkward collection of hypotheses in this Theorem is precisely the situation encountered in Theorem 5.3.23.

5.1.27. Theorem. Let Y be a continuum such that it is the union of two disjoint, nonempty sets K and E with the following properties: (1) Y is aposyndetic at each point of E, (2) Y is not aposyndetic at any point of K, (3) Y is aposyndetic at each point of K with respect to each point of E, (4) K has a metric d such that (K, d) has the property of Effros, (5) Each homeomorphism h of (K, d) can be extended to a homeoˆ of Y , by defining h(z) ˆ morphism h = z for each z ∈ E, (6) K is connected and open, and (7) dim(E) = 0 (i.e., E is totally disconnected). If G = {T ({x}) | x ∈ Y }, then G is a continuous, terminal decomposition of Y with the following properties: (8) The degenerate elements of G are precisely the points of E, (9) The nondegenerate elements of G are mutually homeomorphic, cell–like, homogeneous, indecomposable continua of the same dimension as Y , (10) The quotient space Y /G is aposyndetic, and (11) The quotient space K/G is homogeneous, where G = {T ({x}) | x ∈ K}. Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Since Y is aposyndetic at every point with respect to at least one other point (by (1) and (3)), the elements of G are proper subcontinua of Y (Theorem 3.1.21). Moreover, if x ∈ K, then T ({x}) ⊂ K since Y is aposyndetic at each point of E, by (1). By (2), the elements of G that are contained in K are nondegenerate. Furthermore, if x ∈ E, then T ({x}) = {x}, since Y is aposyndetic at every point of K with respect to x, by (3). Note that by (4) and by Theorem 4.2.33, K is a homogeneous space. By (5) and Lemma 5.1.1, the elements of G contained in K are mutually homeomorphic and homogeneous. Thus, the elements of G are the points of E and some nondegenerate subcontinua of K, i.e., G = {{x} | x ∈ E}∪{T ({x}) | x ∈ K}. The proof of the fact that G is a decomposition of Y is similar to the one given in Theorem 5.1.2 (using (5) to extend the homeomorphisms defined on K to homeomorphisms defined on Y ). Now, we see that G is continuous. First, we show that G is upper semicontinuous. Let x ∈ E. Then T ({x}) = {x}. Suppose G is not upper semicontinuous at {x}. Then there exists an open subset U of Y such that {x} ⊂ U and for each n ∈ IN, there exists xn ∈ V d1 (x) such n that T ({xn }) \ U = ∅. Note that lim xn = x. Since the hyperspace n→∞

of subcontinua of Y , C(Y ), is compact (Theorem 1.8.5), without loss of generality, we assume that the sequence {T ({xn })}∞ n=1 converges to a subcontinuum Z of Y . Note that Z \ U = ∅. Let z ∈ Z \ U . Since z ∈ Y \ T ({x}), there exists a subcontinuum W of Y such that z ∈ Int(W ) ⊂ W ⊂ Y \ {x}. Since lim T ({xn }) = Z, for each n→∞

n ∈ IN, there exists zn ∈ T ({xn }) such that lim zn = z. Hence, n→∞

there exists N ∈ IN such that zn ∈ Int(W ) for every n ≥ N . Since x ∈ Y \ W , and lim xn = x, there exists N ∈ IN such n→∞

that xn ∈ Y \ W for every n ≥ N . Let n = max{N , N }. Then zn ∈ Int(W ) and xn ∈ Y \W . This implies that zn ∈ Y \T ({xn }), a contradiction to the choice of zn . Hence, G is upper semicontinuous at {x}. Let x ∈ K, and let U be an open subset of Y such that T ({x}) ⊂ U . Without loss of generality, we assume that U ⊂ K. By a similar argument to the one given in Theorem 5.1.4, we can find an open subset V of Y such that T ({x}) ⊂ V ⊂ U and if x ∈ Y such Copyright © 2005 Taylor & Francis Group, LLC


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∅, then T ({x }) ⊂ U . Hence, G is upper that T ({x }) ∩ V = semicontinuous at T ({x}). Therefore, G is upper semicontinuous. Next, we prove that G is lower semicontinuous. Let x ∈ E. Then, since T ({x}) = {x}, clearly, G is lower semicontinuous at {x}. Let x ∈ K. Take p, q ∈ T ({x}), and let U be an open subset of Y such that p ∈ U . Without loss of generality, we assume that U ⊂ K. By a similar argument to the one given in Theorem 5.1.4, we can find an open subset V of Y such that q ∈ V and if x ∈ Y such that T ({x }) ∩ V = ∅, then T ({x }) ∩ U = ∅. Hence, G is lower semicontinuous at T ({x}). Therefore, G is a continuous decomposition. Each degenerate element of G is obviously a terminal subcontinuum of Y . A proof of the fact that the nondegenerate elements of G are terminal subcontinua is similar to the one given in Theorem 5.1.19 (only the property of Effros is used). The proof of the fact that the nondegenerate elements of G are indecomposable is similar to the proof given in Theorem 5.1.19. By Theorem 1.7.3, Y /G is a continuum. Let q : Y → → Y /G be the quotient map. Then q is a monotone map. Since K has the property of Effros, q|K : K → → K/G is completely regular. Since the nondegenerate elements of G are terminal, by Theorem 5.1.17, each such element of G is cell–like. A proof of the fact that the elements of G have the same dimension as Y may be found in Corollary 9 of [32]. Since the elements of G are terminal subcontinua, the proof of the fact that Y /G is aposyndetic is similar to the one given in Theorem 5.1.19. Since G is a continuous decomposition, the proof of the fact that K/G is homogeneous is similar to the one given in Theorem 5.1.4. Q.E.D.


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5.2


Detour to Covering Spaces

We present the necessary definitions and results of the theory of covering spaces to state and prove Rogers’s Terminal Decomposition Theorem (Theorem 5.3.28). 5.2.1. Definition. Let X and Y be metric spaces. A map f : X → Y is called a local homeomorphism provided that for each point x ∈ X, there exists an open subset U of X such that x ∈ U , f (U ) is an open subset of Y and f |U : U → → f (U ) is a homeomorphism. 5.2.2. Proposition. Let X and Y be metric spaces. If f : X → →Y −1 is a surjective local homeomorphism, then f (y) is a discrete subset of X for every y ∈ Y . Proof. Let y ∈ Y . Then, by definition, for each point x ∈ f −1 (y) there exists an open subset U of X such that x ∈ U and f |U : U → → f (U ) is a homeomorphism. Then U ∩ f −1 (y) = {x}. Hence, each point of f −1 (y) is an isolated point in f −1 (y). Therefore, f −1 (y) is a discrete subset of X. Q.E.D. 5.2.3. Corollary. Let X and Y be metric spaces. If X is compact and f : X → → Y is a surjective local homeomorphism, then f −1 (y) is finite for every y ∈ Y . 5.2.4. Definition. Let X, Y and Z be metric spaces. If f : X → Y and g : Z → Y are maps, then a lifting of g relative to f is a map g˜ : Z → X such that f ◦ g˜ = g. 5.2.5. Proposition. Let X, Y and Z be metric spaces. Let f : X → Y be a local homeomorphism. If Z is connected and g : Z → Y is a map, then two liftings g˜, gˆ : Z → X of g relative to f such that g˜(z0 ) = gˆ(z0 ), for some point z0 ∈ Z, are equal. Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Let A = {z ∈ Z | g˜(z) = gˆ(z)}. Then, since g˜(z0 ) = gˆ(z0 ), A = ∅. Since X is a metric space, A is a closed subset of Z. To conclude that g˜ = gˆ, it is enough to show that A is an open subset of Z (because Z is connected). Let z ∈ A. Then g˜(z) = gˆ(z). Since f is a local homeomorphism, there exists an open subset U of X such that g˜(z) = gˆ(z) ∈ U and f |U : U → → f (U ) is a homeomorphism. Since g˜ and gˆ are continuous, there exists an open subset V of Z such that z ∈ V and g˜(V ) ∪ gˆ(V ) ⊂ U . Then for each z ∈ V , (f ◦ g˜)(z ) = g(z ) = (f ◦ gˆ)(z ). Hence, for each z ∈ V g˜(z ) = gˆ(z ), since f |U is a homeomorphism. Thus, V ⊂ A. Therefore, g˜ = gˆ. Q.E.D. 5.2.6. Definition. Let X be a metric space. A covering space of . and a map σ : X . → X is a pair consisting of a metric space X →X such that the following condition holds: for each point x ∈ X, there −1 exists an open subset U of X such that x ∈ U , σ (U ) = Vλ , λ∈Λ

. where {Vλ }λ∈Λ is a family of pairwise disjoint open subsets of X, and σ|Vλ : Vλ → → U is a homeomorphism for every λ ∈ Λ. Any such open subset U of X is called an evenly covered subset of X. The map σ is called a covering map.

5.2.7. Remark. Observe that, by definition, every covering map is a local homeomorphism. The converse is not necessarily true as can be easily seen using the interval (0, 1) and the map f : (0, 1) → → S1 given by f (t) = exp(4πt). The next Lemma says that the Cartesian product of covering spaces is a covering space. .1 → → X1 5.2.8. Lemma. Let X1 and X2 be metric spaces. If σ1 : X .2 → . 1 ×X .2 → and σ2 : X → X2 are covering maps, then (σ1 ×σ2 ) : X →X1 × X2 is a covering map. Proof. Let (x1 , x2 ) ∈ X1 ×X2 . Let Uj be an open subset of Xj such that xj ∈ Uj and Uj is evenly covered by σj , j ∈ {1, 2}. Suppose Copyright © 2005 Taylor & Francis Group, LLC

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σ1−1 (U1 ) =

Vλ1 and σ2−1 (U2 ) =

λ∈Λ

γ∈Γ

(σ1 × σ2 )−1 (U1 × U2 ) =

Vγ2 . Then

Vλ1

×

λ∈Λ

Vγ2

.

γ∈Γ

→ U1 × U2 is a homeomorphism Note that (σ1 × σ2 )|Vλ1 ×Vγ2 : Vλ1 × Vγ2 → for every λ ∈ Λ and γ ∈ Γ. Therefore, σ1 × σ2 is a covering map. Q.E.D. 5.2.9. Definition. Let X be a metric space. A covering space . of X is said to be universal if X . is arcwise connected and its X fundamental group is trivial.

. → 5.2.10. Definition. Let σ : X → X be a covering map. A homeomorphism . → . ϕ: X →X is covering homeomorphism of X if σ ◦ ϕ = σ.

5.2.11. Definition. Let X and Y be metric spaces, with metrics d and d , respectively. A map f : X → Y is a local isometry provided that for each point x ∈ X, there exists an open subset U of X such that x ∈ U and d(x , x ) = d (f (x ), f (x )) for every x , x ∈ U . The map f is an isometry if d(x , x ) = d (f (x ), f (x )) for every x , x ∈ X. The following Proposition tells us that a connected covering space of a locally connected continuum has a metric such that the covering map is a local isometry. 5.2.12. Proposition. Let X be a locally connected continuum . → . is conwith metric d. If σ : X → X is a covering map, where X . such that σ is a local nected, then there exists a metric d˜ for X isometry and every covering homeomorphism of X is an isometry. Copyright © 2005 Taylor & Francis Group, LLC

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. such that Proof. First, we show the existence of a metric d˜ for X σ is a local isometry. . is a covering space, for each x ∈ X, there exists an Since X evenly covered open subset Ux of X. Thus, each σ −1 (Ux ) may be written as Vλ (x), σ −1 (Ux ) = λ∈Λx

. and σ|V (x) : Vλ (x) → where each Vλ (x) is an open subset of X → Ux λ is a homeomorphism. Since X is locally connected, without loss of generality, we assume that each Ux is connected. .x is an open subset of X. . Note . Vλ (x). Then U Let Ux = λ∈Λx

.x | x ∈ X} is an open cover of X. . Hence, since X . that U = {U . is connected, by Theorem 3–4 of [12] for any two points x˜, y˜ ∈ X there exists a finite subfamily {U1 , . . . , Un } of U such that x˜ ∈ U1 , y˜ ∈ Un and Uj ∩ Uk = ∅ if and only if |j − k| ≤ 1, j, k ∈ {1, . . . , n}. Let z˜1 = x˜, zñ = y˜ and let z˜j ∈ Uj−1 ∩Uj for each j ∈ {2, . . . , n}. Then z˜1 , . . . , zñ is a chain of points from x˜ to y˜. We define the length of this chain of points as n

d(zj−1 , zj ), where zj = σ(˜ zj ), j ∈ {1, . . . , n}.

j=2

˜ x, y˜) to be the infimum of the lengths of chains of points Define d(˜ ˜ x, y˜) ≥ 0 and from x˜ to y˜. We assert that d˜ is a metric. Clearly, d(˜ ˜ x, y˜) = d(˜ ˜ y , x˜) for every x˜, y˜ ∈ X. . d(˜ . Note that if We show the triangle inequality. Let x˜, y˜, z˜ ∈ X. a ˜1 , . . . , a ñ is a chain of points from x˜ to y˜ and c˜1 . . . , c˜m is a chain of points from y˜ to z˜, then a ˜1 , . . . , a ñ , c˜1 . . . , c˜m is a chain of points from x˜ to z˜. Hence, n m d(aj−1 , aj ) + d(cj−1 , cj ) ∈ j=2

k

j=2

d(zj−1 , zj ) z˜1 , . . . , z˜k is chain of points from x˜ to z˜ .

j=2


246


Since n m d(aj−1 , aj ) + d(cj−1 , cj ) inf j=2

ñ , c˜1 . . . , c˜m is a ˜1 , . . . , a a

j=2

chain of points from x˜ to z˜ inf

n

≤

d(aj−1 , aj ) a ñ is a chain of points from x˜ to y˜ + ˜1 , . . . , a

j=2

inf

m

d(cj−1 , cj ) c˜1 . . . , c˜m is a chain of points from y˜ to z˜ ,

j=2

˜ x, z˜) ≤ d(˜ ˜ x, y˜) + d(˜ ˜ y , z˜). Hence, d˜ satisfies the triangle inequality. d(˜ . Therefore, d˜ is a metric for X. . and let Next, we see that σ is a local isometry. Let x˜ ∈ X, x = σ(˜ x). Let Ux be an open subset of X such that x ∈ Ux and Ux is evenly covered by σ. Let Vx˜ be an open subset of σ −1 (Ux ) such that x˜ ∈ Vx˜ and σ|Vx˜ : Vx˜ → → Ux is a homeomorphism. Take w, ˜ z˜ ∈ Vx˜ , let w = σ(w) ˜ and let z = σ(˜ z ). Observe that w, ˜ z˜ is a chain of points from w˜ to z˜. Hence, d(w, z) is the smallest ˜ w, length of chains of points from w˜ to z˜. Thus, d( ˜ z˜) = d(w, z). Therefore, σ is a local isometry. To finish the proof, we show that each covering homeomorphism . → . be a covering homeomorphism. is an isometry. Let ϕ : X → X . Let z˜1 , . . . , zñ be a chain of points from x˜ to y˜. Let x˜, y˜ ∈ X. Then ϕ(˜ z1 ), . . . , ϕ(˜ zn ) is a chain of points from ϕ(˜ x) to ϕ(˜ y ). Since σ ◦ ϕ = σ, d(σ(˜ zj−1 ), σ(˜ zj )) = d(σ ◦ ϕ(˜ zj−1 ), σ ◦ ϕ(˜ zj )). Hence, n d(σ(˜ zj−1 ), σ(˜ zj )) z˜1 , . . ., zñ is a chain of points from x˜ to y˜ j=2


5.2. DETOUR TO COVERING SPACES ⊂

n

247

d(σ(w˜j−1 ), σ(w˜j )) w˜1 , . . ., wñ is a chain of points from

j=2

ϕ(˜ x) to ϕ(˜ y) . Since ϕ is a covering homeomorphism, it is easy to see that ϕ−1 is also a covering homeomorphism. Thus, the reverse inclusion is ˜ x, y˜) = d(ϕ(˜ ˜ x), ϕ(˜ also true. Consequently, d(˜ y )). Therefore, ϕ is an isometry. Q.E.D. The next Theorem says that, in certain cases, the property of Effros can be lifted to a covering space. 5.2.13. Theorem. Let X be a compact and connected absolute .→ neighborhood retract, and let σ : X → X be a covering map, where . X is connected. If M is a homogeneous subcontinuum of X, then / = σ −1 (M ) has the property of Effros. M Proof. Since X is an absolute neighborhood retract, X is locally connected (by (iii) (p. 339) of [16]). Hence, by Proposition 5.2.12, . has a metric d˜ such that σ is a local isometry we assume that X and every covering homeomorphism is an isometry. Let U be a finite open cover of X by connected and evenly covered subsets of X. Let ε > 0 such that 2ε is a Lebesgue number for U (Theorem 1.6.6). Since X is a compact absolute neighborhood retract, by Theorem 1.1 (p. 111) of [13], there exists β > 0 such that for any . → X such that d(k(˜ two maps k, k : X x), k (˜ x)) < β for every . . x˜ ∈ X, there exists a homotopy F : X × [0, 1] → X such that F ((˜ x, 0)) = k(˜ x), F ((˜ x, 1)) = k (˜ x), and d(F ((˜ x, t)), F ((˜ x, s))) < ε . and every s, t ∈ [0, 1] (such a homotopy is called for every x˜ ∈ X an ε–homotopy). Without loss of generality, we assume that β ≤ ε. Let M be a homogeneous subcontinuum of X. Then, by Theorem 4.2.31, there exists an Effros number δ > 0 for β. We assume ˜ p, q˜) < δ. We show there / such that d(˜ that δ ≤ β. Let p˜, q˜ ∈ M ˜: M ˜ p) = q˜. /→ / such that h(˜ exists an ε–homeomorphism h →M Copyright © 2005 Taylor & Francis Group, LLC

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Let p = σ(˜ p) and let q = σ(˜ q ). Since σ is a local isometry, ˜ d(˜ p, q˜) = d(p, q) (δ < ε and 2ε is a Lebesgue number for U). Hence, there exists a β–homeomorphism h : M → → M such that h(p) = q. Since h is a β–homeomorphism, there exists an ε–homotopy F between h and 1M . Since 1M can be extended to 1X , there exist a map f : X → X such that f |M = h and an ε–homotopy F between f and 1X such that F |M ×[0,1] = F (Theorem 2.2 (p. 117) of [13]). . × [0, 1] → Let G : X → X be given by G = F ◦ (σ × 1[0,1] ). Then G is a homotopy between f ◦ σ and σ because: G((˜ x, 0)) = F ((σ(˜ x), 0)) = f (σ(˜ x)) = f ◦ σ(˜ x) and G((˜ x, 1)) = F ((σ(˜ x), 1)) = σ(˜ x). Note that 1X. is a lifting of σ relative to σ. By Proposi¸cão 9 (p. .: X . × [0, 1] → X . between 132) of [17], there exists a homotopy G . →X . with σ ◦ f˜ = f ◦ σ such that σ ◦ G . = G. 1X. and a map f˜: X ˜ = f˜| / . We show that h ˜ is desired homeomorphism. Note Let h M ˜ M /) = σ ◦ f˜(M /) = f ◦ σ(M /) = f ◦ σ(σ −1 (M )) = f (M ) = that σ ◦ h( ˜ M /) ⊂ M /. h(M ). Hence, h( ˜ moves no point of M / more than ε. Let Now, we see that h / x˜ ∈ M . Observe that G|M / ×[0,1] is an ε–homotopy. Hence, we have that d(G((˜ x, t)), G((˜ x, s))) < ε for every s, t ∈ [0, 1]. Thus, G({˜ x} × . [0, 1]) is contained in an element of U. Since σ(G({˜ x} × [0, 1])) = ˜ . . x, s))) < ε G({˜ x} × [0, 1]) and σ is a local isometry, d(G((˜ x, t)), G((˜ . x, 0)) = x˜ = 1 / (˜ . x, 1)) = for every s, t ∈ [0, 1]. Since G((˜ M x) and G((˜ ˜ x), d(˜ ˜ x, h(˜ ˜ x)) < ε. Therefore, h ˜ moves no point of M / f˜(˜ x) = h(˜ more than ε. ˜ is a homeomorphism, follow the procedure of the To see that h /→M / of the homeomorabove paragraphs to construct a lift k : M −1 ˜ z , k(˜ /. Let phism h : M → → M satisfying d(˜ z )) < ε for all z˜ ∈ M −1 −1 ˜ x) = h ◦ h ◦ σ(˜ ˜ x) = h ◦ σ ◦ h(˜ /. Then σ ◦ k ◦ h(˜ x) = σ(˜ x), x˜ ∈ M −1 ˜ and σ ◦ h ◦ k(˜ x) = h ◦ σ ◦ k(˜ s) = h ◦ h ◦ σ(˜ x) = σ(˜ x). Hence, −1 ˜ ˜ ˜ ˜ k ◦ h(˜ x), h ◦ k(˜ x) ∈ σ (σ(˜ x)). Since neither k ◦ h nor h ◦ k moves a ˜ x) = x˜ and h ˜ ◦ k(˜ point more than 2ε, it follows that k ◦ h(˜ x) = x˜. ˜ Therefore, h is an ε–homeomorphism. Copyright © 2005 Taylor & Francis Group, LLC

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Since h(p) = q, f is an extension of h and f˜ is a lift of f ◦ σ, it ˜ p, h(˜ ˜ p)) < ε, ˜ follows that h(p) ∈ σ −1 (q). Since d(˜ ˜ q , h(˜ ˜ p)) ≤ d(˜ ˜ q , p˜) + d(˜ ˜ p, h(˜ ˜ p)) < δ + ε ≤ 2ε. d(˜ ˜ p) = q˜. Therefore, M / has the property of Effros. Hence, h(˜ Q.E.D.

5.3

Rogers’s Theorem

We use covering spaces techniques due to James T. Rogers, Jr. to study homogeneous continua. We begin with a discussion of Poincaré model of the hyperbolic plane IH ([3], [9] and [36]). Let IH be the interior of the closed unit disk D in IR2 , and let S 1 be its boundary. 5.3.1. Definition. A geodesic in IH is the intersection of IH and a circle C in IR2 that intersects S 1 orthogonally (straight lines through the origin are considered circles centered at ∞).

5.3.2. Definition. A reflection in a geodesic C ∩ IH is a Euclidean inversion in the circle. A hyperbolic isometry of IH is a composition of reflections in geodesics.

5.3.3. Definition. The set IH with the family of isometries is the Poincaré model of the hyperbolic plane. The boundary S 1 of IH, which is not in IH, is called the circle at ∞. Copyright © 2005 Taylor & Francis Group, LLC

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Let IF be a double torus. The universal covering space of IF may be chosen to be IH ([3] and [36]), and the group of covering homeomorphisms to be a subgroup of the orientation–preserving isometries of IH. Each covering homeomorphism (except the identity map 1IH ) is a hyperbolic isometry. The pertinent property for us is that a hyperbolic isometry, when extended to the circle at ∞, has exactly two fixed points in S 1 and none in IH (Theorem 9–3, (p. 132) of [3] and p. 410 of [36]). The universal covering map σ : IH → → IF may be chosen to be a local isometry. 5.3.4. Definition. A geodesic in IF is the image under σ of a geodesic in IH. A geodesic is simple if it has no transverse intersection.

5.3.5. Definition. A simple closed geodesic is a geodesic in IF that is a simple closed curve. A simple closed curve in IF is essential if it does not bound a disk. Each essential simple closed curve in IF is isotopic to a unique simple closed geodesic (p. 339 of [33]). Assume the universal covering space (IH, σ) of IF is constructed with the following properties. The geodesic IH ∩ x–axis maps to a simple closed geodesic C1 under σ. The geodesic IH ∩ y–axis maps to a simple closed geodesic C2 under σ. Furthermore, C1 ∪ C2 is a figure eight W , and C1 ∩ C2 = {v}. A proof of the following Theorem may be found in Theorem 5 of [28]. 5.3.6. Theorem. Let σ : IH → → IF be the universal covering map of IF. Let U be a finite cover of IF by evenly covered sets, and let 2ε be a Lebesgue number for U. If IK is a compact subset of IH and if ϕ : IH → → IH is a covering homeomorphism different from the ˜ x, ϕ(IK)) > ε. identity map 1IH , then there exists x˜ ∈ IK such that d(˜



251

5.3.7. Theorem. Let σ : IH → → IF be the universal covering map / = σ −1 (M ). Suppose of IF. Let M be a subspace of IF, and let M /. If IL is compact, then σ|IL is one–to–one. IL is a component of M Proof. Suppose there exist x ∈ M , and two points x˜1 , x˜2 ∈ IL ∩ σ −1 (x). Let ϕ : IH → → IH be a covering homeomorphism such that ϕ(˜ x1 ) = x˜2 (Proposi¸cao 8 (p. 162) of [17]). Since ϕ = 1IH , by Theorem 5.3.6, there exists z˜ ∈ ϕ(IL) \ IL. Hence, IL ∪ ϕ(IL) is a / which properly contains IL, a contradiction to subcontinuum of M /. Therefore, σ|IL is one–to–one. the fact that IL is a component of M Q.E.D. 5.3.8. Theorem. Let σ : IH → → IF be the universal covering map / = σ −1 (M ). Suppose IL of IF. Let M be a subspace of IF, and let M /. If M is arcwise connected, then σ(IL) = M . is a component of M Proof. Let z ∈ σ(IL), and let z ∈ M . Since M is arcwise connected, there exists map α : [0, 1] → M such that α(0) = z and α(1) = z . Let z˜ ∈ IL such that σ(˜ z ) = z. By a proof similar to the one given in Theorem 1.3.31, it is shown that there exists / such that α a map α ˜ : [0, 1] → M ˜ (0) = z˜ and σ ◦ α ˜ = α. Since /, α α ˜ ([0, 1]) ∩ IL = ∅ and IL is a component of M ˜ ([0, 1]) ⊂ IL. Hence, σ(˜ α(1)) = z . Therefore, σ(IL) = M . Q.E.D. 5.3.9. Definition. Let Q be the Hilbert cube, and let σ ×1Q : IH× Q → → IF × Q be the universal covering map of IF × Q. Let X be a continuum essentially embedded in W × Q (this means that the embedding is not homotopic to a constant map; note that this eliminates some continua from consideration). Let f : X → → W be −1 ˜ . . → / the projection map. Let X = (σ × 1Q ) (X), and let f : X →W . then the set be the projection map. If IK is a component of X, 1 IE(IK) = {z ∈ S | z is a (Euclidean) limit point of f˜(IK)} is called the set of ends of IK. / is the universal covering space of / = σ −1 (W ). Then W Let W W , the “infinite snowflake” pictured on top of page 253. The set / ) ∩ S 1 is a Cantor set; call it Z (p. 341 of [33]). Cl(W


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5.3.10. Lemma. Let X be a continuum. If : X → W is a map which is not homotopic to a constant map, then X can be essentially embedded in W × Q. Proof. By Theorem 1.1.16, there exists an embedding g : X → Q. Let h : X → W × Q be given by h(x) = ((x), g(x)). Then h is an embedding of X into W × Q which is not homotopic to a constant map. Q.E.D. A proof of the following Theorem may be modeled from the proof of Theorem 8 of [28], using the results in [1] to obtain X as the remainder of a compactification of [0, 1). 5.3.11. Theorem. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF×Q. Let X be a continuum essentially embedded . = (σ × 1Q )−1 (X), then no component of X . is in W × Q. If X compact. 5.3.12. Theorem. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF×Q. Let X be a continuum essentially embedded . = (σ × 1Q )−1 (X). Then there exists δ > 0 in W × Q, and let X ˜ x, x˜ ) < δ (where IK and IK are such that if x˜ ∈ IK, x˜ ∈ IK and d(˜ . then IE(IK) = IE(IK ). components of X), Proof. This is an immediate consequence of Theorem 5.2.13 and . preserves ends. the fact that each bounded homeomorphism of X Q.E.D. 5.3.13. Notation. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF×Q. Let X be a continuum essentially embedded . = (σ × 1Q )−1 (X), and let IK be a component in W × Q. Let X . Assume each of the geodesics C1 and C2 has length greater of X. 1 than one and that 2δ is an Effros number for ε = . Cover {v} × Q 2 with a finite collection B of open δ–balls. Since IK is unbounded (Theorem 5.3.11) and locally compact, there exist a ball B ∈ B, .0 and B .m of B and a subcontinuum M of W / meeting two liftings B .0 and B .m . Let ϕ × 1Q : IH × Q → both B → IH × Q be a covering . .m . homeomorphism such that (ϕ × 1Q )(B0 ) = B Copyright © 2005 Taylor & Francis Group, LLC


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σ

C1

C2

a

b

5.3.14. Theorem. If IK and ϕ×1Q are as in Notation 5.3.13, then IE((ϕ × 1Q )(IK)) = IE(IK). .0 ∩ M and let x˜2 ∈ B .m ∩ M . Hence, Proof. Let x˜1 ∈ B ˜ x2 , (ϕ × 1Q )(˜ x1 )) < 2δ. d(˜ By Theorem 5.3.12, IE(IK) = IE((ϕ × 1Q )(IK)). Q.E.D. Compactify IH×Q with S 1 ×Q. Shrink each set of the form {z}× Q, where z ∈ S 1 , to a point to obtain another compactification of IH×Q. This time the remainder is S 1 . Let π : (IH×Q)∪S 1 → → IH∪S 1 be the map of this latter compactification onto the disk D obtained by naturally extending the projection map. Note that the restriction . is just f˜. (If x˜ ∈ X, . then σ ◦ f˜(˜ of π to X x) = f ◦ (σ × 1Q )(˜ x) ∈ W . / .) Hence, f˜(˜ x) ∈ W 5.3.15. Definition. Let X be a continuum, with metric d, and let f: X → → X be a surjective map. A point x ∈ X is a fixed attracting Copyright © 2005 Taylor & Francis Group, LLC

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point provided that f (x) = x and there exists a neighborhood U of x in X such that lim d(x, f n (y)) = 0 for every y ∈ U . n→∞

X U f(y)

2

f (y) 3

f (y)

y x

Fixed attracting point

5.3.16. Definition. Let X be a continuum, with metric d, and let f: X → → X be a surjective map. A point x ∈ X is a fixed repelling point provided that f (x) = x and there exists a neighborhood U of x in X for each y ∈ U \ {x}, there exists n ∈ IN such that f n (y) ∈ X \ U . X U f(y) y

x

2

f (y)

3

f (y)

Fixed repelling point

5.3.17. Remark. It is known that a hyperbolic isometry of IH has an attractive fixed point and a repelling fixed point on S 1 .

5.3.18. Theorem. If IK and ϕ×1Q are as in Notation 5.3.13, then the attracting point of ϕ belongs to IE(IK). Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Let IKn = Cl(IH×Q)∪S 1 ((ϕn × 1Q )(IK)). Since the hyperspace of subcontinua of (IH × Q) ∪ S 1 , C((IH × Q) ∪ S 1 ) is compact (Theorem 1.8.5), without loss of generality, we assume that the sequence 1 {IKn }∞ n=1 converges to IP ∈ C((IH × Q) ∪ S ). This implies that {π(IKn )}∞ n=1 converges to π(IP). Since π(IKn ) ∩ S 1 = IE((ϕn × 1Q )(IK)) = IE(IK) for all n ∈ IN, by 1 Theorem 5.3.14, and since {π(IKn ) ∩ S 1 }∞ n=1 converges to π(IP) ∩ S , it follows that π(IP) ∩ S 1 = IE(IK). If y˜ ∈ π(IK) and z˜ is the attracting point of ϕ, then {ϕn (˜ y )}∞ ˜ in IH ∪ S 1 . n=1 converges to z Since y˜ ∈ π(IK), it follows that ϕn (˜ y ) ∈ π ((ϕn × 1Q )(IK)), and so z˜ ∈ π(IP). Hence, z˜ ∈ IE(IK). Q.E.D. 5.3.19. Theorem. If IK and ϕ×1Q are as in Notation 5.3.13, then the repelling point of ϕ belongs to IE(IK). Proof. The repelling point of ϕ is the attracting point of ϕ−1 . Hence, the Theorem follows from Theorem 5.3.18. Q.E.D. 5.3.20. Theorem. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF × Q. Let X be a homogeneous continuum . = (σ × 1Q )−1 (X) and IK is essentially embedded in W × Q. If X . then IE(IK) is either a two–point set or a Cana component of X, tor set. Furthermore, IE(IK) contains a dense subset each point of which is a fixed point of a hyperbolic isometry ϕ such that ϕ × 1Q is a covering homeomorphism. Proof. Suppose IE(IK) has more than two points. Since IE((ϕn × 1Q )(IK)) = IE(IK) for each n ∈ IN (Theorem 5.3.14), and ϕ only fixes two points of S 1 (p. 419 of [36]), it follows that IE(IK) is infinite. In fact, there exists a sequence of points {˜ zm }∞ m=1 in IE(IK) that converges to the attracting point, z˜, of ϕ. Furthermore, if z˜ ∈ IE(IK), then arbitrarily close to z˜ there is a point z˜ ∈ IE(IK) such that z˜ is the attracting fixed point of a hyperbolic isometry ϕ : IH → → IH such that ϕ × 1Q is a covering homeomorphism. Such a hyperbolic isometry is constructed in the same way as ϕ was (Notation 5.3.13). In particular, ϕ has the Copyright © 2005 Taylor & Francis Group, LLC

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same properties as ϕ. It follows that IE(IK) is perfect. Therefore, IE(IK) is a Cantor set. Q.E.D. A proof of the following Theorem may be found in Theorem 4.1 of [34]. → IF × Q be the universal 5.3.21. Theorem. Let σ × 1Q : IH × Q → covering map of IF × Q. Let X be a nonaposyndetic homogeneous . = (σ × 1Q )−1 (X) continuum essentially embedded in W × Q. Let X . Let Y = Cl(IH×Q)∪S 1 (IK). Then and let IK be a component of X. Y \ IK = IE(IK), and Y is connected im kleinen at each point of IE(IK).

5.3.22. Corollary. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF × Q. Let X be a nonaposyndetic homogeneous . = (σ × 1Q )−1 (X) continuum essentially embedded in W × Q. Let X . Let Y = Cl(IH×Q)∪S 1 (IK). Then Y and let IK be a component of X. is aposyndetic at each point of IE(IK).

5.3.23. Theorem. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF × Q. Let X be a nonaposyndetic homogeneous . = (σ × 1Q )−1 (X) continuum essentially embedded in W × Q. Let X . Let Y = Cl(IH×Q)∪S 1 (IK). Then and let IK be a component of X. the collection G = {TY({˜ x}) | x˜ ∈ Y} is a continuous, terminal decomposition of Y with the following properties: (1) The quotient space Y/G is aposyndetic. x}) is degenerate if x˜ ∈ Y \ IK. (2) TY({˜ (3) The nondegenerate elements of G are mutually homeomorphic, cell–like, indecomposable homogeneous continua of the same dimension as Y. (4) IK/G is homogeneous, where G = {TY({˜ x}) | x˜ ∈ IK}. Proof. Note that Y = IK ∪ IE(IK), where IK is open and connected. By Theorem 5.3.20, dim(IE(IK)) = 0 and each homeomorphism h Copyright © 2005 Taylor & Francis Group, LLC


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ˆ: Y → of IK can be extended to a homeomorphism h → Y such that ˆh(˜ z ) = z˜ for each z˜ ∈ IE(IK). By Corollary 5.3.22, Y is aposyndetic at each point of IE(IK). By Proposition 5.2.12, (IH × Q) ∪ S 1 has a metric d˜ such that . has a metric σ ×1Q is a local isometry. Thus, by Theorem 5.2.13, X . has the property of Effros. Hence, IK has a metric d˜ d˜ such that X such that IK has the property of Effros. It can be shown that each point of IE(IK) has a local basis of open sets of Y whose complements are connected. Hence, Y is aposyndetic at each point of IK with respect to each point of IE(IK). Since X is not aposyndetic and homogeneous, X is not aposyndetic at any point with respect to any other. Let x˜ ∈ IK and suppose Y is aposyndetic at x˜ with respect to x˜ ∈ IK. Thus, there exists a subcontinuum W of Y such that x˜ ∈ IntY(W) ⊂ W ⊂ Y \ {˜ x }; without loss of generality, we assume that W ⊂ IK and (σ×1Q )(˜ x) = (σ × 1Q )(˜ x ). Hence, since σ × 1Q is a local isometry, (σ × 1Q )(W) satisfies that (σ × 1Q )(˜ x) ∈ IntX ((σ × 1Q )(W)) ⊂ (σ × 1Q )(W) ⊂ X \ {(σ × 1Q )(˜ x )}, a contradiction. Therefore, Y is not aposyndetic at any point of IK. Therefore, the Theorem follows from Theorem 5.1.27. Q.E.D. 5.3.24. Theorem. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF × Q. Let X be a nonaposyndetic homogeneous . = (σ × 1Q )−1 (X) continuum essentially embedded in W × Q. Let X . Let Y = Cl(IH×Q)∪S 1 (IK). If x˜ ∈ IK, and let IK be a component of X. then (σ × 1Q )|TY({˜x}) is one–to–one. x}) such that Proof. Let x˜1 and x˜2 be distinct points of TY({˜ (σ × 1Q )(˜ x1 ) = (σ × 1Q )(˜ x2 ). By Proposi¸cão 9 (p. 132) of [17], there exists a covering homeomorphism ϕ : IH × Q → → IH × Q such that ϕ(˜ x1 ) = x˜2 . Since ϕ does not fix (setwise) any compact set (Theorem 5.3.6), it follows that ϕ(TY({˜ x})) ∩ (Y \ TY({˜ x})) = ∅ and TY({˜ x}) ∩ (Y \ ϕ(TY({˜ x}))) = ∅. Since ϕ(TY({˜ x})) ∩ TY({˜ x}) = ∅, this contradicts the fact that the decomposition, G, of Y is terminal (Theorem 5.3.23). Q.E.D.


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→ IF × Q be the universal 5.3.25. Theorem. Let σ × 1Q : IH × Q → covering map of IF × Q. Let X be a nonaposyndetic homogeneous . = (σ × 1Q )−1 (X) continuum essentially embedded in W × Q. Let X . Let Yj = Cl(IH×Q)∪S 1 (IKj ) and let IK1 and IK2 be components of X. j ∈ {1, 2}. If x˜1 ∈ IK1 and x˜2 ∈ IK2 such that (σ × 1Q )(˜ x1 ) = (σ × 1Q )(˜ x2 ), then TY1 ({˜ x1 }) is homeomorphic to TY2 ({˜ x2 }), and (σ × 1Q )(TY1 ({˜ x1 })) = (σ × 1Q )(TY2 ({˜ x2 })). Proof. Let ϕ : IH×Q → → IH×Q be a covering homeomorphism such that ϕ(˜ x1 ) = x˜2 (Proposi¸cão 9 (p. 132) of [17]). By Lemma 5.1.1, ϕ(TY1 ({˜ x1 })) = TY2 ({ϕ(˜ x1 )}) = TY2 ({˜ x2 }). Hence, TY1 ({˜ x1 }) is homeomorphic to TY2 ({˜ x2 }), and (σ × 1Q )(TY1 ({˜ x1 })) = (σ × 1Q )(TY2 ({˜ x2 })). Q.E.D. 5.3.26. Lemma. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF × Q. Let X be a nonaposyndetic homogeneous . = (σ ×1Q )−1 (X). continuum essentially embedded in W ×Q. Let X . such that (σ × 1Q )| . is one–to–one, If Z. is a subcontinuum of X Z . of IH × Q such that Z. ⊂ U . and then there exists an open subset U (σ × 1Q )|U. is one–to–one. Proof. Suppose the lemma is not true. Then for each n ∈ IN, ˜ . there exist xñ , yñ ∈ V d1 (Z) such that xñ = yñ and (σ × 1Q )(˜ xn ) = n yn ). Since IF × Q is compact and σ × 1Q is a covering map, (σ × 1Q )(˜ without loss of generality, we assume that the sequences {˜ xn }∞ n=1 and {˜ yn }∞ ˜0 and y˜0 , respectively. Note that x0 , y0 ∈ n=1 converge to x . Z. xn )}∞ Since σ × 1Q is continuous, the sequences {(σ × 1Q )(˜ n=1 and {(σ × 1Q )(˜ yn )}∞ converge to (σ × 1 )(˜ x ) and (σ × 1 )(˜ y Q 0 Q 0 ), n=1 respectively. Since for each n ∈ IN, (σ × 1Q )(˜ xn ) = (σ × 1Q )(˜ yn ), we have that (σ × 1Q )(˜ x0 ) = (σ × 1Q )(˜ y0 ). Hence, x˜0 = y˜0 ((σ × 1Q )|Z. is one–to–one). Let V be an evenly covered open subset of IF × Q such that (σ × 1Q )(˜ x0 ) ∈ V . Let V. be the open subset of (σ × 1Q )−1 (V ) such Copyright © 2005 Taylor & Francis Group, LLC


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→ V is a homeomorphism. Then that x˜0 ∈ V. and (σ × 1Q )|V. : V. → xn }∞ V. is an open subset of IH × Q containing x˜0 . Since {˜ n=1 and ∞ {˜ yn }n=1 converge to x˜0 , there exists N ∈ IN such that xñ , yñ ∈ V. for every n ≥ N . Hence, (σ × 1Q )(˜ xn ) = (σ × 1Q )(˜ yn ) for each n ≥ N . ((σ × 1Q )|V. : V → → V is a homeomorphism), a contradiction to the choices of xñ and yñ . . of IH × Q such that Therefore, there exists an open subset U . and (σ × 1Q )| . is one–to–one. Z. ⊂ U U Q.E.D. 5.3.27. Theorem. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF × Q. Let X be a nonaposyndetic homogeneous . = (σ ×1Q )−1 (X), continuum essentially embedded in W ×Q. Let X . Let Y = Cl(IH×Q)∪S 1 (IK). If and let IK be a component of X. x˜ ∈ IK, then (σ × 1Q )(TY({˜ x})) is a maximal terminal proper, cell– like subcontinuum of X. Proof. Let x˜ ∈ IK. Recall that TY({˜ x}) is cell–like ((3) of Theorem 5.3.23). It follows, from Theorem 5.3.24, that (σ×1Q )(TY({˜ x})) is cell–like. Hence, (σ×1Q )(TY({˜ x})) is a proper subcontinuum of X (there exists a map f : X → → W that is not homotopic to a constant map). Since (σ×1Q )|TY({˜x}) is one–to–one (Theorem 5.3.24), there exists . and (σ × 1Q )| . is one–to– . such that TY({˜ x}) ⊂ U an open set U U . → . ) is a → (σ × 1Q )(U one, by Lemma 5.3.26. Thus, (σ × 1Q )|U. : U homeomorphism. If M is a subcontinuum of X such that M ∩(σ×1Q )(TY({˜ x})) = ∅ and M ∩(X \(σ×1Q )(TY({˜ x}))) = ∅, then we may assume, by taking . ). Hence, M has a subcontinuum if necessary, that M ⊂ (σ × 1Q )(U /∩(Y\TY({˜ / such that M /∩TY({˜ x}) = ∅ and M x})) = ∅. Since a lift M / x}) is terminal, TY({˜ x}) ⊂ M . Thus, (σ × 1Q )(TY({˜ x})) ⊂ M . TY({˜ Hence, (σ × 1Q )(TY({˜ x})) is a terminal subcontinuum of X. Now, we show that if T is a proper, terminal, cell–like subcontinuum of X such that (σ × 1Q )(TY({˜ x})) ⊂ T , then T = (σ × 1Q )(TY({˜ x})). Since T is cell–like, the inclusion map i : T → IF × Q lifts to a one–to–one map ˜ı : T → IH × Q such that ˜ı((σ × 1Q )(˜ x)) = . x˜. Hence, ˜ı(T ) ∩ TY({˜ x}) = ∅. Let T = ˜ı(T ). Copyright © 2005 Taylor & Francis Group, LLC

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Since ˜ı is one–to–one, (σ × 1Q )|T. is one–to–one. Thus, there exists an open set V. such that T. ⊂ V. and (σ × 1Q )|V. is one– to–one, by Lemma 5.3.26. Since TY({˜ x}) is a maximal terminal . . subcontinuum of Y, either T is not terminal or T. = TY({˜ x}). Let N . ∩ T. = ∅ and N . ∩Y\ T. = be a proper subcontinuum of Y such that N . ⊂ V. . Hence, (σ × 1Q )(N . ) ∩ T = ∅ and ∅. Again, we assume that N . ) ∩ (X \ T ) = ∅. Since T is terminal, T ⊂ (σ × 1Q )(N . ). (σ × 1Q )(N . , and so T. is terminal. Hence, T. = TY({˜ Thus, T. ⊂ N x}), and T = (σ × 1Q )(TY({˜ x})). Q.E.D. We are ready to state and prove Rogers’s Terminal Decomposition Theorem. 5.3.28. Theorem. Let σ × 1Q : IH × Q → → IF × Q be the universal covering map of IF × Q. Let X be a homogeneous continuum that admits a map into the figure eight W that is not homotopic to a constant map. Hence, we may consider X essentially embedded in . = (σ × 1Q )−1 (X). If G = {(σ × 1Q )(TY({˜ IF × Q. Let X x})) | x˜ ∈ . 1 IK, where IK is a component of X, and Y = Cl(IH×Q)∪S (IK)}, then G is a continuous decomposition such that the following hold: (1) G is a monotone and terminal decomposition of X. (2) The elements of G are mutually homeomorphic, indecomposable, cell–like terminal, homogeneous continua. (3) The quotient space, X/G, is a homogeneous continuum. (4) X/G does not contain any proper, nondegenerate terminal subcontinuum. (5) If X is decomposable, then X/G is an aposyndetic continuum; in fact, the decomposition G is Jones’s decomposition. (6) If the elements of G are nondegenerate, then they have the same dimension as X and X/G is one–dimensional. Proof. Note that, by Theorem 5.3.27, G is a collection of maximal proper terminal cell–like subcontinua of X. . such We show that G is a decomposition of X. Let x˜1 , x˜2 ∈ X that (σ × 1Q )(TY1 ({˜ x1 })) ∩ (σ × 1Q )(TY2 ({˜ x2 })) = ∅ (IKj is the . component of X such that x˜j ∈ IKj and Yj = Cl(IH×Q)∪S 1 (IKj ), Copyright © 2005 Taylor & Francis Group, LLC


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x1 }))∩(σ×1Q )(TY2 ({˜ x2 })). Then j ∈ {1, 2}). Let z ∈ (σ×1Q )(TY1 ({˜ there exists z˜j ∈ TYj ({˜ xj }) such that (σ × 1Q )(˜ zj ) = z, j ∈ {1, 2}. Hence, by Theorem 5.3.25, TY1 ({˜ x1 }) is homeomorphic to TY2 ({˜ x2 }) and (σ × 1Q )(TY1 ({˜ x1 })) = (σ × 1Q )(TY2 ({˜ x2 })). Since σ × 1Q is a covering map, X = {(σ × 1Q )(TY({˜ x})) | x˜ ∈ IK, where IK is a component of . and Y = Cl(IH×Q)∪S 1 (IK)}. X, Therefore, G is a decomposition of X. Now, we prove that the homeomorphism group of X, H(X), respects G, and then, apply Theorem 5.1.4 to conclude that G is continuous. . such that Let h ∈ H(X), and let x˜1 , x˜2 ∈ X h((σ × 1Q )(TY1 ({˜ x1 }))) ∩ (σ × 1Q )(TY2 ({˜ x2 })) = ∅. x1 })) is terminal, h((σ × 1Q )(TY1 ({˜ x1 }))) is Since (σ × 1Q )(TY1 ({˜ terminal (Corollary 5.1.12). Thus, since (σ × 1Q )(TY2 ({˜ x2 })) is a maximal terminal subcontinuum, h((σ × 1Q )(TY1 ({˜ x1 }))) ⊂ (σ × 1Q )(TY2 ({˜ x2 })). x1 })) ⊂ h−1 ((σ × 1Q )(TY2 ({˜ x2 }))). This implies that (σ × 1Q )(TY1 ({˜ −1 Hence, (σ × 1Q )(TY1 ({˜ x1 })) = h ((σ × 1Q )(TY2 ({˜ x2 }))). Consequently, h((σ × 1Q )(TY1 ({˜ x1 }))) = (σ × 1Q )(TY2 ({˜ x2 })). Therefore, H(X) respects G. Since H(X) respects G, by Theorem 5.1.4, the elements of G are mutually homeomorphic homogeneous continua, and the quotient space X/G is homogeneous. The fact that the elements of G have the same dimension as X follows from Theorem 8 of [32]. The proof of the indecomposability of the elements of G is similar to the one given in Theorem 5.1.19. The fact that X/G is one–dimensional may be found in Theorem 6 of [35]. Next, we show that X/G does not contain nondegenerate proper terminal subcontinua. To this end, let Γ be a nondegenerate terminal proper subcontinuum of X/G. Let q : X → → X/G be the quotient −1 map. Then q is a monotone map. Thus, q (Γ) is a subcontinuum of X (Lemma 2.1.12). Note that x})), | x˜ ∈ IK, where IK is a component q −1 (Γ) = {(σ×1Q )(TY({˜ Copyright © 2005 Taylor & Francis Group, LLC

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CHAPTER 5. DECOMPOSITION THEOREMS . Y = Cl(IH×Q)∪S 1 (IK) and q((σ × 1Q )(TY({˜ x}))) ∈ Γ}. of X,

Since Γ is a nondegenerate proper subcontinuum of X/G, q −1 (Γ) is a proper subcontinuum of X and if q((σ × 1Q )(TY({˜ x}))) ∈ Γ, −1 −1 (σ × 1Q )(TY({˜ x})) q (Γ). Hence, q (Γ) is not a terminal subcontinuum of X (each element (σ × 1Q )(TY({˜ x})) of G is a maximal terminal proper subcontinuum of X). Thus, there exists a subcontinuum Z of X such that Z ∩ q −1 (Γ) = ∅, Z \ q −1 (Γ) = ∅ and q −1 (Γ) \ Z = ∅. Since q(Z ∩ q −1 (Γ)) = q(Z) ∩ Γ, q(Z) is a subcontinuum of X/G such that q(Z) ∩ Γ = ∅. Since Γ is a terminal subcontinuum of X/G, either q(Z) ⊂ Γ or Γ ⊂ q(Z). If q(Z) ⊂ Γ, then q −1 (q(Z)) ⊂ q −1 (Γ). Hence, Z ⊂ q −1 (Γ), a contradiction. Suppose, then, that Γ ⊂ q(Z). This implies that q −1 (Γ) ⊂ q −1 (q(Z)). Thus, for each x ∈ q −1 (Γ), there exists z ∈ Z such that q(z) = q(x). Hence, q −1 (q(x))∩Z = ∅. Since q −1 (q(x)) is a terminal subcontinuum of X, either Z ⊂ q −1 (q(x)) or q −1 (q(x)) ⊂ Z. If Z ⊂ q −1 (q(x)), then Z ⊂ q −1 (Γ), a contradiction. Thus, q −1 (q(x)) ⊂ Z. Since x was an arbitrary point of q −1 (Γ), q −1 (Γ) ⊂ Z, a contradiction. Therefore, X/G does not contain nondegenerate terminal proper subcontinua. Finally, suppose X is decomposable. Let G = {TX ({x}) | x ∈ X} be Jones’s decomposition (Theorem 5.1.19). We show that G = G . Thus, by Theorem 5.1.19, X/G is aposyndetic. x})) ∈ G. Let x ∈ (σ × 1Q )(TY({˜ x})).Then Let (σ × 1Q )(TY({˜ (σ × 1Q )(TY({˜ x})) ∩ TX ({x}) = ∅. Since both of these continua are maximal terminal subcontinua of X (Corollary 5.1.20), (σ × 1Q )(TY({˜ x})) = TX ({x}). Thus, (σ × 1Q )(TY({˜ x})) ∈ G . Hence, G ⊂ G . Let TX ({x}) ∈ G . Since σ × 1Q is a covering map, there exists . such that (σ × 1Q )(˜ . x˜ ∈ X x) = x. Let IK be the component of X such that x˜ ∈ IK. Then (σ × 1Q )(TY({˜ x})) ∩ TX ({x}) = ∅. Since both of these continua are maximal terminal subcontinua of X, (σ × 1Q )(TY({˜ x})) = TX ({x}). Thus, TX ({x}) ∈ G. Hence, G ⊂ G. Therefore, G = G . Q.E.D.


5.4. CASE AND MINC–ROGERS CONTINUA

5.4

263

Case and Minc–Rogers Continua

Aposyndetic homogeneous continua of dimension greater than one are very easy to construct, since the product of homogeneous continua is homogeneous and aposyndetic (Corollary 3.3.9). It is difficult to find one–dimensional aposyndetic homogeneous continua. In fact, R. D. Anderson showed that the simple closed curve and the Menger universal curve are the only one–dimensional locally connected homogeneous continua (Theorem XIII of [5]). Hence, if the only one–dimensional aposyndetic homogeneous continua were locally connected, then Theorem 5.1.19 would imply that it would be enough to study one–dimensional indecomposable homogeneous continua. We present a collection of one–dimensional aposyndetic homogeneous continua which are not locally connected. We use inverse limits to construct such continua. First, we present the construction of J. T. Rogers, Jr. [29] of a continuum originally constructed by J. H. Case [7]. Afterwards, we give a sketch of the generalization of Rogers’s construction made by P. Minc and J. T. Rogers, Jr. [23]. The Sierpi´ nski universal plane curve, S, can be constructed by taking the unit square with boundary B, deleting the interior of the middle–ninth of that square (leaving the boundary of such a square, C), then deleting the interiors of the middle–ninths of each of the eight squares remaining. Let us continue in this manner step by step. The points which have not been removed constitute the required curve. The next Theorem gives a characterization of S; a proof of it may be found in Theorem 4 of [37]:

5.4.1. Theorem. In order that a plane one–dimensional locally connected continuum Z be the Sierpi´ nski universal plane curve S it is necessary and sufficient that for each open connected subset U of Z and each point z ∈ U , U \ {z} is connected. Copyright © 2005 Taylor & Francis Group, LLC

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The Menger universal curve, M, can be described as the set of all points of the unit cube, [0, 1]3 , that project, in each of the x, y and z directions, onto Sierpi´ nski universal plane curves on the faces of the cube. 5.4.2. Lemma. Each point of the Menger universal curve has arbitrarily small neighborhoods with connected boundary. Proof. Note that the point (1, 0, 0) ∈ M has arbitrarily small neighborhoods homeomorphic to M. Since M is homogeneous (Theorem III of [4]), this is true for all points of M. Q.E.D. The next Theorem presents a characterization of M; a proof of it may be found in Theorem XII of [5]:

5.4.3. Theorem. In order that a one–dimensional locally connected continuum Z be the Menger universal curve M it is necessary and sufficient that for each open connected subset U of Z, U is not embeddable in the plane, and for each point z ∈ U , U \ {z} is connected. Copyright © 2005 Taylor & Francis Group, LLC


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. → 5.4.4. Definition. A covering map σ : X → X is regular provided that for each point x ∈ X and each pair of points x˜1 , x˜2 ∈ . → . such σ −1 (x), there exists a covering homeomorphism ϕ : X → X that ϕ(˜ x1 ) = x˜2 .

5.4.5. Definition. A solenoidal sequence is an inverse sequence {Xn , fnn+1 } of continua such that each bonding map fnm : Xm → → Xn is a regular covering map. The inverse limit of a solenoidal sequence is called a solenoidal space. ∞ 5.4.6. Theorem. If the points p = (pn )∞ n=1 and q = (qn )n=1 of the solenoidal space X∞ = lim{Xn , fnn+1 } have the same first coordi←− nate, then there exists a homeomorphism of X∞ onto itself mapping p to q.

Proof. For each n ∈ IN, we construct a homeomorphism hn : Xn → → Xn such that hn (pn ) = qn and fnn+1 ◦ hn+1 = hn ◦ fnn+1 . Then Theorems 2.1.46, 2.1.47 and 2.1.48 imply that h = lim{hn } is a ←−

homeomorphism of X∞ onto itself such that h(p) = q. Since p1 = q1 , we take h1 = 1X1 . Note that p2 , q2 ∈ (f12 )−1 (p1 ). Since f12 is a regular covering map, there exists a covering homeomorphism h2 : X2 → → X2 such that h2 (p2 ) = q2 and f12 ◦ h2 = f12 = h1 ◦ f12 . Copyright © 2005 Taylor & Francis Group, LLC

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Note that p3 , q3 ∈ (f13 )−1 (p1 ). Since f13 is a regular covering map, there exists a covering homeomorphism h3 : X3 → → X3 such 3 3 3 that h3 (p3 ) = q3 and f1 ◦ h3 = f1 = h1 ◦ f1 . Observe that both h2 ◦ f23 and f23 ◦ h3 are liftings of f13 such that h2 ◦ f23 (p3 ) = h2 (p2 ) = q2 and f23 ◦ h3 (p3 ) = f23 (q3 ) = q2 . Hence, h2 ◦ f23 = f23 ◦ h3 , by Proposition 5.2.5. → Suppose we have defined a covering homeomorphism hk : Xk → k k k Xk for f1 : Xk → → X1 such that hk (pk ) = qk , f1 ◦ hk = f1 and k k hk−1 ◦ fk−1 = fk−1 ◦ hk for each k ∈ {1, . . . , n − 1}. Since pn , qn ∈ (f1n )−1 (p1 ) and f1n is a regular covering map, there exists a covering homeomorphism hn : Xn → → Xn such that hn (pn ) = n n qn and f1n ◦ hn = f1n . Again, both hn−1 ◦ fn−1 and fn−1 ◦ hn are n n liftings of f1 such that hn−1 ◦ fn−1 (pn ) = hn−1 (pn−1 ) = qn−1 and n n n n fn−1 ◦ hn (pn ) = fn−1 (qn ) = qn−1 . Hence, hn−1 ◦ fn−1 = fn−1 ◦ hn (Proposition 5.2.5). Q.E.D.

5.4.7. Definition. Let V be an open subset of a metric space Y . Call Y stably homogeneous over V if for each pair of points p, q ∈ V , there exists a homeomorphism h : Y → → Y such that h(p) = q and h|(Y \V ) = 1(Y \V ) .

5.4.8. Definition. A metric space Y is strongly locally homogeneous if for each point y ∈ Y and each open subset U of Y such that y ∈ U , there exists an open subset V of Y such that y ∈ V ⊂ U and such that Y is stably homogeneous over V .

5.4.9. Definition. Let {Xn , fnn+1 } be an inverse sequence of continua. If X∞ = lim{Xn , fnn+1 }, we say that X∞ is locally a product ←− of an open subset of X1 and the Cantor set, C, provided that for each point x ∈ X∞ , there exists an open subset U1 of X1 such that x ∈ f1−1 (U1 ) and f1−1 (U1 ) is homeomorphic to U1 × C.



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5.4.10. Theorem. Let {Xn , fnn+1 } be a solenoidal sequence such that X1 is strongly locally homogeneous. If X∞ = lim{Xn , fnn+1 } ←− and X∞ is locally a product of an open subset of X1 and the Cantor set, C, then X∞ is homogeneous. Proof. Let p = (pn )∞ n=1 ∈ X∞ . Let H be the set of points q1 ∈ X1 such that there exists a homeomorphism of X∞ onto itself such that takes p to a point whose first coordinate is q1 . We show that H is both open and closed. Let q1 ∈ H. Then there exists a homeomorphism g : X∞ → → X∞ such that f1 (g(p)) = q1 (recall that f1 : X∞ → X1 is the projection map). Since X∞ is locally a product of an open subset of X1 and the Cantor set, there exists an open set U1 of X1 containing q1 such that f1−1 (U1 ) is homeomorphic to U1 × C. Since X1 is strongly locally homogeneous, there exists an open set V of X1 such that q1 ∈ V ⊂ U1 and X1 is stably homogeneous over V . If r1 ∈ V , then there exists a homeomorphism h1 : X1 → → X1 such that h1 (q1 ) = r1 and h1 |(X1 \V ) = 1(X1 \V ) . Define the map h : X∞ → → X∞ as follows: h|(X∞ \f1−1 (V )) = 1(X∞ \f1−1 (V )) and h|f1−1 (V ) = h1 × 1C . Then h is a homeomorphism, and f1 (h(g(p))) = r1 . Hence, r1 ∈ H. Thus, H is open. A similar argument shows that H is closed. Hence, H = X1 . Then, by Theorem 5.4.6, X∞ is homogeneous. Q.E.D. 5.4.11. Lemma. View the Sierpi´ nski universal plane curve S as a subset of an annulus D with boundary B ∪ C. If g : D → → D is a double–covering map, i.e., g is a covering map such that its fibres have exactly two points, such that g(B) = B and g(C) = C, then g −1 (S) is homeomorphic to S. Proof. First, we show that g −1 (S) is connected. Let p2 ∈ g −1 (S), and let p1 ∈ S such that g(p2 ) = p1 . Let A be an arc in S from p1 Copyright © 2005 Taylor & Francis Group, LLC

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. in g −1 (S) from p2 to a to a point of B. The arc A lifts to an arc A point of B, by a proof similar to the one given in Theorem 1.3.31. Hence, g −1 (S) is arcwise connected. Since g is a covering map, g preserves local properties. Hence, by Theorem 5.4.1, g −1 (S) is homeomorphic to S. Q.E.D. 5.4.12. Lemma. View the Menger universal curve M as a subset of D × [0, 1], where D is an annulus with boundary B ∪ C. If f : D × [0, 1] → → D × [0, 1] is given by f = g × 1[0,1] , where g : D → → D is a double–covering map such that g(B) = B and g(C) = C, then f −1 (M) is homeomorphic to M. Proof. A proof similar to that of Lemma 5.4.11 shows that f −1 (M) is arcwise connected. Since f is a covering map, f preserves local properties. Hence, by Theorem 5.4.3, f −1 (M) is homeomorphic to M. Q.E.D. The following Theorem is a consequence of Theorem 5.6 of [22]:

5.4.13. Theorem. If {Xn , fnn+1 } is a solenoidal sequence of manifolds, then X∞ = lim{Xn , fnn+1 } is locally a product of an open ←− subset of X1 and the Cantor set C.

5.4.14. Definition. A continuum X is said to be colocally connected at x ∈ X provided that for each open subset U of X such that x ∈ U , there exists an open subset V of X such that x ∈ V ⊂ U and X \ V is connected The continuum X is colocally connected if it is colocally connected at each of its points.

5.4.15. Remark. Note that each colocally connected continuum is aposyndetic. Copyright © 2005 Taylor & Francis Group, LLC


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The continuum, C, constructed in the following Theorem is known as the Case continuum. 5.4.16. Theorem. Let f : D × [0, 1] → → D × [0, 1] be the covering map of Lemma 5.4.12. If X1 = M, X2 = f −1 (M), . . . , Xn = (f n )−1 (M), and if for each n ∈ IN, fnn+1 = f |Xn+1 , then {Xn , fnn+1 } is a solenoidal sequence of Menger curves and C = lim{Xn , fnn+1 } ←− is an aposyndetic, homogeneous, one–dimensional solenoidal space that is not locally connected. Proof. Since the solenoidal sequence {Xn , fnn+1 } is obtained as the restriction of a solenoidal sequence of manifolds, C is locally homeomorphic to the product of an open subset of X1 = M and the Cantor set (Theorem 5.4.13). Hence, C is not locally connected. Since M is one–dimensional, C is one–dimensional (15.6 of [25]). By Theorem XVI of [5], M is strongly locally homogeneous. Hence, C is homogeneous, by Theorem 5.4.10. To see C is aposyndetic, we show that C is colocally connected. Since C is locally homeomorphic to the product of an open subset of X1 = M and the Cantor set, by Proposition 2.1.9 and Lemma 5.4.2, each point of C has arbitrarily small neighborhoods of the form U = fn−1 (Un ), where Un is an open subset of Xn with connected boundary Bn , and ClC(U ) is homeomorphic to ClXn (Un )×C. Let U = fn−1 (Un ) be one of such neighborhoods, and suppose C \ U is not connected. Then there exist two disjoint closed subsets C and D of C such that C \ U = C ∪ D. Let Bn be the boundary of Un . Note that no component of fn−1 (Bn ) intersects both C and D. Define C to be the union of all the components K of U such that ClC(K) ∩ C = ∅. Define D in a similar way. Note that C ∪ C and D ∪ D are disjoint closed subsets of C whose union is C. Since C is connected, either C ∪ C or D ∪ D is empty. Hence, C \ U is connected. Q.E.D. To finish this section, we sketch the construction of the Minc– Rogers continua. Fix n ∈ IN such that n ≥ 3. Let T

n

=

n k=1


S 1 be the n–

270


dimensional torus.

Let Z =

−1

{e} × S × 1

k=1

n

{e}, for ∈

k=+1

{1, . . . , n}, where e = (1, 0). One can assume that the Menger universal curve M is embedded in Tn in such a way that Z ⊂ M for every ∈ {1, . . . , n}. For each α ∈ ZZn , α = (α1 , . . . , αn ), define fα : Tn → Tn by fα((z1 , . . . , zn )) = (z1α1 , . . . , znαn ).

The proof of the next Lemma is similar to the proof given for Lemma 5.4.12. −1 (M) is homeomor5.4.17. Lemma. With the above notation, fα phic to M containing Z for every ∈ {1, . . . , n}.

The proof of the following Theorem is similar to the one given in Theorem 5.4.16. The spaces obtained from Theorem 5.4.18 are known as Minc–Rogers continua. 5.4.18. Theorem. With the above notation. Let Λ = {αk }∞ k=1 be n k k k a sequence of elements of ZZ . Assume that α = (α1 , . . . , αn ). Let Λ Λ −1 MΛ1 = M, and MΛk = fα = lim{MΛk , fαk |MΛk+1 }, then k (Mk−1 ). If M ←−

MΛ is a one–dimensional colocally connected homogeneous continuum.

5.5

Covering Spaces of Some Homogeneous Continua

We study covering spaces of certain homogeneous continua. Copyright © 2005 Taylor & Francis Group, LLC

5.5. COVERING SPACES

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Let us note first that, by Corollary 1.3.30, we have that IR is a covering space of S 1 since the exponential map is a covering map. It is easy to see that, for a given n ∈ IN, the map p : S 1 → → S1 n 1 given by p(z) = z , where S is considered as a subset of the set of complex numbers CI, is a covering map. Hence, S 1 is a covering space of itself in a nontrivial way. Let X be a continuum for which we can define a map fIH : X → W , where W is the figure eight C1 ∪ C2 that is not homotopic to a constant map. Let g : X → Q be an embedding of X into the Hilbert cube (Theorem 1.1.16). Then the map (fIH , g) : X → W × Q given by (fIH , g)(x) = (fIH (x), g(x)) is an embedding that is not homotopic to a constant map. Let /×Q → σ × 1Q : W → W × Q be the universal covering map of W × Q, . . IH for and define X IH = (σ × 1Q )−1 (X). We are going to study X several continua. 5.5.1. Definition. A simple triod is a continuum which is the union of three arcs having only one end point in common. 5.5.2. Theorem. Let n= {nk }∞ k=1 be a sequence of positive integers. Let fIH : Σ → W be a map that is not homotopic to a constant map, where Σ is the . IH is homeomorphic to IR × C ∗ , where C ∗ is the n–solenoid. Then Σ Cantor set minus one point. Proof. Let fIH : Σ → W be an essential map. By Theorem of [2], . IH is homeomorphic to IKIH × IB . IH , where IKIH is a we know that Σ . IH and IB . IH is a zero–dimensional locally compact component of Σ homogeneous space. Hence, IBIH is homeomorphic to C ∗ . We divide the proof in eight steps. . IH does not contain simple triods. Step 1. Σ Copyright © 2005 Taylor & Francis Group, LLC

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It is known that each proper subcontinuum of Σ is an arc (Theorem 2 of [11]). Hence, Σ does not contain simple triods. Since σ×1Q . IH does not contain simple triods. is a local homeomorphism, Σ . IH does not contain simple closed curves. Step 2. Σ . IH contains a simple closed curve C. . Then (σ ×1Q )(C) . Suppose Σ . is an arc A (Theorem is a subcontinuum of Σ. Hence, (σ × 1Q )(C) 2 of [11]). Let a be one of the end points of A. Let U be an evenly covered open set of Σ containing a. Let x ∈ U \ A be in the same arc component of U containing a; so there is an arc xa from x to a . be such that contained in U , such that xa ∩ A = {a}. Let a ˜∈C . be an open set of Σ . IH containing a (σ × 1Q )(˜ a) = a, and let U ˜ so . . that (σ × 1Q )|U. : U → → (σ × 1Q )(U ) = U is a homeomorphism. Let −1 . x˜ ∈ U ∩ (σ × 1Q ) (x). Then the arc xa can be lifted to an arc xã ˜, . . and xã ˜ ∩ C = {˜ a}. This implies that ΣIH contains a simple triod, . IH does not contain a simple a contradiction to Step 1. Therefore, Σ closed curve. . IH , the Given two points p˜ and q˜ in the same arc component of Σ . IH that have p˜ as an end point and contain union of all the arcs in Σ q˜ is called a ray starting at p˜. . in Σ . IH is the union of a countable number of Step 3. A ray R arcs. . and let {˜ Let p˜ be the starting point of R, p n }∞ n=1 be a count. . able dense subset of R. We assert that R is the union of the arcs ∞ .\ {˜ ppñ }∞ ˜ in R p˜pñ . Consider n=1 . Suppose there exists a point r n=1

the arc p˜r˜. Now, r˜ must be in the interior of an arc. Thus, p˜r˜ may be extended to an arc p˜s˜ so that r˜ is contained in the relative . IH contains no simple triods, each pñ belongs interior of p˜s˜. Since Σ . to the arc p˜r˜. This implies that {˜ pn }∞ n=1 is not dense in R, since no ∞ .= pñ is near s˜, a contradiction. Therefore, R p˜pñ . n=1

. IH , A. is the Step 4. For each point p˜ of an arc component A. of Σ .1 and R .2 starting at p˜ such that R .1 ∩ R .2 = {˜ union of two rays R p}.



273

. IH does We know that p˜ is in the interior of an arc a ˜˜b. Since Σ . not contain simple triods, we have that A is the union of two rays . IH does starting at p˜ going through a˜ and ˜b, respectively. Since Σ not contain simple closed curves, these rays intersect only at p˜. . IH has uncountably many arc components. Step 5. Σ . IH has only countably many arc components. Then Suppose Σ . IH is the union of countably many arcs. By Theorem 1.5.12, one Σ . IH , a contradiction to the fact of these arcs contains an open set of Σ . IH is not locally connected. that Σ . IH are unbounded. Step 6. The arc components of Σ . We . IH , and let p˜ ∈ Cl . (A). Let A. be an arc component of Σ ΣIH . = Cl . (A.p˜), where A.p˜ is the arc component show that ClΣ. (A) ΣIH IH . of ΣIH containing p˜. Suppose there exists a point q˜ ∈ Cl . (A.p˜) \ ΣIH

˜ q , Cl . (A)) . Let ε ≤ d(˜ . such that each ε–homeomorphism ClΣ. (A). ΣIH IH . IH (Theorem 5.2.13). Hence, by Theorem 4.2.31, may be lifted to Σ there exists an Effros number δ > 0 for this ε. Let a ˜ ∈ A. such that ˜ ˜: Σ . IH → . IH d(˜ a, p˜) < δ. Hence, there exists an ε–homeomorphism h →Σ ˜ p) = a ˜ q ) ∈ Cl . (A); . this is a contradicsuch that h(˜ ˜. Note that h(˜ ΣIH . tion. Thus, Cl . (A.p˜) ⊂ Cl . (A). ΣIH

ΣIH

. ⊂ Cl . (A.p˜). A symmetric argument shows that ClΣ. (A) ΣIH IH . = Cl . (A.p˜). Therefore, ClΣ. (A) ΣIH IH Let . | A. is an arc component of Σ . IH }. G = {ClΣ. (A) IH

The above argument shows that the elements of G are pairwise disjoint. . IH is bounded. Then all of Suppose that one arc component of Σ them are. Hence, G is an uncountable collection of continua (Step 5). Since each arc component of Σ is dense, σ × 1Q sends each element of G onto Σ, and this contradicts the fact that the fibres of σ × 1Q are countable. . IH . IH is a closed subset of Σ Step 7. Each arc component of Σ which is homeomorphic to IR. Copyright © 2005 Taylor & Francis Group, LLC

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Since Σ does not contain simple closed curves, by Corollary 8 of [31], there is a one–to–one map j : IR → Σ such that j(IR) is a . IH , then let dense arc component of Σ. If IKIH is a component of Σ j˜: IR → IKIH be a lift of j so that j˜(IR) ⊂ IKIH (Lemma 79.1 of [24]). Let ε > 0 be a positive number such that each ε–homeomorphism . IH (Theorem 5.2.13), and let of Σ lifts to an ε–homeomorphism of Σ 2δ > 0 be an Effros number for this ε (Theorem 4.2.31). Let v ∈ W be the common point of C1 and C2 . Cover {v} × Q with a finite collection B of δ–balls. Since the arc components of .0 , . IH are unbounded (Step 6) and B is finite, we can find an arc A Σ .0 contained in j˜(IR), whose end points are in two different liftings, B .1 , of the same element B of B. and B .k , k ∈ {0, 1}. Let (ϕ × 1Q ) : IH × .0 = a ˜0 a ˜1 , where a ˜k ∈ B Let A .0 onto B .1 . Q → IH×Q be the covering homeomorphism that maps B ˜ . . . Since (ϕ×1Q )(˜ a0 ) ∈ B1 , there is an ε–homeomorphism h : ΣIH → ΣIH ˜ ˜ .1 = A .0 ∪ h◦(ϕ×1)( .0 ) is an such that h◦(ϕ×1Q )(˜ a0 ) = a ˜1 . Then A A ˜ .0 ∩ h◦(ϕ×1)( .0 ) and Σ . IH does not contain either arc, because a ˜1 ∈ A A ˜ simple triods or simple closed curves. If we call a ˜2 = h◦(ϕ×1 a1 ), Q )(˜ .1 = a we may write A ˜0 a ˜1 a ˜2 . ˜ ◦ (ϕ × ˜ .1 ) is an arc, and A .1 ∩ h Now observe that h ◦ (ϕ × 1Q )(A ˜ ◦ (ϕ × 1Q )(A .1 ) = a .2 = A .1 ∪ h .1 ) is 1Q )(A ˜1 a ˜2 . Thus, we have that A ˜ ◦ (ϕ × 1Q )(˜ .2 = a an arc. If a ˜3 = h a2 ) we may write A ˜0 a ˜1 a ˜2 a ˜3 . ∞ .n }n=1 of arcs If we continue in this way, we obtain a sequence {A so that the “right” end points of the arcs tend to the attracting point z + of ϕ × 1Q . −1 ˜ Similarly, using h ◦ (ϕ × 1Q ) , we can construct a sequence .−n }∞ of arcs having their “left” end points tending to the re{A n=1 pelling point z − of ϕ × 1Q . Thus, j˜(IR) is closed, and each point of j˜(IR) has a neighborhood homeomorphic to an open interval. Therefore j˜(IR) is homeomorphic to IR. . IH are homeomorphic to IR. Step 8. The components of Σ . IH . . IH is a component of Σ We prove that each arc component of Σ By Corollary of [2], we only need to show that each arc component is a quasicomponent, and for this, it is enough to prove that if A.1 Copyright © 2005 Taylor & Francis Group, LLC


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. IH , then there is a closed and A.2 are two distinct arc components of Σ . . and open subset of ΣIH containing A1 which is disjoint from A.2 . . IH . Let Let A.1 and A.2 be two different arc components of Σ . . a ˜1 ∈ A1 . Since ΣIH is locally homeomorphic to Σ, we can find a . IH of the form [r, s] × C, such that (r, s) × C is an compact set of Σ . IH containing a˜1 . Let U . be a closed and open subset open subset of Σ . ∩ A.2 = ∅. of (r, s) × C containing a˜1 such that U . IH such that Let V. be the union of all the arc components A. of Σ . . . . . . A ∩ U = ∅. Clearly A1 ⊂ V and A2 ∩ V = ∅. We show that V. is a . IH . closed and open subset of Σ . First, we show that V is open. Let x˜ ∈ V. and let A.x˜ be the . IH containing x˜. Let x˜1 ∈ A.x˜ ∩ U . . Take ε > 0 arc component of Σ . IH (Thesuch that each ε–homeomorphism of Σ may be lifted to Σ d˜ . . Let δ > 0 be an Effros number x1 ) ⊂ U orem 5.2.13), and Vε (˜ ˜ x). Then there is an for this ε (Theorem 4.2.31). Let x˜2 ∈ Vδd (˜ ˜ ˜ x) = x˜2 . Hence, . . ε–homeomorphism h : ΣIH → → ΣIH such that h(˜ ˜ x) ∈ A.x˜ , A.x˜ being the arc component of Σ . IH containing x˜2 . h(˜ 2 2 ˜ x1 ), x˜1 < ε. Thus, A.x˜ ∩ U . = ∅. Consequently, Therefore, d˜ h(˜ 2 x˜2 ∈ V. . Therefore, V. is open. Now, we prove that V. is closed. Let x˜ ∈ ClΣ. (V. ), and let A.x˜ be IH . IH containing x˜. For every n ∈ IN, let δn > 0 the arc component of Σ 1 ˜ x). For each n ∈ IN, be an Effros number for . Let xñ ∈ V. ∩ Vδdn (˜ n 1 ˜ n (˜ ñ : Σ . IH → . IH so that h there is a –homeomorphism h →Σ xn ) = x˜. n . IH containing xñ . Let xñ ∈ Let A.xñ be the arc component of Σ ˜ ˜ ˜ . . . Axñ ∩ U . Then for each n ∈ IN, hn (˜ x ) ∈ Ax˜ , and d x˜ , hn (˜ x ) < n

n

n

1 . n

. is compact, without loss of generality, we assume that Since U . . Let ε > 0 be the sequence, {˜ xn }∞ ˜ , where x˜ ∈ U n=1 , converges to x ε ε 1 xn , x˜ ) < . Then given, and let n ∈ IN be such that < and d(˜ n 2 2 ε 1 ˜ n (˜ ˜ n (˜ d˜ x˜ , h xn ) ≤ d(˜ x , xñ ) + d˜ xñ , h xn ) < + < ε. 2 n ˜ Hence, for every ε > 0, Vεd (˜ x )∩A.x˜ = ∅. Therefore, x˜ ∈ ClΣ. (A.x˜ ) = IH


276


. = ∅. Thus x˜ ∈ V. , and V. is closed. A.x˜ (Step 7), and A.x˜ ∩ U . IH is homeomorphic to IR × C ∗ . Therefore, Σ Q.E.D. 5.5.3. Theorem. Let M be the Menger universal curve. If fIH : M . IH → W is a map that is not homotopic to a constant map, then M has countably many components, and for each component IKIH of . IH , YIK = Cl(IH×Q)∪S 1 (IKIH ) is homeomorphic to M. M IH . IH is homeomorphic to IKIH × IB . IH . Proof. By Theorem of [2], M . IH is locally homeomorphic to M, and M is locally connected, Since M . IH is an at most countable and discrete space. Therefore, IKIH is IB locally homeomorphic to M. Thus, IKIH is connected, locally arcwise connected (hence arcwise connected), and locally compact. Note . of IKIH and each point z˜ ∈ U ., that for each connected subset U . U \ {˜ z } is connected, and no open set of IKIH can be embedded in the plane. By Theorem 5.3.21, YIKIH is connected im kleinen at each point of IE(IKIH ) = YIKIH \ IKIH . Hence YIKIH is locally connected. Now, we show that for each point z˜ of point of IE(IKIH ) and each . of YIK , such that z˜ ∈ U ., U . \ {˜ connected subset U z } is connected. IH . Let z˜ ∈ IE(IKIH ) and let U be a connected open set of YIKIH . is arcwise connected. Let U . be an containing z˜. Observe that U . ) ⊂ open connected subset of YIKIH containing z˜ such that ClYIK (U IH . . The projection of U . \IE(IKIH ) into W / contains an infinite number U / contained / . Let V. be the set of all vertices v˜ of W of vertices of W . / . , in the projection of U \ IE(IKIH ) into W for which Qv˜ ∩ IKIH ⊂ U where Qv˜ = {˜ v } × Q. Let v ∈ W be the common point of C1 and C2 . Cover ({v} × Q) ∩ M with a finite open cover B, where each element of B is evenly covered and arcwise connected. We take the elements of B small enough such that for each v˜ ∈ V. , the lifting Bv˜ of B, covering .. Qv˜ ∩ IKIH , is such that Bv˜ ⊂ U . \ {˜ . \ {˜ Suppose that U z } is not connected. Then U z } can be . . . . .. written as U1 ∪ U2 , where U1 and U2 are disjoint open sets of U Copyright © 2005 Taylor & Francis Group, LLC


277

We assert that there is an infinite subset V. of V. with the prop.1 or in U .2 . erty that if v˜ ∈ V. , then Bv˜ is totally contained in U To show this, let v˜ ∈ V. , and suppose that Bv˜ = {Bv˜ ,1 , . . . , Bv˜ , }. Let x˜v˜ ,j ∈ Bv˜ ,j . Since IKIH is arcwise connected, for each j ∈ {1, . . . , − 1}, there is an arc αj : [0, 1] → IKIH such that αj (0) = −1 x˜v˜ ,1 and αj (1) = x˜v˜ ,j+1 . Since αj ([0, 1]) is compact, there j=1

is a covering (ϕ × 1Q ) : IKIH → IKIH such that

−1 homeomorphism .. (ϕ × 1Q ) αj ([0, 1]) ⊂ U j=1

v ) = v˜. Since Let (ϕ × 1Q )(˜

Bv˜ ∪ (ϕ × 1Q )

−1

αj ([0, 1])

.1 or in U .2 . Hence is connected, it is contained in U Bv˜ is totally .1 or in U .2 . Since this was true for every v˜ ∈ V. , such contained in U a subset V. of V. exists. .2 . .1 or in U Bv˜ is contained in U Now, we assert that j=1

v˜∈V.

.1 and Bv˜1 ⊂ U To see this, let v˜1 and v˜2 in V. and suppose that .2 . Let x˜k ∈ Bv˜2 ⊂ U Bv˜k , k ∈ {1, 2}, then x˜k = (w˜k , qk ), where / and qk ∈ Q. Now, w ˜k and z˜ determine a unique arc [w˜k , z˜] w˜k ∈ W / in ClIH (W ), from w˜k to z˜ (p. 284 of [34]). Now [w˜1 , z˜] ∩ [w˜2 , z˜] . contains an element v˜ of V . Without loss of generality, we assume .2 . Since U . is arcwise connected, there that Bv˜ is contained in U . such that α(0) = x˜1 and α(1) = x˜2 . Since exists an α : [0, 1] → U .k , there is a t ∈ [0, 1] such that α(t) = z˜. On the other x˜k ∈ U hand, any arc from x˜1 to z˜ must Qv˜ ∩ IKIH , which is a intersect .1 or in U .2 , Bv˜ is contained in U contradiction. Therefore, v˜∈V.

.1 . say U . \ {˜ .2 is empty. Hence, U z } is A similar argument shows that U connected. Therefore, by Theorem 5.4.3, YIKIH is homeomorphic to M. Q.E.D. Copyright © 2005 Taylor & Francis Group, LLC

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In the following Theorem we construct a covering space of the Case continuum whose components are locally connected. 5.5.4. Theorem. The Case continuum C has a covering space with locally connected components. Proof. Note that, by Theorem 5.4.16, C ⊂ Σ × [0, 1]2 , where Σ is the dyadic solenoid. First, we use inverse limits to construct a covering space of Σ × [0, 1]2 . Let Yn = S 1 × [0, 1]2 , and let Y.n = IR × [0, 1]2 × {2n−1 th roots of unity}. Define the bonding maps fnn+1 : Yn+1 → → Yn and fñn+1 : Y.n+1 → → Y.n by fnn+1 ((z, x)) = (z 2 , x), and fñn+1 ((r, x, t)) = (2r, x, t2 ), respectively. Let qn : Y.n → Yn be the covering map given by qn ((r, x, t)) = (t exp(2πr), x). Thus we can consider the following infinite ladder: fn

n+1

n−1 . fn . · · · ←− Y.n−1 ⏐ ←− Y⏐n ←− Yn+1 ⏐ ←− · · · ⏐qn−1 ⏐qn ⏐qn+1 & & & · · · ←− Yn−1 ←− Yn ←− Yn+1 ←− · · · n+1 n fñ−1

fñ

:⏐ Y.∞ ⏐q∞ & : Y∞

where Y.∞ = lim{Y.n , fñn+1 }, Y∞ = lim{Yn , fnn+1 } and q∞ = lim{qn }. ←−

←−

←−

Note that Y.∞ is homeomorphic to IR × [0, 1]2 × C and Y∞ is homeomorphic to Σ × [0, 1]2 . Let (IR × [0, 1]2 × {t})n+1 be a component of Y.n+1 . Note that the restriction of the map fñn+1 to this component, fñn+1 |(IR×[0,1]2 ×{t})n+1 : → (IR × [0, 1]2 × {t2 })n ⊂ Y.n , is (IR × [0, 1]2 × {t})n+1 ⊂ Y.n+1 → n a homeomorphism, since the map gn+1 : (IR × [0, 1]2 × {t2 })n ⊂ n Y.n → → (IR × [0, 1]2 × {t})n+1 ⊂ Y.n+1 given by gn+1 ((r, x, t2 )) = ( 12 r, x, t) is its inverse. 2 ˜ Let t˜ = (tn )∞ n=1 ∈ C, and let (IR × [0, 1] × {t})∞ be a component of Y.∞ . Consider the map fñ |(IR×[0,1]2 ×{t˜})∞ : (IR × [0, 1]2 × {t˜})∞ ⊂ → (IR × [0, 1]2 × {tn })n ⊂ Y.n . We claim that this map is a Y.∞ → homeomorphism. To see this, let us define the map g n : (IR × [0, 1]2 × {tn })n ⊂ Y.n → → (IR × [0, 1]2 × {t˜})∞ ⊂ Y.∞ by



279

n ((r, x, tn )), (r, x, tn ), g n ((r, x, tn )) = (f1n ((r, x, tn )), . . . , fn−1 n n gn+1 ((r, x, tn )), . . . , gn+m ((r, x, tn )), . . . ) n+m−1 n n = gn+m ◦ · · · ◦ gn+1 . Since g n ◦ fñ |(IR×[0,1]2 ×{t˜})∞ = where gn+m 1(IR×[0,1]2 ×{t˜})∞ , and fñ |(IR×[0,1]2 ×{t˜})∞ ◦ g n = 1(IR×[0,1]2 ×{tn })n , we have that fñ |(IR×[0,1]2 ×{t˜})∞ is a homeomorphism. Now, assume that the Menger curve M is embedded in Y1 , as described in Lemma 5.4.12. Since (f12 )−1 (M) is homeomorphic to M, setting X1 = M, X2 = (f12 )−1 (M), . . . , Xn = (f1n )−1 (M), we obtain that C = lim{Xn , fnn+1 }. ←−

.n is a locally compact, metric .n = q −1 (Xn ). Note that X Let X n n−1 space, with 2 components, and each component is locally homeomorphic to M. . = lim{X . is a locally compact, metric .n , fñn+1 }. Then C Let C ←−

. To see that IK is locally space. Let IK be a component of C. connected, note that IK is contained in one of the components of Y.∞ . Let (IR × [0, 1]2 × {t˜})∞ be the component of Y.∞ containing IK. Since f˜1 |(IR×[0,1]2 ×{t˜})∞ is a homeomorphism, we have that f˜1 |IK .1 . Since X .1 is also a homeomorphism onto f˜1 (IK), but f˜1 (IK) = X is locally homeomorphic to M, IK is locally homeomorphic to M. Thus IK is locally connected, because M is locally connected. Q.E.D. To finish this chapter, we show two different covering spaces of one of the Minc–Rogers continua (Theorem 5.4.18). 5.5.5. Theorem. Let Λ be the constant sequence {(2, 2, 1)}. If MR = MΛ , then MR has a covering space whose components are not locally connected. Proof. Note that, by Theorem 5.4.18, MR ⊂ Σ × Σ × S 1 , where Σ is the dyadic solenoid. First, we construct a covering space for Σ × Σ × S 1 using inverse limits. Copyright © 2005 Taylor & Francis Group, LLC

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Let Yn = S 1 × S 1 × S 1 , and Y.n = IR × S 1 × S 1 × {2n−1 th roots of unity}. Define the bonding maps fnn+1 : Yn+1 → → Yn and fñn+1 : Y.n+1 → → Y.n by fnn+1 ((z1 , z2 , z3 )) = (z12 , z22 , z3 ) and fñn+1 ((r, z2 , z3 , t)) = (2r, z22 , z3 , t2 ), respectively. Let qn : Y.n → → Yn be the map given by qn ((r, z2 , z3 , t)) = (t exp(2πr), z2 , z3 ). Note that qn is a covering map. Hence, we have the following infinite ladder: fn

n+1


fñ

:⏐ Y.∞ ⏐q∞ & : Y∞

where Y.∞ = lim{Y.n , fñn+1 }, Y∞ = lim{Yn , fnn+1 } and q∞ = lim{qn }. ←−

←−

←−

Note that Y.∞ is homeomorphic to IR × Σ × S 1 × C and Y∞ is homeomorphic to Σ × Σ × S 1 . 1 ˜ Let t˜ = (tn )∞ n=1 ∈ C, and let (IR × Σ × S × {t})∞ be a com. ponent of Y∞ . We assert that q∞ |(IR×Σ×S 1 ×{t˜})∞ is one–to–one. To see this, let (˜ r, s˜, x˜, t˜) = ((rn , sn , xn , tn ))∞ r , s˜ , x˜ , t˜) = n=1 and (˜ 1 ˜ ((rn , sn , xn , tn ))∞ n=1 be two points of (IR × Σ × S × {t})∞ such that q∞ ((˜ r, s˜, x˜, t˜)) = q∞ ((˜ r , s˜ , x˜ , t˜)). This equality implies that, for every n ∈ IN, qn ((rn , sn , xn , tn )) = qn ((rn , sn , xn , tn )). Then (tn exp(2πrn ), sn , xn ) = (tn exp(2πrn ), sn , xn ). Thus, for each n ∈ IN, sn = sn , xn = xn and tn exp(2πrn ) = tn exp(2πrn ). From this last equality we obtain that exp(2πrn ) = exp(2πrn ). Hence, for every n ∈ IN, there is an mn ∈ ZZ such that 2πrn +2πmn = 2πrn , from here, rn + mn = rn . For n = 1, we have that r1 = r1 + m1 . Then r2 = 12 r1 = 12 r1 + 12 m1 , r3 = 12 r2 = 14 r1 + 14 m1 , etc. Now, if m1 = 0, there is a k ∈ ZZ such that 21k m1 ∈ ZZ, and, on the other hand, rk+1 = 21k r1 + 21k m1 , a contradiction. Hence, m1 = 0. Similarly, mn = 0 for all n ∈ IN. Thus, rn = rn , and (˜ r, s˜, x˜, t˜) = (˜ r , s˜ , x˜ , t˜). Note that if (˜ r, s˜, x˜, t˜) ∈ (IR×Σ×S 1 ×C)∞ , then q∞ ((˜ r, s˜, x˜, t˜)) = (s , s˜, x˜), i.e., q∞ sends the second and third coordinates of a point in IR × Σ × S 1 × C identically to the second and third coordinates of its image in Σ × Σ × S 1 . Copyright © 2005 Taylor & Francis Group, LLC


281

˜ × C ) × (˜ . = (˜ e, f˜) × C be a basic open set of Let U a, ˜b) × ((˜ c, d) ˜ × C ) × (˜ . ) = ((a, b) × C ) × ((˜ Y.∞ . Then q∞ (U c, d) e, f˜). Hence, if ˜ ˜ ˜ . Uc = (˜ a, b) × ((˜ c, d) × C ) × (˜ e, f ) × {t˜} is an open set of a component ˜ × C ) × (˜ . . of Y∞ , then q∞ (Uc ) = ((a, b) × {t}) × ((˜ c, d) e, f˜). Assume that the Menger curve M is embedded in Y1 in such a way that {e} × S 1 × {e} ⊂ M and S 1 × {e} × {e} ⊂ M (Lemma 5.4.17). Since (f12 )−1 (M) is homeomorphic to M, taking X1 = M, X2 = (f12 )−1 (M), . . . , Xn = (f1n )−1 (M), we obtain that MR = lim{Xn , fnn+1 }. ←−

.n = qn−1 (Xn ). Observe that X .n is a locally compact, metric Let X n−1 space, with 2 components, each of which is locally homeomor˜ × 0 .n , fñn+1 }. Let U . = (˜ phic to M. Let MR = lim{X a, ˜b) × ((˜ c, d) ←−

0 is a . ∩ MR e, f˜) × C be a basic open set of Y.∞ , then U C ) × (˜ 0 We can take U . such that q∞ | . is a homebasic open set of MR. U −1 0 . 0 omorphism. Since # MR = q∞ (MR), we have that q∞$(U ∩ MR) = ˜ × C ) × (˜ . ) ∩ MR = ((a, b) × C ) × ((˜ q∞ (U c, d) e, f˜) ∩ MR. But # $ ˜ ˜ this is homeomorphic to (a, b) × (˜ c, d) × (˜ e, f ) × C × C ∩ MR. Since MR is locally homeomorphic to the product of an open set of # M and a Cantor set, there is $ an open set V of MR so that ˜ ˜ (a, b) × (˜ c, d) × (˜ e, f ) × C × C ∩ MR is homeomorphic to V × ˜ × C ) × (˜ 0 is a, ˜b) × ((˜ c, d) e, f˜) × {t˜}] ∩ MR) C × C . Hence q∞ ([(˜ 0 are homeomorphic to V × C × {t}. Then the components of MR locally homeomorphic to V × C . In particular they are not locally connected. Q.E.D. 5.5.6. Theorem. Let Λ be the constant sequence {(2, 2, 1)}. If MR = MΛ , then MR has a covering space whose components are locally connected. Proof. Let the inverse sequences {Yn , fnn+1 } and {Xn , fnn+1 } be as in the proof of Theorem 5.5.5. Hence, MR = lim{Xn , fnn+1 }. ←−

Let Y.n = IR×IR×S 1 ×{2n−1 th roots of unity}, and let fñn+1: Y.n+1 → Y.n be given by fñn+1 ((r1 , r2 , z3 , t)) = (2r1 , 2r2 , z3 , t2 ). Define the Copyright © 2005 Taylor & Francis Group, LLC

282


covering map qn : Y.n → Yn by qn ((r1 , r2 , z3 , t)) = (t exp(2πr1 ), t exp(2πr2 ), z3 , t). Thus, we may construct the following infinite and commutative ladder: fn

n+1


fñ

:⏐ Y.∞ ⏐q∞ & : Y∞

where Y.∞ = lim{Y.n , fñn+1 } and q∞ = lim{qn }. Note that Y.∞ is ←−

←−

homeomorphic to IR × IR × S 1 × C. Let (IR×IR×S 1 ×{t})n+1 be a component of Y.n+1 . Note that the restriction of bonding maps fñn+1 |(IR×IR×S 1 ×{t})n+1 : (IR × IR × S 1 × {t})n+1 ⊂ Y.n+1 → (IR × IR × S 1 × {t2 })n ⊂ Y.n is a homeomorphism, n since the following map gn+1 : (IR×IR×S 1 ×{t2 })n ⊂ Y.n → (IR×IR× n ((r1 , r2 , z3 , t2 )) = ( 12 r1 , 12 r2 , z3 , t) S 1 × {t})n+1 ⊂ Y.n+1 given by gn+1 is its inverse. n+m−1 n n Define gn+m = gn+m ◦ · · · ◦ gn+1 . Let t˜ = (tn )∞ n=1 ∈ C, and let 1 . ˜ (IR × IR × S × {t})∞ be a component of Y∞ . Consider the map fñ |(IR×IR×S 1 ×{t˜})∞ : (IR × IR × S 1 × {t˜})∞ ⊂ Y.∞ → (IR × IR × S 1 × {tn })n ⊂ Y.n . We claim that this map is a homeomorphism. To this end, let g n : (IR×IR×S 1 ×{tn })n ⊂ Y.n → (IR×IR×S 1 ×{t˜})∞ ⊂ Y.∞ be given by n ((r1 , r2 , z3 , tn )), g n ((r1 , r2 , z3 , tn )) = (f1n ((r1 , r2 , z3 , tn )), . . . , fn−1 n n ((r1 , r2 , z3 , tn )), . . . , gn+m ((r1 , r2 , z3 , tn )), . . . ). (r1 , r2 , z3 , tn ), gn+1

Since it is easy to see that g n ◦fñ |(IR×IR×S 1 ×{t˜})∞ = 1(IR×IR×S 1 ×{t˜})∞ and fñ |(IR×IR×S 1 ×{t˜})∞ ◦ g n = 1(IR×IR×S 1 ×{tn })n , fñ |(IR×IR×S 1 ×{t˜})∞ is a homeomorphism. .n = qn−1 (Xn ), then, X .n is a locally compact, metric space Let X n−1 with 2 components, each of which is locally homeomorphic to 0 0 .n , fñ+1 }, and let IK be a component of MR. M. Let MR = lim{X n ←−

IK is contained in a component of Y.∞ . Let (IR × IR × S 1 × {t˜})∞ be such a component. Since f˜1 |(IR×IR×S 1 ×{t˜})∞ is a homeomorphism, Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

283

.1 . Since X .1 is locally then f˜1 |IK is a homeomorphism onto f˜1 (IK) = X homeomorphic to M, IK is locally homeomorphic to M. Therefore, IK is locally connected. Q.E.D.

REFERENCES

[1] J. M. Aarts and P. Van Emde Boas, Continua as Remainders in Compact Extensions, Nieuw Archief voor Wiskunde 3 (1967), 34–37. [2] J. M. Aarts and L. G. Oversteegen, The Product Structure of Homogeneous Spaces, Indag. Math., 1 (1990), 1–5. [3] L. V. Ahlfors, Conformal Invariants, Topics in Geometric Function Theory, Series in Higher Mathematics, McGraw–Hill Book Co., New York, 1973. [4] R. D. Anderson, A Characterization of the Universal Curve and a Proof of its Homogeneity, Ann. Math., 67 (1958), 313– 324. [5] R. D. Anderson, One–dimensional Continuous Curves and a Homogeneity Theorem, Ann. Math., 68 (1958), 1–16. [6] D. P. Bellamy and L. Lum, The Cyclic Connectivity of Homogeneous Arcwise Connected Continua, Trans. Amer. Math. Soc., 266 (1981), 389–396. [7] J. H. Case, Another 1–dimensional Homogeneous Continuum Which Contains an Arc, Pacific J. Math., 11(1961), 455–469. [8] C. O. Christenson and W. L. Voxman, Aspects of Topology, Monographs and Textbooks in Pure and Applied Math., Vol. 39, Marcel Dekker, Inc., New York, Basel, 1977. [9] R. Fenn, What is the Geometry of a Surface?, The Amer. Math. Monthly, 90 (1983), 87–98. Copyright © 2005 Taylor & Francis Group, LLC

284

REFERENCES

[10] J. Grispolakis and E. D. Tymchatyn, On Confluent Mappings and Essential Mappings–A Survey, Rocky Mountain J. Math., 11 (1981), 131–153. [11] C. L. Hagopian, A Characterization of Solenoids, Pacific J. Math., 68 (1977), 425–435. [12] J. G. Hocking and G. S. Young, Topology, Dover Publications, Inc., New York, 1988. [13] S. T. Hu, Theory of Retracts, Wayne State University Press, Detroit, 1965. [14] J. Krasinkiewicz, On One–point Union of Two Circles, Houston J. Math., 2 (1976), 91–95. [15] J. Krasinkiewicz, On Two Theorems of Dyer, Colloq. Math., 50 (1986), 201–208. [16] K. Kuratowski, Topology, Vol. II, Academic Press, New York, N. Y., 1968. [17] E. L. Lima, Grupo Fundamental e Espa¸cos de Recobrimento, Instituto de Matemática Pura e Aplicada, CNPq, (Projeto Euclides), 1993. (Portuguese) [18] J. C. Mac´ıas, El Teorema de Descomposición Terminal de Rogers, Tesis de Maestr´ıa, Facultad de Ciencias F´ısico Matemáticas, B. U. A. P., 1999. (Spanish) [19] S. Mac´ıas, Covering Spaces of Homogeneous Continua, Topology Appl., (1994), 157–177. [20] S. Mac´ıas, Homogeneous Continua for Which the Set Function T is Continuous, to appear in Houston Journal of Mathematics. [21] T. Maćkowiak and E. D. Tymchatyn, Continuous Mappings on Continua II, Dissertationes Math., 225 (1984), 1–57. [22] M. C. McCord, Inverse Limit Sequences with Covering Maps, Trans. Amer. Math. Soc., 114 (1965), 197–209. Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

285

[23] P. Minc and J. T. Rogers, Jr., Some New Examples of Homogeneous Curves, Topology Proc., 10 (1985), 347–356. [24] J. Munkres, Topology, second edition, Prentice Hall, Upper Saddle River, NJ, 2000. [25] S. B. Nadler, Jr., Dimension Theory: An Introduction with Exercises, Aportaciones Matemáticas, Serie Textos # 18, Sociedad Matemática Mexicana, 2002. [26] J. T. Rogers, Jr., Homogeneous Separating Plane Continua are Decomposable, Michigan J. Math., 28 (1981), 317–321. [27] J. T. Rogers, Jr., Decompositions of Homogeneous Continua, Pacific J. Math., 99 (1982), 137–144. [28] J. T. Rogers, Jr., Homogeneous Hereditarily Indecomposable Continua Are Tree–like, Houston J. Math., 8, (1982), 421–428. [29] J. T. Rogers, Jr., An Aposyndetic Homogeneous Curve That is not Locally Connected, Houston J. Math., 9 (1983), 433– 440. [30] J. T. Rogers, Jr., Cell–like Decompositions of Homogeneous Continua, Proc. Amer. Math. Soc., 87 (1983), 375–377. [31] J. T. Rogers, Jr., Homogeneous Curves That Contain Arcs, Topology Appl., 21 (1985), 95–101. [32] J. T. Rogers, Jr., Orbits of Higher–dimensional Hereditarily Indecomposable Continua, Proc. Amer. Math. Soc., 95 (1985), 483–486. [33] J. T. Rogers, Jr., Hyperbolic Ends And Continua, Michigan Math. J., 34 (1987), 337–347. [34] J. T. Rogers, Jr., Decompositions of Continua Over the Hyperbolic Plane, Trans. Amer. Math. Soc., 310 (1988), 277–291. [35] J. T. Rogers, Jr., Higher Dimensional Aposyndetic Decompositions, Proc. Amer. Math. Soc., 131 (2003), 3285–3288. Copyright © 2005 Taylor & Francis Group, LLC

286

REFERENCES

[36] P. Scott, The Geometries of 3–manifolds, Bull. London Math. Soc., 15 (1983), 401–487. [37] G. T. Whyburn, Topological Characterization of the Sierpi´ nski Curve, Fund. Math., 45 (1958), 320–324.


Chapter 6 n–FOLD HYPERSPACES

We give a brief overview of n–fold hyperspaces. Other hyperspaces have been studied extensively in [40] and [21]. Throughout this chapter, σ denotes the union map and H denotes the Hausdorff metric.

6.1

General Properties

We present general properties of n–fold hyperspaces.

6.1.1. Lemma. Let n ∈ IN, and let a continuum. If A is X be a connected subset of Cn (X), then A = {A | A ∈ A} has at most n components. Proof. Suppose the result is not true. Then there exists a connected subset A of Cn (X) such that A has at least n + 1 components. Thus, we can find n + 1 pairwise separated subsets, C1 , . . . , Copyright © 2005 Taylor & Francis Group, LLC

287

CHAPTER 6. n–FOLD HYPERSPACES

288

Cn+1 , of X such that

A=

n+1

Cj . Let

j=1

B=

n Cj A∈AA⊂

j=1

and D = {A ∈ A | A ∩ Cn+1 = ∅}. Then B and D are separated subsets of Cn (X) and A = B ∪ D, a contradiction. Therefore, A has at most n components. Q.E.D. 6.1.2. Corollary. Let n ∈ IN. If X is a continuum and A ∈ C (Cn (X)), then σ (A) ∈ Cn (X). Proof. By Lemma 1.8.11, σ (A) is a closed subset of X. By Lemma 6.1.1, σ (A) has at most n components. Therefore, σ (A) ∈ Cn (X). Q.E.D. 6.1.3. Notation. Let X be a continuum. To simplify notation, we write: U1 , . . . , Um n , to denote the intersection of the open set U1 , . . . , Um , of the Vietoris Topology, with Cn (X). 6.1.4. Theorem. Let n ∈ IN. Then the continuum X is locally connected if and only if Cn (X) is locally connected. Proof. If X is locally connected, then Wojdyslawski has shown that Cn (X) is an absolute retract (Théorème IIm of [47]). Hence, it is locally connected ((2.6) (p. 101) of [5]). Suppose Cn (X) is locally connected. Let x be a point in X and let U be an open subset of X containing x. Since Cn (X) is locally connected, there exists a connected open subset V of Cn (X) such that {x} ∈ V ⊂ Cl(V) ⊂ U n . Let V1 , . . . , Vk n be a basic open k set of Cn (X) such that {x} ∈ V1 , . . . , Vk n ⊂ V. Let V = Vj . j=1


6.1. GENERAL PROPERTIES

289

Now, if y ∈ V , then y ∈ σ(Cl(V)). Since {x} ∈ Cl(V), we have that σ(Cl(V)) ∈ C(X) ((1.49) of [40]). Thus, both x and y belong to σ(Cl(V)) ⊂ U . Hence, X is connected im kleinen at x. Since x was an arbitrary point of X, we have that X is locally connected (Theorem 1.7.12). Q.E.D. 6.1.5. Definition. A free arc in a metric space X is an arc α such that α \ {end points} is an open subset of X. A proof of the following Theorem may be found in Theorem 7.1 of [29]. 6.1.6. Theorem. If X is a locally connected continuum without free arcs, then Cn (X) is homeomorphic to the Hilbert cube for each n ∈ IN.

6.1.7. Theorem. Let X be a continuum. If n ∈ IN, then Cn (X) is nowhere dense in Cn+1 (X). In particular, Cn (X) is nowhere dense in 2X . Proof. Let n ∈ IN. Suppose IntCn+1 (X) (Cn (X)) = ∅. Recall that F(X) is dense in 2X (see the proof of Corollary 1.8.9). Hence, X does not belong to IntCn+1 (X) (Cn (X)). Let A ∈ IntCn+1 (X) (Cn (X) \ {X}), and suppose that A1 , . . . , Ak are the components of A. Thus, k ≤ n. Then there exists ε > 0 such that VεH (A) ∩ Cn+1 (X) ⊂ IntCn+1 (X) (Cn (X)). Without loss of generality, we assume that ε is small enough that Vεd (Aj )∩Vεd (A ) = ∅ if and only if j = and j, ∈ {1, . . . , k}. Suppose that n−k = m and let x1 , . . . , xm+1 ∈ Vεd (A1 )\ A1 be m+1 distinct points. Let B = A∪{x1 , . . . , xm+1 }. Then B ∈ Cn+1 (X) \ Cn (X) and H(A, B) < ε. Therefore IntCn+1 (X) (Cn (X)) = ∅. Similarly, Int2X (Cn (X)) = ∅. Q.E.D.


290


6.1.8. Lemma. Let X be a continuum, with metric d. If n ∈ IN, then there exist n pairwise disjoint nondegenerate subcontinua of X. Proof. Let n ∈ IN, and let x1 , . . . , xn be n distinct points of X. Let η = min{d(xj , xk ) | j, k ∈ {1, . . . , n} and j = k}. Then η > 0 and Vηd (xj ) ∩ Vηd (xk ) = ∅ for each j, k ∈ {1, . . . , n} and j = k. By Corollary 1.7.23, there exists a subcontinuum Aj of X such that {xj } Aj ⊂ Vηd (xj ) for every j ∈ {1, . . . , n}. Therefore, A1 , . . . , An are n pairwise disjoint nondegenerate subcontinua of X. Q.E.D. 6.1.9. Theorem. Let X be a continuum. If n ∈ IN, then Cn (X) contains an n–cell. Proof. Let A1 , . . . , An be n pairwise disjoint nondegenerate subcontinua of X (Lemma 6.1.8). For each j ∈ {1, . . . , n}, let aj ∈ Aj , and let αj : [0, 1] → C(X) be an order arc such that αj (0) = {aj } and αj (1) = Aj (Theorem 1.8.20). Then the map ξ : [0, 1]n → Cn (X) given by ξ((t1 , . . . , tn )) = α1 (t1 ) ∪ · · · ∪ αn (tn ) is an embedding of [0, 1]n in Cn (X). Q.E.D. If the continuum X contains decomposable continua, more can be said. 6.1.10. Theorem. Let X be a continuum and let n ∈ IN. If X contains k pairwise disjoint decomposable subcontinua (k ≤ n), then Cn (X) contains a (k + n)–cell. Proof. First suppose k < n. Let M1 , . . . , Mk be k pairwise disjoint decomposable subcontinua of X. Suppose that Mj = Aj ∪Bj , where Aj and Bj are continua, for each j ∈ {1, . . . , k}. By the proof of (1.145) of [40], we may assume that for each j ∈ {1, . . . , k}, Aj ∩ Bj is connected, Aj \ (Aj ∩ Bj ) = ∅, Bj \ (Aj ∩ Bj ) = ∅, and [Aj \ (Aj ∩ Bj )] ∩ [Bj \ (Aj ∩ Bj )] = ∅. Let Ck+1 , . . . , Cn be n − k pairwise disjoint nondegenerate subcontinua of X such that Mj ∩ C = ∅ for every j ∈ {1, . . . , k} and every ∈ {k + 1, . . . , n}. For each j ∈ {1, . . . , k}, let αj : [0, 1] → C(Aj ) and βj : [0, 1] → C(Bj ) be Copyright © 2005 Taylor & Francis Group, LLC


291

order arcs such that αj (0) = Aj ∩ Bj , αj (1) = Aj , βj (0) = Aj ∩ Bj , and βj (1) = Bj (Theorem 1.8.20). For each ∈ {k + 1, . . . , n}, let x ∈ C . Let γ : [0, 1] → C(C ) be an order arc such that γ (0) = {x } and γ (1) = C , ∈ {k + 1, . . . , n}. Since [0, 1]k+n is homeomorphic to [0, 1]2k × [0, 1]n−k , we need an embedding of [0, 1]2k ×[0, 1]n−k into Cn (X). The map ξ : [0, 1]2k ×[0, 1]n−k → Cn (X) given by ξ((t1 , . . . , t2k ), (t1 , . . . , tn−k )) = n−k

k (αj (t2j−1 ) ∪ βj (t2j )) ∪ γk+ (t ) j=1

=1

is an embedding of [0, 1]k+n in Cn (X). When k = n, repeat the argument without using the γ’s. Q.E.D. As a consequence of Theorem 6.1.10, we have the following two Corollaries: 6.1.11. Corollary. If X is a continuum which is not hereditarily indecomposable, then dim(Cn (X)) ≥ n + 1. 6.1.12. Corollary. Let X be a continuum and let n ∈ IN. If X contains n pairwise disjoint decomposable subcontinua, then Cn (X) contains a 2n–cell.

6.1.13. Corollary. If X is a continuum containing an arc, then Cn (X) contains a 2n–cell for each n ∈ IN. 6.1.14. Theorem. Let X be a continuum. If n ∈ IN, then the following are equivalent: (1) 2X is contractible; (2) Cn (X) is contractible; (3) C(X) is contractible. Copyright © 2005 Taylor & Francis Group, LLC

292


Proof. Suppose 2X is contractible. Then there exists a map H : 2X × [0, 1] → 2X such that for each A ∈ 2X , H ((A, 0)) = A and H ((A, 1)) = X. Let H : 2X × [0, 1] → 2X be the segment homotopy associated with H defined by H((A, t)) =

{H (A, s) | 0 ≤ s ≤ t}.

Then H is continuous ((16.3) of [40]). Observe that for each A ∈ 2X , H((A, 0)) = A, H((A, 1)) = X, and H({A} × [0, 1]) is an order arc from A to X. Note that if A ∈ Cn (X) and B ∈ H({A}×[0, 1]), then B ∈ Cn (X) (Theorem 1.8.20). Therefore, G = H|Cn (X)×[0,1] : Cn (X)× [0, 1] → Cn (X) is a continuous function such that for each A ∈ Cn (X), G((A, 0)) = A and G((A, 1)) = X. Hence, Cn (X) is contractible. A similar argument shows that if Cn (X) is contractible, then C(X) is contractible. The other implication is contained in the proof of (16.7) of [40]. Q.E.D.

6.1.15. Definition. A continuum X is said to have the property of Kelley provided that given any ε > 0, there exists δ > 0 such that if a, b ∈ X, d(a, b) < δ, and a ∈ A ∈ C(X), then there exists B ∈ C(X) such that b ∈ B and H(A, B) < ε. This number δ is called a Kelley number for the given ε.

6.1.16. Corollary. If X is a continuum having the property of Kelley, then Cn (X) is contractible for each n ∈ IN. Proof. If X is a continuum having the property of Kelley, then 2X and C(X) are contractible ((16.15) of [40]). Hence, the result follows from Theorem 6.1.14. Q.E.D. 6.1.17. Lemma. Let n ∈ IN, and let X be a continuum having the property of Kelley. If W is a subcontinuum of Cn (X) having nonempty interior, then σ (W) ∈ Cn (X) and IntX (σ (W)) = ∅. Copyright © 2005 Taylor & Francis Group, LLC


293

Proof. By Corollary 6.1.2, σ (W) ∈ Cn (X). Let A ∈ IntCn (X) (W). Let A1 , . . . , Ak be the components of A. Then k ≤ n. Let ε > 0 be given such that VεH (A) ∩ Cn (X) ⊂ W, and such that Vεd (Aj ) ∩ Vεd (A ) = ∅ when j = . Let δ > 0 be a Kelley number for the given ε. Clearly, A ⊂ σ (W). For each j ∈ {1, . . . , k}, let aj ∈ Aj . For every j ∈ {1, . . . , k}, let xj ∈ Vδd (aj ). Since X has the property of Kelley, there exists a subcontinuum Bj of X such that xj ∈ Bj k and H(Aj , Bj ) < ε for each j ∈ {1, . . . , k}. Let B = Bj ; then B ∈ Cn (X). Note that Vεd (A) =

k

j=1 k

Vεd (Aj ), Vεd (B) =

j=1

Vεd (Bj ),

j=1

and H(Aj , Bj ) < ε for each j ∈ {1, . . . , k}; hence, H(A, B) < ε. k Thus, B ∈ W, and B ⊂ σ (W). Therefore, Vδd (aj ) ⊂ σ (W), and IntX (σ (W)) = ∅.

j=1

Q.E.D. 6.1.18. Theorem. Let n ∈ IN. If X is an indecomposable continuum having the property of Kelley, then X is the only point at which Cn (X) is locally connected. Proof. Using order arcs (Theorem 1.8.20), it is easy to see that for each ε > 0, VεH (X) ∩ Cn (X) is arcwise connected. Suppose that Cn (X) is locally connected at the point A = X. Since Cn (X) is locally connected at A, there exists a subcontinuum W of Cn (X) such that A ∈ IntCn (X) (W) and σ (W) = X. By Lemma 6.1.17, σ (W) ∈ Cn (X) and IntX (σ (W)) = ∅. Since σ (W) has finitely many components and IntX (σ (W)) = ∅, it follows that at least one of the components of σ (W) has nonempty interior, which is impossible because σ (W) = X and X is an indecomposable continuum (Corollary 1.7.21). Q.E.D. As a consequence of the previous Theorem, we have the following result:



294

6.1.19. Theorem. Let n ∈ IN. If X is a hereditarily indecomposable continuum, then X is the only point at which Cn (X) is locally connected. Proof. Since hereditarily indecomposable continua have the property of Kelley ((16.27) of [40]), we have that the result follows from Theorem 6.1.18. Q.E.D. The following result is easy to establish. 6.1.20. Lemma. Let n ∈ IN, n ≥ 2, and let X be a continuum. Let A be a point of Cn (X) \ Cn−1 (X), and suppose A1 , . . . , An are d d the components of A. Let ε > 0 be such that V2ε (Aj ) ∩ V2ε (Ak ) = ∅ if and only if j = k and j, k ∈ {1, . . . , n}. Let B be a point of Cn (X) \ Cn−1 (X), and suppose B1 , . . . , Bn are the components of B. Then H(A, B) < ε if and only if H2 ({A1 , . . . , An }, {B1 , . . . , Bn }) < ε, where H2 is the Hausdorff metric on Fn (C(X)) induced by H. As a consequence of Lemma 6.1.20 , we have the following result. 6.1.21. Theorem. Let n ∈ IN, n ≥ 2. If X is a continuum, then the function fn : Cn (X) \ Cn−1 (X) → Fn (C(X)) given by fn (A) = {K | K is a component of A} is an embedding. The next Corollary says that given a continuum X, if C(X) is of finite dimension, then all the n–fold hyperspaces of X are of finite dimension too. 6.1.22. Corollary. If X is a continuum such that C(X) is of finite dimension then for each n ≥ 2, dim(Cn (X)) ≤ n(dim(C(X))). Proof. Let n ≥ 2. Then dim(Fn (C(X))) ≤ n(dim(C(X))) (Lemma 3.1 of [9]). Also, dim(Cn (X)) ≤dim(Fn (C(X))) (Theorem 6.1.21 and 7.3 of [42]). Therefore, dim(Cn (X)) ≤ n(dim(C(X))). Q.E.D.


6.2. UNICOHERENCE

295

6.1.23. Corollary. If X is a continuum such that all its nondegenerate proper subcontinua are arcs, then dim(Cn (X)) = 2n for every n ∈ IN. Proof. Since the hyperspace of subcontinua of an arc is a 2–cell (5.1 of [21]), dim(C(Y )) < 3 for every proper subcontinuum Y of X. Hence, by Theorem 2.2 of [43], dim(C(X)) < 3. Also, by 22.18 of [21], dim(C(X)) ≥ 2. Thus, dim(C(X)) = 2. Therefore, by Corollaries 6.1.13 and 6.1.22, dim(Cn (X)) = 2n for every n ∈ IN. Q.E.D.

6.2

Unicoherence

We show that Cn (X) has trivial shape and it is unicoherent for every n ∈ IN. 6.2.1. Definition. A compactum X has trivial shape if X is homeomorphic to lim{Xn , fnn+1 }, where each Xn is an absolute neighbor←− hood retract, and for each n ∈ IN, there exists m ≥ n such that fnm is homotopic to a constant map.

6.2.2. Theorem. If X is a continuum, then Cn (X) has trivial shape for each n ∈ IN. Proof. Let n ∈ IN. Note that by Theorem 2.1.52, X is homeomorm+1 phic to lim{Xm , fm }, where each Xm is a polyhedron. Hence, by ←−

Théorème IIm of [47], Cn (Xm ) is an absolute retract. Also, by Theom+1 rem 2.3.4, Cn (X) is homeomorphic to lim{Cn (Xm ), Cn (fm )}. Since ←−

absolute retracts are contractible (1.6.7 of [37]), by Theorem 1.3.12, all the maps Cn (fkm ) are homotopic to a constant map. Therefore, Cn (X) has trivial shape. Q.E.D.


296


ˇ 1 (X; ZZ) denotes the first 6.2.3. Remark. Given a continuum X, H ˇ Cech cohomology group of X with integer coefficients. It is known ˇ 1 (X; ZZ) = {0} if and only if each map from X into S 1 is that H homotopic to a constant map (8.1 of [10]). This implies, by 19.7 of ˇ 1 (X; ZZ) = {0}, then X is unicoherent. [21], that if H 6.2.4. Theorem. If X is a continuum, then for every n ∈ IN, each map from Cn (X) to the unit circle, S 1 , is homotopic to a constant map. In particular, we have that Cn (X) is unicoherent. Proof. By Theorem 2.1.52, X is homeomorphic to the inverse limit of an inverse sequence of compact and connected polyhedra, m+1 {Xm , fm }. Hence, Cn (Xm ) is a continuum which is an absolute retract (Théorème IIm of [47]) for each n ∈ IN. By 2.4 (p. 86) of [5] and 8.1 of [10], each map from Cn (Xm ) into S 1 is homotopic to a constant map. Then we obtain that ˇ 1 (Cn (Xm ); ZZ) = {0} (8.1 of [10]). By the continuity theorem H ˇ ˇ 1 (Cn (X), ZZ) = {0}. for Cech cohomology (Theorem 7–7 of [44]), H Thus, every map from Cn (X) into S 1 is homotopic to a constant map (8.1 of [10]). Therefore, by Remark 6.2.3, we have that Cn (X) is unicoherent for each n ∈ IN. Q.E.D.

6.3

Aposyndesis

We prove that for every n ∈ IN, Cn (X) is colocally connected and finitely aposyndetic. 6.3.1. Theorem. Let X be a continuum, with metric d, and let n ∈ IN be given. Then Cn (X) is colocally connected. Proof. Let A be a point of Cn (X). We consider two cases. First assume that A ∈ Cn (X) \ Fn (X). Let ε > 0 be given such that (VεH (A) ∩ Cn (X)) ∩ Fn (X) = ∅. To see that Cn (X) \ Copyright © 2005 Taylor & Francis Group, LLC

6.3. APOSYNDESIS

297

(VεH (A) ∩ Cn (X)) is connected, let B ∈ Cn (X) \ (VεH (A) ∩ Cn (X)), and let B1 , . . . , B be the components of B. Then ≤ n. Since B ∈ Cn (X) \ (VεH (A) ∩ Cn (X)), we have that H(A, B) ≥ ε. Hence, either A ⊂ Vεd (B) or B ⊂ Vεd (A). If A ⊂ Vεd (B), then there exists a point a in A such that a ∈ Vεd (B). Thus, for every b ∈ B, d(a, b) ≥ ε. For each j ∈ {1, . . . , }, let bj ∈ Bj . Let α : [0, 1] → Cn (X) be an order arc from {b1 , . . . , b } to B (Theorem 1.8.20). Since for each t ∈ [0, 1], α(t) ⊂ B, and for every b ∈ B, d(a, b) ≥ ε, we have that α(t) ∈ VεH (A) ∩ Cn (X) for any t ∈ [0, 1]. Thus, α([0, 1]) ∪ Fn (X) ⊂ Cn (X) \ (VεH (A) ∩ Cn (X)). If B ⊂ Vεd (A), then there exists a point b in B such that b ∈ d Vε (A). Thus, for every point a of A, d(b, a) ≥ ε. Without loss of generality, we assume that b ∈ B1 . For each j ∈ {2, . . . , }, let bj be any point of Bj . Let β : [0, 1] → Cn (X) be an order arc from {b, b2 , . . . , b } to B. Since for each t ∈ [0, 1], b ∈ β(t) ⊂ B and d(b, a) ≥ ε for any a ∈ A, we have that β(t) ∈ VεH (A) ∩ Cn (X) for any t ∈ [0, 1]. Thus, β([0, 1]) ∪ Fn (X) ⊂ Cn (X) \ (VεH (A) ∩ Cn (X)). Therefore, if A ∈ Cn (X) \ Fn (X) and ε > 0 is given such that H (Vε (A) ∩ Cn (X)) ∩ Fn (X) = ∅, then each element B of Cn (X) \ (VεH (A) ∩ Cn (X)) can be joined with Fn (X) by an order arc completely contained in Cn (X) \ (VεH (A) ∩ Cn (X)). Now, assume that A ∈ Fn (X). Suppose A = {a1 , . . . , ak } and let ε > 0 be given such that Vεd (aj ) ∩ Vεd (am ) = ∅ if and only if j = m and j, m ∈ {1, . . . , k}. Let B ∈ Cn (X) \ (VεH (A) ∩ Cn (X)) and let B1 , . . . , B be the components of B. Then ≤ n. Hence, H(A, B) ≥ ε. Thus, we have that either A ⊂ Vεd (B) or B ⊂ Vεd (A). If A ⊂ Vεd (B), then there exists a point a in A such that for each point b of B, d(a, b) ≥ ε. Without loss of generality, we assume that a = a1 . Hence, B ∩ Vεd (a1 ) = ∅. Let α : [0, 1] → Cn (X) be an order arc from B to X. We claim that for each t ∈ [0, 1], H(α(t), A) ≥ ε. To show this, suppose it is not true. Then there exists a point t0 in [0, 1] such that H(α(t0 ), A) < ε. Thus, we have that A ⊂ Vεd (α(t0 )) and α(t0 ) ⊂ Vεd (A). Since A ⊂ Vεd (α(t0 )), we have that for each j ∈ {1, . . . , k}, α(t0 ) ∩ Vεd (aj ) = ∅. On the other k d hand, since α(t0 ) ⊂ Vε (A) = Vεd (aj ), and these balls are pairwise j=1

disjoint, we have that each component of α(t0 ) is contained in one such ball. In particular, there is a component of α(t0 ) contained Copyright © 2005 Taylor & Francis Group, LLC

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in Vεd (a1 ). Hence, B ∩ Vεd (a1 ) = ∅, a contradiction. Therefore, α([0, 1]) ⊂ Cn (X) \ (VεH (A) ∩ Cn (X)). If B ⊂ Vεd (A), then given an order arc β : [0, 1] → Cn (X) from B to X, we have that β([0, 1]) ⊂ Cn (X) \ (VεH (A) ∩ Cn (X)). Therefore if A ∈ Fn (X), where A = {a1 , . . . , ak }, and ε > 0 is given such that Vεd (aj ) ∩ Vεd (am ) = ∅ if and only if j = m and j, m ∈ {1, . . . , k}, then each element B ∈ Cn (X) \ (VεH (A) ∩ Cn (X)) can be joined with {X} by an order arc contained in Cn (X) \ (VεH (A) ∩ Cn (X)). Therefore Cn (X) is colocally connected. Q.E.D. 6.3.2. Definition. A continuum X is said to be finitely aposyndetic provided that for each finite subset K of X, T (K) = K.

6.3.3. Corollary. If X is a continuum and n ∈ IN, then Cn (X) is aposyndetic and finitely aposyndetic. Proof. Clearly, any colocally connected continuum is aposyndetic. Given a continuum X and a positive integer n, since Cn (X) is unicoherent (Theorem 6.2.4) and aposyndetic, we have that Cn (X) is finitely aposyndetic (Corollary 1 of [3]). Q.E.D.

6.4

Arcwise Accessibility

We present results concerning arcwise accessibility of points of the n–fold symmetric product from the n–fold hyperspace of a given continuum. 6.4.1. Definition. Let X be a continuum. Let Σ1 and Σ2 be two arcwise connected closed subsets of 2X , such that Σ2 ⊂ Σ1 . A Copyright © 2005 Taylor & Francis Group, LLC

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member A of Σ2 is said to be arcwise accessible from Σ1 \ Σ2 beginning with K if and only if there is an arc α : [0, 1] → Σ1 such that α(0) = K, α(1) = A and α(t) ∈ Σ1 \ Σ2 for all t < 1. X

Σ1 K A

Σ2

We begin observing that in the proof of (2.2) of [39], Nadler actually showed the following: 6.4.2. Theorem. Let X be a continuum. If A is a nondegenerate subcontinuum of X and q is any point of A, then there exist a point p ∈ A \ {q} and an order arc α : [0, 1] → C(A) such that α(0) = {q}, α(1) = A and p ∈ α(t) for any t < 1.

6.4.3. Theorem. Let n ∈ IN, n ≥ 2. Let X be a continuum and let A be an element of Cn (X) having exactly n components and at least one of them is nondegenerate. Then A is arcwise accessible from Cn+1 (X) \ Cn (X), beginning with an element in Fn+1 (X) \ Fn (X). Proof. Suppose A1 , . . . , An are the components of A and that An is not degenerate. For each j ∈ {1, . . . , n}, let qj ∈ Aj . By Theorem 6.4.2, there exist qn+1 ∈ An and an order arc αn : [0, 1] → C(An ) such that αn (0) = {qn }, αn (1) = An and qn+1 ∈ αn (t) for any t < 1. For each j ∈ {1, . . . , n − 1}, let αj : [0, 1] → C(Aj ) be an order arc such that αj (0) = {qj }, αj (1) = Aj (Theorem 1.8.20). Let γ : [0, 1] → Cn+1 (X) be given by γ(t) = α1 (t) ∪ . . . ∪ αn (t) ∪ {qn+1 }. Since for each j ∈ {1, . . . , n}, αj is continuous and the union is also continuous (Lemma 1.8.11), then γ is continuous. Also, we have that γ(0) = {q1 , . . . , qn+1 }, γ(1) = A, and γ(t) ∈ Cn+1 (X)\ Copyright © 2005 Taylor & Francis Group, LLC

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Cn (X) for each t < 1. Therefore, A is arcwise accessible from Cn+1 (X) \ Cn (X), beginning with an element in Fn+1 (X) \ Fn (X). Q.E.D. 6.4.4. Corollary. Let n ∈ IN, n ≥ 3. Let X be a continuum. If x is a point of X such that {x} is arcwise accessible from C2 (X)\C(X) with an arc α : [0, 1] → C2 (X) such that α([0, 1])∩(C2 (X) \ F2 (X)) = ∅, then {x} is arcwise accessible from Cn (X) \ C(X).

6.4.5. Theorem. Let n ∈ IN, n ≥ 3. Let X be a continuum, and let a be a point of X such that {a} is arcwise accessible from 2X \ C(X). Then each element A of Fn (X) \ Fn−1 (X) containing a is arcwise accessible from 2X \ Cn (X). Proof. Let {a1 , . . . , an } ∈ Fn (X)\Fn−1 (X) containing a. Without loss of generality, we assume that a = an . Let U be an open set of X such that an ∈ U , and for each j ∈ {1, . . . , n − 1}, aj ∈ U . Since {an } is arcwise accessible from 2X \ C(X), there exists an arc α : [0, 1] → 2X such that α(1) = {an } and α(t) ∈ 2X \ C(X) for each t < 1. By continuity, there exists t1 ∈ [0, 1) such that α(t) ∈ U for each t ≥ t1 . Let β : [0, 1] → 2X be given by β(s) = {a1 , . . . , an−1 } ∪ α((1 − s)t1 + s). Since α is continuous and the union function is also continuous (Lemma 1.8.11), we have that β is continuous. By construction, we also have that β(1) = {a1 , . . . , an }, and β(t) ∈ 2X \ Cn (X) for each t < 1. Therefore, {a1 , . . . , an } is arcwise accessible from 2X \ Cn (X). Q.E.D. From the proof of Theorem 6.4.5 we have the following: 6.4.6. Corollary. Let n, m ∈ IN, n ≥ 3. Let X be a continuum and let a be a point of X such that {a} is arcwise accessible from Cm (X) \ C(X). Then each element A of Fn (X) \ Fn−1 (X) such that a ∈ A is arcwise accessible from Cm+n−1 (X) \ Cn (X).


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6.4.7. Theorem. Let n ∈ IN, n ≥ 2. Let X be a continuum with metric d. Suppose that for each j ∈ {1, . . . , n}, aj is a point of X such that {aj } is not arcwise accessible from 2X \ C(X). Then {a1 , . . . , an } is not arcwise accessible from 2X \ Cn (X). Proof. Suppose that {a1 , . . . , an } is arcwise accessible from 2X \ Cn (X). Then there exists an arc α : [0, 1] → 2X such that α(1) = {a1 , . . . , an } and α(t) ∈ 2X \ Cn (X) for each t < 1. Let U1 , . . . , Un be open sets of X such that for each j ∈ {1, . . . , n}, aj ∈ Uj , and Cl(Uj ) ∩ Cl(Uk ) = ∅ if and only if j, k ∈ {1, . . . , n} and j = k. By continuity, there exists t1 ∈ [0, 1) such that α(t) ∈ U1 , . . . , Un for each t ≥ t1 . For each j ∈ {1, . . . , n}, let βj : [0, 1] → 2X be given by βj (s) = α((1 − s)t1 + s) ∩ Uj . Since the family {U1 , . . . , Un } of open sets has pairwise disjoint closures, βj is well defined for each j ∈ {1, . . . , n}. To see that each βj is continuous, let j ∈ {1, . . . , n} and ε > 0 such that d(Cl(Uk ), Cl(U )) > ε, for each k, ∈ {1, . . . , n} and k = . Let s0 ∈ [0, 1]. By continuity, there exists δ > 0 such that if |s0 − s1 | < δ, then H(α((1 − s0 )t1 + s0 ), α((1 − s1 )t1 + s1 )) < ε. Let x ∈ α((1 − s0 )t1 + s0 ) ∩ Uj . Since H(α((1 − s0 )t1 + s0 ), α((1 − s1 )t1 + s1 )) < ε, there exists y ∈ α((1 − s1 )t1 + s1 ) such that d(x, y) < ε. n Since y ∈ Uk , there exists ∈ {1, . . . , n} such that y ∈ U . Since k=1

d(x, y) < ε and d(Cl(Uk ), Cl(U )) > ε for each k ∈ {1, . . . , n} and k = , we have that = j and y ∈ Uj . Therefore, y ∈ α((1 − s1 )t1 + s1 ) ∩ Uj , and α((1 − s0 )t1 + s0 ) ∩ Uj ⊂ Vεd (α((1 − s1 )t1 + s1 ) ∩ Uj ). Similarly α((1 − s1 )t1 + s1 ) ∩ Uj ⊂ Vεd (α((1 − s0 )t1 + s0 ) ∩ Uj ). Thus, H(α((1−s0 )t1 +s0 )∩Uj , α((1−s1 )t1 +s1 )∩Uj ) = H(βj (s0 ), βj (s1 )) < ε. Hence, βj is continuous. Note that for each j ∈ {1, . . . , n}, βj (1) = {aj }. Since, for each j ∈ {1, . . . , n}, {aj } is not arcwise accessible from 2X \ C(X), we have that, for each j ∈ {1, . . . , n}, there exists sj ∈ [0, 1) such that βj (s) ∈ C(X) for every s ≥ sj . Let s∗ = max{s1 , . . . , sn }. Then for each j ∈ {1, . . . , n}, βj (s) ∈ C(X) for every s ≥ s∗ . n n βj (s) = α((1 − s)t1 + s) ∩ Uj = Let s ≥ s∗ . Then α((1 − s)t1 + s) ∩

n

j=1

j=1

Uj = α((1 − s)t1 + s). Also, if s ≥ s∗ , then

j=1


302 n


βj (s) ∈ Cn (X). Thus, for each s ≥ s∗ , α((1 − s)t1 + s) ∈ Cn (X),

j=1

a contradiction. Therefore, {a1 , . . . , an } is not arcwise accessible from 2X \ Cn (X). Q.E.D. 6.4.8. Corollary. Let n ∈ IN, n ≥ 2. If X is a hereditarily indecomposable continuum, then {a1 , . . . , an } is not arcwise accessible from 2X \ Cn (X) for any {a1 , . . . , an } ∈ Fn (X). Proof. Since X is hereditarily indecomposable, no singleton is arcwise accessible from 2X \ C(X) ((3.4) of [39]). Thus, we have the result follows from Theorem 6.4.7. Q.E.D.

6.5

Points that Arcwise Disconnect

We consider when a point arcwise disconnects Cn (X). We also study the arc components of Cn (X) \ {X}, where X is an indecomposable continuum. 6.5.1. Lemma. Let X be a continuum and let n ∈ IN be given. If A is a proper subcontinuum of X, then Cn (X) \ Cn (A) is arcwise connected. Proof. Each element of Cn (X) \ Cn (A) can be joined with X with an order arc contained in Cn (X) \ Cn (A) (Theorem 1.8.20). Q.E.D. The proof of the following Theorem is similar to the one given in (11.3) of [40]. 6.5.2. Theorem. Let X be a continuum and let n ∈ IN be given. If A ∈ Cn (X) is such that Cn (X) \ {A} is not arcwise connected, then A is connected. Copyright © 2005 Taylor & Francis Group, LLC

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The following Theorem characterizes indecomposable continua.

6.5.3. Theorem. A nondegenerate continuum X is indecomposable if and only if for each n ∈ IN, Cn (X) \ {X} is not arcwise connected. Proof. Suppose X is decomposable. We show that for each n ∈ IN, Cn (X) \ {X} is arcwise connected. The proof is done by induction. The result is known for n = 1 ((1.51) of [40]). Let n = 2, and let A and B be two elements of C2 (X) \ {X}. If both A and B are connected then there exists an arc in C(X) \ {X} joining A and B ((1.51) of [40]). Suppose A has two components, say A1 and A2 . Since X is decomposable, there exist two proper subcontinua H and K of X such that X = K ∪ H. We show there exists an arc in C2 (X) \ {X} joining A and an element of C(X). We have to consider several cases. If either A ⊂ H or A ⊂ K, then there exists an order arc joining A with H or K (Theorem 1.8.20), and we are done. If A2 = K, then let x ∈ H ∩ K. Let α : [0, 1] → C2 (X) be an order arc joining A1 ∪ {x} and A1 ∪ K = A. Let β : [0, 1] → C2 (X) be an order arc joining A1 ∪ {x} and H. Then α([0, 1]) ∪ β([0, 1]) contains an arc having A and H as its end points, contained in C2 (X) \ {X}. If A1 ⊂ H, A2 ∩ (H ∩ K) = ∅, and A2 = K, then H ∪ A2 is a proper subcontinuum of X, and we can take an order arc joining A to H ∪ A2 . If A1 ∩ (H ∩ K) = ∅ and A2 ∩ (H ∩ K) = ∅, then let aj ∈ Aj ∩ H for j ∈ {1, 2}. Let α : [0, 1] → C2 (X) be an order arc joining {a1 , a2 } and A, and let β : [0, 1] → C2 (X) be an order arc having {a1 , a2 } and H as its end points. Then α([0, 1]) ∪ β([0, 1]) is contained in C2 (X) \ {X} and contains an arc joining A and H. If A1 ⊂ H \ K and A2 ⊂ K \ H, then let x ∈ H ∩ K. Let α : [0, 1] → C2 (X) be an order arc from A to H ∪ A2 . Let β[0, 1] → C2 (X) be an order arc from {x}∪A2 to H∪A2 . Let γ : [0, 1] → C2 (X) be an order arc from {x} ∪ A2 to K. Then α([0, 1]) ∪ β([0, 1]) ∪ γ([0, 1]) contains an arc joining A and K contained in C2 (X) \ {X}. The rest of the cases are similar to the ones treated. Copyright © 2005 Taylor & Francis Group, LLC

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Now, let n ≥ 3 and suppose Cn (X) \ {X} is arcwise connected. We show that Cn+1 (X) \ {X} is arcwise connected. Let A and B be two points in Cn+1 (X) \ {X}. If both A and B belong to Cn (X), by induction hypothesis we can find an arc in Cn (X)\{X} ⊂ Cn+1 (X)\{X} having A and B as its end points. Hence, assume that A has n + 1 components. Let A1 , . . . , An+1 be the components of A. Since X is decomposable, there exist two proper subcontinua H and K of it such that X = H ∪ K. Since n ≥ 3, at least two components of A intersect either H or K; suppose that two components of A intersect H. Without loss of generality, we assume that An ∩ H = ∅ and An+1 ∩ H = ∅. For each j ∈ {1, . . . , n − 1}, let aj ∈ Aj . Take an ∈ An ∩ H and an+1 ∈ An+1 ∩ H. Let α : [0, 1] → Cn+1 (X) be an order arc from {a1 , . . . , an+1 } to A. Let β : [0, 1] → C2 (X) be an order arc from {an , an+1 } to H. Let γ : [0, 1] → Cn+1 (X) be given by γ(t) = {a1 , . . . , an−1 } ∪ β(t). Then γ is continuous, γ(0) = {a1 , . . . , an−1 } ∪ β(0) = {a1 . . . , an+1 } and γ(1) = {a1 , . . . , an−1 } ∪ β(1) = {a1 , . . . , an−1 }∪H ∈ Cn (X)\{X}. Hence, α([0, 1])∪γ([0, 1]) is contained in Cn+1 (X) \ {X} and contains an arc having A and γ(1) as its end points. Similarly, if B has n + 1 components, we can find an arc in Cn+1 (X) \ {X} having B and an element of Cn (X) as its end points. Thus by induction hypothesis, we are done. The proof of the reverse implication is similar to the one given in (11.4) of [40]. Q.E.D. The next Theorem tells us more about what type of subcontinua may arcwise disconnect the hyperspaces. 6.5.4. Theorem. Let X be a continuum, and let E be a nondegenerate proper subcontinuum of X. Consider the following statements: (1) E is a terminal subcontinuum of X. (2) 2X \ {E} is not arcwise connected. (3) For each n ∈ IN, Cn (X) \ {E} is not arcwise connected. (4) C(X) \ {E} is not arcwise connected. Then (1) implies (2), (3) and (4). Furthermore, if E is decomposable then all four statements are equivalent. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. The proof of (1) implies (2) and (4) are given in (11.5) of [40]. The proof of (1) implies (3) is similar to the one given in (11.5) of [40]. Now suppose E is a decomposable nondegenerate subcontinuum of X. The equivalence between (1), (2) and (4) is given in (11.5) of [40]. Suppose E is not terminal, then E = X. We show that for each n ∈ IN, Cn (X) \ {E} is arcwise connected. This is done by induction. For n = 1 the result is known. Suppose Cn (X) \ {E} is arcwise connected. To show that Cn+1 (X) \ {E} is arcwise connected, let A and B be two points in Cn+1 (X) \ {E}. If both A and B belong to Cn+1 (X) \ Cn+1 (E), then there exists an arc having A and B as its end points and contained in Cn+1 (X)\{E} (Lemma 6.5.1). If both A and B belong to Cn+1 (E), then by Theorem 6.5.3, there exists an arc joining A and B and contained in Cn+1 (E) \ {E} ⊂ Cn+1 (X) \ {E}. Thus, suppose, without loss of generality, that A ∈ Cn+1 (E) \ {E} and B ∈ Cn+1 (X) \ Cn (E). If A has less than n + 1 components then, by induction hypothesis, there exists an arc in Cn (X) \ {E} joining A and X. So, assume A has exactly n + 1 components. Since E is decomposable, by Theorem 6.5.3, there exists an arc joining A and an element A of Cn (E) \ {E}. By induction hypothesis, there exists an arc in Cn (X) \ {E} joining A and X. Hence, there exists an arc in Cn+1 (X) \ {E} joining A and X. Since B ∈ Cn+1 (X) \ Cn+1 (E), by Lemma 6.5.1, there exits an arc having B and X as its end points. Therefore, there exists an arc in Cn+1 (X) \ {E} joining A and B. Q.E.D. 6.5.5. Theorem. If X is a continuum, then for any E ∈ 2X , the following are equivalent: (1) 2X \ {E} is not arcwise connected. (2) C(X) \ {E} is not arcwise connected. (3) For each n ∈ IN, Cn (X) \ {E} is not arcwise connected. Proof. The proof of the equivalence of (1) and (2) is given in (11.8) of [40]. Clearly (3) implies (2). We show that (2) implies (3). Copyright © 2005 Taylor & Francis Group, LLC

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Let E ∈ 2X and suppose C(X) \ {E} is not arcwise connected. Then E ∈ C(X) ((11.3) of [40]). Let n ≥ 2. If E = X, then, by Theorem 6.5.3, X is indecomposable and Cn (X)\{E} is not arcwise connected. Suppose E = X. If E is decomposable, then, by Theorem 6.5.4, E is a terminal subcontinuum of X and Cn (X) \ {E} is not arcwise connected. Thus, assume E is indecomposable. Let B ∈ C(E) ⊂ Cn (X) and let A ∈ Cn (X) \ Cn (E). Let α : [0, 1] → Cn (X) be an arc such that α(0) = B and α(1) = A. Let β : [0, 1] → Cn (X) be given by β(t) = σ(α([0, t])). Then β is an order arc from B to σ(α([0, 1])). Hence, β([0, 1]) ⊂ C(X) ((1.11) of [40]). In particular, σ(α([0, 1])) is connected. Since β(0) = B ⊂ E, β(1) = σ(α([0, 1])), (σ(α([0, 1])))∩ (C(X)\C(E)) = ∅, and C(X)\{E} is not arcwise connected, we have that there exists t0 ∈ [0, 1] such that β(t0 ) = E. Let t1 = min{t ∈ [0, 1] | β(t) = E}. Then t1 > 0 and β(t1 ) = E. Since for each t < t1 , β(t) is a proper subcontinuum of E and E is indecomposable, we have that β(t) is nowhere dense in E (Corollary 1.7.21). Note that, for every t < t1 , E = β(t) ∪ (σ(α([t, t1 ]))), and σ(α([t, t1 ])) is compact. Hence, E = σ(α([t, t1 ])). By continuity, we have that E = α(t1 ). Therefore, Cn (X) \ {E} is not arcwise connected. Q.E.D. 6.5.6. Theorem. If E, A and B are subcontinua of the continuum X and n ∈ IN, then the following are equivalent: (1) If Γ is an arc in C(X) such that A, B ∈ Γ, then E ∈ Γ. (2) If Γ is an arc in Cn (X) such that A, B ∈ Γ, then E ∈ Γ. (3) If Γ is an arc in 2X such that A, B ∈ Γ, then E ∈ Γ. Proof. Clearly (3) implies (2) and (2) implies (1). The proof of the implication from (1) to (3) is in (11.13) of [40]. Q.E.D. 6.5.7. Corollary. Let E be a subcontinuum of the continuum X. If Λ is an arc component of 2X \ {E} and Λ ∩ Cn (X) = ∅, for some n ∈ IN, then Λ ∩ Cn (X) is an arc component of Cn (X) \ {E}. Copyright © 2005 Taylor & Francis Group, LLC


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6.5.8. Theorem. If X is a continuum, then the following are equivalent: (1) X is hereditarily indecomposable. (2) For each nondegenerate subcontinuum E of X, 2X \ {E} is not arcwise connected. (3) For each nondegenerate subcontinuum E of X, Cn (X) \ {E} is not arcwise connected, for each n ∈ IN. (4) For each nondegenerate subcontinuum E of X, C(X) \ {E} is not arcwise connected. Proof. By Theorem 6.5.5, we have that (2), (3) and (4) are equivalent. The proof of the equivalence between (4) and (1) is given in (11.15) of [40]. Q.E.D. 6.5.9. Definition. Let X be a continuum, and let p ∈ X. The composant of p in X is the union of all proper subcontinua of X containing p.

6.5.10. Notation. Given a continuum X, a positive integer n and a composant κ of X, let Cn (κ) denote the set Cn (κ) = {A ∈ Cn (X) | A ⊂ κ}. In the next two Theorems we describe the arc components of Cn (X) \ {X}, where X is an indecomposable continuum. 6.5.11. Theorem. Let n be an integer greater than one. If κ is a composant of an indecomposable continuum X, then Cn (κ) is an arc component of Cn (X) \ {X}. Proof. By Theorem 6.5.3, Cn (X) \ {X} is not arcwise connected. First, observe that Cn (κ) is arcwise connected. To see this, let A ∈ Cn (κ). Then, since A has finitely many components and X is indecomposable, it is easy to show that there exists a proper subcontinuum B of X containing A. Thus, there exists an order Copyright © 2005 Taylor & Francis Group, LLC


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arc from A to B (Theorem 1.8.20). Since C(κ) is arcwise connected ((1.52.1) of [40]), we have that Cn (κ) is arcwise connected. Let A be the arc component of Cn (X) \ {X} containing Cn (κ), and suppose there exists B ∈ A \ Cn (κ). Let A ∈ C(κ). Since A and B belong to A, there is an arc α : [0, 1] → A such that α(0) = A and α(1) = B. Let β : [0, 1] → Cn (X) be given by β(t) = σ(α([0, t])). Then β is well defined (Corollary 6.1.2), β is an order arc and β(0) = α(0) = A. Hence, β(t) ∈ C(X) for each t ∈ [0, 1] ((1.11) of [40]), and B ⊂ β(1). Since B is not contained in κ and B ⊂ β(1), we have that β(1) is a subcontinuum of X intersecting two different composants of it. Thus, β(1) = X. Let t0 = min{t ∈ [0, 1] | β(t) = X}. Then β(t0 ) = X and t0 > 0. Observe that if 0 ≤ t < t0 , then β(t) is a nowhere dense subset of X (Corollary 1.7.21). Note that for each 0 < t < t0 , X = β(t0 ) = β(t) ∪ (σ(α ([t, t0 ]))). Since β(t) is nowhere dense in X, we have that X = σ(α ([t, t0 ])) for each t < t0 . By continuity, α(t0 ) = X, a contradiction. Therefore, Cn (κ) = A. Q.E.D. 6.5.12. Theorem. Let n be an integer greater than one, and let X be an indecomposable continuum. If A is an arc component of Cn (X)\{X}, which is not of the form Cn (κ), where κ is a composant of X, then there exist finitely many composants κ1 , . . . , κ of X and positive integers m1 , . . . , m such that there exists a one–to–one map from Cmj (κj ) (with the “max” metric ρ1 ) onto A. j=1

Proof. Let A0 be a point of A, and let κ1 , . . . , κ be the composants of X which intersect A0 . Since A is not of the form Cn (κ), ≥ 2. First, we show that every element of A intersects each κj for each j ∈ {1, . . . , }. To see this, suppose that there is a point B of A such that B ∩ κj = ∅, for some j ∈ {1, . . . , }. Since A0 and B belong to A, there exists an arc α : [0, 1] → A such that α(0) = A0 and α(1) = B. Let β : [0, 1] → Cn (X) be given by β(t) = σ(α([0, t])). Then β is well defined (Corollary 6.1.2), β is an order arc, β(0) = α(0) = A0 , and β(1) = σ(α([0, 1])) is an element of Cn (X) intersecting κj and containing B. On the other hand, the map γ : [0, 1] → Cn (X) given by γ(t) = σ(α([1 − t, 1])) is also well defined, γ is an order arc, γ(0) = α(1) = B and γ(1) = σ(α([0, 1])) = β(1). Thus, there Copyright © 2005 Taylor & Francis Group, LLC


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exists an order arc in Cn (X) from B to β(1), B ∩ κj = ∅ and β(1) ∩ κj = ∅; this contradicts Theorem 1.8.20 if σ(α([0, 1])) is a proper subset of X. Otherwise, a similar argument to the one given in Theorem 6.5.11 shows that there exists t0 ∈ [0, 1] such that α(t0 ) = X, which is also a contradiction. Therefore, every element of A intersects each κj , j ∈ {1, . . . , }. A similar argument proves that if A ∈ A and A ∩ κ = ∅, for some composant κ of X, then κ ∈ {κ1 , . . . , κ }. For each j ∈ {1, . . . , }, let mj = max{number of components of A contained in κj | A ∈ A}. Let us observe that

mj = n. Let f :

j=1

Cmj (κj ) → A be given

j=1

by f ((A1 , . . . , A )) =

Aj .

j=1

Let us see first that f is well defined. Clearly, f ((A1 , . . . , A )) ∈ Cn (X). On the other hand, for each j ∈ {1, . . . , }, Aj and A0 ∩ κj both belong to Cmj (κj ). Since Cmj (κj ) is arcwise connected, there is an arc αj : [0, 1] → Cmj (κj ) such that αj (0) = A0 ∩ κj and αj (1) = αj (t) is a Aj . Hence, α : [0, 1] → Cn (X) \ {X} given by α(t) = j=1

path joining A0 and

Aj . Therefore,

j=1

Aj ∈ A.

j=1

If A is a point of A, then (A ∩ κ1 , . . . , A ∩ κ ) is an element of Cmj (κj ) such that f ((A ∩ κ1 , . . . , A ∩ κ )) = (A ∩ κj ) = A.

j=1

j=1

Thus, f is surjective. Let (A1 , . . . , A ) and (B1 , . . . , B ) be two different points of Cmj (κj ). Then Aj0 = Bj0 for some j0 ∈ {1, . . . , }. Hence, j=1 j=1

Aj =

Bj , being a disjoint union. Therefore, f is one–to–one.

j=1



310

To see that f is continuous, let ε > 0 be given. Let (A1 , . . . , A ) Cmj (κj ) such that and (B1 , . . . , B ) be two points of j=1

ε ρ1 ((A1 , . . . , A ), (B1 , . . . , B )) < . 2 Then for each j ∈ {1, . . . , }, H(Aj , Bj ) < 2ε . Hence, for every

Br and Bj ⊂ V dε (Aj ) ⊂ j ∈ {1, . . . , }, Aj ⊂ V dε (Bj ) ⊂ V dε 2 2 2 r=1

d d Vε Ar . Thus, we have that Aj ⊂ V ε Bj and Bj 2

2

r=1

⊂ V dε

2

j=1

j=1

j=1

Aj . This implies that

j=1

H

Aj ,

j=1

j=1

Bj

≤

ε < ε. 2

Therefore, f is continuous. Q.E.D. 6.5.13. Theorem. Let n be an integer greater than one. Let X be an indecomposable continuum. If A is an arc component of Cn (X)\{X}, which is not of the form Cn (κ), where κ is a composant of X, then for any arc α : [0, 1] → A, σ(α([0, 1])) is not connected. Proof. Let α : [0, 1] → A be an arc and suppose σ(α([0, 1])) is connected. Observe that α(0) is a nonconnected subset of σ(α([0, 1])) intersecting at least two composants of X. Hence, σ(α([0, 1])) = X. An argument similar to the one given in the proof of Theorem 6.5.11 shows that there exists a point t0 in [0, 1] such that α(t0 ) = X, which is not possible. Therefore, σ(α([0, 1])) is not connected. Q.E.D. We end this section showing that hereditarily indecomposable continua have unique n–fold hyperspaces.



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6.5.14. Theorem. Let n, m ∈ IN, m ≥ 2. Let X be a hereditarily indecomposable continuum. If Y is a continuum such that Cm (Y ) is homeomorphic to Cn (X), then Y is homeomorphic to X. Proof. Let h : Cn (X) → → Cm (Y ) be a homeomorphism. We consider first the case n = 1. Since X is hereditarily indecomposable, C(X) is uniquely arcwise connected ((1.61) of [40]). Hence, Cm (Y ) is uniquely arcwise connected. Then, by Theorem 6.1.9, m = 1. A contradiction. Suppose now that n ≥ 2. Since X is hereditarily indecomposable, X is the only point at which Cn (X) is locally connected (Theorem 6.1.19). Hence, h(X) = Y . Now, we show that h(C(X)) ⊂ C(Y ). To see this, let A ∈ C(X)\ Fn (X), then Cn (X) \ {A} is not arcwise connected (Theorem 6.5.8). Thus, Cm (Y ) \ {h(A)} is not arcwise connected. Hence, h(A) ∈ C(Y ) (Theorem 6.5.2). Since C(Y ) is closed in Cm (Y ) and F1 (X) ⊂ ClCn (X) (C(X) \ F1 (X)), we have that h(C(X)) ⊂ C(Y ). Next, we prove that h(F1 (X)) ⊂ F1 (Y ). To this end, suppose there exists a point {x} in F1 (X) such that h({x}) ∈ C(Y ) \ F1 (Y ). Then there exists an order arc α : [0, 1] → C2 (Y ) such that α(0) ∈ F2 (Y ), α(1) = {x}, and α([0, 1)) ⊂ C2 (Y ) \ C(Y ) (see (2.2) of [39]). Hence, h−1 ◦ α : [0, 1] → Cn (X) is an arc such that h−1 ◦ α(1) = {x} and h−1 ◦α([0, 1)) ⊂ Cn (X)\C(X). This contradicts the fact that in 2X , such arcs do not exist (i.e., singletons are not arcwise accessible from 2X \ C(X)) ((3.4) of [39]). Therefore, h(F1 (X)) ⊂ F1 (Y ). Let Y ∈ C(Y ) be such that F1 (Y ) = h(F1 (X)). Note that C(Y ) ⊂ C(Y ) and h−1 (C(Y )) is an arcwise connected subcontinuum of Cn (X). Thus, C(X) ∩ h−1 (C(Y )) is arcwise connected ((5.2) of [39]), and F1 (X) ⊂ C(X) ∩ h−1 (C(Y )). We claim that h−1 (Y ) ∈ C(X). Suppose, to the contrary, that h−1 (Y ) ∈ Cn (X) \ C(X). Let x1 and x2 be two points in different composants of X. Let β1 , β2 : [0, 1] → h−1 (C(Y )) be two arcs such that βj (0) = {xj } and βj (1) = h−1 (Y ), j ∈ {1, 2}. By (3.4) of [39], we have that βj ([0, 1])∩ C(X) = {{xj }}, j ∈ {1, 2}. Hence, (β1 ([0, 1]) ∪ β2 ([0, 1])) ∩ C(X) contains an arc from {x1 } to {x2 } ((5.2) of [39]). Since X is indecomposable and x1 and x2 are in different composants of X, we have that X = σ([(β1 ([0, 1]) ∪ β2 ([0, 1])) ∩ C(X)]) ((1.51) of [40]). Then X ∈ (β1 ([0, 1]) ∪ β2 ([0, 1])) ∩ C(X) ((1.50) of [40]). Thus, X ∈ h−1 (C(Y )). It follows that Y = h(X) ∈ C(Y ) and Y = Y , a Copyright © 2005 Taylor & Francis Group, LLC

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contradiction. Therefore, h−1 (Y ) ∈ C(X). We assert that h−1 (Y ) = X. To see this, let x1 and x2 be two points in different composants of X. Since C(X) ∩ h−1 (C(Y )) is arcwise connected, there exist two arcs α1 , α2 : [0, 1] → C(X) ∩ h−1 (C(Y )) such that αj (0) = {xj } and αj (1) = h−1 (Y ), j ∈ {1, 2}. Then α1 ([0, 1]) ∪ α2 ([0, 1]) contains an arc from {x1 } to {x2 }. Since X is indecomposable, we have that X = σ(α1 ([0, 1]) ∩ α2 ([0, 1])) ((1.51) of [40]). Thus, X ∈ α1 ([0, 1]) ∪ α2 ([0, 1]) ((1.50) of [40]). Then h(X) ∈ C(Y ). Hence, since h(X) = Y , Y ∈ C(Y ). Hence, Y = Y . Thus, by definition of Y , h (F1 (X)) = F1 (Y ). Therefore, Y is homeomorphic to X. Q.E.D. 6.5.15. Corollary. Let n ∈ IN, and let X be a hereditarily indecomposable continuum. If Y is a continuum such that Cn (Y ) is homeomorphic to Cn (X), then X and Y are homeomorphic. Proof. By Theorem 6.5.14, we only have to consider the case when n = 1. Since X is hereditarily indecomposable, C(X) is uniquely arcwise connected ((1.61) of [40]). Hence, C(Y ) is uniquely arcwise connected. Thus, by (1.61) of [40], Y is hereditarily indecomposable. Therefore, Y is homeomorphic to X ((0.60) of [40]). Q.E.D.

6.6

Cn∗ –smoothness

We study the continuity of taking n–fold hyperspaces. 6.6.1. Definition. Let n ∈ IN. A continuum X is Cn∗ –smooth at A ∈ Cn (X), provided that for any sequence {Ak }∞ k=1 of elements of Cn (X) converging to A, the sequence {Cn (Ak )}∞ k=1 of hyX ∗ perspaces converges to Cn (A); i.e., the map Cn : Cn (X) → 22 given by Cn∗ (A) = Cn (A) is continuous at A. A continuum X is Cn∗ –smooth if it is Cn∗ –smooth at each element of Cn (X); i.e., Cn∗ is continuous. Copyright © 2005 Taylor & Francis Group, LLC

6.6. Cn∗ –SMOOTHNESS

313

6.6.2. Remark. For n = 1, Cn∗ –smoothness is called C ∗ –smoothness instead of C1∗ –smoothness. C ∗ –smoothness was introduced in Chapter XV of [40].

6.6.3. Theorem. If X is a C ∗ –smooth homogeneous continuum, then X is indecomposable. Moreover, if X is a C ∗ –smooth homogeneous plane continuum, then X is hereditarily indecomposable. Proof. Let X be a C ∗ –smooth homogeneous continuum. Then X is hereditarily unicoherent ((3.4) of [13]). By Theorem 1 of [22], X is indecomposable. If X is a C ∗ –smooth homogeneous plane continuum, we have that X is indecomposable. Since any indecomposable homogeneous plane continuum is hereditarily indecomposable (Theorem 1 of [15]), X is hereditarily indecomposable. Q.E.D. The following result is easy to prove. 6.6.4. Lemma. Let n ∈ IN. Let X be a continuum and let {Ak }∞ k=1 be a sequence in Cn (X) converging to A. If lim Cn (Ak ) exists, then lim Cn (Ak ) ⊂ Cn (A).

k→∞

k→∞

6.6.5. Theorem. Let X be a continuum. If A is a subcontinuum of X, then the following are equivalent: (1) X is C ∗ –smooth at A; (2) Cn∗ |C(X) is continuous at A for all n ∈ IN; (3) Cn∗ |C(X) is continuous at A for some n ∈ IN. Proof. Assume X is C ∗ –smooth at A, and let n ≥ 2. We prove Cn∗ |C(X) is continuous at A. Let {Ak }∞ k=1 be a sequence of subcontinua of X converging to A. Let B be any element of Cn (A). Let B1 , . . . , B ( ≤ n) be the components of B. Hence, each Bj is a subcontinuum of A, j ∈ {1, . . . , }. Since X is C ∗ –smooth at A, there exist subcontinua Bk1 , . . . , Bk of Ak for each k ∈ IN such that Copyright © 2005 Taylor & Francis Group, LLC


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lim

k→∞

Bkj

= Bj for each j ∈ {1, . . . , }. Hence, Bk =

Bkj is an ele-

j=1

ment of Cn (Ak ) for each k ∈ IN, and lim Bk = B (Lemma 1.8.11). k→∞

Therefore, Cn (A) ⊂ lim Cn (Ak ). By Lemma 6.6.4, we may conclude k→∞

that lim Cn (Ak ) = Cn (A). Therefore, Cn∗ |C(X) is continuous at A. k→∞

Assume Cn∗ |C(X) is continuous at A for some n ≥ 1 and some element A of C(X). We prove X is C ∗ –smooth. Let {Ak }∞ k=1 be a sequence of subcontinua of X converging to A. Let B be a nondegenerate proper subcontinuum of A. Let x1 , . . . , xn−1 be n − 1 distinct points in A\B. Let D = B∪{x1 , . . . , xn−1 }. Since Cn∗ |C(X) is continuous at A, there exists Dk ∈ Cn (Ak ) for each k ∈ IN such that the sequence {Dk }∞ k=1 converges to D. Since D has n components, we assume without loss of generality that Dk also has n components for any k ∈ IN. Since n is the maximum number of components we allow, there exists a component Dk1 of Dk such that {Dk1 }∞ k=1 converges to B. Therefore, C(A) ⊂ lim C(Ak ). By Lemma 6.6.4, we conclude that lim C(Ak ) = C(A).

k→∞

k→∞

The fact that (2) implies (3) is obvious. Q.E.D. 6.6.6. Definition. A continuum X is absolutely C ∗ –smooth, provided that for any continuum Z in which X can be embedded and for each sequence {Ak }∞ k=1 of elements of C(Z) converging to X, the ∞ sequence {C(Ak )}k=1 of hyperspaces converges to C(X). With a proof similar to the one given for Theorem 6.6.5, we have the following result: 6.6.7. Theorem. Let X be a continuum. Then the following statements are equivalent: (1) X is absolutely C ∗ –smooth; (2) for any continuum Z in which X is embedded, Cn∗ |C(Z) is continuous at X for all n ∈ IN; (3) for any continuum Z in which X is embedded, Cn∗ |C(Z) is continuous at X for some n ∈ IN. Copyright © 2005 Taylor & Francis Group, LLC


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6.6.8. Lemma. Let n ∈ IN. Let X be a continuum, with metric d, let A be an indecomposable subcontinuum of X and let {Bm }∞ m=1 be a sequence of elements of Cn (X) converging to A. Then there exists ∞ a subsequence {Bmk }∞ k=1 of {Bm }m=1 such that for each k, there exists a component Dk of Bmk such that the sequence {Dk }∞ k=1 of continua converges to A. Proof. Since A is an indecomposable continuum, A has uncountably many mutually disjoint composants (11.15 and 11.17 of [41]). Let a1 , . . . , an+1 be n + 1 points in n + 1 distinct composants of A. We may assume that V d1 (ai ) ∩ V d1 (aj ) = ∅ if i = j for each positive integer . Since {Bm }∞ m=1 converges to A, for each , there exists an integer 1 m such that H(A, Bm ) < . Thus, Bm ∩ V d1 (aj ) = ∅ for each j ∈ {1, . . . , n + 1}. Since Bm has at most n components, we have that at least one of the components of Bm intersects two of the balls V d1 (aj ), j ∈ {1, . . . , n + 1}.

Since we only have n + 1 balls, there exist j0 , j1 ∈ {1, . . . , n + 1} such that for infinitely many indices k, Bmk has a component Dk such that Dk ∩ V d1 (aj0 ) = ∅ and Dk ∩ V d1 (aj1 ) = ∅ for each k. k k Since C(X) is compact (Theorem 1.8.5), we assume without loss of generality that the sequence {Dk }∞ k=1 converges to a subcontinuum D of A. Since aj0 and aj1 belong to D and they are in different composants of A, we conclude that D = A. Q.E.D.

6.6.9. Remark. The converse of Lemma 6.6.8 is false, as can be seen from the argument of Example 3.4 of [34].

6.6.10. Lemma. Let X be a decomposable continuum, with metric d, and let A and B be nondegenerate proper subcontinua of X such that X = A ∪ B. Assume that there exist two order arcs α, β : [0, 1] → C(X) with the following properties: α(0) ∈ F1 (A), α(1) = A, β(0) ∈ F1 (B), β(1) = B and (A ∩ B) ∩ (α(t) ∪ β(t)) = ∅ for each t ∈ [0, 1). Then X is not Cn∗ –smooth at X for any n ≥ 2. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Suppose X is Cn∗ –smooth at X. Let {tm }∞ m=1 be an increasing sequence of numbers in [0, 1) converging to 1. For each m ∈ IN, let Dm = α(tm )∪β(tm ). For each m ∈ IN, (A∩B)∩(α(tm )∪β(tm )) = ∅, hence, Dm ∈ C2 (X) \ C(X). Let R be a component of A ∩ B. Let H and K be proper subcontinua of A and B, respectively, such that they properly contain R (Corollary 1.7.23). Let x1 , . . . , xn−1 be n − 1 distinct points of X \ (H ∪ K). Let L = {x1 , . . . , xn−1 } ∪ (H ∪ K). Let ε > 0 be such that the following hold: d d V2ε (xi ) ∩ V2ε (xj ) = ∅ if and only if i = j, d (H ∪ K) = ∅, {x1 , . . . , xn−1 } ∩ V2ε n−1

d V2ε (xj ) ∩ (H ∪ K) = ∅,

j=1 d d (K) = ∅, and K \ V2ε (H) = ∅. H \ V2ε

Since X is Cn∗ –smooth at X, there exists m0 ∈ IN such that if m ≥ m0 , then there exists Em ∈ Cn (D

m ) such that H(Em , L) < ε. n−1 Vεd (xj ) ∪ Vεd (H ∪ K), Let m ≥ m0 . Then Em ⊂ Vεd (L) = j=1

Em ∩Vεd (xj ) = ∅ for each j ∈ {1, . . . , n−1}, and Em ∩Vεd (H ∪K) = ∅. Hence, Em has exactly n components. Let G1 , . . . , Gn be the components of Em . Since the ε–balls about each x1 , . . . , xn−1 and H ∪ K are pairwise disjoint we assume, without loss of generality, that Gj ⊂ Vεd (xj ) for each j ∈ {1, . . . , n − 1} and Gn ⊂ Vεd (H ∪ K). Since Gn is a subcontinuum of Dm , Gn is contained either in α(tm ) or in β(sm ). Suppose that Gn is contained in α(tm ). Let x ∈ d K \ V2ε (H). Then for each point z of Em , d(y, z) ≥ ε. This is a contradiction; therefore, X is not Cn∗ –smooth at X. Q.E.D. The following result characterizes the class of continua for which the map Cn∗ is continuous for n ≥ 2. 6.6.11. Theorem. A continuum X is Cn∗ –smooth for some n ≥ 2 if and only if X is hereditarily indecomposable. Copyright © 2005 Taylor & Francis Group, LLC


317

Proof. Suppose X is hereditarily indecomposable. Then X is Cn∗ – smooth by Lemma 6.6.8 and (1.207.8) of [40]. Suppose that X is Cn∗ –smooth for some integer n ≥ 2. Then condition (3) of Theorem 6.6.7 is satisfied. Hence, X is C ∗ –smooth by Theorem 6.6.7. Since X is C ∗ –smooth, X is hereditarily unicoherent ((3.4) of [13]). Suppose X is decomposable. Then there exist two proper subcontinua A and B of X such that X = A ∪ B. Let a ∈ A \ B and b ∈ B \ A. Let α, β : [0, 1] → C(X) be order arcs such that α(0) = {a}, α(1) = A, β(0) = {b} and β(1) = B (Theorem 1.8.20). Let t0 and s0 be points of [0, 1] such that α(t0 ) ∩ β(s0 ) = ∅ and such that for each t < t0 and each s < s0 , α(t) ∩ β(s) = ∅. Note that t0 > 0 and s0 > 0. Let {tk }∞ k=1 and {sk }∞ be increasing sequences in [0, 1] converging to t 0 and s0 , k=1 respectively. Let Y = α(t0 ) ∪ β(s0 ). Then Y is a subcontinuum of X. Then, by Lemma 6.6.10, X is not Cn∗ –smooth at Y , a contradiction. Therefore, X is indecomposable. A similar argument shows that each subcontinuum of X is indecomposable. Q.E.D. We now present some results about the points at which a continuum X is Cn∗ –smooth. 6.6.12. Theorem. Let X be a continuum and let A be an element of Cn (X) for some n ≥ 2. If X is Cn∗ –smooth at A, then X is C ∗ – smooth at each component of A. Proof. Let A be an element of Cn (X) and suppose X is Cn∗ –smooth at A. Observe that if A is connected, then X is C ∗ –smooth at A by Theorem 6.6.5. Suppose A has at least two components. Let A1 , . . . , Ak be the components of A. We show that X is C ∗ –smooth at A1 . Let {Km }∞ m=1 be a sequence of subcontinua of X converging

k to A1 . Without loss of generality, we assume that Km ∩ Aj = ∅ for each m ∈ IN. Let L be a subcontinuum of A1 . Copyright © 2005 Taylor & Francis Group, LLC

j=2


318

Let α : [0, 1] → C(X) be an order arc such that α(0) ∈ F1 (A2 ) and α(1) = A2 (Theorem 1.8.20). Let {tm }∞ m=1 be an increasing sequence of numbers in [0, 1) converging to 1. For each m ∈ IN, let (1) (n−k) pm , . . . , pm be n − k distinct points in A2 \ α(tm ). For each m ∈ IN, let

k (n−k) Fm = Km ∪ α(tm ) ∪ Aj ∪ {p(1) }. m , . . . , pm j=3

Then lim Fm = A. Since X is Cn∗ –smooth at A, for every m ∈ IN, m→∞

there exists an element Dm of Cn (Fm ) such that lim Dm = L ∪ m→∞

k (1) (n−k) α(t1 ) ∪ Aj ∪ {p1 , . . . , p1 }. j=3

For each m ∈ IN, let Lm = Dm ∩ Km . Then Lm is a subcontinuum of Km and lim Lm = L. Therefore, X is C ∗ –smooth at A1 . m→∞ Similarly, X is C ∗ –smooth at the other components of A. Q.E.D. 6.6.13. Lemma. Let C be a closed subset of a space Z. Let A = Cl(Z \ C) and let B = Cl(Z \ A). Then A = Cl(Z \ B). Proof. Since A is closed in Z, Cl(Int(A)) ⊂ A; thus, since A = Cl(Z\C) = Cl(Int(Z\C)) ⊂ Cl(Int(Cl(Z\C))) = Cl(Int(A)), we have that A = Cl(Int(A)). Therefore, since Int(A) = Z \Cl(Z \ A) = Z \ B, A = Cl(Z \ B). Q.E.D. 6.6.14. Theorem. If X is an irreducible continuum such that X is Cn∗ –smooth at X for some n ≥ 2, then X is indecomposable. Proof. Assume that a and b are points about which X is irreducible. Suppose X is decomposable. Let C be a nondegenerate proper subcontinuum of X, with nonempty interior, containing b. Let A = Cl(X \ C) and B = Cl(X \ A). Then A and B are subcontinua of X (Theorem 1.7.26) containing a and b, respectively. Note that A = Cl(X\B), by Lemma 6.6.13. Copyright © 2005 Taylor & Francis Group, LLC


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Since A ∩ B = Bd(A) = Bd(B) and since B = Cl(X \ A) (and A = Cl(X \ B)), A (and B, respectively) is irreducible between a (b, respectively) and any point of A ∩ B (11.42 of [41]). Let α, β : [0, 1] → C(X) be order arcs such that α(0) = {a}, α(1) = A, β(0) = {b} and β(1) = B (Theorem 1.8.20). Notice that for any t ∈ [0, 1), (A ∩ B) ∩ α(t) = ∅ and (A ∩ B) ∩ β(t) = ∅. By Lemma 6.6.10, X is not Cn∗ –smooth at X, a contradiction. Therefore, X is indecomposable. Q.E.D. 6.6.15. Theorem. Let X be a continuum and let A be an element of Cn (X) with exactly n components, n ≥ 2. Then X is Cn∗ –smooth at A if and only if X is C ∗ –smooth at each component of A. Proof. Let A ∈ Cn (X) \ Cn−1 (X). If X is Cn∗ –smooth at A, then X is C ∗ –smooth at each component of A by Theorem 6.6.12. Let A be an element of Cn (X) with n components A1 , . . . , An . Suppose X is C ∗ –smooth at each Aj for each j ∈ {1, . . . , n}. Let {Bk }∞ k=1 be a sequence of elements of Cn (X) converging to A. Since A has n components, without loss of generality, we assume that Bk has n components, Bk1 , . . . , Bkn , for each k ∈ IN. In fact, we may suppose that lim Bkj = Aj for each j ∈ {1, . . . , n}. k→∞

Let C be an element of Cn (A). Let Aj1 , . . . , Aj be the compo nents of A intersecting C, i.e., C = (Aji ∩ C). Let Cji = Aji ∩ C i=1

for each i ∈ {1, . . . , }. Since X is C ∗ –smooth at Aji , there exists a subcontinuum Dkji of Bkji for each i ∈ {1, . . . , } such that lim Dkji = Cji . For k ∈ IN, let Dk = Dkji . Hence, Dk ∈ Cn (Bk ) k→∞

i=1

and lim Dk = C. k→∞

Therefore, X is Cn∗ –smooth at A by Lemma 6.6.4. Q.E.D. 6.6.16. Theorem. Let X be a continuum. If A is an element of Cn (X) for some n ≥ 2 such that all the components of A are indecomposable and X is C ∗ –smooth at each component of A, then X is Cn∗ –smooth at A. Copyright © 2005 Taylor & Francis Group, LLC


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Proof. Let A be an element of Cn (X). Let A1 , . . . , A ( ≤ n) be the components of A. Suppose Aj is an indecomposable continuum and X is C ∗ –smooth at Aj for each j ∈ {1, . . . , }. Let {Bk }∞ k=1 be a sequence of elements of Cn (X) converging to A. Let C be an element of Cn (A). Let Aj1 , . . . , Ajs be the components s (Aji ∩ C). Let Cji = Aji ∩ C for of A intersecting C, i.e., C = i=1

each i ∈ {1, . . . , s}. In what follows, k ∈ IN and i ∈ {1, . . . , s}. Since each Aji is indecomposable, by a similar argument to the one given in Lemma 6.6.8, there are components Bkji of Bk such that lim Bkji = Aji . Since X k→∞

is C ∗ –smooth at each Aji , there are subcontinua Dkji of Bkji such s ji Dkji . Then Dk ∈ Cn (Bk ) and that lim Dk = Cji . Let Dk = k→∞

i=1

lim Dk = C. Therefore, X is Cn∗ –smooth at A by Lemma 6.6.4. k→∞ Q.E.D. 6.6.17. Corollary. Let X be a continuum and let n ≥ 2. If A is an element of Cn (X) such that all the components of A are hereditarily indecomposable, then X is Cn∗ –smooth at A. Proof. The corollary follows from Theorem 6.6.16 and the fact that hereditarily indecomposable continua are absolutely C ∗ –smooth continua (by (14.14.1) of [40] and 3.2 of [14]). Q.E.D.

6.7

Retractions

We present results about retractions between the hyperspaces of locally connected continua. 6.7.1. Lemma. Let n ∈ IN. If X is a continuum containing an open set with uncountably many components, then Cn (X) contains an open set with uncountably many components. Copyright © 2005 Taylor & Francis Group, LLC

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Proof. Let U be an open subset of X with uncountably many components. Let Γ = U n . Then Γ is an open subset of C n (X), and Γ ⊂ U . Since for each x ∈ U , {x} ∈ Γ, we have that Γ = U. By Lemma 6.1.1, if Λ is a component of Γ, then Λ has at most n components. Since U has uncountably many components, Γ=U and for each component Λ of Γ, Λ has at most n components, we have that Γ has uncountably many components. Q.E.D. We begin considering maps between some of the hyperspaces. 6.7.2. Theorem. Let n ∈ IN. Then for any continuum X, there exists a map of 2X onto Cn (X). Proof. If X is locally connected, then 2X and Cn (X) are locally connected continua (Théorème II and Théorème IIm of [47]). Hence, the result follows from 8.19 of [41]. Suppose that X is not locally connected. Then there exists a map f of 2X onto the cone over the Cantor set ((1.39) of [40]). Also, there exists a map g of the cone over the Cantor set onto Cn (X) (Remark (p. 29) and Theorem 2.7 of [23]). Thus g ◦ f is a map of 2X onto Cn (X). Q.E.D. 6.7.3. Theorem. Let n ∈ IN, n ≥ 2. If X is a continuum containing an open set with uncountably many components, then there exists a map of Cn (X) onto C(X). Proof. Since X contains an open subset with uncountably many components, Cn (X) contains an open subset with uncountably many components (Lemma 6.7.1). Thus, there exists a map f of Cn (X) onto the cone over the Cantor set (Theorem II of [2]). Also, there exists a map g of the cone over the Cantor set onto C(X) (Theorem 2.7 of [23]). Hence, g ◦ f is a map of Cn (X) onto C(X). Q.E.D.


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6.7.4. Definition. Let Z be a metric space. By a deformation we mean a map H : Z × [0, 1] → Z such that for each z ∈ Z, H((z, 1)) = z. Let A = {H((z, 0)) | z ∈ Z}. If the map h : Z → A given by h(z) = H((z, 0)) is a retraction from Z onto A, then H is a deformation retraction from Z onto A. If H is a deformation retraction from Z onto A such that for each z ∈ A and each t ∈ [0, 1], H((z, t)) = z, then H is a strong deformation retraction from Z onto A. The set A is called deformation retract (strong deformation retract, respectively). The next Lemma provides a sufficient condition for a continuum X for the nonexistence of a deformation of the m–fold hyperspace of X onto a subset of the n–fold hyperspace of X, m > n. 6.7.5. Lemma. Let n, m ∈ IN and let X be a nonlocally connected continuum, with metric d. Let p be a point of X at which X is not connected im kleinen, and let A ⊂ Cn (X) such that {p} ∈ A. If m > n, then there does not exist a deformation, H, of Cm (X) onto A such that H(({p}, t)) = {p} for each t ∈ [0, 1]. Proof. Since X is not connected im kleinen at p, there exists a neighborhood U of p and a sequence {Kj }∞ j=1 of components of ClX (U ) converging to a continuum K ⊂ ClX (U ) such that p ∈ K and such that for each j ∈ IN, Kj ∩ K = ∅ ((12.1) (p. 18) of [45]). Let {pj }∞ j=1 be a sequence in ClX (U ) converging to p and such that pj ∈ Kj for each j ∈ IN. → Cm (X) Suppose there exists a deformation H : Cm (X) × [0, 1] → from Cm (X) onto A such that for each t ∈ [0, 1], H(({p}, t)) = {p}. Let H be the segment homotopy associated with H ((16.3) of [40]), given by H ((A, t)) = σ({H((A, s)) | 0 ≤ s ≤ t}). Observe that for each t ∈ [0, 1], H (({p}, t)) = {p} and for each A ∈ Cm (X), H ((A, 0)) = H((A, 0)) ∈ Cn (X). Let ε > 0 such that Vεd (p) ⊂ U . Then there exists δ > 0 such that if A ∈ Cm (X) and H(A, {p}) < δ, then for each t ∈ [0, 1], H(H ((A, t)), H (({p}, t))) = H(H ((A, t)), {p}) < ε. For each j ∈ IN, let Aj = {p} ∪ {pj+ }m−1 =1 . Hence, there exists j0 ∈ IN such that if j ≥ j0 , then H(Aj , {p}) < δ. Choose j ≥ j0 . Then {H (Aj , t) | t ∈ [0, 1]} is a connected subset of VεH ({p}) such that H ((Aj , 0)) ∈ Cn (X). Thus σ({H ((Aj , t)) | t ∈ Copyright © 2005 Taylor & Francis Group, LLC

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[0, 1]}) is a closed subset of X having at most n components (by a similar argument to the one given in Lemma 6.1.1). Since Aj ⊂ σ({H ((Aj , t)) | t ∈ [0, 1]}) ⊂ Vεd (p) ⊂ U and since Aj ∩ K = ∅ and Aj ∩Kj+ = ∅ for each ∈ {1, . . . , m−1}, we obtain a contradiction. We conclude that there does not exist a deformation H of Cm (X) onto A such that for each t ∈ [0, 1], H(({p}, t)) = {p}. Q.E.D. 6.7.6. Corollary. Let m, n ∈ IN. If X is a nonlocally connected continuum and m > n, then there does not exist a strong deformation retraction of Cm (X) onto Fn (X).

6.7.7. Theorem. Let n ∈ IN, n ≥ 2, and let X be a continuum. If F1 (X) is a deformation retract of 2X , then F1 (X) is a deformation retract of Cn (X). → 2X be a deformation retraction of 2X Proof. Let H : 2X × [0, 1] → onto F1 (X). Let r(A) = H((A, 0)), for each A ∈ 2X . → Cn (X) by Define G : Cn (X) × [0, 1] →

⎧ 1 ⎪ ⎪ if t ∈ 0, ; ⎨σ({H((A, s)) | 0 ≤ s ≤ 2t}) 2

G((A, t)) = 1 ⎪ ⎪ ⎩σ({H((A, s)) | 2t − 1 ≤ s ≤ 1}) if t ∈ , 1 . 2 1 , then both definitions of G give the value 2

1 σ({H((A, s)) | 0 ≤ s ≤ 1}). Let t ∈ 0, . Since H is continuous, 2 {H((A, s)) | 0 ≤ s ≤ 2t} is a closed connected subset of 2X that contains the element H((A, 0)) = A and A ∈ Cn (X). By similar argument to the one given in Lemma 6.1.1, G((A, t)) ∈ Cn (X) for each A ∈ Cn (X). Similarly, since H((A, 1)) ∈ F1 (X), G((A, t)) ∈ Cn (X) 1 for each t ∈ , 1 and each A ∈ Cn (X). Hence, G is well defined. 2 Since H and σ (Lemma 1.8.11) are continuous, it follows that G is continuous also. Note that if t =


324


Note that G((A, 0)) = r(A) and G((A, 1)) = A. Therefore, F1 (X) is a deformation retract of Cn (X). Q.E.D. The proof of the following Theorem is similar to the one given in Theorem 6.7.7. 6.7.8. Theorem. Let n ∈ IN, n ≥ 2, and let X be a continuum. If F1 (X) is a deformation retract of Cn (X), then F1 (X) is a deformation retract of C(X).

6.7.9. Definition. A metric ρ for a continuum X is said to be convex provided that given two points x and y of X there exists a ρ(x, y) point z in X such that ρ(x, z) = = ρ(z, y). 2 The following result was proved by R. H. Bing [4] and E. E. Moise [38], independently. 6.7.10. Theorem. Every locally connected continuum admits a convex metric.

6.7.11. Definition. If X is a locally connected continuum with a convex metric ρ, let Kρ : [0, ∞) × 2X → 2X be given by Kρ ((t, A)) = {x ∈ X | ρ(x, y) ≤ t for some y ∈ A}.

6.7.12. Remark. If X is a locally connected continuum with a convex metric ρ, then given two points x and y of X, there exists an arc γ : [0, ρ(x, y)] → X such that γ is an isometry ((0.65.3)(a) of [40]). This implies that the set Kρ ((t, A)) has at most as many components as A for each t ∈ [0, ∞). Also, observe that if t ≥ diam(X), then Kρ ((t, A)) = X for each A ∈ 2X .


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6.7.13. Theorem. Let n ∈ IN and let X be a locally connected continuum with a convex metric ρ. Then the function αρn : 2X → IR given by αρn (A) = inf{t ≥ 0 | Kρ ((t, A)) ∈ Cn (X)} satisfies the following: (a) Kρ ((αρn (A), A)) belongs to Cn (X) for each A ∈ 2X . (b) If A and B belong to 2X , t ≥ 0, Kρ ((t, A)) ∈ Cn (X) and Hρ (A, B) ≤ η, then Kρ ((t + η, B)) ∈ Cn (X). (c) αρn is continuous. Proof. Since X is locally connected with a convex metric, Kρ is continuous ((0.65.3)(f) of [40]). To see Kρ ((αρn (A), A)) belongs to Cn (X) for each A ∈ 2X , observe that, by definition of αρn , there exists n a decreasing sequence, {tj }∞ j=1 , of real numbers converging to αρ (A) such that Kρ ((tj , A)) belongs to Cn (X) for each j ∈ IN. By the continuity of Kρ , we have that lim Kρ ((tj , A)) = Kρ ((αρn (A), A)). Since j→∞

Cn (X) is compact (Theorem 1.8.5), we have that Kρ ((αρn (A), A)) ∈ Cn (X). Now, suppose A and B belong to 2X , t ≥ 0, Kρ ((t, A)) ∈ Cn (X) and Hρ (A, B) ≤ η. Observe first that A ⊂ Kρ ((t, A)) ⊂ Kρ ((t + η, B)). Let x ∈ Kρ ((t + η, B)), then there exists b ∈ B such that ρ(x, b) ≤ t + η. Since b ∈ B and Hρ (A, B) ≤ η, there exists a ∈ A such that ρ(a, b) ≤ η. Since ρ is a convex metric, there exist arcs γ1 : [0, ρ(x, b)] → X and γ2 : [0, ρ(b, a)] → X such that γ1 (0) = x, γ1 (ρ(x, b)) = b, γ2 (0) = b, γ2 (ρ(b, a)) = a, and both γ1 and γ2 are isometries ((0.65.3)(a) of [40]). Let Gx = γ1 ([0, ρ(x, b)]) ∪ γ2 ([0, ρ(b, a)]). Then Gx is a connected subset of Kρ ((t + η, B)) containing x and intersecting Kρ ((t, A)). Hence, Kρ ((t + η, B)) has at most as many components as Kρ ((t, A)). Therefore, Kρ ((t + η, B)) ∈ Cn (X). To show αρn is continuous, let η > 0. Let A and B be two elements of 2X such that Hρ (A, B) ≤ η. By (a), Kρ ((αρn (A), A)) belongs to Cn (X). Hence, by (b), Kρ ((αρn (A) + η, B)) also belongs to Cn (X). By definition of αρn (B), we have that αρn (B) ≤ αρn (A) + η. Interchanging the roles of A and B, in the above argument, we obtain αρn (A) ≤ αρn (B) + η. Hence, |αρn (A) − αρn (B)| ≤ η. Therefore, αρn is continuous. Q.E.D. Copyright © 2005 Taylor & Francis Group, LLC

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The next five results give conditions for the existence of retractions between the hyperspaces of locally connected continua. 6.7.14. Theorem. Let n, m ∈ IN, m > n, and let X be a continuum. Then Cn (X) is a strong deformation retract of Cm (X) if and only if X is locally connected. Proof. If X is not locally connected, then there is not a strong deformation retraction from Cm (X) onto Cn (X) (Lemma 6.7.5). If X is a locally connected continuum, then the map H : Cm (X)× [0, 1] → → Cm (X) given by H((A, t)) = Kρ ((1 − t)αρn (A), A) is a strong deformation retraction from Cm (X) onto Cn (X) (Remark 6.7.12 and Theorem 6.7.13). Q.E.D. The proof of the following Theorem is similar to the one given for Theorem 6.7.14. 6.7.15. Theorem. Let n ∈ IN and let X be a continuum. Then Cn (X) is a strong deformation retract of 2X if and only if X is locally connected.

6.7.16. Lemma. Let n ∈ IN. If X is a locally connected continuum, then F1 (X) is a retract of Cn (X) if and only if X is an absolute retract. Proof. Since X is a locally connected continuum, Cn (X) is an absolute retract (Théorème IIm of [47]). If F1 (X) is a retract of Cn (X), then F1 (X) is an absolute retract (see (2.2) (p. 86) of [5]). Since X is homeomorphic to F1 (X), X is an absolute retract. If X is an absolute retract, then F1 (X) is a retract of Cn (X) (1.5.2 of [37]). Q.E.D.


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327

6.7.17. Theorem. Let n ∈ IN. If X is a locally connected continuum, then the following are equivalent: (1) X is an absolute retract. (2) F1 (X) is a retract of Cn (X). (3) F1 (X) is a deformation retract of Cn (X). (4) F1 (X) is a strong deformation retract of Cn (X). Proof. Clearly (4) implies (3), and (3) implies (2). Note that (2) implies (1) by Lemma 6.7.16. Suppose now, X is an absolute retract. Then Cn (X) is an absolute retract (Théorème IIm of [47]). Hence, there exists a retraction rn : 2X → Cn (X). By 2.5 of [12], there exists a strong deformation retraction G : 2X × [0, 1] → 2X , from 2X onto F1 (X). Thus, rn ◦ G|Cn (X)×[0,1] is a strong deformation retraction from Cn (X) onto F1 (X). Therefore, (1) implies (4). Q.E.D.

6.7.18. Theorem. Let n ∈ IN. If X is a locally connected continuum, then the following hold: (1) Cn (X) is a retract of 2X . (2) Cn (X) is a deformation retract of 2X . (3) Cn (X) is a strong deformation retract of 2X . (4) C(X) is a retract of Cn (X). (5) C(X) is a deformation retract of Cn (X). (6) C(X) is a strong deformation retract of Cn (X). Proof. By Théorème IIm of [47], Cn (X) is an absolute retract. Hence, (1) and (4) hold. Since X is locally connected, X has the property of Kelley. Thus, Cn (X) is contractible (Corollary 6.1.16) and 2X is contractible too (Theorem 6.1.14). This implies, by 32E.4 of [46], the equivalence of (1) and (2) and of (4) and (5). By Theorems 6.7.15 and 6.7.14, (3) and (6) are equivalent to the fact that X is locally connected. Q.E.D.



328

6.8

Graphs

We present results about the n–fold hyperspaces of graphs. 6.8.1. Definition. A graph is a continuum which can be written as the union of finitely many arcs, any two of which are either disjoint or intersect only in one or both of their end points.

6.8.2. Definition. Let X be a graph and let A be a subset of X. Let β be a cardinal number. We say that A is of order less than or equal to β in X, written ord(A, X) ≤ β, provided that for each open subset U of X containing A, there exists an open subset V of X such that A ⊂ V ⊂ U and BdX (V ) has cardinality less than or equal to β. We say that A is of order β in X, written ord(A, X) = β, provided that ord(A, X) ≤ β and ord(A, X) ≤ α for any cardinal number α < β. A point x of a graph X is a ramification point of X if and only if ord({x}, X) ≥ 3. A point a of a graph X is a end point of X if and only if ord({a}, X) = 1. The next Theorem characterizes graphs as the class of locally connected continua X having finite dimensional hyperspaces Cn (X).

6.8.3. Theorem. A locally connected continuum X is a graph if and only if for each n ∈ IN, Cn (X) is of finite dimension. Proof. Suppose X is a graph. Then dim(C(X)) < ∞ (Lemma 5.2 of [23]). Hence, given n ∈ IN, by Corollary 6.1.22, we have that dim(Cn (X)) ≤ n(dim(C(X)) < ∞. Suppose X is a locally connected continuum such that for each n ∈ IN, dim(Cn (X)) < ∞. Then we have that dim(C(X)) < ∞. Thus, X is a graph (Lemma 5.2 of [23]). Q.E.D.


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329

6.8.4. Theorem. Let n ∈ IN, n ≥ 2. Then Cn ([0, 1]) \ Cn−1 ([0, 1]) is embeddable in IR2n . Proof. Given A ∈ Cn ([0, 1]) \ Cn−1 ([0, 1]), without loss of generality, we assume that [a1 , a1 ], . . . , [an , an ] are the components of A and a1 ≤ a1 < a2 ≤ a2 < . . . < an ≤ an . Define ξ : Cn ([0, 1]) \ Cn−1 ([0, 1]) → IR2n by ξ(A) = (a1 , a1 , . . . , an , an ). Clearly, ξ is a one–to–one function. To see its continuity, let ε > 0 and let B = n [bj , bj ] be a point of Cn ([0, 1])\Cn−1 ([0, 1]) such that H(A, B) < ε. j=1

Then for each j ∈ {1, . . . , n}, |aj − bj | < ε and |aj − bj | < ε. Hence, D(ξ(A), ξ(B)) = max{|aj −bj |, |aj −bj | | j ∈ {1, . . . , n}} < ε. Thus, ξ is continuous. Note that ξ VεH (A) ∩ Cn ([0, 1]) \ Cn−1 ([0, 1]) = 2n VεIR (ξ(A)) ∩ ξ (Cn ([0, 1]) \ Cn−1 ([0, 1])). Therefore, ξ is an embedding. Q.E.D. 6.8.5. Theorem. Let n ∈ IN, and let En = {(x1 , . . . , x2n ) ∈ IR2n | 0 ≤ x1 ≤ . . . ≤ x2n ≤ 1}. Then the function f : En → Cn ([0, 1]) given by n f ((x1 , . . . , x2n )) = [x2j−1 , x2j ] j=1

is continuous and surjective. Furthermore, En /Gf is homeomorphic to Cn ([0, 1]). Proof. Let ε > 0 be given. If (x1 , . . . , x2n ) and (x1 , . . . , x2n ) belong to En and D((x1 , . . . , x2n ), (x1 , . . . , x2n )) = max{|xj − xj | | j ∈ {1, . . . , 2n}} < ε, then we have that H(f ((x1 , . . . , x2n )), f ((x1 , . . . , x2n ))) ≤ max{|xj − xj | | j ∈ {1, . . . , 2n}} < ε. Thus, f is continuous. k [xj , yj ], where k ≤ n, then (x1 , y1 , . . . , xk , yk , yk , . . . , yk ) If A = j=1

belongs to En and f ((x1 , y1 , . . . , xk , yk , yk , . . . , yk )) = A. Then f is Copyright © 2005 Taylor & Francis Group, LLC


330

surjective. The fact that En /Gf is homeomorphic to Cn ([0, 1]) follows from Theorem 1.2.10. Q.E.D. 6.8.6. Notation. Let S 1 denote the unit circle in the plane. If θ ∈ S 1 and t ∈ [0, 1), then let A(θ, t) be the arc (possibly degenerate) contained in S 1 having θ as mid point and length 2πt. For t = 1, A(θ, 1) denotes S 1 . The following Lemma is easily established. 6.8.7. Lemma. Let ε > 0 be given. If θ and ϕ belong to S 1 , t and ε s belong to [0, 1] and satisfy that ||θ − ϕ|| < ε and |t − s| < , 2π then H(A(θ, t), A(ϕ, s)) < ε.

6.8.8. Notation. Let n ∈ IN. If n > 1, let Tn = 1S 1 × ·23 · · × S41 . If n times

n = 1, then T1 = S 1 .

6.8.9. Theorem. Let n ∈ IN. If f : Tn × [0, 1]n → Cn (S 1 ) is a function given by f ((θ1 , . . . , θn , t1 , . . . , tn )) =

n

A(θj , tj ),

j=1

then f is continuous and surjective. Furthermore, (Tn × [0, 1]n )/Gf is homeomorphic to Cn (S 1 ). Proof. Let ε > 0 be given. If (θ1 , . . . , θn , t1 , . . . , tn ), (ϕ1 , . . . , ϕn , s1 , . . . , sn ) ∈ Tn × [0, 1]n and D((θ1 , . . . , θn , t1 , . . . , tn ), (ϕ1 , . . . , ϕn , ε ε ε s1 , . . . , sn )) < , then |θj − ϕj | < < ε and |tj − sj | < for 2π 2π 2π n each j ∈ {1, . . . , n}. By Lemma 6.8.7, it follows that A(θj , tj ) ⊂ j=1


6.8. GRAPHS

Vεd H

n

331

A(ϕj , sj )

j=1 n

A(θj , tj ),

j=1

and

n j=1

n

A(ϕj , sj ) ⊂

A(ϕj , sj )

Vεd

n

A(θj , tj ) . Thus,

j=1

< ε. Therefore, f is continuous.

j=1

Let A ∈ Cn (S 1 ), and suppose A1 , . . . , Ak be the components of A, with k ≤ n. For each j ∈ {1, . . . ,

k}, let θj be the midpoint of Aj t1 and let tj be the length of Aj . Then θ1 , . . . , θk , θk , . . . , θk , , . . . , 1 23 4 2π n−k times tk tk tk , ,... , is an element of Tn × [0, 1]n whose image under 2π 12π 23 2π4 n−k times

f is A. Thus, f is surjective. The fact that (Tn ×[0, 1]n )/Gf is homeomorphic to Cn (S 1 ) follows from Theorem 1.2.10. Q.E.D. As a consequence of Corollary 6.1.23 we have the following Theorem: 6.8.10. Theorem. dim(Cn ([0, 1])) = dim(Cn (S 1 )) = 2n for every n ∈ IN. The proof of the following Theorem is due to R. Schori: 6.8.11. Theorem. C2 ([0, 1]) is homeomorphic to [0, 1]4 . Proof. Let D1 = {A ∈ C2 ([0, 1]) | 1 ∈ A} and D01 = {A ∈ C2 ([0, 1]) | {0, 1} ⊂ A}. We divide the proof in three steps. Step 1. C2 ([0, 1]) is homeomorphic to K(D1 ) (the cone over 1 D ). Let f : K(D1 ) → → C2 ([0, 1]) be given by f ((A, t)) = (1 − t)A = {(1 − t)a | a ∈ A}. Since f ((A, 1)) = {0} for every A ∈ C2 ([0, 1]), f is a well defined map. Let A, B ∈ D1 and let s, t ∈ [0, 1] such that f ((A, t)) = f ((B, s)). Since 1 ∈ A ∩ B, 1 − t = max(1 − t)A and 1 − s = max(1 − s)B. This implies that t = s. If t = s = 1, then (A, 1) and (B, 1) both represent the vertex of K(D1 ). Copyright © 2005 Taylor & Francis Group, LLC

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Hence, we assume that t < 1. Since (1 − t)A = (1 − t)B, A = B. Therefore, f is one–to–one. Now,let A ∈ C2 ([0, 1]) \ {{0}}. Take 1 1 t = 1 − max A. Then max A = 1. Thus, A ∈ D1 1 − t 1 − t 1 and f A, t = A. Hence, f is surjective. Therefore, 1−t C2 ([0, 1]) is homeomorphic to K(D1 ). Step 2. D1 is homeomorphic to K(D01 ). → D1 be given by g((A, t)) = t + (1 − t)A = Let g : K(D01 ) → {t + (1 − t)a | a ∈ A}. Proceeding as in Step 1, it can be seen that g is a homeomorphism. Step 3. D01 is homeomorphic to [0, 1]2 . Let T = {(a, b) ∈ IR2 | 0 ≤ a ≤ b ≤ 1}, and let S = T /∆, where ∆ = {(a, b) ∈ T | a = b}. Note that S is homeomorphic to [0, 1]2 . Let h : T → → D01 be given by h((a, b)) = [0, a] ∪ [b, 1]. Then h is continuous. Note that h((a, b)) = h((c, d)) if and only if a = c and b = d. Hence, by Theorem 1.2.10, S and D01 are homeomorphic. Q.E.D. Recall that for any continuum X and any n ∈ IN, Cn (X) is unicoherent (Theorem 6.2.4). The next Theorem, whose proof is contained in Lemma 2.3 of [17], says that unicoherence of these hyperspaces may be destroyed by removing a point. 6.8.12. Theorem. C2 (S 1 ) \ {S 1 } is not unicoherent. The proof of the following Theorem is the content of [20]. 6.8.13. Theorem. C2 (S 1 ) is homeomorphic to the cone over the solid torus. The proof of the following Theorem is the content of [17] and [19].


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6.8.14. Theorem. Let X be a graph and let Y be a continuum such that Cn (Y ) is homeomorphic to Cn (X). Then: (1) if n = 1 and X is neither an arc nor a simple closed curve, then Y is homeomorphic to X, (2) if n ≥ 2, then Y is homeomorphic to X.

6.9

Cones

Recall that given a space Z, K(Z) denotes the cone over Z. Theorem 6.8.11 gives an example of a continuum whose 2–fold hyperspace is a cone. Corollary 6.9.4 extends this example. Note that Theorem 6.8.13 also gives such an example. We need the following definitions. 6.9.1. Definition. A dendroid is an arcwise connected and hereditarily unicoherent continuum. A fan is a dendroid with exactly one ramification point (i.e., with only one point which is the common part of three otherwise disjoint arcs) [8]. The unique ramification point of a fan F is called the top of F ; τ always denotes the top of a fan. By an end point of a fan F we mean an end point in the classical sense, that is, a point e of F which is a nonseparating point of any arc in F that contains e; E(F ) denotes the set of all end points of a fan F . A leg of a fan F is the unique arc in F from τ to some end point of F . Given two points x and y of a fan F , xy denotes the unique arc in F joining x and y. Given an m ∈ IN, an m–od is a fan for which E(F ) has exactly m elements.

6.9.2. Definition. A fan F is said to be smooth provided that whenever {xi }∞ i=1 is a sequence in F converging to a point x of F , then the sequence of arcs {τ xi }∞ i=1 converges to the arc τ x. Copyright © 2005 Taylor & Francis Group, LLC

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Given a fan F , let G(F ) denote either of the hyperspaces 2F , Cn (F ), or Fn (F ), for n ∈ IN. 6.9.3. Theorem. If F is a fan, with metric d, which is homeomorphic to the cone over a compact metric space, then G(F ) is homeomorphic to the cone over a continuum. Proof. Let F be a fan which is a cone. Then F is smooth. We assume by (Corollary 4 (p. 90) of [11] and Theorem 9 (p. 27) of [8]) that F is embedded in IR2 , τ = (0, 0) is the top of F and the legs of F are convex arcs of length one (4.2 of [35]). Given two points a and b of IR2 , [a, b] denotes the convex arc in IR2 whose end points are a and b, and ||a|| denotes the norm of a in IR2 . Given an element A of G(F ) and r ≥ 0, rA = {ra | a ∈ A}. Note that for r = 0, rA = {(0, 0)} = {τ }. Let E(F ) = {eλ }λ∈Λ . Then F = K(E(F )) by 4.2 of [35]. Note that this equality implies that E(F ) is closed in F . Hence, E(F ) is a compactum. Let B = {{A ∈ G(F ) | eλ ∈ A} | λ ∈ Λ}. Let ϕ : B × I → G(F ) be given by ϕ((A, t)) = (1 − t)A. Clearly, ϕ is well defined. Observe that if t ∈ [0, 1) and A ∈ B, then {τ } ∈ ϕ((A, t)) if and only if τ ∈ A. We show that ϕ is ε continuous. Let ε > 0 be given and let δ = . Let A, B ∈ G(F ) and 2 let t, s ∈ [0, 1] such that H(A, B) < δ and |t − s| < δ. Let a ∈ A. Then there exists b ∈ B such that ||a − b|| < δ. Note that ||(1 − t)a − (1 − s)b|| ≤ ||(1 − t)a − (1 − t)b|| + ||(1 − t)b − (1 − s)b|| ≤ (1 − t)||a − b|| + |s − t|||b|| ≤ ||a − b|| + |s − t| < 2δ = ε. Thus, ϕ((A, t)) ⊂ Vεd (ϕ((B, s))). Similarly, we have that ϕ((B, s)) ⊂ Vεd (ϕ((A, t))). Therefore, H(ϕ((A, t)), ϕ((B, s))) < ε and ϕ is continuous. We show that ϕ is one–to–one on B × [0, 1). Let t, s ∈ [0, 1), and let A, B ∈ B. Suppose that ϕ((A, t)) = ϕ((B, s)). Since A, B ∈ B, there exist eλ , eλ ∈ E(F ) such that eλ ∈ A and eλ ∈ B. Copyright © 2005 Taylor & Francis Group, LLC

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Case (1). τ ∈ A. Then τ ∈ B. Let [a, eλ ] be the component of A containing eλ . Let [b, eλ ] be the component of B containing eλ . Since ϕ((A, t)) = ϕ((B, s)), there exist [bλ , cλ ] ⊂ A and [bλ , cλ ] ⊂ B such that [(1 − t)a, (1 − t)eλ ] = [(1 − s)bλ , (1 − s)cλ ] and [(1 − s)b, (1 − s)eλ ] = [(1 − t)bλ , (1 − t)cλ ]. From the first equality we obtain that (1 − t)eλ = (1 − s)cλ , which implies that 1 − t = (1 − s)||cλ || ≤ 1 − s. From the second inequality we obtain that (1 − s)eλ = (1 − t)cλ , which implies that 1 − s = (1 − t)||cλ || ≤ 1 − t. Therefore, t = s. Hence, A = B. Case (2). τ ∈ A. Then τ ∈ B. Let us observe that either [τ, eλ ] ⊂ A or there exists a ∈ A such that [a, eλ ] ⊂ A. In either case, as in Case (1), we conclude that (1 − t)eλ = (1 − s)cλ , for some cλ ∈ E(B), and that (1 − s)eλ = (1 − t)cλ , for some cλ ∈ E(A). These two equalities imply that t = s. Hence, A = B. We show that ϕ is surjective. Let B ∈ G(F ). If B = {τ }, then ϕ((A, 1)) = {τ } for any A ∈ B. Thus, assume B = {τ }. If B ∩ E(F ) = ∅, then ϕ((B, 0)) = B. Suppose B ∩ E(F ) = ∅. Let t = inf{||b − eλ || | b ∈ B and eλ ∈ E(F )}. Since B = {τ }, t = 1. Then there exists λ0 ∈ Λ such that ||bλ0 − eλ0 || = t, where bλ0 ∈ B ∩ [τ, eλ0 ]. 1 1 B. Note that for λ0 , bλ ∈ A ∩ [τ, eλ0 ]. Let A = 1−t 1−t 0 1 bλ = eλ0 , we have that eλ0 ∈ A. Hence, A ∈ B, and Since 1−t 0 1−t B = B. ϕ((A, t)) = 1−t By Theorem 1.2.10, the hyperspace G(F ) is homeomorphic to K(B). Since no point of G(F ) arcwise disconnects 2F ((11.5) of [40]), we have that B is a continuum. Q.E.D. Note that a similar proof to the one given for Theorem 6.9.3 shows:


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6.9.4. Corollary. If G([0, 1]) ∈ {2[0,1] , Cn ([0, 1]), Fn ([0, 1])}, then G([0, 1]) is homeomorphic to the cone over a continuum. Since, clearly, a simple m–od is a fan homeomorphic to the cone over a finite set, we have the following:

6.9.5. Corollary. Let m and n be positive integers. If F is a simple m–od, and G(F ) ∈ {Fn (F ), Cn (F )}, then G(F ) is homeomorphic to the cone over a finite–dimensional continuum. The next Theorem shows that for n ≥ 2, no n–fold hyperspace of a finite–dimensional continuum is homeomorphic to its cone.

6.9.6. Theorem. Let X be a finite–dimensional continuum. Then for each integer n ≥ 2, Cn (X) is not homeomorphic to K(X). Proof. Let n ≥ 2 and suppose Cn (X) is homeomorphic to K(X). Since X is of finite dimension, K(X) is of finite dimension too. In fact dim(K(X)) = dim(X) + 1 ((8.0) of [40]). Since C(X) ⊂ Cn (X) and Cn (X) is homeomorphic to K(X), we have that dim(C(X)) < ∞. Hence, by the dimension theorem, dim(X) = 1 (Theorem 2.1 of [26]). Thus, dim(K(X)) = 2, and dim(Cn (X)) = 2. Therefore, since Cn (X) contains an n–cell (Theorem 6.1.9), n = 2. We consider two cases. Case (1). X contains a proper decomposable subcontinuum. Then, by Theorem 6.1.10, C2 (X) contains a 3–cell, a contradiction to the fact that dim(C2 (X)) = 2. Case (2). All proper subcontinua of X are indecomposable. Then, by Lemma 1.7.24, X is hereditarily indecomposable. Hence, K(X) is uniquely arcwise connected. On the other hand, since C2 (X) contains 2–cells (Theorem 6.1.9), C2 (X) is not uniquely arcwise connected. Therefore, Cn (X) is not homeomorphic to K(X). Q.E.D.


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6.9.7. Lemma. Let X be a continuum such that C(X) is finite dimensional. If A is a nondegenerate indecomposable proper subcontinuum of X, then at most a finite number of composants of A have the property that some subcontinuum of X contains a point of X \ A and a point of the composant but does not contain A. Also, Cn (X) \ {A} has uncountably many arc components. Proof. The first part follows by the proof of (∗) (p. 312) of [40]. By the proof of (v) (p. 312) of [40], C(X) \ {A} has uncountably many arc components. Hence, Cn (X)\{A} is not arcwise connected (Theorem 6.5.5) and has uncountably many arc components by 11.15 of [41] and Theorem 6.5.11. Q.E.D. 6.9.8. Theorem. Let X be a continuum and let n ≥ 2 be an integer. If Z is a finite–dimensional continuum such that K(Z) is homeomorphic to Cn (X), then dim(X) = 1 and X contains at most one nondegenerate indecomposable continuum. Hence, X is not hereditarily indecomposable. Proof. Let n ≥ 2. Let h : Cn (X) → K(Z) be a homeomorphism. The proof of the fact that dim(X) = 1 is similar to the one given in Theorem 6.9.6. Suppose X contains two nondegenerate indecomposable continua, A and B. Then Cn (X) \ {A} and Cn (X) \ {B} have infinitely many arc components (Lemma 6.9.7). Hence, K(Z) \ {h(A)} and K(Z) \ {h(B)} both have infinitely many arc components. On the other hand, for each p ∈ K(Z) \ {νZ }, K(Z) \ {p} has at most two arc components. Therefore, X contains at most one nondegenerate indecomposable subcontinuum. Q.E.D. 6.9.9. Theorem. Let X be a continuum and let n ≥ 2. Then every 2–cell in Cn (X) is nowhere dense. Proof. First, suppose n ≥ 3. Let U be any nonempty open set in Cn (X). Then there exists A ∈ U such that A has exactly n components, A1 , . . . , An (Theorem 6.1.7). By Corollary 1.7.23, for Copyright © 2005 Taylor & Francis Group, LLC


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each j ∈ {1, . . . , n}, there is a subcontinuum Bj of X such that Bj n contains Aj properly, Bj ∩ B = ∅ if j = and B = Bj ∈ U. j=1

For each j ∈ {1, . . . , n}, let αj : [0, 1] → C(X) be an order arc such that αj (0) = Aj and αj (1) = Bj (Theorem 1.8.20). Define n γ((t1 , . . . , tn )) = αj (tj ). Then γ ([ 0, 1]n ) is an n–cell contained j=1

in U. Since n ≥ 3, U cannot be a 2–cell. Next, suppose n = 2. Suppose that D is a 2–cell in C2 (X) with nonempty interior in C2 (X). Then there exists D ∈ IntC2 (X) (D) such that D has two nondegenerate components, D1 and D2 (Theorem 6.1.7 and Corollary 1.7.23). Let ε > 0 such that if E ∈ C2 (X) and H(E, D) < ε, then E ∈ IntC2 (X) (D). Let µ : C(D2 ) → [0, 1] be a Whitney map. Let ϕ : C(D2 ) → C2 (X) be given by ϕ(B) = D1 ∪ B. Then ϕ is an embedding of C(D2 ) into {G ∈ C2 (X) | G ⊂ D}. Let t0 ∈ (0, 1) such that if µ(B) ≥ t0 , then H(ϕ(B), D) < ε. Let B = {ϕ(B) | t0 < µ(B) < 1}. Then B is a 2–dimensional subset of IntC2 (X) (D), dim(B) = 2 is seen using the fact that no zero–dimensional set separates C(D2 ) by ((2.15) of [40]). Hence, IntC2 (X) (B) = ∅ (10.2 of [42]). Thus, letting B0 ∈ C(D2 ) such that ϕ(B0 ) ∈ IntC2 (X) (B), we have that ϕ(B0 ) is not arcwise accessible from C2 (X) \ B. However, let β : [0, 1] → C(D1 ) be an order arc such that β(0) ∈ F1 (D1 ) and β(1) = D1 (Theorem 1.8.20). Then β(s) ∪ B0 ∈ B for any s ∈ [0, 1) and β(1) ∪ B0 = D1 ∪ B0 = ϕ(B0 ) ∈ B, a contradiction. Therefore, every 2–cell in Cn (X) is nowhere dense. Q.E.D. We end this chapter with the following Theorem, which gives conditions on an indecomposable continuum X in order to have its n–fold hyperspaces homeomorphic to a cone over a finite–dimensional continuum. 6.9.10. Theorem. Let X be a continuum containing a nondegenerate indecomposable subcontinuum A. Let n ≥ 2 be an integer, and let Z be a finite–dimensional continuum such that K(Z) is homeomorphic to Cn (X). If h : Cn (X) → K(Z) is a homeomorphism, then Copyright © 2005 Taylor & Francis Group, LLC

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(1) h(A) = νZ ; (2) Z has uncountably many arc components. In particular, Z is not locally connected; (3) dim(Cn (X)) ≥ 2n and dim(Z) ≥ 2n − 1; (4) Each point z of Z is contained in an arc in Z and some points of Z belong to locally connected subcontinua of Z whose dimension is at least 2n − 1; (5) No point of K(Z) \ {νZ } arcwise disconnects K(Z); (6) If A = X, then X does not contain a nondegenerate proper terminal subcontinuum; (7) Z is not irreducible. In particular, Z is decomposable. Proof. (1) The proof is similar to the proof of Theorem 6.9.8. (2) By Lemma 6.9.7, Cn (X) \ {A} has uncountably many arc components. Since h(A) = νZ (by (1)), K(Z) \ {νZ } has uncountably many arc components. Thus, since K(Z) \ {νZ } is homeomorphic to Z × [0, 1), we conclude that Z has uncountably many arc components. (3) By Theorem 6.9.8, each subcontinuum of X, distinct from A, is decomposable. Thus, Cn (X) contains a 2n–cell (Corollary 6.1.12). Hence, dim(Cn (X)) ≥ 2n. Since K(Z) is homeomorphic to Cn (X) and dim(K(Z)) = dim(Z)+1 ((8.0) of [40]), we have that dim(Z) ≥ 2n − 1. (4) Let z be any point of Z. Let π : K(Z) \ {νZ } → → Z be the projection map. We consider two cases. First, suppose there exists t0 ∈ [0, 1) such that h−1 ((z, t0 )) ∈ Cn (X) \ C(X). Let B = h−1 ((z, t0 )) and let B1 , . . . , Bk be the components of B, where k ∈ {2, . . . , n}. By Corollary 1.7.23, for each j ∈ {1, . . . , k}, there exists a subcontinuum Cj of X containing Bj properly. We assume, without loss of generality, that Cj ∩C = ∅ if j = . For each j ∈ {1, . . . , k}, let αj : [0, 1] → C(X) be an order arc (Theorem 1.8.20) such that αj (0) = Bj and αj (1) = Cj . Let k k α : [0, 1] → Cn (X) be given by α((t1 , . . . , tk )) = αj (tj ). Let

j=1

D = α [0, 1] . Then D is a k–cell such that B ∈ D and A ∈ D. k


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Thus, h(D) is a k–cell containing the point (z, t0 ) and not containing νZ . Hence, π(h(D)) is a locally connected subcontinuum of Z containing z. Since k ≥ 2, π(h(D)) is nondegenerate. Thus, z is contained in an arc by (8.23 of [41]). Next, suppose that h−1 ((z, t)) ∈ C(X) for each t ∈ [0, 1). Since h(A) = νZ , there exists t ∈ [0, 1) such that h−1 ((z, t )) = A and h−1 ((z, t )) ∈ F1 (X). Let E = h−1 ((z, t )). Since E = A and E is nondegenerate, E is a decomposable continuum (by Theorem 6.9.8). Hence, there are two proper subcontinua K and H of E such that E = H ∪ K. Suppose, first, that A is not contained in E. Take x1 ∈ H \ K and x2 ∈ K \ H. Let βj : [0, 1] → C(X) be an order arc such that βj (0) = {xj }, j ∈ {1, 2}, β1 (1) = H and β2 (1) = K. Let β : [0, 1]2 → Cn (X) be given by β((t1 , t2 )) = β1 (t1 ) ∪ β2 (t2 ). Let G = β ([0, 1]2 ). Then G is a locally connected subcontinuum of Cn (X) such that G contains a 2–cell and such that E ∈ G and A ∈ G. Thus, h(G) is a locally connected subcontinuum of K(Z) containing a 2–cell, such that (z, t ) ∈ h(G) and νZ ∈ h(G). Hence, π(h(G)) is a nondegenerate locally connected subcontinuum of Z containing z. Thus, z is in an arc by (8.23 of [41]). Suppose next A is contained in E. Since A is indecomposable, E = A. Hence, there exists a point x1 ∈ E \ A. Suppose that x1 ∈ H. Choose a point x2 ∈ K \ {x1 }. Then we just repeat the argument in the preceding paragraph to construct a nondegenerate locally connected subcontinuum of Z containing z. This completes the proof of the first part of (4). We prove the second part of (4) as follows: By Corollary 6.1.12, there exists a 2n–cell E in Cn (X). We may choose E such that A ∈ E. Let B ∈ E. Then h(E) is a 2n–cell such that h(B) ∈ E. Hence, π(h(E)) is a locally connected subcontinuum of Z containing π(h(B)), and dim(π(h(E))) ≥ 2n − 1 (by 20.10 [42] since π(h(E))×[0, 1) contains h(E) and, thus, has dimension at least 2n). (5) By (4), each point of K(Z) lies in the cone over an arc. Hence, (5) follows easily. (6) This is a consequence of Theorem 6.9.8, Theorem 6.5.4 and part (5) of this theorem. (7) Suppose there exist two points z1 and z2 of Z such that Z is Copyright © 2005 Taylor & Francis Group, LLC

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irreducible between them. First, we prove that both z1 and z2 belong to π(h(Fn (X))). Suppose this is not true. Note that Fn (X) intersects all the arc components of Cn (X) \ {A}. Hence, there are arcs α1 and α2 in Z such that one end point of αj is zj and the other point of αj is in π(h(Fn (X))), for j ∈ {1, 2}. Hence, by irreducibility, Z = α1 ∪ α2 ∪ π(h(Fn (X))). On the other hand, Z cannot contain free arcs (otherwise, Cn (X) contains 2–cells with nonempty interior, which contradicts Theorem 6.9.9). Thus, we have proved that z1 and z2 belong to π(h(Fn (X))). Let t1 and t2 be points of [0, 1) such that (z1 , t1 ) and (z2 , t2 ) belong to h(Fn (X)). Let B1 and B2 be the elements of Fn (X) such that h(B1 ) = (z1 , t1 ) and h(B2 ) = (z2 , t2 ). Take x1 ∈ B1 and x2 ∈ B2 . Let B1 = {{x1 } ∪ B | B ∈ Fn−1 (X)} and B2 = {{x2 } ∪ B | B ∈ Fn−1 (X)}. Then B1 and B2 are subcontinua of Cn (X) containing B1 and B2 , respectively; also, B1 ∩ Fn−1 (X) = ∅ and B2 ∩ Fn−1 (X) = ∅. Hence, B1 ∪ B2 ∪ Fn−1 (X) is a subcontinuum of Cn (X) which does not intersect all the arc components of Cn (X) \ {A}, which we prove as follows: By Lemma 6.9.7, Cn (X) \ {A} has uncountably many arc components. Let a1 , . . . , an be n points of A in n distinct comn posants, κ1 , . . . , κn , of A such that {x1 , x2 } ∩ κj = ∅; if A = X, j=1

by Lemma 6.9.7, we may take n composants that are not accessible from X \ A. Let G be the arc component of Cn (X) \ {A} containing {a1 , . . . , an }. Then G ∩ (B1 ∪ B2 ∪ Fn−1 (X)) = ∅. Since B1 ∪ B2 ∪ Fn−1 (X) does not intersect all the arc components of Cn (X) \ {A}, we have that h (B1 ∪ B2 ∪ Fn−1 (X)) is a subcontinuum of K(Z) containing both (z1 , t1 ) and (z2 , t2 ), which does not intersect all the arc components of K(Z) \ {νZ }. Then π(h (B1 ∪ B2 ∪ Fn−1 (X))) is a proper subcontinuum of Z containing z1 and z2 , a contradiction. Therefore, Z is not irreducible. Q.E.D.


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REFERENCES REFERENCES

[1] G. Acosta, Continua with Unique Hyperspace, in Continuum Theory: Proceedings of the Special Session in Honor of Professor Sam B. Nadler, Jr.’s 60th Birthday. Lecture Notes in Pure and Applied Mathematics Series, Vol. 230, Marcel Dekker, Inc., New York, Basel, 2002. pp. 33–49. (eds.: Alejandro Illanes, Ira Wayne Lewis and Sergio Mac´ıas.) [2] D. P. Bellamy, The Cone Over the Cantor Set–continuous Maps From Both Directions, Proc. Topology Conference (Emory University, Atlanta, Ga., 1970), J. W. Rogers, Jr., ed., 8–25. [3] D. E. Bennett, Aposyndetic Properties of Unicoherent Continua, Pacific J. Math., 37 (1971), 585–589. [4] R. H. Bing, Partitioning a Set, Bull. Amer. Math. Soc., 55 (1949), 1101–1110. [5] K. Borsuk, Theory of Retracts, Monografie Mat. Vol. 44, PWN (Polish Scientific Publishers), Warszawa, 1967. [6] E. Casta˜ neda, A Unicoherent Continuum for Which its Second Symmetric Product is not Unicoherent, Topology Proceedings, 23 (1998), 61–67. [7] E. Casta˜ neda, Productos Simétricos, Tesis Doctoral, Facultad de Ciencias, U. N. A. M., 2003. [8] J. J. Charatonik, On fans, Dissertationes Math. (Rozprawy Mat.), 54 (1967), 1–37. [9] D. Curtis and N. T. Nhu, Hyperspaces of Finite Subsets Which are Homeomorphic to ℵ0 –dimensional Linear Metric Spaces, Topology Appl., 19 (1985), 251–260. [10] C. H. Dowker, Mapping Theorems for Non–compact Spaces, Amer. J. Math., 69 (1947), 200–242. [11] C. A. Eberhart, A Note on Smooth Fans, Colloq. Math. 20 (1969), 89–90. Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

343

[12] J. T. Goodykoontz, Some Retractions and Deformation Retractions on 2X and C(X), Topology Appl., 21 (1985), 121– 133. [13] J. Grispolakis, S. B. Nadler, Jr. and E. D. Tymchatyn, Some Properties of Hyperspaces with Applications to Continua Theory, Can. J. Math., 31 (1979), 197–210. [14] J. Grispolakis and E. D. Tymchatyn, Weakly Confluent Mappings and the Covering Property of Hyperspaces, Proc. Amer. Math. Soc., 74 (1979), 177–182. [15] C. L. Hagopian, Indecomposable Homogeneous Plane Continua are Hereditarily Indecomposable, Trans. Amer. Math. Soc., 224 (1976), 339–350. [16] J. G. Hocking and G. S. Young, Topology, Dover Publications, Inc., New York, 1988. [17] A. Illanes, The Hyperspace C2 (X) for a Finite Graph X is Unique, Glansnik Mat., 37(57) (2002), 347–363. [18] A. Illanes, Comparing n–fold and m–fold hyperspaces, Topology Appl., 133 (2003), 179–198. [19] A. Illanes, Finite Graphs Have Unique Hyperspaces Cn (X), Topology Proc., 27 (2003), 179–188. [20] A. Illanes, A Model for the Hyperspace C2 (S 1 ), Q. & A. in General Topology, 22 (2004), 117–130. [21] A. Illanes and S. B. Nadler, Jr., Hyperspaces: Fundamentals and Recent Advances, Monographs and Textbooks in Pure and Applied Math., Vol. 216, Marcel Dekker, New York, Basel, 1999. [22] F. B. Jones, Certain Homogeneous Unicoherent Indecomposable Continua, Proc. Amer. Math. Soc., 2 (1951), 855–859. [23] J. L. Kelley, Hyperspaces of a Continuum, Trans. Amer. Math. Soc., 52 (1942), 22–36. Copyright © 2005 Taylor & Francis Group, LLC

344

REFERENCES

[24] J. Krasinkiewicz, Curves Which Are Continuous Images of Tree–like Continua Are Movable, Fun. Math., 89 (1975), 233– 260. [25] K. Kuratowski, Topology, Vol. II, Academic Press, New York, N. Y., 1968. [26] M. Levin and Y. Sternfeld, The Space of Subcontinua of a 2–dimensional Continuum is Infinitely Dimensional, Proc. Amer. Math. Soc., 125 (1997), 2771–2775. [27] S. Mac´ıas, On Symmetric Products of Continua, Topology Appl., 92 (1999), 173–182. [28] S. Mac´ıas, Aposyndetic Properties of Symmetric Products of Continua, Topology Proc., 22 (1997), 281–296. [29] S. Mac´ıas, On the Hyperspaces Cn (X) of a Continuum X, Topology Appl., 109 (2001), 237–256. [30] S. Mac´ıas, On the Hyperspaces Cn (X) of a Continuum X, II, Topology Proc., 25 (2000), 255–276. [31] S. Mac´ıas, On Arcwise Accessibility in Hyperspaces, Topology Proc., 26 (2001–2002), 247–254. [32] S. Mac´ıas, Fans Whose Hyperspaces Are Cones, Topology Proc., 27 (2003), 217–222. [33] S. Mac´ıas and S. B. Nadler, Jr., n–fold Hyperspaces, Cones and Products, Topology Proc., 26 (2001–2002), 255–270. [34] S. Mac´ıas and S. B. Nadler, Jr., Smoothness in n–fold Hyperspaces, Glasnik Mat., 37(57)(2002), 365–373. [35] S. Mac´ıas and S. B. Nadler, Jr., Fans Whose Hyperspace of Subcontinua are Cones, Topology Appl., 126 (2002), 29–36. [36] S. Mardeˇsić and J. Segal, ε–mappings onto Polyhedra, Trans. Amer. Math. Soc., 109 (1963), 146–164. [37] J. van Mill, Infinite–Dimensional Topology, North Holland, Amsterdam, 1989. Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

345

[38] E. E. Moise, Grille Decomposition and Convexification Theorems for Compact Locally Connected Continua, Bull. Amer. Math. Soc., 55 (1949), 1111–1121. [39] S. B. Nadler, Jr., Arcwise Accessibility in Hyperspaces, Dissertationes Math., 138 (1976), 1–29. [40] S. B. Nadler, Jr., Hyperspaces of Sets, Monographs and Textbooks in Pure and Applied Math., Vol. 49, Marcel Dekker, New York, Basel, 1978. [41] S. B. Nadler, Jr., Continuum Theory: An Introduction, Monographs and Textbooks in Pure and Applied Math., Vol. 158, Marcel Dekker, New York, Basel, Hong Kong, 1992. [42] S. B. Nadler, Jr., Dimension Theory: An Introduction with Exercises, Aportaciones Matemáticas, Textos # 18, Sociedad Matemática Mexicana, 2002. [43] J. T. Rogers, Jr., Dimension and the Whitney Subcontinua of C(X), Gen. Top. and its Applications, 6 (1976), 91–100. [44] A. H. Wallace, Algebraic Topology, Homology and Cohomology, W. A. Benjamin Inc., 1970. [45] G. T. Whyburn, Analytic Topology, Amer. Math. Soc. Colloq. Publ., vol. 28, Amer. Math. Soc., Providence, R. I., 1942. [46] S. Willard, General Topology, Addison–Wesley Publishing Co., 1970. [47] M. Wojdislawski, Rétractes Absolus et Hyperespaces des Continus, Fund. Math., 32 (1939), 184–192.


Chapter 7 QUESTIONS

In this small chapter we include some open questions related to the topics of the rest of the book.

7.1

Inverse Limits

7.1.1. Question. Is it true that X is the inverse limit of graphs with monotone simplicial retractions as bonding maps if and only if X is locally connected and each cyclic element of X is a graph? (B. B. Epps) Comment: If M is a semi–locally connected continuum a cyclic element of M is either a cut point of M , an end point of M or a nondegenerate subset of M which is maximal with respect to being a connected subset without cut points. A map f : X → Y between graphs is simplicial if it is a map between the set of vertixes of the graph and it is a linear extension on the edges. 7.1.2. Question. Let {Xn , fnn+1 } be an inverse sequence of polyhedra with bonding maps fnn+1 : Xn+1 → Xn such that the inverse limit is a hereditarily indecomposable continuum. Let gn : S 2 → Xn Copyright © 2005 Taylor & Francis Group, LLC

347

348

CHAPTER 7. QUESTIONS

be a map such that gn is homotopic to fnn+1 ◦ gn+1 for each n ∈ IN. Is g1 homotopic to a constant map? (J. Krasinkiewicz) 7.1.3. Question. If X and Y are continua and X ×Y is disk–like, then must X be arc–like? (C. L. Hagopian) 7.1.4. Question. If a product of two continua is disk–like, must it be embeddable in IR3 ? (C. L. Hagopian) 7.1.5. Question. Does every disk–like continuum have the fixed point property? (S. Mardeˇsić and C. L. Hagopian, independently) Comment: An affirmative answer to this question would imply that every nonseparating plane continuum has the fixed point property. (C. L. Hagopian) 7.1.6. Question. Does each tree–like homogeneous continuum have the fixed point property? (C. L. Hagopian) Comment: For the plane, Oversteegen and Tymchatyn have shown that this question has a positive answer. (W. Lewis) 7.1.7. Question. Does every triod–like continuum have the fixed point property? (D. Bellamy) 7.1.8. Question. Is there a map f : [0, 1] → [0, 1] such that f has a periodic orbit of odd period larger than one, f has no periodic orbit of period three and lim{[0, 1], f } is homeomorphic to the pseudo– ←−

arc? (L. Block, J. Keesling and V. V. Uspenskij) Comment: Let f : X → X be a map. A point x ∈ X is said to be periodic provided that there exists m ∈ IN such that f m (x) = x. Now, x is said to have period n if n = min{m ∈ IN | f m (x) = x}. We say that f has a periodic orbit of period n provided that there exists a point x ∈ X such that x has period n.



349

1 7.1.9. Question. Given λ ∈ , 1 , define Tλ : [0, 1] → [0, 1] by 2

⎧ 1 ⎪ ⎪ if x ∈ 0, ; ⎨2λx 2

Tλ (x) = 1 ⎪ ⎪ ⎩2λ(1 − x) if x ∈ , 1 . 2

1 < λ1 < λ2 ≤ 1, then is lim{[0, 1], Tλ1 } topologically different ←− 2 from lim{[0, 1], Tλ2 }? (W. T. Ingram) If

←−

7.1.10. Question. Suppose n ≥ 4 is a positive integer and H and K are chainable continua having the property that each is an indecomposable continuum with only n end points and every nondegenerate proper subcontinuum is an arc. Are H and K homeomorphic? (W. T. Ingram) Comment: This question has been solved by Barge and Diamond for n = 5. Kailhoffer and Raines have further partial solutions. (W. T. Ingram) ˇ Comment: Sonja Stimac has announced at the meeting Geometric Topology II, celebrated at the Inter–University Centre, Dubrovnik, Croatia (September 29 through October 5, 2002), that in her dissertation (in Croatian) she has a positive answer to this question. No paper has appeared yet. (S. Mac´ıas)

7.2

The Set Function T

7.2.1. Question. If the set function T is continuous for the continuum X, is it true that X is T –additive? (D. Bellamy) Comment: A bushel of Extra Fancy Stayman Winesap apples for the solution. (D. Bellamy)


350


7.2.2. Question. If T is continuous for the continuum X, is it true that the collection {T ({p}) | p ∈ X} is a continuous decomposition of X such that the quotient space is locally connected? (D. Bellamy) Comment: A partial answer to this question is presented in Corollary 5.1.24. (S. Mac´ıas) 7.2.3. Question. If T is continuous for X and there is a point p in X such that T ({p}) has nonempty interior, is X indecomposable? (D. Bellamy)

7.2.4. Question. If X and Y are indecomposable continua, is T idempotent on X × Y ? Even for only closed sets in X × Y ? (D. Bellamy)

7.2.5. Question. If T is continuous for X and X is a decomposable continuum, is it true that for each p ∈ X, Int(T ({p})) = ∅? (D. Bellamy)

7.2.6. Question. Let X be a continuum. If X/T denotes the finest decomposition space of X which shrinks each T ({p}) to a point, is X/T locally connected? (D. Bellamy) Comment: The answer is affirmative if it is assumed that X is T –additive. 7.2.7. Question. If X is an indecomposable continuum and W is a subcontinuum of X ×X with nonempty interior, is T (W ) = X ×X? (F. B. Jones) Comment: C. L. Hagopian has shown that this question has an affirmative answer if X is chainable. (S. Mac´ıas)



351

7.2.8. Question. If X is an atriodic continuum (or X contains no uncountable collection of pairwise disjoint triods) and X does not have a weak cut point, then is there a continuum W ⊂ X such that Int(W ) = ∅ and T (W ) = X? (H. Cook) Comment: Let X be a continuum and let x be a point of X. We say that x is a weak cut point of X provided that there exist two points y and z in X such that if W is a subcontinuum of X and {y, z} ⊂ W , then x ∈ W . 7.2.9. Question. If T is continuous for the continuum X and f: X → → Z is a continuous and monotone surjection, is T continuous for Z also? (D. Bellamy)

7.2.10. Question. If X is a strictly point T –asymmetric dendroid, then is X smooth? (D. Bellamy) Comment: A continuum X is strictly point T –asymmetric if for any two distinct points p and q of X with p ∈ T ({q}), we have that q ∈ T ({p}).

7.2.11. Question. Let X be a continuum. Suppose the restriction of T to the hyperspace of subcontinua of X is continuous. Does this imply that T is continuous for X? (D. Bellamy) Comment: A partial answer to this question is presented in Theorem 3.1.32. (S. Mac´ıas)

7.2.12. Question. Do open maps preserve T –additivity? T –symmetry? (D. Bellamy)


352

7.3


Homogeneous Continua

7.3.1. Question. Is every homogeneous, hereditarily indecomposable, nondegenerate continuum a pseudo–arc? (F. B. Jones) Comment: The pseudo–arc is the only nondegenerate, hereditarily indecomposable chainable continuum. Comment: J. T. Rogers, Jr. has shown that it must be tree–like.

7.3.2. Question. Is each nondegenerate, homogeneous, nonseparating plane continuum a pseudo–arc? (F. B. Jones) Comment: F. B. Jones and C. L. Hagopian have shown that it must be hereditarily indecomposable. J. T. Rogers, Jr., has shown that it must be tree–like. 7.3.3. Question. If a homogeneous continuum X contains an arc must it contain a solenoid or a simple closed curve?(C. L. Hagopian)

7.3.4. Question. Is the simple closed curve the only nondegenerate, homogeneous, hereditarily decomposable continuum? (P. Minc)

7.3.5. Question. Is every decomposable, homogeneous continuum of dimension greater than one aposyndetic? (J. T. Rogers, Jr.)

7.3.6. Question. Is every nondegenerate, homogeneous, indecomposable continuum one–dimensional? (F. B. Jones)

7.3.7. Question. Is each atriodic, homogeneous continuum circle– like? (C. L. Hagopian) Copyright © 2005 Taylor & Francis Group, LLC

7.3. HOMOGENEOUS CONTINUA

353

7.3.8. Question. Is each aposyndetic, nonlocally connected, one– dimensional, homogeneous continuum an inverse limit of Menger curves? Menger curves and covering maps? (J. T. Rogers, Jr.)

7.3.9. Question. Let X be an arcwise connected homogeneous continuum which is not S 1 . Must X contain a simple closed curve of arbitrary small diameter? (L. Lum)

7.3.10. Question. Let X be a homogeneous arcwise connected continuum which is not S 1 . Let U be an open set in X and let M be an arc component of U . Is M cyclicly connected? (D. P. Bellamy) Comment: An arcwise connected space Z is cyclicly connected provided that each pair of points of Z lie on a simple closed curve. 7.3.11. Question. Let X be a homogeneous arcwise connected continuum which is not a simple closed curve. Is every arcwise connected open subset of X also cyclicly connected? (D. P. Bellamy)

7.3.12. Question. Can Jones’s Aposyndetic Decomposition Theorem be strengthened to give decomposition elements which are hereditarily indecomposable? (J. T. Rogers, Jr.)

7.3.13. Question. If X is an arcwise connected homogeneous continuum other than a simple closed curve, must each pair of points of X be the vertices of a θ–curve in X? (D. Bellamy) Comment: D. Bellamy and L. Lum have shown that each pair of points of X must lie on a simple closed curve. 7.3.14. Question. Does each finite subset of a nondegenerate arcwise connected homogeneous continuum lie on a simple closed curve? (D. Bellamy) Copyright © 2005 Taylor & Francis Group, LLC

354


7.3.15. Question. Does there exist a homogeneous one–dimensional continuum with no nondegenerate chainable subcontinuum? (W. Lewis) Comment: If there exists a nondegenerate, homogeneous, hereditarily indecomposable continuum other than the pseudo–arc, the answer is yes. 7.3.16. Question. Is every continuum a continuous image of a homogeneous continuum? In particular, is the spiral around the triod such an image? (L. Fearnley) 7.3.17. Question. Is there a nondegenerate homogeneous plane continuum X not homeomorphic to a simple closed curve, the pseudo–arc or the circle of pseudo–arcs?

7.4

n–fold Hyperspaces

7.4.1. Question. Let X be a continuum. If Cn (X) does not contain (n+1)–cells, then is X hereditarily indecomposable? (S. Mac´ıas) 7.4.2. Question. Is C3 ([0, 1]) homeomorphic to [0, 1]6 ? (R. Schori) 7.4.3. Question. If n ≥ 2 is an integer and X is a continuum, then is Cn (X) countable closed aposyndetic? 0–dimensional closed aposyndetic? (S. Mac´ıas) 7.4.4. Question. Let n be an integer greater than two. Let X be a continuum and let x ∈ X. If {x} is arcwise accessible from Cn (X) \ C(X), then is {x} arcwise accessible from C2 (X) \ C(X)? (S. Mac´ıas) Copyright © 2005 Taylor & Francis Group, LLC

REFERENCES

355

7.4.5. Question. Does there exist an indecomposable continuum X such that Cn (X) is homeomorphic to the cone over a finite– dimensional continuum for some integer n ≥ 2? (S. Mac´ıas and S. B. Nadler, Jr.)

7.4.6. Question. Does there exist a hereditarily decomposable continuum X that is neither an arc nor a simple m–od such that Cn (X) is homeomorphic to the cone over a finite–dimensional continuum for some n ≥ 3? (S. Mac´ıas and S. B. Nadler, Jr.)

7.4.7. Question. Find a geometric model for C2 (X), where X is a simple triod. (A. Illanes)

REFERENCES

[1] D. P. Bellamy, Questions In and Out of Context, talk given in the VI Joint Meeting AMS–SMM, Houston, TX, May 13–15, 2004 (LATEX edition by Jonathan Hatch). [2] L. Block, J. Keesling and V. V. Uspenskij, Inverse Limits Which are the Pseudoarc, Houston J. Math., 26 (2000), 629– 638. [3] H. Cook, W. T. Ingram and A. Lelek, A List of Problems Known as Houston Problem Book, in Continua with the Houston Problem Book, Lectures Notes in Pure and Applied Mathematics, Vol. 170, Marcel Dekker, Inc., New York, Basel, Hong Kong, 1995. (eds. H. Cook, W. T. Ingram, K. T. Kuperberg, A. Lelek, and P. Minc) [4] C. L. Hagopian, Mutual Aposyndesis, Proc. Amer. Math. Soc., 23 (1969), 615–622. Copyright © 2005 Taylor & Francis Group, LLC

356

REFERENCES

[5] C. L. Hagopian, Disk–like Products of λ Connected Continua, I, Proc. Amer. Math. Soc., 51 (1975), 448–452. [6] C. L. Hagopian, Disk–like Products of λ Connected Continua, II, Proc. Amer. Math. Soc., 52 (1975), 479–484. [7] A. Illanes, A Model for the Hyperspace C2 (S 1 ), Q. & A. in General Topology, 22 (2004), 117–130. [8] W. T. Ingram, Inverse Limits on [0, 1] Using Tent Maps and Certain Other Piecewise Linear Bonding Maps, in Continua with the Houston Problem Book, Lectures Notes in Pure and Applied Mathematics, Vol. 170, Marcel Dekker, Inc., New York, Basel, Hong Kong, 1995, pp. 253–258 (eds. H. Cook, W. T. Ingram, K. T. Kuperberg, A. Lelek, and P. Minc) [9] W. Lewis, Continuum Theory Problems, Topology Proc., 8 (1983), 361–394. [10] W. Lewis, The Classification of Homogeneous Continua, Soochow J. of Math., 18 (1992), 85–121. [11] S. Mac´ıas, On the Hyperspaces Cn (X) of a Continuum X, Topology Appl., 109 (2001), 237–256. [12] S. Mac´ıas, Fans Whose Hyperspaces Are Cones, Topology Proc., 27 (2003), 217–222. [13] S. Mac´ıas and S. B. Nadler, Jr., n–fold Hyperspaces, Cones and Products, Topology Proc., 26 (2001–2002), 255–270. [14] S. Mardeˇsić, Mappings of Inverse Systems, Glasnik Mat. Fiz. Astronom. Ser. II, 18 (1963), 195–205.


Topics on Continua (Chapman & Hall CRC Pure and Applied Mathematics)

CRC Pure and Applied Mathematics)

Shock Waves & Explosions (Chapman and Hall Crc Monographs and Surveys in Pure and Applied Mathematics)

Modular Lie Algebras and Their Representations (Chapman & Hall CRC Pure and Applied Mathematics)

An Introduction to Functional Analysis (Chapman & Hall CRC Pure and Applied Mathematics)

Matrix Theory: From Generalized Inverses to Jordan Form (Chapman & Hall CRC Pure and Applied Mathematics)

Algebraic Combinatorics (Chapman Hall Crc Mathematics Series)

Algebraic Combinatorics (Chapman Hall Crc Mathematics Series)

Engineering BGM (Chapman & Hall Crc Financial Mathematics Series)

Residuals and Influence in Regression (Chapman & Hall CRC Monographs on Statistics & Applied Probability 18)

Multivariate Models and Multivariate Dependence Concepts (Chapman & Hall CRC Monographs on Statistics & Applied Probability)

Expansions and Asymptotics for Statistics (Chapman & Hall CRC Monographs on Statistics & Applied Probability)

Smoothing Splines: Methods and Applications (Chapman & Hall CRC Monographs on Statistics & Applied Probability)

Introduction to non-Kerr Law Optical Solitons (Chapman & Hall CRC Applied Mathematics & Nonlinear Science)

Integral and Discrete Transforms with Applications and Error Analysis (Chapman & Hall CRC Pure and Applied Mathematics)

Statistics in the 21st Century (Chapman & Hall CRC Monographs on Statistics & Applied Probability)

Antedependence Models for Longitudinal Data (Chapman & Hall CRC Monographs on Statistics & Applied Probability)

Topology (Pure & Applied Mathematics)

Introduction to Time Series Modeling (Chapman & Hall CRC Monographs on Statistics & Applied Probability)

Functional Analysis: A Primer (Chapman & Hall Pure and Applied Mathematics)