Foundations of Analysis: A Straightforward Introduction: Book 2, Topological Ideas

THE FOUNDATIONS OF ANALYSIS: A STRAIGHTFORWARD INTRODUCTION BOOK 2 TOPOLOGICAL IDEAS

THE FOUNDATIONS OF ANALYSIS: A STRAIGHTFORWARD INTRODUCTION BOOK 2 TOPOLOGICAL IDEAS

K. G. BINMORE Professor of Mathematics London School of Economics and Political Science

CAMBRIDGE UNIVERSITY PRESS Cambridge London New York New Rochelle Melbourne Sydney

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www. Cambridge. org Information on this title: www.cambridge.org/9780521233507 ©Cambridge University Press 1981 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1981 Re-issued in this digitally printed version 2008 A catalogue record for this publication is available from the British Library ISBN 978-0-521-23350-7 hardback ISBN 978-0-521-29930-5 paperback

CONTENTS

Book 1: Logic, Sets and Numbers Introduction 1

Proofs

2

Logic (I)

3

Logic (II)

4

Set operations

5

Relations

6

Functions

7

Real numbers (I)

8

Principle of induction

9

Real numbers (II)

iot

Construction of the number systems

lit

Number theory

12

Cardinality

Book 2: Topological Ideas Introduction 13 13.1 13.3 13.5 13.9 13.11 13.14

Distance The space U" Length and angle in Un Some inequalities Modulus Distance Euclidean geometry and Un

xi 1 1 4 5 7 7 8

vi

Contents

13.17| 13.18 13.19| 13.20

Normed vector spaces Metric space Non-Euclidean geometry Distance between a point and a set

14 15 16 17

14 14.1 14.2 14.3 14.7 14.15

Open and closed sets (I)

21 21 21 22 25 29

15 15.1 15.4 15.8| 15.11

Open and closed sets (II)

16 16.1 16.2 16.7 16.13

Continuity

16.17t 17 17.1 17.2 17.6 17.8 17.15 17.18 17.21| 17.25| 18 18.1 18.4| 18.9| 19 19.1

Introduction Boundary of a set Open balls Open and closed sets Open and closed sets in Un Interior and closure Closure properties Interior properties Contiguous sets Introduction Continuous functions The continuity of algebraic operations Rational functions Complex-valued functions Connected sets

Introduction Connected sets Connected sets in U1 Continuity and connected sets Curves Pathwise connected sets Components Structure of open sets in Un Cluster points

Cluster points Properties of cluster points The Cantor set Compact sets (I)

Introduction

31 31 32 33 34 39 39 39 41 44 46 47 47 47 49 50 52 53 56 57 60 60 62 63 66 66

|This material is more advanced than the remaining material and can be omitted at a first reading.

Contents

vii

19.2 19.5 19.12

Chinese boxes Compact sets and cluster points Compactness and continuity

67 69 73

20f 20.lt 20.2t 20.4| 20.7f 20.15| 20.16| 20.20|

Compact sets (II) Introduction Open coverings Compact sets Compactness in Un Completeness Compactness in general metric spaces A spherical cube

77 77 77 78 78 83 84 86

21 21.1 21.2 21.3 21.4 21.6 21.8 21.9| 21.15f 21.18f

Topology Topological equivalence Maps Homeomorphisms between intervals Circles and spheres Continuous functions and open sets Topologies Relative topologies Introduction to topological spaces Product topologies

89 89 90 91 92 94 95 95 100 102

22 22.1 22.2 22.3 22.4 22.9 22.12 22.15 22.16 22.19 22.23 22.25

Limits and continuity (I) Introduction Open sets and the word 'near' Limits Limits and continuity Limits and distance Right and left hand limits Some notation Monotone functions Inverse functions Roots Combining limits Complex functions

106 106 109 109 110 113 116 118 119 121 123 125 128

Limits and continuity (II) Double limits Double limits (continued) Repeated limits Uniform convergence Distance between functions Uniform continuity

130 130 132 132 136 138 145

22.34t 23f 23.lt 23.3t 23.5t 23.11t 23.12f 23.20t

viii

Contents

24 24.1 24.2 24.3 24.4 24.5 24.8 24.111 24.12|

Points at infinity Introduction One-point compactification of the reals The Riemann sphere and the Gaussian plane Two-point compactification of the reals Convergence and divergence Combination theorems Complex functions Product spaces

149 149 150 153 154 157 163 165 165

25 25.1 25.2 25.7 25.12 25.18 25.23

Sequences Introduction Convergence of sequences Convergence of functions and sequences Sequences and closure Subsequences Sequences and compactness

169 169 170 172 175 176 179

26 26.1 26.2 26.5| 26.11t

Oscillation Divergence Limit points Oscillating functions Lim sup and lim inf

181 181 182 184 186

27 27.1 27.2 27.8 27.13| 27.16| 27.18|

Completeness Cauchy sequences Completeness Some complete spaces Incomplete spaces Completion of metric spaces Completeness and the continuum axiom

190 190 190 193 195 197 199

28 28.1 28.7 28.8 28.11| 28.15| 28.19| 28.26|

Series Convergence of series Absolute convergence Power series Uniform convergence of series Series in function spaces Continuous operators Applications to power series

201 201 204 204 208 209 212 215

29|

Infinite sums Commutative and associative laws Infinite sums

218 218 220

29.lt 29.2t

Contents

ix

29.4| 29.9| 29.17| 29.23f

Infinite sums and series Complete spaces and the associative law Absolute sums Repeated series

221 223 226 230

30t

Separation in Un Introduction Separation Separating hyperplanes Norms and topologies in (1 Curves and continua Simple curves Simply connected regions

234 234 235 235 239 241 242 243

Notation

245

Index

246

30.lt 30.2| 30.4f 3O.8t 30.11f 30.12| 30.14|

The diagram on p. x illustrates the logical structure of the books. Broken lines enclosing a chapter heading indicate more advanced material which can be omitted at a first reading. The second book depends only to a limited extent on the first. The broken arrows indicate the extent of this dependence. It will be apparent that those with some previous knowledge of elementary abstract algebra will be in a position to tackle the second book without necessarily having read the first.

-»»-j 2. Logic (I)

Proofs

n

3. Logic (II)

±

-j4. Set operations

112. Cardinality L L J T

5. Relations

7. Real numbers (I)

6. Functions

8. Principle of induction

Book I Logic, Sets and Numbers

I

I 9. Real ] numbers ( i

T

1

110. Construction I of number systems]

—

I

I |

14. Open and closed sets (I)

11. Number theory

13. Distance

JL

1

15. Open and I closed sets (II)

16. Continuity

17. Connected sets

H

30. Separation I in W

18. Cluster points

i

Book 2 Topological

]20. Compact 1 I sets (II) !

P

19. Compact sets (I)

Ideas

—j 2 1 . T o p o l o g y (-*•

22. Limits and continuity (I)

24. Points at infinity

I

i 23. Limits and] j continuity (II) i

c

25. Sequences U H 26. Oscillation

- - T " ^ •427. Completeness!—*H I

I

I

28. Series

|—•{i 29. Infinite iI iI sums

I

i

,

I i

INTRODUCTION

This book is intended to bridge the gap between introductory texts in mathematical analysis and more advanced texts dealing with real and complex analysis, functional analysis and general topology. The discontinuity in the level of sophistication adopted in the introductory books as compared with the more advanced works can often represent a serious handicap to students of the subject especially if their grasp of the elementary material is not as firm as perhaps it might be. In this volume, considerable pains have been taken to introduce new ideas slowly and systematically and to relate these ideas carefully to earlier work in the knowledge that this earlier work will not always have been fully assimilated. The object is therefore not only to cover new ground in readiness for more advanced work but also to illuminate and to unify the work which will have been covered already. Topological ideas readily admit a succinct and elegant abstract exposition. But I have found it wiser to adopt a more prosaic and leisurely approach firmly wedded to applications in the space Un. The idea of a relative topology, for example, is one which always seems to cause distress if introduced prematurely. The first nine chapters of this book are concerned with open and closed sets, continuity, compactness and connectedness in metric spaces (with some fleeting references to topological spaces) but virtually all examples are drawn from Un. These ideas are developed independently of the notion of a limit so that this can then be subsequently introduced at a fairly high level of generality. My experience is that all students appreciate the rest from 'epsilonese' made possible by this arrangement and that many students who do not fully understand the significance of a limiting process as first explained find the presentation of the same concept in a fairly abstract setting very illuminating provided that some effort is taken to relate the abstract definition to the more concrete examples they have met before. The notion of a limit is, of course, the single most important concept in mathematical analysis. The remainder of the volume is therefore largely devoted to the application of this idea in various important special cases. Much of the content of this book will be accessible to undergraduate students during the second half of their first year of study. This material has xi

xii

Introduction

been indicated by the use of a larger typeface than that used for the more advanced material (which has been further distinguished by the use of the symbol t)- There can be few institutions, however, with sufficient teaching time available to allow all the material theoretically accessible to first year students actually to be taught in their first year. Most students will therefore encounter the bulk of the work presented in this volume in their second or later years of study. Those reading the book independently of a taught course would be wise to leave the more advanced sections (smaller typeface and marked with a | ) for a second reading. This applies also to those who read the book during the long vacation separating their first and second years at an institute of higher education. Note, incidentally, that the exercises are intended as an integral part of the text. In general there is little point in seeking to read a mathematics book unless one simultaneously attempts a substantial number of the exercises given. This is the second of two books with the common umbrella title Foundations of Analysis: A Straightforward Introduction. The first of these two books, subtitled Logic, Sets and Numbers covers the set theoretic and algebraic foundations of the subject. But those with some knowledge of elementary abstract algebra will find that Topological Ideas can be read without the need for a preliminary reading of Logic, Sets and Numbers (although I hope that most readers will think it worthwhile to acquire both). A suitable preparation for both books is the author's introductory text, Mathematical Analysis: A Straightforward Approach. There is a small overlap in content between this introductory book and Topological Ideas in order that the latter work may be read without reference to the former. Finally, I would like to express my gratitude to Mimi Bell for typing the manuscript with such indefatigable patience. My thanks also go to the students of L.S.E. on whom I have experimented with various types of exposition over the years. I have always found them to be a lively and appreciative audience and this book owes a good deal to their contributions. June 1980

K. G. BINMORE

13 13.1

DISTANCE

The space R"

Those readers who know a little linear algebra will find the first half of this chapter very elementary and may therefore prefer to skip forward to §13.18. The objects in the set Un are the rc-tuples

in which x19 x2,..., xn are real numbers. We usually use a single symbol x for the n-tuple and write

The real numbers xv x2,...,xn are called the co-ordinates or the components of x. It is often convenient to refer to an object x in R" as a vector. When doing so, ordinary real numbers are called scalars. If x = (x1, x 2 , . . . , xn) and y = (yu yii--) yn) a r e vectors and a is a scalar, we define 'vector addition' and 'scalar multiplication' by

ax = (ocxv OLX29. .., a x j .

These definitions have a simple geometric interpretation which we shall illustrate in the case n = 2. An object xe[R2 may be thought of as a point in the plane referred to rectangular Cartesian axes. Alternatively, we can think of x as an arrow with its blunt end at the origin and its sharp end at the point (x19 x2).

(xltx2)

x as a point

x as an arrow

Distance

Vector addition and scalar multiplication can then be illustrated as in the diagrams below. For obvious reasons, the rule for adding two vectors is called the parallelogram law.

x+y

ccx2

x2 y\

The parallelogram law is the reason that the navigators of small boats draw little parallelograms all over their charts. Suppose a boat is at 0 and the navigator wishes to reach point P. Assuming that the boat can proceed at 10 knots in any direction and that the tide is moving at 5 knots in a south-easterly direction, what course should be set? N

/ y / /

/io /

\

/ /

\

\\\

\

\\ /

^p

y

/

o Is

7/ / /

The vector x represents that path of the boat if it drifted on the tide for an hour (distances measured in nautical miles). The vector y represents the path of the boat if there were no tide and it sailed the course indicated for

Distance

3

an hour. The vector x + y represents the path of the boat (over the sea bed) if both influences act together. The scalar t is the time it will take to reach P.

13.2

Example Let x = (l, 2, 3) and y = (2, 0, 5). Then = ( l , 2 , 3) + (2,0,5) = (3,2, 8) 2x = 2(l,2, 3) = (2,4,6).

It is very easy to check U" is a commutative group under vector addition. (See §6.6.) This simply means that the usual rules for addition and subtraction are true. The zero vector is, of course,

The diagram below illustrates the vector y — x = (yx — xv y2 — x 2 , •. •, yn — xn) in the case n = 2.

y-x

---

It is natural to ask about the multiplication of vectors. Is it possible to define the product of two vectors x and y as another vector z in a satisfactory way? There is no problem when n = 1 since we can then identify U * with U. Nor is there a problem when n = 2 since we can then identify U 2 with C (§10.20). If n ^ 3 , however, there is no entirely satisfactory way of defining multiplication in Un. Instead we define a number of different types of 'product' none of which has all the properties which we would like a product to have. Scalar multiplication, for example, tells us how to multiply a scalar and a vector. It does not help in multiplying two vectors. The 'inner product', which we shall meet in §13.3, tells how two vectors can be 'multiplied' to produce a scalar. In R 3 , one can introduce the 'outer product' or 'vector product' of two vectors x and y. This is a vector denoted by x A y or x x y. Unfortunately, x A y = — y A x .

Distance

4

Multiplication is therefore something which does not work very well with vectors. Division is almost always meaningless. 13.3

Length and angle in Un The Euclidean norm of a vector x in Un is defined by

We think of ||x|| as the length of the vector x. This interpretation is justified in U2 by Pythagoras' theorem (13.15).

The inner product of two vectors x and y in R" is defined by <x,y> = x 1 >/ 1 +x 2 ); 2 + ... + xnyn. It is easy to check the following properties: (i) <x,x> = ||x||2 (ii) <x,y> = (iii) The geometric significance of the inner product can be discussed using the cosine rule (i.e. c2 = a2 4- b2 — 2ab cos y) in the diagram below. y-x

O

Rewriting the cosine rule in terms of the vectors introduced in the right-

Distance hand diagram, we obtain that

But, llx -y|| 2 = <x - y , x - y > = <x, x - y > - = <x,x>-2<x,y> + = ||x||2 + ||y|| 2 -2<x,y> It follows that <x,y> = ||x||.||y||cosy. Of course, this argument does not prove anything. It simply indicates why it is helpful to think of

<x,y> IMNIyll as the cosine of the angle between x and y.

13.4 Example Find the lengths of and the cosine of the angle between the vectors x = (l, 2, 3) and y = (2, 0, 5) in R 3 . We have that

||y|| = {2 2 + 0 2 -f- 5 2 } 1 / 2 = V 2 9 ' <x, y> 1-2 + 2 - 0 + 3-5 17 ^14x29'

13.5

Some inequalities

In the previous section y was the angle between x and y. The fact that |cos y\ < 1 translates into the following theorem.

13.6

Theorem (Cauchy-Schwarz inequality) If xeR" and yeR", then

Proof Let aeR. Then ay||2 = < x - a y , x - a y > = ||x|| 2 -2a<x,y> + a2||y||2.

6

Distance

It follows that the quadratic equation ||x||2 —2a<x, y> + a 2 ||y|| 2 has at most one real root (§10.10). Hence ' b 2 - 4 a c g 0 ' - i.e. 4<x,y> 2 -4||x|| 2 ||y|| 2 ^0.

It is a familiar fact in Euclidean geometry that one side of a triangle is shorter than the sum of the lengths of the other two sides. x+y

llyll

llxll

This geometric idea translates into the following theorem.

13.7

Theorem (Triangle inequality) If xeU" ajid yeR", then ||x + y||g||x|| + ||y||. Proof 2

13.8

, x + y> + 2<x, y> + ||y||2 + 2||x||. ||y|| + ||y||2

(theorem 13.6)

Corollary If xeR" and yeUn, then

Proof It follows from the triangle inequality that

Distance 13.9

7

Modulus

The modulus \x\ of a real number x coincides with the Euclidean norm of x thought of as a vector in R 1 . We have that

One has to remember that y1/2 represents the non-negative number whose square is y (§9.13). The modulus \z\ of a complex number z = x + iy is identified with the Euclidean norm of (x, y) thought of as a vector in R2. We have that = \\{x9 y)\\. 13.10 Then

Theorem Suppose that u and v are real or complex numbers. \uv\ = \u\-\v\.

Proof We need only consider the complex case. If u = a + ib and v = c + id, then \uv\2 - \u\2\v\2 = (ac - bd)2 + (be + ad)2 - (a2 + b2)(c2 + d2) = 0.

13.11

Distance The distance d(x, y) between two vectors x and y in R" is defined

by In interpreting this idea in Un, it is better to think of x and y as the points at the end of the arrows rather than the arrows themselves.

x-y

8

Distance

13.12

Examples

(i) The distance between the vectors x = (l, 2, 3) and y = (2, 0, 5) in IR3 is

(ii) The distance between 3 and 7 (regarded as vectors in R1) is |3-7| = | - 4 | = 4 .

-,

13.13

i-

Exercise

(1) Let x = (0, 1, 0) and y = (1, 1, 0) be vectors in U3. Calculate the quantities (i) x + y (iv) ||x||

(ii) x - y (v) | | x - y | |

(iii) 2x (vi) <x,y>.

What is the length of the vector x? What are the distance and the angle between x and y? (2) Calculate the moduli of the following real and complex numbers: (i) - 3 (ii) 0 (iii) 4 (iv) 3 + 4j (v) 4 - 3 i (vi) I (3) If a and b are any real numbers, prove that |a|f (S) convex. f(ll) Prove that a function/: IR"-^ 1 is linear if and only if there exists an eeR" such that/(x) = <x, e> for each xelR". t(12) Anticipating the definition of § 13.17, prove that a normed vector space % (with real scalars) admits an inner product satisfying (i), (ii) and (iii) of § 13.3 if and only if ||x + y||2 + ||x-y|| 2 = 2||x||2 + 2||y||2

for each xe % and each ye %. [Hint. Use exercises 13.16(1) and 13.13(5).]

13.17|

Normed vector spaces

The space Un is an example of a normed vector space. For a vector space, one first needs a commutative group % (§ 6.6) to serve as the vectors. The group operation is then called vector addition. Next one needs a field to serve as the scalars. A meaning then has to be assigned to scalar multiplication in such a way that the following rules are satisfied: (i) (ii) (a + £)x = ax + fix (iii) (a0)x = a(£x) (iv) 0x = 0; lx = x. For a normed vector space, we usually insist that the field of scalars be R or C. A norm function from X to R is required which satisfies the following properties: (i)

(ii) ||x||=0~x = 0 (iii) ||ax|| = |a|||x|| (iv) ||x + y|| ^ ||xj| + ||y||

(triangle inequality)

Distance

15

for all xe X, ye Z and all scalars. (Here ||x|| denotes the image of x under the norm function.) The chief reason for giving the requirements for a normed vector space at this stage is to provide a summary of the properties of U" which will be important in the next few chapters. Other normed vector spaces will be mentioned hardly at all. It should be noted, however, that the properties of infinite-dimensional normed vector spaces are often counter-intuitive. For example, a hyperplane in an infinitedimensional normed vector space does not, in general, split the space into two separate half-spaces.

13.18

Metric space A metric space is a set % and a function d: %x Z-+U which

satisfies (i) (ii) (iii) (iv)

d(x, y ) ^ 0 d(x, y) = 0 ^ x = y d(x,y) = d(y,x) d(x, z)^d(x, y) + d(y, z).

(triangle inequality)

The function d is said to be the metric for the metric space and we interpret d(x, y) as the distance between x and y. The metric space in which we shall be most interested is the space Un. As we have seen in §13.11, the metric d2 :Un x Un'-+U in R" is defined by

Strictly speaking, one should refer to d as the Euclidean metric. It is trivial to check that the Euclidean metric satisfies the requirements for a metric given above. In particular, the triangle inequality for a metric follows from theorem 13.7 since

Any subset £ of Un is also a metric space provided that we continue to use the Euclidean metric in £. We then say that £ is a metric subspace ofMn. For example, the interval (0, 1] is a metric subspace of IR1 provided that the metric d: (0, 1] x (0, 1]->R is defined by d(x9 y) = \x-y\.

d(x,y)

of More exotic examples of metric spaces can be obtained by starting with any normed vector space % and defining d: % x Z-+M by

16

Distance

But there are also many interesting metric spaces which have no vector space structure at all. Since we have been discussing the fact that U 2 (with the Euclidean metric) is a model for Euclidean geometry (§13.14), we shall give Poincare's model for a non-Euclidean geometry as an example of such a metric space.

13.19|

Non-Euclidean geometry

Perhaps the most well-known of Euclid's postulates is the parallel postulate (quoted in exercise 13.16(4)). This was always felt to be less satisfactory than Euclid's other assumptions in that it is less 'intuitively obvious' than the others. Very considerable efforts were therefore made to deduce it as a theorem from the other axioms. All these efforts were unsuccessful. Finally, it was realised that the task is impossible and Gauss, Lobachevski and Bolyai independently began to study a geometry in which the parallel postulate is false but all the other assumptions of Euclidean geometry are true. Gauss did not publish his work and Lobachevski published before Bolyai. Hence the non-Euclidean geometry they studied will be called Lobachevskian geometry. (Sometimes it is called 'hyperbolic geometry'.) This Lobachevski, incidentally, is the Nikolai Ivanovich Lobachevski of the immortal Tom Lehrer song but the scandalous suggestions made in this song are totally unfounded! In Lobachevskian geometry there are many parallel lines through a given point parallel to a given line. This may seem intuitively implausible as a hypothesis about the 'real world' because we have been trained from early childhood to think of space as Euclidean. In fact, Einsteinian physics assures us that, in the vicinity of a gravitating body, space is very definitely not Euclidean. The purpose of this long pre-amble is to explain the interest of the following metric space which was introduced by the great French mathematician Poincare. This space provides a model for Lobachevskian geometry. Its existence therefore demonstrates that the axioms of Lobachevskian geometry are consistent. What is more, since the parallel postulate is true in U2 but false in the Poincare model, it must be independent of the other axioms of Euclidean geometry. In particular, it cannot be deduced from them. The set % in Poincare's metric space is the set 3T={(x, y):x2 + y2 0, there exists an xeS such that d(£, x)0 is a lower bound for D i.e. not 3e>0 VxeS (d( But this is equivalent to

as required (§3.10).

It is important to take note of the fact that d(£, S) = 0 does not imply that

13.23

Examples

(i) Consider the point 1 and the set S= (0, 1) in R. For each £>0, we can find an xeS such that x> 1 — s.

Since 1—e<x 0, we can find an neN such that n> 1/e (because N is unbounded above). But then d(0, l/n) = l/n<e and hence d(0, S) = 0.

points of S

20

Distance

13.24

Exercise

(1) Find d(£, S) for the following points £ and sets S in (i) (ii) (iii) (iv) (v) (vi) (vii)

£ = 1; S = (0,2) £ = 0 ; S = [l,2] £=±; S=N £=0; S = (0,l) { = 1; S = (2,3) £ = 2; S = (0,l)u(3,4) £ = y/2; S = Q. [tfmr: Use theorem 9.20.]

(2) Find d($, S) for the following points $ and sets S in R 2 . (i) $=(1,1); S = {(x,y (ii) $ = (2,1); S = {(x,y (iii) $ = (0,0); S = {(x,>; (3) Let S be a non-empty set of real numbers which is bounded above. Prove that d(sup S, S) = 0. (4) Let $ be a point in a metric space X and let S and T be non-empty sets in X. Prove that (i) (ii) (iii) 9 S) = 0 is meaningless and hence cannot be true. Thus the empty set has no boundary points - i.e. d0 = 0 . Similarly, d% = 0.

14.3

Open balls

The open ball B with centre £, and radius r > 0 in a metric space % is defined by In U3, an open ball is the inside of a sphere. In U 2, an open ball is the inside of a circle. In U1, an open ball is an open interval.

%

14.4 Theorem Let S be a set in a metric space % and let £ be a point in %. Then £ is a boundary point of S if and only if each open ball B with centre ^ contains a point of S and a point of C S.

Open and closed sets (I) point of S

23 point of 6 S

Proof The fact that each open ball B with centre ^ and radius r > 0 contains a point of S and a point of CS is equivalent to the assertion that, for each r > 0 , there exists xeS and yeCS such that d(£, x ) < r and d(£, y) = 0}. Suppose that d(y, H) = 0. We propose to use theorem 14.10 and therefore seek to prove that y e / / . By theorem 13.22, given any e>0, there exists an xeH such that d(y, x) = ||y —x||<e. But since xeH,

Hence, using the Cauchy-Schwarz inequality (13.6),

30

Open and closed sets (I)

Since this is true for every a>0, it follows that =0 and so yeH. Thus H is closed. 14.17

Exercise

(1) Determine which of the sets of exercise 14.6(1) and (2) are open and which are closed. (2) Prove that the only sets in Un which are both open and closed are 0 and

or. (3) Let S be the set of exercise 14.6(5). Find a point ££S such that d(E>, S) = 0. Deduce that S is not closed. (4) Prove that open balls in a metric space % as defined in §14.3 are open sets as defined in §14.7. (5) Prove that a set consisting of a single point in a metric space % is closed. Deduce that any finite set in a metric space % is closed. (6) Let Sv S2,...,Sn be closed sets in U1. Prove that S = S1xS2x ... xSn is a closed set in Un. [Hint: Use exercise 13.24(6).] (7) Prove that the types of set in GT said in §14.15 to be closed actually are closed. (8) Aflat (or affine set) in Un may be defined to be the intersection of a collection of hyperplanes. Prove that a flat is always closed. (9) The sets S and T in or are defined by (where u # 0 and c are constant). We call S a closed half-space of R" and T an open half-space. Prove that S is closed and T is open in the sense of §14.7. |(10) Give examples to show that a convex set in U" may be (i) open (ii) closed (iii) neither open nor closed (iv) both open and closed. |(11) Prove that any open set G in a metric space % is the union of the collection of all open balls it contains. |(12) Find a sequence of open intervals and a sequence <Jk> of closed intervals in U such that, if

then / is not open and J is not closed.

15

15.1

OPEN AND CLOSED SETS (II)

Interior and closure

Let £ be a set in a metric space %. The interior E of the set E is defined by E = E\dE. The closure E is defined by

The interior E is obtained from E by removing all the boundary points of E which happen to be elements of E. The closure E is obtained from E putting in all the boundary points of E which happen not to be elements of E.

15.2

Example Consider the set S in R 2 defined by

This set is neither open nor closed. Its interior S is the set $={(x,y):0 ^ c} (where u ^ 0) be half-spaces in Un. Let / / be the hyperplane H = {x: <x, u> =c}. Prove that (i) dS = H (iv) 8T=H

(ii) 5 = T (v) T = T

(iii) S = S (vi) f=S.

[Hint: For (iv), (v) and (vi), recall question (3).] t(6) Let B1 = {x: d& x)0) be balls in Un. Prove that (i)B 1 =B 1 (ii) ^7 = B2 (Hi) B^B, [Hint: Use exercise 14.6(4).]

(iv) B2 = B2.

15.4

Closure properties

15.5

Theorem For any set £ in a metric space X, xeE d(x, E) = 0

Proof (i) If xeE = E\jdE, then x e £ or xedE. In the former case d(x, E) = d(x, x) = 0. In the latter case, we have by definition that d(x, E) = 0 and d(x, GE) = 0. Thus xeE =>d(x9 E) = 0. (ii) If d(x, E) = 0, then either xeEaE or else xeCE. In the latter case d(x, CE) = d(x9 x) = 0 and so xedEcE. Thus d(x, £) = 0 => xeE.

Open and closed sets (II) 15.6

33

Corollary Let S and T be sets in a metric space X. Then

Proof Let xeS. By theorem 15.5, d(x, S) = 0. But S c T implies , S)^d{\, T). Hence d(x, T) = 0. Thus x e T .

15.7f Theorem Let £ be a set in a metric space £. Then £ is the smallest closed set containing E. Proof To show that £ is closed, we use theorem 14.10. We therefore assume that dfe, £) = 0 and seek to deduce that £e£. By theorem 13.22, given any £>0, we ean find an xe£ such that d(£, x)<e. But xe£ d(x, E) = 0 by theorem 15.5. Hence £) = d(§, x) (exercise 13.24(5)). Thus Since this is true for any e>0, we conclude that dfe, E) = 0 and hence ^eE by theorem 15.5. Now suppose that F is any closed set containing E. Since £czF, we have from corollary 15.6 that EaF. But F is closed and therefore F = F (exercise 15.3(lii)). It follows that EaF for each closed set F containing E. Thus E is the smallest closed set containing E.

15.8|

Interior properties From exercise 15.3(31) we know that

Thus any result about closures leads to a corresponding result about interiors. For example,

es^eT^Js^Jf=> e(es)c:e(ef) and so 5c=T => $czt. 15.9| Theorem Let £ be a set in a metric space X. Then £ is the largest open set contained in £. Proof Note first that E=C(CE) and hence is open because it is the complement of a closed set. If Gc£, then GczE. If G is also open it follows that

34

Open and closed sets (II)

GczE (because then G = (5). This argument shows that £ is the largest open set contained in E.

15.10|

Exercise

(1) Aflat (or affine set) F is the intersection of a collection of hyperplanes in U". Prove that (i) F = F

(ii) / = 0 .

Show that the same results hold if F is a half-line or a closed line segment. (2) Let £ be a set in a metric space %. Prove that xeE if and only if there exists an open ball B with centre £ such that BaE. (3) Let £ be a set in a metric space £. Prove that dE is closed. [Hint. d£ = £nC(£).] (4) Let £ be a set in a metric space %. Prove that (i) £ = H

£

(ii) E=\J

EczF F closed

G.

GczE G open

(5) Let A and £ be sets in a metric space %. Prove that (i)

0

)

o

(ii) AczB => AczS;AczB =>Aa3 o

(iii)

i^

Give examples to show that c cannot be replaced in general by = in (iii). (6) Let £ be a set in a metric space %. Prove that d(dE) = dE. Show also that: (i) d(E)adE (ii) d(£)czdE. Give examples to show that c cannot be replaced in general by = in either case. (7) Let C be a convex set in U" with a non-empty interior. Prove that (i) C=£

(ii) f~1(C) and f~l{D) separated.

Our first theorem records the unremarkable fact that, if one continuous transformation is followed by another, then the result will be a continuous transformation.

16.3 Theorem Let X, % and £ be metric spaces and let S c £ , Suppose that g: S^y, and / : T-»£ are continuous on S and T respectively and that g(S)czT. Then the composite function/°g: £-•£ is continuous on S.

Proof The diagram renders the proof obvious.

16.4 Theorem Let % and y, be metric spaces and let S c l Then / : S-+y, is continuous on the set S if and only if, for each xeS and each £ c S , d(x,£) = 0=>d(/(x), /(£)) = 0.

Continuity

41

Proof (i) Suppose that / is continuous on S. If d(x, E) = 0, then {x} and E are contiguous by theorem 15.12. Hence {/(x)} and f(E) are contiguous. Thus d(/(x), f(E)) = 0 by theorem 15.12 again. (ii) Suppose that d(\, E) = 0=>d(f(x)J(E)) = 0. Let A and B be contiguous subsets of S. Then a point of one set is at zero distance from the other by theorem 15.12. Suppose aeA and d(a, B) = 0. Then d(/(a), /(£)) = 0 and hence f(A) a n d / ( £ ) are contiguous by theorem 15.12 yet again.

If/: S->R", the formula

defines n real-valued functions/^ S->R,/ 2 : S—•IR,... ,/„: S-*R. We call these functions the component functions o f / a n d write

f=(fvf2,.. -./„)• 16.5 Theorem Let £ be a metric space and let Sa%. A function / : S^Un is continuous on the set S if and only if each of its component functions is continuous on S. Proof This follows immediately from theorem 16.4 and exercise 13.24(6).

16.6

Corollary The projection function Pk: tR"-*[R defined by Pk(xv

x 2 , . . . , xn) = xk

is continuous on Un. Proof The identity function / : [Rn->[Rn defined by /(x) = x is obviously continuous. But I = (PV P 2 , . . . , Pn) and hence Pk is continuous by theorem 16.5.

16.7

The continuity of algebraic operations

It is easy to see that constant functions and identity functions are continuous. We shall obtain some slightly less trivial examples of continuous functions from these by using the operations of addition, sub-

42

Continuity

traction, multiplication and division. But first we must prove that these operations are themselves continuous. We begin with a simple lemma.

16.8 Lemma Let % and y, be metric spaces and let Sa %. Then/: S-+y, is continuous on S provided that for each xeS there exists a y > 0 and a c > 0 such that

d(x,y)d(f(x)J(y))^cd(x,y) for each yeS. Proof Suppose that d(x, E) = 0. Given any ^>0, we shall prove that there exists a y e £ such that d(/(x),/(y))<e. Thus d(f(x),f{E))^=0 by theorem 13.22. Choose eo = min {sc'1, y}. Then eo>O and hence there exists yeE such that d(x, y)<e 0 by theorem 13.22. But so^y. It follows that , /(y)) S cd(x, y) < cs0 g s as required. 16.9 Theorem Let a and b be real numbers and let L: (R"xR"M: U x R ->R and D:Ux(U\ {0})^>U be defined by (i) L(x 1 ? x 2 ) = ax 1 (ii) M(x 1 ? x 2 ) = x 1 x 2

(iii) D(xv x2) = xjx2. Then L, M and D are continuous on their domains of definition. Proof (i) We have that || Xl - y ^ l K x ^ x 2 ) - ( y 1 , y2)|| and I|x 2 -y 2 ||^||(x 1 , x 2 )-(y 1 ? y2)|| because ||(x1? x 2 )-(y 1 ? Hence ||L(x1? x 2 )-L(y 1 ?

^{\a\ +

\b\}-\\(xvx2)-(y19y2)\\.

Thus Lis continuous on [R2" by lemma 16.8. (ii) We use lemma 16.8 with y = l. Assume that \\(xvx2)-(yvy2)\\R defined by P(x, y) = x 2 + 3xy + x 2 y 3 + 1

is a polynomial. The function R: R 2 \ {(0, 0)} ->R defined by

Continuity

45

is a rational function.

16.15 Theorem All rational functions are continuous on their domains of definition. Proof Constant functions and the identity function are continuous. So are the projection functions (corollary 16.6). The fact that rational functions are continuous therefore follows from corollaries 16.10, 16.11 and 16.12.

16.16

Exercise

(1) For each of the following functions/: R -»R, find non-empty contiguous sets A and B in R, for which f(A) and f(B) are not contiguous.

Deduce that in neither case is / continuous. (2) Let eeR". Prove that the functions/:R"-*R 1 and g-.W-^M1 defined by (i) /(x) = ||x|| (ii) (see §13.3) are continuous. (3) A function/: R"->Rm is said to be linear if and only if /(ax + jSy) = a/(x) + j8/(y) for all real a and /? and all x and y in U". Prove that a linear function f{. IR"-^ 1 satisfies / 1 (x) = <x, e> for some eeUn. Hence show that any linear function/: [R"->lRm is continuous. (4) Let % be a metric space and let S be a non-empty set in %. Prove that the function / : 5E-+R defined by /(x) = d(x, S) is continuous on %. [Hint: Begin by showing that \d(x, S) — d(y, S)\^d(x, y) using exercise 13.24(5).] (5) Let X and % be metric spaces and let ScTaZ. If/: T-^% and g: T->% satisfy f(x) = g(x) for each xeS, prove that g continuous on T implies / continuous on S. t(6) Prove theorem 16.9 with U replaced everywhere by C.

46

Continuity

16.171

Complex-valued functions

In the above discussion we have confined our attention to real polynomials and real rational functions. However, it is worth noting that all of the results of this chapter remain valid if U is replaced throughout by C and the word 'real' replaced by 'complex'. No other change in the proofs is necessary.

17

17.1

CONNECTED SETS

Introduction

The characteristic feature of an interval in IR* is that it is 'all in one lump'. A little man standing on any point of an interval / would be able to walk to any other point of / without having to cross any points of the complement of /.

What sets in IR" have the same property? One can think, for example, of a set S in IR 2 as a country entirely surrounded by water. When does S consist of a single island and when does it consist of a whole archipelago of islands? Before trying to answer this question, we must of course first express it in precise mathematical terms.

17.2

Connected sets

A set S in a metric space % is connected if and only if S cannot be split into two non-empty separated subsets. This means that if A and B are any two non-empty subsets of S satisfying AuB = S, then A and B are contiguous.

connected set

disconnected set

If one thinks of a connected set S in IR 2 as a country entirely surrounded by water, the definition says that, however one divides S into two provinces 47

48

Connected sets

A and B, these two provinces will be 'joined up'. A little man will be able to go from A to B without getting his feet wet.

17.3 Theorem A metric space X is connected if and only if the only sets in X which are both open and closed are 0 and X. Proof Observe that

(i) AKJB=% and AnB = 0 => AuB=X and AnB = 0 => B=CA => AuB=X. (ii) AuB=X and Ar\B = 0 => A\jB=X and Ar\B = 0 => B=6A Taking these two results together we obtain that AKJB=%

and AnB = 0 o B = B and B=6A o B is closed and A = CB.

It follows that X can be split into two disjoint, separated sets A and B if and only if A = C B and B is simultaneously open and closed. The theorem is an immediate consequence.

17.4

Corollary The metric space R" is connected. Proof See exercise 14.17(2).

17.5 Theorem Let W be a collection of connected sets in a metric space X all of which contain a common point j;. Then

r=U s Sew

is connected. Proo/ Let A and 5 be non-empty subsets of T such that AuB= T. Either £e,4 or ^eB, Suppose that ^ e A Then all of the sets SnA with SeW are non-empty. Suppose that SnB = 0 for all S e ^ Then

S = BnT=B Seiu

Seut

because BaT. This is a contradiction because B^0.

Hence there exists an

Connected sets SxeW such that S^nB^^.

49 Also S1nAj=0

and

Because S1 is connected, it follows that S1nA and S1nB are contiguous. Thus A and B are contiguous by theorem 15.15.

17.6

Connected sets in U1

An interval I in U* is a set with the property that, if ael and then

aR 2 cannot be continuous on the set A if t(6) Let if be a hyperplane in Un. Prove that Un\H has two components and that these are the open half-spaces determined by H. (See exercise 14.17(9).) Explain why any line segment which joins two points from different half-spaces contains a point of if. (See §13.14.)

18

18.1

CLUSTER POINTS

Cluster points A cluster point of a set £ in a metric space % is a point £e X such

that

A cluster point of a set E may or may not be an element of E. Equally, an element of E may or may not be a cluster point of E. Those elements of E which are not cluster points of E are called isolated points of E.

isolated points ofE

cluster points ofE

Cluster points are often referred to as 'points of accumulation' or as 'limit points' of the set. The latter usage is unfortunate and we prefer not to employ the word 'limit' except when discussing convergence. (See §26.2.) The tables below list a number of equivalent assertions about cluster points and isolated points. The proofs of these equivalences provide useful practice in the techniques of chapters 14 and 15 and so are the subject of one of the problems in the next set of exercises.

60

Cluster points TABLE I

(i) ^ is a cluster point of E. (ii) \eE but is not an isolated point of E. (iii) dfe, E \{£}) = 0. (iv) Each open ball with centre % contains a point of E other than £. (v) Each open ball with centre ^ contains an infinite subset of E.

61 TABLE II

(i) £ is an isolated point of £. (ii) £e£ but is not a cluster point of E.

P) .

-

_

(iv) There exists an open ball containing no point of E except £.

To deduce Iv from Iiv, consider a sequence of smaller and smaller open balls with centre £. Note that Iv implies that every point of a. finite set is an isolated point.

18.2

Examples

(i) The set S = {1/n: neN} in U* has a single cluster point, namely 0. Note that 0£S. Every point of S is an isolated point of S. (ii) The set N in U * has no cluster points. All its points are isolated, (iii) The cluster points of the set Tin U2 defined by are the points of {(x, y): x2 + y2 ^ 1}. The point (2, 2) is an isolated point of T. (2,2)

18.3

Exercise

(1) Find the cluster points and the isolated points of the following sets in

62

Cluster points (i) (0,1) (ii) [0,1) (iii) ( - o o , 0 ] (iv) N (vi) U (vii) 0 (viii) {1, 2, 3} (ix) [0, l]u{2}

(v) Q (x) [0, l ] u ( l , 2]

(2) Repeat the above question for the following sets in U2. (i) (ii) (iii) (iv) (v) (vi) (vii)

A = {(x9y): B = [(x,y):x^0} C = {(x,y):x2 + y2 of closed boxes each of which contains an infinite subset of E. By the Chinese box theorem, there exists a £ which belongs to each of these boxes. Let B denote any open ball with centre !;. Since the dimensions of Sk are 2~k times those of S, a closed box SK will be a subset of B provided that K is sufficiently large. Thus B contains an infinite subset of E and so ^ is a cluster point of E (§18.1(Iv)).

• •

• •

• • • •

•

•

•

*

19.7 Note It is of great importance to remember that the BolzanoWeierstrass theorem (and hence the Heine-Borel theorem which follows) are FALSE in a general metric space %. This point is taken up again in chapter 20. 19.8 Theorem (Heine-Borel theorem) A set X in R" is compact if and only if it is closed and bounded. Proof The proof that a compact set in 1R" is closed and bounded is quite easy and, since we prove the same result for a general metric space later on (theorems 20.12 and 20.13), we shall therefore consider only the deeper part of the proof - i.e. that a closed, bounded set in tR" is compact. Suppose that K is a closed, bounded set in Un and let E be an infinite subset of K. From the Bolzano-Weierstrass theorem it follows that E has a cluster point £. By theorem 18.7, the fact that K is closed implies that Thus K is compact.

19.9 Theorem (Cantor intersection theorem) Let be a nested sequence of non-empty closed subsets of a compact set K in a metric

72

Compact sets (I)

space X. Then

n Proof If one of the sets Fn is finite, the result is trivial. Otherwise we can construct an infinite subset E of K consisting of one point from each of the sets F . points ofE

Since E is an infinite subset of a compact set K it has a cluster point \. Since all but a finite number of points of E belong to each Fn, £ must be a cluster point of each of the sets Fn. But each Fn is closed. Hence, by theorem 18.7, £>eFn for each nef^J. Thus

19.10 Example The sequence of sets constructed in §18.9 is a nested sequence of non-empty closed subsets of the compact set [0, 1]. By the Cantor intersection theorem, its intersection F is non-empty. In fact, as is explained in §18.9, F is uncountable.

19.11

Exercise

(1) Which of the sets given in exercise 18.3(1) are compact? (2) Which of the sets given in exercise 18.3(2) are compact?

Compact sets (I)

73

(3) Give examples of a nested sequence of non-empty closed sets in U* and a nested sequence of non-empty bounded sets in Ul which have empty intersections.

19.12

Compactness and continuity

19.13 Theorem Let X and % be metric spaces and let Scz X. If/: S^y, is continuous on the set S, then S compact =>f(S) compact. Proof Suppose that S is compact. Let E be an infinite subset of f(S). We need to show that E has a cluster point in f(S). Define g:f(S)-+S so that/(gf(y)) = y (see example 6.9). Then D = g(E) is an infinite subset of S. Since S is compact, it follows that D has a cluster point Now let TJ =/(£). Then r\ef(S). Also since/is continuous on <S, d& D \ &}) = 0 => d(i,, f(D \ ft})) = 0 by theorem 16.4. B u t / ( D \ { ^ } ) = £ \{i|} and so i| is a cluster point of E.

The next theorem is of fundamental importance in optimisation theory. It asserts that any real-valued continuous function achieves a maximum and minimum value on a compact set K. It is worth noting that this theorem need not be true if K is not compact. Consider, for example, the function / : R-+R defined by f(x) = x. This does not achieve a maximum nor does it achieve a minimum on the open set (0, 1).

74

Compact sets (I)

19.14 Theorem Suppose that K is a non-empty compact set in a metric space % and t h a t / : K-+M is continuous on the set K. T h e n / achieves a maximum and a minimum value on the set K - i.e. there exists ^eK and H\GK such that, for any xeK,

/(I)

By theorem 19.13, f(K) is compact. A compact set in R 1 is closed and bounded by the Heine-Borel theorem (19.8). Because/(X) is a closed, non-empty set of real numbers which is bounded above it has a maximum (corollary 14.11). For similar reasons, f(K) has a minimum.

19.15 Corollary Suppose that K is a non-empty compact set in a metric space %. and that ye %. Then there exists a ^ e K such that

Compact sets (I)

75

Proof The function/: Z-+U defined by/(x) = d(y, x) is continuous on % by exercise 16.16(4). Hence it achieves a minimum on K by theorem 19.14. 19.16 Corollary Suppose that F is a non-empty closed set in Un and n that yeU . Then there exists a ^ e F such that

Proof Let B be the closed ball with centre y and radius d(y, F ) + 1 . Then BnF is a non-empty, closed and bounded set in Mn. It follows from the Heine-Borel theorem (19.8) that BnF is compact. The result therefore follows from corollary 19.15. (Note that the result does not hold in a general metric space.)

19.17

Exercise

(1) The sets A and B in R 2 are defined by Explain why a function / : R 2 -»R 2 cannot be continuous on A if (2) Let / be an interval in (R1 and l e t / : I R 1 - ^ 1 be continuous on /. Give counter-examples to each of the following (false) propositions: (i) / closed =>/(/) closed (ii) / closed =>/(/) bounded (iii) / open =>/(/) open (iv) / bounded =>/(/) bounded (v) / open and bounded => / ( / ) has no maximum. (3) Which of the propositions above are necessarily true when/: [ R 1 - ^ 1 is continuous on a compact interval J containing /. t(4) The distance d(S, T) between two non-empty sets S and Tin a metric space % is defined by d(S, T) = inf {d(x, y):xeS and ye'T}.

76

Compact sets (I)

Show that, if S and Tare compact, then there exist ssS and t e T s u c h that d(S, T) = d(s, t). Show that the same result holds when X = U" if 5 is compact but Tis only closed. Give examples of closed sets S and Tin !R 2 for which the result is false. f(5) Let SczU and suppose t h a t / : S-+M is continuous on the set S. The graph of/is the set Tin 1R2 defined by

Prove that (i) S connected => T connected (ii) S compact => T compact. Give examples of non-continuous functions/: S-+M for which (i) and (ii) are true. t(6) L e t / : U^-U be a bounded function whose graph is a closed set in R2. Prove t h a t / is continuous on U.

20f

20. If

COMPACT SETS (II)

Introduction

In this chapter we give the 'proper definition' of a compact set (see § 19.5). Most of the chapter is then concerned with proving that this definition is equivalent to that of the previous chapter for a metric space Z. This is quite a lengthy piece of work and many readers will prefer to skip this chapter for the moment. The book has been written with this possibility in mind.

20.2t

Open coverings A collection U of sets is said to cover a set E if and only if

SeU

sets in U

20.3

Example

Let S: [0,1] ->(0, oo). Then the collection

covers the set £ = [0, 1].

77

78

Compact sets (II)

20.4|

Compact sets A set K in a metric space X is said to be compact if any collection U of open sets which covers K has a / m t e subcollection 5 cz U which covers X.

20.5 by

Example

Suppose the function (0, oo) of example 20.3 is given

*

(

x

)

.

x+2 Then a finite subcollection 5 of U which covers [0, 1] is

- i.e. one needs only the intervals (x — S{x% x + d(x)) with x = 0 and x = l.

20.6|

Exercise

(1) A function/: R-*R has the property that, for each xe[0, 1], there exists a 0 such t h a t / is either increasing or else decreasing on (x — 0 such that Bo a G (theorem 14.12). Since D is dense in £, there exists a deD such that d(£, d)
However, if just one point is removed from C we obtain a set which is topologically equivalent to a line. The homeomorphism which is usually used to show this is called the stereographic projection. This is illustrated in the diagram below. The point deleted from C (labelled N in the diagram) is referred to as the 'north pole' of the projection.

Similarly, a 'punctured sphere' is R 3 (i.e. a sphere with one point removed) is topologically equivalent to a plane. Again, the stereographic projection provides a suitable homeomorphism.

Topology

93

Let S be the sphere in R 3 with centre (0, 0, j) and radius j . Take as the 'north pole' the point N = (0, 0, 1) and let P be the plane {(x, y, z): z = 0}. Then the stereographic projection/: P-+S \ {N} is given by

fix, y, OH

-

a n d / " 1 : S \ {iV}->P is given by

Both functions are continuous by theorems 16.5 and 16.15.

21.5

Exercise

(1) Prove that topological equivalence as defined in §21.1 is an equivalence relation as defined in §5.5. (2) Let 5 be any set in R" which is topologically equivalent to a closed ball B. Assuming Brouwer's fixed point theorem as quoted in note 17.14, show that for any function/: S-+S which is continuous on S there exists a $eS such that/(§) = £. (3) Let 3 denote the collection of intervals in R1 which have at least two points. Let U denote the collection of all open intervals in 5 , let V denote the collection of all compact intervals in 3 and let W denote the remaining intervals in 3 .Show that U, V andW are the equivalence classes into which 3 is split by the relation of topological equivalence. (4) Let L be a closed line segment in U2 and let S be the closed box [0, 1] x [0, 1]. Prove that the removal of any three points from L produces a disconnected set but that the same is not true of S. Deduce that L and S are not topologically equivalent. t(5) Check the formula given in §21.4 for the stereographic projection between the plane P and the punctured sphere S\{N}. t(6) A function F: C->S\{N} is denned by 3mz

\z\2 \

W'TW/ Prove that F is a homeomorphism between the complex plane C and the punctured sphere S \ {N} (where S \ {N} is regarded as a metric subspace of IR3). Show that

94

Topology

21.6

Continuous functions and open sets

21.7 Theorem Let X and % be metric spaces and let/: X-+%. Then the following statements are all equivalent. (i) / is continuous on X. (ii) For any xe X and any Sf"1(F) closed, (v) For any G c ^, G open =>/~ 1 (G) open. Proo/ We shall show that (i)=>(ii)=>(iii)=>(iv)=^(v)=>(i). (i)=>(ii). This is just theorem 16.4. (ii) => (Hi). Assume (ii) and suppose that yef(S). Then y =/(x) where xeS. Since d(x, S) = 0 it follows that d(f(x), f(S)) = 0. Thus y=/(x)e/(S). (iii)=>(iv). Assume (iii) and suppose that Fczy, is closed. Since

Hence f~\F)czf-\F). T h e r e f o r e / " 1 ^ ) is closed. (iv)=>(v). Assume (iv) and suppose that G(i). Assume (v) and suppose that C and D are separated sets in y,. Then disjoint, open sets G and H exist with CaG and DczH by theorem 15.16. The s e t s / ' ^ G ) a n d / - 1 ( i J ) are then disjoint, open sets in £ by (v). But then/" 1 (C)cz/- 1 (G) and/'~ 1 (Z))c:/- 1 (/J) are separated sets in 2 . Thus C and D separated implies / - 1 ( Q a n d f~1{D) separated. Hence / is continuous (§16.2).

The most significant of the equivalences in the above theorem is that which asserts that a function/: X-*% is continuous if and only if, for each

Topology

95

G c ^ , G open implies/ 1(G) open. The next few sections are devoted to exploring the consequences of this result.

21.8

Topologies

As we know from the previous section, two metric spaces X and y are topologically equivalent if and only if there exists a bijection/: X^y such that, for each Gay, G open f~\G)

open.

(1)

Thus, if we wish to determine whether or not X and y are topologically equivalent, the only question we need to ask about X and y is: what are their open sets? Any bijection/: X-^y can then be tested to see whether it satisfies (1). Other information about X and y may be helpful or interesting but it is not strictly necessary. Thus we know all that there is to know about the topological structure of a space if we have a list of all its open sets. For this reason the collection of all open sets in a space is called its topology.

21.9|

Relative topologies

Notice that theorem 21.7 refers to a function/: Z-^y, where % and y, are metric spaces. Various equivalent conditions are then given for/ to be continuous on the whole space X. In contrast, the theorems of chapter 16 always involved a subset £ of X and were concerned with the continuity of a function/: Z^% on the set Z.

If £ is a subset of X, then it is, of course, true that Z is itself a metric space provided that we use the same definition for distance in Z as is used in %. We say that £ is a metric subspace of %.

In chapter 16, we carefully chose a definition for a function /: Z^% to be continuous on the set £ which makes it quite irrelevant whether we regard our

96

Topology

underlying metric space to be X or whether we simply throw out all the points of X which are not in £ and take our underlying metric space to be £. This is because two subsets A and B of £ are contiguous in the metric space X if and only if they are contiguous in the metric space £. It follows that theorem 21.7 applies equally well in respect of a function/: Z-+% which is continuous on a subset £ of the metric space X. One simply takes note of the fact that £ is a metric subspace of X and replaces X at each occurrence by £. This leaves the meaning of item (i) of theorem 21.7 unaltered and the same is true of item (ii). However, very considerable care is necessary with items (iii), (iv) and (v) of theorem 21.7 when the underlying metric space is switched from X to £. Such a switch affects the meaning of the words 'open', 'closed', 'closure' etc. This is because these ideas all depend on the notion of a boundary point. Recall that \e X is a boundary point of S if and only if 0 in the metric space £ is defined by B = {z: d(z, £)% is a continuous bijection with a continuous inverse/" 1 : y,-+%. If a homeomorphism/: X^% exists, we say that the topological spaces X and y, are topologically equivalent (or homeomorphic). From the topological point of view, two homeomorphic spaces are essentially the same (see the discussion of §9.21). Let £ be a subset of a topological space X with topology S. Then

is a collection of subsets of £ which satisfies the conditions for a topology on £. We call V the topology on £ relative to X. Thus theorem 21.11 for a metric space becomes a definition for a topological space. The set £ with the topology V is called a topological subspace of X. The introduction of the idea of a relative topology means that we do not need a separate definition for continuity on a subset £ of a topological space X. We use the definition given above but with £ (regarded as a topological subspace of X) replacing X. Similarly, we need only provide definitions of a connected topological space and a compact topological space. A connected topological space X is one in which the only sets which are both open and closed are 0 and X. Theorem 17.3 for a metric space therefore becomes a

102

Topology

definition for a topological space. It is important that theorem 17.9 remains true in a topological space. With our new definitions, the proof is even easier.

21.16f Theorem Let X and y be topological spaces and let f:X^y continuous surjection. Then

be a

X connected => y connected. Proof Suppose that E is a set in y which is both open and closed. Since f:X-+y is c o n t i n u o u s , / " 1 ^ ) is both open and closed in X. But % is connected. Thus, / " 1 ( £ ) = 0 or f~1(E)= X. Because / is a surjection, it follows that E = 0 or E= X. Thus y, is connected.

A compact topological space X is one with the property that any collection U of open sets which covers X has a finite subcollection which covers X. This definition is identical with that of §20.4. It is important that theorem 19.13 remains true in a topological space and, again, with our new definitions, the proof is even easier.

21.17f Theorem Let X and y be topological spaces and let f\X-+y continuous surjection. Then

be a

X compact =>y compact. Proof Let V be a collection of open sets which covers y. Then

is a collection of open sets which covers X. Since X is compact, a finite subcollection £ covers X. Let

$={f(G):Ge£}. Then 5 is a finite subcollection of V which covers y. Hence y is compact.

21.18t

Product topologies In the space U2 the Euclidean metric d: U2-^U is defined by

We use this metric because it corresponds to the notion of the distance between two points as understood in Euclidean geometry. However, the Euclidean metric is not the only possible metric which can be used in R 2 . The function m: [R2—>IR defined by m(x, y) = max{|x 1 - t y 1 |,

\x2-y2\}

is an example of an alternative metric. So is the function /:R2->[R defined by /(x, y) = | x 1 -

103

Topology

To verify that these functions satisfy the requirements for a metric given in §13.1 is very easy. Of course, an open ball with respect to one of the metrics / and m looks very different from a Euclidean open ball as the diagrams below illustrate.

£, x)R defined by d(x, y) =

{(d1(xl9

x2i y2)):, 211/2

(1)

and m(x, y) =

2,

y2)}

(2)

are metrics on. X^ x X2 which generate the same topology on Xi x X2. Why should this topology be more useful than the various other topologies which one might impose on Xx x £ 2 ? The significant fact about the metrics d and m is that they generate a topology on XxxX2 which makes the projection functions Pl: Xxx X2^-X1 and P2: X1 x X2^X2 continuous. For example, if we use the metric d in Xi x X2, then

and hence P1 is continuous on X±x X2 by lemma 16.8.

104

Topology

P 2 (x)

X = (Xj, X 2 )

•X,

This observation makes it natural to define the product topology on Zl x X2 to be the weakest topology (i.e. the topology with the fewest open sets) with respect to which the projection functions P x : %l x X2~*Xl and P2:Xlx X2-+X2 are continuous. It follows that, if Gx is an open set in Xu then Pi~1(G1) must be an open set in the product topology of X± x X 2. Similarly, if G2 is an open set in X2, then P2~ 1{G2) must be an open set in the product topology of Xx x X2.

Since the intersection of a finite collection of open sets must be open, it follows that

must be an open set in the product topology of X^ x X2. Finally, the union of any collection of open sets must be open. Thus, if W is any collection of sets of the form Gx x G2 (where Gj is open in Xx and G2 is open in X2\ then

5= KJ

(3)

must be an open set in the product topology of Xi x X2. The collection 5 of all sets 5 of the form (3) satisfies the requirements for a topology on Xtx X2 given in §21.15. It follows that 3 is the product topology on Xi x X2.

Topology

105

S=

Now return to the case in which %1 and £ 2 a r e metric spaces with metrics dx and d2 respectively. If we use the metric m: Xx x £ 2 ->R defined by (2), then the open balls B in %l x ft2 are of the form B = BX xB2 where JBJ is an open ball in %x and 5 2 is an open ball in %2. It is therefore apparent from the preceding discussion, that the open sets in %x x %2 generated by the metric m are precisely those which lie in the product topology 3'. The same is therefore true of the metric d: %x x SC2-+U defined by (1) and of numerous other metrics. From the topological point of view it does not matter which of these metrics we choose to use in %x x %2 since they all generate the same topology. When discussing topological matters, we therefore work with the metric in %1 x %2 which happens to be most convenient for the problem in hand. In U2 (or U") this is usually the Euclidean metric. When Zx and %2 are more general metric spaces, however, the metric m is often much less cumbersome.

22

22.1

LIMITS AND CONTINUITY (I)

Introduction

->^ and suppose that £eX and \\e%. In this chapter, we shall study the meaning of the statement as Sometimes this is written in the equivalent form

We say that 4/(x) tends to r\ as x tends to £' or that '/(x) converges to the limit r\ as x approaches £'. The diagrams below illustrate the idea we are trying to capture. The first diagram is of a function/: R 2 ->R 2 for which/(x)->77 as x-»£ It shows x approaching £ along a path. As x describes this path,/(x) approaches r\. We shall of course want the same to be true however x approaches £.

The next diagram shows the graph of a function/: as x-ȣ.

for which f(x)-+r]

Limits and continuity (I)

107

y

Note that in both diagrams it is false that r\ =f(Z>). It is not the case that one can always find the value of

by replacing x in the formula for /(x) by £. This point is of some importance in calculus. The derivative of a function f:M->U at the point £ is defined by

/'(«) = li /eX and x\ey,. Then we say that as through the set S if and only if For any open set G containing t|, there exists an open set H containing £ such that

xeHnS=>f(x)eG provided

110

Limits and continuity (I)

One thinks of the choice of G as specifying how 'near' we want/(x) to be to r\. Since we are interested only in the limit through the set S, we restrict attention to xeS. We also exclude the case x = £. With these provisos, the definition says that, if we take x 'sufficiently near' to ^ - i.e. x e H - then/(x) will be as 'near' to r\ as we specified - i.e. /(x)eG. We now define the statement as

This should mean that /(x) approaches r\ however x approaches appropriate definition is therefore that:

The

For any open set G containing r\, there exists an open set H containing £ such that xeH =>/(x)eG provided In order that this definition makes sense, it is necessary that/be defined on some open set S containing £ (except possibly at £ itself). Such an open set S contains all points of Z which are 'sufficiently near' to £. Indeed, our definition of the statement '/(x)-n] as x->^' is equivalent to the assertion that/(x)^i| asx-^^ through the set S for some open set S containing £. 22.4

Limits and continuity

If/(x)--•/(£) as x->^, we say that/is continuous at the point ^. The next theorem relates this terminology to our previous work on continuity. 22.5

Theorem Let % and y. be metric (or topological) spaces and let

Limits and continuity (I)

111

f: Z-*y. where S is a set in Z. Then/ is continuous on the set S if and only if, for each as through the set S. Proof We give the simplest proof. This depends on the results of chapter 21. An alternative proof using only the ideas of chapter 16 is suggested in exercise 22.24(4). (i) Suppose that / is continuous on the set S. Let \ be any point of S and let G be an open set containing/^). Then/~ 1 (G) is an open set relative to S by theorem 21.7. From theorem 21.11, it follows that there exists an open set if in Z such that/" 1 (G) = if nS. We then have that xeHnS=>/(x)eG. This shows that/(x)->/(!;) as x->^ through the set S. (ii) Suppose that / is not continuous on the set S. Then there exists an open set G in y, such that/ - 1 (G) is not open relative to S (theorem 21.7). Let iief~1{G) be a boundary point of/ - 1 (G) relative to S. Then any open set H in Z which contains \ has the property that HnS contains a point

Since zeHnS but/(z)R defined by

By corollary 22.6, f(x, y ) - 3

as

(x, y)->(l, 1).

(ii) Let S = U \ { 0 } and consider the rational function F: S-+U defined by

We know from corollary 22.6 that F(x)->1 as x-> — 1 but corollary 22.6 does not help immediately in evaluating lim F(x) x->0

because F(0) is not defined. Note, however, that for x # 0, _ . l + 2x + x 2 - l F(x) = x If G: IR->R is the polynomial defined by G(x) = 2 + x, it follows that F(x) = G(x) unless x = 0. Since we explicitly ignore what happens when x equals 0 when evaluating a limit as x approaches 0, it follows that lim F(x) = lim G(x) = lim (2 + x) = 2. x->0

x~*0

x-*0

22.8 Exercise (1) Evaluate the following limits

(i)

lim

y2-x2 -= 2

l.D y

(ii)

^

m

v2-x2

(2) Let K = {(x, y): O ^ x ^ l and O^ygl} and let/: [R2-IR be defined by fl

(x,y)eK

Limits and continuity (I)

113

Prove that/(x, y)-+\ as (x, y)->(l, 1) through the set K and/(x, y)-+0 as (x, y)->(l, 1) through the set e K. (3) Suppose that 2; is not a cluster point of S. Explain why it is true that f(x)-+r\ as x-»£ through the set 5 for all r\. If £ is a cluster point of S, prove that there exists at most one r\ such that/(x)->Tj as x->£ through the set S provided that, for each pair of distinct points r\1 and r\2 in yy there exist disjoint open sets Gx and G2 such that r\ieG1 and r\2eG2. Show that such disjoint open sets always exist when y is a metric space. (4) Suppose that S ^ L ^ ^ . Prove that/(x)-nj as x->£ through 5 if and only if/(x)-*Tj as x->^ through Sx and through S2. Deduce that, if/ is the function of question 2, then there is no r\eU for which/(x, y)->r\ as (x, JO->(1, 1).

(5) Suppose that T is a subset of 5. If f(x)-+r\ as x-ȣ through the set S9 prove that/(x)-^i| as x->^ through the set T. (6) Suppose t h a t / : R-+R and g: U-^U are continuous on the set U and /(x) = gf(x) for each xeO. Prove that f = g. [Hint: Let J be irrational. Then f is a cluster point of Q (why?). Also/(x)-^/((^) a s i ^ through Q and g(x)-^g(^) as x ^ ^ through O.]

22.9

Limits and distance

In a metric space % a set G is open if and only if each xeG is the centre of an open ball B which is entirely contained in G. This fact means that, if X and y, are metric spaces, then we can rewrite the definition of a limit given in §22.3 in terms of open balls rather than open sets. Let % and y. be metric spaces and let/: S^y where S is a set in X. Then /(x)->i| a s x ^ through the set S if and only if, for each open ball E with centre t|, there exists an open ball A with centre \ such that xeAnS =>/(X)G£

provided that

114

Limits and continuity (I)

If we take the radius of the open ball E to be e and that of A to be 3, the definition assumes the following familiar form: For any e>0, there exists a 3>0 such that for each xeS,

It is sometimes helpful to think of/(x) as an approximation to r\. The quantity d(r\, /(x)) is then the error in approximating to r\ by /(x). The definition then asserts that this error can be made as small as we choose by taking x sufficiently close to £. (Here, of course, it is understood that values of x outside S \ {^} are to be ignored.) 22.10 Example Consider the function/: [R2->[R2 defined by (u,v)=f(x,y) where u=x+y) v = x-y.) The fact that/(x, y)->(2, 0) as (x, y)->(l, 1) follows from theorem 22.5 because/is continuous on U2. In this example we give an alternative proof using the criterion for convergence given in §22.9. Proof Given any e>0, we shall demonstrate the existence of a d > 0 such that We begin by observing that \\(x,y)-(l

l)\\ = {(x-l)2

and y, x-y)-(2,

Given e > 0 we therefore have to choose S > 0 such that 0 0, there exists a 3 > 0 such that 0rj as

as x->b —

->fc through the set (a, b).

/ ti + E -

Limits and continuity (I)

117

It is useful to note that f(x)^rj as x-»a + if and only if, for any 8>0, there exists a 8>0 such that a<x \f(x) — rj\0, there exists 3. S>0 such that b-S<x \f(x)-rj\<e. 22.13 Theorem Let / be an open interval containing £ and suppose that /:/\{£}-»R. Then /(x)->77 as x-+f if and only if/(x)->77 as x-»£— andf(x)^rj as x-xi; + . Proo/ This follows immediately from exercise 22.8(4). 22.14

Example Let /: U ->R be denned by 2x

Then/(*)-• 0 as x->l— and/(x)->2 as x->l+. It follows from theorem 22.13 that lim f(x) does not exist. In particular, / is not continuous at the point 1.

y=fW

Note that it is not necessary to use an argument involving e and 6 to show that/(*)-> 2 as x->l +. The function g: U -*[R defined by g(x) = 2x is a polynomial and hence is continuous everywhere. A\sof(x) = g(x) when x> 1.

118

Limits and continuity (I)

In calculating a limit as x-*l + we ignore what happens when x ^ 1. Hence lim f(x)= lim g(x)= lim 2x = 2. X-*\+

X-+1+

X-+1 +

Similarly for/(x)—•() as x->l —. 22.15

Some notation

When the limits exist, the notation - ) = lim/(x);

= lim f(x)

can often be useful. However, it is important not to confuse /(£ —) or/(£ + ) with/(£). These three quantities are equal only when/is continuous at the point £. If / ( £ - ) = /(£) we say that / is continuous on the left at £. If/(
/«)=/«+) neither / ( £ —) nor + ) is equal to

119

Limits and continuity (I) 22.16

Monotone functions

Suppose that S is a non-empty subset of U. We say that / : Sincreases on S if and only if

for each x and y in S. Similarly, / : S->R decreases on S if and only if

for each x and y in 5. We say t h a t / i s strictly increasing on S if and only if

xf(x)f(y).

strictly increasing function

decreasing function

A function which is either increasing or else decreasing on S is said to be monotone on S. (A function can be both increasing and decreasing on S. But then it must be constant.) A function which is either strictly increasing or strictly decreasing is said to be strictly monotone.

22.17 Theorem Suppose that I = (a, b) is an open interval in U and that / : I-+R is increasing on /. (i) If/is bounded above with supremum L on /, then/(x)-»L as x^b — . (ii) If / is bounded below with infimum / on /, then /(x)->/ as x-+a +. Proof We prove only (i). From §22.12, we know that, given any e > 0, we have to demonstrate the existence of a 8 > 0 such that

b-5<x\f(x)-L\<e. But \f(x) — L\<s

— 8
Limits and continuity (I)

121

Proof The function / is bounded above on the interval (a, £) by By theorem 22.17, the smallest upper bound of/on (a, £) is/(£ — ). It follows that, for any xe(a, £), A similar argument for the interval (£, b) yields the other inequalities. 22.19

Inverse functions

Recall that a function /: S-»T admits an inverse function f'1: T^>S if and only if/ is bijective. A function is bijective if and only if it is both surjective and injective. Surjective means that f(S) = T and injective means that f(x1) = f(x2) xx =x2.

In this section, we are interested in the case when S and Tare intervals in U and / is continuous on S. It then seems 'intuitively obvious' that / is bijective if and only if T=f(S) and / is strictly increasing or strictly decreasing on S. Furthermore, it seems equally clear that the inverse function/" 1 : T^S will be continuous on T.

x=fl(y)

*~ y

122

Limits and continuity (I)

But, like so many 'intuitively obvious' theorems, the proof requires some deep results. 22.20 Proposition Let / be an interval in R and suppose that/: J->R is continuous on L Then/is injective if and only if/is either strictly increasing* or strictly decreasing on /. Proof The fact that a strictly increasing or decreasing function is injective is trivial. The proof of the other half of the theorem is the content of exercise 17.28(3). It requires a simple application of the intermediate value theorem (17.11). The diagram illustrates a function/: I-+R which is continuous on / but neither increasing nor decreasing on /. If aT| through B\ A limit specified in this manner will be called a double limit. It is important to be aware of the fact that statements of this sort about double limits can be highly ambiguous and that it is therefore always necessary to examine the context carefully in order to determine precisely what the author means. The most straightforward case is that in which x and y are intended to be 'independent variables'. In this case the double limit statement simply means that /(x,y)->£

as

(x, y)->& i|)

through the set S = AxB. However, in order to use the definition of a limit given in §22.3 we need to know what sets are open in Xxfy. We therefore augment our definition of a double limit in this case by observing that the open sets of X x y, are to be those in the product topology of X xy, (§21.18).

130

Limits and continuity (II)

131

As explained in §21.18, when % and y are metric spaces with metrics dx and d2 respectively, there are various metrics which we can introduce into % x y which generate the product topology of % x y and the choice of which of these metrics to use in % x y is largely a matter of convenience. When % = y. = M and so % x y = U2, we usually use the Euclidean metric but, in many cases, it is easier to use the metric m: Zxy-*M defined by 1,

), (x 2 ,

x 2 ), d2(y1, y2)}.

The use of this metric in the formulation of the definition of a limit given in §22.9 yields the following criterion for the existence of a double limit: For any s>0, there exists a 3>0 such that for each (x, y)eS = Ax B, ),

(x, y))0, there exists a 3>0 such that for any xeA and any yeB, 9 x)£ as (x, y)->(£, rj). If we use the Euclidean metric in U 2, the definition of this statement can be expressed in the form: For any g>0, there exists a 3>0 such that 0£ as x->£ through A and y-n] through B\ In the discussion above we considered its meaning when x and y were to be understood as 'independent variables'. However, the statement is often used in circumstances when x and y are not 'independent variables' - i.e. some relation between x and y must be satisfied. Sometimes one is told explicitly what this relation is. On other occasions one has to guess the relation from the context. The existence of such a relation means that there is a set R in Zxy, with the property that (x, y) not in R are to be disregarded. Our double limit statement then means that /(x,y)->§ as (x,"y)->&ii) through the set Rr\S (where S — AxB topology is used in Z x y).

and it is understood that the product

23A Example Consider the function/: [R2->[R of exercise 22.8(2). If one is asked to consider the double limit £ defined by the statement/(x, y)->C as x-»l and y->l where x + y>l, then one should interpret this as meaning that f(x9 JO-C

as (JC, y M l , 1)

through the set J = {(x, y):x + y>l}. Since J& I])

although the latter notation is somewhat less precise in that it takes for granted the use of the product topology in Zxy,. It is important not to confuse either of these pieces of notation with repeated limits lim I lim /(x, y)); lim (lim /(x, y) I. x-*

\y-ii

/

y-i

\x-*

/

The reasons why one needs to be careful in dealing with repeated limits are best explained with the help of some examples.

Limits and continuity (II) 23.6

133

Example Observe that r

x2

-y2

hm —-z

1 T

= 1

and hence lim (lim-T-S-1 = 1. 2 2 x^o\y-,o x + y j On the other hand, a similar argument shows that lim (lim —z

T|

= — 1.

The order in which the limits appear cannot therefore be reversed in general without changing the result. 23.7

Example Observe that, for each xeU,

It follows that (

*y

lim I lim —

\ 1 = 0.

Similarly, lim f lim —z. j I = 0. However, the double limit nm

z

IT

does not exist (for the same reason that the similar limit in example 22.11 does not exist; see also example 23.15). 23.8

Example Consider the function h: U2^>M defined by h(x,y) = x

We have that \h(x, y)\ = \x\£\\(x, y)\\

134

Limits and continuity (II)

and so it is easily shown that lim

h(x, y) = 0.

(x, y)-(0, 0)

On the other hand, we have that h{x, y)-*x as y-»0 + and /i(x, y)—• — x as y—•0 —. It follows that lim h(x9 y)

(1)

y-0

does not exist unless x = 0 and hence that the repeated limit lim (lim h(x, y) J does not exist. The results of examples 23.6 and 23.7 are not particularly surprising. We have already seen a number of examples in which /(x, y) tends to different limits as (x, y)->(0,0) along different paths and, as the diagrams below indicate, one can think of taking a repeated limit as another special way of allowing (x, y) to approach (0, 0). lim Aim f(x, y)\

x - + 0 \ v -*•()

I

(x.y)

—»

t

1

(0,0)

(0,0)

lim | lim fix, y) Example 23.8 requires a little more thought. The problem is that the existence of the double limit as (x, y)->(0, 0) does not guarantee that h(x, y) tends to a limit as (x, y)-+ (X, 0) along the line x = X (unless X = 0). As the diagrams below indicate, there is no reason why matters should be otherwise.

(x,y)

(0,0)

(1,0)

135

Limits and continuity (II)

However, if we remove this pitfall by restricting our attention to those functions for which the limit (1) does exist, then the existence of the double limit implies the existence of the repeated limit and the two are equal. This result is the content of the next theorem.

23.9| Theorem Let %, fy and £ be metric spaces and let ^e %, \\e% and Suppose that A is a set in X with cluster point £, and that B is a set in y. Let / : S\(£, !))-•£ where S = A x B. Suppose that /(x,y)->£

as

(x,

(2)

/(x, y)->Z(y) as

(3)

through the set S and that, for each yeB,

through the set A where /: £->£. Then /(y)-»£

as y

(4)

through the set B.

B4

Proof Let e>0 be given. By (2) there exists a £ > 0 such that for each xeA and each yeB & x) < d and d2(t!, y) < 5 => ^ , /(x, y)) < e/2 provided (x, xeA,

x\). By (3), for each yeB there exists a A y >0 such that, for each ), f(x, y))<e/2.

136

Limits and continuity (II)

For each ye£, choose xyeA so that d^, 0£

/

To proceed any further with this topic we need to introduce the subject of uniform convergence. In particular, we shall show that, if both sides of (5) exist, then the two sides are equal provided that one of the inner limits is a uniform limit.

23.1 If

Uniform convergence

Suppose that % and £ are metric spaces and that/:X xB^Z where A is a set in %. Let £E % and suppose that /:B->£ has the property that, for each yeB, /(x,yH/(y)

as

x^£

through the set A. Expressed in full detail, this assertion means that Vyefl V £ >0 3(5>O VxeX, (1) We know from chapter 3 that the order in which the quantifiers V and 3 occur in a statement is significant to its meaning. In particular, if P(u, v) is a predicate, it is in

Limits and continuity (II)

137

general false that Vw

3vP(u9v)o3vVuP(u9v).

It follows that, if we move the term 'Vyel?' from the beginning of our list of quantifiers to the end, we obtain a statement which means something different from the original statement - i.e. the statement V£>0

VxeA VyeB, (2)

does not mean the same as (1).

pointwise convergence

uniform convergence

138

Limits and continuity (II)

The difference in the meaning of the two statements is that, in statement (1), the value of S which is asserted to exist may depend both on the value of e and on the value of y. In statement (2), the value of S depends only on the value of e - i.e. the same value of S works uniformly for all values of yeB. If (2) holds, we say that /(x,y)->/(y)

as

x-+^

through the set A uniformly for yeB. It is sometimes necessary to emphasise that this statement is stronger than (1). When (1) holds, we therefore say /(x,yH/(y)

as

x->^

through the set A pointwise for yeB. Note that it is obvious that uniform convergence implies pointwise convergence. But fhe converse is very definitely false.

23.12f

Distance between functions With the notation of the previous section, suppose that /(x,y)-/(y)

as

x->$

(1)

through the set A uniformly for yeB. Suppose that £ > 0 is given. Then e/2 > 0. Hence there exists a S > 0 such that for each xeA and yeB 0 < dft, x) < d => d(l(yl /(x, y)) < e/2. Since the final inequality holds for all yeB we may conclude that, for any e > 0 there exists a d>0 such that, for each xeA, 0 < d& x)<S=> sup d(l(y\ f{x, y)) < e yeB

i.e. sup $

(2)

yeB

through the set A. We have shown that (1)=>(2; and it is even easier to show that (2)=>(1). This observe ion suggests the following definition. Suppose that F: B^Z G: B->£. Then we define the uniform distance between F and G by II(F, G) = supd(F(y), G(y)) yeB

where this quantity exists.

and

(3)

Limits and continuity (II)

139

23.13 Example Consider the function F: [0,1]->R defined by F(y) = $y2 + \ and the function G: [0, 1]->R defined by G(y) = 6_y. The uniform distance between F and G is given by u(F,G)= sup

= max \Sy2-6y + l\. The maximum of H(y) = \8y2 — 6y + 1| is either attained at y = 0 or y = 1 or else at a point 7/e(0, 1). In the latter case, differentiation yields that 16/7-6 = 0

Since tf(0) = l,

= 3 and if(|) = i, it follows that M(F,

G) = 3.

140

Limits and continuity (II)

In terms of the uniform distance notation, statement (2) assumes the less clumsy form, u(/,/(x,#)H0

as x ^

(4)

through the set A, where /(x,»): B-+Z is the function whose value at yeB is /(x, y).

B

If F and G are bounded on B then w(F, G) defined by (3) always exists and the set of all bounded functions on B is a metric space with u as metric. (See §13.18.) When dealing with bounded functions we may therefore rewrite (4) in the simple form: as

(5)

through the set A. It is often helpful to 'forget' that/(x,«) and / are functions and to think of them instead as 'points' in the metric space of bounded functions on B.

space of bounded functions on B

Limits and continuity (II)

141

It is important, however, not to forget that the distance between the 'points' /(x,») and / is defined by p(yJ(x,y)).

(6)

yeB

The above discussion is intended to indicate why uniform convergence is in some respects a more straightforward concept than pointwise convergence although it seems at first sight as though the opposite were the case. However, the discussion also has some practical benefits in that it is often easiest when seeking to establish uniform convergence to begin by calculating or estimating (6).

23.14

Example

Let ^ = R \ { 0 } and define g: AxU->U by

We begin by observing that, for each yeU, g(x, y)-+y as x->0 - i.e. g(x, y)->y as x-»0 pointwise for yeU. It follows that, if g(x, y)-*l(y) as x->0 uniformly for yeU, then it must be the case that l{y) = y. Consider u(l, g(x,m)) = sup \l(y)-g(x, y)\. yen

Taking l(y) = y, we obtain that, for each xeA,

\Ky)-g(x,y)\ =

y— |x|3

Hence, for each xeA, Ixl 3 yeU X

\y

Since |x|—>0 as x-^0, it follows that g(x, y)-*y as x-»0 uniformly for yeU. The diagram over illustrates the set of (x, y) for which it is true that I y — g(x, y)\ < £• Observe that the choice 5 = s ensures that o < |x| < d =>| y — g(x, y)\ < e for all yeU.

142

Limits and continuity (II)

23.15

Example Let A = R\{0} and define h: AxU-+U by h{x, y)=~2

xv

j•

xz + y2 We begin by observing that, for each yeU, h(x, y)-*0 as x-+0 - i.e. h(x, y)->0 as x->0 pointwise for yeU. It follows that, if h(x, y)—>l{y) as x->0 uniformly for yelR, then /(y) = 0. Observe that, for each xeA, 0-yen

(The geometric-arithmetic mean inequality gives \xy\i^\{x2 + y2) and equality is attained when x = y) Since i/(0, h(x9 •))-f>0 as x^>0, it follows that it is not true that h(x, y)^>0 as x->0 uniformly for yeU. The diagram below illustrates the set of (x, y) for which it is true that |/z(x, y)\ <s when 0 < £ < ^ . (See exercise 22.24(3).)

143

Limits and continuity (II)

-4s2)}'1

\h(x,y)\0 such that for all xeA and all

yen If y 7^0, the largest value of d>0 for which it is true that 0ri

(8)

through the set B. Then

/(x,y)-S as (x,yHfen) through the set S. Proof Let £ > 0 be given. By (7) there exists a ^ > 0 such that, for each xeA and each yeB,

By (8), there exists a S2 >0 such that for each y

144

Limits and continuity (II)

Choose ^ = min{^1? S2}. Then, if 0 < ^ 1 ( ^ , x)£ as x-»£ through the set A. Proof The theorem is simply a combination of theorems 23.9 and 23.16.

23.18

Example

Let # be defined as in example 23.14. We have that

(i) g(x, y)->y as x->0 uniformly for yeU. (ii) y-»0 as y-»0. (iii) #(x, y)->x as y^O for xelR \ { 0 } . From theorem 23.17 we may conclude that lim lim g(x, y) = \im lim g{x, y). y-*0 x->0

x->0 >;-+0

This is easily verified directly.

23.19*}"

Exercise

(1) Consider the function/: U2 \{(0, 0)}->[R defined by J \x-> y)—~~i

2'

Prove the following: (i) / ( x , y)^l as (x, y)->(0, 0) along the line y = Q. (ii) / ( x , y)-> — 1 as (x, y)-*(0, 0) along the line x = 0. (iii) / ( x , y)-*0 as (x, y)->(0, 0) along the line x = y.

Limits and continuity (II)

145

Deduce that/(x, y) does not tend to a limit when x->0 and y->0 independently. (2) Let / be as in question 1. If 0 < e < l , prove that |/(x, y)\<s if and only if 1-e 1+e


R given below, decide whether or not it is true that lim (lim f(x, y) I = lim ( lim/(x, y) I. 2

2

(i) f(x, y)= * ~[

2 4-

2

(ii) /(x, y) = 4 ^ T

l + x z + y2

x2 + y

(4) In each case considered in question 3, find a function /: IR ->R such that /(x, y)^/(y)

as x->0

pointwise for (5) In each case considered in question 3, decide whether or not it is true that /(x, y)-*l(y)

as x->0

uniformly for (a) yeU and (b) ye[ — l, 1]. (6) A function/: R2-»[R1 has the property that (x1^x2

and y i ^ ^ ) ^ / ^ ! . y i ) ^ / f e , y2)-

Prove that lim I lim /(x, y) 1= lim ( lim /(x, y) 1. £ \ } \Z J

23.20|

Uniform continuity

Let % and y, be metric spaces and suppose that/:S->^ where 5 is a set in %. To say that / is continuous on S is equivalent to the assertion that, for each / ( x ) - , / © as through the set S. This means, in turn, that

x ^

146

Limits and continuity (II)

In this statement the value of 3>0 which is asserted to exist may depend both on the value of s>0 and on the value of ^eS. If, given e>0, a value of S>0 can be found which works for all ^eS simultaneously, then we say that / is uniformly continuous on the set S. Thus / is uniformly continuous on the set S if and only if V £ >0 35>O VxeS

23.21

Example The function /: [0, oo)->[0, oo) defined by

is uniformly continuous on [0, oo). Proof It is evident from the diagram that \y/x — y/^\ is largest for x^O, £^0 and \x-£\^d when £ = 0 and x = S. (This is easily checked analytically by proving that Jb-^Ja^{b-a)112 when Ogagfr.)

=

It follows that, given any £>0, there exists a any x^O and any y^O,

23.22

y/x

(namely S = s2) such that, for

Example The function/: (0, oo)->(0, oo) defined by

is not uniformly continuous on (0, oo).

Limits and continuity (II)

147

y

Proof Given e > 0, the largest value of d > 0 for which

satisfies

1

1

-S

I

1

1

x

t

= e.

Hence Since this expression tends to zero as £—>0 + it follows that no d>0 can work for all

23.23| Theorem Let X and y be metric spaces and suppose that / : K^y, where K is a compact set in X. If / is continuous on K, then / is uniformly continuous on K. Proof Since/is continuous on K,/(x)-»/(y) as x->y through K for each yeK. Let e > 0 be given. Then ^e>0 and so, for each yeK, there exists a (5(y)>0 such that, for each xeK, d(y9 x)«5(y) => d(f(y),

(1)

Let 1L denote the collection of all open balls with centre yeK and radius ^S(y). Then U covers K. From the definition of a compact set (§20.4) it follows that a finite subcollection § of It covers K. Let d>0 be the radius of the open ball of minimum radius in the subcollection $. Now suppose that x and £, are any points of K which satisfy d(£, x) (provided x ^ ± 00). x/ - 00 = 0 j Perhaps the most significant feature of these definitions are those items which are omitted. Observe that (+00) + ( — 00), ( + 00) —(+00), 0(+oo),

Points at infinity

157

0(— oo), +00/ +00 and other expressions are not defined. There is no sensible way to attach a meaning to these collections of symbols and we do not attempt to do so. This fact means that there is little point in attempting algebraic manipulations involving + 00 and — 00. Definitions (i)-(iv) should therefore be regarded as handy conventions rather than as a basis for a serious mathematical theory. Note also that division by zero remains unacceptable. As an example of the use to which these conventions may be put, consider two non-empty sets 5 and T of real numbers and two non-empty sets P and Q of positive real numbers. We then have that (i) sup ( — x)= —I inf x xeS

(ii)

V xeS

sup (x + y) = ( sup x l + l sup y (x,y)eSxT

(iii)

\ xeS

)

\ yeT

sup (xy) = I sup x 11 sup y {x,y)ePxQ

\xeP

) \

yeQ

without any need for assumptions about the boundedness of the sets concerned provided that the suprema and infima are allowed the values + 00 or —00 (See exercise 9.10(9).) Note, however, that some vigilance is necessary if one or more of the sets is empty. For example, if S — 0 and T is unbounded above, the right-hand side of (ii) becomes meaningless. The system [—00, +00] with the structure described in this section is called the extended real number system. It should be pointed out that some authors use IR * (which we have used for the one-point compactification) to stand for [— 00, + 00]. Since other authors use U * for yet other purposes, some element of confusion is inevitable in any case.

24.5

Convergence and divergence

Let S be a set of real numbers. Usually it would be convenient to regard S as a set in the space IR but, in what fallows, we shall wish to regard S as a set in the space [—00, +00]. Next consider a function / : S-*[— 00, +00] for which f(S)aM. Again, one would normally regard / as taking values in IR but here it is convenient to regard / as taking values in [— 00, + 00].

158

Points at infinity •

+00

+00

If ^e[— 00, + 00] and ne\_— 00, + 00], it makes sense to ask whether or not/(x)->77 as x->£ through the set S. The definition is the same as always. (See §22.3.) For any open set G in [—00, +00] containing rj, there must exist an open set H in [—00, +00] containing £ such that xeHnS =>f(x)eG provided that x # £ . As we found in §22.4, it is more helpful for technical purposes to rephrase this definition in terms of 'open balls' in [—00, +00]. We obtain that f(x)-+n as x-+£ through 5 if and only if: For each 'open ball' E in [— 00, + 00] with centre n, there exists an 'open ball' A i n [ — 0 0 , +00] with centre £ such that xeAnS =>f(x)eE provided x ^ £ . If £ and n are real numbers, this definition is exactly the same as the familiar definition of §22.4. But if £ or n are + 00 or — 00 then something new is obtained. If we insert the relevant definition for an 'open ball' given in §24.4 and bear in mind that/takes only real values, we obtain the criteria listed in the table below. In this table / and a are to be understood as real numbers. There is little point in committing these criteria to memory since it is easy to work out in any particular case what the form of the definition must be. There is some point, however, in acquiring a feeling for the intuitive meaning of the different statements. Consider, for example, the statement '/(x)-W as x-> + 00'. One can think of f(x) as an approximation to /. Then |/(x) —/| is the error involved in using

159

Points at infinity

f(x) as an approximation for /. The statement '/(x)-W as x-> + oo' then tells us that this error can be made as small as we choose (i.e. less than e) by making x sufficiently large (i.e. greater than X). Similarly, '/(x)-» —oo as x-> —oo' means that we can make/(x) as negative as we choose (i.e. less than Y) by making x sufficiently negative (i.e. less than X). /(x)->/ as n through S

V £ >0 3^>0 VxeS, 0 1./

/(x)-W as x-i through S

V£>0 3X VxeS,

/(x)->/ as x-» — oo through S /(x)-» + oo as x->a through S

VX 3^>0 VxeS,

/(x)—• — oo as x->a through S

VY 3(S>0 VxeS,

/(x)-» + oo as x-> 4- oo through S

VX 3YVxeS, x VX

/(x)-* + ooas x ^ — oo through S /(x)-> — oo as x-»- + oo through S

VY3X VxeS,

f(x)-> — oo as x-> — oo through S

VY3X VxeS, >/(x)x

24.6 Example Consider the /(x) = l/x. We have that (i) l/x->0 as x ^ + oo (ii) 1/x-^ + oo as x->0 +

=*/(x) — oo as x-+0 — (iv) l/x->0 a s x - ^ - oo. To prove (i) we have to show that, for each e > 0, there exists an X such that x>X

0 such that x Any value of S>0 is adequate for this purpose when Y^0. When Y/ as x->0+ and no real number m for which f(x)-+m as x->0 —. If we restrict our background space to be the real number system R, it is therefore untrue to say that f(x) converges asx->0+ or x->0 —. We next consider a set S in Un# and a function/: S->Rm#. (Recall that U is the one-point compactification of Un.) We are interested in the case when SczM" and f(S)cRm. If ^eUn# and r\eMm#, we have that/(x)->Ti as x->£ through S if and only if: For each 'open ball' E in R m# with centre TJ, there exists an 'open ball' A in R" # with centre \ such that n#

xeAnS=>/(x)e£ provided x ^ ^ .

Points at infinity

161

If we insert the relevant definition for an 'open ball' given in §24.2 and bear in mind that/takes values only in Um, we obtain the criteria listed in the table below. In this table, I and a are to be understood as vectors in Um and Un respectively. /(x)->lasx->a through S /(x)-»lasx->oo through 5 /(x)->oo as x->a through S /(x)->oo as x->oo through S

24.7

Ve>0 3S>0 VxeS 0X=>||/(x)-l||<e VX 3O VxeS 0X VX 37 VxeS ||x||>y=>||/(x)||>X

Example Consider the function/: [R2\{(0, 0)}->[R2 defined by /

x

—y

We have that as

as

Proof To prove (i), we have to show that, for any e > 0, there exists an X such that

x 1 The choice X = 1/e clearly suffices. To prove (ii) we have to show that, for any X, there exists a 5>0 such that If X ^ 0 any value of d > 0 is adequate. Otherwise we require a 8 > 0 such that 2 0<x 2 +y 2 / 2>X x +y

162

Points at infinity

and the choice 6 = 1/X clearly suffices. The diagram illustrates (i). The function value /(x, y) lies in the disc with centre (0, 0) and radius s>0 provided (x, y) lies outside the disc with centre (0, 0) and radius X.

This example, incidentally, is easier if complex notation is used. Since 1 x + iy

1 (x — iy) x — iy (x + iy) (x — iy) x2 + y2'

we can identify the given function with the function/: C\{0}-»C defined b y / ( z ) = l / z . To prove (i) we then have to show that, for any e>0, there exists an X such that

When working with U1 there is sometimes room for confusion over whether the one-point or two-point compactification is in use especially since it is not uncommon for authors to write oo where we have been writing +oo. Indeed, in earlier chapters, we have been using the notation (a, co) = {x: x>0} rather than (a, + oo) since the latter notation seems painfully pedantic. In those exceptional cases when one wishes to make use of the one-point compactification of U it is therefore as well to employ notation of the type as |x| >oo X) as or in order to avoid any possible confusion. Similar notation is also sometimes useful when working in R" in order to emphasise the use of the onepoint compactification.

Points at infinity 24.8

163

Combination theorems

In the previous section we have explained how some problems concerning the divergence of functions / : R"->[Rm can be replaced by convergence problems by replacing U by [—00, +00] and Un by Un#. A natural next question is to ask to what extent the combination theorems of §22.25 carry over to the new situation. There is no problem at all when [— 00, + 00] or R" # replaces the metric space X in these theorems. Thus, for example, theorem 22.30 shows that, if /j(*)->/! as x-> — 00 and/ 2 (x)-»/ 2 as x-> — 00, then as

x-> — 00.

However, there are a number of cases not covered by the theorems of §22.25 The most important of these cases is the subject of the next result.

24.9 Proposition Let X be a metric (or a topological) space and let j \ : S-*[— 00, +00] and/ 2 : S-*[—00, +00], where S is a set in X. Let £e X and suppose that Mx^rji x

and

/2( )-^2

as

x->£

as

x

~ȣ

through the set S. Then, if a and b are real numbers (i) a (ii) (iii) as x ^ ^ through the set S, in each case for which the right-hand side is a meaningful expression in the sense of §24.4.

It is sometimes useful in addition to note that, if 00 for xeS and n2 = 0, then /i(x)// 2 (xH + oo

as

x->^

through S. Similarly, if — 00 ^nx < 0 , / 2 ( x ) > 0 for xeS and n2 = 0, then /2W// 2 (x)-^-co

as

x-ȣ

through S. It is dangerous to assume the truth of results other than those explicitly mentioned above. For example, it is tempting to assume that, if/^x)-* + 00 as x->0+ and /2(x)--> + oo as x->0 + , then fi(x)/f2(x)-+l as x->0 + . However, as the examples below indicate, this result is false in general.

(i)f1(x) = x2, f2(x) = x

164

Points at infinity

(ii)/ 1 (x) = 2x, f2(x) = x (iii)/ 1 (x) = x, / 2 (x) = x 2 .

24.10 Exercise (1) Prove the following results: (i) sup (2x + 3) = + oo (ii) sup (2x + 3) = — oo x 2 0

(iii)

inf

(xy) = — oo (iv)

x>0,y0,y>0

(2) L e t / : U \{1}-+R be denned by

Prove the following results: (i) /(x)->l (ii) /(x)->l (iii) /(x)->l (iv) /(x)-> + oo (v)/(x)-> —oo (vi) |/(x)|->oo (vii) x/(x)-> + oo (viii) x/(x)-> — oo (ix) |x/(x)|-»oo

as x-> + oo as x - > - o o as |x|->oo as x-»l + asx-»l — as x->l as x-> + oo as x-> — oo as |x|-^oo.

(3) (i) Let / : R2->R be defined by / ( x , y)=(l + x 2 + ^ 2 ) " 1 . Prove that /(x, y ) ^ 0 as (x, y)^co. Consider also the function g: U2^U defined by g(x, y) = (l+(x — y)2)~1. Prove that g(x, y)-+l as (x, .y)-^oo along the line x = y. Does g(x, y) approach a limit as (x, y)->oo? (ii) The function h: (0, oo)->R2 given by h(t) = (l/t, t) defines a curve in U2. Sketch this curve. Prove that (a) h(t)^oo as £-•() + (b) h(t)-+co as r-> + oo. (4) A function f:M-+M has the property that f{x)->l as x-> + oo and f(x)^m as x-* — oo where / and m are real numbers. Define g: [^-oo, +oo]->[R by

(f(x) g(x)=\ I [ m

(xeU) (x=+oo) ( x = — oo).

I f / i s continuous on M, explain why g is continuous on [— oo, oo]. Deduce the following results: (i) The set /(R)u{/, m} is compact, (ii) The function / is bounded on U. (iii) If there exists anxeIR such that /(x) ^ / and /(x) ^ m,

Points at infinity

165

then / achieves a maximum on U. t(iv) The function / i s uniformly continuous on U (§23.20). (5) The metric space R # with metric c: U*-*M was introduced in §24.2. Explain why a set G in U * is open if and only if, for each £eG, there exists an 'open ball' B with centre £ such that BaG. Here 'open ball' has the meaning assigned in §24.2. (6) Prove proposition 24.9 Suppose that /i(x)-> + oo as x-» — oo and / 2 (x)-^ + oo as x-+ —oo. Give examples to show that neither of the following statements need be true. (i)/i(x)-/2(x)->0 (ii) / i (x)//2(x) -»1

24.11|

as

as

*->-oo x - • - oo.

Complex functions

We have studied the one-point compactification of U 2 and observed that it should always be assumed that this compactification is in use when U2 is identified with the system C of complex numbers. Since C has a rich algebraic structure (i.e. C is a field), it is natural to seek to extend some of this structure to C * just as we extended some of the algebraic structure of U to [— oo, + oo]. There is no point, of course, in seeking to extend the order structure of C since C has no order structure. (See §10.20.) In this context, we make the following definitions: (i) (ii) x • oo = oo • x = oo (iii) x/oo=0 (iv) x/0=oo

(x^O) (x^oo) (x#0).

With these definitions proposition 24.9 remains valid when C * replaces [ — oo, +oo]. Note that oo4-oo, 0*oo, oo/oo and 0/0 are not defined. In particular, we do not define 00 + 00 = 00. For example, z->oo as z—»oo and — z—•oo as z—•oo but z + ( — z)-^0 as z-*oo. Even more than in §24.4, it is necessary to warn against attempting algebraic manipulations on the basis of definitions (i)—(iv). Item (iv), for example, does not represent a genuine 'division by zero' but merely indicates a handy convention.

24.12|

Product spaces

The one-point compactification of R" is not the only possible compactification of Un. An obvious and important alternative is to regard Un as sitting inside the space [—00, +oo] n . In the case of IR2 this amounts to thinking of the plane as squeezed into the interior of a square and regarding the edges of the square as the 'points at infinity'.

166

Points at infinity oo, + oo)


oo

171

Sequences 25.4

Example Consider the sequence ^cos k sin k

of points in U2. We have that cos k

as fc-+oo

1 k

sin k 0 k

and

as fc->oo.

Hence, by theorem 25.3, xfc-•((), 0) as

^^"^ X

r X

\ \ \

/

\

I \ \

\ X

c

5

"\

N

\ i /

y /

3

25.5 Theorem Suppose that <xfe> and are sequences of points in Un and that and

xfc->^ as /c->oo yk—>i| as fc->oo.

Then, for any real numbers a and b, as /c-^oo. Proo/ Theorem 22.30.

25.6 Theorem Suppose that and that

and are sequences of real numbers as k-+oo

and Then;

as

/c->oo.

172

Sequences

(i) xkyk-+£rj as /c-»oo and, provided that x T| as x->£ through S; (2) For each sequence <xfc> of points o

are equivalent. Proof (i) (1)=>(2). Suppose that -(1) holds. Then, given any e>0, there exists a £ > 0 such that for any xeS \{^} (3) Now suppose that <xk> is a sequence of points of S \ { ^ } such that

>^ as

k-KX>. Then there exists a K such that (4)

Sequences

173

Combining (3) and (4), we obtain that and hence /(x k )-n| as k-+co. (ii) not (1) => not (2) (see §2.10). Suppose that it is false that f(x)-+x\ as x->^ through S. If it is false that Ve>0 then it is true that 3 £ >0 V(5>0 3xeS\{£}, (d($, x) 0 and all S>0. Given 8 = l//c (fc = 1, 2, 3,...), we can therefore find x k eS\{^} such that d& x k )ooand, from the second, it follows that/(xk)-Ar| as k-*co. 25.9 Corollary Let % and y, be metric spaces and suppose that/: S->^ where 5 is a set in %. Then / is continuous on S if and only if, for each sequence <xk) of points of S which converges to a point of S,

limf(xk)=fUmx\ 25.10 Example In §9.2 we explained how Archimedes trapped the area of a quarter-circle of radius 1 between two sequences {An} and of rational numbers. The terms of these sequences satisfy A1=^, B1 = l and (ii)

B;+\=2{A;+\+B;1}.

Using the fact that the geometric and harmonic mean of two real numbers lies between these numbers, it is trivial to prove that From this it follows that and ^AnSAn+t^Bn+1g>BnR defined by f(x) = y/x is continuous on [0, oo). We

174

Sequences

have that AnBn-+AB as rc->oo and hence, by corollary 25.9, as

Since An + l=^/(AnBn), is obtained again.

25.11

it follows that A = J(AB) and hence the result A = B

Exercise

(1) Determine whether or not the sequence <xfc> converges in U2 in each of the following cases. 'fc + 1 fc + 2\

(iii) xfc = f-, k) \k )

_

/

1

(iv) xk = (cos /c, sin k)>

(2) Write down definitions for the statements 'xfc-» + oo as /c-*oo' and '**-* — oo as /c->oo' in the case when <xfc> is a sequence of real numbers. Show that /c2-> + oo as /c->oo and that — ^Jk-+ — oo as /c->oo. Discuss the sequences {{ — If} and . (3) Let <xk> be a sequence of points in 1R". Write down a definition for the statement cxk—XX) as fe->oo' based on the one-point compactification of Un. For which of the sequences of question 1 is it true that xfc->oo a? /c->oo? (4) A sequence <xk> of real numbers is said to increase if and only if x x k^ k+i f° r e a c n keM. Prove that an increasing sequence <xk> of real numbers is unbounded above if and only if x k -* + oo as /c->oo. Prove that an increasing sequence of <xk> of real numbers is bounded above if and only if xfc->x as /c->oo and that the limit x is the supremum of the sequence. What are the corresponding results for decreasing sequences? (5) A sequence <xk> is defined by x1 = 1 and xk + 1 =f{xk) where/:[0, oo)->lR is defined by J \x) —

TTT •

Prove that 0<xk+1^xk (fc = l, 2, ...) and explain why it follows that <xk> converges. If xk->x as /c-^oo, justify the conclusion that x=f(x) and hence show that

Sequences

175

(6) Prove that the sequence <xk> of rational numbers defined by Xj = 2 and (

2

decreases and is bounded below. Deduce that <xfc> converges and prove that the limit is yjl.

25.12

Sequences and closure We begin with the following lemma.

25.13 Lemma Let S be a non-empty set in a metric space % and let ^G Z. Then d(^, S) = 0 if and only if there exists a sequence {x^) of points of S such that xfc->£ as /c-+oo.

Proof (i) Suppose that d(^ S) = 0. By theorem 13.22, given any e>0, there exists an xeS such that d{£,, x)0, it follows that there exists an xkeS such that d(£, x k ) of points of S such that x f e -^ as /C-KX). Then, given any a>0, there exists a K such that k>K=> dfe, xk)<s. In particular, dfe, xK+1)<s. It follows from theorem 13.22 that

25.14 Theorem Let 5 be a set in a metric space X and let ^ e £ . Then ^eS if and only if there exists a sequence <xfc) of points of S such that xfc—•I; as fc-»oo.

176

Sequences

ProofBy theorem 15.5, ^eSd(^9 S) = 0 and so the theorem follows immediately from lemma 25.13.

25.15 Corollary A set S in a metric space X is closed if and only if each convergent sequence of points of 5 converges to a point of S. Proof By exercise 15.3(1), S is closed if and only if S = S.

25.16"}* Note It is of some importance to take note of the fact that theorem 25.14 and corollary 25.15 are false for a general topological space X. Analogues of these results are valid but the notion of a sequence must be replaced by the more general notion of a net.

25.17

Exercise

(1) If <xk> is a sequence of points in Rn such that xfc->£ as /c-*oo, explain why (i) <xfc, u>->

(ii) llxJMISH as

as

/C->GO

fc-oo.

[Hint: Use corollary 25.9.] Deduce from corollary 25.15 that the sets S = {x: <x, u>^c} and T={x: ||x||^r} are closed. (2) Let A and B be two non-empty sets in a metric space X. Prove that A and B are contiguous if and only if a sequence of points in one of the sets converges to a point of the other. Use this result and corollary 25.9 to prove that, if 5 is a connected set in X and /: S-+y, is continuous on S, then f(S) is connected. (See theorem 17.9.) (3) Let S be a set in a metric space X and let ^e X. Prove that £ is a cluster point of S if and only if there exists a sequence <xk> of distinct points of <S such that xk-+Z> as /c->oo. (Distinct means that x} = xk j = /c.)

25.18

Subsequences

Suppose that g: N-*N is strictly increasing. Then the sequence fOg:N -+y. is said to be a subsequence of the sequence/: N -+y,. If we think of the subsequence illustrated in the diagram as a list, then its first few terms are

177

Sequences 1

1

2

2

3

3

4

4

Z7^ —

5

V

6 7

Note that this is obtained from the original sequence

by omitting some of the terms. If (yky is a sequence and is a strictly increasing sequence of natural numbers, then

denotes a subsequence of . (Iff(k) = y a n d g(l) = kh then For example, if = and = , then (yki} is . This subsequence is obtained from the sequence by deleting terms as indicated below: /, 5, jtf, 17, ^ , 37, jtf, 65, ^ , 101, ...

25.19 Theorem Let be a sequence of points in a metric space y, with the property that y k - m as k-+oo. If is any subsequence of , then as

yjt - ^

/->oo.

Proof Let / : f^J->^ be defined by f(k) = yk and let gf: M-^l^l be defined by g{l) = kt. We have that /(/c)-ni as /c-> + oo through N and that 6f(/)^ + oo as /-> + oo through l\l (see exercise 25.22(1)). It follows from exercise 22.33(3ii) that ii

as

/-^ + oo

through N - i.e. yk -*r\ as /->oo.

25.20

Example We have that l/k->0 as k-*oo. From theorem 25.19 it

178

Sequences

follows without further calculation that 1/2*^0

as /

because is a subsequence of . 25.21 Example Consider the sequence of real numbers. If k ( - l ) ^ / a s fc->oo, then (-1) 2 *-Was /c-^oo and ( - l ) 2 k + 1-W as /c->oo. But (-l) 2 / c = l and ( - l ) 2 k + 1 = - l . Hence / = 1 = - 1 . From this contradiction we deduce that diverges.

25.22

Exercise

(1) Suppose that is a strictly increasing sequence of natural numbers. Prove by induction that kt ^ / for all le N and deduce that fcj-> + oo

as

/->oo.

(2) Show that, for any mef^J, the sequence of real numbers

is increasing and bounded above. Deduce that the sequence converges to a real number (m). By considering an appropriate subsequence, prove that

(3) Let be a sequence of points in a metric space y.. If r\ is a cluster point of £ = {yk: /CGN}, prove that T] as h-> oo, prove that £U{TJ} is closed. If has no convergent subsequences, prove that E is closed. (5) Let (xky be a sequence of real numbers. If each set En = {xk: k> n] has a maximum, prove that <xk> has a decreasing subsequence. If at least one of the sets En has no maximum, prove that <xk> has an increasing subsequence. Deduce that any bounded sequence of real numbers has a convergent subsequence. [Hint: Use exercise 25.11(5).] (6) Let {(xk,yk)} be a bounded sequence of points in U2. Prove that has a convergent subsequence. [Hint. Begin with a convergent subsequence <xk > of <xk> and consider

Sequences 25.23

179

Sequences and compactness

The Bolzano-Weierstrass theorem (19.6) was central to our discussion of compact sets in chapter 19. We therefore begin by giving a version of the Bolzano-Weierstrass theorem for sequences.

25.24 Theorem {Bolzano-Weierstrass theorem) Any bounded sequence <xfc) of points in Un has a convergent subsequence. Proof This may be deduced from theorem 19.6 as follows. If E = {xk: he N} is infinite, then theorem 19.6 asserts that E has a cluster point £. This cluster point must be the limit of a subsequence of <xfe> (exercise 25.22(3)). If E is finite, then <xk> has a subsequence all of whose terms are equal. This subsequence therefore converges. Alternatively, the theorem may be proved directly as indicated in exercise 25.22(5) and (6).

In chapters 19 and 20, we considered a number of different definitions of compactness and the above version of the Bolzano-Weierstrass theorem suggests yet another. We say that a set K in a metric space % is sequentially compact if and only if every sequence of points in K has a subsequence which converges to a point of K.

25.25 Theorem A set X in a metric space X is compact if and only if each sequence <xk> of points of K has a subsequence which converges to a point of K - i.e. in a metric space, compactness and sequential compactness are the same. Proof We use the definition of compactness given in §19.5 - i.e. a set K in a metric space % is compact if and only if each infinite subset E of S has a cluster point ^eS. (i) Let K be compact and let <xfc> be a sequence of points of K. If E = [\k: he N} is finite, then <xk> has a subsequence <xk ) all of whose terms are equal to one of the terms of <xfc>. Hence <xkj>" converges to a point of K. If E is infinite, then E has a cluster point ^eX. By exercise 25.22(3), <xfc> has a subsequence which converges to £,. (ii) Let K be sequentially compact and let E be an infinite subset of K. Since E is infinite, there exists a sequence <xk> of distinct points of E. Because K is sequentially compact, <xk> has a convergent subsequence whose limit % lies in K. But, by exercise 25.17(3), ^ is a cluster point of E.

180

Sequences

25.26| Note It is important to bear in mind that sequential compactness is not the same as compactness in a general topological space. It is true that a countably compact set (§20.7) in a topological space is sequentially compact but even the converse of this assertion is false without some subsidiary assumptions.

25.27 Example Suppose that X and % are metric spaces and that S c l If/: S-+y, is continuous on the set S, then S compact =>f(S) compact. We have already seen two proofs of this very important theorem. (See theorems 19.13 and 21.17.) A third proof may be based on theorem 25.25. Let be any sequence of points in/(S). We seek to show that has a subsequence which converges to a point of f(S). Write yk =/(x k ). Since S is compact, <x k ) has a subsequence which converges to a point §eS. Suppose xk/-+£ as /->oo. By corollary 25.9, /(xki)->/(!;) as /->oo. It follows that the subsequence converges to the point/(£)e/(S).

25.28

Exercise

(1) Let be a sequence of points in a metric space % and let E = {yk: keN}. If yk->il, prove that Eu{r\] is compact. t(2) Let % and y. be metric spaces and let/: K->y, be continuous on the compact set K in %. Use theorem 25.25 to show that/ is uniformly continuous on K (see §23.20). [Hint: The contradictory of the statement that / is uniformly continuous on K begins with ' 3s > 0 V S > 0\ Use this assertion with d = l//c for each ke N.] t(3) Given an example of a bounded sequence in Z00 (see §20.20) which has no convergent subsequence.

26

26.1

OSCILLATION

Divergence

Suppose that % and y, are metric spaces and t h a t / : S-+y. where S is a set in %. Let tje^/ and let ^ be a cluster point of S.

g3 oscillates 'finitely' as x->-0+

181

I g4 oscillates 'infinitely' as x->0 +

182

Oscillation

If/(x)->i| as x->£, through the set S, we say that the function converges as x approaches £, through S. If the function does not converge as x approaches ^ through S, then it is said to diverge. Divergent functions can exhibit a variety of different behaviour as the diagrams on p. 181 illustrate. We saw in chapter 24 that the behaviour of the functions g1 and g2 can be treated by the same machinery as one uses for convergent functions. Although these functions do not converge as x approaches 0 from the right when regarded, as mappings to the space U, they do converge when regarded as mappings to the space [— oo, + oo]. The functions g3 and g4 are more interesting. These are examples of oscillating functions. To discuss these, we require the notion of a limit point.

26.2

Limit points

Suppose that X and y, are metric spaces and t h a t / : S^y, where S is a set in X. Let x\e% and let £ be a cluster point of S. Suppose it is true that, for some as through the set T where T is a subset of S for which ^ is a cluster point. Then we say that A is a limit point o f / a s x approaches £ through S.

Observe that there is a distinction between a 'limit' and a 'limit point'. A limit exists only when /(x) converges as x approaches £ through S. But limit points may exist even though /(x) diverges as x approaches \ through S.

Oscillation 26.3

183

Examples (i) Consider the function/: U->U defined by 1

(x^O)

This function diverges as x approaches 0 but /(x)-»l as x-»0 + and /(x)-» — 1 as x-»0 —. Hence the function has two limit points as x approaches 0. These are 1 and — 1. (ii) Consider the function/: [ R 2 - ^ 1 of example 22.11. We have seen that as

along the line y = ax. Examining each aeU in turn, we find that every point of [0, 1) is a limit point of /(x, y) as (x, y) approaches (0, 0). Also Ax,y)->1

as

(XJH0,0)

along the line x = 0. Hence 1 is also a limit point. Since — l:g/(x, y ) ^ l for all (x,'y)eU2, the function can have no other limit points. It follows that the set of limit points of/(x, y) as (x, y)->(0, 0) is equal to [—1, 1].

26.4 Exercise (1) Suppose that % and y are metric spaces and t h a t / : S^y where 5 is a set in % for which ^ is a cluster point. Prove that 'key, is a limit point o f / a s x approaches ^ through the set S if and only if there exists a sequence <xfc) of points of S such that x f c 7^ (/c = 1, 2,...) and x k -^^ as /c->oo for which /(x k )-»i,

as

/C-KX).

[Hint: Use theorem 25.8]. (2) Find all real limit points of the following sequences of real numbers: (i) (iv)

(ii) (V)

(iii) (Vi)

These sequences may also be regarded as taking values in [— oo, + oo]. Find all limit points from [— oo, +oo] of the given sequences. (3) Find all real limit points as x->0+ / : (0, oo)-*R:

of the following functions

(i) fix) = sin — x

(ii) fix) = x sin — x

(iii) /(x) = - sin -

(iv) /(x) = 1 - sin -

XX

X

184

Oscillation

(v) f(x) =

sin X

X

(vi) /(x) = - j — sin - . X

X

X

X

These functions may also be regarded as taking values in [ — oo, -f oo]. Find all limit points from [—00, + 00] of the given functions as x->0 + . (4) Find all real limit points as (x, y) approaches (0, 0) of the function / of exercise 22.24(3). (5) Give examples of functions/: (0, oo)->[R for which the set L of real limit points as x->0 + is as indicated below: (i) L = 0 (iv) L = [ 0 , 00) (vii) L = {0} but

(ii) L = U (iii) L = [ 0 , 1] ( v ) L = { - l , 1} (vi) L=N f(x)±0 as x->0 + .

Show that in the case of each of your examples except (iii) and (v), + 00 or — 00 is also a limit point. (6) Suppose t h a t / : (0, oo)->IR is continuous on (0, 00). Let L be the set of real limit points as x->0 + . Prove that L is connected.

26.5f

Oscillating functions

26.61 Theorem Suppose that % and y, are metric spaces and let/: S-*y. where S is a set in X for which \ is a cluster point. Then the set L of limit points of/ as x approaches \ through S is given by L = O f(SnAk) where the set Ak is obtained by removing £ from the open ball with centre £ and radius l//c.

Proof (i) Suppose that AeL. Then there exists a sequence <xfc> of points of 5 \{£} satisfying x k ->^ as /c->oo such that/(x k )-*A as /c->oo (exercise 26.4(1)). Given any J, there exists a X such that xkeAjnS => f(xk)ef(AjnS). Hence Xef(AjnS). It follows that f(AknS). (ii) Suppose that Aef(AknS) for each /ce^J. Then for each fceN, there exists a sequence <xM> of points of AknS such that f{xkj)->'k as /->oo. For each keN, choose lk so that d(X,f(xktlJ£ as /c->oo and/(xMfc)->X, as fc->oo.

185

Oscillation Thus XeL and so f(AknS)czL.

26.7| Theorem Let ft and ^ be metric spaces and suppose that/: S^y. where 5 is a set in % for which £ is a cluster point. Then the set L of limit points of/ as x approaches £ through S is closed. Proof By the previous theorem, L is the intersection of a collection of closed sets. Thus L is closed by theorem 14.14. 26.8| Theorem In addition to the assumptions of the previous theorem, suppose that/(S) lies in a compact subset K of y. Then the set L of limit points of/ as x approaches \ through S is non-empty. Proof The sets SnAk are non-empty for each keN because £ is a cluster point of S. It follows that/(SnAfc) is non-empty for each keN. Thus (f(SnAk)} is a nested sequence of non-empty, closed subsets of a compact set K. It follows from the Cantor intersection theorem (19.9) that the sequence has a non-empty intersection i.e. 26.9f

Theorem With the assumptions of the previous theorem, /(x)-*t| as

through S if and only if L= {r\}:

x-^

186

Oscillation

Proof If /(X)-»TI as x->£ through S, then it is trivial to prove that L={TI}. If it is false that / ( x ) - n j as x->E, through S, then there exists a sequence <xfc> of points of S \ { £ } such that xk->£ as /c->oo but/(x k ) -/m as /c->oo. We deduce the existence of an so>0 and a subsequence <xki> for which d(r\, f(\k())^e0 But has a convergent subsequence by theorem 25.25. Hence L l )

The assumption that f(S) be a subset of a compact set Kay which is made in theorems 26.8 and 26.9 is not so restrictive as it may seem at first sight. For example, if y. = U, one can always begin by replacing U by the compact space [— oo, + oo].

26.10 that

Example

Consider the sequence oo --oo

as

k-+ao.

The sequence therefore has only one real limit point namely 0. However, one cannot conclude from theorem 26.9 that the sequence converges because the range of the sequence does not lie in a compact subset of U. Note that, if we regard the sequence as taking values in the compact space [— oo, +oo], we find that the set L of limit points contains two elements namely 0 and — oo. Thus again theorem 26.9 does not apply.

Suppose that X is a metric space and t h a t / : S->R where S is a set in X for which £ is a cluster point. Let L denote the set of limit points from [— oo, +oo] as x approaches £ through S. Since [— oo, + oo] is compact, we know from theorem 26.8 that L^0. If L consists of a single point, then either/converges as x approaches § through S or else /diverges to + oo or diverges to - oo as x approaches. E, through S. If L consists of more than one point, we say that / oscillates as x approaches % through S. If/oscillates as x approaches £ through S and L is a compact subset of R, it is customary to say that / 'oscillates finitely'. (Note that, since L is closed, it follows that L is a compact subset of U if and only if L is a subset of R which is bounded in U.) Otherwise/is said to 'oscillate infinitely'.

26.1 It

Lim sup and lim inf

Let X be a metric space and suppose t h a t / : S-+M where S is a set in X for which £ is a cluster point. The set L of limit points from [ — oo, + o o ] a s x approaches % through S is nonempty and closed by theorems 26.7 and 26.8. It follows that L has a maximum element A and a minimum element X (corollary 14.11). We call A the limit superior (or lim sup) o f / a s x approaches % through S. We call k the limit inferior (or lim inf). Sometimes A is referred to as the upper limit and X as the lower limit.

Oscillation

187

The function g3 illustrated in §26.1 is an example of a function for which both X and A are finite (i.e. X and A are elements of U). The function g 4 of §26.1 is an example of a function for which X is finite but A = + oo.

26.12| Theorem Let % be a metric space and suppose t h a t / : S->R where S is a set in % for which ^ is a cluster point. Then the function / oscillates as x approaches £, through S if and only if X # A. Moreover, given any ^G[—oo, +oo], f(x)-+rj

as

x-+£

through iS if and only if A = rj = A. Proof This is an immediate consequence of theorem 26.9.

26.13| Proposition With the assumptions of the preceding theorem, A is the limit superior o f / a s x approaches \ through S if and only if (i) V/>A 3(5>OVxeS, o/(x)[— oo, + oo] defined by F(d)= sup /(x) increases on (0, oo). This guarantees the existence of the limit as (5->0 + . The identification of the limit with A is then achieved with the help of proposition 26.13. 26.15| Exercise (1) Determine the lim sup and lim inf for each of the examples given in exercises 26.4(2), (3) and (4). (2) Suppose that is any sequence of real numbers. Prove that Ae[— oo, + oo] is the lim sup of the sequence if and only if (i) For any subsequence of , ak -+rj as /—• oo => rj ^ A and

(ii) There exists a subsequence of such that ak —>A as /-•oo.

What is the corresponding result for the lim inf of a sequence? (3) Suppose that is any sequence of real numbers. Prove that A e [ — oo, + oo] is the lim sup of the sequence if and only if and

(i) VL>A, {k:ak^L) is finite (ii) V / < A, {k: ak ^ /} is infinite.

What is the corresponding result for the lim inf of a sequence? (4) Suppose that (ak} and (bk} are any sequences of real numbers. Prove that lim sup (ak + bk) ^ (lim sup ak I + (lim sup bk I fc->oo

\

fc-oo

/

\

/e-+oo

/

whenever the right-hand side makes sense. Give an example for which the lefthand side is — oo and the right-hand side is H-oo. Show that the two sides are equal whenever is a convergent sequence of real numbers. What are the corresponding results for the lim infs of the sequences? (5) Suppose that {ak} and {bk} are any bounded sequences of real numbers. Prove that lim inf (ak + bk) ^ (lim inf ak I + ( lim sup bk j ^ lim sup (ak + bk). k~*oo

\

/c-»oo

/

\

k-+ao

J

k-*oo

Show also that lim sup (— ak) = — ( lim inf ak I. Deduce that, for any bounded sequence of real numbers, lim inf (ck — c f e + 1 )^0^1im sup (ck — /c-+oo

k-+oo

Oscillation

189

(6) Suppose that is any sequence of positive real numbers. If there exists a K such that ak + 1^rak for any /c>K, prove that there exists an H such that ak < Hrk for all keN. Deduce that lim sup allk-^\\m sup ——.

27

27.1

COMPLETENESS

Cauchy sequences

A Cauchy sequence <xfc> of points in a metric space % is a sequence with the property that, for any e > 0, there exists an open ball E of radius e and a K such that k>K

x

>xkeE.

i X

2

The definition of a convergent sequence asserts that the terms of the sequence <xk> can be forced as close to the limit \ as we choose by taking k sufficiently large. The definition of a Cauchy sequence asserts that the terms of the sequence <xk> can be forced as close to each other as we choose by taking k sufficiently large. This latter point is more evident if the definition is rewritten in the form:

(k>K and 1>K) => d(xk, \t)<e.

27.2

Completeness

A complete metric space % is a metric space in which every Cauchy sequence converges. Our first task is to link this notion with the ideas introduced in §20.15 by showing that a complete metric space is one in which a suitable analogue of the BolzanoWeierstrass theorem (19.6) is true. 190

Completeness

191

27.3")" Theorem A metric space % is complete if and only if every totally bounded set in X has the Bolzano-Weierstrass property. Proof (i) Suppose that X is complete and that S is a totally bounded set in X. Let E be an infinite subset of S and let <xfc> be a sequence of distinct points of E. The definition of a Cauchy sequence is equivalent to the assertion that the range of the sequence is totally bounded. It follows that <xfc> is a Cauchy sequence and hence converges. Its limit is then a cluster point of E (exercise 25.17(3)). (ii) Suppose that every totally bounded set in X has the Bolzano-Weierstrass property. Let <xfc> be a Cauchy sequence. Then its range is totally bounded and hence is either finite or else possesses a cluster point. In either case <xfc> has a convergent subsequence (exercise 25.17(3)) and therefore converges (exercise 27.7(3)).

Not only is an analogue of the Bolzano-Weierstrass theorem true in a complete metric space, but we also have the following analogues of the Heine-Borel theorem and the Chinese box theorem.

27.4f Theorem A set K in a complete metric space X is compact if and only if it is closed and totally bounded. Proof See §20.16.

27.5f Theorem Let be a nested sequence of non-empty closed balls in a complete metric space X whose radii tend to zero. Then 00

P i Bk k=l

is non-empty (and in fact consists of a single point). Proof Note that a closed ball need not be compact in a general complete metric space X. (See §20.20.) The Cantor intersection theorem (19.9) therefore does not apply. Let x k e£ k . Given any e>0, let BK be the first ball with radius less than e. Then

k>K=>xkeBK and hence <xk> is a Cauchy sequence. Since % is complete, <xk> therefore converges. But each of the sets Bk are closed and the limit of <xk> therefore belongs to each of these sets (corollary 25.15).

Theorem 23.17 was concerned with the conditions under which one can reverse the limiting operations in a repeated limit. If £ is a complete metric space, this theorem takes a more satisfactory form.

192

Completeness

27.6f Theorem Let X, % and £ be metric spaces and suppose that £ is complete. Let £e X and Tje^. Suppose that A is a set in X for which £ is a cluster point and that B is a set in %. Let/: S \(£, f|)->£ where S = A x £. Let l:B-+Z and letm:,4->£. Suppose that (i) /(x, y)->/(y) as x->£ through the set A uniformly for yeB, and (ii) /(x, y)->m(x) as y—m through the set B pointwise for xeA. Then there exists a £e£ such that (iii) /(y)-»£ as y->n through the set B, and (iv) m(x)-+£ as x->^ through the set A.

Proof Let s>0 be given. From (i) we have that there exists a, d>0 such that for each xeA and each yeB,

From (ii) we have that, for each xeB there exists a Ax such that 0«i(y, 11) < AX => d(m(x)J(x, y))<e/4. It follows that if xeA satisfies 0 which satisfies.

d(xki,xkiJS2-1

(ZeN).

(3) Prove that any Cauchy sequence in a metric space X which has a convergent subsequence is itself convergent. t(4) Let y. be a set in a metric space X. Then y may be regarded as a metric subspace of X. Prove the following: (i) If X is complete, then y is complete if and only if y is closed in X. (ii) If y is compact in X, then y is complete. t(5) Let X be a metric space for which it is true that every nested sequence of nonempty closed balls whose radii tend to zero has a non-empty intersection. Prove that X is complete. (See theorem 27.5.) t(6) If X is a metric space, a function/: X-+X is called a contraction if and only if there exists a real number a < 1 such that

for each xe X and each yeX. Let/: X-+X be a contraction. Prove that: (i) / is continuous on X. (ii) Any sequence <xfc> of points of X which satisfies xk + j =/(x fc ) is a Cauchy sequence. (iii) If X is complete, t h e n / has a fixed point - i.e. for some £e5C,/(£) = £. (See example 17.13.)

27.8

Some complete spaces

The results of the previous section show that complete spaces are of some interest. In this section we shall consider a few specific examples of complete spaces. Since the Bolzano-Weierstrass theorem holds in Un9 it follows from theorem 27.3 that R" is complete. The proof of the next theorem establishes this result in a more direct fashion.

27.9

Theorem The metric space U" is complete.

194

Completeness

Proof Let <xfc> be a Cauchy sequence in Un. By exercise 27.7(1), <xk> is bounded and hence, by theorem 25.24, <xfc> has a convergent subsequence. The conclusion therefore follows from exercise 27.7(3).

Let S be a non-empty set and let y be a metric space. We shall denote the set of* all bounded functions/: S-^y by 53 (S, %\ The set $ (S, %) becomes a metric space if we define the distance between two 'points' / and goffB (S, y) by u(fg)= sup d(f(x\g(x)). xeS

We call u the uniform metric on & (S, %).

This is not a new notion. We introduced the same idea in §23.12 in connection with uniform convergence and in §20.20 we studied the special case /°° = 5

27.10|

Theorem If y is complete, then so is the metric space $ (S, %).

Proof Let be a Cauchy sequence in 3 (S, ^). Then, for each xeS, is a Cauchy sequence in y,. Since ^ is complete, it follows that, for each xeS, converges - i.e. there exists a function/: S-+y, such that/fc(x)->/(x) as k-^oo. In the language of §23.11 and §23.12, we have shown that converges pointwise for xeS. What we need to show is that converges in the space $ {S, %) - i.e. converges uniformly for xeS. Given any e>0, there exists a K such that for any k>K and any

It follows that, for each xeS, But, for each xeS,/fc(x)->/(x) as /c->oo. Thus, for any rf(/(x),/(x))^/2. (See exercise 16.16(4).) We deduce that, for any 1>K u(fbf)^s/2<s. It follows that/->/ as /->oo in the space S (S, ^). 27.1 If Theorem Suppose that y, is a complete metric space. Then the set C of all continuous functions in $ (5, ^) is closed in 3? (S, ^). Proof Suppose that is a sequence of functions in C such that/ k ->/ as /c-»oo. To prove that C is closed, we need to show that feC (corollary 25.15). Let £>0 be given. Then there exists a K such that k>K=>u(fJk)<s/3.

Completeness

195

Let ^e 5 and let J > K. Since/; is continuous on S, there exists a S > 0 such that for each xeS

It follows that, if d(&, x) < S, then

fMUx)) + d(Ux)J(x)) and so / is continuous on S - i.e. fe C. We shall use the notation C (5, y) to denote the space of all continuous functions / : S^-y.. Such functions need not in general be bounded. However, if S is a compact set in a metric space %, then a continuous function / : S-> ^/ is bounded and, moreover,

xeS

To obtain this result we apply theorem 19.14 to the continuous function F:

defined by F = do{f,g). 27.12"}" Theorem Suppose that ^ is a complete metric space and S is compact. Then d (S, %) is complete. Proof This follows from theorem 27.11 and exercise 27.7(4i).

27.13|

Incomplete spaces

It is by no means the case that all metric spaces are complete. As a typical example of an incomplete space we shall consider the system Q of rational numbers (regarded as a metric subspace of U). Some more examples are given in exercise 27.15. The fact that Q is not complete follows from the fact that Q is not a closed subset of U (see exercise 27.7(4i)). It is instructive, however, to consider some specific examples for which the properties of a complete space fail to hold in Q.

27.14 yt=j

Examples

Consider the sequences <xfc> and defined by

x1=2,

and

1/

2\

1/

1

These are both sequences of rational numbers. The sequence <xk> decreases and converges (in the space U) to ^2. (See exercise 25.11(6).) Similarly, the sequence increases and converges (in the space U) to ^J2.

196

Completeness

1—v

V2

(i) Since <xfc> converges in the space R, it is a Cauchy sequence in IR (exercise 27.7(1)). It follows that <xfc> is a Cauchy sequence in the space Q. But <xfc> does not converge if it is regarded as a sequence of points in Q. There is no rational number £ such that xk—•£ as k—>oo. In fact, xk-*^/2 as k-+oo and yjl is irrational. (ii) Let Ik = {r: reQ and xk^r^yk}. Then Ik = \_xk, > J n Q and hence is closed in the metric space Q (corollary 21.12). It follows that is a nested sequence of closed boxes in Q (§ 19.2) but 00

n ik=o - i.e. the Chinese box theorem fails in the space Q. (iii) Consider the set S = {r: reQ and O ^ r ^ ^ / 2 } . This set is closed in the metric space Q. It is also totally bounded. But it is not compact in O. For example, the continuous function/: Q-+R defined by/(x) = x does not achieve a maximum on S.

27.15f

Exercise

(1) Find the uniform distance between the functions/: R -•R and g: R -•R defined by l+x2'

"

w

l+x2

where these functions are regarded as points in the metric space 3 (IR, IR). (2) Let % be the set of all continuous functions/: [«,fr]-*(R.Let the metric in Z be defined by

/, g)= f

d(f,g)=\ \f(x)-g(x)\dx. Ja

Prove that X is not complete. (3) Let S be a set in a metric space % and let £e Z. Let ^ be a complete metric space and suppose that L^(S, y) is the subset of 9$ (S, y) consisting of all bounded functions / : S->y. such that there exists an x\ey. for which /(x)->i] as x-+£ through S. Prove that L^(5, y) is closed in SB (S, y). [Hmt: Use theorem 27.6.] (4) Explain why the open interval (0, 1) (regarded as a metric subspace of IR) is not complete. Give an example of a Cauchy sequence in (0, 1) which does not converge in (0, 1). Let/: R xR-»(0, 1) be defined by/(x, y) = xy. Show that/(x, y)-*y as x->l through (0, 1) uniformly for ye(0, 1) and that/(x, y)-+x as y-+l through (0, 1) pointwise for xe (0, 1). Explain the relevance of this result to the assumption in theorem 27.6 that £ is complete. (5) Let Pn: [0, 1]->R be defined by P()

l

- + - + ...+-.

Completeness

197

Prove that converges in the space S([0, 1], U). (6) Let 9 denote the metric space of all real polynomials where the metric is defined by

u(P, Q)=max \P(x)-Q(x)\. 0£x£l

Prove that 9 is not complete. [Hint. See the previous question.]

27.16|

Completion of metric spaces

When a space lacks a certain desirable property, a natural mathematical response is to seek to fit the space inside a larger space which does have the desirable property. In particular, if X is an incomplete metric space, can we fit X inside a larger complete metric snace? In this section we answer this question in the affirmative by constructing such a complete metric space. We shall, in fact, construct the smallest complete metric space inside which X can be fitted. This is called the completion of X and denoted by %*. The construction is very simple. The defect in X is that it has Cauchy sequences <xk> which do not converge. We therefore need to invent a limit for each such Cauchy sequence. We must be careful, however, not to invent too many objects. In particular, if two Cauchy sequences <xk> and satisfy d(\k, yk)->0

as

/c-^oo

(1)

then we shall want both sequences to converge to the same limit. We therefore begin by introducing an equivalence relation ~ on the set 5 of Cauchy sequences of points of X by writing

if and only if (1) holds. The set ft* is defined to be the set of all equivalence classes of S defined by this equivalence relation. We shall use the notation [<xfc>] to denote the equivalence class containing the Cauchy sequence <xfc>. The original space X is fitted inside the new space X* by identifying an element x of X with the element [<x>] of Z*. We need to show that X* is a complete metric space. First it is necessary to introduce a metric into X* which is consistent with that of X. We therefore define d: £*->R'by ], [])=lim

d(xk,yk).

k-+oo

Note that the limit on the right-hand side exists because and are Cauchy sequences in X. (See exercise 27.20(1).) But every Cauchy sequence of real numbers converges because U is complete (theorem 27.9).

198 27.17|

Completeness Theorem

For any metric space X, the metric space X* is complete.

Proof We have to prove that every Cauchy sequence in X* converges to a point of X*. We proceed by showing that, if <Xy> is a sequence in X* satisfying d(Xp Xj+1) is a sequence in X which satisfies

and hence d(xku\ x ^ ) < 2 - ( / c + 1)

(3)

provided that l^k. Now (2) asserts that lim d(xku\ x ^ + We may therefore deduce the existence of a strictly increasing sequence of natural numbers such that, for any l^kp d{xt{j\ x ^ + 1 ) ) < 2 - ( j + 1).

(4)

From (3) we also have that d(xkjij+1\

xkj+u+1)) is a Cauchy sequence in X. This fact follows from exercise 27.7(2) because

by (4) and (5). It remains to show that X^X

as /->oo. From (4) we have that, for each 7 ^ / ,

Completeness

199

But Xj = [<xk.(Z)>] and thus it follows that d(Xl9 X)2~~l-*0

27.18|

as

/->oo.

Completeness and the continuum axiom

After the previous discussion it is natural to consider first the completion Q * of the rational number system. Is Q * the same as the real number system R? Before seeking to answer this question, we should remind ourselves that Q and U are not just metric spaces. Both are also ordered fields. The question is therefore only meaningful if we can extend the algebraic structure of Q to Q * (as well as the metric structure). In particular, we need to be able to add, multiply and to order the equivalence classes which constitute the elements of Q>*. The appropriate definitions are the obvious ones. We define (ii)

(iii) l<xkyi>L(ykyi *>3K(k>K => xk>yk). With these definitions Q* becomes an ordered field and ], C< yk>]) = l To prove this is a somewhat tiresome but essentially trivial task and so we shall omit the details. Recall that Q is the 'smallest' ordered field. Thus O* is the 'smallest' complete ordered field. We know from theorem 27.9 that U is complete and thus K*c:[R. (These remarks, of course, take for granted that structures isomorphic to Q, Q * or U are to be identified with Q, Q* or U respectively - see §9.21.) The next theorem shows that the system Q * and the system U are the same - i.e. the real number system is the completion of the rational number system. Recall that U is an ordered field which satisfies the continuum axiom. This asserts that every non-empty set which is bound above has a smallest upper bound.

27.19|

Theorem

The system Q* satisfies the continuum axiom.

Proof Let S be a non-empty set in Q* which is bounded above. Let axeS and let b1 be an upper bound of S. We construct an increasing sequence of points of S and a decreasing sequence (bk} of upper bounds of S such that

The construction is inductive. If ck=^{ak + bk) is an upper bound of 5, we take bk + 1=ck and ak + 1=ak. If ck is not an upper bound we take bk + 1=bk and choose ak + i G S so that ak +1 ^ ck. The sequences and and be Cauchy sequences in a metric space X. Prove that is a Cauchy sequence of real numbers and hence converges. (2) Check that Q * is an ordered field with the definitions of addition, multiplication and an ordering given in §27.18. (3) Every ordered field X contains the system N of natural numbers (or else a system isomorphic to N). We say that X is Archimedean if N is unbounded above in X. (See §9.16.) The system U is Archimedean (theorem 9.17) and hence so are Q and Q*. Prove that an Archimedean ordered field X is complete if and only if it satisfies the continuum axiom (i.e. %=M). (Note that by the assertion that the Archimedean ordered field X is complete we mean that X is 'complete' with respect to the 'metric' d: X x X -+ X defined by

28

28.1

SERIES

Convergence of series

Suppose that is a sequence of points in a normed vector space Z. (See §13.17.) The Kth partial sum of the sequence is defined by

If the limit of the sequence <sK> exists, then we say that the series 00

converges and we write 00

s= I ak if and only if sK->s as X->oo. If the sequence of partial sums does not converge, we say that the series diverges.

28.2

Example If |x| < 1, then

To prove this result we observe that K

k=0

=

l-xK+1 1-x

•

1 1-x

as

K-»oo.

If |x| ^ 1, then the sequence of partial sums diverges and so the infinite series does not exist. 201

202

Series

28.3 Theorem (Comparison test) Suppose that % is a complete normed vector space and that

(i)

£ K is a convergent series of non-negative real numbers. If Hak||^fck

(fceN),

then the series

converges. Proof The partial sums of the sequence (1) are a Cauchy sequence (exercise 27.7(1)). Given any e>0, it follows that there exists an N such that, for any K>J>N, K

where tK denotes the Kth partial sum of the series (1). It follows that, for

any K>J>N, K

bk<s k=J+\

k=J+l

and so <sk> is a Cauchy sequence. Since % is complete, we may conclude that <sk> converges.

Many analysis textbooks are stuffed full with tests for establishing the convergence of series. Most of these tests, however, are useful only for somewhat obscure series and we shall therefore restrict our attention to two of the more important.

28.4 Proposition (Ratio test) Let denote a sequence of positive real numbers. Then the series

k=l

Series

203

converges if (i)

I

i

fc->oo

^ bk

and diverges if (ii) liminf ^ ^ > 1 . k-+oo

Ok

Proof If (i) holds, there exists a p < 1 and an H such that bk < Hpk (fceN). The convergence of the series then follows from the comparison test. If (ii) holds, then bfe-/»0 as /c-»oo and so the series diverges.

28.5 Proposition (Root test) Let denote a sequence of nonnegative real numbers. Then the series

k=l

converges if (i) limsup

bk1/kl.

fc^oo

Proof The same remarks apply as in the proof of proposition 28.4.

28.6

Example Consider the case bk = k~a. Then l i m - 4 ±1 = l; k-+cc

Dk

lim bki/k = l. fc^oo

It follows that neither the ratio test nor the root test are helpful in determining the convergence or divergence of the series

I1i—i La k= 1 K

As it happens, the series converges for a > l and diverges for a ^ l . (See exercise 28.10(3).)

204

Series

28.7

Absolute convergence A series GO

fc=l

converges absolutely if and only if

t Ikll fc=l

converges. The comparison test shows that in a complete space any absolutely convergent series converges. The ratio test and the root test can therefore be used withfcjt= ||ak|| to establish the convergence of certain series in a complete space. But note that both tests (and the comparison test) are only able to establish absolute convergence. However, series exist which converge but do not converge absolutely. (See exercise 28.10(4).)

28.8

Power series

If is a sequence of real (or complex) numbers and £ is a real (or complex) number, then the series

k=l

is called a power series (about the point £). For what values of z does this power series converge? The root test is useful here. Suppose that lim sup \ak\1/k = fc^oo

Then

If 0 < p < + o o , it follows that the power series converges when \z — £| < 1/p. If p = 0, the power series always converges and, if p = +oo, it converges only when z = £. This argument shows that there exists an Re[0, +oo] such that the power series converges when \z — £ | < # and diverges when |z — £|>K. It follows that the set S of values of z for which the power series converges satisfies

Series

205

Thus, if z is a real variable, S is an interval with midpoint £. We say that S is the interval of convergence of the power series. If z is a complex variable, S is a disc with centre (. We call S the disc of convergence of the power series. For obvious reasons, R is called the radius of convergence of the power series.

////////////A 7777777777777

interval of convergence

disc of convergence

A given boundary point £ of S may or may not be an element of S - i.e. the power series may or may not converge at £. The above theory yields no information about this question. Note finally that we have shown that a power series converges absolutely at interior points of S (i.e. for \z — (|<JR). If it converges at a boundary point of 5, however, there is no reason why the convergence should necessarily be absolute.

28.9

Example The power series oo

I (has radius of convergence 1. We have that lim

1/fc

(-1)*

= 1.

The series obtained by writing z = 1 converges but not absolutely. The series obtained by writing z = — 1 diverges.

28.10

Exercise

(1) Suppose that is a sequence of points in a normed vector space % and that the series 00

k= 1

converges. Prove that ak—•() as fc->oo. Give an example of a sequence

206

Series (bky of real numbers for which bk^0

as /c->oo but the series

k=l

diverges. (2) Suppose that {ak} is a sequence of non-negative real numbers. Prove that the infinite series

exists if and only if the sequence of partial sums is bounded above. (3) Prove that

K

k=l

diverges by demonstrating that

where sK denotes the Kth partial sum. Prove that, if <x> 1, then "

1

1

1

converges by demonstrating that

where tK denotes the Kth partial sum. (4) If is any decreasing sequence of non-negative real numbers, prove that K+L

aK-aK+1S\

I (-l) fc ak \^aK. k=K

Show that, if also afc->0 as /c-»oo, then

converges. [Hint. Show that the sequence of partial sums is Cauchy.] Deduce that the series - (-If'1

111

Series

207

converges and that j ^ s r g l . Prove also that 3

1 1 1 1 1 1 1 1

S

2~

3 2

5 7 4

9

11 6

[/fmt: If rK denotes the Xth partial sum of the second series, then 1 1

1

1

1

(5) Suppose that y, where y,is a. normed vector space. We say that I /k(x)

(1)

fc=i

converges uniformly for xeS if and only if the sequence of partial sums converges uniformly for xeS. Similarly, (1) converges pointwise if and only if the sequence of partial sums converges pointwise for xeS. 28.12*1* Theorem Suppose that % is a metric space and that y. is a complete normed vector space. Let SaX and suppose .that, for each keN, the functions fk:S-^% satisfy /*(*)-•% as x->£ through the set S and that

k=l

converges uniformly for xeS. Then 00

00

k=\

k=l

£ AW-* E % as x -^ through the set S. Proof This is an immediate consequence of theorem 27.6. 28.13f Corollary Suppose that % is a metric space and that y. is a complete normed vector space. Let Scz% and suppose that, for each keN, the function fk: S-+y, is continuous on S. Then the function/: S-+y. defined by

/(x)= I Mx) k=l

is continuous on S provided the series converges uniformly for xeS.

Series 28.14

Example

209 The series

converges pointwise for each xe[0, 00). For x > 0 , / ( x ) may be evaluated using the formula for a geometric progression. We obtain that

l-^r1

(x>0)

Observe that/(x)->l as x->0 + but/(0) = 0. Thus / is not continuous on [0, oo) and hence the series does not converge uniformly for xe[0, OO).

28.15f

Series in function spaces

Recall from §27.8 that $ (5, %) denotes the space of all bounded functions F : S-+y with metric u defined by u(F, G) = supd(F(x), G(x)). xeS

If y, is a normed vector space, then we may regard $ (S, y) as a normed vector space by defining vector addition and scalar multiplication in $ (S, y) by (F + G)(x) = F(x) + G(x) (aF)(x) = aF(x). The norm on fS (S, y) is, of course, defined by ||F|| = sup||F(x)|| xeS

so that u(F, G) = \\F-G\\. We proved in theorem 27.10 that, if y is complete, then so is $ (S, y). In particular, if y is a complete normed vector space, then so is $ (S, y). Suppose that y is a normed vector space and that, for each keN, fk:S-+y is bounded - i.e. is a sequence of 'points' in the normed vector space ® (S, y). Then, as explained in §23.12, to say that the series

converges uniformly for xeS is the same as saying that the series OO

converges in the space 3$ (S, y). This is a useful observation because it allows us to extend many of the preceding

210

Series

results of this chapter to the case of uniformly convergent series. In particular, we obtain the following version of the comparison test. 28.16| Theorem (Weierstrass 'M test) Suppose that, for each keN,fk: S-+y., where y, is a complete normed vector space. Suppose also that is a sequence of non-negative real numbers for which 00

IK k=l

converges and that, for each keN and each xeS,

IL/iMII^*.

(i)

Then the series

I k=l

converges uniformly for xeS. Proof Condition (1) implies that \\fk\\£bk Hence the series 00

I/* converges in the space $ (S, %) by the comparison test. 28.17 Example Consider the function fk: [0, oo)-»R defined by fk(x) = x2e~kx. This achieves a maximum at x = 2k~1. Hence, for each xe[0, oo) and each keN,

It follows from the Weierstrass 'M test' that the series

k=\

converges uniformly for xe[0, oo). From corollary 28.13 we may conclude that / is continuous on [0, oo). In fact, for each xe[0, oo), we may evaluate f(x) as in example 28.14 to obtain

l-rr1

(x>0) (x = 0).

Observe that /(x)-*/(0) as x->0 +.

Series 28.18J

211

Exercise

(1) Prove that the series

converges point wise for x^O. Show that / is not continuous on [0, oo) and deduce that the series does not converge uniformly for x^O. Does the series converge uniformly for (a) O g x ^ l , (b) l ^ x ^ 2 ? (2) Prove that the series

/ and are sequences of real numbers and let K

U = y

uk

(K>J).

Prove that K k=J+l

K-l k=J+l

(Abel's 'partial summation' formula). (4) Suppose that (ak) is a decreasing sequence of positive real numbers which tends to zero and that 4>k: S-+M satisfies K

£ (t>k(x) ^H lc=l

for all KeN and all XES. Prove that

converges uniformly for xeS. [Hint: Use question 3.] (5) Establish the identity . ^ . cos \x-cos{n+j)x sin x + sin 2x + ... + sin nx= :—'-. 2 sin ^x and deduce that the series

/w= Z k=i

sin kx

kK

converges uniformly for xe[(5, 2n — S'] provided that 0. [Hint: Use question 4.] Deduce that / is continuous on (0, n).

212

Series

(6) Suppose that the real power series

f(x)= £ akxk k=O

has interval of convergence / and let J be any compact subinterval of /. Prove that the series converges uniformly for xel. [Hint: To show that the series converges uniformly on [0, X]ciJ, take uk = akXk and vk = (xX~1)k in question 3.] Deduce that the function F: [ — 1, 1)—•IR defined by oo

k

is continuous on [ — 1, 1).

28.19|

Continuous operators

Suppose that % and y. are metric spaces and that T: %-*% is continuous on %. Then we know from corollary 25.9 that r / l i m XfcJ = lim T{Xk) \k-*co

I

k-*ao

provided that the left-hand side exists. It follows that

provided that the left-hand side converges. This is a useful result, particularly in the case when % is a function space. We then usually call the continuous function T: Z-^y, a. continuous operator. We shall illustrate the usefulness of the result by examining the case in which % is the space of all continuous functions/: [a, b]-*R (usually denoted by C[a, bj) and T is the 'integration operator' on C[a, b~\.

28.20| Theorem Let C[a, fr] denote the space of all continuous functions / : [a, b~\-+U and define the operator T: C[a, &]-*R by

Then T is continuous on C[a, b~\. Proof We have that

\T(f)-T(g)\=

- | | g(t)dt

213

Series

S\ \f(t)-g(t)\dt £(b-a)

max \f(t)-g(t)\

= (b-a)\\f-g\\. It follows that T(/)->T(g) as f-*g and hence that T is a continuous operator on Cla, b].

28.21"]" Corollary suppose that

For each /ceN, l e t / t : [a, £>]->IR be continuous on [a, b~\ and

(1) converges uniformly for xe[a, 6]. Then (2) Proof To say that (1) converges uniformly for xe[a, i>] is the same as saying that

converges in C\_a, b]. The result therefore follows immediately from theorem 28.20.

28.22 Example It is worth noting that, without the uniformity condition, corollary 28.21 does not hold. Consider, for example, the function Fk: [0, 1]->R illustrated below.

We have that, for each xe[0, 1], Ffe(x)-»0 as /c->oo - i.e. converges pointwise to

214

Series

the zero function. But

f

Fk(x)dx = l

It follows that the series whose sequence of partial sums is (Fk(x)} fails to satisfy (2) of corollary 28.21.

28.23f Theorem Suppose that X and y, are metric spaces and that T: X-+y, is continuous and bijective. Then 00

\

=i

/

00

) k=i

provided that the right-hand side exists. Proof Since T is continuous

I yt. k=l

Hence 00

X r- 1 (yj=r- 1 k=l

We illustrate this result by taking X = C[a, b] and % = C£a, b] where the latter notation denotes the set of all F: [a, b]-+U which can be written in the form

F(x) =

f{t)dt

{a^x^b)

where/is a function in C[a, b]. The role of the operator T:X-*y, will be assumed by the 'integration operator' /: C[a, b] -> C\[a, b] defined by / ( / ) = F. The fundamental theorem of calculus asserts that the 'differentiation operator' D:C£a, b]->C[a, b~\ defined by D ( F ) = / where

f(x) = F(x)

(a<xR is defined by

28.25

Example

Consider the function Gk: [0,1]->R illustrated below.

We have that GkeC[a, b] and that converges uniformly to the zero function for xe[0, 1]. But

It follows that the series whose sequence of partial sums is fails to satisfy (3) of theorem 28.24.

28.26t

Applications to power series

A trivial consequence of the Weierstrass 'M test' (28.16) is that a real (or complex) power series

converges uniformly for \z — Q ^ r provided that the constant r is chosen so that r < R where R is the radius of convergence of the power series. Exercise 28.18(6) is a more subtle result. This demonstrates that a real power series converges uniformly on any

216

Series

compact subinterval of its interval of convergence. We may deduce the following results.

28.27| Proposition (Abel's theorem) The sum of any real power series is continuous on its interval of convergence.

28.28*f" Proposition real power series

If c and d are any points in the interval of convergence of a

/(*)= £ ak(x-Z)\ then fd

00

fd

f(x)dx= X flk (x-tfdx.

28.29| Proposition real power series

If y is any interior point of the interval of convergence of a oo

f(x)= £ ak(x-0\ k=0

then/is differentiate at y and 00

f\y)= Z

ka

k(y-Ok~l-

The second and third propositions are true also of complex power series (provided we replace 'interval of convergence' by 'disc of convergence'). Abel's theorem, however, is false for complex power series. The diagram below represents the disc of convergence of a complex power series. If the power series converges at the point y, then it is true that f(z)-+f(y) as z-»y along curves like Cl but it is in general false that/(z)->/(>>) as z->y along 'tangential' curves like C 2 .

Series 28.30

217

Example If |x| < 1, we have that 00

1

k=o

1+x

By proposition 28.28, if |y| < 1, then fy

fy / °°

A

\

=Jo T^=\ ( I (-l)VW 1 + x Jo \k = o J i)fc+1

00

= Z (-D

+1

The latter power series has interval of convergence ( — 1, 1] and, by Abel's theorem (28.27), its sum/(y) is therefore continuous on ( — 1, 1]. But/(^) = log(l +y) for ye(-l, 1). It follows that/(I) = log 2 - i.e. log 2 = 1 1 1 * 2 3 4 5

28.31|

...

Exercise

(1) Prove that, for |x| < 1,

(-i)kx2k

1+x 2

Hence obtain a power series expansion for arctan x which is valid for - 1 <xS 1. Deduce that n

1 1 1 1 1

+

(2) Obtain the general binomial theorem by differentiating

(3) The power series

(z)=% bk(z-ok k=0

both converge for \z — Q0, there exists a. finite set Fez I such that, for each finite set G,

A vector s defined in this way will be called an infinite sum. (Note incidentally that the convergence notion introduced here differs somewhat from those we have met earlier. It is an example of the convergence of a net.) Since the definition takes no account of any ordering which may exist on the index set /, it is clear that infinite sums must satisfy the commutative law. The next theorem puts the proposition in formal terms.

29.3| Theorem Suppose that Z is a normed vector space and that a: I- Z. If 0 is a permutation on /, then

iel

provided one side or the other of the equation exists. Proof We shall suppose that it is the left-hand side which exists. (Otherwise replace a by b = a O and (j> by (f)'1.) Let £ > 0 be given. Then there exists a finite set F a I such that, for any finite set G,

iel

Hence, for any finite set H,

Infinite sums

221 Hl£a(O-

iel

Z fl(OIIZ and

Z isl

exists. Then, given any £>0, there exists a finite set Fez I such that, for e a c h ; e / ,

JtF=>\\aU)\\<e. Proof Let £ > 0 be given. Then E / 2 > 0 and hence there exists a finite set Fez I such that for any finite set G

LQtjeI\F. Then

)II = II Z a ( 0 -

2 >

ieFv{j}

ieF

ieF

222

Infinite

29.6| Corollary suppose that

sums Suppose that Z is a normed vector space. Let a: I-+Z and

iel

exists. Then the set H = {i: a(i) # 0} is countable. Proof Let fceN. By the previous theorem the set Gk = {i: ||fl(0|| = l/fc} is finite. But

and so if is the countable union of finite sets. It follows that H is countable by theorem 12.12.

The definition of an infinite sum given in §29.2 seems to be one of considerable generality in that the index set may be any set whatsoever while the index set in the case of an infinite series is restricted to be N. But the last result shows that this generality is largely illusory. An infinite sum of an uncountable collection of non-zero objects never exists. In future, we shall therefore always assume that the index set / is countable. Indeed, it will often be convenient to assume that I = N. This assumption can be made without loss of generality when / is countable because there then exists a bijection <j>: N-+I. Thus / may be replaced by N provided that a: /-> Z is replaced by the sequence b:N-+% where b = a O <j>.

29.7| Theorem Suppose that ft is a normed vector space and that (bky is a sequence of points in Z. Then

Z K= £ K keN

(l)

k=l

provided that the left-hand side exists. Proof Let e > 0 be given. If the left-hand side of (1) exists, then there exists a finite set Fez I such that for any finite set G,

2> fceN

keG

Let X 0 = max F. Then, provided that K>K0, the set G = {1, 2, 3, . . . , K} satisfies ^J. Hence

and the equation (1) follows.

Infinite sums

223

Theorem 29.7 says that, if an infinite sum exists, then so does the corresponding series and the two are equal. Of course, as we know from §29.1 it is in general false that the convergence of a series implies the existence of the corresponding infinite sum. There is, however, an important special case for which the convergence of a series does imply the existence of the corresponding infinite sum. This is the case of a series with non-negative terms.

29.8| Proposition bers. Then

Suppose that (bk} is a sequence of non-negative real num-

keN

k= 1

provided that the right-hand side converges.

The proof follows from exercise 28.10(5). An immediate consequence is that series of non-negative real numbers satisfy the commutative law although, as we have seen, this is certainly not the case of series in general.

29.9|

Complete spaces and the associative law

To proceed any further with the study of infinite sums, we need to restrict our attention to the case when % is a complete normed vector space. (Such a space is called a Banach space.) This is no surprise since most of the results of the previous chapter required completeness as well. Recall that a complete metric space is one in which every Cauchy sequence converges (§27.2). Alternatively, it is a metric space in which every totally bounded set has the Bolzano - Weierstrass property. We shall wish to use the fact that a version of the Heine-Borel theorem holds in a complete space. This asserts that every closed and totally bounded set is compact (see theorem 27.4). Some examples of complete normed vector spaces were discussed in §27.8. The most notable example of such a space is, of course, the space Un.

29.10| Theorem Suppose that Z is a complete normed vector space and that is the collection of all finite subsets of/. Let a: I-+Z and suppose that

exists. Then the set

is totally bounded and hence its closure is compact.

224

Infinite sums

Proof We have to show that, for each £>0, there exists a. finite collection of open balls of radius £ which covers S. Let £ > 0 be given. Then there exists a finite set Fez I such that, for each finite set

Define the set C by

Then C is a finite set. The collection of all open balls of radius s whose centres are elements of C is therefore finite. It covers S because, for each He 5,

||2>(0-S+ X 4011 = 11 I 0(O-S|| 0, there exists a finite set F a I such that, for any finite set if, e.

(2)

Proof It is quite easy to prove that the existence of (1) implies criterion (2). We therefore leave this as an exercise. There then remains the problem of showing that criterion (2) implies the existence of (1). With the notation of the previous theorem, let

£ 40: He 5.and £c= ieH

Then {SE: Ee 5} is a collection of closed subsets of the compact set S. Since this collection has the finite intersection property (exercise 20.6(3)) it follows that it has a non-empty intersection. Let seSE for each Ee S. Let £ > 0 be given. Then, by (2), there exists a finite set Foa:I such that, for any finite set H,

ieH

Also, since SGS FQ , there exists a finite set F satisfying F0F. Then G\Fcil\F0

and so

||s-£a(OI|:£||s-£ 4011 + 11 I 40 ieG

ieF

ieG\F

Hence (1) exists.

Theorem 29.11 is a variant of the result which says that, in a complete space X, every Cauchy sequence converges. It is not surprising therefore that it has some useful corollaries.

29.12"}* Corollary Suppose that X is a complete normed vector space and that a: / - • X. If J c /, then the existence of the infinite sum

£40 iel

implies the existence of

140. ieJ

29.13f Corollary Suppose that X is a complete normed vector space and that a: J-* X. If J and K are disjoint sets with union /, then

£40 = 140+140 iel

ieJ

ieK

provided that one side or other of the equation exists.

29.14*}* Corollary Suppose that X is a complete normed vector space and that a: I->Z. Then the existence of the infinite sum

iel

implies that, for any £>0, there exists a finite set Fez I such that for any set G,

||£fl(0-£fl(0ll = || E 40ll<e. iel

ieG

ieI\G

226

Infinite sums

The proofs of the corollaries are easy and we give them as exercises (29.22(1), (2) and (3)). Having obtained these results, we are now in a position to prove a version of the associative law in complete spaces.

29.15| Theorem Suppose that % is a complete normed vector space and that a : I-+X. Let W denote any collection of disjoint sets with union /. Then

I«(0= I jz««} iel

JeW- lieJ

(3)

)

provided the left-hand side exists. Proof Let e > 0 be given. Since the left-hand side of (3) exists, it follows from corollary 29.14 that there exists a finite set Fal such that, for any set G,

iel

ieG

Let 5 — {J:JeW and FnJ^ 0} and suppose that Q is a finite collection of sets satisfying S a QaW. Then the set G=U J JeQ

satisfies Fa Gal

and therefore, using corollary 29.13,

iel

JeQ LieJ

)

iel

ieG

Equation (3) follows.

Theorem 29.15 asserts that, if an infinite sum exists, then one can bracket its terms in any manner whatsoever without altering its value. The insertion of brackets in infinite sums therefore creates no problem. But the problem of removing brackets remains - i.e. the existence of the right-hand side of (3) does not in general guarantee the existence of the left-hand side. The infinite sum

10 certainly exists. But, if we write 0 = (1 — 1) in this sum and then remove the brackets, we obtain

which does not exist. We noted at the end of §29.4 that, for the special case of series of non-negative real numbers, the commutative law does not fail. It is also true that the removal of brackets from infinite sums of non-negative real numbers creates no problems.

Infinite sums

227

29.16f Proposition Suppose that U. Suppose that the infinite sum

I MO exists and that \\a(i)\\^b(i)

(iel).

228

Infinite

sums

Then the infinite sum ieE

exists. Proof This is the same as the proof of theorem 28.3 except that theorem 29.11 is used instead of the fact that Cauchy sequences converge.

29.19"}* Corollary Suppose that % is a complete normed vector space and that a: I-^Z. Then the existence of

El won iel

implies the existence of iel

Proof Take b(i) = \\a(i)\\ in proposition 29.18.

29.20 Corollary Suppose that % is a complete normed vector space and that a: I-^Z. Let W denote any collection of disjoint sets with union /. Then the existence of

(2)

I il Jem

U

J

implies the existence of the infinite sum

(3)

Z«(0iel

Proof By proposition 29.16, the existence of (2) implies that (3) exists absolutely.

An immediate application of this result is to the problem of removing brackets from infinite sums.

29.211 Corollary Suppose that % is a complete normed vector space and that a: I^Z. Let W denote any collection of disjoint sets with union /. Then

2X0= £ {z«(»)} iel

Jew. UeJ

provided the ng/zt-hand side exists absolutely.

)

Infinite sums

229

Proof This follows immediately from corollary 29.20 and theorem 29.15.

We shall use this result in §29.23 to study the problem of reversing the order of summation in repeated series.

29.22|

Exercise

(1) Suppose that % is any normed vector space and that a: I-+%. If J and K are disjoint sets with union /, prove that

iel

ieJ

provided the right-hand side exists. (2) Show that the equation of question 1 holds when % is a complete normed vector space provided the left-hand side exists (i.e. prove corollaries 29.12 and 29.13). (3) Suppose that % is a complete normed vector space and that a: / - » £ . Show that.

iel

exists if and only if, for any e > 0, there exists a finite set G such that for any set

HczI\G ieH

(4) Let be a sequence of real numbers. Write a k + = m a x ak~ = max{ — ak, 0}. Prove that the infinite sum

{ak, 0} and

exists absolutely if and only if any two of the series 00

00

(i) £ ak (ii) £ ak+ k=l

00

(iii) £ ak~

k=l

k=l

converge. If (i) converges but neither (ii) nor (iii) converge, then the series (i) is said to converge conditionally. Prove that, in this case, given any £eM there exists a permutation <j)\ N -• N such that

(The latter result is called Riemann's theorem.) (5) If a: /->[R, prove that the infinite sum

230

Infinite sums

exists if and only if it exists absolutely. [Hint: Use the previous question.] Obtain the same result when a: /->R". Let be the sequence in Z00 (see §20.20) whose terms are all zero except for the /cth which is equal to l//c. Prove that

exists but not absolutely. Deduce that the converse of corollary 29.19 is false in general (although, as question 4 shows, the converse is true for Z = Un). (6) Prove that a normed vector space % is complete if and only if, for every a: / - • £, the existence of

I IMOI iel

implies the existence of

I a(Q. iel

29.23|

Repeated series One sometimes has to deal with repeated series of the form

I

I *A-

(1)

It follows from corollary 29.21 that if 00

I

f

00

)

I IM

7=1 U=i

(2) J

exists, then o o f o o

0=1

This is, of course, a special case of a much stronger result. If (2) exists, then any method of adding up the terms of (1) will yield the same answer.

29.24

Example

It follows from the binomial theorem (exercise 28.31(2)) that ( l - z ) - 2 = l + 2 z + 3z2 + 4z3 + ...

(3)

provided that |z| < 1. It is instructive to see how this result may be deduced from the theory developed in this chapter. We begin with the formula for a geometric progression (l-z)-1

which is valid for | z | < l . Thus

= l + z + z2 + z3 + ...

(4)

Infinite sums

231

and so

a-*r2=E i ^

(5)

provided that | z | < l . But we know that the power series (4) exists absolutely for \z\ < 1. Thus (5) exists absolutely and so we can reverse the order of summation and obtain oo

I

oo

I

(1-*)-*= E I z ' = I z < I 1. 1=0 j=0

Z=0

j=0

Thus

provided that \z\ < 1. Some explanation of these calculations may be helpful. Given that oo

j=0

oo

k=0

exists, the same result will be obtained regardless of whether the rows or columns in the array below are summed first.

+ r2

= a20

+ a21 +

a22

+ r

i

=

a

io

+

a

n

+

II

a2l

a22

^23

+

+

4

+ al3

a

n

+

4

r0

^20

+

= a 00 + fl01 + a 02 II

c

o

+

+

^02

fl

n

II

4 c.

+ C2

O3 II

In (5) we are concerned with the special case in which the terms in the array above the main diagonal are all zero as on page 232.

232

Infinite sums

0

0

0

0

0

("22

0

("11

"00

+ "01

(733

0

0

•f

"23

0

0

+ "12 "f

"13

0

("n

+

"02

•f

"12

+

"ool

"03

0

"01

"02

Then 00

00

= Z rj= Z

00

Z

s

j=0

00

a

j= 0 1 = j

j= 0 1 = 0

00

00

1=0

00

00

Ji= Z Z aji 00

i

1 = 0 j=0

1 = 0 j=0

But summing an array by first adding the rows or columns is not the only way in which one can proceed. One can also sum, for example, by diagonals as indicated below.

a

30\.

5

"31

=

"32

"33

"22

"23

d{

Observe that

This method is particularly suitable for multiplying power series. We have that

= f Z «,vi+* [=o

j+fc=i

00

= Z *' Z «A

Infinite sums

233 00

I

Zl

= Z

a

Z m=0

1=0

mbl-m

provided that |z| and to denote these alternative norms. In chapter 23 we found that the first of these alternative norms is sometimes more convenient to use than the Euclidean norm. The reason, of course, that this norm could be substituted for the Euclidean norm in this context is that both norms generate the same topology in 1R" and, when considering limits, it is only the topologies in the spaces concerned which matter. (See §21.18 and §23.1.) There is an infinite collection of norms which are compatible with the vector space structure of R" and it is natural to ask whether all these different norms generate the same topology on Un. The next proposition answers this question in the affirmative. Note, however, that the corresponding result is very definitely false in infinite-dimensional spaces.

30.9|

Proposition

All norms on Un generate the same topology on Un.

Proof We shall use the usual notation for the Euclidean norm on Un. We shall suppose that we are also given an alternative norm on Un and use the notation |||x||| to denote this norm. As indicated in §21.18, it is enough if we can show that the set

contains an open Euclidean ball Br = {\: ||x|| 0 and this is a contradiction. Hence S is bounded and so there exists an R for which

240

Separation

in Un

Next suppose that 0 is a boundary point of S (with respect to the Euclidean metric). Since S is convex, it follows from theorem 30.6 that there exists an X^O such that <X, s > ^ 0 for all seS. But, if aXeS and a > 0 , then <X, aX> = a||X|| 2 >0. This is a contradiction. We conclude that 0 is an interior point of S (with respect to the Euclidean metric) and hence there exists an r > 0 such that Br(^S.

<X,x> =

30.1 Of

Exercise

(1) Prove that any hypersphere (§13.14) separates U" into two components. (2) Prove that any ball, box or line segment S (§13.14) has the property that U" \S is connected. (3) Find hyperplanes which separate the following sets in U2: (i) A = {(x, y): x2 + y 2 g l } ; B = {(x, y): y^l} (ii) A = {{x, y): x>0 and xy> 1}; B = {(x, y): x>0 and xy< - 1 } . (4) Find two non-empty, convex sets A and B in R 2 with AnB = 0 which cannot be separated by a hyperplane. (5) Let S be a convex set in (RM. (i) If ^edS, prove that ^ed(eS). Give an example of a non-convex set S for which this result is false. (ii) If S is unbounded, prove that S contains a half-line. Give an example of a convex set S in Z00 for which this result is false. (6) Prove that in a Hausdorff space (§30.2) every compact set is closed. Prove that Rn is a Hausdorff space.

Separation in Rn 30.1 If

241

Curves and continua

In the remainder of this chapter, we focus our attention on !R2. We have seen that lines and circles separate !R2 into two distinct components. What can be said about other curves? First, something needs to be said about curves in general. In §17.15, we defined a curve in R 2 to be the image/(/) of a compact interval / under a continuous function / : I-+U2. As we remarked at the time, this definition is not very satisfactory from the intuitive point of view since it fails to classify as a curve various 'one-dimensional' objects such as the example of Brouwer's (§17.18) consisting of a spiral wrapped around a circle.

Brouwer's set E is compact and connected. Such a set is called a continuum (although this usage is not consistent with the familiar description of U as 'the continuum'). The fact that E fails to qualify as a curve is something to which one can accommodate oneself. But much worse things can happen than this as Peano discovered in 1890. We describe a version of his construction given by Hilbert. Let / = [0, 1] and let S be the square [0, 1] x [0, 1]. For each neN, Hilbert constructed a continuous function fn: I->S. The curves ft(I), j'2(I) and / 3 (/) are illustrated in the diagram below.

I i —U i

The sequence of functions converges uniformly to a continuous function / : / - » £ . As is evident from the construction, the curve/(/) passes through every point of S i.e. / ( / ) = 5. We say that / ( / ) is a 'space-filling curve'.

242

Separation in Un

This construction shows that 'two-dimensional' objects can be curves according to the definition given at the head of this section. This is, of course, highly counterintuitive. (Even worse, the function/: 7-+S has the property that it is continuous on I but differentiate at no point of /.) Peano's discovery is by no means an oddity. The Mazurkiewicz-Moore theorem asserts that every locally-connected continuum is a curve. (A set E in U2 is locally. connected if for each eeE and each £>0 there exists a 0 such that for each xeE satisfying ||e — x\\C. A curve C will be called a Jordan curve (or simple 'closed' curve) if it is topologically equivalent to the unit circle U in U2.

The word 'simple' is used to indicate that we are discussing curves which do not 'cross themselves'. Hilbert's space-filling curve is therefore very definitely not simple. Simple arcs and Jordan curves also admit characterisations similar to that provided by the Mazurkiewicz-Moore theorem for general curves. Thus a simple arc is a continuum with the property that the removal of any point with two exceptions produces a disconnected set. (The exceptions are the endpoints.) A Jordan curve is a continuum with the property that the removal of any single point produces a connected set but the removal of any two distinct points produces a disconnected set. It is clear what the separation properties of simple arcs and Jordan curves 'ought' to be. If A is a simple arc, then U2 \ A should be connected while if J is a Jordan curve, then U 2 \ J should have precisely two components. Both results are true but neither is easily proved. The latter result is the more important and we quote it as the Jordan curve theorem.

Separation

in Un

243

3O.13| Proposition Let J be a Jordan curve in U2. Then U2\J has two components and J is the boundary of both components. One of the components is bounded and the other is unbounded. (The bounded component is called the 'inside' of J and the unbounded component is called the 'outside' of J.)

outside

Note that the Jordan curve theorem would be trivial if it were known that there exists a homeomorphism F: U2-^U2 such that F(U) = J. But we are only given that there exists a homeomorphism/: U->J. It is a result of Schonflies that each such homeomorphism/: U-*J can be extended to a homeomorphism F: IR2->R2 but the proof of this useful result is harder than the proof of the Jordan curve theorem itself.

30.14J

Simply connected regions

Recall that the Riemann sphere (or Gaussian plane) U 2 * is the one-point compactification of M2. This notion is particularly important in those cases when one is identifying U 2 with the system C of complex numbers. We begin by observing that the Jordan curve theorem remains true with U2 replaced by U 2 * (except that, if oo e J, neither component is bounded in U 2). In complex analysis, one is often concerned with open, connected subsets of IR2#. Such sets are called regions (or domains). Of particular importance, are simply connected regions. These are regions which have no 'holes'. More precisely, a region R in IR 2 * is simply connected if and only if IR 2 * \ R is connected.

simply connected region

multiply connected region

An important alternative characterisation of a simply connected region R is the fact that if R contains a Jordan curve J, then R contains one of the two components of IR 2 * \ J. In particular, if oo ^R, then R contains the bounded component of IR 2 * \ J.

244

Separation

in Un

\

30.15f Exercise (1) Explain why Brouwer's set E (§30.11) separates U2 into two components. Explain also why E is not a locally connected continuum. (2) Consider the set R = {(x, y):y>x2} in U2. Explain why R is simply connected. (3) Suppose that % and y. are topological spaces and that y, is Hausdorff (§ 30.2 and exercise 30.10(6)). If X is compact and / : %-+% is continuous and bijective, prove that / is a homeomorphism. (4) If / : /->R 2 and g: U-+M2 (where / = [0, 1] and U is the unit circle in U2) are continuous and bijective, prove t h a t / ( / ) and g(U) are respectively a simple arc and a Jordan curve. Explain why Hilbert's space-filling curve is not simple. (5) Suppose that J is a Jordan curve in IR2 and that ^eJ. Let S and T denote the inside and the outside of J respectively. Let a e S and be T. Prove that there exists a simple arc joining a and % which lies entirely inside S and that there exists a simple arc joining b and \ which lies entirely inside T. [Hint: Use the result of Schonflies quoted in §30.12.]. (6) Explain why the result of the previous question is false when J is replaced by Brouwer's set E.

NOTATION*

=> e

{x:P(x)}

0

V 3 AaB

es

AuB AnB A\B

(§2.4) (§2.10) (§3.1) (§3.1) (§3.1) (§3.1) (§3.4) (§3.4) (§4.1) (§4.4) (§4.7) (§4.7) (§4.9)

7 9 14 14 15 15 16 16 21 22 23 23 23

(§7.13) 52 (§7.13) 52 52 (§7.13) 52 (§7.13) 52 (§7.13) 52 (§7.13) 54 (§8.2) 56 (§8.7) 60 (§8.13) 60 (§8-14) (§9.7, §24.4) 68,156 (§9.7, §24.4) 65,156 69 (§9.10)

(-oo, b)

fob]

fo oo) (-oo, b\

fob)

]a,bl N

I Q

sup S inf S

sup f(x) xeS

U S

(§4.10)

24

Se WL

n s

(§4.10)

24

Se W

(a,b) AxB A2 Un

9 (A) f:A->B AS)

f~\T) f-u.B->A <xky fog

u uU_+ (a,b) (a, oo)

inf/W

(§9.10)

(§5.1, §7.13) 28,52 28 (§5.2) 29 (§5.2) (§5.2, §13.1) 29 32 (§5.9) 33 (§6.1) 35 (§6.2) 35 (§6.2) 36 (§6.2) (§6.3, §25.1) 59,169 (§6.5) 39 (§7.2) 46 50 (§7.10) 50 (§7.10) (§7.13, §5.1) 52,28 52 (§7.13)

max S min 5 00

Q

+

C i AB X

(xvx2, ..., 0

INI <x,y> |x|

d(x, y) d&S) dS

t

E

(§9.7) 68 (§9.7) 68 (§9.16, §24.2) 73,151 84 (§10.11) (§10.20) 92 93 (§10.20) (§12.22) 122 1 (§13.1) 1 •*„) (§13.1) 3 (§13.1) (§13.3, §13.17) 4, 14 4 (§13.3) 7 (§13.9) (§13.9, §13.18) 7, 15 18 (§13.20) 21 (§14.2) 31 (§15.1) 31 (§15.1)

* Numbers in italics refer to pages in Book 1: Logic, Sets and Numbers.

245

69

xeS

246

Notation

r I2 m(x, y) /(x, y)

(§20.20) (§20.21) (§21.18). (§21.18)

86 87 102 102

[ - o o , -hoc] - o o , +oo [-oo, +oo]" ( + oo, +oo)

(§24.4) (§24.4) (§24.12) (§24.12)

155 155 165 166

/ ( X ) - T I as x-+§

(§22.1)

106

lim/(x)

(§22.1)

106

lim sup/(x)

(§26.11)

187

(§22.12) (§22.12) (§22.15)

116 116 118

liminf/(x)

(§26.11)

187

%(S, %)

(§27.8, §28.15)

(§23.1)

130

(§23.5)

132

6(S, y.) Z*

(§27.8) (§27.16)

194, 209 195 197

(§28.1)

201

(§23.5)

132 (§28.19) (§28.19) (§28.19)

212 214 214

(§29.2)

220

/(x)-» I as x-+a + /(x)-» •/ as x-+bf(i+) • , / ( £ - ) /(x,y; and y lim

/(x, y)

I ak ', y) y-^n 1

c , G) * * *

(§23.12) 138 (§24.2) 151 (§24.3) 153 (§24.3, §24.11) 154,165

C[fl, C[«, b] C [a,

Z 40 ie/

fc] fe]

INDEX*

Abel's theorem (28.27) 216 absolute convergence (§28.7) 204 absolute sums (§29.17) 226 accumulation point (§18.1) 60 affine set (14.17) 30 angle (§13.3) 4 Appel(§21.2) 91 Archimedean ordered field (27.20) 200 Archimedes (§9.2, §9.16) 63, 74 associative law (§6.6) 41 axiom of the continuum (§9.5, §27.18) 67, 199 ball (§13.14, §14.3) 11,22 bijection (§6.2) 36 binomial theorem (28.31) 217 Bolyai (§13.19) 16 Bolzano-Weierstrass property (§20.7, 27.3) 78, 191 Bolzano-Weierstrass theorem (19.6, 25.24, 27.3) 70, 179, 191 boundary (§14.2) 21 boundary point (§14.2) 21 bounded sets (§9.3, 19.1) 66, 66 Brouwer (17.14, §17.18) 52, 53 Cantor (§12.20, §12.23, §24.1) 121,123, 150 Cantor intersection theorem (19.9, 20.6) 71, 78 Cantor set (§18.9) 63 cardinality (§12.2) 110 cardinal numbers (§12.25) 126 Cartesian product (§5.2) 28 Cauchy-Schwarz inequality (13.6) 5 Cauchy sequence (§27.1) 190 Chinese box property (§20.7) 78 Chinese box theorem (19.3, 27.5) 67, 191 circle (§13.14) 8 closed interval (§7.13) 57 closed set (§14.7, §21.15) 25, 100 closure (§15.1, §21.15) 31,100 cluster point (§18.1) 60

commutative law (§6.6) 41 compactification (§24.2, §24.4) 150, 154 compact interval (§7.13) 57 compact set (§19.5, §20.4, §21.15, §25.23) 69, 78, 100,179 comparison test (28.3) 202 complement (§4.4) 22 complete metric space (§20.15, §27.2) 83, 190 completion of a metric space (§27.1.6) 197 complex number (§10.20) 92 component function (§16.2) 41 components of a set (§17.21) 56 components of a vector (§13.1) 1 composition (§6.5) 39 co-ordinates (§13.1) 1 cone (13.16) 14 connected set (§17.2, §21.15) 47, 101 consistent (§10.1, §13.14) 78, 9 contiguous sets (§15.11) }A continuous (§16.2, §21.15, §22.4) 39, 101, 110 continuous at a point (§22.4) 110 continuous on the left (§22.15) 118 continuous on the right (§22.15) 118 continuous operator (§28.19) 212 continuum (§8.14, §30.11) 62, 241 continuum axiom (§9.5, §2.18) 67, 199 continuum ordered field (§9.5) 66 contraction (27.7) 193 contradiction (§2.9) 9 contradictory (§2.9) 9 cosine rule (§13.3) 4 countable set (§12.4) 112 countably compact set (§20.7) 78 convergence (§22.1) 106 convex set (§13.14) 12 cover (§20.2) 77 curve (§17.15) 52 decreasing function (§22.16) dense (§9.19, §20.7) 75, 80 disc (§13.14) 11

* Numbers in italics refer to pages in Book 1: Logic, Sets and Numbers.

247

119

248

Index

disc of convergence (§28.8) 205 discrete topology (21.14) 100 disjoint (§4.9) 23 distance (§13.11, §13.18, §23.12) 7, 15, 138 divergence (§26.1) 181 domain (§6.2, §30.14) 35, 243 double limit (§23.1) 130 element (§3.1) 14 empty set (§3.1) 15 endpoint, of interval (§7.13) 52 equivalence class (§5.5) 31 equivalence relation (§5.5) 30 Euclid (§1.8, §7.2, §8.14, §10.1, §11.1, §13.4) 5, 45, 61, 78, 99, 9 Euclidean geometry (§13.14) 8 Euclidean metric (§13.18) 15 Euclidean norm (§13.3) 4 extended real number system (§24.4) 154 extension, of a function (§6.4) 39 field (§7.3) 46 finite (§12.1) 107 finite intersection property (20.6) four colour problem (§21.2) 90 function (§6.1) 33 Gauss (§13.19) 16 Gaussian plane (§24.3) graph (§6.1) 34

interval (§7.13, §17.6) 57, 49 interval of convergence (§28.8) 205 inverse function (§6.2) 36 irrational number (§8.15, §11.1) 61, 98 isolated point (§18.1) 60 isomorphic (§9.21) 75 Jordan curve (§30.12) 242 Jordan curve theorem (30.13)

243

left-hand limit (§22.12) 116 length (§13.3) 4 limit (§22.3) 109 limit inferior (§26.11) 186 limit point (§18.1, §26.2) 60, 182 limit superior (§26.11) 186 Lindelof property (§20.7) 79 Lindelof theorem (20.10) 81 line (§13.14) 9 linear (13.16) 14 Lobachevski (§13.19) 16 locally connected set (§30.11) 242 lower limit (§26.11) 186

78

154

Haken (§21.2) 91 half-line (§13.14) 10 Hausdorff space (§30.2) 235 Heine-Borel theorem (19.8, 27.4) 71, 191 Hilbert (§13.14) 9 homeomorphic spaces (§21.15) 101 homeomorphism (§21.1, §21.15) 89, 101 homomorphism (§21.1) 89 hyperbolic geometry (§13.19) 16 hyperplane (§13.14) 13 hypersphere (§13.14) 11 identity function (§6.5) 39 image (§6.2) 34 implies (§2.10) 10 increasing function (§22.16) 119 independent (§12.26, §13.19) 127, 16 index set (§29.2) 220 inductive definition (§8.7) 56 infinite set (§12.1) 107 infinite sum (§29.2) 220 infinity (§9.16, §24.1) 73, 149 injection (§6.2) 36 inside (§30.12) 242 integers (§8.13, §11.2) 60, 100 interior (§15.1, §21.15) 31, 101 intersection (§4.7) 23

mapping (§6.1) 34 Mazurkiewicz-Moore theorem (§30.11) metric (§13.18) 15 metric space (§13.18) 15 metric subspace (§13.18, §21.9) 15, 95 monotone function (§22.16) 119

242

natural number (§8.2, §10.3) 54, 79 nested sequence (§19.2) 67 non-Euclidean geometry (§13.19) 16 norm (§13.17) 14 normal (§13.14) 12 normed vector space (§13.17) 14 north pole (§21.4) 92 nowhere dense set (§18.9) 64 one-point compactification (§24.2) 150 open ball (§14.3, §24.2) 22, 153 open covering (§20.2) 77 open interval (§7.13) 52 open set (§14.7, §21.15) 25, 100 operator (§28.19) 212 ordered field (§8.1) 54 orthogonal (§13.14) 10 orthogonal projection (§30.4) 237 oscillating function (§26.5) 186 outer product (§13.1) 3 outside (§30.12) 242 parallelogram (§13.1) 2 parallel postulate (§10.1, 13.16, §13.19) 13, 16 partial sum (§28.1) 201 Pasch's theorem (13.16) 14

78,

249

Index path (§17.15) 53 pathological (§18.9) 63 pathwise connected set (§17.18) 53 Peano (§30.11) 241 perfect set (§18.9) 63 permutation (§29.1) 218 perpendicular (§13.14) 10 plane (§13.14) 12 Poincare (§13.19) 16 point (§13.14) 9 point of accumulation (§18.1) 60 pointwise convergence (23.11) 138 polynomial (§10.23, §16.13) 96, 44 power series (§28.8) 204 product topology (§21.18) 104 projection function (16.6, §21.18) 41, 103 projection theorem (30.5) 236 punctured sphere (§21.4) 92 Pythagoras' theorem (13.15) 10

separating hyperplane (§30.4) 238 separation (§30.2) 235 sequence (§6.3, §25.1) 36, 169 sequentially compact set (§25.23) 179 series (§28.1) 201 set (§3.1) 14 simple arc (§30.12) 242 simple curve (§30.12) 242 simpy connected (§30.14) 243 space-filling curve (§30.11) 241 sphere (§13.14) 11 stereographic projection (§21.4, §24.2) 92, 151 strictly decreasing function (§22.16) 119 strictly increasing function (§22.16) 119 subsequence (§25.18) 176 subset (§4.1) 21 supporting hyperplane (§30.4) 237 surjection (§6.2) 36

quantifier (§3.4) 16

term, of a sequence (§25.1) 169 topological equivalnce (§21.1, §21.15) 89, 101 topological space (§21.15) 100 topological subspace (§21.15) 101 topology (§21.8, §21,15) 95, 100 totally bounded set (§20.15) 83 totally disconnected set (17.24) 57 triangle inequality (13.7, §13.18) 6, 15 two-point compactification (§24.4) 154

radius of convergence (§28.8) 204 range (§6.2) 35 rational function (§16.13) 44 rational number (§8.14, §11.13) 60,102 ratio test (28.4) 202 ray (§13.14) 10 real number (§7.2) 44 re-arrangement (§29.1) 218 region (§30.14) 243 relative topology (§21.9, §21.15) 96, 101 repeated limit (§23.5) 132 repeated series (§29.23) 230 Riemann sphere (§24.3) 154 Riemann's theorem (29.22) 229 right angle (§13.14) 10 right-hand limit (§22.12) 116 root (§9.13, §22.23) 72, 123 root test (28.5) 203 scalar (§13.1, §13.17) 1, 14 scalar multiplication (§13.1, §13.17) Schonflies (§30.12) 243 separated sets (§15.11) 35

1, 14

uncountable set (§12.14) 118 uniform continuity (§23.20) 145 uniform convergence (§23.11, §28.11) 138, 208 uniform distance (§23.12) 138 uniform metric (§23.12, §27.8) 138, 194 union (§4.7) 23 upper limit (§26.11) 186 vector (§13.1, §13.17) 1 vector addition (§13.1, §13.17) vector space (§13.17) 14 Weierstrass M test (28.16)

210

1, 14