Norman R. Howes
MODERN ANALYSIS AN'O
TOPOLOGY
t
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Norman R. Howes
Modern Analysis and Topology
Springer-Verlag New York Berlin Heidelberg London Paris
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Norman R. Howes Institute for Defense Analyses 180 IN. 1801 N. Beauregard Street Alexandria, VA 22311-1772 USA
Editorial Board (North America): J.H. Ewing Department of Mathematics Indiana University Bloomington, IN 47405 USA
F. W. Gehring Department of Mathematics University of Michigan Ann Arbor, MI 48109 USA
P.R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA
54D20, 54060, 54D60, 54E15, Mathematics Subject Classifications (1991): 26-02, 54-02, 54020, 28CIO, 46Exx 28A05, 28CI0, Library of Congress Cataloging-in-Publication Data Howes, Norman R. topology!/ Norman R. Howes. Modern analysis and topology em. - (Universitext) p. cm. Includes bibliographical references (p. - ) and index. ISBN 0-387-97986-7 (softcover : acid-free) . 1. Mathematical analysis. 2. Topology I. Title. QA300.H69 1995 95-3995 515' .I3-dc20 .13-dc20 Printed on acid-free paper.
© 1995 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Production managed by Bill Imbornoni; manufacturing supervised by Joe Quatela. Camera-ready copy prepared by the author. Printed and bound by R.R. Donnelley & & Sons, Harrisonburg, VA. Printed in the United States of America. 987654321 ISBN 0-387-97986-7 Springer-Verlag New York Berlin Heidelberg
In Memory of Hisahiro Tamano
Preface
The purpose of this book is to provide an integrated development of modern analysis and topology through the integrating vehicle of uniform spaces. It is intended that the material be accessible to a reader of modest background. An advanced calculus course and an introductory topology course should be adequate. But it is also intended that this book be able to take the reader from that state to the frontiers of modern analysis and topology in-so-far as they can be done within the framework of uniform spaces. Modern analysis is usually developed in the setting of metric spaces although a great deal of harmonic analysis is done on topological groups and much of offimctional functional analysis is done on various topological algebraic structures. All of these spaces are special cases of uniform spaces. Modern topology often involves spaces that are more general than uniform spaces, but the uniform spaces provide a setting general enough to investigate many of the most important ideas in modem modern topology, including the theories of Stone-Cech compactification, Hewitt Real-compactification and Tamano-Morita Paracompactification, together with the theory of rings of continuous functions, while at the same time retaining a structure rich enough to support modern modem analysis.
There is probably more material in this book than can comfortably be covered in a one year course. It is intended that a subset of the book could be used for an upper-level undergraduate course, whereas much of the full text would be suitable for a one year graduate class. The more advanced chapters are suitable for seminar work and contain many central unsolved problems suitable for a research program. The book emphasizes theory as opposed to application and does not stray far from the setting of uniform spaces, although some of the classical development of convergence theory and measure theory is done in a more abstract setting. The uniform structure is what is needed to give topological spaces enough structure to support modern analysis. The lack of coverage of more general topological spaces that are currently popular in set theoretic and geometric topology and the minimal coverage of topological algebraic structures (like topological groups, vector spaces, etc.) that are more specialized will no doubt make this book inappropriate for some types of analysis or topology courses. But for a middle-of-the-road approach that covers the central theories of modern analysis and topology, the uniform space setting has a lot to offer. It alloWS us to cut a path through this literal forest of subjects that starts out at also allows first principles, follows a common thread and arrives at the frontier, albeit in an area that not many researchers venture into. Part of the reason for this is that analysis on uniform spaces is not as easy to develop as on metric spaces, and
VIII V111
Preface
part of the reason is that uniform spaces are widely perceived to be harder than they really are. We will attempt to change that perception in this book. The book is roughly divided into two parts. The first seven chapters are essentially topology and the last five are mostly analysis with some more topology added where needed to support the development of analysis. It is not necessary to cover the chapters in order. A strictly analysis course could cover Chapters 1, I, 2 and 4 and then skip to Chapter 8 and work as far as time or choice permitted. A strictly topology course could cover just the first seven chapters. Chapters 3,6,7, 3, 6, 7, 10, and 12 could be skipped for an undergraduate introduction to modern analysis and topology, while Chapters 6, 7, 9, 10 and 12 contain advanced material suitable for research seminars.
form. There Over half the material in this text has never appeared in book fonn. is much from the recent literature of the 1980s and 1990s including answers to some unsolved unifonn uniform space questions from the 1960s, an extension of the concept of Haar measure to the Borel sets of an isogeneous unifonn uniform space (which generalizes and corrects work done during the 1940s through the 1970s), a necessary and sufficient condition for a locally compact space to have a topological group structure, and a development of uniform measures and uniform differentiation. There is also much from the theory of unifonn uniform spaces unifonn uniform paracompactness from the 1960s and 1970s including the concepts of unifonn and paracompactifications. It is the intent that there be much in this book to interest the expert as well as the novice. An attempt has been made to document the history of all the central ideas and references so historical notes are embedded in the text. These can lead the interested reader to the foundational sources where these ideas emerged. The author is indebted to Prof. Arthur Stone whose encouragement through the years has helped make this work possible, to Prof. John Mack whose discussions have shed light on several problems presented herein and to Prof. Hisahiro Tamano who taught the author most of what he knows and transferred to him the uniform space viewpoint.
Alexandria, Virginia 1995
Norman Nonnan R. Howes
CONTENTS
PREFACE INTRODUCTION: TOPOLOGICAL BACKGROUND
Vii vii xvii
CHAPTER 1: METRIC SPACES 1.1 Metric and Pseudo-Metric Spaces Distance Functions, Spheres, Topology of Pseudo-Metric Spaces, Functions. Spheres. Spaces. The Ring C*(X), Space, The Distance from a Point to a Set. Set, C*(X). Real Hilbert Space. Partitions of Unity
6
1.2 Stone's Theorem Refinements. Star Refinements and lI-Refinements, Refinements, ~-Refinements, Full Normality, Paracompactness, Shrinkable Coverings, Stone's Theorem 1.3 The Metrization ProblelTI Problem
13
Functions That Can Distinguish Points from Sets, a-Local Finiteness, Urysohn's Metrization Theorem, The Nagata-Smirnov Metrization Arhangel'skii's Theorem, Local Starrings, Arhangel' skii' s Metrization Theorem
20
1.4 Topology of Metric Spaces Complete Normality and Perfect Normality, First and Second Countable Spaces, Separable Spaces, The Diameter of a Set, The Lebesgue Number, Precompact Spaces, Countably Compact and Sequentially Compact Spaces
25
1.5 Uniform Continuity and Uniform Convergence Uniform Continuity, Uniform Homeomorphisms and Isomorphisms, Isometric Functions, Uniform Convergence
28
1.6 Completeness Convergence and Clustering of Sequences, Cauchy Sequences and Cofinally Cauchy, Sequences, Complete and Cofinally Complete Spaces, The Lebesgue Property, Borel Compactness, Regularly Bounded Metric Spaces 1.7 Completions
.
38
The Completion of a Metric Space, Uniformly Continuous Extensions
CHAPTER 2: UNIFORMITIES
43
2.1 Covering Unifonnities Uniformities
43
Normal Sequences of Coverings, Bases and Subbases Uniform Spaces, Nonna! for Uniformities, Nonna! Normal Coverings, Uniform Topology
x
Contents
Unifonn Continuity 2.2 Uniform
48
Unifonn Continuity, Uniform Unifonn Homeomorphisms, Pseudo-Metrics Uniform Determined by Normal Nonnal Sequences
2.3 Uniformizability and Complete Regularity
52
Uniformizable Unifonnizable Spaces, The Equivalence of Uniformizability and Complete Regularity, Regularly Open Sets and Coverings, Open and Closed Bases of Uniformities, Regularly Open Bases of Uniformities, Unifonnities, Universal or Fine Uniformities Unifonnities Nonnal Coverings 2.4 Normal
56
The Unique Uniformity of a Compact Hausdorff Space, Tukey's Nonnal Spaces, Star-Finite Coverings, Precise Characterization of Normal Refinements, Some Results of K. Morita, Some Corrections of Tukey' s Theorems by Morita
CHAPTER 3: TRANSFINITE SEQUENCES
62
3.1 Background
62
Unifonn Spaces 3.2 Transfinite Sequences in Uniform
63
Cauchy and Cofinally Cauchy Transfinite Sequences, A Characterization of Paracompactness in Terms Tenns of Transfinite Sequences, Unifonnity, Some Characterizations of the Lindelof LindelOf Shirota's e Uniformity, Property in Terms Tenns of Transfinite Sequences, The B p Uniformity, Unifonnity, A Characterization of Compactness in Terms Tenns of Transfinite Sequences 3.3 Transfinite Sequences and Topologies
75
Characterizations of Open and Closed Sets in Terms Tenns of Transfinite Sequences, A Characterization of the Hausdorff Property in Terms of Transfinite Sequences, Cluster Classes and the Characterization of Topologies, A Characterization of Continuity in Terms of Transfinite Sequences
CHAPTER 4: COMPLETENESS, COFINAL COMPLETENESS AND P ARACOMP ACTNESS UNIFORM PARACOMPACTNESS
83
4.1 Introduction
83
4.2 Nets
84
Convergence and Clustering of Nets, Characterizations of Open and Closed Sets in Terms Tenns of Nets, A Characterization of the Hausdorff Tenns of Nets, Subnets, A Characterization of Property in Terms
Contents
Xl
Compactness in Terms of Nets, A Characterization of Continuity in Terms of Nets, Convergence Classes and the Characterization of Topologies, Universal Nets, Characterizations of Paracompactness, the LindelOf Property and Compactness in Terms of Nets Lindelof 4.3 Completeness, Cofinal Completeness and Uniform Paracompactness
92
Cauchy and Cofinally Cauchy Nets, Completeness and Cofinal Completeness, The Lebesgue Property, Precompactness, Uniform Paracompactness 4.4 The Completion of a Uniform Space
97
Fundamental Nets, Completeness in Terms of Fundamental Nets, The Construction of the Completion with Fundamental Nets, The Uniqueness of the Completion 4.5 The Cofinal Completion or Uniform Paracompactification
103
The Topological Completion, Preparacompactness, Countable Boundedness and the LindelOf Lindelof Property, A Necessary and Sufficient Condition for a Uniform Space to Have a Paracompact Completion, A Necessary and Sufficient Condition for a Uniform Space to Have a Lindelof LindelOf Completion, The Existence of the Cofinal Completion, A Characterization of Preparacompactness
Chapter 5: FUNDAMENTAL CONSTRUCTIONS
110
5.1 Introduction
110
5.2 Limit Uniformities
111
Infimum and Supremum Topologies, Infimum and Supremum Uniformities, Projective and Inductive Limit Topologies, Projective and Inductive Limit Uniformities 5.3 Subspaces, Sums, Products and Quotients
114
Uniform Product Spaces, Uniform Subspaces, Quotient Uniform Spaces, The Uniform Sum 5.4 Hyperspaces The Hyperspace of a Uniform Space, Supercompleteness, Burdick's Characterization of Supercompleteness, Other Characterizations of Supercompleteness, Supercompleteness and Cofinal Completeness, Paracompactness and Supercompleteness
119
xii
Contents
5.5 Inverse Limits and Spectra
126
Inverse Limit Sequences, Inverse Limit Systems, Inverse Limit ofUnifonn Systems of Uniform Spaces, Morita's Weak Completion, The Spectrum Uniform Spaces, Morita's and Pasynkov's of Weakly Complete Unifonn Characterizations of Closed Subsets of Products of Metric Spaces 5.6 The Locally Fine Coreflection
133
Unifonnly Locally Uniform Unifonn Coverings, Locally Fine Unifonn Uniformly Uniform Spaces, The Derivative of a Unifonnity, Uniformity, Partially Cauchy Nets, Injective Uniform Spaces, Subfine Unifonn Uniform Spaces, The Subfine Coreflection Unifonn 5.7 Categories and Functors
146
Concrete Categories, Objects, Morphisms, Covariant Functors, Isomorphisms, Monomorphisms, Duality, Subcategories, Reflection, Coreflection
P ARACOMPACTIFICATIONS CHAPTER 6: PARACOMPACTIFICATIONS
156
6.1 Introduction
156
Some Problems ofK. Morita and H. Tamano, Topological Completion, Paracompactifications, P aracompactifications, Compactifications, Samuel Compactifications, The Stone-Cech Compactification, Uniform Paracompactifications, Tamano's Paracompactification Problem 6.2 Compactifications
159
Extensions of Open Sets, Extensions of Coverings, The Extent of a Covering, Stable Coverings, Star-Finite Partitions of Unity 6.3 Tamano's Completeness Theorem
171
The Radical of a Unifonn Uniform Space, Tamano's Completeness Theorem, Necessary and Sufficient Conditions for Topological Completeness 6.4 Points at Infinity and Tamano's Tamano' s Theorem
178
Points and Sets at Infinity, Some Characterizations of Paracompactness by Tamano, Tamano's Theorem 6.5 Paracompactifications
pX, A Solution of Completions of Unifonn Uniform Spaces as Subsets of ~X, Tamano's Paracompactification Problem, The Tamano-Morita ParaLindelOf compactification, Characterizations of Paracompactness, the Lindelof Property and Compactness in Tenns Terms of Supercompleteness, Another Necessary and Sufficient Condition for a Uniform Unifonn Space to Have a Paracompact Completion, Another Necessary and Sufficient Condition
182
Contents
Xlll xiii
LindelOf COlnpletion, Completion, The Definition for a Uniform Unifonn Space to Have a Lindelof and Existence of the Supercolnpletion Supercompletion 6.6 The Spectrum of ~X PX
192
Spectrum of ~X, pX, The Spectruln Spectrum of uX, Morita's Weak Completion The Spectruln 6.7 The Tamano-Morita Paracolnpactification Paracompactification
197
M-spaces, Perfect and Quasi-perfect Mappings, The Topological Completion of an M-space, The Tamano-Morita Paracompactification of an M-space
CHAPTER 7: REALCOMPACTIFICATIONS
202
7.1 Introduction
202
PX, Q-spaces, CZ-maximal CZ-maxilnal Families Another Characterization of pX, 7.2 Realcompact Spaces
203
Realcompact Spaces, The Hewitt Realcompactification, Characterizations of Realcompactness, Properties of Realcompact Spaces, Pseudo-metric Uniformities, The c and c* Uniformities 7.3 Realcompactifications
210
Realcompactifications, The Equivalence of uX and eX, The Uniqueness of the Hewitt Realcompactification, Characterizations of uX, Realcompactness Properties of uX, Hereditary Realcolnpactness 7.4 Realcompact Spaces and Lindelof LindelOf Spaces
217
Tamano's Characterization of Realcompact Spaces, A Necessary and LindelOf, Sufficient Condition for the Realcompactification to be Lindelof, Tamano's Characterization of Lindelof LindelOf Spaces 7.5 Shirota's Theorem
221
Measurable Cardinals, {O, I} Measures, The Relationship of Non-Zero {O,l} Measures and CZ-maximal Families, A Necessary and Sufficient {0,1} Condition for Discrete Spaces to be Realcompact, Closed Classes of Cardinals, Shirota's Theorem
CHAPTER 8: MEASURE AND INTEGRATION
229
8.1 Introduction
229
Riemann Integration, Lebesgue Integration, Measures, Invariant Integrals
xiv
Contents
8.2 Measure Rings and Algebras
230
Rings, Algebras, a-Rings, a-Algebras, Borel Sets, Baire Sets, Measures, Measure Rings, Measurable Sets, Measure Algebras, Measure Spaces, Complete Measures, The Completion of a Measure, Borel Measures, Lebesgue Measure, Baire Measures, The Lebesgue Ring, Lebesgue Measurable Sets, Finite Measures, Infinite Measures 8.3 Properties of Measures
235
Monotone Collections, Continuous from Below, Continuous from Above 8.4 Outer Measures
238
Hereditary Collections, Outer Measures, Extensions of Measures, 11* -Measurability J.l*-Measurability 8.5 Measurable Functions
243
Measurable Spaces, Spaces. Measurable Sets, Measurable Functions, Borel Functions, Limits Superior, Limits Inferior, Point-wise Limits of Functions, Simple Functions, Simple Measurable Functions 8.6 The Lebesgue Integral
249
Development of the Lebesgue Integral
8.7 Negligible Sets
256
Negligible Sets, Almost Everywhere, Complete Measures, Completion ofa Measure
8.8 Linear Functionals and Integrals
257
Linear Functionals, Positive Linear Functionals, Lower Semi-continuous, Upper Semi-continuous, Outer Regularity, Inner Regularity, Regular Measures, Almost Regular Measures, The Riesz Representation Theorem
CHAPTER 9: HAAR MEASURE IN UNIFORM SPACES
264
9.1 Introduction
264
Isogeneous Uniform Spaces, Isomorphisms, Homogeneous Spaces, Translations, Rotations, Reflections, Haar Integral, Haar Measure 9.2 Haar Integrals and Measures Development of the Haar integral on Locally Compact Isogeneous Uniform Spaces
267
Contents
9.3 Topological Groups and Uniqueness of Haar Measures
xv 271
Topological Groups, Abelian Topological Groups, Open at 0, Right Uniformity, Left Uniformity, Right Coset, Left Coset, Quotient of a Topological Group, A Necessary and Sufficient Condition for a Locally Compact Space to Have a Topological Group Structure
CHAPTER 10: UNIFORM MEASURES
284
10.1 Introduction
284
Uniform Measures, The Congruence Axiom, Loomis Contents 10.2 Prerings and Loomis Contents
285
Prerings, Hereditary Open Prerings, Loomis Contents, Uniformly Separated, Left Continuity, Invariant Loomis Contents, Zero-boundary Sets
10.3 The Haar Functions
292
The Haar Covering Function, The Haar Function, Extension of Loomis Contents to Finitely Additive Measures 10.4 Invariance and Uniqueness of Loomis Contents and Haar Measures
299
Invariance with Respect to a Uniform Covering, Invariance on Compact Spheres, Development of Loomis Contents on Suitably Restricted Uniform Spaces. 10.5 Local Compactness and Uniform Measures
304
Almost Uniform Measures, Uniform Measures, Jordan Contents, Monotone Sequences of Sets, Monotone Classes, Development of Uniform Measures on Suitably Restricted Uniform Spaces
CHAPTER II: 11: SPACES OF FUNCTIONS
317
11.1 U LP -spaces
317
LP -norm, The Essential Supremum, Essentially Conjugate Exponents, U Inequalit~ "-I6Ider's Bounded, Minkowski's Inequalit: -folder's Inequality, The Supremum LP -norm Norm, The Completion of CK(X) with Respect to the U
11.2 The Space L 2\11) (fJ-) and Hilbert Spaces Square Integrable Functions, Inner Product, Schwarz Inequality, CombinHilbert Space, Orthogonality, Orthogonal Projections, Linear COlnbinations, Linear Independence, Span, Basis of a Vector Space,
326
Contents
XVI XVi
Orthononnal Orthononnal Bases, Bessel's Inequality, RieszOrthonormal Sets, Orthonormal Fischer Theorem, Hilbert Space Isomorphism LP (Jl) (J.!) and Banach Spaces 11.3 The Space U
340
Nonned Linear Space, Banach Space, Linear Operators, Kernel of a Normed Linear Operator, Bounded Linear Operators, Dual Spaces, HahnBanach Theorem, Second Dual Space, Baire's Category Theorem, Nowhere Dense Sets, Open Mapping Theorem, Closed Graph Theorem, Uniform Boundedness Principle, Banach-Steinhaus Theorem 11.4 Unifonn Uniform Function Spaces
355
Uniformity of Pointwise Convergence, Unifonnity of Uniform Unifonn Convergence, Joint Continuity, Uniformity Unifonnity of Unifonn Convergence on Compacta, Topology of Compact Convergence, Compact-Open Topology, Joint Continuity on Compacta, Ascoli Theorem, Equicontinuity
CHAPTER 12: UNIFORM DIFFERENTIATION
370
12.1 Complex Measures
370
Measure, Total Variation, Absolute Continuity, Complex fvleasure, Concentration of a Measure on a Subset, Orthogonality of Measures 12.2 The Radon-Nikodym Derivative
373
Radon-Nikodym Derivative and its Applications 12.3 Decompositions of Measures and Complex Integration
380
Polar Decomposition, Lebesgue Decomposition, Complex Integration 12.4 The Riesz Representation Theorem
386
Regular and Almost Regular Complex Measures, The Riesz Representation Theorem
Unifonn Derivatives of Measures 12.5 Uniform
389
Differentiation of a Measure at a Point, Differentiable Measures, Measures, Unifonnly Differentiable Measures, Fubini's Theorem
L 1I -differentiable
INDEX
394
Introduction TOPOLOGICAL BACKGROUND
This book is intended for readers with a basic understanding of general topological spaces and advanced calculus on Euclidean spaces who perhaps know little or nothing about the unifonn structure these spaces may possess. uniform structure but a broad class of them, Not all topological spaces have a unifonn known as the Tychonoff spaces, do. A unifonn structure in a space can be used to measure the "nearness" of one point to another in much the same way a metric is used to measure the distance of one point to another in a metric space. Consequently, many theorems about metric spaces have their counterparts in the theory of unifonn spaces and much of the analysis perfonned on metric spaces can be generalized to unifonn spaces. The concept of a unifonnity on a space is far less restrictive than the concept of a metric as there are many unifonn spaces that are not metric. On the other hand, all metric spaces are uniform spaces. One of the purposes of this book is to indicate to the reader how much of the mathematical analysis of metric spaces can be extended to uniform spaces. In this introduction, the topological background the reader is assumed to possess will be reviewed. All the needed definitions are here but the propositions and theorems are stated without proof. Most of the proofs are elementary and can be deduced with very little effort. A few are hard. Those proofs can be found in the elegant little book Introduction to General Topology by S. T. Hu published in 1966. The reader is also assumed to have some knowledge of Set Theory including cardinal arithmetic and the most widely known equivalents of the Axiom of Choice. A development such as found in of K. Kunen's Kunen' s Set Theory, published in 1983, or the appendix of the first chapter ofK. J. L. Kelly's General Topology, published in 1955, is sufficient.
Topological Spaces There are several ways to approach the subject of topological spaces. In our approach we will use our axioms to characterize the behavior of the so-called "open" subsets of a given set X. Toward this end we let X be a set and define a topology in X to be a collection ~'t of subsets of X that satisfy the following axioms:
Introd uction Introduction
XVIll xviii
(1) (2) (3) (4)
(2) belongs to 1, The empty set 0 t, The set X belongs to 1, t, The union of a family of members of t1 belongs to 1't and 'to The intersection of finitely many members of 1't belongs to 1.
When working with topological spaces it is customary to call the members of X points and the members of 1't open sets. A set X is said to be topologized (X,1) is called a topological space or simply a by the topology 't1 and the pair (X,t) space if the topology is understood. If we wish to be as concise as possible, we can define a topology in X without using axioms 1 and 2. Both of them follow from axioms 3 and 4 since the empty union of subsets of X is 0 (2) and the empty intersection is X. But it is customary to include axioms 1 and 2 in the definition of a topology. Since topologies are sets of subsets of a given set X, we can compare topologies in the same set X by means of the inclusion relation. Let (j 0 and 1't be 0 C c 1't we say 0' 0 is coarser than 1 0. 't and that 1 t is finer than 0'. topologies in X. If (j A basis for a topology 1t is a subcollection ~ of 1't such that each open set U in 1:'t is a union of members of~. In other words, for each U in 't1 and point p E U, there is a V in ~ with p EVe E V C U. The members of ~ are called basic open sets. A sub-basis for a topology 1't in X is a subcollection 0' 0 of 1't such that the finite intersections of members of 0' 'to In other words, for each 0 form a basis for 1. V I ... Vnn such that 0', say VI· .. V U E 't1 and p E U there are finitely many members of 0, pE n ... p E VI V In' .. nVncU. n Vn cU.
The members of 0' 0 will be called sub-basic open sets. Clearly, a topology is completely determined by any given basis or sub-basis. A space is said to satisfy the second axiom of countability or simply be second countable, denoted 22', if it has a countable basis. From the definition of sub-basis, it is easily seen that a space is second countable if and only if it has a countable subbasis. 0
,
Let X be a space and p E X. N c X is said to be a neighborhood of p if there is an open set U such that p E U c N. It is an easy exercise to show: PROPOSITION 0.1 A set U in a space X is open if and only if it contains a neighborhood of each of its points.
For any point p in a space X, the collection N of all neighborhoods of p is called the neighborhood system for p. The neighborhood system of p behaves in accordance with the following proposition: PROPOSITION 0.2 Finite intersections of members of N belong to N and each set in X containing a member ofN of N belongs to N.
xix
Introduction
By a neighborhood basis or local basis for a point p we mean a collection B of neighborhoods of p such that every neighborhood of p contains a member of B. The members of B will be referred to as basic neighborhoods of p. A space is said to satisfy the first axiom of countability or simply be first countable, denoted 1I', if it has a countable local basis at each of its points. 0
,
PROPOSITION 0.3 If a space is second countable then it is also first countable.
The converse of Proposition 0.3 is clearly false since every uncountable discrete space fails to satisfy the second axiom of countability. A point is said to be an interior point of a set A if there is a neighborhood of p contained in A. p is an exterior point of A if there is a neighborhood of p that contains no point of A. p is said to be a boundary point of A if every neighborhood of p contains a point of A and a point not in A. The set Int(A) of all interior points of A is called the interior of A while the set Ext(A) of all exterior points of A is called the exterior of A. The boundary of A, denoted aA, dA, is the set of all boundary points of A. PROPOSITION 0.4 The interior of A is the largest open set contained inA.
COROLLARY 0.1 A set is open boundary points.
if and only if it contains none of its
A set in a space X is said to be closed if it contains all of its boundary points. Of course, in most spaces there is an abundance of sets that contain some, but not all of their boundary points, and consequently are neither open or closed. PROPOSITION 0.5 A set is closed
If if and only If if its complement is
open.
As one might expect, since closed sets are merely complements of open sets, rules for the behavior of closed sets under union and intersection (similar to the defining axioms for a topological space) ought to exist. The next proposition enumerates them. PROPOSITION 0.6 The closed sets of a space satisfy the following four conditions: (1) ( 1) The empty set is closed, (i.e., the underlying set) is closed, (2) The space itself itselj(i.e., (3) The intersection of afamiiy a family of closed sets is closed and (4) Any finite union of closed sets is closed.
xx
Introduction
Notice that for a space X, 0 and X are both open and closed at the same time. Again, as might be expected, the four conditions of Proposition 0.6 could have been taken as the axioms for closed sets and then the axioms for open sets could have been proved as a proposition. Consequently, one has a choice of viewpoints and approaches when dealing with topological spaces. Yet another important approach to topological spaces is by means of the so-called "closure operators." If A ec X, the closure of A, denoted CI(A), is defined to be the operators." smallest closed set containing A. Condition 3 of Proposition 0.6 guarantees the Cl(A) is the existence of a smallest closed set containing A and we see that CI(A) intersection of all closed sets containing A. CI(A) = dA = dA. PROPOSITION 0.7 For A ce X we have Cl(A) =AAu u aA =Int(A) uU aA. PROPOSITION 0.8 For any A ec X we have: (1) A is open if and only ifA if A = Int(A); =Cl(A). CI(A). (2) A is closed if and only if A =
By a closure operator in X we mean a function f that assigns to each subset A of X a subset /(A) f(A) of X such that the following four conditions are satisfied: (1) (2) (3) (4)
't
f(0) = /(0) = 0, A e/(A) cf(A) for each A eX, c X, f(f(A)) = /(A) f(A) for each A ce X and f(A u f(B) for each A, B c X. /(A U B) = f(A) u /(B) Be
f be a closure operator on the space X X and let PROPOSITION 0.9 Let Letf that f(X - A) = - A. Then: be the collection of subsets A of X such thatf(X =XX-A. (1) 't is a topology in X; of A in X with respect to 'to (2) For each A ec X,f(A) is the closure ofA
Yet another approach to topology is via the "limit point." This approach is rooted in classical analysis and will be expounded in Chapter 1 on metric spaces. For the time being, we only introduce the concept of a limit point. CI(A) PROPOSITION 0.10 A point p is in Cl(A) of p meets A. neighborhood ofp
if and only if each
CI(A) The concept of a limit point is motivated by the above proposition. If p E Cl(A) but p does not belong to A, each neighborhood of p must meet A, aA. A. so p E dA. Moreover, since p does not belong to A, each neighborhood of p must contain a point of A distinct from p. p is said to be a limit point of A if each neighborhood of p contains a point of A distinct from p.
xxi
Introduction
0.111 A set is closed PROPOSITION 0.1 limit points.
if and only if it contains all of its
As the name "limit point" implies, limit points also have to do with a limiting process similar to the process of taking limits of sequences of real numbers. These ideas will be pursued in detail in Chapters 1 and 3. A set is said to be dense in X if Cl(A) CI(A) = = X. PROPOSITION 0.12 A is dense in X X if and only that every nonempty basic open set meets A.
if XX has a
basis such
A space is said to be separable if it contains a countable subset that is dense in X. PROPOSITION 0.13 Every second countable space is separable.
Mappings Let f:X ~ Y be a function from a space X into a space Y. f is said to be U in Y containingf(p) there continuous at the point p E X if for each open set V Y in X containing p such thatf(V) thatf(Y) E V. is an open set V U. fis said to be continuous if it is continuous at every point of X. Continuous functions will also be called map,ings map!lings or maps. X into a PROPOSITION 0.14 If f:X ~ Y is a function from a space X space Y then the following statements are equivalent: (1) f:X ~ Y is a mapping, (2) lfV if U is a neighborhood off(p) thenf-1(V) then l (U) is a neighborhood ofp, of p, (V) is open in X, (3) if V U is a basic open set in Y then f- 1l (U) (4) if V U is a sub-basic open set in Y then f- 1l (V) (U) is open in X, (5) ifF if F is closed in Y then f- 1l (F) is closed in X, f(CI(A)) c Cl(f(A))for CI(f(A))for each A c X and (6) I(Cl(A)) (7) f- 1l (CI(A)) (Cl(A)) contains Cl(r Cl(f- 1l (A))for each A c Y.
r
r
r
r
r
A mapping f: X ~ Y is said to be open if the image feU) U in X f:X f(V) of each open set V Similarly,fis closed iff(F) is closed in Y for each closed F in X. If is open in Y. Similarly,/is is sometimes called bijective if it is one-to-one and surjective if it is onto. Consequently, bijective mappings have well-defined inverse functions defined on the image f(X) c Y.
Introduction
xxii XXII
PROPOSITION 0.15 For any bijective mappingf:X mappinglX ~ Y, thefollowing the following statments are equivalent: (1) f- 11 .f(X) I(X) c Y ~ X is continuous, (2) fis f is open and (3) ffis is closed.FR
r
If the mapping f:X ~ Y has a continuous inverse rl:y f- 1 :y ~ X, thenfis said to be a homeomorphism. This is equivalent to f being both bijective and surjective and satisfying one of the equivalent properties of Proposition 0.14. If h:X ~ Y is a homeomorphism then the spaces X and Y are said to be homeomorphic or topologically equivalent. A property that is preserved under homeomorphisms (i.e., if P is a property such that whenever X has property P then Y = heX) h(X) also has property P) is said to be a topological property. PROPOSITION 0.16 Iff:X ~ Y Y and g:Y ~ Z are continuous. then the composition g © f:X ~ Z is also continuous. PROPOSITION 0.17 If f:X ~ Y and g:Y then so is the composition g © f:X ~ Z.
~
Z are homeomorphisms
Letf:X ~ Y be a function from a set X into the set Y. If If'tt is a topology in X we can define a collection cr of subsets of Y as follows: U c Y is in cr if fr 11 (U) E 'to 'to It is easily verified that cr is a topology in Y. In fact, cr is the finest thatf topology in Y such that f is continuous. cr is called the topology induced by f and't and 't or simply the induced topology iff and 't are understood.
Conversely, if ~ is a topology in Y, then we can define a collection ~ in X by U E ~ if U =rf- 11 (V) for some V E ~. Then ~ is the coarsest topology in X that makes f continuous. Similarly, ~ is said to be the topology induced by f and ~ and is denotedr denotedf- 11 (~). PROPOSITION 0.18 For any mapping f:X ~ Y Y the following two statements are equivalent: (1) A c Y Y is open in Y if and only if f- 11 (A) is open in X; (2) A c Y is closed in Y if and only if f- 11 (A) is closed in X.
r
r
A continuous surjectionf:X ~ Y is said to be an identification if it satisfies the equivalent conditions (1) and (2) of Proposition 0.18.
PROPOSITION 0.19 Iff:X thenf is an identification.
~
Y is either an open or closed surjection
PROPOSITION 0.20 Iff:X ~ Y is an identification and g:Y ~ Z is a SUfficient condition for the function from Y into Z, then a necessary and sufficient continuity of g is that of the composition g © f
Introduction
xxiii XXlll
Iff:X ~ Y is a surjective function from the space X onto the set Y then as already discussed, f induces a topology in Y. When f is a surjection, this induced topology is usually called the identification topology in Y with respect tofbecause when Y Y is topologized in this manner,fbecomes an identification. On the other hand, a continuous bijection f:X ~ Y is said to be an imbedding if it satisfies the equivalent conditions (1) and (2) of Proposition 0.18. In this case, X is said to be an imbedding of X Y and f(X) is said to X into Y be imbedded in Y. Let A be a subset of a topological space (X;t). (X,t). The inclusion mapping i:A ~ X induces a topology i-I (t) in A as defined above. This topology in A is
t. When A is topologized in to.. called the relative topology in A with respect to this manner, A will be referred to as a subspace of X. It follows from the definition of i-I Cr) (.) that a set U c A is open in A if and only if there is an open set V in X with U = i-I (V) = A (nl V. Similarly, F c A is closed in A if and only if there is a closed set K in X with F = A (nl K.
PROPOSITION 0.21 A subspace S of a space X is open in X if and only if the inclusion mapping i:S ~ X X is an open mapping. Consequently, every open subset of an open subspace is open in X. PROPOSITION 0.22 A subspace S of a space X X is closed if and only if the inclusion mapping i:S ~ X is closed. Hence, every closed subset of a closed subspace S is closed in X. For any mapping f:X ~ Y the composition mapping g =f © i:S ~ Y is called the restriction off onto the subspace S of X denoted by g =II fl S or g =fs. fIi(x)] = f(x). Consequently, g is merely For every point XES we have g(x) = f1i(x)] 'cut down" to the subspace S. On the other hand, if g:S ~ Y, an the function ff'"cut = g. extension of g over X is a mappingf:X ~ Y such thatfl S =
Sums, Products and Quotients of Spaces Let F = = {X IX a Ia EE A} be a family of topological spaces such that if a =1;"# B ~ then X a (nl X J3j3 = 0. Then the family F is said to be disjoint. Let S be the union of the sets X a in F. Define a collection •t of subsets of S as follows: U c S E A. Clearly Clearly.'t satisfies the belongs to •'t if U (nl X a is open in X a for each a E (S,.) is called the defining axioms 1 through 4 of a topology. The space (S,'t) topological sum of the family F and is usually denoted by S = LX a. =IX a' Most topological sums of interest are formed from disjoint families of spaces and these are often called disjoint topological sums. In other instances, families of spaces that are not disjoint are considered to be formally disjoint for the purpose of fonning forming the disjoint diSjoint topological sum.
xxiv
Introd uction Introduction
PROPOSITION 0.23 For each a E AI a --) A, the inclusion function ia:X ia:Xa---7 F is disjoint, each i a is both open and closed. S is continuous. If IfF COROLLARY 0.2 If F is disjoint, each i a is an imbedding of X a as an open and closed subspace of S.
I X aa Ia E A}. Then P consists Let P denote the Cartesian product set P = 0 {X of all functions fA f:A --) ---7 S such that f (a) E X a for each a E A. Note that members of P are actually choice functions, so the existence of the product of an arbitrary collection of sets is dependent on the Axiom of Choice. The product topology in P is defined by defining which subsets of P are the subbasic open sets. For any a E A and open set Va V a of X a, the set V~= {fE liE Plf(a)E plf(a)E Va} will be called a sub-basic open set in P. Let cr be the collection of all sub-basic open sets in P and let 't be the smallest topology in P containing cr. Then 't is called the product topology in P. The space (P,'t) is known as the topological product of F, often denoted by nx a' a. Clearly cr is a sub-base for't. When all X then the topological product is denoted the members of F are the same space X A XA. • by X In order that the topological product of a family F be non-trivial we always P --) assume that for some a E A that X a :t= "# 0. For each a E A let p a: a:P ---7 X aa denote the function defined by p a if) (f) = f (a) for each f E P. Then p a is a surjection for each a E A. p aa will be called the canonical projection or simply the projection of P onto its a-th coordinate space X a' a. Notice that V~ = p~l (Va). P;,} (Va)' PROPOSITION 0.24 The projection P a:P --) Pa:P ---7 X a is an open mapping from P into X afor each a E A. PROPOSITION 0.25 A function f:Y --) ---7 P from a space Y into the topological product P is continuous if and only if if the composition Pa p a © f is continuousfor continuous for each a E A.
= II a:X aa --) ---7 Y aa I Now let {fa:X I a E A} be a family of mappings and let f = Ofa:P a: P --) ,fa(a ---7 Y be the Cartesian product function (Le., (i.e., f(a) = = (... ( ... ,f a(a a), ... ) where P is the topological product of the spaces X X aa and Y is the topological product of the spaces Y a. a. PROPOSITION 0.26 The Cartesian product of a family of continuous functions is continuous with respect to the product topologies.
Introduction
xxv
By a decomposition of a space X we mean a disjoint collection D of subsets of X that covers X. Then each member of X belongs to one and only one member of the decomposition. The function p:X ~ D defined by p (x) = d where d is the unique member of D containing x is called the canonical projection of X onto D. If we gi ve D the identification topology with respect to give p then D is called a decomposition space of X. If an equivalence relation R in the space X is given, then the equivalence classes of R are a decomposition of X. Let Q denote the disjoint collection of equivalence classes of R. Then the decomposition space Q obtained by topologizing Q with the identification topology with respect to the canonical Q is called the quotient space of X with respect to R and is projection of X onto Q denoted by X/R. Conversely, if we are given some identification j:X ~ Y we can define an equivalence relation R in X by defining xRy in X if j (x) = j (y) for any pair of points x,y E X. It is left as an exercise to show that R is indeed an equivalence Q be the collection of equivalence classes with respect relation in X. If we let Q to R then the quotient space Q = X/R is the decomposition space formed by taking the identification topology with respect to the canonical projection p:X = Q since the ~ R. Here, p:X ~ R is the same as the projection p:X ~ X/R = members of X/R are the decomposition (partition) induced by R. Note that the members of Q Q are precisely the sets r j-Il (y) for the points y Q by k (y) = r Y. We can define a one-to-one function k: Y ~ Q j-Il (y) for each y Y. Since both j and p are identifications, the triangle
E E
Q Q is commutative so by Proposition 0.20, k is a homeomorphism. Consequently, every identification mapping is essentially a canonical projection of a space onto its quotient space.
Separation Axioms One of the first separation properties to be studied extensively, and perhaps the most natural separation property to arise, is the so-called Hausdorff property. Two points x and y in a space X are said to be separated if there exist disjoint open sets U and V in X such that x E U and y E V. In this case the sets U and V are said to separate the points x and y. X is said to be Hausdorff if each pair of
Introduction
XXVI
distinct points can be separated. Hausdorff spaces are often said to satisfy the T 2 separation axiom and are often called T 2 spaces or simply said to be T 2. PROPOSITION 0.27 D.27 Hausdorff·
Every subspace of a Hausdorff space is
PROPOSITION 0.28 D.28 The topological sum of a disjoint collection of Hausdorff spaces is Hausdorff. PROPOSITION 0.29 D.29 The topological product of a collection of Hausdorff spaces is Hausdorff. It is easily shown that in a Hausdorff space every point is a closed set. Spaces that satisfy this property are said to be Frechet spaces or to satisfy the T 1 separation axiom. The T 1 separation axiom is usually stated in the following equivalent form: if x and y are distinct points of X, then there exist two open subsets U and V of X such that x E U and y E V and such that x does not belong to V V and y does not belong to U. Since points are closed sets in Hausdorff spaces, a natural strengthening of the T 2 separation axiom is the following: given two disjoint closed sets E and F in X, there exist disjoint open sets U and V in X such that E c U and F c V. This separation axiom is called the T 4 separation axiom, and spaces satisfying it are said to be normal. Normal spaces are said to be able to separate disjoint closed sets.
In the special case where one of the closed sets is a single point we obtain a separation axiom that lies between the T 2 and the T 4 axioms as follows: given a closed set F and a point p E X - F there exist disjoint open sets U and V in X such that p E U and F c V. This is called the T 3 separation axiom and spaces that satisfy it are said to be regular. Regular spaces are said to be able to separate points from closed sets. D.3D Every regular Frechet space is Hausdorff. PROPOSITION 0.30 THEOREM 0.1 D.1 (Urysohn's Lemma, 1925) A space X is normal if and only if if whenever E and F are disjoint closed sets in X there exists a continuous functionf:X ~ [0,11 such thatf(E) = --7 fD,l} = D0 andf(F) == 1. THEOREM 0.2 (Tietze's Extension Theorem) A space X is normal if and only if whenever E is a closed subset of X, every continuousfunctionf:E ~ continuous function f:E --7 [0,1] can be extended to a continuousfunctionf*:X --7 ~ [0,1] (i.e., afunctionf* such thatf*(x) =/(x)for =f(x) for eachxE each x E E).
On the surface, normality ~ppears to be a very natural topological property and Theorems 0.1 and 0.2 prOVide prOVIde powerful tools with which to work in nonnal normal spaces. However, even more powerful tools will not help solve some of the
Introduction
XXVll xxvii
fundamental problems involving normal nonnal spaces. It appears that normality nonnality is a property that is so "set theoretic" in nature that additional axioms for set theory are needed to settle these questions. In recent years, some fundamental questions regarding whether normal spaces satisfy certain topological properties or not have been shown to be independent of current set theoretic axioms.
J. W. Tukey in his 1940 Annals of Mathematics Studies (Princeton) monograph titled Convergence and uniformity in general topology concluded that normality nonnality might not prove to be as important a property as full normality (paracompactness). Essentially, Tukey's prediction has proven true, as we will discuss in later chapters.
By far the most useful separation axiom has proved to be an axiom that lies between regularity and normality called complete regularity although sometimes it is referred to by the rather clumsy name of T 3-l/2. 3-l/2' The so-called T 3-112 3-l/2 separation axiom is stated as follows: given a point p and a closed set E ---t [0,1] in X such that x E X - E then there exists a continuous function f:X -'t such that f(P) = 1 and f(E) = O. Completely regular T 1I spaces are called Tychonoff spaces. It is this class of spaces we will be most interested in.
Covering Properties A covering of a space X is a collection of subsets U of X such that X = u {u II u U}. An open covering is a covering consisting of open subsets and a closed covering is one consisting of closed sets. X is said to be compact if every open covering has a finite subcovering. X is said to be locally compact if each point of X is contained in a compact neighborhood. A c X is said to be covered by a collection V of subsets of X if A c u { V I V E V}. A is said to be a compact subset of X if each collection V of open subsets of X possesses a finite subcollection that also covers A. An equivalent definition of compactness of a subset A is that A is compact in the subspace topology when it is considered as a subspace of X. E
PROPOSITION 0.31 A closed subset of a compact space is compact. PROPOSITION 0.32 A compact Hausdorff space is normal.
Two useful generalizations of compactness are countable compactness and LindelOf property. A space is said to be countably compact if each the Lindelof countable open covering has a finite subcovering. It is said to be Lindelof LindelOf if each open covering has a countable subcovering. PROPOSITION 0.33 A space is compact if and only if it is both Linde/oj LindeLOf and countably compact.
XXVlll xxviii
Introduction
PROPOSITION 0.34 Every second countable space is LindelOj. Lindel6f. A family F of sets is said to have the finite intersection property if the intersection of the members of each finite subfamily of F is nonempty.
PROPOSITION 0.35 A space is compact if if and only if if every family of closed sets with the finite intersection property has a non-void intersection.
Part I: Topology
Chapter 1 METRIC SPACES
The study of metric spaces preceded the study of topological spaces. The emerging awareness of the significance of the so-called open sets in metric spaces led to the concept of a topological space. Although many properties of topological spaces that have been studied extensively are motivated by our understanding of metric spaces, there are many topological spaces that are quite different from metric spaces. When we relax the conditions on a space so that we no longer have a metric we may get some surprising (and unpleasant) properties.
Metric spaces have a wonderful uniformity or homogeneity about them. Each point in a metric space has a local base with the :~:ame arne structure as every other point in the space. Also there is a uniform measure of nearness throughout metric spaces. In subsequent chapters, we will study a class of topological spaces (namely the uniformizable spaces) that retain much of the uniform or homogeneous nature of metric spaces while at the same time being considerably more general than metric spaces.
1.1 Metric and Pseudo-Metric Spaces The concept of a pseudo-metric space is a minor abstraction of a metric space and since there are as many important examples of pseudo-metric spaces in modem analysis and topology as there are metric spaces, we may as well include them in our study since they share all the uniform properties of metric spaces. In fact, pseudo-metric spaces behave exactly like metric spaces except for the fact that they need not be Hausdorff.
By a pseudo-metric on a set X we mean a real valued function d:X x X ~ --t R such that for any points x,y,z E X we have: (1) d(x, z) $~ d(x, y) + dey, d(y, z), (2) d(x, x) = = 0, and (3) d(x, y) = =d(y, dey, x).
Pseudo-metrics are often referred to as distance functions and property (1) is called the triangle inequality for pseudo-metrics.
2
1. Metric Spaces
LEMMA 1.1 If d is a distance function on X then d(x, y) pair x, y EEX. X.
~
0 for each
Proof: By property (1), d(x, x) + d(x, y) ~ d(x, y) which implies d(x, x) ~ 2d(x,y). Since d(x, x) = 0 we have 0 ~ 2d(x, y) so that d(x, y) ~ O. •By a metric on a set X we mean a pseudo-metric d on X that satisfies an additional property:
X,d(p,q)=Oimpliesp=q. (4) foreachpairp,qE for each pair p, q E X, dCp, q) = 0 implies p = q. many interesting examples of metric spaces that arise naturally, the There are Inany most familiar of which is probably the usual metric defined on the real numbers R by d(x,y) = = I1x - y I. But it is also instructive to consider certain less interesting pseudo-metrics such as the trivial pseudo-metric d defined on an arbitrary set X by d(x,y) = 0 for every pair of points x,y E E X or the discrete = y. metric defined by d(x,y) = = 1 if xx:#:=t y and d(x,y) == 0 if x = If d:X x X ---t ~ R is a distance function on the set X then for any p in X and positive number £ > 0 the set S(p, SCp, £) called the sphere about p of radius £ or the £-sphere about p is defined as: SCp,£) S(p, £)
=
{XE XldCp,x) 0 such that S(p, SCp, £) cC U and let "t't be the collection of all open sets in X. Then
PROPOSITION 1.1 The collection "t't of open sets in a metric space X is a topology for X. 'to To show that Proof: Clearly 0 and X are open by the above definition of "t. unions of open sets are open let {U u I1 a E E A} c"t c 't and put U == uU u. u' Let p E E U. Then p EE U ~ C U for some ~ E E "t't implies the existence of an ££ > E A. Now U ~ E o such that S(p, U. But then U EE "t. SCp, £) cC U l3~ ccU. 'to
To show finite intersections of open sets are open, notice that it suffices to Un V E 't since if it can be shown for two elements, U show that if U,V E "t't then UnV and V, the result can be extended via induction to finitely many elements. To Un V E 't let p E UnV. Un V. Since U,V E 't there exists an £1 £, > 0 and an £2 > 0 show UnV with S(p, SCp, £1) E,) c U and S(p, SCp, £2) E2) c = min {£1' {EJ, £2 E2}. SCp, £) E) c C UnV Un V so C V. Let £E = }. Then S(p, UnV E 't. • 't.By a pseudo-metric space we mean a set X and a topology 't on X that is 2 defined by some pseudo-~etric pseudo-~etric d:X 2 ~ R. Such a pseudo-metric space is Instead of (X,'t). usually denoted by (X, d) instead (X;t). (X, d) will be called a metric space if d is a metric.
1.1 Metric and Pseudo-Metric Spaces
3
PROPOSITION 1.2 A pseudo-metric space is a metric space if if and only if if it is T 1• I. Proof: Suppose the pseudo-metric space (X, d) is T I. 1. Let p and q be distinct {q I c X is closed which implies X - {q} {q I is open. Since p E points of X. Then {q} X - {q} E) c C X - {q}. (q I there is an £E > 0 with S(p, £) (q I. Since q is not contained in S(P,£), d(p, q) ~ £ > O. Therefore p #::1= q implies d(p, q) > 0 so d is a metric. S(p,E), Conversely, assume (X, d) is a metric space. Let p E X. For each q in X Eq > 0 such distinct from p we have d(p, q) > O. Consequently there exists an £q Eq ). Clearly X that 0 < E£qq < d(p, q) which implies p is not contained in Seq, £q). {p} = u{S(q, £q) Eq ) Iq # p} {p 1= q::l= P I and hence X - {p} {p I is open which implies {p} {p I is closed. I. •Therefore X is T 1. There are many important examples of metric spaces (other than Euclidean n-space with which the reader is no doubt familiar). Examples of two of these spaces will be considered here. The first is the so called ring of real-valued bounded continuous functions on a topological space X which is denoted by C*(X). If f:X ~ --7 R and g:X ~ --7 R are bounded and continuous then the function f+g defined by [f+ g l(x) ](x) = =f(x) + g(x) is bounded and continuous and hence f+ g belongs to C*(X). Moreover, for any pair of real numbers a and b, it is known that the function af+bg defined by [af+bg lex) = = af(x) + bg(x) and the function fg defined by [fg l(x) ](x) = f(x)g(x) are bounded and continuous so that C*(X) does form a ring with respect to these operations of "functional addition" indeed fonn and "functional multiplication." If f and g belong to C*(X), the function f - g defined by [f - g](x) gl(x) = f(x) ~ R by g(x) is bounded so we can define the function d:C*(X)xC*(X) --7
d(f, g)
= sup{ If(x) - g(x) Ilx E X} XI
for each pair f,g E C*(X). Then for any three functionsf,g,h in C*(X) and any point x in X we have: d(g ,h) ~ If(x) - g(x) I + Ig(x) - hex) I ~ If(X) If(x) - hex) I. d(f, g) + d(g,h) Since this holds for all x in X we have
d(J,g)+d(g,h) X} = d(f,g) + d(g,h) ~ sup{lf(x)-h(x)llxE XI = d(f,h)
so that d satisfies the triangle inequality for a distance function. Clearly d also satisfies properties (2), (3) and (4) of the definition of a metric. Therefore (C*(X), d) is a metric space.
4
1. Metric Spaces
The next example is one we will meet again in later chapters. It is known as the real Hilbert space H and it has many important applications in modern analysis and topology. Let H denote the set of all sequences x = {x n } of real Ix~ converges. The number Xi in {x n } is called the numbers such that the series u~ i th Ih coordinate of x. Let x and y be any two members of H. Then for any pair of real numbers a and b and positive integers m and n, the Schwarz inequality yields:
which can be used to show that the sequences ax + by = {ax n + by n} belongs to H. If we put a = 1 and b = -1 then L(Xn - Yn)2 is a convergent series so we can don define a function d on H x H by d(x,y) = -/L(X ~L(Xnn - Yn? Yn)2. For any points x, y, Zz E H we have
because of the triangle inequality that holds for the Euclidean metric in Rn. Rn • Since this inequality holds for all positive integers n we have d(x,y) + d(y,z) ~ d(x,z) and hence the triangle inequality holds for d. Finally, we note that properties (2), (3) and (4) in the definition of a metric follow directly from the definition of d so d can be seen to be a metric on H.
turn our attention to the fact that distance functions are Next we tum continuous in the topologies they generate and certain frequently used consequences of this fact. As seen from Proposition 1.1, a distance function d:X 22 ~ R generates a topology 't on X. If we denote the product topology on X x X by 1t the statement that d is continuous in the metric topology will mean that d is continuous with respect to 1t. PROPOSITION 1.3 J.3 A distance function d:X x X continuous with respect to the topology 1t.
~
R is
form H(z) = {x E R I x < z} and Proof: Since the collection of all sets of the fonn K(z) = {x E R II z < x} is known to be a sub-basis for the topology for R it will suffice to show that d- 1i (H(z» and d- 1i (K(z» are open in X x X for any z in R. To show d- 1i (H(z» is open first note that if z ::s; ~ 0 then d- 1i (H(z» = = 0 which is 1 (p, q) in d- i (H(z». Then d(p, q) = = 8 < z for open in X so assume z > O. Pick (P, £) x Seq, e) £) which is open in X xX. x X. some 8 ~ O. Put e£ = (z - 8)/2 and let U = S(p, e) ::s; d(x, p) + d(p, q) + d(q, y) < 8 + 2£ = 8 + (z For each (x,y) in U we have d(x, y) ~ - 8) = = z. But then d(x, y) < z which implies (x,y) E d- 1i (H(z». Consequently U c d- 1i (H(z». We conclude that d- 1i (H(z» (H(z» is open in X x X. The proof that d- 1i (K(z» (K(z» is open in X x X is similar. -
1.1 Metric and Pseudo-Metric Spaces
5
When d:X 22 ~ R is a distance function and (p, q) is a point in X x X, X.. d(p,q) is called the distance between p and q. If Hand K are two non-void subsets of X the distance between Hand K is defined to be
d(H, K)
= inf{d(x, y) Ix E
Hand y E K}.
If H = {p} then d(H, K) is simply denoted by d(p, K) and is called the distance from the point p to the set K.
PROPOSITION 1.4 If d:X x X ~ R is a distance function and E is a non-void subset of X, X. then the function dE:X ~ R defined by dE(x) = d(x, d(x. E) for each x E X is continuous. Proof: Analogously to the proof of Proposition 1.3, it suffices to show that the sets dEl (H(z» (K(z» are open for each z E R. To show dEl (H(z» (H(z» and dEl (K(z» (H(z» is open (H(z» = = 0 which is open. Consequently we may ~ 0 then dEl (H(z» note that if z :S: b)/2. Since assume z > O. Let p E dEl (H(z». Then dE(P) = b 0 < z. Put ££ = (z - 0)/2. dEep) E. Let U = dE(P) = d(p, E) = b 0 there exists a point q E E with d(p,q) < 0b + £. S(P,£). Then x E U implies that S(P,E).
= d(x, d(x,E):S:d(x,q):s: d(x,p)+d(P,q) dE(x) = E) ~ d(x, q) ~ d(x,p) + d(p, q) < E£+0+£ +b+E
= z.
dEl (H(z» so that U c dEl (H(z». Consequently dE(x) < z which implies x E dEI(H(z» similar.Therefore dEl (H(z» is open. The proof that dEl (K(z» is open is similar. •
=
PROPOSITION 1.5 Let d:X x X ~ R. Then for any E c X we have d(x. E) = = O} with respect to the topology t"t generated by d. xl d(x,
CI(E) = {x E Cl(E)
Proof: By Proposition 1.4 it is clear that the set F = {x E X II d(x, E) = O} is C/(E) c closed. Since d(x, E) = = 0 for each x E E it follows that E c F and hence Cl(E) "t. F. Conversely, suppose x E F. Let N be a neighborhood of x with respect to t. Since d generates "t, t, there exists an £E > 0 such that Sex, £) E) c N. Now x in F E. But then y implies d(x, E) = 0 < E£ so there exists a y in E with d(x,y) < £. belongs to N. Consequently NnE 7= oF 0 for each neighborhood N of x. C/(E) which implies F c CI(E). C/(E).Therefore x E Cl(E) • Let (X,t) (X,"t) be a space and let P
= {fa Ia
A} be a family of mappings from X there is a neighborhood N of x such that for all but a finite number of the fa's, fa' s, fa (X - N) = = 0 and such that L {fa (x) Ia E A} = 1 then PPiscalledapartition is called a partition of unity. If U is a covering of L{fu(x)laE Al = X, P is said to be subordinate to U (denoted P < U) if for each a E A the 0 I is contained in some member of U. support of fa defined by (y {y E X IIf faa (y) > O} If for each p E X there exists a neighborhood of p that meets only finitely many members of U then U is said to be locally finite.
X to the unit interval [0,1]. If for each x
E
E
1. Metric Spaces
6
PROPOSITION 1.6 J.6 For every locally finite open covering U of a partitIOn of unity P = {Iv Ifu II U E U} on X pseudo-metric space, there exists a partition such that the support offv of ff) is precisely the open set U E U. Proof: Let U be a locally finite open covering of the pseudo-metric space fu by (X,d). For each U E U define Iv fu(x) !u(x)
=
d(x,X-U) L{d(x,X-V) I VE U) .
I
Since the denominator of the above expression is positive for each x E X, fv is X,fu L Ifu (x) I U E U} = 1I for each x E X and well defined for each U E U. Clearly L{fv(x) fu(x) > 0 if and only if x E U. Therefore P = {Iv Ifu I U E U} is the desired Iv(x) partition of unity subordinate to U.-
EXERCISES 1. Show that d(CI(A),CI(B» d(C/(A),C/(B)) pseudo-metric space (X, d).
= dCA,
B) for any pair of subsets A A and B of the
2. Show that if A A and B are cOlnpact compact subsets of the pseudo-metric space (X, d) bE = dCa, b). that there exists points a,b E X such that a E A, b E Band dCA, B) =
3. Let (X, d) be a pseudo-metric space and {Xi Ii = 1I ... n} be a finite subset of d(xj,x n ) ~ Li~ld(Xi,Xi+l). L7~id(Xi,Xi+d. X. Show that d(Xl,X 4. Two pseudo-metrics d and d' on a set X are said to be equivalent if both define the same topology for X. For each x E X let Sex, e) e) be the £) and S'(x, c:) spheres about x of radius e£ with respect to d and d' respectively. Show that d and d' are equivalent if and only if for each e c: > 0 there exists positive numbers 8 and 8' such that Sex, 8) c S'(x, £) e) and S'(x, 8') c Sex, c:). e). 5. Let (X, d) be a pseudo-metric space and x E X. Show that if e£ > 0 the closed sphere of radius e£ about x defined by {y E X I d(x,y) ~ e} c:} is a closed set in the topology generated by d. 6. A space is said to be locally compact if each point is contained in a compact neighborhood. Show that real Hilbert space H is not locally compact.
1.2 Stone's Theorem The study of topology was fairly well developed by the mid forties. A topologist of that day probably would not have guessed the development that was to take place in the next two decades. The beginning of this extremely
1.2 Stone's Theorem
7
productive era was marked by a theorem due to A. H. Stone (1948) which provided the insight and techniques for much of this major advance. The concept of full normality was studied by J. W. Tukey, who showed in 1940 that every metric space is fully normal. In 1944, J. Dieudonne introduced the important concept of paracompactness. Stone's achievement consisted of showing that full normality and paracompactness are the same thing and consequently every metric space is paracompact. Stone's Theorem elevated the importance of paracompactness. Originally investigated by Dieudonne as a generalization of compactness, para- compactness was now seen to be a silTIultaneous simultaneous generalization of both metric spaces and compact spaces. Hmetrization problem" One of the key open questions at the time was the "metrization i.e., the problem of characterizing which topological spaces have topologies Le., that are generated by a metric (which spaces are metrizable). Stone's technique ~ ~ decomposing" an open covering into a produced the important concept of "decomposing" countable family of locally finite sets, which led to the simultaneous solution of the metrization problem by J. Nagata and Y. M. Smirnov in 1951 as acknowledged by Nagata in his book Modern Dimension Theory published in 1964. It was also the basis for the proof of Shirota's celebrated theorem for uniform spaces in 1951. Furthermore, according to Nagata, "[Stone's Theorem] made an epoch not only for modern general topology but for modern dimension theory. On the foundation of the developed covering theory for metric spaces, M. Katetov in 1952 and K. Morita in 1954 independently succeeded in extending the principle spaces ... ". results of the classical dimension theory to general metric spaces...". In this section, the original version of Stone's Theorem will be given. It is customary in topology books to obtain the proof of this theorem using arguments that rely on results that were discovered later, thereby not only disguising the historical development, but more importantly, camouflaging some of the strength of full normality, which is the side of paracompactness that plays an important role in uniform spaces. Since the emphasis of this book is on concepts rather than conciseness, the original approach will be the preferred one in this particular case. A covering U of a set X is called a refinement of a covering V if for each U E U there is a V E V such that U c V. In this case we write U < V and 'proceeds to observe that < is a partial ordering (Note: in this case the ordering' ordering "proceeds the left" in the sense that U is a successor of V if U < V). If HeX, the star of H with respect to U is the union of all elements of U that meet H and is denoted by Star (H, U). Formally,
=
Star(H,U) = U{UE UIUnH#0}.
In the event H is a single point say H = {x} then Star (H, U) is simply written as
8
1. Metric Spaces
Star(x,U). The covering U* = {Star(U,U) (Star(U,U) I U E U} is called the star of U. If U* refines V then U is called a star refinement of V in which case we write U Then let Va = Va for each a> 1. Clearly U 1 = = {Valaa}. Vau{V~I~>a}. Ifb 0 a transfinite sequence {H net } of closed sets by (1.7)
H na = X - Star(X - [Va - u{Hn~ I ~ < a}], Un) eVa.
(1.8)
To show that if a"# H net and H a;t: 8 () then no Un in Un can meet both Hncr Hn/i' () < nc ' suppose 8 a and UnnH net ;t: 0 for some Un in Un. Pick X x in Un nH nHnn"a .. By (1.1) ncr "# XE
HnetimpliesxE
{XE
XIStar(x,U n) c Va-u{Hn~I~ 0 with Star (x, Un) cC Va. belong to H na nr:J.'. From (1.1) and (1.8), Star (x, Un) contains a point y such that y Vu - u{Hn~ I ~ < a}. But then y EE u{Hn~ I ~ < a} which does not belong to Va Star (Hn~ ,Un) cC V~ V~ which implies y EE Hn~ for some ~ < a. Therefore x EE Star(Hn~,Un) contradicts a being the first member of A with x in V Va. u' Thus H covers X. For each a < y and positive integer n put
(1.9)
Encr CI(Enet Clearly Hncr H net cC E ncr ) c G na ncr and since net c Cl(E meet both H we have: Hn" Hn/i na and H nB (1.10)
a"# () a;t: 0
a;t: b () a"#
implies no Un in Un can
implies no Un Gn/i' n +2 in U n n +2 meets both G net ncr and G nB .
). For each n put F n = a) Ia < y}. = u{ Cl (En (Ene.> y). To show FFnn is closed let x EE Cl (Fnn).
12
1. Metric Spaces
Then every neighborhood N of x meets some Cl (En a) and hence meets E (En,) Ena' na . Suppose N c Star(x,U n++2 ). By (1.10) N can meet at most one G na and hence at most one E Enu' Ena na which na . Therefore each neighborhood of x must meet E implies x E Cl(E CI(Ena Fn. Fnn is closed. nu ) c F n. But then F
yand positive integer n, define W Wna For each a < 'Yand na = G na - (F 1 uF 2U . . . ) where W la lu is considered to be G la. lu' Then each W Wna na is open. It will be shown that W = {W na} } is the desired locally finite open refinement of U. For na Hna CI(Ena ) for some a < 'yand this let x E X. Then x E H Y and some positive na c Cl(E integer n. Let m be the smallest positive integer for which there exists a Cl(E CI(Em~) Gm~ mB and x is not contained in F 1uF 2U . . . mB ) containing x. Then x E G uFm m --I1 so that x E W mB m ~.. Hence W covers X. Furthermore, uFn n -_ l
so W is a refinement of U. It only remains to show that W is locally finite. Hna yand positive integer n. Let x E X. Again, we have x E H na for some a < 'Yand Therefore Star(x, U n+ +3 C E c F . Consequently Star(x,U =0 ) Ena Fn. Star(x,Un+3)nWk~ n + 3 )nWkB = n 3 na k > n. Furthermore, for a given k ~ n, Star (x, U n + + + for each k> ) C U C U + n 3 k 2 for n 2 (LlO) some Un +2 in U n+2 and Uk +2 in U k+2 • Then by (1.9) and (1.10) Star(x,U + )nG i:0 for most one P in A. But then Star(x,Un+3)nWk~ i:Star(x,Un+3)nGk~ to at B Star(x,Un+3)nWk~ to n 3 kB for at most one B p in A. Hence Star (x, U n+3) is a neighborhood of x that meets at most n of the sets of W so W is locally finite. This shows X is paracompact.
o
Conversely, assume X is paracompact and U = {U a I a E A} is a locally finite open covering of X. It will suffice to show that U has a ~-refinement. By Lemma 1.2, X is normal and by Lemma 1.3, the covering U is shrinkable (since = {Val a E A} be a V = a locally finite covering is clearly point finite). Let V CI(V a ) C U a for each a E A and V is an open covering of shrink of U. Then Cl(V X. By hypothesis each x E X has an open neighborhood N(x) meeting only 0). Let B(x) = {a E finitely many of the U a's. Put A(x) = {a < 'yl 11 U anNex) i:-to 0}. CI(V a a)}. A(x) Ix E U a} and C(x) = {a E A(x) Ix is not contained in Cl(V )}. Clearly B(x)uC(x) = A(x). Now put
=
W(x) = N(x)n[n{ U a Ia E B(x)} ]n[n{ X - Cl(V a ) I a E C(x)}] N(x)n[n{UalaE B(x)}]n[n{X-CI(Va)laE
for each x E X and let W = {W(x) II x EX}. Since W(x) is an open set containing x for each x E X, W is an open covering of X. To show that W is a ~-refinement of U let Y y E X. Then there exists a V 13~ containing y. If y ~ W(x) (V~) to 0 so ~ E A(x) but is not contaIned in for some x E X then W(x)nCI (V 13) i:C(x). Thus ~ E B(x) which implies W(x) c U 13. Therefore Star(y, W) c U 13~ U~.
1.3 The Metrization Problem
13
which shows that W is a Ll-refinement ~-refinement of U. •The concept of full normality was studied by 1. J. W. Tukey in 1940 in his Annals of Mathematics Studies (Princeton) monograph titled Convergence and uniformity in general topology. In that work he introduced the notion of a normal open covering as follows: Let {Un} be a sequence of open coverings such that U n+ 1 star refines Un for each positive integer n. Then {Un} is said to be a normal sequence of open coverings. If U is an open covering such that there exists a normal sequence {V n} of open coverings with V VI1 refining U then U is said to be a normal covering. (1. W. Tukey, Tukey. 1940) Every pseudo-metric space isfully is fully THEOREM 1.2 (J.
normal. Proof: Let (X, d) be a pseudo-metric space and let U = = {U a II a < y} be an open E) c C U a for covering of X. For each x E X there exists an eE > 0 such that Sex, e) some a < y. Choose e(x) E(X) and a(x) such that 0 < £(x) E(X) < 1 and Sex, 4e(x)) 4E(X» cC U a(x)' Then V = (Sex, {S(x, £(x)) E(X» Ix EX} is an open covering of X. It will be shown that V ~-refines U. For this let y E X and consider H = (x {x lyE # Ll-refines lYE Sex, £(x))}. E(X»}. Now H ::j; since y E E H such that e(z) E(Z) > 2/3sup{£(x) 2/3sup{E(X) Ix E E H. Therefore choose ZZ E E H}.
o
Then if x E H we have:
S (x, £(x)) cC S (Y, 2e(x)) cC S (Y, 3£(z)) cC S (z, 4£(z)). S(X,E(X») S(y,2E(X)) S(y,3E(Z» S(Z,4E(Z»). Therefore Star (Y, V) c S (z, 4£(z)) Star(y, 4E(Z» c U a(z)' a(z)' Consequently V ~ Ll U. Let U 1 = = U. By successive applications of the above argument it is possible to construct a sequence {Un} of open coverings of X such that U n+ 1 Ll each n. But ~ Un for ea~h then {U 2n } is a normal sequence and hence U is normal. Therefore every open covering in a pseudo-metric space is normal which means every pseudo-metric normal.space is fully normal. • COROLLARY 1.1 paracompact.
(A.
H. Stone) Every pseudo-metric space is
EXERCISE 1. Let {U a} be a covering of a space X. Another covering {Va} with the same Vaa C U a fqr each a. index set is said to be a precise refinement of {U a} if V U ex} Show that if {{U a} has a point finite (locally finite) refinement {W {w P ~ }~, I"~ th'en there U a}. a }. exists a precise point finite (locally finite) refinement {Vex} {Va} of {{U
1.3 The Metrization Problem The metrization problem is the problem of detennining determining which topological spaces have topologies that are generated by metrics (pseudo-metrics). It dates
14
1. Metric Spaces
back to the earliest studies of topology. One of the early researchers of the problem was P. Urysohn whose metrization theorem will be proved in this section. Urysohn' Urysohn'ss Metrization Theorem states that regular second countable spaces are metrizable. Urysohn's Urysohn' s Metrization Theorem is not a solution to the metrization problem because it is not both necessary and sufficient. Although there are a number of known solutions to the mctrization problem, many topologers consider the problem to be "essentially solved" solved" by Nagata-Smimov the Nagata-Smimov Nagata-Smirnov Metrization Theorem. Certainly the Nagata-Smirnov solution is the most successful and widely known solution. None-the-Iess, new solutions appear from time to time. There is, however, something unsatisfying 'defines" the about the Nagata-Smirnov topology "defines" Nagata-Smimov solution. Since a basis for a topology' topology, placing a metrization requirements on the basis is closely akin to defining the space to be metric. One would like a characterization of metrizable metrizablc terms of something like a covering or separation property that does spaces in tenns not explicitly involve a basis. In Chapter 2 it will be shown that uniform spaces can be characterized by a separation property. The purpose of this section is not to give a comprehensive treatment of the metrization problem but to give a representative sample that will provide motivation for our study of unifonn uniform spaces in later chapters. Following the historical precedent we will first establish the Urysohn Metrization Theorem. Two lemmas are needed for this. LEMMA 1.4 A space is metrizable (pseudo-metrizable) if and only there exists an imbedding of the space into a metric (pseudo-metric) space.
if
Proof: Assume X is metrizable and that d is a metric that generates the topology of X. Let Y denote this metric space. Then the identity mapping i:X ~ Y is a homeomorphism and hence an imbedding of X into the metric space Y.
Conversely, let f:X ~ Y be an imbedding of the space X into the metric space Y. Let d denote the metric for Y. Define the function d':X x X ~ R by d'(x,y) = = d[f(x)f(y)] d[f(x)j(y)] for each pair of points x and y in X. That d' is a metric that generates the topology of X is left as an exercise (Exercise 1).COROLLARY 1.2 Every subspace of a metric (pseudo-metric) space is metrizable (pseudo-metrizable). In particular, the closed unit interval I = [0,1] is metrizable. LEMMA 1.5 The topological product of a countable family of metric (pseudo-metric) spaces is metrizable (pseudo-metrizable). Proof: It will be shown that the topological product of a finite number of metric x ..., .· X XXnn is metrizable. For each i == 1I ... n let dd;i be the spaces X =X 1I X X 2 X
1.3 The Metrization Problem
15
metric for Xi. Define d:X x X
~
R by
(1.11) for each pair of points x = (x 1 ••• xn) x n) and Yy = (y 1 ••• Yn) in X. That d is a metric that generates the topology of X is left as an exercise (Exercise 2(a)). 2(a». This shows X is metrizable. Next it will be shown that the topological product X of a sequence {X {Xn} n } of metric spaces is metrizable. For each positive integer n let d n denote the metric for X n • First we want to show that for each n it is possible to construct a new Ynn in X n. that In (xn,y n) ~ S; 1 for each pair of points X Xnn and Y metric In from dn such thatln(xn,Yn) For this let
(1.12) for each pair of points XXnn and Yn in X n. Thatlnn is a metric for X Xnn that generates Xnn is left as an exercise (Exercise 2(b». 2(b)). We can now define a the topology of X metric for X by
(1.13) for any pair of points x = = (x 11,, X 2, . . . ) and Y == (y 11,, Y 2, . . . ) of X. That d is a metric that defines the topology of X is left as an exercise (Exercise 2(c».The subspace I( (J) w) of real Hilbert space H (defined in Section 1.1) consists {xn} S; X Xnn S; } of all points {x such that 0 ~ ~ lin is known as the Hilbert Cube. It can be n shown (Exercise 3) that I(co) I(w) is homeomorphic to the countable product
°
/N=n{/ n ln=I,2,3 ... } andl n =[O,l] foreachn of unit intervals under the homeomorphism 1:I(w) 1:I(co) {Xj, X2,X3 .•• ... }. {Xl,X2,X3 }.
~
IN defined by I( {x n }) I({x
=
COROLLARY 1.3 J.3 The Hilbert Cube is metrizable.
Let F be a family {fa:X {I a:X ~ Y aa Ia < y} of mappings from a space X into spaces Y a for each a < y. Let Y denote the topological product of the Y a}· X ~ Y denote the product mapping defined by famil y {{Y family f:X a }. Let f:
= If a(x) for each x E
[f(X)]a =
X and a < y.
Then If is a continuous mapping from X into Y. The family F is able to distinguish points of X if for any two distinct points x and Y of X there exists a (x) =I: #- ffa(y). a(y). F is able to distinguish points from closed sets if an a < y with Ifa(x)
16
1. Metric Spaces
(P) is for any closed set K in X and point p in X - K there is an aex < y such that ffau(p) a(K». not contained in Cl(f CI(fu(K».
LEMMA 1.6 If the family r can distinguish points of X and can ~ Y is an distinguish points from closed sets, then the product mapping f:X -7 imbedding. Proof: Assume F can distinguish points and can distinguish points from closed "F ffau(y) sets. If x and y are distinct points of X then there is an aex < y with ffau(x) =t= (y) which implies f(x) "F =t= f(Y). f(y). Therefore f is a one-to-one mapping. Let U be an open subset of X. It will be shown that feU) is open in Y. For this let p EE U and let q = f(P). Since X - U is closed and p does not belong to X - U, there is an aex such thatf a(P) does not belong to Cl(f a(X - U». Let < yysuch thatfu(P) Cl(fu(X
V = = {y (y EE Y I Ya Yu is not contained in Cl(fa(X u(X - U»}. Then V is a basic open set in Y which implies Vnf(X) is open in f(X). Now q = = f (P) EE Vnf(X) c feU). It follows that feU) is open in f(X). Therefore f is a one-to-one continuous open mapping and hence an imbedding. •THEOREM 1.3 J.3 (Urysohn's Imbedding Theorem) A regular T T]1 space with a countable basis can be imbedded as a subspace of the Hilbert Cube.
I(w) is homeomorphic to IN, it Proof: In view of Lemma 1.6 and the fact that 1(00) will suffice to show the existence of a countable family F of continuous functions from X into the unit interval I = = [0,1] which can distinguish points from closed sets. For this let B be a countable basis for X and let C be the C/(V) cU. subset of B x B which consists of pairs of open sets (U,V) such that CI(V) c U. Then C is countable. By Urysohn's Lemma (Theorem 0.1) we can then obtain a = {fcu,v):X {f(U,v):X -7 ~ I I (U,V) EE C} of continuous functions which countable family F = map CI(V) into zero and X - U into one. To show that F can distinguish points from closed sets let p EE X - K where K is closed in X. Since X is regular and B is a basis, it is possible to find U and V in B such thatp that p EVe CI(V) C I(V) cUe X - K. Thenfcu,v)(P) Then f(U,v) (P) = = 0 andf(u,v)(K) and f(U,vl (K) = = 1. Consequently F can distinguish points from closed sets. -
°
T]1 THEOREM 1.4 J.4 (Urysohn's Metrization Theorem) Every regular T space with a countable basis is metrizable metrizable.. Proof: By Theorem 1.3, X can be imbedded as a subspace of the Hilbert Cube. lA, X is By Corollary 1.3 the Hilbert Cube is metrizable. Hence by Lemma 1.4, metrizable. •~etrization Theorem was discovered independently The Nagata-Smirnov Nagata-Smimov ~etrization by J. Nagata and Y. M. Smrrnov. The first to appear was Nagata's paper in the
1.3 The Metrization Problem
17
Journal of the Polytechnical Poly technical Institute at Osaka City University in 1950, Volume 1, pp. 93-100. Smirnov's paper appeared in 1951 in Doklady Akad. Nauk S.S.S.R.N.S. Volume 77, pp. 197-200. In the proof of Stone's Theorem it was shown that if an open covering U is normal then an associated normal sequence {Un} could be used to decompose U into a countable sequence of families {G n a.rxl a < y} for n = 1, 2, 3 ... where {G mrx } is locally finite. A family F of for each positive integer m, the family {Gma.} X is said to be a-locally finite if it is the union of countably subsets of a space X many locally finite subfamilies. Using this terminology, the proof of Stone's Theorem shows that full normality implies each open covering has a a-locally finite open refinement. . 1
With this in mind, it is natural to wonder what a space would be like if it had a a-locally finite basis. Clearly metric spaces have a a-locally finite basis Bnn = {S(x, 2- nn)lx E X} is the covering for if (X, d) is a metric space and B nn consisting of the spheres of radius 2- for some positive integer n, then since X X is paracompact (Corollary 1.1) there is a locally finite open refinement Un of Bn. u{U B Un}. n }. Then U is a a-locally finite collection of open sets and n. Put U = u{ } since B = u{B is a basis, so is U. The Nagata-Smirnov Theorem shows that u{Bn} n among regular T 1 spaces, the spaces with the a-locally finite bases are precisely the metrizable ones.
THEOREM 1.5 J.5 (Nagata-Smirnov Metrization Theorem) A T 1 space is metrizable if if and only if if it is regular and has a a-locally finite basis. Proof: Let X X be a regular space with a a-locally finite basis B = {B {Bn} n } where each B K be disjoint closed Bnn is locally finite. To show X is normal let H and K subsets of X. Since X is regular, if p E H there is an open set W(P) in B such CI(W(P)) c X-H. Define the open coverings U and V V of H that p E W(P) c Cl(W(P)) (W(P) Ip Ip E H} and V = {W(q) (W(q) IIq E K}. For each and K respectively by U = {W(P) UnBnn and Vn positive integer n put Un = UnB Vn = VnB n. Since Un and Vn Vn are locally finite collections we have: Cl(U Cl(Un) n)
= u{ Cl(U) I U E u{CI(U)IU
Un}
C
X-K K and
Cl(V Cl(Vn) = u{Cl(V)IVE u{ CI(V) IV E V nn}} eX-H. C X-H. n) =
For each positive integer n, define the open sets
WmnVnn is empty whenever m ~ n it follows that WmnG nn = Since WmnV = 0 whenever m ~ n we have WmnG nn = 0 m ~ n. Similarly, since UmnG n = 0 whenever m::; n = W mnGn = whenever m::; m ~ n. Hence WmnGn = 0 for all positive integers m and n. Now put
18
1. Metric Spaces
=
=
=
}. ClearlyHcW,KcGandWnG=0. W=u{Wn}andG=u{G W U{W n } andG u{G nn }. Clearly HeW, KcGand WnG 0. ThusXis Thus X is
normal. nonnal. Next it will be shown that there exists a metric for X that generates the topology t of X. For this let N denote the positive integers and A = N x N. Let Bnn I CI(V) c V}. a = (m,n) be a member of A. For each V U in Bm B m let W = u{V E B U}. Since B U. By Urysohn's Bnn is locally finite CI(W) c V. Vrysohn's Lemma (Theorem 0.1) there exists a continuous!u:X that!u(X - V) U) = 0 andfu(CI(W)) = continuousfu:X ~ [0,1] such thatfu(X = 1. Next, define the pseudo-metric d aa for X by
Bm for any two points x and y of X. Since B m is locally finite the summation is finite, so d a is well defined. For each a < y, (X, d a) a ) is a pseudo-metric space a:(X,"t) ~ which generates some topology t"ta for X. The identity function i a:(X,'t) (X,"ta ) can be shown to be continuous by showing that S (x 0,£) 0 ,f) E 't t for any x 0 in (X,t =fu(xo),gu(Y) = = leo Ico -!u(y) -fu(y)1I andF(y)=L{gu(y)IUE andF(y)=L{gu(y)IVE B Bm}· X. For this put Co =!u(Xo),gu(Y) m }· Since fu is continuous for each V Bm Since!u U in B m, so is guo Since B m is locally finite F:X ~ R is also continuous. Now S(xo, f) £) = £}. Consequently = {y E YIF(y) yIF(y) < f}. 0 ,f) E t which shows that ii a is continuous. Next let Sex 0,£)
Clearly F can distinguish points of X. To show that F can distinguish points from closed sets let K be a closed subset of X and p a point in X - K. Since X is regular, there are basic open sets V U and V such that p EVe CI(V) eVe cUe X K. Let m and n be positive integers such that V UE B Bmm and V E B Bn. = n • Then a = (m,n) E A and da(P, K) ~ 1. By Proposition 1.5, ia(P) is not in the closure of i a(K). Thus F can distinguish points from closed sets.
By Lemma 1.6, the Cartesian product n:X ~ n(X, d a ) of the family F is an imbedding of X into the topological product of the (X, da)'s. Since A is countable, it follows from Lemma 1.5 that this product is pseudo-metrizable and by Corollary 1.2 this implies X is pseudo-metrizable. Since X is T 1 it is metrizable. It has already been shown that if X is metrizable then it has a a-locally finite basis. That X is regular follows from the fact that every metric space is paracompact (Corollary 1.1) and hence normal (Lemma 1.2).-
COROLLARY 1.4 A space is pseudo-metrizable regular and has a a-locally finite basis.
if and only if it is
The remainder of this section will be devoted to another solution to the metrization problem that does not require regularity in the hypothesis and does not state the metrization criteria in terms tenns of a basis. Let U be an open covering
1.3 The Metrization Problem
19
of the space X. A sequence {Un} of open coverings of X is said to be locally starring for U if for each x E X there is a neighborhood N of x and a positive integer n such that Star (N, Un) C U for some U E U. i) A T 1 space X is metrizable if and THEOREM 1.6 J.6 (A. V Arhangel' ski ski[) only if if there exists a sequence {Vn} of open coverings that is locally starring for all open coverings ofX.
Proof: First suppose (X, d) is a metric space. Let n be a positive integer and n) Ix EX}. Then u {B n} is a basis for the topology generated put B n = {S(x, 2-n by d. Let U be an open covering of X and pick U E U such that x E U. Then there is a positive integer m such that Sex, 2- mm ) c U. Now
so that {B n} is locally starring for all open coverings of X. Conversely, assume X X is T 1 and there exists a sequence {V n} of open coverings of X that is locally starring for all open coverings of X. It will first be shown that X is paracompact. Let {Un} be the sequence of open coverings U 1 = VI and Un = {V In ... nV nVnn IVi E Vi for i = 1I ... n}. Clearly defined by U {Un} is also locally starring for all open coverings of X X and U n+ 1 refines Un for each positive integer n. Let U be an open covering of X and put
= {W c X IW is open and for some positive integer n, W cC Un for some W = Un E Un and Star(W, Un) C U for some U E U}. We For each W E E W let neW) be the least positive integer such that W c Un(W) for some Un(W) in Un(w)andStar(W, Un(W) and Star(W, Un(w)cUforsomeUE Un(W) c U for some U E U. Since {Un} is someUn(W) locally starring for U, W is an open covering of X. It will be shown that W Ll-refines U. Let x EE X X and put n(y) = min {n(W) lYE W E W} and let V E W d-refines V E such that n(V) = n(x). For each W E W containing x it is clear that neW) ~ n(x). Thus
Star(x,W) c u{Star(x,U u{Star(x, Un) n) I n~n(y)}. But since Unn+ 1 refines Un for each positive integer n, we have
Star (x, W) c Star(x,Un(x) Star(x,Un(x) c Star(V, Un(V)
c U
for some U E U. Therefore W ~-refines Ll-refines U and hence X is fully nonnal. normal. By Stone's Theorem, X is paracompact. Bnn be a locally finite open refinement of Un. It will be For each n let B shown that B = u{B u{Bn} n } is a basis for the topology of X. For this let A be an open subset of X and let x E A. Let W be an open set such that x EWe Cl(W) CI(W) c A
1. Metric Spaces
20
and put B = X - CI(W). Since {U} {U I is locally starring for the covering {A, B} BI and x is not contained in B, there is a neighborhood N of x and a positive integer n such that Star(N, Un) C A. But since Bn B n refines Un we have: X E
Star(x, B n) c Star(x, Un)
C
Star(N, Un)
C
A.
Bnn is a locally finite Therefore B is a basis for the topology of X. Since each B collection, B is a a-locally finite basis. Consequently, by the Nagata-Smirnov Theorem X is metrizable. •-
EXERCISES 1. Complete the proof of Lemma 1.4 by showing d' is a metric. 2. Complete the proof of Lemma 1.5 by showing:
= (a) d defined by (1.11) is a metric that generates the topology of X = Xl x ... xX , xXn' n functionfn (b) the function in defined by (1.12) is a metric that generates the topology of X Xnn and (c) the function d defined by (1.13) is a metric that generates the product topology of X. 3. Show that the Hilbert Cube l(w) I(w) is homeomorphic to a countable product of unit intervals.
1.4 Topology of Metric Spaces This section is devoted to the behavior of metric (pseudo-metric) topologies. From Section 1.2 we know metric spaces are paracompact. We also know paracompact spaces are normal. Metric spaces also satisfy a variety of other separation properties such as being completely normal, perfectly normal and collectionwise collection wise normal. The verification that this is the case is left as an exercise (Exercise 2(a) through 2(c». The definitions of these properties are given below. A space is said to be completely normal if every subspace is normal. An alternate definition is that for each pair of subsets A and B such that A nel nCI (B) (8) = (A )nB there exist open sets U and V containing A and B respectively o == Cl (A)nB with Un V = 0. That these definitions are equivalent is left as an exercise H is called aGo (pronounced "G-delta") if it is the (Exercise 1). A set H intersection of a countable collection of open sets. A space is said to be normal and every closed subset is aGo. A collection F perfectly normal if it is nonnal = = {F a Ia E A} of subsets is said to be discrete if each point of the space is con-
1.4 Topology of Metric Spaces
21
tained in a neighborhood that meets at most one of the members of F. A space is said to be collectionwise normal if for each discrete collection F F = {F a II aex E = {G a Iaex E A} of closed sets with uF being closed, there exists a collection G = A} of disjoint open sets with F Faa eGa for each a ex E A. Clearly all pseudo-metrizable spaces are first countable since for each n point x, the sequence {Sex, 22~n)} )} is a local basis for x. Recall from Section 1.1 that discrete spaces are metrizable and consequently there exist metric spaces that are not second countable or separable. One of the distinct features of metric spaces is that the concepts of second countability and separability are equivalent in metric spaces. THEOREM 1.7 J.7 A metric (pseudo-metric) space is second countable if and only if If it is separable.
Proof: Assume (X, d) is a second countable metric space and let B = {B {Bn} n } be a countable basis for X. For each positive integer n pick Xn Xn E B Bn. n • Clearly the sequence {x n } is countable and dense in X. Therefore X is separable. Conversely, assume (X, d) is separable and let K be a countable dense subset. Put B = {Sex, {S(x, 2- nn ))lx Ix E K and n is a positive integer}. Then B is countable. It will be shown that B is a basis for the topology of X. For this let U be open in X and let p E U. Then there exists an E > 0 such that S(p, E) c U. m 2(2~m)) < E. Since K is dense in X there Let m be a positive integer such that 0 < 2(2m m Seq, 2~m). is a q E K such that q E S(P,2- mm).). Then p E S(q,2). To show Seq, 2- m)) c U m m m m 2~m). let x E Seq, 2). Then d(p, x) ~ d(p, q) + d(q, x) < 2- m + 2- m = 2(2- m)) < E. m m Therefore x EE S(P, C U which shows S(p, E) CC U. Consequently Seq, 2- )) cC S(P, S(p, £) E) c that B is a basis for the topology of X.By Exercise 6 of Section 1.1 we see that a metric space may fail to be locally compact. This is one way in which an arbitrary metric space differs from Euclidean n-space. However, many of the properties of En are preserved in arbitrary metric spaces. For a non-empty subset E of a metric space (X, d) we define the diameter 8(E) B(E) by
8(E) SeE)
= sup {d(x,y) Ix,y E
E}.
If the diameter 8(E) B(E) of E is finite, E is said to be bounded. A subset K of En is compact if and only if it is closed and bounded. In an arbitrary metric space we have: THEOREM J.8 Every compact subset of a metric space is closed and
bounded. Proof: Let K be a compact subset of the metric space (X, d). Suppose p E X X-K. Then for each q E K there exists open sets U(q) and V(q) containing p and q
22
1. Metric Spaces
respectively such that U(q)n U(q)nV(q) V(q) = 0. Then V = {V(q) II q E K} is a covering of K and therefore has a finite subcovering {V(qJ {V(qi) II i = 1 ... n}. Then U = ,,7=1 U(qi) is a neighborhood of p such that UnK == 0. Therefore p cannot be a "£=1 limit point of K which implies K is closed. {Sex'!) Ix Ix E K}. Then U is To show K is bounded consider the family U = {S(x,l) 1, 1) an open covering of K and as such has a finite subcovering say {S(x I, l) ... l)}. Let k = = max{d(xi, xj)li,j:S; mi. If x,y E K then there exist i and j S(x m , 1)}. Xj) Ii,j ~ m}. S(Xj,!). Then by the triangle inequality we have such that x E S(xi,l) and y E S(xj,l). d(x,y) :s; d(x,xI) + d(Xi,Xj) + d(xj,y) :s; 1 + k + 1 = k + 2. ~ k + 2. Therefore Since x and y were chosen arbitrarily, we conclude that S(K) 3(K) :s; K is bounded. -
THEOREM 1.9 (Lebesgue Covering Theorem) For each open covering U of a compact metric (pseudo-metric) space (X, d) there exists a positive A. such that every subset E of X X with 3(E) 'A. is contained in some number 'A S(E) < A member of U. Proof: Since X is compact there is a finite subcovering {V 1 ..• • •• V n} n } of U. For each i = fi(x) = UJ. By = 1 ... n define a function fi:X ~ R by Ji(x) = d(x, X - Vi)' Proposition 1.4, each fi is continuous and hence the function f:X ~ R defined L£=1fi(x) is continuous. Moreover, f(x) > 0 for each x. Since X is by f(x) = L7=IJi(X) compact, so is f(X). Therefore there is a positive number p such thatf(x) that f(x) ~ p for A. = L7=1!i(Y) = = n 'A. A.. each x E X. Let 'A = p In and let y E X. Then L'!=lfi(Y) = f(y) ~ p = Therefore there is an i such that !i(Y) fi(y) ~ 'A. VJ ~ 'A A.. But then d(y, X - Vi) A. which A.) C Vi. Vi' Let E cC X such that S(E) 3(E) < A. 'A.. Pick e E E and let x be an implies S(y, 'A) arbitrary point of E. S(E) 3(E) < 'A A. implies d(e, dee, x) < A. 'A.. Thus x E S(e, See, 'A) A.) which Ui for some i = implies E c S(e, See, 'A) A.) c Vi = 1 ... n.-
A. in Theorem 1.9 is said to be a Lebesgue number The positive number 'A for the covering U. A subset K of a metric space is said to be precompact if for each £ > 0 there is a finite F c K such that K c u{S(x,£) II x E F}. The proofs of the following propositions are left as exercises (Exercises 6 and 7). PROPOSITION 1.8 Every compact subset of a metric (pseudo-metric) compact subset is bounded. space is precompact and every pre precompact J.9 A precompact PROPOSITION 1.9 pre compact metric (pseudo-metric) space is separable. COROLLARY 1.5 J.5 separable.
A compact metric (pseudo-metric) space is
1.4 Topology of Metric Spaces
23
There are several generalizations of compactness besides paracompactness and local compactness that have been studied extensively such as countable compactness, sequential compactness, precompactness and pseudocompactness. Some of these (e.g., countable compactness and sequential compactness) are equivalent to compactness in metric (pseudo-metric) spaces. A space X is said to be countably compact if every countable open covering has a finite subcovering. It is said to be sequentially compact if every sequence has a convergent subsequence. Sequential compactness is not a generalization of compactness as the name seems to imply, for there are compact spaces that fail to be sequentially compact. THEOREM 1.10 For an arbitrary metric space X X the following statements are equivalent: (1) X is compact, (2) X is countably compact and (3) X is sequentially compact. Proof: Clearly (a) ~ (b). To show (b) ~ (c) let (X, d) be a countably compact metric space and let {x {x nn I} has no limit point. {xnn I} be a sequence in X. Suppose IX } } Then {x is so X-{x is open. For each Xi E {x IX n I closed X-Ixn I {Xn}, n n }, Xi is not a limit n point of {X {Xnn I} so there is an open set U(xJ U(Xi) containing Xi but no other point of {x n }. Then U = (X-{x IX (X-{xnn })u{U(xJ} })ulU(xJ I is a countable open covering of X with no n I. finite subcovering, which is a contradiction, so IX {x n I} has a limit point after all.
Let p be a limit point of IX {x nn I. }. We will construct a subsequence of IX {x n n I} that converges to p. Since X is metric, the family {S(P, 2-nn ))}I is a local basis for p. Let x m1 {xnn I} contained in S(p, 2- 11). For each positive m1 be the first element of {x {xnn I} greater than XXmk integer k > 1 let x mk mk be the first element of {x mk _ 1 1 that belongs to S(p, 2- kk ). From this definition of {x mn } it is easily seen that ) {x mn }) converges mn mn to p. Consequently X is sequentially compact. ~ (a) let (X, d) be a sequentially compact metric space. We To show (c) -7 first show X is countably compact. Suppose it is not. Then there is a countable open covering U = I{ Un} Un I with no finite subcovering. For each positive integer n Um I m ~ n }. ). Let p E X. Since U pick XXnn E X such that XXnn is not contained in u {{ Urn covers X there is a positive integer n such that thatpp E Un. Un" By the definition of the sequence IX {x nn I} it is clear that Un can contain at most finitely many members of {X {x n IXnn }. I. Hence no subsequence of IX n I} can converge to p. But this means X is not sequentially compact which is a contradiction. Therefore X is countably compact.
Next we will show X is precompact. Suppose it is not. Then there is an E£ £)Ix > 0 such that no finite F c X satisfies X = = uu{S(x, {S(x, E) I X E Fl. F}. Therefore it is possible to construct a sequence {x {xnn I} in X of distinct points such that IXn = {Xi} {Xi I for each positive integer i. Define the open set G and the {x }nS(Xi' £) E) = n InS(xi, closed set H by bv
24
1. Metric Spaces
= u{S(x,E/2)lxE u{S(x,£/2)lxE H}. H =X-u{S(x =X-u{S(Xn,E)} n,£)} and G = Then G = Gu{S(xnn,, £)} is a countable open covering with no finite subcovering which is a contradiction. Therefore X must be precompact. By Proposition 1.9, X is separable and hence by Theorem 1.7, second countable. But then X is Lindelof Lindelbf and a countably compact Lindelof Lindelbf space is compact so (c) ~ (a).Theorem 1.10 hints at the role sequences play in metric and pseudo-metric spaces. From it we see that a metric space is compact if and only if each sequence has a convergent subsequence. What is more striking is the fact that all topological properties in metric spaces can be determined by the behavior of the sequences in the space. In fact, the topology itself can be characterized in terms of which sequences converge to which points. PROPOSITION 1.10 A set U in a pseudo-metric space (X, d) is open if and only if no sequence in X - U converges to a point of U. Proof: Assume U is open. If {x {xn} {xn} n } is a sequence in X - U then {x n } cannot be eventually in U and consequently cannot converge to any point in U since U is open. Conversely, assume no sequence in X - U can converge to a point of U and suppose U is not open. Then there is a p in U such that each neighborhood n S(p,2-n)n(X-U). of p meets X - U . For each positive integer n let XXnn E S(p,2)n(X-U). Then n n {x n } C X - U. But clearly {x {x n } {xn} {xn} n } is eventually in S(p, 2- )) for each n so {xn} converges to p which is a contradiction. Therefore U must be open after all. -
PROPOSITION 1.11 1.1 1 A point p in a pseudo-metric space (X, d) belongs to the closure of a set F c X if and only if if there is a sequence in F that converges to p. The proof of this proposition and the next are left as exercises (Exercises 9 and 10).
COROLLARY 1.6 A set F in a pseudo-metric space (X, d) is closed if and only iffor lffor each ppin in F there is a sequence in F that converges to p. PROPOSITION 1.12 A function f:X ~ Y from a pseudo-metric space (X, d) to a pseudo-metric space (Y, p) is continuous if if and only if if for each sequence {x )} in Y {xn} (f(x nn )} n } C X that converges to some p E X, the sequence {f(x converges to f(p). EXERCISES 1. Show that a space X is completely normal if and only if for any pair of Ct(B) and Bn Cl(A) Ct(A) are both empty, there exist subsets A and B such that An Cl(B) disjoint open sets U and V with A c U and B c V.
1.5 Uniform Continuity and Uniform Convergence
25
2. Show that pseudo-metric spaces are: (a) completely normal, nonnal, nonnal, and (b) perfectly normal, collection wise normal. nonnal. (c) collectionwise 3. Show that the product of uncountably many unit intervals is not perfectly nonnal and hence not metrizable. normal 4. A subset of a space is said to be an F crG (pronounced "F-sigma") if it is the nonnal space is union of a countable collection of closed sets. Show that a normal nonnal if and only if every open set is an F cr. G' Hence open sets in perfectly normal G's. metric spaces are F cr' s. 5. If an open set U in a metric space (X, d) contains a compact subset K, show that d(K, X - U) > 0 if both K and X - U are non-empty. 6. Prove Proposition 1.8. 7. Prove Proposition 1.9.
8. Show that precompactness is a hereditary property in metric spaces; i.e., show that every subspace of a precompact metric space is precompact. 9. Prove Proposition 1.11.
10. Prove Proposition 1.12. 11. Show that a sequence {x {xn} E n } in a metric space (X, d) converges to a point p E X if and only if limn~ood(p, limn ->~d(p, xxn) n) = O. 12. Show that the real Hilbert space is not locally compact. Hence a metrizable space may fail to be locally compact.
1.5 Uniform Continuity and Uniform Convergence Recall from elementary analysis that a real-valued function f:R --7 -7 R from the reals R is continuous at a point p if for each £ > 0 there exists a () > 0 such that for each q E R with IIp-ql p - qI < If(P) - f(q) I < E. Then f is defined to foreachqE Rwith 0 there is a 8 (5 then p(j(p),f(q)) p(f(p),f(q)) < £. E. d(p,q) < 8
The proof of this proposition is left as an exercise (Exercise 1). In the (5 that is assumed to exist depends on both E £ above definition of continuity, the 8 (5 > 0 such that for and the point p. If for each E£ > 0 it is possible to find a single 8 (5 then If(P) - f(q) I < £E thenf is said to any pair of points p, q E R with Ip - q I < 8 be uniformly continuous. This definition is extended to arbitrary metric spaces X --7 f:X .~ Y from a metric space (X, d) to a metric space (Y, as follows: a function f: (5 > 0 such p) is said to be uniformly continuous if for each £ > 0 there exists a b that whenever p, q E X with d(p, q) < 8 (5 then p(f(P), f(q» < £. E. Clearly, a uniformly continuous function is continuous.
°
°
An example of a continuous function that is not uniformly continuous ~j R defined by f(x) = = l/x for each x E (0,1). However, the function f:(O,I) f:(O.1) we define f on [0,1] [OJ] instead, we find that f is now uniformly continuous. It impossible to find a continuous real valued function on a closed interval that not uniformly continuous. In fact for arbitrary metric spaces we have:
is if is is
PROPOSITION 1.14 Every continuous function from a compact metric space into a metric space is uniformly continuous.
Proof: Let f:X ~j Y be a continuous function from a compact metric space (Y,p) E > 0. O. Then U = {S(y, £/2») E/2» lyE lYE Y} is an (X,d) into a metric space (Y ,p) and let £ 1 open covering of Y so f- 1l (U) = = {f{f-I (S(y, £/2» E/2» lyE lYE Y} is an open covering of (5 > 0 for this X. Since X is compact, by Theorem 1.9, is a Lebesgue number b (5. Then the set {p,q} covering. Let p and q be two points in X with d(p, q) < 8. (5 must lie entirely in one of the sets f- 1l (S(y, £/2)). E/2». Thus f(P) having diameter < 8 E/2) which implies p(f(P), f(q» f(q» < E. £. This and f(q) both belong to S(y, £/2) establishes the uniform continuity of f. •
r
°
r
If f:X ~j Y is a uniformly continuous one-to-one function from X onto Y then the inverse image function f- 1l is well defined. If f- 1l is continuous then, of course, X and Y are homeomorphic topologically. If in addition f- 1l is uniformly continuous thenfis said to be a uniform homeomorphism.
r
r
r
Properties preserved by homeomorphisms are said to be topological properties whereas properties preserved by uniform homeomorphisms (and not by homeomorphisms) are called uniform properties. There is a special type of uniform homeomorphism that preserves distance; that is, for each pair of points p(f(P),f(q». p, q E X d(p, q) = = p(f(P), f(q». Such a uniform homeomorphism is called an isomorphism. Properties preserved by isomorphisms are called metric properties. Clearly the concepts of uniform continuity, uniform homeomorphism and isomorphism also apply to pseudo-metric spaces.
1.5 Uniform Continuity and Uniform Convergence
27
There exist uniformly continuous functions that preserve distance but are not isomorphisms. Functions that preserve distance are called isometric functions. An isometric function may fail to be an isomorphism for the following reasons: the inverse function does not exist (in case f is not one-toone) or the inverse function is not uniformly continuous. An important example of an isometric mapping is the following. Let (X, d) be a pseudo-metric space X by x-y if d(x, y) = O. Let Y Y be the and define an equivalence relation - in X quotient space X/- and p:X ~ Y the canonical projection that maps each point of X onto the equivalence class that contains it. Define the metric p in Y by
for each pair a, bEY. It is easily shown that p is indeed a metric and p defines the quotient topology on Y. (Y, p) is called the metric space associated with (X, d). If X is a metric space it is clear that X == Y and p is the identity mapping. p(y» = d(x, y) for each pair x, y E X we see that the canonical Since p(P(x), p(y» projection is an isometric mapping. Topologically, isometric mappings are very well behaved as can be seen by the following: J.J 1J If f:X ~ Y is an isometric function from the metric THEOREM 1.1 space X to the metric space Y thenfis an imbedding ofX of X into Y.
that f Proof: Let d denote the metric in X and p the metric in Y. We first show thatf is an open mapping of X onto f(X). For this let V U be an open set in X and put V f(V) c f(X). Pick p E V =feU) U and let q = = f(P). Choose £E > 0 such that S(p, £) cC U. Let y E Seq, £)nf(X). Then there exists an x E X with f(x) = y. Since f is £. Consequently x E S(p, isometric we have d(p, x) = p(q, y) < E. S(P, £) E) C V U so Y E V. Therefore Seq, £)nf(X) c V so V is open in f(X). This establishes that f is an open mapping. It remains to show that f is a one-to-one one-lo-one mapping which is left as an exercise (Exercise 2). •-
In calculus, another "uniform" "uniform" concept of significance is that of uniform convergence. A sequence {fn} {fn) of real valued functions from the reals is said to converge uniformly to a function f if for every £ > 0 there is a positive integer m such that for each x E X and each k > m, l!k(x) Ifk(X) - f(x) II < E. £. Furthermore, the following classical theorem can be proved. THEOREM 1.12 A function f:R ~ R is continuous if there is a sequence {fn} {fn) of continuous functions that converges uniformly to f.
Proof: Let x 0a E R and £ > O. Put £' = = £/4. Since {fn} {fn) converges uniformly to f there is a positive integer k such that if m ~ k then Ifm(x) - f(x) I < £' for each x - f(xo) I < £'. Since!kiscontinuousthereisa 0 E R. In particular Ifk(XO) Ifk(XO)-!(xo)1 0 there is a positive d(fk(x),f(x» integer m such that for each x E X and each kk> > m, d(fk(x), f(x)) < £e where d is the metric on Y. Furthermore, the method of proof of Theorem 1.12 can be modified to prove: ~ Y from a topological space X to a THEOREM 1.13 A function f:X --7 metric (pseudo-metric) space Y is continuous if there is a sequence {fn} Un} of continuous function that converges uniformly to f.
EXERCISES 1. Prove Proposition 1.13 2. Finish the proof of Theorem 1.11; i.e., that f is one-to-one. 3. Prove that an isometric function f:X itself is an onto function.
~ --7
X of a compact metric space into
4. Prove the following: (a) compositions of isometric functions are isometric, (b) compositions of isomorphisms are isomorphisms and (c) the inverse of an isomorphism is an isomorphism.
1.6 Completeness The concept of completeness occupies a central role in the theory of metric spaces and as we shall see, it is equally significant in the theory of uniform spaces. We devote this section to its study. We will also spend some time on a stronger form of completeness that has not previously been investigated in form of completeness, referred to as cofinal metric spaces. This stronger fonn completeness, arose in the study of paracompactness in unifonn uniform spaces. Because of its importance in the following chapters it seems appropriate to introduce it in the setting of metric spaces where it takes on a simpler form.
1,6 1.6 Completeness
29
To facilitate our study of completeness, we first introduce some tenninology terminology that will help streamline the discussion. Let (X'l (X, d) be a metric {xn} {xn} space and A a subset of X. If {x n } is a sequence of points in X we say {x n } is eventually in A if there exists a positive integer m such that XXnn E A whenever n C: m. Also, we say {x {xn} ~ n } is frequently in A if for each positive integer m there is Xnn E A. Clearly if {x {xn} a positive integer n C: ~ m such that X n } is eventually in A it } is frequently in A if and must be frequently in A. It is also easily seen that {x {xn} n {xn} } is eventually in A. only if some subsequence of {x n In Chapters 3 and 4, more general objects than sequences will be introduced for the purpose of studying completeness in uniform spaces. They will be seen to have certain interesting subsets called cofinal subsets. Although we will not need this concept for our study of metric spaces, it will be seen in these later chapters that a subsequence of a sequence will satisfy the definition of a cofinal subset in the special case of sequences. For this reason we say vate our subsequences are cofinal in sequences. This terrninology terminology will moti motivate terminology of cofinal completeness a little later in this section. tenninology We can restate the classic definitions of convergence and clustering that one encounters in elementary analysis in terms of eventually and frequently defined above. We say a sequence {x n } converges to a point p if it is eventually {xnn }} clusters to p or that p is a cluster point in each neighborhood of p and that {x {xn} {xn} of {x } if {x } is frequently in each neighborhood of p. By now the reader has n n no doubt observed that in a metric space, one neighborhood base for a point p consists of the sequence {S(P, 2-nn )} of spheres about p of radius 2-nn for each positive integer n. Consequently, {x {xn} n } converges to p if and only if for each positive integer n there exists a positive integer m such that d(p, Xk) < 2- nn for each positive integer k ~ C: m. {xn} In elementary analysis, the sequence {x n } is said to be Cauchy if for each C: < E£ for each m ~C: k and n ~ k. The following proposition is easily verifiable and left as an exercise. E> xn) E > 0 there is a positive integer k such that d(x m m ,, x n)
J .15 A sequence {x {xn} if and only iffor each PROPOSITION 1,15 n } is Cauchy if
> 0, {x {xn} E. n } is eventually in some sphere of radius E,
£ E>
The concept of a cofinally Cauchy sequence can be obtained by generalizing the property in Proposition 1.15. A sequence {x {xn} n } is said to be {xn} } cofinally Cauchy if for each E > 0, {x is frequently in some sphere of radius n {xn} each E > } E. But this is equivalent to defining {x to be cofinally Cauchy if for n 0, {x E. As {xn} n } has a subsequence that is eventually in some sphere of radius e. mentioned previously, subsequences are cofinal subsets of sequences. Hence the name cofinally Cauchy.
PROPOSITION 1.16 A convergent sequence in a metric (pseudometric) space is Cauchy.
30
1. Metric Spaces
{xn} Proof: Assume the sequence {x n } converges to the point p in the metric space (X, d). Then {x {xn} n } is eventually in each neighborhood of p. Let £ > O. Then {xn} {xn} Cauchy.{x n } is eventually in S(p, E). By Proposition 1.15 this implies {x n } is Cauchy.
The converse of Proposition 1.16 is not true in general. For example. the sequence {x {xn} Xnn = l/n lin in the open interval (0,1) of real numbers n } defined by X with the usual metric is Cauchy but does not converge to a point of (0,1). This example motivates the concept of completeness" completeness. A metric (pseudo-metric) space is said to be complete if each Cauchy sequence converges.
PROPOSITION 1.17 A sequence that clusters in a metric (pseudometric) space is cofinally cofi,nally Cauchy. The proof of Proposition 1.17 is similar to the proof of Proposition 1.16. The example above used to show the existence of a Cauchy sequence that does not converge is also an example of a cofinally Cauchy sequence that does not cluster. This is due to the fact that by definition all Cauchy sequences are cofinally Cauchy and also the following:
PROPOSITION 1.18 If a Cauchy sequence clusters to a point p then it also converges to p. Proof: Assume the Cauchy sequence {x {xn} n } clusters to p in the metric space (X,d). Let £ > O. Then {xn} {x n } is frequently in S(p, £/3). Since {xn} {x n } is Cauchy {xn} in } there is a q in X with {x eventually Seq, £/3). Let k be a positive integer n such that XXmm E Seq, £/3) for each m ~ k. Then there exists a positive integer j such that k < jj and Xj E S(P, S(p, £/3). But then Xj E S(p, £/3)nS(q, £/3). Let m be a positive integer such that m ~ k. Then
Therefore XXmm E S(P, S(p, £) which implies {x {xn} n } is eventually in each neighborhood of p so {x converges to p.} {xn} n A metric (pseudo-metric) space will be called cofinally complete if each cofinally Cauchy sequence clusters. Since, as noted above, all Cauchy sequences are cofinally Cauchy we have the following:
cofi,nal/y complete metric (pseudo-metric) space is COROLLARY 1.7 A cofinally complete.
PROPOSITION 1.19 A sequence in a metric space clusters to a point p if and only if it has a subsequence that converges to p. Proof: Let {x {xn} n } be a sequence in the metric space (X, d) that clusters to p
E
X.
1.6 Completeness
31
The sequence B = {S(P, (S(P, 2- nn )} is a neighborhood base for p so {x {xn} n } is frequently m m in S(p, 2- ) for each positive integer m. For each positive integer n let Xkn Xk n {Xk nn }I is denote the first element of XXnn such that knn > n and Xkn Xk n E S(p,2-n). S(P,2- n ). Then {Xk a subsequence of {x {xn} n } and {Xk nn } clearly converges to p. Conversely, assume {x {xn} n } has a subsequence {Xk nn }} that converges to some p {Xk nn }} is eventually in each neighborhood of p. As previously noted, in X. Then {Xk this means {x {xn} {xnn }} clusters n } is frequently in each neighborhood of p and hence {x top. •COROLLARY 1.8 J.8 A metric (pseudo-metric) space is sequentially compact if if and only if if each sequence clusters.
There is another property that can be sandwiched in between the properties of compactness and cofinal completeness. Using the Lebesgue Covering Theorem for motivation, we define a metric space to have the Lebesgue property if for each open covering U, there is an E > 0 such that U can be refined by the covering consisting of spheres of radius E. By the Lebesgue Covering Theorem it is clear that compact metric spaces have the Lebesgue property. PROPOSITION J1.20 .20 A metric space with the Lebesgue property is cofinally complete. Proof: Assume the metric space (X, d) has the Lebesgue property and suppose X is not cofinally complete. Then there exists a cofinally Cauchy sequence {x {xn} n } that does not cluster. For each x E X pick an open set U(x) containing x such {xn} X - Vex) U(x) and put U = = {V(x) (U(x) I x EX}. Since (X, d) has that {x n } is eventually in X the Lebesgue property, there exists an E> E > 0 such that (S(x, {S(x, E) II x EX} refines U. {Xn} But then {x n n }} is eventually in X - Sex, E) for each x E X which implies {x n } is not cofinally Cauchy which is a contradiction. We conclude (X, d) is cofinally complete. •-
These results yield the following implication diagram for metric (pseudometric) spaces: compact
Lebesgue property
cofinally complete
J-t complete
1. Metric Spaces
32
PROPOSITION 1.21 A sequence in a metric space has a limit point p
if and only if it clusters to p. Proof: Assume the sequence {x n } in the metric space (X, d) has a limit point p. m )) contains a point of {x n }. In fact, Then for each positive integer m, S(p, 2- m m m S(p,2-m)) contains infinitely many points of {x n }. If S(p, 2- m S(P,2)) only contained finitely many points of {x {xn} {Xk I ... . . . Xkj} then we could put e = mint min (d(P, £ = d(p, n } say {Xkl b = 1I ... j} and pick a positive integer b such that 2- bb < E. £. Then S(p, 2- b ) Xk,) II i = would contain no points of {xn} {x n } which would be a contradiction. For each positive integer m let kmm denote the first positive integer greater than m such m S(p~ 2- m). that Xk Xkmm EE S(p. ). Then {Xk nn } is a subsequence of {x {xn} n } that converges to p. {xn} } By Proposition 1.19 {x clusters to p. n The converse is obvious since if a sequence clusters to a point, it is frequently in each neighborhood of the point. •The Lebesgue property in a metric space is very close to the property of compactness. In what follows it will be shown that if a metric space has the Lebesgue property then it is the union of a compact set and a discrete set.
THEOREM 1.14 (Kasahara (Kasahara, 1956) If (X, d) is a metric space with the Lebesgue property such that S(x, e) £) contains at least two distinct points for each E e > 0 and x in X then every sequence in X clusters. 1
Proof: Assume {x {xn} n } is a sequence in X that does not cluster. By Proposition 1.21, {x n } has no limit point. Therefore V = = X - {x n } is an open set. Moreover, em > 0 such that S(x m Xnn Em) contains no X for each positive integer m there is an Em m ,, em) for each n 7; of. m. Furthermore, it is possible to pick ()m b m > 0 such that: m m (1) S(x m bm ) is a proper subset of S(x m m,, ()m) m ,, 2- )) and
(2) ()m bm
< em. Em.
Put U = Vu {S(x n , b n )}. Then U is an open covering of X. Since (X, d) has the U=Vu{S(xn,()n)}' UisanopencoveringofX. (X,d) (S(x, E) e) II x EX} refines U. Let k Lebesgue property there is an Ee > 0 such that {Sex, k I x EX} be a positive integer such that 2-kk < e. E. Then {Sex, 2- k))lx E X} refines U. k S(Xb 2- k) is not contained in V. Suppose n is a positive integer such that Clearly S(Xk' k of. k. Since Xk does not belong to S(xn,b S(xn,()n) ()nn we see that S(Xk, S(Xb 2- kk) n 7; ~b n) and 22k < Ee::; is not a subset of S(xn,b S(xn,()n) of. k. n) for each n 7; S(Xb 2- kk) is not a subset of S(xk,b S(Xb()k) S(xkod Finally, S(Xk' k) is a k) since by (1) above, S(xk,B proper subset of S(Xk' S(Xko 2-kk). Consequently S(Xk' S(Xko 2- kk) is not a subset of U for each (S(x, 2- k) II x EX} does not refine U which is a U E U which implies {S(x, {xn} contradiction. We conclude that {x n } must cluster. -
COROLLARY 1.9 (Kasahara) If a metric space has the Lebesgue property then it is the union of a compact set and a discrete set.
1.6 Completeness
33
Proof: Let (X, d) be a metric space having the Lebesgue property and let Y be p such that S(p, £) contains at least the subspace of X consisting of all points P two distinct points for each £ > O. Then (Y, d) is a metric space satisfying the
hypothesis of Theorem 1.14 and so every sequence in Y clusters. By Corollary 1.8 (Y, d) is sequentially compact so by Theorem 1.10, Y is compact. Now X Y must consist of discrete points by the definition of Y. Therefore X is the union of a compact set and a discrete set. •We now consider some examples of complete metric spaces. Consider first the real numbers R with the usual metric. It should come as no surprise that R is complete. In fact R is cofinally complete. Since R is not a union of a compact set and a discrete set, R cannot have the Lebesgue property by Corollary 1.8. To show R is cofinally complete, let {x (xn} n } be a cofinally Cauchy sequence {Xk nn }) of (xn) {x n } that is eventually in S(p,l) S(P,l) for in R. Then there is a subsequence (Xk some P in R. Since CI(S(P,I» Cl(S(P,l» is compact in R, pin R. by Corollary 1.8 and Theorem 1.10, {Xk (Xk nn }} clusters. By Proposition 1.18, {Xk (Xk nn }} has a limit point which implies {x n } has a limit point so {x {xn} {xn} n } clusters. Next consider the real Hilbert space H introduced in Section 1.1. Let h = p/H -7 R be a Cauchy sequence in H. For each positive integer j let Pj:H denote the canonical projection of H onto its ph jth coordinate subspace defined by p/x) = = Xj where x = = {xn} = {p/h Pj(x) {x n } E H. For each positive integer j let Yj = {pj(h n )}. To show Yj is a Cauchy sequence of real numbers let E£ > O. Since h is Cauchy in H d(hm,hn) £. Since there is a positive integer N such that m,n > N implies d(hm,h n ) < E. {h nn
}
we conclude that Ipj(h mm)) -- pj(h p/h nn)) I1 = I1hmj mj -- hnj nj I1 < £ so that Yj is Cauchy. We (rj). = {rj}' have just seen that R is complete so Yj converges to some rj E R. Let r = H and {h (h n }) converges to r. Since (h We will show r E Hand {h n }) is Cauchy in H, if £E > othere exists a positive integer N such that m,n > N implies
for each positive integer k. If we hold nand k fixed and let n increase we have {Pj(h {pih mm)} )} converges to rj. Therefore
2 :s; ~LJ=l 1I rj - Pj(h n ) 12 1 ~ E£ for each positive integer k. But then Consequently ,Jr.J=1 ~Lj=l I1rj - pj(h /2 :s; ~ eE for each n > N. If r EE H we see that this implies ,Jr.;=1 p/hn) n ) 12
I, Metric Spaces 1.
34
for each n > N. This shows h converges to r if r E H. Therefore it only remains to show that r E H. For this note that
»2
rJ r; = == (rj - pj(h p/h nn)) + Pj(h pj(h nn»2 = == (rj - Pj(h p/h nn »)2 }y2 + 2[rj - pj(h pj(hn)][p/h }] + [Pj(h [p/hn}f n )] [Pj(h n n )] n )]2 ~ 2«rj - Pj(h p/h nn»2 }}2 + [Pj(h [p/h nn )]2) }]2}
y2 holds for each pair of real numbers x and because the inequality 2xy ~ x 22 + y2 y. Therefore
for each integer k > 0 and n > N. Since h n E H the series on the far right converges. Therefore the series consisting of the terms {r;} {rJ} converges to r E H. Consequently, the real Hilbert space is complete. However, unlike our previous example R, H is not cofinally complete. To demonstrate this we will exhibit a cofinally Cauchy sequence that does not cluster. For each positive == {xi} x? = == n and xJ xi = == 0 for each j > 1. Then for each n, integer n let hn = {xJ} where Xl Knn = == {y'!} {yf} where y'! yj = == {z)'} {Z}'} and h nn E H. Next, for each positive integer n let K where Z'!I }
= xl!} + 2- n 12 if i =}' or
Z'!l }
= xl!} if i # }'.
yi E H for each positive integer i. Put Y = == uKn • Then Y is countable. Then y,/ Well order Y by < such that each initial interval of Y has cardinality less than Y. Then Y can be represented as some sequence {aj} c H. For each positive integer i
Cn - 1» for each positive integer n. Thus countably many yi E S(h nn,, 22-(n-l)} Therefore, y,/ C 2-(n-l)} members of {aj} are contained in S(h nn,, 2n-l) for each positive integer n. Hence {aj} is cofinally Cauchy. Cn - 1» In 2-(n-l)} u{S(h n is a positive integer}. Then if p is a It is clear that Y c u{ S(h n n ,, 21 Cn S(hn' 2-(n-l» limit point of Y, P E S(h - » for some positive integer n. Now for each n , 2pair of positive integers i,j with i of. # j 1
I nj nj 1 2 n L m=l I n n 12 d(y d(Y?,Yi} == ~oo :E;;;=llz~-z;;;12 = ~m=l Zm - Zm -== i ,Yjn) -_== :E;;;=IIY?m-YJJ2 Yim - Yim 00
zni 1Z?i I
I
n12 2 12 ++ 1Z?j znj -_ Z'lj znj 122 ++ 1xi Xn + + 22-n12 -_ xl! 122 ++ 1xi X(l -_ xl! ~ XJ XJ -_ 22-ni2 12 ~ J J
2 znj -_ Z?j 1 I
I
I
1
I
I
1
I
I}
1
35
1.6 Completeness
2-(n+l) can contain at most one yJ which implies P is not a limit point Then S(p, 2-(n+l)) {ai} does not cluster so H is not cofinally complete. after all. Therefore {aj}
At this point it may be interesting to consider under what conditions cofinal completeness and completeness are identical in metric spaces. Theorem 1.8 states that compact subsets of metric spaces are closed and bounded. Some metric spaces have the property that all closed and bounded subsets are compact. We will call such metric spaces Borel compact. Borel compact metric spaces are cofinally complete. In fact, if {x {xn} n } is a cofinally Cauchy sequence in a Borel compact metric space (X, d) then there exists a 11 subsequence {x mn mn }} of {x n n }} that is contained in S(p, 2- ) for some P E X. Since 11 Cl(S(P, CI(S(P, 2- ) is compact in X, {x mn mn }} has a convergent subsequence. But then {x n } has a convergent subsequence so {xn} {x n } clusters. {xn}
»
The Borel compactness property can be "'factored" "factored" into the properties of cofinal completeness and regularly bounded; i.e., into a pair of properties that together imply Borel compactness and are both implied by Borel compactness. The property of being regularly bounded is defined by the rule that every bounded closed subset is totally bounded. Regularly bounded metric spaces are also of interest because in them, the concepts of completeness and cofinal completeness are identical. The proofs of these assertions are left as exercises.
PROPOSITION JJ.22 .22 A metric space is precompact if if and only if if each sequence has a Cauchy subsequence. Proof: Assume each sequence in the metric space (X, d) has a Cauchy subsequence and suppose X is not precompact. Then there exists an £e > 0 such {Sex, e) e)lx E X} has no finite subcovering. Pick Pickpl,P2 E X such that thatp2 that {S(x, Ix EX} PI, P 2 E P2 does not belong to S(p 1, e). Next assume {Pi Ii I i = 1 ... n} has been defined such that for each pair i,j i,} with i O. Since (y:" {y~}I converges to Zi Zi there is d;(Zi,yj) < -VEin -If In for each integer j > k i. Put k = a positive integer k i such that di(Zi'y)) max{k 1l •• .. •. kknl. n }. Then
(xmm I} converges to z. We conclude X is complete.for every j > k. Therefore {x complete. -
EXERCISES 1. Show that Borel compact metric spaces are regularly bounded. 2. Show that a metric space is regularly bounded if and only if each bounded sequence has a Cauchy subsequence. 3. Show that a metric space is Borel compact if and only if it is both regularly bounded and complete. 4. Show that a metric space is Borel compact if and only if each bounded sequence clusters.
5. Show that the metric space C*(X) of all bounded real valued continuous functions on the space X (introduced in Section 1.1) is complete.
1. Metric Spaces
38
6. Show that in a regularly bounded metric space that completeness and cofinal completeness are equivalent.
1.7 Completions complete. Certain metric spaces, although not complete, are "almost" "almost" cOlnpleteo E2 by Consider the metric space M constructed from the Euclidean plane £2 = £2 E2 - (0.0). (0, 0). Let d be the Euclidean removing the point at the origin. Then M = E2. Let {x {xn} metric on M considered as a subspace of £2. n } be the sequence of points in M such that XXnn = = (2- nn, 0). Clearly {x {xn} {xn} n } is Cauchy with respect to d, but {x n} does not converge in M. Of course {xn} {x n } converges to (0, 0) in £2 E2 but since we have removed (0. (0, 0) from £2 E2 to get M, M. {x {xn} n } cannot converge to a point of M. This is a special case of a completion of a metric space. Here M is a metric space that is not complete, complete. E £22 is a metric space that is complete, complete. M is a 2 E2 and CI(M) = E E2. E2 is a completion of M. More subspace of £2 • We say that £2 generally, generally. we define the completion of a metric space as follows: A metric space is said to be isometrically im bedded in a metric space Y if there exists imbedded an isomorphism h:X ~ Y. If X is isomorphically imbedded in a complete metric space Y and heX) is dense in Y, then Y is called a completion of X. Notice that E E22 satisfies this definition of a completion. In this case the E22 is the identity mapping. We can construct infinitely isomorphism h:M ~ E £22 that are not complete but for which E E22 is the many other subsets of E fact. if we let A denote the set of all points in £E22 that have at completion. In fact, least one irrational coordinate, coordinate. then X = = E£22 -- A is a metric subspace of £2 E2 that is not complete. X is the set R x R where R represents the rationals and is E2. Again, the identity therefore countable. However, we still have CI(X) = £2. £22 is a completion of X. mapping is the required isomorphism h:X ~ £2 so that E Notice that in this case the cardinality of the completion is greater than the cardinality of the original metric space X. The following theorem is of central importance in the theory of metric spaces. It not only guarantees the existence of a completion, but its constructive proof shows that this completion is essentially constructed by appending all the "missing" "missing" limit points of Cauchy sequences that do not converge in the original metric space.
THEOREM 1.15 J.1 5 Every metric space has a completion. Proof: Let (X, (X. d) be a metric space. Two Cauchy sequences in X will be said to be equivalent, denoted {x {xn} - {Yn} if limn~ood(xn,Yn) limn-'>~d(xn,Yn) = o. O. Clearly ~ - is an n} ~ X* be the set of all equivalence classes with respect equivalence relation. Let x* to~. to -. We will denote the members of x* X* by {x {xn}* {xn} n } is any member of n }* where {x {x n }*. A metric for X* can be defined by
39
1.7 Completions
where {x {Yn}*. To show this Ixnl IYnl are any representatives of {x Ixnl* n } and {Yn} n }* and {ynl*. definition is independent of the choice of the representati ves {x n }, let representatives IX n IYn), n I} and {y I{w Wnn I} be another representative of {x {xnn I}* and {zn} another representative of {Yn (Yn }*. 1*. Then
Hence limn~c>od(xn,Yn) limn-->~d(xn,Yn) = = limn~c>od(wn,zn). limn-->~d(wn,zn)' The verification that d* is indeed a metric on X* is left as an exercise (Exercise 1). Next, define the mapping h that carries p E X onto the class {x }* E X* IX n 1* where {x }*"# IX n I} converges to p. {x IXn 1* ::f- 0 since the sequence {Yn} IYn} defined by Yn = p for each n belongs to Ix ----7 X* is a one-to-one mapping onto {x n } *. Clearly h:X ~ heX) c X*. That h is an isomorphism is left as an exercise. X* is complete, let {I Ix;;'}* To show x* {x~ } *m} be a Cauchy sequence in X*. Then we have a sequence of sequences
xl, x~, xL xL X~, xL ••• .. .
XI, x~, x~, . . . xI. XI, xi, xj, xl. xL .. .
e
For the k thh sequence there is, by definition, an integer nk such that d(X~k' x~) < 11k for each i > nk. nk' For each positive integer k we can define the constant sequence {y~} Iy~} by y~ = = X~k for each n. From the definition of d* we have d*( {x~}* ,{y~ }*) < 11k. d*(lx~}*,{y~}*)
Hence limn~c>od*({x~}*,{y~}*) limn-->~d*({x~}*,{Y~I*) = 0 so {y~} IY~} - {x~} in X*. Let E £ > O. Since } *k} = {{ x~ }*k} we see that {{ y~ } *k} is Cauchy in x* I{ {y~ {y~}*d {IX~}*k} {{y~}*d X* so there exists a positive integer j such that d*({{y~}*d,{{y~}*d)
j.
But then d(X~k' X~l) X~/) < £ whenever k,l > j. Consequently, the sequence {X~k} is to show that {I {x;;' {x~ } *m} C X* converges to {X~k }} *. Now Cauchy in X. We want wantto
40
1. Metric Spaces
{x;;' } *m }} converges to {x~" {X~k }* so X* is complete. By Exercise 11 of Section 1.4 { {x'; To show h(X) heX) is dense in x* X* let {x nn }* E x* X* and let £ > O. Since {xn} {x n } is Cauchy in X there is a positive integer k such that d(x mm,, xxn) ) < £/2 whenever m,n > k. n Select a particular m > k and let p = = X m • Let {Yn} be the constant sequence defined by Yn = P for each n. Then {Yn}* = h(P). Now d(Yn' d(Yn, xxn) n) = d(x mm,, xxn) n) < £/2 whenever n > k. Thus
S({x {Xn} Therefore h(P) = {Yn}* E S( {x n }*, f). E). Since {x n } ** and E£ were chosen heX) is dense in X* so X* is a completion of X. •arbitrarily we conclude h(X) Theorem 1.15 as well as the rest of the theorems that follow in this chapter were the motivating results in the early study of uniform spaces. Each of these theorems will have its counterpart in the theory of uniform spaces. THEOREM 1.16 Iff is a unlformly uniformly continuous function on a subset A of a metric space X into a complete metric space Y then f has a unique CI(A). uniformly continuous extensionf* to Cl(A).
Proof: Let d represent the metric of X and p the metric of Y. For each x E A let {x {xn} (xnn = x for each n) that converges to x and for n } be the constant sequence (x each x E Cl(A) CI(A) - A pick a sequence {x {xn} n } in A that converges to x. This is of course possible by Proposition 1.11, and by Proposition 1.16 {xn} {x n } is Cauchy for each x. Now let E £ > O. By the uniform continuity of f there is a 8 0 > 0 such that whenever a,b E A with d(a,b) < 8 f(b» < £. Since each {x n } is 0 then p(f(a), feb»~ Cauchy there exists a k(x) such that if m,n > k(x) then d(x d(xm' xn) 0 which m, x n) < 8 implies P(f(xm),f(x p(f(xm ), f(x n» ») < E. )} C Y is Cauchy in Y £. But then the sequence {f(x {f(xn)} n CI(A). Since Y is complete {f(x n)} converges to some point x*. for each x E Cl(A). Define !*:Cl(A) f*:CI(A) ~ Y by !*(x) f*(x) = = x* for each x E Cl(A). CI(A). Clearly!*(a) Clearly f*(a) = =f(a) for A. each a EEA. To show the definition of f* is independent of the choice of sequences {xn} X* i= :t:- y*. Then {x n } let {Yn} be another sequence converging to x. Suppose x* £ > 0 with S(x*, £)nS(y*, £)"S(y*, £) there exists an E E) = 0. Again, by the uniform 0 > 0 such that d(a,b) < 8 0 implies p(f(a), feb»~ continuity of f there is a 8 f(b») < £. Since both {x {xn} n } and {Yn} converge to x there is a positive integer k such that n > d(xn,yn) < 8 0 which in turn implies p(f(xn), f(yn» < £. But this is k implies d(xn,Yn) {f(xn)} impossible since {f(x E) and {f(yn)} is eventually in n)} is eventually in S(x*, £) f). Hence x* = y* so f* is well defined. S(y*, e). To show f* is unifonnly uniformly continuous on Cl(A) CI(A) let E£ > 0 and pick 0 > 0 such that whenever a,b E A with d(a,b) < a 0 then p(f(a),f(b» < e/3~ £/3. Letx,y Let x,Y E Cl(A) CI(A) with d(x,y) < 0/2. Since {x {xn} x, {Yn} converges to y, {f(x (f(xn)} n )} n } converges to X, converges to x* and {f(yn)} converges to y* it is possible to choose a positive
1.7 Completions
41
x} < 8/4, d(Yn' dCyn, Y) integer k such that whenever n > k we have d(x n, x) y) < 8/4, p(fCyn),Y*) < £/3. Then d(xn,Yn) < 8 and hence p(f(x n), p(x* J(x j(x n» < £/3 and p(f(yn),Y*) fCyn» < £/3. Consequently !(Yn»
Thus f* is uniformly continuous on Cl(A) and therefore a uniformly continuous f' is another extension of f from A to Cl(A). To show f* is unique, suppose [' uniformly continuous extension of ffrom A to Cl(A) and let x E Cl(A) - A. Then {xn) {x n } c A converges to x. Since both [' f' and f* are continuous it follows from )} and f'(x) and f* (x) Proposition 1.12 that {f'(x {['(xn}) {f* (x (xn») n n)} converge to [,(x) respectively. But since {x } c A we have {f'(x )} = {f(x )} = {f* (x {xn) {['(x (xn»). n n n n)}. Since a sequence in a metric space cannot converge to two distinct points it follows that [,(x) f'(x) = = f* (x) so thatf* that f* is unique.unique. The theorem (Theorem 1.17) will show that completions of metric spaces are essentially unique. To prove it we will make use of the following pair of lemmas.
LEMMA 1.7 Let X and Y be metric spaces. Let A be a subset of X and ---7 Y be a continuous function with a continuous extensionf* extensionf*:Cl(A) ---7 Y. letf:A -7 :Cl(A) -7 Iffis Iff is isometric then so isf*. {Yn) in A Proof: Let x and Y be two points of Cl(A). Pick sequences {x n }) and {Yn} that converge to x and Y respectively. Let dx denote the metric in X and d yy the (x n)} converges tof*(x) to f* (x) and {f* (Yn)} metric in Y. By Proposition 1.12, {f* {f*(xn») {f*Cyn») Cy). Consequently, the sequence {(f*(x n), 1*(Yn»} f*Cyn») in Y x Y converges to f* (y). (f*(x), f*Cy» in YxY. Since by Proposition 1.3, d dyy is converges to the point (j*(x),1*(Y» continuous in the product topology we have:
= dy(f(x since Xn,Yn E A for each n. Since f is isometric dx(xn,Yn) = dy(f(xn), fCyn» and n), f(yn» since d x is also continuous we have:
Hence dy(j*(x),j*(y» dy(f*(x),f*(y»
= dx(x, y) so 1* f* is isometric.-
LEMMA 1.8 An isometric image of a complete metric space is complete.
Proof: Let f:X -7 ---7 Y be an isometric function from a complete metric space X onto a metric space Y. Let {Yn} be a Cauchy sequence in Y. Since f is onto, f(xn) {x n } in X with Yn = = f(x there exists a sequence {xn} n) for each n. Let dx denote the
42
1. Metric Spaces
metric in X and d y the metric in Yo Since f is isometric we have:
{xn} for each pair of positive integers m,n. Since {Yn} is Cauchy, this implies {x n } is Cauchy in X. Therefore {x converges to p E X. Put q f (p). Then } {xn} some = (P). n
{xn} for each pOSItIve positive integer n. Since {x {Yn} n } converges to p, this implies {yn} converges to q. Hence Y is complete. •-? Y denote a completion of a metric space X THEOREM 1.17 Let h:X -t and let j:X -t -? W be another completion of X. Then there exists an isomorphism i:W -t -? Y such that i ©j= h.
j:j(X) -t -? Y by few) = Proof: Define a function f:j(X) = hU- 11(w)) for each w E j(X). By -? Y. Since unifonnly continuous extension i:W -t Theorem 1.16,fhas a unique uniformly h and j are isomorphisms, f is an isomorphism of jeW) onto heX) [Exercise 4, Section 1.5]. By Lemma 1.7, i is isometric. By Lemma 1.8, i(W) is complete. By Proposition 1.21, i(W) is closed in Y.
SincefU(X)) andfU(X)) Since fU(X)) c i(W) and fU(X)) = = hU-1U(X)) hU- 1 U(X)) = heX) we have that heX) c i(W). Since heX) is dense in Y and i(W) is closed in Y we have i(W) = Y. -? Y is a unifonnly uniformly continuous isometric mapping of W onto Y. By Hence i:W -t Theorem 1.11, i is one-to-one and hence an isomorphism. Since i is an h.= h we have i © j = h.· extension offandf© j =
EXERCISES 1. Complete the details of the proof of Theorem 1.15. 2. A metric space X is said to be absolutely closed if every isometric image of X into a metric space Y is closed in Y. Show that a metric space is complete if and only if it is absolutely closed.
Chapter 2 UNIFORMITIES
2.1 Covering Uniformities The concept of a metric space leads naturally to the concept of a uniform space, especially if one approaches the topic from the point of view of covering fes uniformities. The first development of uniform spaces by A. Weil titled Sur /es espaces a structure uniforme et sur /a fa top%gie topofogie generale published in Actualities Sci. Ind. 551, Paris, 1937 took a different approach. It involved a family of pseudo-metrics that generate the topology of the space as opposed to a family of coverings. Weil's original approach was rather unwieldy and was soon replaced by two others. In 1940 1. J. W. Tukey, in his Annals of Mathematics Studies (Princeton) monograph titled Convergence and uniformity in general topology, presented an elegant equivalent development of uniform spaces based on covering uniformities. The covering approach was popularized by K. Morita and T. Shirota in the 1950s and by 1. J. R. Isbell in an American Mathematical Society publication of 1964 titled Uniform Spaces. Also, in 1940, N. Bourbaki advanced a development of uniform spaces based on entourages which are subsets of the product space of the given uniform space. The entourage approach is still favored by many. An English translation of Bourbaki's Uniform Spaces was published by Addison-Wesley in 1966. In what follows uniform spaces will be developed from the covering point of view as being a more natural generalization of a metric space. Covering uniformities will also be much easier for us to use for the material we will be developing. In a metric space M, the metric function is the measure of distance, e but in most proofs, one is usually concerned with the spheres S (x, e) E) of radius E > 0 about some point x. Furthermore, the uniform coverings of the form U = {S(x, M}, for each positive integer n, provide a uniform measure of {Sex, 2- nn ) Ix EEM}, nearness throughout the entire space. A broad class of topological spaces admit a uniform structure that provides a "uniform measure of nearness" throughout the entire space. Even in the absence of a metric, this structure allows one to apply many metric space techniques to non-metric topological spaces. A family 11 f.l of coverings of a set X is called a Hausdorff uniformity if it satisfies the following conditions:
44
2. Uniformities
(1) (2) (3) (4)
If U, V VEil E Jl there is aWE 11 Jl with W < U and W < V, if U Eiland E Jl and U < V V then V E f.l, VEil, every element of 11 Jl has a star refinement in 11 f.l and if x =t y E X there is aU f.l with Star (x, U)nStar(y, a U E 11 U)nStar (y, U)
*
= 0.
The set X together with the Hausdorff uniformity 11 f.l is called a Hausdorff uniform space. Then members of 11 f.l are called uniform coverings. We will V instead of U 1. If U = = {Va}, {U a}, a T 1 space a. E E A, is a star-finite open covering of a aT U a C Starn(U 13, U) X, we can define an equivalence relation ~ - on A by a a. ~- P ~ if Va Starn(V~, for some positive integer n. Let D be the collection of equivalence classes with -. If C E D, put Xc = uaEc uaEcVa. to~. respect to U a. Clearly X = UCEDX CC•' Also, if C, C' E D with C "# =Ie C', then xcnX XcnXc = 0 for otherwise, there exists some P E XcnX xcnXc E c' = c' which implies p E V U arlU anV 13~ for some a a. E C and p ~ E C' which in turn tum implies aa. -- p "# C'. For each C E D, Xc is ~ which contradicts the assumption that C =Ie = UaEA-C obviously open. Also, X - Xc = UaEA-CVa U a is open, so Xc is closed. Therefore,
U»
2.4 Normal Coverings
59
Xc is a component of X. Now {Val a E C} C I must be a countable open covering of Xc. We record these observations as:
THEOREM 2.15 If U = {Va}, {Va). a E A, is a star-finite open covering of a T 1 space X, then A can be decomposed into a collection of equivalence classes D such thatfor XCnX B that for each C E D, Xc == uaEcVa is a component of X, XCnXB = 0 ifC"# ifC:F- Band {Val a E C} CI is countable. THEOREM 2.16 (K. Morita, 1948) If U = = {Vi} {Vi I is a countable open covering of a T 1 space X and V = {F iffor {Fii I} is a precise closed refinement, and if for each i there exists a real valued continuous function fi on X such that fieF;) fi(F;) = =0 and fi(X J, then U admits a countable open star-finite !1-refinement. ~-refinement. f;(X - Vi) V;) = 1, Proof: Using the continuous functions fi, we can construct open sets V7 Vi for n = = + 1, ... such that F Fii cC Vi V7 c CI(Vi) Cl(V?) c Vi+ V7+ 11 for each n = i, i + 1, ... Then if i, i + we put X CI(X n) c Xn+1 X n+1 for each positive Xnn = Ui=l u7=1 Vi we have X = U;;'=lX U;;'=IXn and Cl(X Kn = ) integer fl. n. Next put H Hnn ==X Xnn - CI(X Cl(X n_3 ) and Kn = CI(X Cl(Xn) n X n-_1 1 for each positive -I = = 0. Then K Knn c H n+1l , u;;'=IK =X X-I =X0 = u';=lKn == X, integer n, where we define X -2 = and HmnH Ui=l (KnnCl(Vi», HmnHnn = 0 if 1m - nnlI ~ 3. Since K Knn = u7=I(K nCl(Ui)), the collection n {KnnCI(Vi) {KnnCl(Vj) IIi == 1 ... nand n = = 1,2, ... I} is a closed covering of X and hence {H Ii== 1I ... nandn= isanopencoveringofX. {Hn+1 nV7+ 11 Ii nand n = 1,2, ... }I is an open covering of X. This latter n+1 nVi+ HmnHnn = = 0 for 1m 1m -- n I ~ 3. Then if we construct covering is star-finite since HmnH {Hn+1 nV7+ 11 ,, X - (KnnCl(Vj) (KnnCI(Ui)}I the intersection W of all the binary coverings {H n+1 nVi+ = 1 ... n and each positive integer n, for i = not by Lemma 2.2, W is a countable ~-refinement of U.· open star-finite !1-refinement (1. W. Tukey and K. Morita, 1948) A TI THEOREM 2.17 (l. T 1 space X is normal if and only if every star-finite open covering is normal.
= {Va}, {Val, a E A, is a star-finite open Proof: Assume X is nonnal and suppose U = covering of X. By Theorem 2.15, A can be decomposed into a collection D of = uaEcVa is a component of equivalence classes such that for each C E D, Xc = X, XCnX XCnXBB = 0 if C "# :F- B, and {Val a E C I is countable for each C E D. Since {U a I C} {Val C) is point finite for each C E D. Therefore, by U is star-finite, {U a I a E C} {Val C I is shrinkable to some covering (Val Lemma 1.3, {U a I a E C} {Val a E C C}I so Cl(Va) there covering Q> Cl(V C Va for each a E C. Consequently, exists a closed = = ) a {F a IIa E A} Al with F Faa C U Vaa for each a E A (simply put Fa = Cl(V for each a ) Fa a E A). Now, by Theorem 2.14, there exists a star-finite open !1-refinement ~-refinement Vof V of U such that UuV is also star-finite. Then we can inductively construct a {Un I of open coverings of X such that Un+11 is a !1-refinement ~-refinement of Un sequence {Un} for each nand U 11 is a !1-refinement ~-refinement of U. Hence U is normal. The proof that each star-finite open covering is normal implies X is nonnal is similar to the proof in Theorem 2.13 that each finite open covering is nonnal implies X is nonnal. •
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2. Uniformities
Morita's paper also contained another theorem that will be needed in later chapters, so we present it here.
PROPOSITION 2.2 (K. Morita, 1948) A countable open covering in a normal space is normal if and only if it has a precise closed refinement. {Un} Proof: Assume X is normal and suppose U = { Un I is a countable open covering with a precise closed refinement n }. Since X is normal, there = {F IFni. exists a continuous functionfn:X ~ 4 [0, 1] such thatfn(F thatfn(Fn) = n) = 0 andfn(X - Un) = 1I for each n. Then by Theorem 2.16, U admits a countable open star-finite L1-refinement ~-refinement V. By Theorem 2.17, V is normal, so U is normal. Conversely, suppose the countable open covering U = I{ Un} is normal. Then U has an open L1-refinement Un, V) form a ~-refinement say V. By Lemma 2.3, the sets Fn Fn = X - Star(X - Un' U.precise closed refinement of U.· COROLLARY 2.6 A countable open covering of a normal space is normal if it is point finite. Proof: Assume X is normal and suppose U = {Un} is a countable open covering of X that is point finite. By Lemma 1.3, U is shrinkable to an open covering V = {V {Vn} n } such that CI(Vn) C Un for each n. Then by Theorem 2.18, U is normal.· normal. PROPOSITION 2.3 A countable open covering of a normal space is normal if and only if it admits a countable star-finite star-jtnite open refinement.
= {I Un} is a countable open covering of X Proof: Assume X is normal and U = that admits a countable star-finite open refinement V. Then V is point finite, so by Corollary 2.6, V is normal. Hence U is normal. Conversely, suppose U is ~-refinement say V. By Lemma 2.3, the sets normal. Then U admits an open L1-refinement Fn = = X - Star(X-Un,V) Star (X -Un' V) form a precise closed refinement of U. Since X is normal, for each n there exists a continuous function fn:X 4~ [0, 1] such that fn(F fn(Fn) n) = 0 and fn(X - Un) = 1. Then by Theorem 2.16, U admits a countable ~-refinement.· open star-finite L1-refinement. Tukey also presented a theorem stating that a star-finite normal covering of a T 1I space has a star-finite L1-refinement ~-refinement V such that V is also normal, and UuV is star-finite, but the proof depended on the same erroneous theorem (Theorem 2.5) that invalidated his proof of our Theorem 2.17. Morita pointed out that this theorem also follows from his results that we have just recorded.
THEOREM 2.18 (J. (1. W. Tukey and K. Morita, 1948) If U is a star-finite normal covering of a T 1I space X, then there exists a star-finite ~-refinement L1-refinement V V of U such that V is a normal covering and UuV is also star-finite.
2.4 Normal Coverings
61
The proof of Theorem 2.18 depends on two lemmas whose proofs are left as exercises (Exercise 1 and 2) and on Lemmas 2.2 and 2.3.
EXERCISES normal covering of a T 1I space X, then the binary 1. Show that if U is a nonnal U») is nonnal normal for any open VeX. covering {V, Star(X - V, U)} 2. Besides the assumption of Lemma 2.2, let us assume further that the binary coverings {U a' ex, X - F ex} are all normal. Show that the covering V defined Fa) there is also normal.
3. Prove Theorem 2.18. 2.l8. STAR REFINEMENTS REFINEl\1ENTS OF COVERINGS 4. Assume U and V are coverings of a set X such that V* < U. Show that there exists a covering W W of X with V < W* < U such that:
[1. Tukey, 1940], (a) if U is finite, so is W [J. (b) if U is point finite, so is W [J. [1. Isbell, 1959], [1. Tukey, 1940]. (c) if U is star finite, so is W [J. a} and V = [Hint: Let U = = {U a) = {V i3~ }.). Let y be a subset of U such that the members of y have a point in common, and let r be the family of all such y. For y,o EE r put W yoyB == Uu {{ V ~i3 IV ~i3 c U a for each U a EE Yyand and Vr; V~ c U a for each pair y,b each U a EO). Let W = {W yo Iy,o E r}.] fl.] eachUaE b}.Let w={WyBly,bE 5. [J. [1. Isbell, 1959] Assume U and V are coverings of a set X such that V** < U. Show that if U is countable, then there exists a countable covering W such that V < W* < U.
Chapter 3 TRANSFINITE SEQUENCES
3.1 Background In the theory of metric spaces, sequences playa fundamental role. Recall that a function from one metric space to another is continuous if it preserves convergent sequences (Proposition 1.12) and that a metric space is compact if each sequence has a convergent subsequence (Theorem 1.10). Furthennore, Furthermore, it is possible to characterize the topology in metric spaces by means of convergent sequences (e.g., Proposition 1.10 and Corollary 1.6).
It was shown in Chapter 2 that for a much broader class of spaces (namely, the completely regular spaces) a structure called a unifonnity uniformity exists that uniform measure of nearness in much the same manner as a metric. provides a unifonn One might hope that sequences would play a fundamental role in uniform unifonn spaces but unfortunately this is not the case. However, as we will see, this does not prohibit the existence of a theory of convergence in unifonn uniform spaces complete with the Cauchy concept and the existence of a suitable completion for each unifonn uniform space. The basic problem with sequences is that they are countable. In a metric space each point p has a neighborhood base consisting of the sequence n {S(p,2-n)}. {S(P,2)}. It is the fact that both sequences and neighborhood bases in metric spaces have the same order structure that makes the theory of convergence of sequences so successful in metric spaces. In general, unifonn uniform spaces do not have countable neighborhood bases. In fact, as was seen in Chapter 2 (Exercise unform space has a countable basis, then the unifonnity uniformity is equivalent to 6), if a unfonn uniformity. a metric unifonnity. In this and the following chapters, different objects will be investigated as possible replacements for sequences. The simplest of these is the transfinite sequence. A transfinite sequence is simply a function :y -7 4 X be a is said to be eventually in A if there transfinite sequence and let A c X. Then (R) c A and frequently in A if there is a cofinal C c y is a residual R c y with ~. Since V (If(F y) "# # 0 there is an x E E F y with f(x) E E u. U". If ()8 = = E (F Y' Yob = xX so fey f(y b) 0) E V. v. But (F p, z) < () 8 so {f(yp)} r' x) then Y (F~, {f(yI3)} is frequently in each (lCI[J(F a)] = = 0. For each neighborhood of p which is a contradiction. Hence (lC/fi(F Va CI[J(F a)] and let V = {Va}. Since M is metric and index a put V a = M - Clfi(F therefore paracompact, there is a locally finite open refinement W of V and (W) IW I W E W} is locally finite in X. If W E W there is a hence f- 1l (W) = = {f-l {f- 1 (w) V c V a = M - Clfi(F Vaa E V with W eVa CI[J(F a)] so that
r
r
paracompact.· • Therefore f- 1l (W) refines {U a} so that X is paracompact. Note that the property that each transfinite sequence that is cofinally Cauchy with respect to u clusters is a topological property (i.e., one that is preserved under homeomorphisms) whereas this same property with respect to any other uniformity unifonnity is a uniform property (i.e., one that is preserved under unifonn uniform homeomorphisms). Let I.. A be the family of all countable normal coverings of a completely regular space X. Xo In 1952 a paper appeared in the Osaka Mathematical Journal (Volume 4, pp. 23-40) titled A Class of Topological Spaces showing that the uniformity for X. This e family e of all covering refined by members of I.. A is a unifonnity unifonnity uniformity plays a fundamental role in realcompact spaces as will be seen in Chapter 7. OUf Our present interest in the e unifonnity uniformity is stated in Theorem 3.2.
LEMMA 3.5 (T. Shirota, 1952) The countable normal coverings of a completely regular space form a basis for a uniformity e that generates the topology.
X is completely regular and U = = { { Un} is a countable normal Proof: Assume X
3. Transfinite Sequences
68
covering of X. We first show there is a countable normal covering V with V* < Z* < U. By Lemma 2.3, Fnn = =X U. Let Z be a normal covering with z* 2.3~ the sets F Star(X - Un, Z) form a precise closed refinement of U. Then for each n there exists a continuous function fn:X ~ [0,1] [0~1] such thatfn(F = that fn(Fn) = 1 andfn(X - Un) = n) = O. To see this recall that by Corollary 2.2 there is a continuous function f:X fX ~ M where M is a metric space such that the inverse image of every set of diameter less than 1 is a subset of an element of Z. Let d be the metric on M d([(Fn),f(x». By Proposition 1.4 g is continuous. and for each x E E X put g(x) = = d(f(Fn),f(x». Now put gn(x) = max{g(x),1} max{g(x)~l} and fn(x) = 1 - gn(x). Then fn is the desired fW where function. Next we define a continuous function h:X ~ /UJ fW
=
= = TI{ln n{/nln= 1,2,3,0... .. }) andl n == [0,1] 1n 1,2,3,
for each index n. For this let x E E X and put hex) = (fl(X),!2(X),!3(X), ([I (x),h(x),!3(x), ... ). Vnn = {h(x) E E Y Ifn(x) > O}. To show Then let Y = heX) and for each index nn put V {Vn} Fnn containing x {V n} is an open covering of Y, note that if x EE X there is an F which impliesfn(x) > 0 and therefore hex) E Vn. To show each V Vnn is open note that V Vnn = =p~I p~1 (O,I]n Y and p~I p~1 (0,1] is open in /UJ fW where Pn denotes the canonical I projection of /UJ fW onto its nth nIh coordinate subspace. Clearly h- I (Vnn)) C Un for each n. f W is a separable metric To complete the argument we use the fact that /UJ space. Clearly /fWUJ is separable since it is a countable product of separable spaces. That /fWUJ is a metric space follows from Lemma 1.5. l.5. For each index n let UJ {OJ Wnn = p~l(O,l]. p~l(O,1]. Then V Vnn = WnnY and {W W {W.n} n } is an open covering of JfW - {O} UJ UJ where 0 is the point of JfW having all coordinates equal to zero. Since JfW - {OJ {O} is this). Since /fWUJ -open in JfWUJ it is also separable (it is left as an exercise to show this)o l.2 it is fully normal. Therefore there are normal {O} is metric, by Theorem 1.2 refinements WI and W 2 of {W nn }} with W~ < W; < {W {W.n}' n }. Let A be a countable dense subset of /UJ WI fW - {Ole {OJ. For each a E E A pick W(a) E E containing a. ao Then let W Waa E W 2 such that Star(W(a),W I ) cC Wac Wa'
To show {Wa {W a Ia {O} and let W E WI la EE A} Al covers IfWUJ -- {O} {OJ let q EE /fWUJ -- {OJ containing q. Then W contains some a E E A which implies W C c Star(W(a),W I ) c W {W a}I is a countable normal star refinement of Waa and hence q E W Wa' a • Clearly {Wa {W {W.n}. n }. For each a E A put Va = WanY. Then {Va} is a countable normal 1 {h-Ieva)}' refinement of {Vnl. {Vn }. Finally, put V = {h(Va)}' Then V is the desired countable normal refinement of U. Next we show that the set A of all countable normal coverings is a basis for a uniformity. From Theorem 2.1 we already know A is a sub-basis for a uniformity. Therefore it only remains to show that if U, V E A there is aWE A with W < U and W < V. Since U and V are both countable so it UnV. If {{Un} Un} and {Vn} are normal sequences of coverings such that U = U 1I and V = = VI then Un+lnVn+! Un+InV n+1 1 = V Unn - [ui~l [u7~l CZ(V CI(Vii )] and B Bnn = =V Vnn - [Ui=l [U7=1 CZ(V CI(U;)]. u;;'=lA nn put An = = u;=lA i )]. Then put A = u;;'=lB n • It is easily seen that H c A and K c B and that A and B are and B = = u;=lB open sets.
= 0, suppose p E AnB. Let j be the least positive To see that AnB = integer such that p E Ai Aj and let k be the least positive integer such that p E B k • U j -- [u{:/ CI(Ui)]' Then pP E Vi [u{:I CI(V CZ(Vii )] and p E V k - [u7=] [U~=l Cl(V i )]. There are two cases to U j and p does not belong to Vi U j which consider: first, U $~ k) which implies p E Vi is a contradiction, and second, U > k) which implies p E Vk and p does not belong to Vk which is also a contradiction. Hence AnB = = 0 so X is normal. Next we show that each open covering has a star-finite refinement. For this let U be an open covering of X. For each p E X let V(P) be an open neighborhood of P p such that CI(Vp) Linde16f, CZ(Vp ) c U for some U E U. Since X is Lindelof, = U;=l u;;'=] V(pn). For each n, we can find a countable set of points {Pn} such that X = let V Unn E U such that Cl(V(pn» CI(V(Pn» C Un. Since X is normal, for each n, there exists fn(Cl(V(pn») = 0 and fn(X - Un) = 1. a continuous function fn:X ~ [0,1] with fn(CI(V(Pn))) Then by Theorem 2.16, {Un} admits an open star-finite refinement V. But then V is a locally finite refinement of U so X is paracompact. •-
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3. Transfinite Sequences
Howes. 1969) A completely regular space X is THEOREM 3.2 (N. Howes, Lindelof LindelOf if and only if each transfinite sequence that is cofinally Cauchy with respect to the e uniformity clusters.
Proof: Assume X is Lindelof Lindel6f and suppose {x Ix a} is a cofinally Cauchy transfinite sequence with respect to e that does not cluster. Then for each x E X there is an Ix a} is eventually in X - Vex). Let U = open Vex) containing x such that {x I{ Vex) Ix EX} EX}. Since X is Lindelof Lindel6f U has a countable subcovering {V(xJ} {V(x,)} and since regular Lindelof Lindel6f spaces are paracompact, {V(Xi)} I V(Xi)} is normal and therefore V(x}) which is a belongs to e. But then {x a} is frequently in some V(x) contradiction. Therefore, {x Ix a} must cluster after all. 1
0
Conversely assume each transfinite sequence that is cofinally Cauchy with respect to e clusters. Then each transfinite sequence that is cofinally Cauchy with respect to u clusters, so by Theorem 3J 301 X is paracompact and therefore {Y131 p ~ < y} countably metacompact. metacoInpact. Next we show that a transfinite sequence {Yf31 with no countable subsequence clusters. Let U E e. Then there is a countable normal covering {V IV n} that refines U. For each index n put En = {~ < 'yl 11 Yy f313 E = {B V nn }} and let E = uEnn .. Suppose En is not cofinal in y for each n. Since {Yi3} {YB} has y such that y 8Ii is no countable subsequence, E is not cofinal in y so there is a ()8 < Ysuch Vnn for each n which is a contradiction. Consequently some Em not contained in V must be cofinal in y so that {y f3} is cofinally Cauchy with respect to e and hence Iy 13} clusters. Finally, we show that a countably metacompact space in which each transfinite sequence with no countable subsequence clusters, is Lindel6f. Lindelof. Let K be the least cardinal such that for some open covering U = = {Val I val a < K} has no countable subcovering. For each a < K put Va Va = U { V f31 ~ a} and let F 131 B ~ :S; Faa = X Faa "# Xaa E F Faa for each index a. If - Va. It is easily shown that each F -:;:. 0 so pick x {x a) has no countable subsequence it clusters which implies nF {xa} /if a = 0 which is impossible since uV uVaa = = X. Therefore assume the existence of a countable cofinal subset {an} of K. But then {Va} I Va} has a countable refinement {Van} I Van} and hence by Lemma 3.2 there is a countable closed refinement {H an} such that {Han} Han Van for each n. Now Un = {V [31 ~ :s; H an C eVart f31 B ~ an} has cardinality less than K and Han' an is closed there is a countable subcollection of Un that covers H an' Since H an Han' covers H an. But since {Han} {H an} is countable and covers X there must be a countable subcovering of U. Therefore X is Lindelof. Lindel6f. COROLLARY 3.1 A regular, countably metacompact space is LindelOf Lindeloj if and only if each transfinite sequence with no countable subsequence clusters.
Let A be the family of all finite normal coverings of a space X. An ~ consisting of all coverings important uniformity for X is the uniformity B refined by members of A. This uniformity gives rise to the celebrated StoneCech Compactification Compactijication that we will study in Chapter 6. Our present interest in
3.2 Transfinite Sequences in Uniform Spaces
71
~ uniformity will be evident in the statement of Theorem 3.3. To show the B a uniformity for X we proceed as follows:
~ is B
slar LEMMA 3.7 Every finite normal covering has a finite normal star refinement.
Proof: Let U be a finite normal covering and V and W be normal coverings V,W E E V put V - W if such that V* < W* < U. For each V~W (I) (1) V and W are contained in the same elements of U and (2) Star(V,V) and Star(W. . V) are contained in the same elements of U. Star(W,V) Clearly - is an equivalence relation on V. Since U is finite, finite. there can be only finitely many equivalence classes. To see this, one can induct on the number of elements of U. Clearly if U has only one member, there can be only finitely many equivalence classes. If it has already been established that when U has n members there are only finitely many equivalence classes and if we now assume U has n+ 1I members, say U 1 ••• U n + 1 ,• and let -n be the equivalence relation (I) and (2) above with respect to {U lUI, defined by (1) 1, ... ••• ,Un} then there are finitely many equivalence classes {E IEj } for some j = = 1I ... k with respect to -n0 For each j, Ejj is partitioned into two distinct sets Hj
IV E = {VE
E j I V and Star(V,V) c U n + 1 } and K j EjIVandStar(V,V)cU
Ej - H j . = Ej-H
But {H IHjj Ij = = 1 ... k}u{K k}ulKjj Ij = = 1I ... k} is merely the collection of equivalence classes with respect to - where - is defined by (1) and (2) above with respect to U. Consequently, if U has n+ 1I members, there are only finitely many equivalence classes, which completes the induction argument. Next, for each equivalence class E with respect to - let Z(E) = = uE and put Z = {Z(E) I Z(E) I E is an equivalence class with respect to -}. Clearly Z is a finite normal covering since V V < Z. We leave it to the reader to show that Z star U.refines U.·
LEMMA 3.8 Each completely regular space admits a uniformity ~ that has a basis A A. consisting of all finite normal coverings. Proof: Let X be a completely regular space and u the universal uniformity on X. Then u consists of all coverings that are normal with respect to the family of all open coverings. So A is the collection of all finite members of u. Let U, V A.. Then the covering Un V is finite. Moreover, Un V E A. V E U by Exercise 2 of Section 2.3 so Un V V E A. Hence A satisfies condition (1) of the definition of a uniformity. By Lemma 3.7 U has a finite uniform star refinement W E U. u. By WE Theorem 2.9, W also has an open uniform refinement say Z. Then
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3. Transfinite Sequences
Z < Int(W) = = {lnt(W) {Int(W) IWE W} which implies Int(W) E u. But Int(W) is finite since W is finite so Int(W) E A. Clearly Int(W) < W* < U. Therefore A satisfies condition (3) of the definition of a uniformity so A is a basis for B. ~. It remains to show that B ~ generates the topology of the space. For this let U be an open neighborhood of p E X. Then there is a uniformly continuous functionf:X functionj:X ~ [0,1] such thatf(X - U) = 0 andf(p) = 1. Let V =f- 11 ([0,1/2». Then {U,V} is a finite open covering of X. For each pair x,y E X put d(x,y) = = if(x) I. Clearly d is a pseudo-metric on X. For each positive integer n put If(x) - fey) I. n)lp E X}. X). As we have seen in previous proofs, U~+l < Un for Un = (S(P, {S(P, 2- n each n. Moreover, if q E X such thatf(q) < 1/4 then
=
S(q,U S(q, U 22 )
C
(pE xllf(P)-f(q)1 1/4) {p E xllf(P) - f(q) I < 1/4}
=
C
(pEE {p
xl
=r
1/2) If(P) I < 1/2}
C
V.
If q E X such that f(q) ~ 1/4 then S(q, {p E X Ilf(P) O} C U. Seq, U2 ) C (p Ilf(P) I1 > 0) (U,V) and Un is an open covering for each n. Therefore Consequently U; < {U,V} W = = {U,V} (U,V) is a finite normal covering and S(P,W) = = U. But then ~ generates the topology of X. •-
THEOREM 3.3 (N. Howes, 1969) A space is compact if and only if each transfinite sequence that is cofinally Cauchy with respect to the ~ uniformity clusters. Proof: Suppose X is compact and (x {x a} a) is a transfinite sequence. For each index a put M a = {x~ {x a }) does not (x~ I ~ > a} a) and let U a = X - Cl(M CI(M a). If (x cluster (U {U a) a} covers X and therefore has a finite subcovering (U {U aa, ). Let () be an index such that ai < () for each i. Then xXli~ is not contained in U aj for each i which is a contradiction. Therefore every transfinite sequence in X clusters. 1
l
}.
Conversely assume each transfinite sequence that is cofinally Cauchy with respect to ~ clusters and suppose X is not compact. Let y be the least infinite cardinal such that there is an open covering (U {U a I1 a < y} y) having no subcovering Va :0:; a}. Then {Va} is of smaller cardinality and for each a < y put V a = = U { UU 131 ~ I~ ~ a well ordered covering of X. For each a < y pick XXaa E X - Va. Then the transfinite sequence {x a} is cofinally Cauchy with respect to ~. In fact, all transfinite sequences in X are cofinally Cauchy with respect to ~ for if U E ~ there is a finite V E ~ with V < U. Since V is finite, {x a} must be frequently in some member of V and hence frequently in some member of U.
Ix a} must cluster to some p E X. According to our original assumption, {x But then p cannot belong to V y which is a contradiction since Vaa for each a < 'Y I{ Va} covers X. Therefore I{U U a} a) must have a subcovering of smaller cardinality. But if each infinite open covering has a subcovering of smaller
3.2 Transfinite Sequences in Uniform Spaces
73
cardinality, each infinite open covering must have a finite subcovering. Consequently X must be compact. -
COROLLARY 3.2 A completely regular TI T 1 space is compact if and only if each transfinite sequence clusters.
EXERCISES 1L~ Show that X is countably paracompact if and only if for each countable descending chain of closed sets {F IFn} = 0 there exists a countable n } with nFIln = descending chain of open sets I{G F n C Gn for each n such that nCI(G G n I} with FIl nCI(Gn) n) =0. 2. Show that X is countably metacompact if and only if for each countable descending chain of closed sets IF {F n} with nFn = 0 there exists a countable descending chain of open sets {G I Gn} with FneGn F neG n for each nn such that nGn nG n = = 0.
LindelOf if and only if 3. Show that a regular countably metacompact space is Lindelof each transfinite sequence with no countable subsequence clusters. (i.e., the 4. Show that a locally finite collection of sets is closure preserving (I.e., closure of a union of members of a locally finite collection is the union of the closures). 5. Prove Lemma 3.2. 6. Mansfield (1957) defined a space to be almost 2-fully normal if for each open covering U there is an open refinement V such that if p E V and q E W for two members V and W of V with VnW = 0, then there is a U EE U sequence", containing both p and q. We define a transfinite sequence 'V to be cofinally d~ Cauchy if for each open covering U of X there is apE X such that 'V U '" is normal T 1 space is frequently in S(p, U). Show that an almost 2-fully nonnal paracomact if and only if each cofinally ~ Cauchy transfinite sequence clusters. M and F is a closed subset 7. If X is M-paracompact for some infinite cardinal M M-paracompact. of X, show that F is M -paracompact.
8. Prove Corollary 3.2.
9. Alexandrov and Urysohn (1929) introduced the concept of final compactness in the sense of complete accumulation points. A space is [a.,p]-compact [a,~]-compact in the sense of complete accumulation points, where aa. and ~P denote cardinals with aa. ~ P, ~, if every subset M of X whose cardinality is regular and lies in the interval
74
3. Transfinite Sequences
a,~] has a point of complete accumulation; i.e., a point p such that if U is an rCl,~] open set containing p then the cardinality of UnM is the same as the cardinality M. A space is finally compact in the sense of complete accumulation of M.A points if it is [a,~]-compact [a,~]-compact in the sense of complete accumulation points for all a. cardinals ~ > Cl.
They then proved the following theorem: A space is [Cl,~]-compact [a,~]-compact in the sense of complete accumulation points if if and only if if every open covering U of X [a,~] has a subcovering U* whose cardinality is regular and lies in the interval [Cl,P] whose cardinality is less than the cardinality of U. We define a space to be linearly Lindelof if each well ordered open covering has a countable subcovering. Prove that the following properties are equivalent:
LindelOf, (1) linearly Lindelof, (2) final compactness in the sense of complete accumulation points, (3) each transfinite sequence with no countable subsequence clusters. 10. RESEARCH PROBLEM Miscenko (1962) exhibited a space that he called R* that is completely regular, 1, finally compact in the sense of complete accumulation points, but not T I,
Lindelof. LindelOf. Later, M. Rudin (1971) showed thatR* is not normal. Question: Does there exist a normal Hausdorff space that is linearly Lindelof LindelOf LindelOf? but not Lindelof? 11. It was long a question as to whether or not the coverings of a given uniformity f..l, J..l, the cardinalities of which are less than a given cardinal number K f..ll( that generates the same topology. In case J..l f..l is form a basis for a uniformity J..lK A = 0). the finest uniformity, Lemma 3.5 gives a positive answer when 'A roo The answer is now know to be positive in other cases: [1. Isbell, 1964] if J..l f..l has a base consisting of point finite coverings,
[G. Vidossich, 1969] if J..l f..l has a base consisting of a-point finite coverings, [A. Kucia, 1973] if we assume the generalized continuum hypothesis (an axiom that is independent of the axioms of ZFC).
J. E. Baumgartner) in 12. [J. [1. Pelant, 1975] There exists a model of ZFC (due to 1. uniform space whose countable uniform coverings do not which there exists a unifonn form the basis for a uniformity. Consequently, this question is independent of the axioms of ZFC.
3.3 Transfinite Sequences and Topologies
75
3.3 Transfinite Sequences and Topologies In this section, the theory of transfinite sequences is presented for arbitrary topological spaces. Although this section is independent of the concept of a uniformity, it is important to the development because it shows that as long as we only consider topological properties (as opposed to uniform properties), transfinite sequences are entirely adequate for characterizing these properties. It is only when uniform properties are considered that the transfinite sequences may be inadequate. We will give examples in later chapters where transfinite sequences cannot be used in the same way as nets and filters to characterize certain uniform properties. This, of course, does not rule out using transfinite sequences in some other way to characterize them. Also, we will show that if we are careful in selecting which uniformities we use to generate a given topology, we can often characterize the uniform properties in which we are interested in terms of transfinite sequences. For instance, the class of relatively fine uniformities for a space, that were introduced in the author's 1994 paper in the journal Questions & Answers in General Topology (Vol. 12) titled Relatively Fine Spaces, is an example of such a class of uniformities. It is interesting to note that the u, e and ~ uniformities all belong to the class of relatively fine uniformities. There are several cases (for example Theorems 3.1 - 3.3) where the behavior of transfinite sequences with respect to a uniformity can be used to characterize topological properties. But this is not the same as being able to characterize uniform properties. At the present time, it is an open problem as to what extent transfinite sequences can be used to characterize uniform properties. As previously mentioned, the success of the theory of convergent sequences in metric spaces is due to the fact that both sequences and neighborhood bases have the same order structure. In fact, in metric spaces, each point has a countable well ordered neighborhood base such that the well ordering is identical with the partial ordering of set inclusion. In more general topological spaces, the existence of a well ordered neighborhood base such that the well ordering is identical to the partial ordering of set inclusion is the exception rather than the rule. Fortunately we can use the Neighborhood Principle (an equivalent form of the Axiom of Choice) to obtain a replacement for this well ordered (by inclusion) neighborhood base. This will enable us to characterize the topology of a space in terms of transfinite sequences. This of course means that every topological property (at least in theory) can be characterized in terms of the behavior of transfinite sequences. The Ordering Lemma (N. Howes, 1968) If (P, Then'HCI" 'Ct" is a closure operator on X and (, (, x) E C if if and only if if of D D which does not follow y let d(B) E~. ~ > y, hence lim~f(B,v) liml3f(~,v) E E U, and since g(~) d(~) we have f © (y, d), then B g(B) > deB) A(~,g) = f(~, g(~» E Uo U. We conclude thatf© 'A 'A(B,g) f(B, g(B» A converges to q.-
u.
Let C be a collection of pairs (q>, (, p) of nets in the space X and points p E X. We say C is a convergence class for X if it satisfies the following conditions: (1) if q> is a constant net such that ( ( a) = p for each a then (q>, (, p) E C, (, p) E E C then so does ('I', subnet 'I' of q>, , (0/, p) for each subnet'V (2) if (q>, (3) if (q>, (. p) does not belong to C then there is a subnet 'V 'I' of q> such that (~, p) does not belong to C for each subnet ~ of 'V 'I' and let E 8 8 be another (4) let D be a directed set and for each bED E Diet
°
~ and P and 'A A be defined as in Theorem 4.1. directed set. Let Land f:"L ~ X for which lim8limJ Then for each function functionf:~ lim8lima! (b,a) (o,a) exists, (j© A,p) EE C.
THEOREM 4.2 Let C be a convergence classfor class for a set X andfor and for each A c X let CI(A) be the set of all p E HCI" is E X X with ('V' ('I'. p) E E C and 0/ 'I' cA. c A. Then "Cl" X and ('V' a closure operator on X ('I'. p) EE C if and only if'V if 'I' converges to p relative to the topology associated with Ct. C/. The proof of Theorem 4.2 is left as an exercise (see Exercise 2). A net in a space X is said to be a universal net if for each A c X the net is eventually in A
90
4. Completeness, Cofinal Completeness and Uniform Compactness
or eventually in X-A. Universal nets have the property that if they are frequently in a set that they must eventually be in the set. Consequently, universal nets converge to each of their cluster points. PROPOSITION 4.9 Every net has a universal subnet.
Proof: Let x = {xa Ix 0.1I a a. ED} be a net in X and let = {F IF c X I1x is eventually in then AnB *i:- 0. Consequently has the two F}. Clearly if A, B E properties: (1) x is frequently in each member of and has the finite intersection property. (2)
Let S be the set of all collections of sets in X that contain and have these two properties. Then S is ordered by set inclusion c. If I{a} o.} is a chain in S with respect to c then it is easily shown that uo. ua also has properties (1) and (2) n above. By Zorn's Lemma (Theorem 0.l.(4», 0.1.(4», there is a maximal collection 0 containing and having properties (1) and (2). Let A c X and assume A does not belong to Q. n. Then either x is eventually in X - A or there is aBE n n such that AnB = 0. If x is eventually in X - A then c O. n. Therefore suppose there is aBE 0n such that AnB = = 0. Then X - A E c Xx must be frequently in X - A or else it could not be frequently in Band B eX A. Therefore, we can append X - A to n n to get a larger collection 0' n' that also n is maximal with respect to properties (1) has properties (1) and (2). But since 0 n' = 0n so (X - A) E Q. n. and (2) above, 0' Since x is frequently in each member of 0nand and 0n has the finite intersection property, by Proposition 4.5 there is a subnet y of x that is eventually in each n then y is eventually in A. If A is not Q. Let A c X. If A E 0 member of n. n then, as shown above, (X - A) E Q n so y is eventually in X-A. X - A. contained in Q Consequently y is a universal subnet of x. •-
THEOREM 4.3 A space is compact if if and only if each universal net nef converges. Proof: Let X be compact and let let x be a universal net in X. By Propositions must also converge. 4.6 and 4.7, x must cluster. But a universal net that clusters tnust Hence x converges. Therefore compactness implies that each universal net converges. Conversely, if every universal net in X converges, then by Proposition 4.9, every net has a universal subnet so every net in X has a convergent subnet. Then by Proposition 4.7, X must be compact. •unifonn space. A net {x a} a.} in X is said to be Cauchy if it is Let (X, f..l.) f.l) be a uniform eventually in some sphere of radius U for each U EE ,..1. f..l.. {x a} a.} is said to be
4.2 Nets
91
E !l. cofinally Cauchy if it is frequently in some sphere of radius U for each U E IJ-. The proofs of the following two propositions are left as exercises (Exercise 3).
uniform space (X,!l) PROPOSITION 4.10 A net {xa} {x a } in a un(form (X, IJ-) is Cauchy if and only if iffor IJ- there is a U E U such that {x a} is eventually in U, } for each U E !l U a a J in (X, !l) PROPOSITION 4.11 A net {x a} J.l) is cofinally Cauchy if and only lffor IJ- there is aU a U E U such that {xa} {x a } isfrequently in iffor each U E !l ill U. U
EXERCISES
OJ be a net in the pseudo-metric space (X, 1. Let {xa {x a I aa. EED} (X~ d). Show that {xa {x a}J X if and only if the net {y a I a ED} ~ defined by Ya = d(p~ converges to p E E {Ya a. E 0 J, d(p, x a) for each aa. EE 0, D~ converges to zero. 2. Prove Theorem 4.2 (see Theorem 3.4). 3. Prove Propositions 4.10 and 4.11.
Let {xa Ia. E D}beanetinR(thereals). O} be a net in R (the reals). Ifa£(8) ~(b) be the first element of E Ebb (with EEl (Q) Let 8 E (S) (S) c Wand q>(S) (S) c WI, so there is a VIE E V;. W** < U implies Thus q>' ' E V;. V~. Similarly, there is a V 2 E W* with 'Jf' ~"E V 1I n V 2 = 0 so V; V~ n V; = 0. Since V; V~ and V; are neighborhoods of q>' ' and 'Jf' ~" respectively in X', X' is Hausdorff. fJ.') is complete. Let {~ {q>~ II U E A} be a fundamental It remains to show (X', Jl') q>~ = i(xu)' Put 'Jf(U) = i(x net in i(X). Then for each U E A, there is an Xu E X with ~ u)' Put'V(U) =Xu' Then 'V:A 'Jf:A ~ X is a fundamental net. It is easily seen that 'V 'Jf is Cauchy. 'Jf', Then there is a UAE fJ.A 'Jf' E JlA with 'V' Let U' be an open set in X' containing 'V'" A Star('Jf', U c U' so there is a VA E U U' with 'V' 'Jf' E VA C 'Jf = {xu} {Xu} is Star('V', C U'. Then 'V q>~ E U'. Thus {~ I converges to 'V'. 'Jf'. eventually in U so ~ A
A A
)
Next let {~ A}I be a fundamental Cauchy net in X'. For each a E A {q>~ II U E A A Raa c A A with {~ {q>~ I U E R Ra) u~ u~ = there is a residual R c V: for some V: E aA. Let V~ = } a A A Star(V:, U = {V:} Star(U~, a A) and put U' {U~} and VA V = {V:}. {V~}. Just as in the proof of the A A A preceding theorem, for each a EE A, if U AE a and {~} {q>~} is eventually in U A, then UAnU~ =F "F 0 so UAc C V:, V~, and if a,b E A with a ~ implies both {x a} and {Ya} converge to x., x a,Ya E U for some U E U which in turn implies f(x a), f(Ya) E V for some V S(x',V) and {f(y {f(Ya)} E V. But this is impossible since {f(x a)} is eventually in S(x'.,V) a)} is S(y',v). Hence x' = y' so f' j' is well defined. eventually in S(y',V). j' is uniformly continuous on CI(A), let W To show f' W E v and pick V E v with V* < W. Since f is uniformly continuous on A? J.l such A, there exists a U E 11 that whenever a,b E A with a,b E U for some U E U, then f(a), f(b) E V for f(a),f(b) Z* < U. Let x.,y X,Y E CI(A) such that x,y X,Y E Z for some V V E V. Let Z E 11 Jl with z* some Z E Z. Since {xa} converges to x., y., {f(x a )} x, {Ya} {Yu} converges to Y, {fey a)} converges to y'., y', it is possible to choose a ~ such converges to x', and {fCY ex > ~ implies x a,x E Z 1, I, and YaS j, Z2 E Z, and x'., x',f(x that a Ya,Y E Z2 for some Z 1, f(x a) a) VI1 and y', y',f(ya) V2 for some VI, Vj, V V2 E V. Then Xa,Ya E Star (Z,Z) c U E V f(ya) E V a), f(Ya) x',y' E for some U E U. Hence f(x a ), flY a) E V for some V E V. But then x',Y' Star(V,V) c W for some W WE Therefore,x,y E W. Therefore, x,y E E Z which impliesj'(x),j'(y) implies f'(x), f'(y) j' is uniformly continuous. E W, so f'
To show f' j' is unique, suppose F is another uniformly continuous extension of ffrom A to CI(A). Let x E CI(A) - A. Then {x a } converges to x. Since both f' f' (x) and {F(x a) } j' and F are continuous, it follows that {f'(x {j'(x a)} {F(xa)} a )} converges to j'(x) {xa} {j'(x a)} )} = = {F(x {F(xa)}' converges to F(x). But {x a} cC A implies {f'(x a) }. Since a )) = {f(x ua)} j'(x) = = F(x), so f' j' is a net in a Hausdorff space has (at most) a unique limit, f'(x) unique. •11') THEOREM 4./0 4.10 For each uniform space (X, 11), IJ), its completion (X', J.l') is unique; i.e., if (X', (X/\, 11') J..L A) is another completion of (X, 11) J.l) then there is a uniform unifornl homeomorphism h:X' ~ -7 X/\ X' that keeps each point of X fixed.
4. Completeness, Cofinal Completeness and Uniform Compactness
102
Proof: Let j:X ~ X X' denote the uniform hOlncolnorphisln homeomorphism that Inaps maps X into the cOlnpletion completion X X'. By Theorem 4.9 there exists a unique unifonnly uniformly continuous extension }':X' j;:X ~ X j" is an extension of j we have Jj = jf © i where i:X X'. Since Since}' ~ X' denotes the unifonn X uniform hOlneolnorphism homeomorphism that Inaps maps X X into its cOlnpletion completion X'. Similarly, there is a unique uniformly continuous extension t:X' iA:X'" ~ X" of i. Then i = (© i'" © j. A i'(j'(x')). Then g' = }' X' is For each x' E X' put g'(x') = i"'(j'(x'». j' © i( so g':X' ~ X X then g'(x) = = i (x) = uniformly continuous. If x E X = {(j'(x» iA(j'(x» = = {(j i"'U (x» = = x so g' is an extension of the identity map i:X ~ X'. But by Theoreln Theorem 4.9, this extension is unique so g' == i' (the identity map on X'). Hence j' © {i = = i' so {i == (j')-l. Hence}' (j'r i . But X' are unifonnly uniformly homeomorphic. •then X' and X'" A
A
r
•
f
A
•
F
•
A
A
A
EXERCISES 1. 1, Prove Proposition 4.20.
2. Show that the cOlnpletion completion of a uniform unifonn space is cOlnpact compact if and only if the precompacr. uniformity is precompacL 3. Show that a complete subspace of a Hausdorff uniform space is closed.
4. [K. Morita, 1951] Let 11* ~ * be the uniformity of~. Jl, of!J.X. For each U = {U a} a I in 11, fJX -- Clf1X(X CI~(X - U a) for each u. Jl} is put U' = {U~} {U~ I where U~ = !J.X a. Then {U" {U'II U E III a basis for 11*. Jl *. CAUCHY FILTERS AND WEAKLY CAUCHY FILTERS A filter F in a uniform space (X, 11) Jl) is said to be Cauchy if for each U E 11, Jl, there exists an F E F such that FeU for some U E U. It is said to be weakly Cauchy if for each U E 11, Jj-, there exists a U E U with UnF t= aU of- 0 for each F E F. 5. Show that (X, 11) Jj-) is complete if and only if each Cauchy filter in X converges. complete if and only if each weakly 6. [N. Howes, 1971] (X, 11) f.!) is cofinally cOlnplete Cauchy filter in (X, (X? u) clusters.
7. [H. Corson, 1958] 19581 X is paracompact if and only if each weakly Cauchy filter in (X, u) clusters.
4.5 The Cofinal Completion or Uniform Paracompactification
103
4.5 The Cofinal Completion or Uniform Paracompactification Since each uniform space has a unique completion. it is natural to define a unifonn space (X~. ~) if (X'. Jl') f..l') to be a cofinal completion of the uniform space (X. f..l) (X'. Jl') f..l') is cofinally complete and (X. Jl) f..l) is uniformly homeomorphic to a dense ~ 'When does (X, uniform subspace of (X', (X'. J.l') f..l') and to ask: "When (X. J.l) f..l) have a cofinal completion. and if a cofinal completion exists, exists. is it unique?" cOlnpletion, unique?" Providing answer~ to these questions is the objective of this section. answers
It turns out these same techniques can be used to provide answers to a variety of other questions asked by K. Morita and H. Tamano. In the late fifties, fifties. '~What is a necessary and sufficient Tamano considered the following question: "What condition for a uniform space to have a paracompact completion?" completion?" Although he did not arrive at a solution, solution. he obtained elegant characterizations of completeness. paracompactness and the structure of the completion by means of cOlnpleteness. a concept called the radical of a uniform space. These results were published in the Journal of the Mathematical Society of Japan in 1960 (Volume 12, No.1, pp. ofJapan 104-117) under the title Some Properties of the Stone-Cech Compactijication. Compactification. We will analyze these results in Chapter 6. In 1970, K. Morita presented a paper titled Topological completions and M-spaces at an international Topology Conference held at the University of Pittsburgh. In Section 7 of that paper, five unsolved problems were listed including a special case of Tamano's question mentioned above and the ~ 'What is a necessary and sufficient condition for a Tychonoff space question: "What to have a Lindelof LindelOf topological cOlnpletion?" completion?" By a topological completion we mean the completion with respect to the finest uniformity. In this section and in later chapters, we provide answers to all these questions. All of the solutions are in terms of the cofinally Cauchy nets and their behavior with respect to various uniformities. Since some of these uniform space has a certain topological questions ask if the completion of a unifonn property, one might be interested in knowing if there are solutions in terms of property. topological properties, or if topological properties exist that are only necessary or only sufficient. sufficiem. Since the literature is rich with characterizations of paracompactness and the Lindelof LindeWf property, one might expect a variety of solutions to these problems. However, at the present time, this area is largely unexplored. We define a uniform space to be preparacompact if each cofinally Cauchy net has a Cauchy subnet. Recall that a uniform space is precompact if every net has a Cauchy subnet. Consequently, preparacompactness is a generalization of precompactness. To continue the parallel. parallel, recall that complete precompact spaces are compact and the completion of a precompact space is compact. We will see that complete preparacompact spaces are paracompact and that the completion of a preparacompact space is paracompact.
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4. Completeness, Cofinal Completeness and Uniform Compactness
unifonn space will be called countably bounded if each uniform unifonn A uniform Jl) is a uniform unifonn space and (X*, Jl J..1 *) covering has a countable subcovering. If (X, fJ,) unifonnity for x* X* and let v be the its completion, let u* denote the universal uniformity unifonnity is called the uniformity uniformity induced on X by u*. The v uniformity derived from u* or simply the derived uniformity. A directed set 0D is said to be ffi C 0 co directed or countably directed if for each countable {d {d;} D there is a d EE i } c D such that d d;i ~ d for each i. A net {x {xaa I a ED} E D} is ffi co directed if 0D is an co o m directed set. LindelOf if and only if each LEMMA 4.1 A completely regular space is Lindelof m co directed net clusters.
m directed net clusters in the completely regular space Proof: Assume that each co Linde16f. Then there is a covering U of X having no X. Suppose X is not Lindelof. countable subcovering. = u {u IU EE V} sUbcovering. For each countable V c U put G(V) = and F(V) = = X - G(V). Let S be the set of all countable subsets of U and for each ~ Xv defines an ffi VE E S pick Xv E E F(V). The assignment V -7 co directed net {xv IV E E S} in X that clusters to some p E E X where S is directed by set inclusion. Let UE E U. Then {xv} is eventually in F( {U} E U such that p E {U})) and hence cannot be frequently in U which is a contradiction. Consequently X is Lindelof. Linde16f.
LindeWf and let {x a II a ED} be an co Conversely assume X is Lindelof m directed net in X. Let U be a member of the e uniformity unifonnity of X. Then there is a countable = {Vi} {Vd such that VEe and V refines U. Suppose {x a} is not subcovering V = D i = {b frequently in some member of V and put D; {O EE 0D Ix 8Ii EE Vi}' V;}. Then there exists a ()i () E i} would be 0; E E 0 D such that ()0 ~ ()i 0; for each bEDi 0 E D; or else {x {Xli8110 E D D;} V;. Since {x a} is ffi CO directed there is abE a 0 E 0D such that ()i 0; ~ ()0 for frequently in Vi' E V each i. Now X8 Xli E Vjj for some j since V covers X. Hence () 0E E D jj which implies ()0< < ()j' OJ. But ()j OJ ~ ()0 which is a contradiction. Therefore xXaa must frequently be in some member of V. Since V was chosen arbitrarily, {xa} is cofinally Cauchy with respect to e and therefore clusters by Theorem 4.4. • The following theorem appeared in an article titled On completeness in the Pacific Journal of Mathematics in 1971 (Volume 38, pp. 431-440). THEOREM 4.11 (N. Howes, 1970) Let (X, Jl) J.1) be a uniform space and v the derived uniformity. Then: Jl) has a paracompact completion if and only if (X,v) is ((1) 1) (X, fJ,) preparacompact and preparacompactand (2) (X, fJ,) Jl) has a Lindelof LindelOf completion if and only if (X,v) is countably bounded and preparacompact, and (3) (X, fJ,) ,V ) is precompact. Jl) has a compact completion if and only if (X (X,v) Proof of (1): Let (X', J..l) Jl') be the completion of (X, J.l) Jl) and let u' be the universal u'). Assume uniformity unifonnity for X'. Then (X,v) is a dense uniform unifonn subspace of (X', U). ",:D -7 (X,v) is preparacompact and that \V:D ~ X' is a cofinally Cauchy net with
4.5 The Cofinal Completion or Uniform Paracompactification
105
respect to u'. Since (X'" (X', Il') complete, so is (X', u'). Let E = D x u' and define Jl') is complete" $ on E by (d, U') s $ (e,v') $ e and V' o'
LEMMA 5.7 If {X {Xn,fn) n, fn} is an inverse limit sequence with onto bonding n1aps {an} of positive integers maps and for some countable counrable set {an) inlegers there is a set {x an an }} such that xxaann E X Xaann for each n and such that if m < fl, n, then fa~m la~m (x an =x am an )) = am '' then eXlsrs a point in limf-X lim.,...X n whose coordinate coordinare in X Xaann is Xxaann for each n. there exists {an) is infinite. For each positive integer m, there exists a Proof: First assume {an} ~ m. If an = m put X Xmm = x an least positive integer an such that an 2:: an '' otherwise let X = f:nn (x ). Then {x } is a point of limf-X such that x E X Xmm f,:: an IXm) lim.,...X n xaann Xaann for each fl. an )· m Ian) is finite. Then there exists a greatest an, say aj' aj. If m < Next, assume {an} (xa). aj, Xm aj put XXmm = = ;:;. t:/z (x ). For each m ~ aj' assume X has already been defined. m aj Xm+! +!) = XXm • Then by Since fm+l is onto, pick XXm+! (X m+l) m+1 such that fm+l (x m+l E X IXm) Ix mm }} E lim~Xn lim.,...Xn and for induction we can complete the sequence {x m } such that {x each n, xxaann E X On an " U
-
Lemma 5.7 illustrates the fundamental property of inverse limit spaces, Xb all elements of limf-X lim"",Xnn having that coordinate have that for any coordinate Xk,
5.5 Inverse Limits and Spectra
127
all other coordinates XXmm with m < kk determined by the inverse limit sequence, sequcncc~ whereas there may be sOlne some room for choice of the coordinates XXmm with m > k. If the bonding maps are not onto, limf-X n may be the empty set. An Xnn are all countable discrete spaces, say Xn = example of this occurs when the X Xn = {x~} and the bonding Inaps {x;:'} maps fn:Xn ~ X n---1 are of the form fn(x~) fn(x;:') = x~~\. x;:,~il' {Xn> Clearly {X n , fn} is an inverse limit sequence, but if we begin with a point x~, it xi and there is only possible to pick the first m coordinates before we reach xT does not exist an X'J'+1 X'J+l such that fm+l (xj+l) Xl. Consequently~ (x'J'+l) = Xl' Consequently, there do not exist any sequences {x {xnl Xnn E X Xnn and fn(x nn )) = = XXn-l n -1 for each n > 1. n } such that X = 0. Under certain conditions, limf-X n ~ l' Therefore, limf-X nn = conditions~ we can assure that lirn~Xn 0. For instance, instance. THEOREM 5.2 If each space Xn X n in the inverse limit sequence {X·fP.fn} /XnJn} Hausdorff space then limf-X n "# i= 0. is a compact Hausdotff Proof: For each positive integer n let Ynn cC OX DXnn be defined by {Yi} E Yn if for each j < n, Yj-l =h(Yj). Jj(y). Then limf-Xn = nY nYn' n, Ynn n. We will show that for each n~ is closed. Suppose p E E ITX Then for some j < n we have fj+l (Pj+l) "# • DXnn -- Y YnJj+l (Pi+l) i= Pj. Pi' n Since X V j containing p Xjj is Hausdorff, there exists disjoint open sets V Ujj and Vj Pjj and h+1 Vj +1 = Jj+l (Pj+l) respectively. Put Vj+l = r;ll J;ll (V (Vi) Upp be a basic open set in j ) and let V nx + 1 as its j/hth and j+ 1 sf factors DXnn containing p and having V Ujj and V Vj+l Fi j Upp since if q = qj+l respectively. Then no point of Yn lies in V , = {qn} E VUp, then qj +1 E p Vjj + Up. Therefore, Yn is closed. + 1 which implies qj E Vj V j so q does not belong to Vp. DXn, nYnn Since {Ynn }} is a decreasing chain of closed sets in the compact space nx n , nY :j; i= 0 so limf-X nn i-: i= 0. •
There are many applications of inverse limit spaces of inverse limit sequences in topology. What we are interested in here is extending the concept to uniform spaces, and extending it in a more general setting. For this first notice that by changing the definition of an inverse limit sequence {Xn.fn} {X n , fn} so Xnn are now uniform spaces and the bonding maps are uniformly that the X limf-Xnn that is a uniform continuous, we get an inverse limit uniform space iin1f-X n0 This follows from the parallel subspace of the uniform product space OX DXno between product topological spaces and product uniform spaces [see Section 5.3]. more general than this. Let (D,~:X/cPll ~:X/)l ~ points x,y E X/A defined by mapping an equivalence class in X/cPj.l X/)l onto the equivalence X/cPA X/A X/)l belongs. class of X/' A to which a member of the equivalence class in X/1l Then we have the following commutative diagram: iJl ill
X
X X)lil
~
~ -7'
.tJ" ·A0'A
II ~
X XA
X/cP X/)lll
.tJ" ~ ~
lj.l t)l
iiAA
X
~ ~
~ ~ ~
X/A X/cPA
where i~ denotes the identity map on X considered as a mapping from X X)lIl onto X A. A' It is left as an exercise [Exercise 3] to show that ~ ~ is uniformly continuous X/)l and X/cPA X/A respectively. Then with respect to the metric uniformities on X/1l {X/A' ~} ~} is an inverse limit system of metrizable spaces. Let veX) = {X/A' = limfX/A and let 1t limrX/A 1tAAdenote the canonical projection of veX) onto X/A. X/A' Then ofv(X) the uniformity of veX) has a basis consisting of the coverings
veX) is called the weak completion of X with respect to v. A uniform space is said to be weakly complete if each (I)-directed co-directed Cauchy net clusters. We will show that (X,v) is uniformly homeomorphic to a dense uniform subspace of veX) and that veX) is weakly complete. The weak completion of a uniform space with respect to a uniformity was introduced by K. Morita in 1970 in a paper titled Topological completions and M-spaces published in Sci. Rep. Tokyo Kyoiku Daigaku 10, No. 271, pp. 271-288. To see that veX) is weakly complete, let \V:D 'I':D ~ veX) be an (I)-directed co-directed Cauchy net. Then for each A E A, 1t A © \V:D A,1t 'I':D ~ X/A is an (I)-directed co-directed Cauchy net in the metric space X/CPA. X/A' The notion of weak completeness is different from the notion of completeness, because every metric space is weakly X/A by complete with respect to its metric uniformity. We prove it for X/CPA showing that 'VA 'l'A = = 1t 1t1..A © 'V 'I' clusters in X/CPA. X/A' For this let n be a positive integer nn 'VA(R n and let R Rnn be residual in D such that 'l'1..(R Xnn EX/CPA. E X/A' n )) c S(x n n ,, 2- )) for some X Then n;;'=I'I'A(R n;=1 'VA (R n ) ;/; 0, so pick Y E n;=1 'VA(R ). Suppose (\ Cl('V'A(R =0 f:. n;;'=I'I'A(R n nIiEDCI('I'" (R 0» = 8ED where R 0 = {y E D I18 0 ~ y}. Then there exists an a E D such that y does not
130
5. Fundamental Constructions
belong to CI('VA(R a». u». Let m be the least positive integer such that m S(y,2-m)nStar(CI('V)...(R a», 2- mm)) = 0. Now for each n > m, y E 'VA(R 'V),JR n) which S(y,2)nStar(CI('VA(R u», m m ) C S(y, 2-so 'VA(Rn)n'VA(R u) = 0 which is a contradiction. implies 'VA(R ) 'V".).., we see that 'VA 'V).. converges to some YA E X/A" X/!!>)... Cauchy in X/A' DAX/!!>).. 'V).. converges to YA y).. for Now {yd {YA} defines a point in n AX/A and since 'VA {yd in each A, we see (by Exercise 2 of Section 5.3) that 'V must converge to {YA} D)..X/!!>)... But since 'V(D) c veX) = lim).. 0AX/Ao limf-X/A and since lim).. limf-X/A is closed in D)..X/!!>A nAX/A by Lemma 5.8, we have that 'V clusters in veX). Consequently, veX) is weakly complete.
To see that (X,v) is uniformly homeomorphic to a dense subspace of veX), (x) I A EX} defines a point of vex) [see notice that for any x E X, {(x) = {q>A(X)} {4>)..(x)} for each x E X. It is easily with A < !-!]. f.l]. Define q>:X shown [see Exercise 4] that (Un)' 1t~! (A(U~+l» (4))..(U~+!)) < (U~) ).. so we have a But 1t A(CI(q>(X») = X/A A(y), A(U~»
contradiction. Therefore, A ~1 (1t ACY» (y» EE qJ, '1', 1t 1tAA(y) X/A and hence a closed set. Let L I: be the collection of countable intersections of members of qJ 'I' and let D= {(A,x)lxE AEE I:}. L}. Definet}. Let ZZ EE (l.m,n=l'V(P(m,n». n.m,n=l'l'(P(m,n». Then jJ.(z) >t(z) = = YJ.l Y>t so Z z is a representative member of the equIvalence eqUivalence class YjJ. Y>t in X. But t) = = YA· YA.
5. Fundamental Constructions
132
= {YA II A E A} lies in veX). We want to show that the net Therefore, the point y = then", © '",:D 11: 0 ~ veX) converges to Y so that if is onto then 'V converges in X (by X is weakly complete with respect to v. Let Proposition 4.8) and hence X S(y,1t~l (A(U~») (A(U~))) be a basic open neighborhood of y in veX). Now S(y,1t~1 A(",(R(A,n») c S(YA' A(U~» A(U~» so A('V(R(A,n»)
n»))) c 1t~1 (S(YA' A(U~») A(U~))) cC S(y, (A(U~))). 1t~1 (A(",(R(A, (A('V(R(A, n»» S(Y, 1t~1 (A(U~»).
:X ~ veX), it is clear that for each x E X, inf(X) in I(X) converging to y. Since g and h are morphisms (uniformly continuous functions in the category of uniform spaces), g © q> f= h © land fand q> c/(X) cf(X) converges to gCy) and h © q> converges to hCy). But g © 1= q>. Since nets converge to unique limits in uniform spaces, implies g © q> = h © 168
(3) ~ = --7 (1). Let A)) " E $ where p E pX ~X - Xo X. For each x EE X and each positi~e positi~e integer n, put Vn(x) = { ( Y EE: xl XI E ~ I,Jx) A (x) -
178
6. Paracompactifications
defined by (Y) (y) = d y for each y E X, is continuous. Therefore, by Lemma 6.13, pX x X. It is clear that d* == 1 on {p} d has a continuous extension d*(z,y) over BX x X and d* = 0 on ~X. L1X. Thus (1) ~ (3).
PX - X and let {p} x X be functionally To show (3) ~ (1), let P E BX L1X by f E C*(~X C*(PX Xx X). Without loss of generality, we may separated from ~X f(L1X) = 0 and f( {p} x X) = 1. For each x E X let fx be defined by fxCz) assume f(I1X) fx(z) pX. Then put d(x,y) = = sUPzefjX SUpzE!3X Ifx(z) IfAz) - fy(z) I. Then d is a = f(z,x) for each Zz E BX. pseudo-metric on X. Let 1't denote the topology of (X, d), and consider the space (X,'t) (X,1) which is paracompact. For each x E X, let Vex) = {y E Xld(x,y) < 1/2}. Then Vex) is open in (X,1). (X,'t). Put U = = {V(x) Ix EX}. Then there exists a partition of unity = {A} {A} that is subordinate to U. Since 1:t is coarser than the original topology, U is an open covering with respect to the original topology is a partition of unity with respect to the original topology. and 1
SUpzE!3X Ifx(z) IfAz) - fy(z) f/z) I1 < 1/2 so Ifx(y) - fy(y) I1 Now d(x,y) < 1/2 implies that sUPze~X < 1/2. Butfy(Y)=f(Y,y)=Oso Ifx(y) I < 1/2. Therefore, fx(z)~ fAz)~ 1/2foreachz pX. On the other hand,fx(P) E Cl 13X[V(x)] ~x[V(x)J since fx is continuous on BX. hand, fx(P) = =f(P, x) = CI ~x[V(x)] [V(x)] 1I for each x E X. It follows that Cl does not contain P for each x E X. 13x is subordinate to U, Cl CI 13x [O(A)] does not contain P for each A 1.. E . Since ~X[O(A)] complete.By Theorem 6.9, this implies X is topologically complete. • 1
EXERCISES
1. Let d be the metric constructed from the supremum norm on C*(X) for a Tychonoff space X. Show that (C*(X), d) is a complete metric space. (f~ ,g 13) ~) = d(f, g) in the proof of Lemma 6.13, for any pair f, g E 2. Show that d ~13 (f13,g C*(X).
6.4 Points at Infinity and Tamano's Theorem Although Tamano's Theorem is not about uniform spaces, the proofs of Tamano's Completeness Theorem and Tamano's Tam an 0 ' s characterization of topological completeness (Theorem 6.9) hold the key to its proof. Also, it is an important example of using points at infinity to prove things about the space itself. If we consider the space R of real numbers, we notice that R has a compactification R* that is constructed by adding two additional points that we denote by -00 _00 and +00. R is sometimes denoted by (-00, +(0) while R* is often referred to as the extended real numbers and denoted [-00, +00]. The point -00 is assumed to precede all members of R and +00 is assumed to follow all members of R. In this way, the natural ordering on R is extended to R*. The topology of R* is formed by taking as a basis, all open intervals in R together with all sets in R* fonned of the fonn form [-00, r) and (r, +00] where [-00, r) = {y E R* Iy 1 y < r} and (r, +00] = = {y ER*ly>r}.
6.4 Points at Infinity and Tamano's Theorem
179
The points -00 and +00 are called points at infinity. In general, if (X, Jl) J.l) is a Jl) is its completion, the set X' - X is called the set of unifonn space and (X', J.l') points at infinity of X. Similarly, if P is a paracompactification of X, PX - X is referred to as the set of points at infinity of X. Points at infinity are always relative to some uniformity. Points at infinity with respect to precompact unifonnities (e.g., ~X - X) play an important role in general topology. If Jl J.l is a unifonnity for X and H cC JlX J.lX - X, then H is called a set at infinity. Compact sets at infinity will be important to us in what follows. Sets at infinity can be very large with respect to the original unifonn space. For example, the space N of positive integers with the usual topology is countable whereas ~N - N is not. We have already seen a characterization of local compactness in tenns of compact sets at infinity, namely, Proposition 6.4. Our next characterization of paracompactness in tenns of compact sets at infinity relies on a well known characterization of paracompactness by E. Michael that appeared in the Proceedings of the American Mathematical Society in 1953 (Volume 4, Number 3, pp. 831-838). LEMMA 6.14 (E. Michael, 1953) The following properties of a regular topological space are equivalent: (1) X is paracompact, X has a locally finite (not necessarily open) (2) every open covering of X refinement and (3) every open covering of X has a locally finite closed refinement. Proof: (1) ~ (2) is obvious. To show (2) ~ (3) let U be an open covering of X. Since X is regular, there exists an open covering V of X such that the closures of elements of V V is a refinement of U. By assumption, there exists a locally finite refinement W of V. Then the closures of elements of W is the desired locally finite closed refinement of U.
To show that (3) ~ (1), let U be an open covering of X. Let V V be a locally ofX. finite refinement of U and let W be a covering of X consisting of closed sets, each one of which intersects only finitely many members of V. Now let Z be a locally finite closed refinement of W. For each V V E V, let V' == X - u{Z E Z IV nZ = (25}. Then V'is = 0}. V' is an open set containing V V such that if Z E Z, then Z VE V intersects V' if and only if Z intersects V. For each V V pick a U v E U such y . Let U' = {V'nUy that V V cC U v· {V' nUv IV V E V}. V). Then U' is an open refinement of U and since each element of a locally finite covering Z intersects only finitely ' , U, is locally finite. many elements of U U', Notice the similarity between statement (2) of the following theorem and the statement of Theorem 6.9. By replacing points at infinity with compact sets at infinity in the hypothesis, the strength of the conclusion is raised from topological completeness to paracompactness.
180
6. Paracompactifications
THEOREM 6.11 (H. Taman 0, 1960) Let BX denote any Tamano. compactification of X. Then the following statements are equivalent: (1) X is paracompact. (2) For each compact set C at infinity, infinity. there exists a partition of unity rJX [O( 1/2}. 1/21 and V = = {y lyE 1/21. Then U and V are disjoint open sets
=
6.5 Paracompactifications
185
in Y such that Z c U and Z' c V. Consequently, Z and Z' have disjoint closures in Y. (3) ---) ~ (4) Clearly Cly(ZnZ') c Cly(Z)nCly(Z'). Cly(Z)nCly(Z,). To show the reverse inclusion, let p E Cly(Z)nCly(Z'). Cly(Z)nCly(Z,). Then there exists a zero set neighborhood V of p in Y, Y. P E Cly(V nZ) and p E Cly(V nZ'). Thus (3) implies (VnZ)n(VnZ') (V nZ)n(V nZ') ~ ~ 0. Therefore, p E Cly(ZnZ'). i= 0 which in tum implies Vn(ZnZ') i= Consequently, Cly(Z)nCly(Z') Cly(Z)nCly(Z,) c Cly(ZnZ'). (4) ---) unifonn subspace ~ (5) We show ~X = ~Y by showing that (X,~) is a uniform of (Y, ~) and appealing to the uniqueness of the completion (Theorem 4.10). To do this, it suffices to show that every finite normal covering U of X can be extended to a finite nonnal = U' nX. If U is a normal covering U' of Y such that U = finite normal covering of X, then it can be shown (Exercise 4) that there exists a nonnal {Un) of finite open coverings of X such that U 1 < U, and for normal sequence {Un} each positive integer nnand and U E Un' X - U is a zero set of X. Next, notice that ••• Zn so we can extend statement (4) of the theorem to any finite collection Z 1 •.. y(n7=1 Z;) that Cl Cly(n7=1 Zi) = n7=1 Cly(Zi). Cly(Zi)' Now UE(Y) is an extension of U into Y for each positive integer n, and U~(Y) is an extension of Un into Y. Furthennore, Furthermore, by the extended version of (4), Un} = n{ Cly(X - V) IV we have for each positive integer n, n{Y - VE(y) IV E Un)
E Un} Cly(0) = 0. Consequently. Consequently, U~(Y) U~(y) is a Un) = Cly(n{ X - V IV E Un}) Un)) = Cly(0) UE(Y) is a covering of Y by covering of Y for each positive integer n, so WrY) Proposition 6.6. Also, by Proposition 6.6, U n+ 1 . By Proposition 6.10, ~AX = ~X so CL.j3,Y(O(:.J) Cl.J3~(O( nJ.!X.'Fhen put '1')" 'VA = )" ':uX '(x) [x] for each x E uX is continuous. Let (M*, d*) be the completion of '(x) = [xl (M,d). Now put 4> == j © i © 4>' ' where i:M ~ -+ M* is the imbedding of M into its I3M* is the imbedding of M* into its Stone-Cech Stone-tech completion and j:M* -+ pM·
187
6.S Paracompactifications 6.5
:uX ~ -7 ~M*. Let rJ ~ be the extension of over BuX ~uX = = compactification. Then :uX ~X. Since ~X and BM* ~M* are both compact, rJ ~ is closed and the inverse image of each compact set in ~M* is compact in ~X. Put Y = (~)-l (~tl (M*) and let ~ ~ = ~ II Y. Then ~:Y ~:Y -7 ~ is rJ ~ M* and since Y is the total inverse image of M*, ~ also closed and the inverse image of each compact set in M* is compact in Y. ~ is a perfect mapping. By Lemma 6.15, Y Y is paracompact. Therefore, ~ Clearly, uX c Y. By a proof similar to the proof of Lemma 6.12, we can show that d*:M* x -7 [0,1]. [OJ]. Set g = = d d~P © M* -7 ~ [0,1] has a continuous extension d~:M* x ~M* ~ (~ x ~). ~). Then g EE C*(Y xX BX) ~X) and g(Y x BX) ~X) c [0,1]. Let (x,y) E X xX X. (~ E X = drJ(~(x),~(y» d~(~(x),~(Y» = = d~(x,y) == d*(x,y) == b(x,y) S(x,y) ~f(x,y). Therefore, Then g(x,y) =
fl X x X
~ g II X x X which implies
fl X x BX ~X
~ g IIX x ~X.
Hence 1 = =f(x, p) ~ g(x, p) for each x EE X. Therefore, 1 ~ g(x, p) for each y EE = b(x, Sex, x) == 0 for each x EE X, g(y,y) == 0 for each y EE Y. Y. Since g(x, x) = Consequently, p does not belong to Y. Hence Y Y is a paracompact subset of ~X {p} {p) containing uX. -
lfrr 1t is the finest uniformity for a space X such that THEOREM 6.17. If rrX is paracompact, then 1t rr = u (the universal uniformityfor uniformity for X). the completion 1tX Il{Y c BX ~xlx paracompact). Clearly PX *Proof: Put PX = = n{Y IX c Y and Y is paracompact}. # 0 rr is finer than ~. Therefore, 1tX rrX c ~X which and since ~X is paracompact, 1t rrX. Now, for each paracompact space Y such that X eYe ~X, implies PX c 1tX. it is easily shown that Y is the completion of X with respect to some uniformity = uX. 'A A that is finer than B. ~. Then by Theorem 6.15, uX c PX. Therefore, PX = ux. rr, namely, u (the finest) and There are two uniformities on X that bound 1t, sup{'AI'AX 'AX is paracompact}, paracompact), which we denote by supA. sup'A. Clearly sup{AI AX c ~X and AX sup'A c 1t rr c u. Now 1tX rrX is paracompact and PX c 1tX. rrX. We claim that the supA sup'AX of X with respect to SUPA sup'A is PX. Since X c PX c 'AX completion supAX AX for each uniformity unifonnity 'A AX is a paracompact subset of ~X, A~ A such that 'AX 'A' (the uniformity of the completion 'AX) AX) induces a uniformity A 'A+ on PX and the A on X. Consequently, supA sup'A++ == sup sup{'A+) uniformity 'A {A+} is a uniformity on PX that induces the unifonnity uniformity SupA sup'A on X. Thus (X, supA) sup'A) is a dense uniform subspace sup'A+) is complete, let 'V \jI be a Cauchy net in of (PX, sup'A+). To see that (PX, supA+) sup'A+. Then 'V \jI is Cauchy in PX with respect to each A 'A+. But PX with respect to SUpA \jI is Cauchy in 'AX \jI converges to some P A A. E E 'AX then 'V AX which implies 'V AX c ~X for each A. = P Jl~ for each pair of uniformities 'A, A, J.l 'A. Since ~X is Hausdorff, p AA. = ~ such AX and J.lX that 'AX ~X are paracompact subsets of ~Xo ~X. Therefore, there exists apE ~X such that 'V \jI converges to p and p E E PX since p = p A.A E AX for each A. Hence (PX, SupA SUPA+) is complete, and since completions are unique, SUpAX = PX.
188
6. Paracompactifications
rrX == uX. Since PX c 1tX, rrX, n: rr induces a uniformity unifonnity n:+ rr+ Finally, we show that 1tX sup/...+ rr is finer than sup'A. sup/.... Then (X, n:) rr) is a dense on PX that is finer than sup 'A+ since n: rr+). Now (PX, n:+) rr+) is complete since (PX, sup'A+) sup/...+) is uniform subspace of (PX, 1t+). rr+ c sup' sup/...+, rrX = PX = uX. complete and n:+ A.+, and since completions are unique, 1tX rr == u. •But then uX is paracompact, so by hypothesis, n: Analogously to the definition of topologically complete, a space will be called topologically preparacompact if it is preparacompact with respect to its unifonnity. finest uniformity.
THEOREM 6.18 (N. Howes, 1992) A necessary and sufficient condition for the existence of the Tamano-Morita paracompactification is topological preparacompactness. Proof: Suppose X is topologically preparacompact and let u be the universal unifonnity on X. Then uX is paracompact which implies that n: rr exists and uniformity equals u. Conversely, if 1t rr exists, then by Theorem 6.17, 1t rr = u which implies that uX is paracompact which in turn implies that u' is cofinally complete. Therefore, u is preparacompact which implies X is topologically preparacompact. •We now show that some of the results in Chapter 4 can be generalized by replacing cofinally Cauchy nets with almost Cauchy nets. Part (1) (I) of the following theorem is due to J. Isbell and appeared in his paper Supercomplete spaces referenced in Chapter 5. (1. Isbell, 1962, N. Howes, 1992) Let X be a THEOREM 6.19 (J. Tychonoff space and let u be the finest uniformity on X. Then (1) X is paracompact if and only if (X, u) is supercomplete, (2) X is Lindelof LindelOf if and only if (X, e) is supercomplete and X is compact if and only if (X, P) (3) X ~) is supercomplete.
Proof: (1) is essentially the equivalence of (1) and (2) in Theorem 5.8. To Linde16f prove (2), we need only show that if (X, e) is supercomplete then X is Lindelof since by Theorem 4.4 and Corollary 5.1, a Lindelof space is supercomplete supercomp1ete with respect to the e uniformity. unifonnity. By Lemma 4.1 it will suffice to show that each (I)-directed (countably directed) net is almost Cauchy with respect to e. For this ro-directed ",:D ~ -7 X be an ro-directed (I)-directed net. Let U E e. Then there exists a countable let o/:D nonnal covering {Vi} {Vii that refines U. For each i put C ij = Vii normal ={d E D I",(d) 0/( d) E Vi} and let E = u{ C ij IC ij is not cofinal in D}. If E = 0 pick any d E D. Then {Cij }} is a collection of cofinal subsets of D such that each d' ;::>: ~ d is in some C ij and for ",(Cij )) cC Vi V j cC U for some U E U. each i, 'V(C If E =F "# 0, then for each i such that C ij is not cofinal in D, there exists a d ij E R(dj)nCij = 0 where R(d R(dj) Dld (I)-directed, D with R(di)nC Idij ~ d}. Since", is ro-directed, i ) = {d E D there exists a d' E D with dij ~ d' for each i such that C ij is not cofinal in D.
6.5 Paracompactifications
189
R(d')ilC ij = = 0 for each i such that Cij is not cofinal in D. Therefore, Hence R(d')nC R(d')nE CjnR(d'). Then {D jj }} is a collection of R(d')ilE = 0. For each j put D jj = CjilR(d'). cofinal subsets of D such that each d+ ~ d' is in some OJ D; and for each J, j, \V(D \jf(D jj )) C Vjj C U for some U E U. Since U was chosen arhitrarily, 'V \jf is almost Cauchy. V
To prove (3), we need only show that if (X, P) ~) is supercoInplete supercomplcte then X is compact since by Theorem 4.4 and Corollary 5.1, a compact space is p. For this, it will suffice to show that each net in supercomplete with respect to to~. to~. \jf:D ~ X be a net in X. Let U E p. ~. X is almost Cauchy with respect to B. Let \V:D Then there exists a finite normal covering {V 1I ... ••• V n} that refines U. For each \jf(d) E V Vi}' i = 1 ... n put Cij = (d {d E D I 'V(d) j }. Then an argument similar to the above can be used to show that there exists a d E D such that the collection {C ij I Cij is cofinal in D} D) is a collection of cofinal subsets of D such that each d' ~ d is in SaIne some Cij and for each i, 'V(C \jf(C;) Vij C U for some U E U. Therefore, each net i) C V ~. • in X is almost Cauchy with respect to B. If instead of Theorem 4.4, we were to use Theorem 6.18 to motivate our definition of preparacompactness, then we would define preparacompactness to be the property that each almost Cauchy net has a Cauchy suhnet. To he almost distinguish between these two notions we define a uniform space to be paracompact if each almost Cauchy net has a Cauchy subnet. Then we can prove:
THEOREM 6.20 (N. Howes, J.l) be a Uniform uniform space and Howes. 1992) Let (X, (X. fJ.) let v be the uniformity on X derivedfrom ]"'hen X derived from J.l. fJ.. Then (X. fJ.) (X.V) is almost (1) (X, J.l) has a paracompact completion if and only if (X,v) paracompact. Lindelof completion if if and only if (X,v) (2) (X, (X. J.l) fJ.) has a LindeLOf (X.V) is countably bounded and almost paracompact. Proof: To prove (1) let (X', fJ.') J.l') be the completion of (X, J.l) fJ.) and let u' be he the finest uniformity for X'. Assume (X,v) is almost paracompact and that \V:D \jf:D ~ X' is an almost Cauchy net with respect to u'. Let E = D xX u' and define < on E (e,V') if d < e and V' ~ e belongs to some £Ea. a' For this choose W' E u' such that W' e then e' = j3 for some index P (d', Z') for some d' > d and Z' ")~}. Recall from {X/Jc, dx)} E uX. It's extension uniform <j>~:~X ~ ~uX ~X can be seen to be the mapping defined by ~(x) = = {~(x)} " show that h is continuous. Since ~:uX following diagram:
X
ff ~
~~ t~
T
t
uX ~ X/' t-! X/c.l>" w = h © 0 for each x E X. Consequently, g-1 defined by g-I(X) g-1 (x) = = l/g(x) is continuous. Therefore, g-1 E C. Since 1t is the product uniformity on P, 1t is the coarsest uniformity on P that f E C, makes all the canonical projections uniformly continuous. But for each IE Pt uniformly continuous. Since for each U E 1t, there is a Z E Z PI = = I,f, so g-1 is unifonnly g -1 is uniformly continuous, then for with Z c U for some U E U, and since g-1 £ > 0, there is a Z E Z such that g-1 (Z) is contained in a set of diameter E. £. each E n++1. 1. Hence there exists a sufficiently large Z(gn), then g(x) < 2- n Now if x E Z(gn)' n with g-I(X) > max{g-I(Y)ly max{g-I(y)ly E Z} for any x E Z(gn). Z(gn)' This implies that Z(gn)nZ = n'/=1 Z(fi) Z(Ji) which implies that Z(gn) E Z. Then Z = 0. But Z(gn) == n7=1 does not satisfy the finite intersection property which is a contradiction. Therefore, Z is a CZ-maximal family. Consequently, by property (2), Z has a pEn non-empty intersection. Let p E n {Z IZ E Z}. Let U E 1t' and choose V E 1t' such that V* < U. Suppose no Z E Z meets S(P,V). Since there exists a Z(V) E Z and a V E V with Z(V) c V, we see that p does not belong to Z(V) which is a contradiction. Therefore, Z(V)nS(p,V) ~ "# 0 which implies there is an open V'nV ~ 0 so there is aaU U E U with V\..jV neighborhood V' of p such that V'nV"# V\..;V cU. But then Z(V) c S(p, U) and Z(V) contains H ua for some index u. a. Hence, {xu} {x a} converges to p so (X, 1t') is complete. This completes the proof of (2) """"* ~ (3).
To show (3) ~ """"* (1), let X be a closed subset of a product P = naR IlaR aa where Ra = = R for each index u. a. Since R aa is complete for each u, a, so is P. Let 1t denote the product uniformity on P and let 1t' denote 1t restricted to X. Since X is closed in P, (X, 1tl 1t) is complete. It will suffice to show that e is a basis for 1t'. For this let U be a basic member of 1t. Then
uniform coverings Uai a, denotes for some finite collection of unifonn ai where P p ai a, of R a, Raj the canonical projection of Ponto R ai for each i = = 1 ... n. Now each UUaiaj has a countable subcovering V R ai is Lindelof. Vaj LindelOf. Furthermore, each V Va, ai since Raj ai is nonnal ai is paracompact. Therefore, p~~ . . np~~ (Van) is a normal since R Raj P~~ (Val)n . ... npu~(Va")
206
7. Realcompactifications Reaicompactifications
member of the e countable normal covering of P that refines U. Hence U is a melnber uniformity for P. But then e is a basis for the n' uniformity on X. Therefore, Therefore~ e = = n', so X is e-complete. •Now that we know realcompact spaces are precisely the e-complete spaces, we can use this fact to derive some interesting properties about realcompact spaces.
THEOREM 7.3 Every Lindelof Lindel6f space is realcompact. reaicompact. Lindelbf spaces are cofinally complete with respect to the e This is because Lindelof uniformity by Theorem 4.4 and hence complete with respect to e hy by Corollary 4.1.
THEOREM 7.4 A closed subspace of a realcompact reaicompact space is realcompact. reaicompact. This is because if Y is a closed subspace of a realcompact space X, then Y is complete with respect to the e uniformity of X relativized to Y by Proposition 4.18. But the e uniformity of X relativized to Y is precisely the e uniformity of Y.
reaicompact spaces is realcompact. reaicompact. THEOREM 7.5 A product of realcompact Proof: Let X = nax a~ let e a be flaX a where each X a is realcompact and for each a, the e uniformity of X a' a. Then for each a, (X a' a~ e a) is complete so X is complete with respect to the product uniformity by Exercise 4 of Section 5.3. Let J.l /-l /-l is of the denote the product uniformity of the e a's. Then a basis element of J.l form
for some finite collection of uniform coverings U a , of e a, where each U a, a , is countable and where p a, aj denotes the canonical projection of X onto X aa,, for each i = 1 ... n. Then U is also countable and therefore belongs to e (the e uniformity of X). Hence J.l (X~ J.l) /-l c e. Since (X, /-l) is complete and e is finer than J.l~ /-l, realcompact.we have that (X, e) is complete so X is realcompact. •
normal sequence of We have already seen in Sections 2.2 and 5.5 how a nonnal open coverings can be used to construct a pseudo-metric. If we consider the famil familyy cI>
A { 0 there exists a d'
E
D
7.2 Realcompact Spaces
207
and a b 8 > 0 such that d'(x,y) < b 8 implies d(x,y) < £e for all x,y E X, then d ED, x -:/- y, there exists ad a d E D with d(x,y) "# -:/- O. (3) whenever x"# where dye is defined by dve(x,y) = = min {d(x,y), e(x,y)} and is called the join of d and e. Property (3) only holds for Hausdorff uniform spaces and property (2) implies that if ddEE D, then rd E D for each r> 0; and if ddEE D and e :::; ~ d then e E D. To show property (l) (1) holds let d 'A, 'J..., deE D and let 0 with d'(x,y) :-::; 8 implies d(x,y) :-::; and £ > 0, there exists aad' ~ () ~ £ for each pair x,y E X, then B is called a base for D. It is left as an exercise (Exercise 4) to show that if S is a subbase for D, then the family B of all finite joins d 1I V ••• vdn such that dd;i E S for each i = = 1 ... n is a base for D. The family C(X) of all real valued continuous functions on X and the family C*(X) of all real valued bounded continuous functions on X can be used C* as follows: for each IE f E to generate two pseudo-metric uniformities C and c* C(X) let dff be defined as
= II(x) If(x) - 1(Y) fCy) II dlx,y) = for each pair x,y E X. It is an easy exercise (Exercise 5) to show that dfr is a (df If E C(X)} and C* = {drllE (df If E C*(X)}. Let c pseudo-metric on X. Let C = {dfllE and c* denote the covering uniformities associated with C and c* C* respectively. Then c and c* have bases of the form
= {u{ b = {U{ I df E C and £ > 0 O}} b* = {U{ {u{ Idf E c* C* and e £ > O}
(Six, £) Ix I x EX} and where Six, e) £) is the sphere respectively, where U{ = {SIx, about x of radius ££ with respect to the pseudo-metric df . We now show that c c e and c* cc~. B. For this, let U E c*. Then U = U{ inf a and sup b for some If E C*(X) and £ > O. Since If E C*(X) there is some inl f(X) c [a,b]. If mayor may not assume the end points a and b. But such that I(X) since If is continuous, it assumes each point P p such that a < P < b. Let 8B = = £/2 f(P 1) I) = f(x) = = a if I(x) = a for some x E X or else pick PIE and pick PIE X such that 1(P f(P 1) I) = a + e. £. If a + £ > b pick any x E (a,b) and put P 1I = x. For X such that 1(P each positive integer k pick Pk+1 Pk+l E X such that l(Pk+l) = f(Pk) + £ if f(Pk) ~ thatf(Pk+l) f(Pd + ££:-::; b. Otherwise put Pk+l = b unless f does not assume the value b in which case Pk+1 = PhI =Pk. Let n be the least positive integer such thatf(Pn+l) = f(Pn). put Pk+l
209
7.2 Realcompact Spaces
Clearly {S/'pi, E) Ii = 1 ... n} is a finite subcovering of U{. It is also clear ufs < U{ and uf, U{; E c. Therefore, every U{ where If E C*(X) has a finite that Ut nonnal normal refinement and consequently belongs to~. This shows that c* c~. This argument can be modified to show that c c e (see Exercise 6). Consequently, if real compact and if X is complete with X is complete with respect to c, then X is realcompact respect to c* then X is compact. In what follows we will show that the completion with respect to c is the Hewitt realcompactification) realcompactification, and the completion with respect to c* is the Stone-Cech compactification, but for now we merely record the following:
PROPOSITION 7.1 II If a Tychonoff space is complete with respect to If it is complete with respect to c*, it is real compact. II the c uniformity, then it is realcompact. compact. uniform continuity into the tenninology terminology of How to translate the concept of unifonn pseudo-metric uniformities follows from the next proposition, whose proof we leave as an exercise (Exercise 7). PROPOSITION 7.2 Let (X, M) and (Y, N) be pseudo-metric uniform spaces and let (X, J.l) J..l) and (Y,v) be their associated covering uniform spaces. -t Y is uniformly continuous with respect to J.l J..l and v if and only if for Then f:X ---., iflor Thenl:X each e E Nand E >0, there exists a d E M and a B 0 > 0 such that if d(x,y) < B 0 e(j(x),f(y)) < E. then e(f(x),I(y)) implication:f:X uniformly Proposition 7.2 has the following useful implication: I:X -t ---., R is unifonnly uniformity D if and only if df E continuous with respect to some pseudo-metric unifonnity D. Consequently,fis Consequently, lis uniformly continuous with respect to c or c* if and only if E C*(X) then df E C* so dff E Cor C or C* respectively. For C* this means that if fIE If is uniformly continuous with respect to c* and hence can be continuously extended to the completion of X with respect to c*. By Proposition 7.1, c*X is compact. But then by Theorem 7.1, c*X is the Stone-Cech compactification of X. Since PX pX has a unique uniformity, we have
THEOREM 7.6 c* =
p. ~.
EXERCISES 1. Let d and d' be two pseudo-metrics on X having the same spheres of radius 2-nn for each positive integer n. Show that d = = d'.
PSEUDO-METRIC PSEUDO-ME1RIC UNIFORMITIES
pseudo-metrIc uniformity on X and let v = {U~ IdE D and E > O} 0I 2. Let D be a pseudo-metric where U~ = = {Sd(X, e) x EX}. E) IIx E Xl. Show that v is the basis for a covering
210
7. Realcompactifications
uniformity that generates the same topology on X as the pseudo-metric uniformity. 3. Show that the intersection of any collection of pseudo-metric uniformities is again a pseudo-metric uniformity. 4. Let S be a subbase for a pseudo-metric uniformity D and let B be the family of all finite joins d 1V . . vd n such that d i E S for each i = = 1 ... n. Show that B 1V . ... is a base for D. 5. Let f E C(X) and define df by dAx,y) dtCx,y) = If(x) - f(y) fey) I for each pair x,y thatfis f is a pseudo-metric on X. Show that
E
X.
6. Show that the e uniformity is finer than the c uniformity.
7.3 Realcompactifications Realcom pactifications By a realcompactification of a Tychonoff space X, we mean a realcompact space Y in which X can be imbedded as a dense subspace. From Theorem 7.2 it is easily seen that a realcompactification Y of X can be realized as the completion of X with respect to some countably bounded uniformity. To see this, note that if Y is realcompact, it is e-complete, and the e uniformity of Y relativized to X is countably bounded. The completion of X with respect to this uniformity is Y. Furthermore, e is the finest countably bounded uniformity on X. For this, first notice that the e uniformity is countably bounded since it has a basis consisting of countable normal coverings. Next, if J..l J...l is a countably bounded uniformity of X and U is an open member of J...l, J..l, then there is a normal sequence {Un} of open members of J..l J...l such that U 1 < U. Let V be a countable subcovering of U and for each positive integer n put Vnn = = Unn V. Then {V n} n} is a normal sequence of open coverings of X such that VI < V C U. Therefore, U E e so e is finer than J...l. J..l. Also, this shows that the countable normal coverings V such that V < U for some U E J...l J..l form a basis for J...l, J..l, so J...l J..l has a basis consisting of (perhaps not all) countable normal coverings. Therefore, the countably bounded uniformities are precisely the ones that have bases consisting of countable normal coverings (i.e., they are the separable uniformities). Shirota showed that if X is a Tychonoff space, that eX is precisely the Hewitt realcompactification uX. He did this by proving the following:
7.3 Realcompactifications Reaicompactifications
211
THEOREM 7.7 (T. Shirota, 1951) Let X be a Tychonoff space. Then eX has the following properties: (1) X is dense in eX, (2) eX is realcompact, eachfEE C(X) can be continuously extended to eX. (3) eachf Also, any space satisfying these three properties is homeomorphic with eX. Clearly eX satisfies property (1) of Theorem 7.7. Shirota states in his proof that it is obvious that eX satisfies property (2) also. This would be the case if we knew, for instance, that e' (the e uniformity of eX) is the e uniformity of eX. Although it is not surprising that this should be so, the proof is not what we usually think of as being obvious. Consequently, we first prove this as a lemma before proving Theorem 7.7.
LEMMA 7.1 Let e' be the uniformity of eX. Then e' is the e uniformity afeX. of ex' Proof: Each U' E e' has a countable normal refinement. To see this, let V' = {v~nX} = {V~} be a closed uniform refinement of U'. Then V = = {V~nX} = {V J3} E e and therefore has a countable uniform refinement say {Vi}. {Vi}' For each Vi there is a VJ3, E V with Vi C VJ3, VJ3 , so VJ3i
(CleX(Vii )} refines V' so W' = since V' is a closed covering. Thus {CleX(V (lntex(CleX(Vii ))} refines U'. Now W' E e'. To see this, note that {Vi} {V;} = {lntex(Clex(V {V;nX} {V~nX} for some uniform covering {V;} {V~ } of eX. Pick y E V~. Then
»}
V;.
Clex(V;) V~ c CleX(V {CleX(Vi )} is a uniform covering of eX. Therefore, V; i ) for each i so (CleX(V But then (Clex(V;)} {CleX(Vi )} has an open refinement in e' which implies W' E e'. Hence U' has a countable normal refinement. It remains to show that all countable normal coverings of eX belong to e'. For this let {W;} {W~} be a countable normal covering of eX. Then there exists a {W~}. But then normal sequence {U~} of open coverings such that CleX(U'I) < {W;}. {U~nX} is a normal sequence of open coverings of X X such that CleX(U'1 )nX refines {W;nX}. Hence, CleX(U'l)nX = Clex(U'dnX E e. But for each VEe, Clex(V) CleX(V) = (CleX(V) {Clex(V) I V E V} E e'. Therefore
Consequently, {W;}
E
e'. Therefore, e' is the e uniformity of eX. -
212
7. Realcompactifications Reaicompactifications
Proof of Theorem 7.7: It remains to show that eX satisfies property (3) and that if Y is another Tychonoff space satisfying properties (1) - (3) then Y and eX are homeomorphic. That eX satisfies property (3) is easily seen from the remarks preceding Proposition 7.1 and those following Proposition 7.2. Iff E C(X) then dfJ E C so that f is uniformly continuous with respect to c. Since c c e, f is uniformly continuous with respect to e and consequently can be uniquely extended to a uniformly continuous functionf':eX functionr:eX ~ --7 R.
To show that if Y is another Tychonoff space satisfying (1) - (3) then Y and eeX X are homeomorphic is a little more difficult. For this we will first show that any Tychonoff space Y satisfying properties (1) - (3) also satisfies the following property: (3') If Zn E Z(X) for each positive integer n, then nCly(Zn) = Cly(nZn). ayE nCly(Zn) For this suppose nCly(Zn) i= :1= Cly(nZn). Then there exists aYE Cly(nZn). For ea~h eaGh positive integer n, let fn E C(X) such that Zn = = Z(Jn) Z(fn) and Cly(nZn), there exists a zero set such that Ifn I ::; ~ 1. Since y does not belong to Cly(nZn)' Z c X such that y E Cly(Z) and Zn(nZn) = 0. LetfE Z(f) Letf E C(X) such that Z = Z(j) L,gnn where gn = = 2-nn( Ifn I + If I). I). Then g is strictly positive, i.e., g(x) and put g == Lg > 0 for each x E X.
Let g' be the extension of g over Y and for each positive integer n let g~ be the extension of gn over Y. Now g'l X = g = L,gn Lg n = Lg~ L,g~ IX. By property (1), X is dense in Y so g' = Lg~. ~ lin I + 1f'1 If' I L,g~. Also, for each positive integer n, g~ ::; where in In is the extension of fn to Y and f' f' is the extension of f to Y. = O. Hence there Consequently, g~(y) == 0 for each positive integer n, so g(y) = exists an h E C(X) such that h(x)g(x) = 1 for each x E X. Let h' be the extension of hover Y. Since (hg)' IX = hg = (h'l X)(g' IX) and since X is dense in Y, we have (hg)' = h'g' = 1. Hence g'(y) = g(y) :1= i= 0 which is a contradiction. Therefore, nCly(Zn) = Cly(nZn). We now use property (3') to show that Y and eX are homeomorphic. For this we show that (Y, e) is the completion of (X, e). We need only show that every countable normal covering U of X can be extended to a countable normal covering U' of Y such that U = UP nX. This will show that (X, e) is a uniform subspace of the complete uniform space (Y, e), and since completions are unique, Y is homeomorphic with eX. Now there exists a normal sequence {Un} of countable open coverings of X such that U 1 < U. U€(y) UE(Y) is an extension of U into Y and for each positive integer n, U~(y) U~(Y) is an extension of Un into Y. By property (3'), for each positive integer n, n{Y - VE(y)
lu E
Un}
= n{Cly(X - U)IU E
Un}
= Cly(n{X - ulu E
Un}) = 0.
U~(Y) is a covering of Y for each positive integer n, so UE(y) UE(y) is a Consequently, U~(y)
7.3 Reaicompactifications Realcompactifications
213
covering of Y. Furthermore, since Un++!1
225
LEMMA 7.3 A non-empty class of cardinals C is closed if and only if, if. whenever ME M E C, then: (1) N E C for each cardinal N < M, M. (2) the sum of any M members of C is in C, M 2M (3) 2
E
C.
Proof: The necessity of conditions (1) - (3) is obvious. Conversely, assume C satisfies conditions (1) - (3). We need to show that C contains products, M so exponentials, suprema and successors. For this let M E C. Then M + 1 0, {Z( g) E Z(Y) 10 "# '# Z(f)nZ c Z( g) for some Z E Z} is a CZ-maximal then Z' = {Z(g) Z(g) family on Y. Proof: It is easily shown that Z' is a non-empty subfamily of Z(Y) with the shOW that Z' is countable intersection property. Therefore, it suffices to show maximal with respect to the finite intersection property in Z(Y). For this suppose Z' is not maximal, i.e., suppose W is a non-empty subfamily of Z(Y) with the finite intersection property that contains Z' as a proper subset. Let Z(g) = Z(g)nZ(j). Then FoE W so F 00'# 0. E W - Z' for some g E C(Y) and put F 0 = NowputF 1 =Y.
*
7.5 Shirota's Theorem
227
Illet (xlf(x) ~ r}, u~ U~ = = For each rational r E [0, 1] let Ur = (xlf(x) {xl/(x) < r}, Zr = {xl/(x) (x Ig(x) ~ r}. Put Gr = UrnU~ and Fr = {x F r = ZrnZ~. Then G r Frr is closed. For any pair of rational numbers r, s E [0, 1] with is open in X and F r < s we have F g'(x) = sup{ r IIx x does not belong Frr c Gs c F Fs. s • For each x E X put lex) o. For to F l E C(X) and Z(g') = rlF nFrr = nr(ZrnZ~) = Z(j)nZ(g) = F Fo. Fr}. r }. Then g' each r, F Frr EE Z(X) and Z(j) c Fr. Therefore, each F Frr E Z which implies nrFr E Z. Hence FOE FoE Z. Then by the definition of Z', Z(g) E Z' which is a contradiction. Therefore, Z' is measurable after all. (x I g(x) < r} and Z~ {x
THEOREM 7.15 (T. Shirota, 1951) Let X be a Tychonoff space such that IX I is non-measurable. Then X is realcompact if and only if if X is complete with respect to u. Proof: Assume X is complete with respect to u. It suffices to show that if Z is a CZ-maximal family of X and U EE u, there is aBE Z and a U E U with B cU. c U. The reason for this is that if this is the case, then a proof similar to (1) ~ (2) of Theorem 7.2 will imply that there is apE X with p E nZ. Thus X will be normal sequence realcompact. For this, observe that U E U implies there is a nonnal {Un} with U 1 1. I. of!-l*, !-l*(A):l!-l(A n >1 Jl(A n) = !-leA) then by Proposition 8.5 !-leA) Jl(A) : a. Therefore, «a, 00]). If x E g-l «a, 00]), then g(x) E (a, 00], ur,/ there exists a positive integer k such that a < hex) fk(X) ~ g(x). Hence hex) fk(X) E (a, 00] -I «a, 00]) cC Uh U]"Il «a, 00]). which implies x E .fkl fkl «a, 00]). Therefore, g g-l l u]"I«a, Conversely, if x E uh «a, 00]) then x E .fkl«a, fkl «a, 00]) for some positive integer k which implies hex) fk(X) E (a, 00] which in tum implies fk(X) fk(x) > a. Since sup{fn(x)}, g(x) = = sup {fn(x) }, we have a < hex) fk(X) ~ g(x) which implies g(x) E (a, 00]. But then x E g-l«a, g-I«a, 00]) so uh1«a, u]"I«a, 00]) Cc g-l«a, g-I«a, 00]). Hence g-l«a, g-I«a, 00]) = U],,1l «a,oo]). Since each fn is measurable, measurable,],,1 Uh hI «a, 00]) is measurable. Since Mis M is 1 a a-algebra, Uh u]"I«a, g-I«a, «a, 00]) E M so g-l «a, 00]) is measurable. Therefore, g is a measurable function.
A similar argument holds with infreplacing sup. Now since h = lim sup fn' for each x E X, h(x) = infk~l {suPn~k{fn(x)}}. For each positive integer k put k hex) = inA~1 {suPn;::dfn(x)}}. k = = sUPn~k{fn(x)}}. sUPn;::dfn(x)}}. Then for each k, k k is measurable as shown above. But h(x) hex) = inh;::1 (k(X)} and so h is measurable. •infk~l {k(X)}
COROLLARY 8.5 The limit of every point-wise convergent sequence of complex measurable functions is measurable. Proof: The proof follows immediately from Proposition 8.17 for the real valued case. Proposition 8.12 can then be used to establish the complex case.· case.COROLLARY 8.6 Iff and g are measurable functions from a space X into [-00,00], then so are max{f, g} and min{f, g}. g).
r r -r·
A special case of Corollary 8.6 is the function f+ = = max{f, O}. f- = = -min {f, O}. f+ is called the positive part of f while f- is negative part of f. Clearly If II =f+ + rand f- and f =f+ - f-·
r
r
r
r
Another is called the
A function s on a measurable space (X, M) whose range is a finite subset of [0, (0) ••• an} is the range of the simple 00) is called a simple function. If {a 11 ... sex) == ai}, ai), then: function s and if Ei Ei = = (x {x E X Is(x)
8.5 Measurable Functions
247
where XE, is the characteristic function of E E;i for each i = = 1 ... n. Clearly, sS is measurable if and only if E E;i is measurable for each i. In this case, S is called a simple measurable function. -7 [0,00] THEOREM 8.7 Iff:X -t [0, ex>] is a measurable function , there exists a X such that sequence {sn} of simple measurable functions on X (l)05, 5, s 22 5:· 5, ..··f, . f, (1)05:ss 1l 5:s {sn(x)) converges to f(x) for each x E X. (2) {sn(x)}
n], put En, = n2n], Proof: For each positive integer n and for each i E [1, n2 n 1 1 f- ([ani ,b n) where an, = (i - 1)/2 1)/2.11n and b nin, = i/2 n and let F ex>]). Then, r1([an"b Fnn = f- 1 ([n, 00]). for each positive integer n, define Sn by
,»
r
By Theorem 8.6.(2), the En; and Fn F n are measurable sets in X. Therefore, the Sn are simple measurable functions. Let m, n be positive integers such that m < n and let x E X. To prove (1), there are three cases to consider: Case 1: I: (f(x) (l(x) sn(x).
E
Fn) Fnn Cc F Fm,f(x) F m' f(x) n) Since F
E
Fm= F m. Therefore, sm(x) = m < n =
(l(x) E Fm Fn) Case 2: (f(x) F m butf(x) but f(x) does not belong to F = m. Let k n) Then sm(x) = be the least positive integer such that x E E nk . Then k > m 22.11n which implies a nk > (m 2.11 - 1)/2.11 (x) ~ (m2n 1)/2n = m - 1/2.11 l/2 n which in turn implies a nk ~ m. Therefore, Sn sn(x) sm(x). Case 3: (f(x) (l(x) does not belong to F m) Let k be the least positive integer such n- m). Hence Emj where jj = k mod (2 (2n-m). that x E E nk . Then x E Em}
5, sn(x). In all of these cases, sm(x) 5: 5, sn(x). To show that for each n, Thus sm(x) 5: 5, f, first note that if f(x) = 00 sn(x) 5, f(x) for each n. Sn 5: ex> for some x E X that Sn (x) 5: Therefore, suppose x E X and f(x) < ex>. 00. Let k be the positive integer such that f(x) E [k, k + 1). I). Then Sk(X) == k 5:f(x). 5,f(x). For each positive integer n > k, there are 2.11 2n subintervals
[(k
+ i - l)/2 n , (k + i)/2n]
I). Let i be the positive integer such that = 1I .. . 2n 2.11 that partition [k, k + 1). where i =
8. Measure and Integration
248
n n ~f(x). Consequently, -1)J2n, [(k + i -1)/2 i)/2 n]. Then sn(x) = (k + i - 1)/2n , (k + i)/2n]. ~ffor ~ffor each positive integer n. This proves (1).
f(x) Sn
E
= 00 then sn(x) == n for each positive To prove (2), first note that if f(x) = integer n so {sn(x)} converges to f(x). Therefore, assume that f(x) < 00. As that f(x) E [k, k + 1), then for each shown above, if k is the positive integer such that/ex) n n 1)/2n, i)/2n] positive integer n > k, there are 2nn subintervals [(k + i - 1)/2 ] that , (k + i)/2 n n 1)J2n,, (k + j)/2 j)/2nj partition [k, k + 1) andf(x) belongs to one of them, say [(k + j - 1)/2 ] n n n l)/2 . Hence If(x) - sn(x) I < 2- . Consequently, the and sn(x) = = (k + j - 1)/2 (f(x) - sn(x)} is Cauchy and therefore converges to 0. O. Therefore, sequence {f(x) (sn(x)} {sn (x)} converges to f(x). This proves (2).(2). ?
COROLLARY 8.7 Sums and products of measurable functions into in£O [0,00] [0.00] are measurable.
Proof: Letfand g be measurable functions from X into [0,00]. By Theorem 8.7, {g n} of simple measurable functions such that 0 there exists sequences {fn} and {gn} ~ fl ~f2 ~ h ~ ... ... ~f, ~ f, 0 ~ g (fn(x)} converges tof(x) to f(x) and {gn(x)} (gn(x)} ~fl glI ~ g2 ~ ... ~ g, {fn(x)} converges to g(x) for each x E X. It is an easy exercise (Exercise 1) to show that sums and products of simple measurable functions are simple measurable functions. Consequently, {fn + gn} and {fngn} are sequences of simple measurable functions. We can also show (Exercise 2) that {[fn ([fn + gn](x)} converges to [f + g](x) and {[fngn](x)} ([fngn](x)} converges to [fg](x) for each x E X. andfg measurable.Hence f + g and fg are measurable. -
°
EXERCISES 1. I. Show that sums and products of simple measurable functions into [0, 00] are simple measurable functions.
[0,00] 2. Show that if {an} and {b n } are sequences in [0, 00] such that 0 ~ a 1I ~ a 2 ~ •• . , 0 ~ b 1 ~ b 22 ~ ", .. . . ~, {an} converges to a and {b n } converges to b, then I an + {an b n } converges to a + b and {anb n } converges to abo o
?
3. Show that the set of points at which a sequence of measurable real valued functions converges is a measurable set. Ix 4. Show that if f is a real valued function on X such that {x f is measurable. measurable for each rational number r, then [is
E
X IIf(x) [(x) ~ r} is
IAn} be a sequence of measurable sets in a a-algebra M and let J.l J..l be a 5. Let {An} measure on M. Define the limit inferior denoted lim in[A infAnn of the sequence
249
8.6 The Lebesgue Integral
(An 1 by lim infA infAnn = = U;=l U;;'=1 (nk=nAn) and define the limit superior denoted lim {An} supAn by n;;'=1 n;;=1 (Uk=nAn). (uk=nAn)' Show that infA n ) ::::: ~ lim infJl(A (a) Jl(lim Il(iim injAn) injll(An) n) and Il(U;;'=1 An) sUPIl(A nn)) ~ ::::: Jl(lim Il(lim supAn). ) < 00 then lim supJl{A (b) if Jl{u;=lA n
8.6 The Lebesgue Integral In the theory of integration we often encounter the concept of infinity and the symbols 00 and -00. We have already defined a measure 11 IJ- to be a set function on a set X into [0, 00]. 00 L In order not to have to make special provisions for dealing with these concepts and symbols in some of the following theorems, we define [-00,00) addition (+) and multiplication (x) on [-00, 00] as follows: a + 00 = = 00 + a == 00 for each a such that -00 < a, a - 00 = -00 + a = -00 for each a such that a < 00, a x 00 = = 00 x a for each a such that < a, = 00 x == 0, x 00 = a x 00 = 00 x a = -00 for each a such that a < 0.
°
°
°
With these definitions, it can be shown that the commutative, commutative. associative and 00). Since -00 + 00 and (-oo) (-00) x (00) are not defined, distributive laws hold for [0, 00]. we cannot extend these laws to [-00, 00], but fortunately, we will not need to. [-00,00], [0,00) The cancellation laws also hold in [0, 00] with the following modifications: a + b = a + c implies b = c if a < 00, = ac implies b = c whenever < a < 00. ab =
°
an 1 c [0, 00) is the If s is a simple measurable function on X where (a {a 1 •••• • • an} Eii = (x Xls(x) ad, and if M M is a range of s and for each i = 1 ... n, E {x E X I sex) = ai}, M. we define the a-algebra on X and 11 J.l is a measure on M, then for E E M, Lebesgue integral of s with respect to 11 f.l as:
fEsdll = 1:7=1 IEsdll L?=1 aill(EnE aill(EnEi). i ). If f:X j:X ~ [0,00] [0,00) is a measurable function, function'l we define the Lebesgue integral of fj with respect to 11 J.l to be the supremum of all simple measurable functions s such i.e., that 0::::: ~ s :::::j, ~ f, I.e.,
°
°: :
IEfdll I Efdll = sup I{IIEsdll EsdlllI0: 0 we call the set B(a, e) = = {xl ai ::;~ Xi < ai + e} 0
0
8.8 Linear Functionals and Integrals
263
the box at a with side £. E. We will also refer to such a box as an E-box. £-box. For each Am be the set of all vectors in R R"n whose coordinates are positive integer m let Am Bmm be the collection of all 22-"n boxes at vectors vector& integral multiples of 2-" 2- n and let B XE An. A".
15. Show the following: R"n lies in one and only one member of B Bm. (1) For each m, each x E R m• (2) If U and U 2 E B where m < k then either U 1 C U 2 or U,1 E 8 Bm Bk leU m k U 1 nU 2 =0. U,nU mn " (3) If U E 8Bm = 2- mn m then v(U) = (4) If m < k and U E BBm then An has precisely 2(k-m)n A" 2(k~m)" vectors in U. m (5) Each non-void open set in R" R n is a countable union of disjoint Bm. boxes, each belonging to some B m• 16. Show that there exists a positive, complete measure m defined on a Min R"n having the following properties: a-algebra M in R (1) m(U) = v(U) for each £-box U. n (2) M contains all the Borel sets in R • R".
(3) E E M if and only if there exists an F cr set F and a G 8Ii set G with FeE c G and meG - F) = =o. (4) m is regular. (5) For each x E R R"n and E E M, m(E + x) = m(E). If J.! is a (6) The property (5) is called translation invariance. If)l positive, translation invariant Borel measure on R R"n with J.!(K) )l(K) < for each ompact set K, then there exists a real )leE) = = rm(E) for each Borel set E in R". Rn • number r with J.!(E) 00
Chapter 9 HAAR MEASURE IN UNIFORM SPACES
9.1 Introduction In 1933, in a paper titled Die Massbegriff Massbegrifj der Theorie der Kontinuierlichen Gruppen published in the Annals of Mathematics (Volume 34, Number 2), A. Haar established the existence of a translation invariant measure in compact, f.l in a separable, topological groups. Translation invariance of a measure J.! f.l(E + x) = J..l(E) f.l(E) for topological group G means that if E is a measurable set then J.!{E {Y E G Iy = G). E + x is called each x E G. Here, E + x = = {y = a + x for some a E G}. the x-translate of E. The transformation T Txx defined on G by Tx(y) = Y + x is called the xx-translation -translation or simply a translation. Topological groups will be defined later in the chapter and these concepts will be developed formally.
In 1934, in a paper titled Zum Haarschen Mass in topologischen Gruppen (Comp. Math., Volume 1), J. von Neumann showed the uniqueness of Haar's measure, and in 1940, A. Weil published L'integration dans les groupes topologiques et ses applications (Hermann Cie, Paris) where Haar's results I. Segal were extended to locally compact topological groups. In 1949, 1. extended Haar's results to certain uniformly locally compact uniform spaces that we will call isogeneous uniform spaces (Journal of the Indian Mathematical Society, Volume 13). In 1958, Y. Mibu, evidently unaware of Segal's work, independently established similar results for this same class of spaces (Mathematical Society of Japan, Volume 10). The Haar measures of both Segal and Mibu were Baire measures rather than Borel measures. Recall from Chapter 8 that Baire measures are defined on a smaller class of sets than Borel measures, namely, on the smallest a-ring containing all the compact sets. In 1972, G. Itzkowitz extended Haar's results to the Borel sets of a class of locally compact uniform spaces (Pacific Journal of Mathematics, Volume 41) that he called equi-homogeneous uniform spaces. That equi-homogeneous uniform spaces are equivalent to isogeneous uniform spaces is the subject of Exercise 2 at the end of this section. Itzkowitz showed the existence of a Haar integral (translation invariant, linear functional on the set of real valued continuous functions with compact support) for locally compact equihomogeneous unifonn uniform spaces. His approach was to show that a locally compact f.l) is homeomorphic to a quotient GIH G/H of equi-homogeneous uniform unifonn space (X, J..1) topological groups, where H is a stability subgroup of G, and then apply Weil's
9.1 Introduction
265
theory of invariant measures on these quotients. quotients~ as recorded in Chapter 3 of L. Integral. (Van Nostrand, New York~ York, 1965) to obtain a Nachbin's book The Haar Integral~ unique Haar measure. Itzkowitz's approach involved showing the modular theorems in Weil Weil's function on H is constant and then appealing to theorelns ~ s theory to show this implies the existence and uniqueness of a Haar measure. In Section 4 of his paper~ paper. he also presented an alternate proof of the existence part of his development of a Haar measure on the Borel subsets of a locally compact equi-homogeneous uniform space. His proof that locally compact equi-homogeneous uniform spaces are quotients of topological groups contains an error as pointed out by the author in a paper titled On Haar Measure in Uniform Spaces (Mathematica Japonica, 1995) with a counterexample to his proof of Lemma 2.1 which is used in an essential manner to prove his Theorem 2.2. However. However~ his alternate existence proof of a Haar measure on the Borel subsets of a locally compact equi-homogeneous uniform space is valid. This leaves his extended theory of Haar measure (on the Borel subsets of a locally compact isogeneous uniform space) incomplete in the sense that the uniqueness part of the proof has not been established. In 1992, the author, unaware of Itzkowitz's work, showed the existence and uniqueness of a Haar measure on the Borel subsets of locally compact isogeneous uniform spaces and presented that development in a series of lectures at the 1992 Topology Workshop at the University of Salerno. The existence part of the author's development is essentially the same as Itzkowitz's alternate existence proof. But the author's uniqueness proof is a uniform space argument rather than an Weil' s theory of invariant measures on quotients of topological appeal to Weil's groups. The author's development, as presented at the Salerno Workshop is given in this chapter. It turns out that Itzkowitz's theorem that locally compact isogeneous uniform spaces are quotients of topological groups is true. We will prove this in Section 3 using a modification of Itzkowitz's approach that allows us to avoid the use of his erroneous Lemma 2.1. We will not nor show that the rest of Itzkowitz's ItZkowitz's approach can be corrected because the topology we get with our new proof is finer than Itzkowitz's topology on the group G and this necessitates additional work to straighten out his approach which is beyond the Scope of this chapter. What we will show is that the converse of this result is true, i.e., i.e .. that quotients of topological groups are isogeneous uniform spaces. This characterizes the locally compact isogeneolls isogeneous uniform spaces as locally compact quotients of topological groups and leads to necessary and sufficient conditions for locally compact uniform spaces to have a topological group structure that generates the uniformity or to have an abelian topological group structure that generates the uniformity.
The Segal-Mibu approach uses a generalization of K. Kodaira's construction given in a paper titled Uber die Beziehung zwischen den Massen und den
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9. Haar Measure in Uniform Spaces
Topologien in einer Gruppe, (Proc. Phys. Math. Soc. Japan, Volume 23, No.3, 1941, pp. 67-119) whereas the Itzkowitz-Howes approach uses a generalization of A. Weil' WeiI'ss technique published in his paper referenced earlier in this section. At the present moment it may appear that the Weil Wei! technique is more powerful in isogeneous uniform spaces in that it can be used to obtain a measure on a larger class of sets. However, it is probable that the two methods are equivalent. If this is the case, we would have a way of constructing the measure directly on the Borel sets using a simple combinatoric method. A uniform space (X, J.!) /.1) is said to be isogeneous if there exists a basis v for J.! /.1 and a collection H of uniform homeomorphisms of X onto itself such that: Sex, U) if and (1) For each i E H, and each pair of points x, y E X, Y E S(x, only if i(y) E S(i(x), U) for each U E v. (2) For each pair x, y E X, there exists an i xxy} E H that carries x onto y.
The members of H are called isomorphisms with respect to v or simply isomorphisms. v is called an isomorphic basis for J.!. /.1. Clearly, if i is an = S(i(x), U) for each x E X. Also, it is isomorphism and U E v, then i(S(x, U» = easily seen that compositions and inverses of isomorphisms are again isomorphisms. A topological space is said to be homogeneous if for each pair of points p, q E X there exists a homeomorphism of X onto itself that carries p onto q. Clearly isogeneous uniform spaces are homogeneous topological spaces. There are various types of isomorphisms. An isomorphism t:X ~ X is called a translation if t has no fixed points. If the isomorphism r:X ~ X has a proper subset F '" 0 of fixed points and F does not separate X - F, then rr is called a rotation. If F separates X - F, r is called a reflection. In what follows, we will show that locally compact isogeneous uniform spaces have a unique integral that is not only translation invariant, but also invariant under rotations and reflections. All topological groups are isogeneous uniform spaces with respect to the classical group uniformities and the classical group translation Tx Tx =Y ++ xx for each xx '" 0 in a topological group satisfies the above defined by Tx(y) = definition of translation.
*
*
Let C(X) denote the ring of real valued continuous functions on X and K(X)' the members of C(X) whose support have compact closures. For any f E CK(X), /; E CK(X). By a Haar CK(X) and isomorphism i:X ~ X, denote f © i by fi X, we mean a positive linear functionall functional! on CK(X) such that Iif;) integral for X~ lUi) = = I(f) for each isomorphism I:X i:X ~ x. X. A Haar measure for X is an almost regular, I(i) m(i(E» = m(E) for each Borel set E and each Borel measure m satisfying meleE»~ isomorphism i:X ~ X. The following lemma is left as an exercise.
/.1) is a locally compact uniform space, then eachf LEMMA 9.1. If (X, Jl) each f E
CK(X) is uniformly continuous.
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267
Isogeneous unifonn spaces were introduced in a series of lectures by the author at the 1992 Topology Workshop, held at the University of Salerno, Italy. The remaining material in this chapter is from the Workshop lecture series.
EXERCISES 1. Prove Lemma 9.1. EQUI-HOMOGENEOUS UNIFORM SPACES
unifonn space. A function !:X f:X ~ X is said to Let (X, U) be an entourage uniform be nonexpansive with respect to a base B for U if for each U E B and (x, y) E (f(x), ICy» f(y» E U also holds. By a B-nonexpansive homeoU, the relation (j(x), of a uniform unifonn space (X, U) onto itself, we mean a homeomorphism If morphism flof of X onto itself such that If is nonexpansive with respect to a base B for the unifonnity U. A uniform unifonn space (X, U) will be called an equi-homogeneous uniformity space if there is a group G of homeomorphisms acting on X such that (i) G is g(P) = q, and (ii) there transitive (i.e., given p, q E X, there is agE G such that g(p) is a base B for U such that G is a group of B-nonexpansive homeomorphisms of unifonn space. the uniform unifonn spaces are precisely the 2. Show that the equi-homogeneous uniform isogeneous unifonn uniform spaces. 3. A collection G of functions from a uniform unifonn space (X, U) to a unifonn space V E V, there is a U E U such (Y,V) is said to be equi-continuous if for each V that for each g E G, [g x g](U) c V. Show that if G is a group of homeomorphisms acting on a unifonn space
unifonnity (X,U), then the following are equivalent: (i) there is a base B for the uniformity such that G is a group of B-nonexpansive homeomorphisms of (X, U), and (ii) unifonn G is an equi-continuous group of unifonn homeomorphisms on the uniform space (X, U). 4. Show that if (X, U) is a locally compact isogeneous unifonn space, then (X,U) is unifonnly locally compact.
9.2 Haar Integrals and Measures
J.I.) is assumed to be a locally compact isogeneous unifonn In this section, (X, J.!) uniform J.I. and H a collection of isomorphisms space. Let v be an isomorphic basis for J.! with respect to v that satisfy condition (2) in the definition of an isogeneous unifonn space. Let g ~ 0 be in CK(X) such that g(x) ~ b for each x in some uniform ~ 0 is in CK(X) CK(X) with f(X - K) = = 0 for some U-sphere S(y, U) where U EE v. If ff~
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9. Haar Measure in Uniform Spaces
compact K c X then there is a finite subset I{xx 1I .••• .. x xn} I ' U) n } of K such that {S(x 1, ... S(X I) = sup {f(x)Ix {f(x) Ix E S(Xj, S(x n Iflj S(Xj' U)} and Qj aj ~ !.t1jlb iflJib for n ,, U)} covers K. If If gXj = g LXi} for each j = 1 ... n, then f(x) ~ L}=l L7=1 ajgXj(x) ajgxi(x) for each x E X where gXj g © LXI} xi onto y. The finite collection {a 1I .. ~X X that carries Xj •• some isomorphism ixjy:X -7 an} is said to dominate fwith respect to g" g. Put [fl g] g 1= inl{ inf{ LjGj Liaj II {aj} dominates ffwith with respect to g}. g }.
satisfies: THEOREM 9.1. The number Ifl [fl g] is non-negative,finite and sati~fies: (1) [fi [Ji Ig] = [fl Ifl g]for each isomorphism i:X ~ -7 X. (2) Ifl [fl + 12 f 2 I g] ~ [fl ,f2 E CK(X). Ifl I g] + [f21 1f21 g]for glfor each pair J~I fl.h CdX). glfor each a> a > O. (3) [afl g] = alfl a[fl g]for 1
(4) fl ~f2 I g] ~ [f21 g]for eachfi ~ 12 f2 E CK(X). (4)fl ~h implies [fl Ifllg] 1f2 lg]foreachfl (5) [fl Ifl h] ~ [fl g][g I h]for each h E CK(X) CdX) with h # O. (6) [h Ij]-I IjJ-I ~ [fl g]/[h I g] ~ [fl h]for hlfor each h E CK(X) with h # O.
'*
'*
Proof: We prove only (1). (2) through (6) are left as an exercise. Now Ji(x) fi(x) = [f©i](x) ~ L)=l aj[gXj © i](x) dominates f with respect to g. L}=laj[gx n(x) for each {aj} {ail that dominatesfwith gxj © i = = g © [iXjY [ix j y © i] where [i xj y © i] is an isomorphism so {aj} {a;} Also, gXj [iXjY , dominatesjiwithrespecttog. HencelJilg]~[flg]. dominatesfi with respect to g. Hence [Ji Ig] ~ [fl g]. Nowjij-l Now fi,-1 =f©i©i= f© i © i-I1 =[ =f [Ji,-J i g] ~ [Ii! [Ji! g]. gJ. and from what has just been proved, we have [fi Vi g] = [fij-l Therefore, [fi Ig] = = [fl g] for each isomorphism i.· i. • j
CK(X) denote the non-negative members of CK(X) and choose sOlne some k Let C"K(X) CK(X). For each g E CK(X), 1~:CK(X) -7 [0, [0,00) [flg]/[klg] E Ck(X). C"K(X), define 19 :C"K(X) ~ 00) by IIg(f) g (f) = = [fl g ]/[k Ig J for each f E CK(X). C"K(X). eachfE
C-;rXJ. COROLLARY 9.1. For each g E Ck(X), (l)lg ~O. (l)lg~O. ~/g(f)~ [flk]. (2) [klj]-I [k IjJ-I ~ 19(j) ~ [fl k]. (3) IIg(fi) g(fi) = IIg(f)for g(f) for each isomorphism i:X -7 -4 X. (4) IIg(rtf) aIg(f)jor each g (oj) = = aJ g (f) fo rea ch a > O. f2) ~ 19(fl) + Ig(f2)' (5) 19(fl Ig(fl + h) l~(h)· Since X is locally compact, for each U E J.l C"K(X) with ~ there exists agE C;(X) = 0 for some U-small compact set K. g(X - K) =
'* '*
C"K(X) fl # 0 i:- 12 f2 andfor CKrX) "Jith with II and lor each e ~ [g(fl f2) + efor Ig(11 + h) elor each g E 0 and such that the support of g is U-small.
LEMMA 9.2. For each 11,12 fI' 12
E
> 0, there exists a U E v such that [g(fl) Ig(fl) + [g(!2) Ig(hJ
'*
Ct:(X) ct;(X) with g #
[fl + h](X letfE Proof: Let e > O. Assume [fl f2](X - K) = 0 for some compact K and letf E CK(X) =II fl + fz f2 + bfand CHX) such thatf(K) that/(K) = = 1. l. For any 8 > 0 put = 8fand for i = = 1,2 put hi = fd q,(x) ~ q>, hi, h 2 EE Ck hI + h 2 $~ 1. /;/ for hi](x) [h;](x) ~ [Ljajgxj(x)]hi(x) [Ljajgx/x)]h;(x) ~ Ljajgxj(x)[h,:(xj) Ljajgx/x)[h,(xj ) + £'l hex) £'1
so Iii [Ii Ig] ~ Ljaj[hi(xj) [Ii Ig] Ljaj[h;(x) + £'] £'l which implies Iii g) + [f21 g] ~ Ljaj(l LjajO + 2£'). Consequently, [f] [12 Ig] ~ [
0 and putf3 =11 9.2~ there exists a + h) + £ for each g E Dwhosesupportis D whose support is U E vsuchthatlg(fl)+lg(f2)~/g(f1 v such that Ig(1l) + Ig(12) ~ Ig(1] +f2)+£foreachgE UE Let/gEE yll\jl(fj)-I(fj)1 forj=I,2,3. U-small. Let Ig£ E {\jIE {\V E yll \V(fj) -I(fj) I l(o.(-j) = - 1(0.(-/)-) = = 0.[1((-j) +) --1«-.1)-)] I ((-j)- )] = = 0.[1(J-) - I er)] = -0.[1(f+) -I(f-)] - I (J-)] = (j). C I early J(fd I (J 1) + a[/«-.I)+) a[l(f-) -Ier)] = -a[l(f+) = ~1 ~J(j). Clearly J(f2) = = l(fl J(f) + h) h'/2 Finally,fl andfi 1(J2) f2) for each fl' f2 E CK(X). Finally, 17 == © i and fi ==f- © i J(f) = l(fl) - 1(Ji) = ler) - l(f-) = = J(j) for each isomorphism i:X --7 ~ X. so 1(h) = lift) I(f;) = uniform integral for X. •Therefore, IJ is a unifonn
r, r
r -r
rr
r
By the Riesz Representation Theorem (Chapter 8), there exists a a-algebra M in X containing all Borel sets, and a unique, positive, regular measure m on M J(j) = Cdx) and such which represents JI in the sense that 1(f) = Jxfdm for each f E CK(X) that m(K) < 00 for each compact K eX. THEOREM 9.3 m is a Haar measure for X.
We first show that if K is compact then m(i(K) = = m(K) for each ~ X. Let U be an open set containing K. Then there is an fv fu isomorphism i:X --7 withfu(K) = = 1 andfv(X andfu(X - U) = = O. Therefore, XK ~fv ~fu ~ Xv Xu so E CK<x) C"K(X) withfv(K)
Proof:
= fxXKdm m(K) = JxXKdm
~
dm fxfudm Jxfu
~
fxXudm = m(U). JxXu dm =
Since m is outer regular, this shows that m(K) = = inf{ infl Jxfvdm Jxfudm IKe 1 K c U and U is = inf{ inftJx[fu ]dmlK c U and U is open}. Thus, open}. But then m(K) = Jx [fv © i-I ]dm IKe m(i(K)
~
1 ]dm Jx[fu © ii-1]dm fx[fu
~
fxX;(U)dm fXXi(U)dm = m(i(U».
Now m(i(K) = Im(V) {m(V) Ii(K) c V and V is open}. For each open V containing i(K), it is possible to find a V E v such that Star(i(K),V) c V. Since Star(i(K),V) = i(Star(K,V», m(i(K) = infl m(i(U» m(i(U) IKe 1K c U and U is open}, so m(i(K» m(i(K) = = = inf{ = m(K). Next suppose E is a Borel set with m(E) < 00 and i is an isomorphism on X. Then there is an ascending sequence {K jj }} of compact subsets of E with m(E) = limj-t~m(Kj). lim) ~oom(Kj). For each positive integer j let XKj be the characteristic function XKJ ~ XKj+l. XKj+l' Since, the sequence {XKj} {XKj } converges almost of K)" Kjo Then for each j, XKj everywhere to XE, {XKj © i-I} r 1 } converges almost everywhere to XE © i-I. By Theorem 8.9, {JxXK.dm} i-I]dm} {fxXK.dm} converges to JxXEdm JxxEdm and {IX[XKj {fX[XKj © rl ]dm} converges J to Ix Ix [XE © i-I ]dm. Therefore, 00
ixXi(E)dm = ix[XE ©i-1]dm © i-I]dm = = limj4-Jx[XKj limj-tJx[XKj ©i-1]dm © i-I]dm m(i(E» = = fxX;(E)dm = fx[XE
= =
9.3 Topological Groups and Uniqueness ofHaar Measures
271
m(i(E» = m(E) for each Borel set E with m(E) < 00, and each Consequently, meleE»~ m(i(E» < 00, then by what has just isomorphism l.i. Finally, suppose m(E) = 00. If meleE»~ been shown, m(E) = = m(l-l mW' (i(E») (i(E))) = = meleE»~, m(i(E», so m(E) < 00 which is a m(i(E» for each contradiction. Therefore, for each Borel set E, m(E) = meleE)~ isomorphism i.l.-
We will defer the uniqueness proof for m until Theorem 10.5.
EXERCISES 1. Prove Theorem 9.1. f.l) be an isogeneous uniform space and let li be an isomorphism of X 2. Let (X, 11) onto itself. Let m be a regular, Borel measure on X. For each Borel set E put m(i(E». Show that mi is an almost regular, Borel measure on X. miCE) = meleE»).
3. Show that the results of this section still hold if CK(X) is replaced by Cc(X) = U:X {f:X ~ C Ifis Cl(O(f) is compact}. Itis continuous and CI(O(/)
Haar Measures 9.3 Topological Groups and Uniqueness of Maar In this section we will introduce the concept of a topological group and see that all topological groups are uniformizable and that locally compact topological groups are cofinally complete (uniformly paracompact) and hence complete. We will also see how the classical existence and uniqueness proofs of the Haar integral in locally compact topological groups follow from the results in the previous section and what form they take in topological groups. This section also contains a solution to the problem of characterizing which uniform spaces have a compatible topological group structure that generates the uniformity for the case when the uniform space is locally compact. The concept of a topological group is simply the combination of the concepts of a group and of a topology on the same underlying set in such a way that the group operation is in some sense compatible with the topology. Precisely, by a topological group, we mean a topological space (X, 't) t) together with a binary operation +:X2 +:X 2 ~ X satisfying the following conditions:
GI. (X, +) is a group. Gl. 2 X2. G2. + is continuous with respect to the product topology on X • G3. The inversion function i:X ~ X defined by i(x) = -x for each x E X is continuous. It is easy to find interesting examples of topological groups. Probably the most
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9. Haar Measure in Uniform Spaces
obvious is the additive group of real numbers (R, +) with the usual top- ology for R. Similarly, coordinatewise addition in R nn makes R nn a topological group with respect to the Euclidean n-space topology. Another important example is the circle group T c C. The unit circle T is defined as the subset of complex numbers consisting of members z such that Iz I == 1. The group operation is usual multiplication of complex numbers and the topology of T is the subspace R22 ". topology of T with respect to the usual topology of R PROPOSITION 9.1 9.i For a group (X, +) with topology 'T, the conditions x, y) = G2 and G3 hold if if and only If if the function f:X 22 ~ X defined by j( f(x, = x - Y isIS continuous. Proof: If G2 and G3 hold then ffis is continuous since f(x, y) = g(x) + i(y) for each pair x, y E X where g is the identity function on X because g x i is continuous X22 by G3 and addition is continuous by G2 and f is the composition of g x i on X and the addition function. = Conversely, if f is continuous then inversion is continuous since i(x) = i(y)) f(O,x) for each x E X, where 0 is the group identity. Also, since x + y = f(x, i(y» X22 it follows that addition is continuous. continuous.for each pair (x, y) E X -
PROPOSITION 9.2 For a topological group (X, 'T, +), the following functions are homeomorphisms o/X of X onto itself: (1) inversion i, =x + a (2) right translation rraa by some a E X X defined by ra(x) = for each x E X, (3) left translation la by some a E X defined by la(x) = a + x for each x E X. Proof: The functions i, rraa and la are clearly continuous for each a E X. Since the compositions i © i, rraa © r -a' and La © La I-a are all the identity mapping for each a E X, it is also clear that i, rraa and La fa are all homeomorphisms for each a E
xX -
S-l is used to denote the set consisting of all For a set SeX, the notation S-1 inverse elements of members of S. S is said to be symmetric in X if S == S-l. If T c X, the notation S + T is used to denote the set consisting of all elements of the form s + t such that s E Sand t E T. PROPOSITION PROPOSiTION 9.3 If A is open and B is closed in the topological group (X, +), and ifC, D c X then: ((1) J) C I(D --11 ) CI(D
= [C I(D )] -1 . [CI(D)]-1
Cl(x+ = x + [CI(D)] + yfor each pair x, x,y (2) Cl(x + D + y) = Y EEX. X. (3)D+A andA+Dareopen. (4) B + Y and y + Bare closedfor each y E X. (5) CI(C) + CI(D) c CI(C + D).
9.3 Topological Groups and Uniqueness ofHaar Measures
273
1 CI(D- 1I )) = Cl(i(D» CIU(D» = i(CI(D» i(CI(D» = [CI(D)r[CI(D)]I. Proof: Since i is a homeomorphism, Cl(D• Similarly, since for each pair x, Y y E X, Lx Ix and rrxx are homeomorphisms,
Cl[x + D + y]
ClOAD + y» = CI(lx(ry(D») CI(lx(ry(D))) = = CI(lx(D
lx(ry(Cl(D») = x + CI(D) + y. IxCry(CI(D»)
This establishes (I) (1) and (2). (3) follows from Proposition 9.2 since for each x E IAA) and D + A = UxED(X + A). Similarly A + D is open. (4) also X, X + A = lx(A) clearly follows from Proposition 9.2. (5) follows from the continuity of + on X CI(C) + CI(D) = = +(Cl(C) +(CI(C) x CI(D» c Cl(+(C CI(+(C x D» D» = = CI(C + D).· D).x X, i.e., Cl(C) CI(D» e PROPOSITION 9.4 If h:X ~ Y is a group homomorphism between topological groups X and Y then: (J) Cl(h(A)) + Cl(h(B)) (1) h(A + B) = h(A) + h(B) e c CI(h(A)) CI(h(B)) ec h(Cl(A h(CI(A + B)) each A, Be X. for eachA, B eX. 1 1 h-I(D) h-1(C D)for (2) h- 1l (C) + h(D) ec h(C + D )for each C, DeY. (3) Cl(h-I(C)) CI(h-I(C)) + Cl(h-1(D)) CI(h-1(D)) ec Cl(h-1(C CI(h-1(C + D))for D))jor each C, DeY. (4) If A is symmetric in X then h(A) and Cl(h(A)) CI(h(A)) are symmetric in Y. (5) If C is symmetric in Y then h- 1l (C) and CI(hCl(h- 1l (C)) are symmetric in X.
The proof of Proposition 9.4 is left as an exercise. exercise, A subgroup H of a group G is said to be normal or invariant if for each x E H, a + x + a-I E H for = b + a for each pair a, bEG. each a E G. G is said to be Abelian if a + b = PROPOSITION 9.5 Let H be a subgroup of a topological group G. Then: (1) Hand Cl(H) CI(H) are topological groups. (J)
(2) If H is normal then CI(H) is normal. (3) If G is Hausdorff and H is Abelian then (hen Cl(H) Cl( H) is Abelian.
(4) If H is open then H = CI(H).
Proof: H is clearly a topological group since the continuous function f:G 22 ~ G H22 ec defined in Proposition 9.1 by f(x, y) = x - y has a continuous restriction to H 2 C l(H) is also a topological group follows from the fact that since H is G2 • That CI(H) [Cl(H)]-l = CI(H) + CI(W CI(H- 1I )) a subgroup of G, H + H = H = HWI1 and so CI(H) + [CI(H)r1 1 c Cl(H CI(H + HWi)) = = CI(H) by Proposition 9.3. Hence CI(H) is also a subgroup of
G. nonnal. Then x + H - x = H for each x E G, so x To show (2), assume H is normal. + CI(H) - x == Cl(x + H - x) == CI(H) for each x E G by Proposition 9.3 which implies Cl(H) CI(H) is normal. nonnal. To show (3), assume G is Hausdorff and H is Abelian. Define g:G 22 ~ G by g(x, y) = = x + y - x - Yy for each pair x, y E G. Then g is continuous since + and i (inversion) are continuous. Since G is Hausdorff, {O}
274
9. Haar Measure in Uniform Spaces HaarMeasure
g-I (0) is is closed where 0 is the group identity element. Since g is continuous, g-l H22 C h- lI (0) since H is Abelian, so CI(H) el(H) x CI(H) el(H) = = CI(H el(H also closed in G. Now H h -Il (0) is closed. Hence CI(H) el(H) is also Abelian. x H) c h-
To show (4), assume H is open in G. Then we have G - H = (G - H) + + H= + H) which is open since x + H is open for each x E G - H by el(H).= CI(H).· Proposition 9.3. Therefore, H is closed in G so H =
UxEG-H(X
Let X and Y be topological groups having neighborhood systems N and B at 0 and e respectively, where 0 is the identity element of X and e is the identity element of Y. A homomorphism h:X ~ 4 Y is open at 0 if for each U E N, there is a V E B with V c h(U). PROPOSITION 9.6 A homomorphism h:X 4~ Y is continuous (open) and only if it is continuous at 0 (respectively open at 0).
if
Proof: Clearly if h is continuous or open, then it is continuous or open respectively at O. Assume first that h is continuous at 0 and let N be the neighborhood system at O. For each x E X and open V containing y = hex), e y is open by Proposition 9.3. Since h is (the identity of Y) is in V - y and V - Y continuous at 0, there is a U E N with h(U) c V - y. Since x E U + + x and h(U+x) = h(U) + hex) c (V - y) + y = V, we have that h is continuous.
Next assume h is open at 0 and let B be the neighborhood system at e. For each x E X and open U containing x, U - x is an open set containing 0, so there exists a V E B with V c h(U - x) = h(U) - hex) which implies V + + y c h(U) y which is open. Therefore, hex) is an where y = hex). Since e E V, Y E V + + Y mapping.• interior point of h(U). Consequently, h is an open mapping. A key feature of topological groups is that knowing the behavior of the neighborhood system of the identity element is equivalent to knowing the behavior of all neighborhood systems at each point of the group. This derives from the fact that if g is any element of the group and U is a neighborhood of the group identity, then both g + U and U + g are neighborhoods of g and if V is g-l + V and V + g-1 g-l are both neighborhoods of the a neighborhood of g, then g-I identity. Consequently, the following results are especially useful in topological groupso groups. PROPOSITION 9.7 If (X, +) is a topological group and N the neighborhood system of the identity 0, O. then: (I) For each V E N there is a V E .N N with V + V cC U. u. (1) V E N N, V-I E IV. N. (2) For each U Nwith[x+V-x]cu. (3) ForeachVE NandxE X,thereisaVE Nwith[x+V-x] cU. (4) Eachfilter N of subsets of X containing 0 and satisfying (J) (1) - (3) determines a unique topology that makes (X, +) a topological group and has N as its neighborhood system at O. 1
ofRaar Measures 9.3 Topological Groups and Uniqueness ofHaar
275
Proof: (1) follows from the continuity of + and (2) from the continuity of I-x and rrxx are homeomorphisms, i-x inversion. Since for each x E X, Lx Lx © rx(V) rx(U) = [(-x) + V U + x] xl E N for each U E N. Hence Rence V c [(-x) + V U + x] xl for some V E N which implies [x +V + (-x)] ccU. V. Now (1) implies the existence of such a V, so (3) follows. let.'t = = {U U, there exists V E N with V + P {V c X I for each p E V? To show (4), let U} .•t is easily shown to be a topology on X. Clearly X, 0 E 'to •. If {Va} Cc .'t c U}. uVaa E •'t since for each x E uV aa ,, X belongs to some V ~ which implies then uV V~ E N with V V ~ + x c V ~ c uV aa . Also, if U V II ... Vnn E •'t then there exists a V . V (")i=1 n?=1 Vi E 't• since for each x E (");=1 n?=1 Vi, there exists a Vi E N for each i with Vi + x C Vi. x E (n;=1 Vi' Since N is a filter, (")i=1 n?=1 Vi E N and andx (n?=1 V;) Vi) + xc x c ni=I n?=1 Vi. Vi' 0
•
°
Clearly •'t has N as its neighborhood system at 0 and if 0cr is another topology for X that makes (X, +) a topological group and has N as its =., • and p E U, neighborhood system at 0, it is easily shown that 0cr = 't, for if U E 't U, so U is a neighborhood of p in the there exists a V E N with V + P c V, cr, i.e., 't• c 0. cr. Similarly 0cr c 'to •. topology 0, It remains to show that that.'t itself makes (X, +) into a topological group. For ~ X as defined in Proposition 9.1 (l(x, (f(x, y) = =x this it suffices to show that f:X 22 -t X22 and let U be a neighborhood of f(P, q) y) is continuous. For this let (p, q) E X = [U - x + y] yl E N. By (1), there exists a V E N with V + V cC = p - q. Then W = W. By (2), V-I E N so A = = V(")VVnV- 1I E N since N is a filter. Then A = = A-I and A + A c V, so A + A-I cC W. Now A + p is a neighborhood of p. Also, there exists aBE N with [p + B - p] c A by (3). Therefore, [p + B - p] + q is a neighborhood of q, so [A + p] x [(P + B - p) + y] is a neighborhood of (p, q). Moreover,f([A p,p bE B) Moreover, f([A + p] x [(P [(p + B - p) + q]) = = {f(a -rT p, p + b -P p + q) I a E A, b E B} a 2 + P - q I a I, I ' a2 a2 E A) = -Ij + (x - y) c W + (x - y) = c {a 1I + P - P - a2 E A} = [A + A A-I] = U. Consequently,f topology.'t makes (X, +) into a topological Consequently, f is continuous, so the topology group. •-
THEOREM 9.4 Let (X, +) be a topological group with topology topology.'t and neighborhood system N at O. For each V E N put Uv == {V + x Ix E X} and let A Nat =[U v IV E N}. Then A is a basis for a uniformity JJf.l that is compatible with •.t. Consequently, every topological group is uniformizable. Proof: To show Ais a basis for a uniformity, it suffices to show that for each pair U v , U w E A, there exists some U u E Athat ,1-refines ~-refines UvnU w . If U v , U w E A, then V, WEN. As shown in the proof of Proposition 9.7, there exists an A EE N ) = {ZE with A + A-II cVnW. c VnW. ThenU=AnAThen U = AnA- 11 EN so S(x, U u )= {z E XIX,ZE xix, Z E U+ U+ withA+AE NsoS(x, yforsomeYE X} = {z+Xlx-y,Z-YE UforsomeYE XI Cc {ZE XI(z-y)Y for some y EX} {z + X Ix - y, Z - Y E U for some y EX} {z E X I(z - y) (x - y) EE U + V-I} U- I } C {z EE X I Z - X EE V} = V + X EE U v . Similarly Sex, S(x, U u ) c W , S(x, + X E UW so sex, U u ) c [V + x]n[W + x] E UvnU w . Therefore, U u is a , w ,1-refinement ~-refinement of UvnU w w.
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9. Haar Measure in Uniform Spaces
To show that J.l J..l is compatible with 't, 1:, by Proposition 9.7 it suffices to show J..l I = No N. For this, first note that if V EE N, then V c 5(0, U v ) that 15(0, {S(o~ U) u) IIU EE J.l} so 5(0, S(O, Uvv ) EE N for each U vy E E A. Since A is a basis for 11, ~, we see that I{S(O,U) S(O,U) II U EE J.l} J..l I c N. Conversely, if V EE N we can choose WEN such that W W-]1 C V so S(O,U w ) = {y Iy E X I 0, yEW YEW + x for some x E X} XI = {y (y E X xly + WI y -- xX W-]1 }) = Iy E X lyE lYE W W + WW-]1 }) Cc V. Therefore, N c {S(O, u) u) II U u E Wand x E W= {y E {S(O, u) E J.l} J..l I which establishes 15(0, U) IU EE J.l} J..ll = = N. •-
Sea,
By now, it has probably occurred to the reader that the unifonnity uniformity J.l J..l of Theorem 9.4 could just as well have been defined by the basis v = = I{W Wvv II V EE N} NI y where W = {x + V I x EX}, and the argument, with minor notational changes, Wv Ix I E X I, would still J..l is called the right uniformity on X stilI be valid. The uniformity J.l whereas the unifonnity J..l' generated by v is the left uniformity on X. It is left uniformity J.l' J..l') uniform spaces (X, J..l) and (X, J..l) as an exercise (Exercise 3) to show that the unifonn are isogeneous uniform unifonn spaces where the collections of homeomorphisms are R E X I and L = {I {Ixx Ix EX} E X I respectively and the isomorphic bases are A = I{rrxx IIx EX} and v (as defined in Theorem 9.4 and above) respectively. Consequently, if (X,+) is a locally compact topological group, then there exist unique Haar measures m and m' for (X, J..l) and (X, J..l') respectively. By definition of Uv and Wvv we then have m(V + x) = = m(V) = = m(V + y) for each pair x, y EE X and each W VE E N (neighborhood system at 0) and similarly, m'(x + V) = = m'(V) == m'(y + V). From Theorem 9.4 it is straightforward to derive a number of useful consequences that are listed in the following proposition and left as an exercise.
PROPOSITION 9.8 If (x. (X, +) is a topological group with neighborhood systemN system N at 0, then: (1) (I) X has a neighborhood base consisting of symmetric open (closed) sets. (2)/nt(Y) {yE Ylv+ycYforsomeVE ylv+ycYforsomeVE N}foreachYcX. (2) Int(Y) = lYE NlforeachYcX. (3)CI(Y)=n{V+Y/VE (3) CI(Y) = nlV + yl V E N}foreachYcX. Nlfor each Y cx. (4)XisT (4)XlsT]l ifandonlyifO=n{VIVE ifandonlyifO=nlVIVE N}. NI. (5) X is locally compact if and only if there exists a compact V E N. Let (X, +) be a topological group and let Z be a subgroup of X. For each x X, the set Z + x is called a right coset of Z in X. Similarly x + Z is called a left coset. Both the right cosets and the left cosets form partitions of X. To show this for the right cosets, define the relation p - q in X by p - q EE Z. It is easily verified that - is an equivalence relation on X. In fact, p - q EE Z if and only if p E E Z + q so that the equivalence class [p] containing P p with respect tois Z + q for each q EE [Pl. Since p EE [P] we see that [P] [p] = Q= =Z + p. Let Q = X/- be the quotient uniform space defined in Chapter 5. In this section the notation for Xj- will be X/Z. It should be noted that X/Z may fail to be a topological group X/if Z is not normal. Never-the-Iess, X/Z will still be referred to as a quotient of topological groups, the reference being to the quotient topological and uniform E E
9.3 Topological Groups and Uniqueness ofHaar Measures
277
structures rather than a quotient group structure. The natural projection n:X X/Z is given by n(x) = Z + x for each x E X.
---7
rAZ) and x + Z = Ix(Z) IAZ) for each x E X. Also, it is easily Clearly Z + x = rx(Z) seen that the above argument could have been carried out had we chosen to define p - q by -p + q EE Z. In this case, the partition would consist of the left cosets (x {x + zlx Xl and we could construct a "left 44left quotient" -IX uniform Z Ix E EX} space. If Z is Abelian, clearly -IX = X/-. Q = .x/z PROPOSITION 9.9 The quotient Q XIZ of a topological group (X, (X. with respect to a subgroup Z is T 1 if and only ifZ if Z is closed in X.
+)
Proof: Assume Q Q is T 1. Then the singleton set {Z} in Q Q is closed. Since the 1 canonical projection n:X ---7 Q ({ Z}) = Q is continuous it follows that 1tn-1({Z}) = Z is Conversely. let Z be closed in X. Let q EE Q. By definition, q is closed in X. Conversely, some right coset Z + p of Z for some p E X. Now Z + P = = rp(Z) and by Proposition 9.2.(2), rr"p is a homeomorphism, so Z + p is closed in X. Since n- 11(q) == Z + p and 1t n is IS an identification mapping, it follows that {q} {q I is a closed subset of Q. Hence Q is T 1. If H is a nonnal subgroup of a group G it can be shown (see Exercise 7) that G/H is also a group with respect to the operation + defined on G/H by (H + x) + (H + y) = H + (x + y). In the case G is also a topological group, it is easily Q= = G/H is a topological group with respect to the verified (see Exercise 8) that Q quotient topology, and as such will be referred to as the quotient topological group of G with respect to H.
Howes, 1992) Let G be a topological group and H THEOREM 9.5 (N (N, Howes~ a subgroup oJ~ of G. Then the quotient G/H GIH with the quotient topology is an uniform space. isogeneous unIform n:G ---7 Proof: Let 1t:G ~X= = G/H be the natural projection defined by neg) == g + HH for each g E G. Then 1t n is continuous and onto X. For each g E G we can define a function g':X -7 ---7 X by g'(j g'(f + H) = (g + f) j) + H for each (j (f + H) E X. It is easily seen that for each g E G, g' gJ' is one-to-one and onto. Furthermore, if x, y E E X, there exists agE G with g'(x) = y.
V is a neighborhood of e (the identity) in G, then for We next show that if V n(f + V) for some f E G, then g'(x), g'(y) E 1t(h n(h + V) for each g E G, if x, y E 1t{f n(f + V) implies some h E G. Now, x, Y E 1t{f g~(x), g'(x), g'(y)
E
gJ'(1t{f g'(n(f + V) V» == {g'(1t(f {g'(n(f + v» v» Iv E V} VI
{g+f+v+HivE {g+!+v+HivE V}. VI.
Put h = g + f. Then, for some VI, V2 E V, g'(x) = h + VI + Hand g'(y) = h + V2 + n(h + V). Let G' be the collection collectIOn of all g' such that g H. But then g'(x), g'(y) E 1t(h E = {f If EE G E G and let v be the collection of all coverings 1t(V) n(V) where V = If + VVi!
278
9. Haar Measure in Uniform Spaces
e). Then each g' E G' is uniformly continuous since and V is a neighborhood of e}. x,y E 1t(j n(J + V) implies g'(x), g'(y) E 1t(h n(h + V) for some h E G. Furthermore, since G is a group, it is easily shown that each g' EGis a uniform
homeomorphism. Finally, we show that each g' E G' is isogeneous with respect to v. For this, let x E X. By what we have already shown, g'(S(x, 1t(V» n(V» c S(g'(x), 1t(V» n(V» n(V) E v. Since G is a group, we also have that (g-l )'(S(g'(x), 1t(V))) n(V») c for each 1t(V) S«g-l )'(g'(x»,, 1t(V» = Sex, 1t(V» S«g-1 r(g'(x» n(V» = n(V» which implies S(g'(x), 1t(V» n(V» c g'(S(x, 1t(V». n(V». Hence g'(S(x, 1t(V» n(V» = S(g'(x),1t(V» S(g'(x),n(V» for each g' E G'. Therefore, g' is an isomorphism with respect to v for each g E Go Consequently, v is an isomorphic basis for a uniformity 11 J..l on X. Therefore, (X, 11) J.!) is an isogeneous uniform space. •
THEOREM 9.6 (N. Howes, 1992) Each locally compact isogeneous uniform space is homeomorphic to a quotient of a topological group. Proof: Let (X, J.l) /l) be an isogeneous uniform space and let v be an isomorphic /l. Let H be the collection of isomorphisms on X with respect to v. If basis for J.l. f, g E H, it is easily shown that f © g and f- 11 are also isomorphisms. Consequently, H can be extended to a group G with respect to the operation of functional composition. Without loss of generality, we may assume G to be the group of all isomorphisms on X with respect to v. For each compact K c X and U E v, put [K, U] =
r
I
{g E G I there exists a V V E v with S(g(x), V) c Sex, U) for each x E K}. K).
Let p ~ = {[K, U] ~ is taken as a system of U]IIKe K c X is compact and U E v}. v). If P neighborhoods of the identity (mapping) i E G, then G is a Hausdorff topological group with respect to the operation of functional composition. To see this, first note that every member of ~Pcontains i. Also, if [K, U], [F, V] E ~, then there exists aWE v with W < U and W < V, so [KuF, W] c P, [K,U]n[F,V]. Consequently, it will suffice to prove:
P,
P P
(1) [F,V]-l1 c [K, U]. 0) For each [K, U] E ~,there there exists [F, V] E ~ with [F,vr (2) For each [K'l E ~,there exists [F,V] [F,V]2 c [K, U]. [K, U] E [F,VJ E ~ with [F.vf (3) For each [K, U) E P ~ and g E G, ~ with G. there exists fF,V] [F,v] EE P g © [F,V] [F,v] © g-l c [K, U]. UJ. (4) For each [K, [K. U] E ~ and g EE [K, [K. U], there exists [F,V] [F,v) EE ~ with g © [F,V] [F,v] c [K, [K. U].
B
W). To prove (1), let [K, U] EE B ~ and pick W E E v with W 0 there exists a closed set F and an open set U E G such that F < A < U and A(U - F) < E. It should be intuitively clear that all of Loomis' results documented in the previous two sections hold for A defined on Z(G). However, to be absolutely sure, one needs to work through all those results replacing T with G and Z with Z( Z(G). G). We leave this as an exercise for the serious reader.
in variance as follows: a Loomis Next we relax Loomis' definition of invariance content l on G is invariant on compact spheres with respect to x E f.l if I(S(P,x)) S(p~ x) and Seq, x) having I(S(P,x» = l(S(q, [(Seq. x)) x» for each pair p, q E X with both S(p, compact closures. I is simply invariant on compact spheres if it is invariant j..l. In this case, we do not necessarily on compact spheres with respect to all x E ,.1. [(S(P, x)) x» for each p E X unless we know that S(p,x) have a common value lex) = l(S(P, X. Again, it should be intuitively clear that has compact closure for each p E X.Again, Loomis' development recorded in the previous two sections can still stilI be sure. one needs to verify accomplished with this restriction, but to be absolutely sure, it for themselves. For this we note that it suffices to prove Lemmas 10.2 and 10.3 using this definition of invariance in variance on compact spheres with respect to an rr
10.4 Invariance and Uniqueness of Loomis Contents and Haar Measures
301
E I.l, invariance is used in the proofs and ~, since these are the only places where invarlance the rest of the development follows from these two lemmas.
THEOREM 10.4 The Loomis content A of Theorem 10.3 (where T is replaced by G) is invariant on compact spheres with respect to all bur hUl countably many indi c es of any interval {y a} c C I.l indices {Ya} ~ * withy with Yaa < Y 13~ whenever a < ~ Pand such that if'Y if y is rational then Yy y E I.l. ~. If lI is any other Loomis content on G that is invariant on compact spheres with respect to all but countably many Band whose indices of any interval {Ya} C J.l*, ~*, with Ya < Y~ Yj3 whenever a < P members with rational indices are in ~, A on /-l, then lI is a constant multiple of A Z(G).
Proof: Let {y a} C I.l ~ ~ * be an interval of indicies with Ya < Y 13~ whenever a < P such that Y Let.p f( a)/g(~o) Yyy E J.l ~ whenever y is rational. Let P E X. The function f(a)/g(Po) given by
in Lemma 10.6, is increasing on [0, 1] so it is continuous at all but an increasing 8 E [0, 1] such that 88 is not in {r n }. If Y sequence {r n} C [0, 1]. Let b Y()c E ~ J.l then c(P), x) = h(y 8(q), c(q), x) for each q in X by A4 so A(y c(P» h(y ()(p), 8(P» = A(y c(q)) 8(q» for each q in X. Thus A is invariant with respect to Y8. y c() is not in ~, J.!, 88 is still a c. If Y f(a)/g(po) a)/g(~o) so there exists an increasing sequence {sn} {s n} of continuity point of f( rationals with Sn ~ b 8 andf(a)/g(~o) andf(a)/g<po) is continuous at each Sn. Then
A(ySn (P» = = A(y A(YsSnn(q» (q» for each q in X since each Sn because, as shown above, A(y Sn (P» sn is f(a)/g(Po). Hence A is invariant rational and because 8 is a continuity point of f(a)/g(~o). Ya's. with respect to all but countably many of the Ya 'so Let lI be another Loomis content on G that is invariant on compact spheres many indices of any interval {Ya} ~** such {y a} C I.l with respect to all but countably Inany y ry E I.l ~ whenever y is rational. Let that Ya < Y ~i3 whenever a < ~Pand such that Y Z/(G) be the ring of zero-boundary sets with WIth respect to I and Z?c(G) ZA(G) the ring of zero-boundary sets with respect to A. If r is a member meinber of I-l f.l** - I-l, f..!, then rr-- in Lemma 10.2 can still be chosen from I-l f.l and similarly in Lemma 10.3. Therefore, both of these lemmas can be generalized to include the possibility I-l*. that r E J.l*. lOA, I satisfies the second Since lI satisfies L5 and L6 of Proposition 10.4, inequalities in (10.2) and (10.3), i.e.,
l(A)/I(r) ~ h(A +, x)/h(r-) and I(A)/I(r-) ~ h(A -, x)/h(r, x) for all sufficiently small x
E
~,if J.l, if lI is invariant on compact spheres with respect
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10. Uniform Measures
to r. By the first inequalities in (10.2) and (10.3), H(A,r) H(A, r)
$; h(A ~, x)/h(r-, x) ~ h(A"',
and H(A, r-)
~ h(A -, x)/h(r, x)
for all sufficiently small x E ,.1. Il. Since I(A)/I(r) and H(A, r) and also I(A)/I(r-) and H(A, r-) play the same roles in these fundamental inequalities from which Lemma 10.6 is eventually proved, it is an easy exercise to show that in Lemma 10.6,f(a)/g(~o) 1O.6,f(a)/g(~o) can be replaced by I (A a) a)11 (r) I (A a)/I a) lim limrr - - - - = II(B~o)ll(r) (B 130 )/ I (r ) II(B~o) (B (30)
since I is invariant on compact spheres with respect to all but countably many Il** with Ya < Y Y~13 whenever a < B ~ and such {y a} C J.l indicies from each interval {Ya} Yyr E Il Jl whenever y is rational. But then that Y (10.14)
A(A a)/(B (30) = leA a)
Aa Z1c(G)nZ/(G). Consequently, I is a constant multiple of A on whenever A a E Z",(G)nZ/(G). Z",(G)nZ/(G). Z1c(G)nZ/(G). It only remains to show that Z(G) = Z",(G) Z1c(G) = Z/(G).
For this let A
E
Z", (G). Then A Z 1c (G).
E {A
a} for some interval of sets with A a
< A 13~ whenever a < ~ and if A = = A ry then y is a continuity point of A(A a). Now Z(G) = = Z",(G)nZ/(G) Z1c(G)nZ/(G) contains all but an increasing sequence {A {Arn} rn }
C {Aa} {A a} (by Z1c(G) and Z,(G» Z/(G» and the remarks preceding Corollary 10.1 as applied to both Z",(G) these are the points of discontinuity of either A(A a) or I(A a). Let k > 0 such = kA(A a) for A a E Z( G). Let E > O. Then there is a 80 > 0 such that that I(A a) = (y-o,'(+o) c Z( G) and for each pair of indicies a, ~ with y - 8 0 < a < y < ~ < y + 8, 0, (y-8,y+8) C Z(G) kA(A~) I/(A~) IkA(A (3) - kA(A a) I < E since kA(A a) is continuous at y. But then I/(A (3) - I(A a) I I(A~13 - C/(Aa» $; I(A I(A~) < E. By the discussion preceding Corollary 10.1, I(A C I(A ~ (3) --/(A~) I(A (3) < E. Hence A Ayr is a zero-boundary set with respect to I so A E Z,(G). Z/(G). By a similar argument, if A E Z/(G) then A E Z",(G). Z1c(G). Hence Z(G) = Z",(G) Z1c(G) = Z/(G).· Z/(G).-
a»
lOA can now be used to prove uniqueness of the Haar measure of Theorem 10.4 Theorem 9.3. THEOREM 10.5 Let (X, v) be a locally compact, isogeneous, uniform space and let Il f..l be an isometric basis for v. Let m be the Haar measure of Theorem 9.3 and suppose h is another Haar measure on (X, v). Then h is a constant multiple of m on (X,V). Proof: Both m and h are defined on G because all open sets are Borel sets. Since m and h are measures they satisfy Ll - L3. Let A E G. Since m and h are IKe A and K = sup{m(K) IKe K is compact} and inner regular on open sets, meA) = sup{h(K) IKe A and KK is compact}. Clearly sup sup{m(V) h(A) = = sup {h(K) IKe {m(V) IV < A and V E
10.4 Invariance and Uniqueness of Loomis Contents and Haar Measures
303
G} ~ meA) m(A) and sup{h(V) sup { h(V) II V < A and V E G} ~ h(A), so we can show m and h to satisfy L4 if we show that for a compact K c A, there exists a V E G with K c V O. Choose a compact C c uU A*(UVn ) -- E. uVnn with A+(C) > A*(uV,,) £. Then there is a finite subsequence {U {Vmn m,,}} of {V nn }} that covers C. Then A*(uV A*(UVnn)) -- E£ < A+(C) :s: ~ A*(UU ) ~ LA*(U ~ A*(uVmJ :s: LA*(VmJ mn mn ) :s: L'A*(Un). Since £E was chosen arbitrarily, A*(uV A*(UVn ) :s: ~ LA*(V LA*(Un). LA*(Un) n ) whenever A*(UVnn )) is finite. N-ow A*(uV Now suppose A*(UU,J A*(uV,,) is infinite. Then for each a> 0 there is a compact C c uU uV"n with A+(C) > u. a. Just as before, there is a finite {V nn }} that covers C so LA*(Umn ) > u. Hence LA*(U subsequence {U {Vm,,} of{V thatcoversCsoLA*(UmJ>a. LA*(Vn) is mn } of infinite. Therefore 1.* also infinite" A* is countably subadditive on open sets. To conclude the proof we must show that 1.* A* is countably subadditive. For this let {En} be a sequence of subsets of X. Note that it is sufficient to show A *(uEn) :s: ~ LA*(E *(En) < 00 for each n, since if some A *(Em ) = = 00 the A*(uEn) LA*(En) A*(En) A*(Em) n) where A proof is triviaL trivial. Let E£ > O. For each positive integer n let Un V" be an open set A*(En) A*(Vn) - £/2n. uEnn c uVn E/2 n. Then uE uUn and A*(uEn) containing En such that A*(E n) > 'A*(U n] = :s: Ln[A*(En) £/2n] = LnA*(En) £. £E was chosen ~ A*(uVn):S: A*(UUn) ~ LA*(V,,) LA*(Un) < Ln[A*(E ) + E/2 LnA*(E ) + E. Since n n A*(uE,,):s: LA*(En).arbitrarily, this implies A *(uEn) ~ LA *(En). PROPOSITION 10.8 If W is an open set with compact closure then A*(W)
= A(W).
= A*(W) A.(W) = = sup{A+(c)lc sup{A+(C)lc c Wand C E Proof: A*(W) = E K} = SUp{A(lnt(C»IC sup{'A(lnt(C»IC c W and C E K} = SUp{A(U) Wand sup{A(U) ICl(U) CI(U) c W and U E ~} 't} since Cl(U) CI(U) E K for each U
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10.5 Local Compactness and Uniform Measures
't with CI(U) c W. Now for each U E 1 't with CI(U) c W there is a V E 1't with E 1 CI(U) eVe CI(V) c W so A*(W) = SUp{A(U) sup{A(U) I U c Wand U E 1} 'tJ = A(W). Therefore A*(W) = A(W).· A*(W) = A(W). -
A*(C)jorr each C COROLLARY 10.2 A+(C) ~ A*(C)fo
E
K.
If x E J.l* /..1* and S(p. 5i(p, x) and S(q. S(q, x) have compact compacc COROLLARY 10.3 !! x)) ~!'A(S(p, x)) = A(S(q, x)r closures then A*(S(p, x)) = A*(S(q, x)) if and only ifA(S(p,
=
=
X are A*-measurable. PROPOSITION 10.9 The open subsets of X Proof: We need to show that for an open set 0 that A *(S) = = A*(OnS) + A*(S-O) A*(S) for any subset S of X. We start by showing that if U, V E 1't and Un V = = 0 then A*(UuV) = A*(U) A*(U) + A*(V). A*(V). For this it is clearly sufficient to consider the case A*(UuV) A*(U) and A*(V) A*(V) are finite. Let E£ > 0 and let C and D be compact where both A*(U) subsets of U and V respectively such that A*(U) - A+(C) ofU A"'(C) < £E and A*(V) - A+(D) A'" (D) < E. That this is possible can be seen from A A*(U) A*(U) c U £. *(U) = A. (U) = sup{A+(K)IK sup (1..+ (K) IKe and K E K} K) ~ A+(C), for any compact C c U, and similarly for a compact D c A+(C) + E£ + 1..+(0) = A+(CuD) + 2E 2£ V. Then CnD = = 0 and A*(U) + A*(V) ~ 'A+(C) A+(D) + £,£ = ~ A*(CuD) A*(CuD) + 2£ (by Corollary 10.2) ~ A*(UuV) + 2£. Since E£ was chosen A*(U) + A*(V) A*(V) ~ A*(UuV). A*(UuV). But then A*(UuV) A*(UuV) = =A *(U) + A*(V). arbitrarily, A*(U) A*(U)
=
=
Next we show that if U and V are open sets then A*(V) A*(V) = A*(UnV) A*(Un V) + A*(V-U). For this let E£ > O. Let C be a compact subset of Un V with A +(C) > UnV A+(C) A.(UnV) CI(W) c A* (UnV) - £E = = A*(UnV) - £.E. Let W be an open set such that C eWe Cl(W) UnV. Then V - U c V - CI(W), so A*(UnV) + 'A*(V A*(V - U) ~ A*(UnV') A*(UnV) + A*(V-CI(W» < A+(C) + E£ + A*(V - CI(W» ~ A*(W) + £ + A*(V - CI(W». Now 'A*(V-Cl(W» W and V - CI(W) are disjoint open sets so as we just showed in the preceding A*(Wu[V - CI(W)]). CI(W) c paragraph. A*(W) + A*(V - CI(W» = A*\WU[V paragraph~ CI(W)1). Also, since Cl(W) V we have Wu[V - Cl(W)] CI(W)] c V so A*(WU[V A*(Wu[V - Cl(W)]) CI(W)]) ~ A*(V). Hence A*(UnV) + A*(V - CI(W» ~ A*(V) + E. arbitrarily, £. Since E £ was chosen arbitrarily. CI(W» :s; A*(UnV) + A*(V - CI(W» ~ A*(V). But then A*(UnV) + A*(V - U) :s; ~ A(V). Lemma 10.7, Conversely, since 1..* A* was shown to be subadditive in the proof of Letnma A*(V) ~ A*(UnV) + A*(V - U). Hence A*(V) = A*(UnV) + A*(V - U). Finally we show that for an open set 0 that A*(S) = = A*(OnS) A*(OnS) + A*(S - 0) A*(OnS) + A*(S - 0) :s; A*(S) so assume for any SeX. If A*(S) is infinite then A*(OnS) ~ A*(S) A*(S) A*(S) is finite. Let E£ > 0 and choose an open set U containing S such that A*(S) > A*(U) - E. O~ so A*(OnS) £. Then OnS c OnU and S - 0 c U - O. A*(O"S) + A*(S - 0) ~ A*(U - 0) = = 'A*(U) A*(U) as we just showed in the preceding paragraph. A*(OnU) + 'A*(U Therefore, A*(OnS) + 'A *(S - 0) ~ A*(S) + E. arbitrarily, A.*(S £. Since E £ was chosen arbitrarily. A.*(OnS) + A*(S A.*(S - 0) $~ 'A*(S). A.*(S). Conversely, since 1.* A.*(5) :s; ~ 'A*(OnS) 'A* is subadditive, A*(S) A.*(OnS) + A*(S A.*(S - 0), so 'A A.*(S)= *(S) = A*(OnS) A.*(On5) + A*(S A.*(S - 0). 0). • _ A We have now shown that 1.* A* is a Borel (and hence Baire) measure on X. This follows from Propositions 10.7 and 10.9 and Theorem 8.4. Next we want
310
10. Uniform Measures
to show that A* is regular on B 0 (the Baire sets). For this we need a number of results that will eventually enable us to conclude that the mere outer regularity of A* on compact sets is sufficient to imply the regularity of A* on all Baire sets. This is important in our development because the outer regularity of A* follows from the definition of A* using the result in the first paragraph of the proof of Proposition 10.7. Consequently, this will demonstrate the regularity of A** on B oo.. The remaining results in this section, up to Theorems 10.10 and A 10.11 date back to the 1940s or earlier.
PROPOSITION 10.10 Every Baire set in X is a-bounded (can be covered by countably many compact sets) and every a-bounded open set is a Baire set. Proof: It is easily shown that the class of all a-bounded sets is a a-ring. Since each compact set is a-bounded, this a-ring contains all compact sets and hence the smallest a-ring (Proposition 8.3) containing the compact sets, namely, B o. Hence every Baire set is a-bounded. The remainder of the proof is left as an exercise. •PROPOSITION 10.11 If~ If ~ is a Baire measure that is outer regular on compact sets, then ~ is outer regular on the difference C - D of any pair of compact sets C and D such that DeC. Proof: Since ~ is outer regular on C, for each E > 0 there exists an open Baire set U with C c U and ~(U) < ~(C) + E. Put V = U - D. Then V is open and C D c V. Moreover, ~(V)
= - ~(C - D) =
~(V
- [C - D]) = =
~(U
= - C) =
~(U)
- ~(C) < E.
Hence for each E > 0 there exists an open V E B 0 that contains C - D such that < ~(C - D) + E, so C - D is outer regular. regular.•
~(V)
We say that an open set is bounded if it is contained in a compact set.
PROPOSITION 10.12 If~ is inner regular on each bounded open set, then ~ is inner regular on the difference C - D of any pair of compact sets C and C such that DeC. The proof of this proposition is similar to the proof of Proposition 10.11 (remember we are assuming X is locally compact) and will be left as an exercise. Moreover, the proof of the next proposition will be left as an exercise for the same reason. PROPOSITION 10.13 If J..1 ~ is inner regular on a finite collection of disjoint sets offinite measure, then J..1 ~ is inner regular on their union.
10.5 Local Compactness and Uniform Measures
311
THEOREM 10.6 Let J.! /-t be a Baire measure and {En} a sequence of the following hold: Baire sets. Then the/allowing If/-t is outer regular on each En' En. then J.l /-t is outer regular on uEn. (1) IfJ.l (2) IfJ.l If/-t is inner regular on each En and {En} is an ascending sequence, sequence. then J.! /-t is inner regular on uEn. (3) IfJ.l If/-t is inner regular on En and J.l(E /-teEn) n. then n) < 00 for each n, J.! /-t is inner regular on nEn. (4) IfJ.l If/-t is outer regular on En and J.!(E /-teEn) n. and if if n) < 00 for each n, {En} is descending, then J.! /-t is outer regular on nEn. 00
00
Proof: To show (1) let E10 > O. For each n there exists an open Baire set Un with En C Un and J.l(U E/2 nn. Put V U = uUn and E = uEn. If J.!(E) /-t(Un) /-teEn) /-teE) = 00 n) < J.l(E n) + e/2 then clearly J.! /-t is outer regular on E. If J..l(E) /-teE) < 00, then 00
00 ..
set,. this shows that J.! /-t is outer regular on uE uEnn ==E. Since V is also an open Baire set. To show (2) put E = uEn. By Proposition 8.6, J.!(E) /-teE) = limnJ.!(E limn/-t(En). n). Let 10E > O. Choose n such that }..l(E) /-teE) - 10E < J.!(E /-teEn). /-t is inner regular on En there exists n). Since J.l a compact C c En with f.l(E) J.!(E) - 10E < J.!(C). /-t(C). But then J.l /-t is inner regular on uE uEnn = E.
enn Cc En with To show (3), let £10 > O. For each n there exists a compact C n J.!(E E/2 . Put C = /-teEn) /-t(e nn)) + e/2n. = ne nen and E == nEn. Then C c E and C is n) < J.!(C compact. Moreover,
Since C is compact, this shows f.l /-t is inner regular on nEn • To show (4), put E = = nEn. By Proposition 8.7, J.l(E) /-teE) = = limnJ.l(E limn/-t(En). n). Let 10E > O. Choose n such that Jl(E < Jl(E) E. Since J.l is outer regular on En there ) /-teEn) /-teE) + £. /-t n exists an open Baire set V such that J.l(E f.l(En) :5: f.l(V) /-teE) e. /-teE) :5: ) ~ }..l(V) < J.l(E) + E. But then J.!(E) ~ n /-t(V) < J..l(E) f.l(E) + 10E so J.! /-t is outer regular on nEnn = = E. •J..l(V)
THEOREM 10.7 A necessary and sufficient condition for a Baire measure J.l /-t to be outer regular on compact sets is that it be inner regular on bounded open sets. Proof: Suppose J..L /-t is outer regular on each compact set and let V be a bounded open set. Let 10£ > O. Let C be a compact set such that V c C. Then C - V is /-t is outer regular on C - V. Therefore, there exists an open compact and hence fJ. Baire set V such that C - V U c V and fJ.(V) E. Since C - V c C J.!(V) < fJ.(C J.!(C - U) + e. = V we have: (C-V) =
J.!(U) - fJ.(C J.!(C - V) = = J.!(U V]) = = J.1(UnV) J.!(Un V) :5: J.!(V - [C - U]). fJ.(U) J.1(U - [C - V) ~ Jl(V
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10. Uniform Measures
/.l(V) - J..l(C /.l(C - U) < e /.l(U) Since the last term on the right is equal to J..l(V) E we have that J.l(U) -- e E < J.l(C /.l(C - V). Since C - V is compact we have shown that J.l /.l is inner regular on U. /.l is inner regular on each bounded open set and let C Conversely, suppose J.l be compact. Let e E > O. Since X is locally compact we can find a bounded open set U containing Co Then, since U - C is a bounded open set, there exists a compact set D c U - C with J.l(U E. Since C = U - (U - C) c U - D /.l(U - C) < J.l(D) /.l(D) + e. we have: J.l(U /.l(U - D) - J.l(C) /.l(C) = = f.l([U /.l([U - C] - D) = = J.l(U /.l(U - C) - J.l(D) /.l(D)
1, Sn s" - Sn-l Sn~1 is a simple 2~"n on the set of points where f - Sn-1 S,,_1 ~ 2" measurable function that equals 22n and n o otherwise. Consequently, 22"0" o n is the characteristic function of this set which we denote by En· E". Then En E" Cc E for each n. Let U be an open set containing E such that CI(U) Cl(U) is compact. Then for each n, there exists a compact set K K"n and n. By Theorem 8.5, an open set Un with K" Kn Cc En C Un C U and J.l(U < e/2 ) J..L(Un - K K,,) £/2". n there exist functions gn E CK(X) with 00:5 ~ gn ~ 1 such that K [I'l" Let g" :5 Knn < gn g" < U",
324
11. 1l. Spaces of Functions
Clearly this series converges unifonnly on X. An easy modification to Theorem Knn we have g = f except on 1.12 implies g is continuous. Since an = 2- nn h nn on K Kn) a measure ) u(U nn -- K which is set of Ineasure less than £. Thus we have proved the n existence part of the theorem for the case where E is compact and 0 s: ~ f < 1. Consequently, it is easily shown that the existence part holds if E is compact andfis and f is merely a bounded complex measurable function. Furthennore, we can remove our compactness assumption, for if Jl(E) < 00, there exists a compact K c E with Jl(E - K) < ££, for any ££, > O. To prove the general case, letf let f be a complex measurable function and for each positive integer n put F Fnn = = {xl If(x) II > n}. F n is measurable and nF (x Ilf(x) n). Then each Fn N n = 0. Consequently, limnJl(F limnJl(Fn) Fn. n) = 0 by Proposition 8.7. For each positive integer n put En = X - F n. Then f is a bounded function on En. For each n define fn by fn(x) = f(x) if x EE (x) = 0 otherwise. Then If is a bounded measurable function on X and En and fn fn(x) Jl( {x If(x) =t;t fn(x)}) ~ Jl(Fn). Now let £ £, > 0 and pick k with Jl(F f.l(F kk)) < £/2. Thenfk Jl«(xlf(x) fn(x))) s: Then/k is a bounded function on X and Jl( {x f(x) =t;t fn (x) }) < £/2. Then there exists a g (x I[f(x) fn(x»)) E CK(X) with Jl( {x ;t f(x)}) (x Ig(x) ;t =t- fn(x)}) fn(x»)) < £/2 so Jl( {x (x I g(x) =tf(x))) < £. This proves the existence part of the proof. To conclude the proof we first observe that if supx If(x) II = 00 then supx Ig(x) I[ ~ s: supx If(x) I so assume supx I/(x) If(x) I < 00. Put b = supx If(x) I and let hex) = x if Ix I s: ~ b and bx/ Ix I otherwise. Then h is a continuous function from C onto CI(S(O, b». If g' E CK(X) CdX) such that f(x) = gJ'(x) g'(x) except on a set of measure less than £,£ and g = h © lgJ' then f(x) = g(x) except on a set of measure less than £,£ and supx Ig(x) I s: ~ supx If(x) I. This concludes the proof.· proof.-
THEOREM 11.4 JJ.4 If X is a locally compact space and Jl is a positive, complete, almost regular Borel measure on X then CK(X) is dense in LP(Jl)for IJ S:p 00 ~ P < 00. 0
Proof: Let SM be the class of complex valued, simple measurable functions on X whose support has finite measure. If s E SM and £ > 0, then by Theorem 11.3, there exists agE CK(X) with g(x) = sex) except on a set of measure less than £ and Igi ~suPxls(x)1 = Isl~. LetE= {xlg(x);ts(x)}. Igl S:suPxls(x)1 (xlg(x)=t-s(x»). Then
Ig-slpp Ig-sl
II s: [fEI2Isl=lpd~]IIP rfEl2lsl~lpdJllllP = 2Isl=E 2Isl~£lIp. = rf[fElg-sIPdJll P. E lg-sIPdJl]IIllPP ~
Therefore, it is possible to find a sequence (gn) {gn} c CK(X) such that {gn} (gn) converges to s with respect to the LP -nonn metric. But then since SM is dense in U(Jl) LP(Jl) by Proposition 11.3, so is CK(X).· CK(X).-
p < 00, U(Jl) Theorems 11.2 and 11.4 together say that for 1 s: ~ P LP (J..l) is the completion of CK(X) with respect to the LP -nonn metric. The case where p 00 is different because the definition of L ~(Jl) (Jl) is essentially different than the
=
00
11.1 LP-spaces LP -spaces
325
definition of LP(f..l) LP (J.l) for p < 00. CK(X) is a metric space with respect to the DO. Now CK(X) L oc ~ -nonn -norm metric. To characterize the completion of CK(X) with respect to this metric, we need the following definition: A complex valued function f on X is said to vanish at infinity if for each £ > 0 there exists a compact set K such that If(x) II < £ for each x E X - K. The class of all continuous, complex valued functions that vanish at infinity is denoted by C ~(X). (X). On CK(X) CK(X) the L ~ -norm coincides with another norm called the supremum norm that is defined by IfI If I = = supx If(x) I. IX)
IX)
THEOREM 11.5 JJ.5 If X is a locally compact space then C ~(X) (X) is the completion ot of CK(X) CdX) with respect to the L = -norm and the supremum norm. 00
IX)
(X) is, indeed, a metric space Proof: It is left as an exercise to show that C =(X) with respect to the L = -nonn. (X) -norm. We need to show that CK(X) is dense in C =(X) and that C =(X) (X) is complete. That CK(X) is dense in C oo(X) =(X) follows from the fact that if f E C =(X) (X) and £ > 0, then there exists a compact set K with if(x) If(x) I < £ outside K and an h E C K(X) with 0 ~ h ~ 1 and h(K) = 1. If we put g = fh then clearly g E CK(X) and If - g I = = = supx If(x) - g(x) I < £. 00
00
IX)
00
00
IX)
To show C ~(X) (X) is complete let {fn} oo(X). For Ifn} be a Cauchy sequence in C =(X). each £ > 0, there exists a positive integer N such that if m, n > N then Ifm - fn I < £ so supx Ifm(x) - fn(x) II < £. Then for each x E X, the sequence {fn(x)} Ifn(x») is a Cauchy sequence in C and hence converges to some point f(x) E C. This = limdn(X) limnin(X) is well defined. Moreover, pointwise limit function defined by f(x) = {fn} Ifn) converges unifonnly uniformly to f since for each £ > 0 there exists a positive integer N such that if m, n > N then supx Ifm(x) - fn(x) II < £/2. Also, for each x E X, there exists a k > N with Ifk(X) fk(X) I < Ih(x) - f(x) I < £/2. But then if m > N, Ifm(x) - hex) £/2 so Ifm(x) - f(x) II < £ for each x E X. Now an easy modification of Theorem thatfis continuous. 8.12 shows that.tis 00
=(X). For this let £ > O. Then there exists an It remains to show that f E C oo(X). If -- In fn I1= < £/2 and there exists a compact set K such that Ifn(x) II < £/2 n with If If(x) II < £ outside K so f E C oo(X).· ~(X).outside K. But then I/(x) IX)
EXERCISES 1. Prove Proposition 11.1. 2. Prove Proposition 11.3.
d(J, g) = = If - g Ip is a pseudo-metric on LP (f..l). 3. Show that d defined by d(j, (J.l). 4. Prove Theorem 11.2 for the case p =
DO. 00.
5. (Jensen's Inequality) If Jl f..l is a positive measure on a a-algebra M in a space
326
11. Spaces of Functions
~(X) = (~) with a < g(x) < b X such that J..l(X) = 1 and if g is a real valued function in L 1I (J..l) iffis for each x E X, and iff is convex on (a, b), show that:
ai > 0 for each i = = 1 ... n such that LUi Lai = = 1 and if Xi is a real number for 6. If Ui e(LaixJ :s;; Laiex, where e is the exponential each i = 1 . . . n, show that e(LuixJ ~ LUiexi function. ~ is a positive, complete, 7. Show that if X is a locally compact space and J..l almost regular Borel measure on X and if the distance d(j, d(f, g) between two d~ then (CK(X), d) is a functions f, f, g E CK(X) is defined by d(f, g) = J Ixx If - g IdJ..l metric space and the completion of (CK(X), d) is precisely the class of Lebesgue integrable functions on X. [Recall that the Lebesgue integrable functions were defined in Exercise 1 of Section 8.6.]
8. Show that if p < q that for some measures J..l, LP (J..l) C L q (J..l) ~, U (~) c (~) whereas for (~) c U (~), and that there are some measures such that other measures L q (J.l) LP (J.l), neither U LP (J.l) (~) or L q (J.l) (~) contains the other. What are the conditions on J.l ~ for which these situations occur?
2 11.2 The Space LL2(~) (J.l) and Hilbert Spaces
(~) is known as the space of square integrable functions. It plays The space L 2 (J..l) a major role in modem physics and in many other mathematical applications. In fact, it is the mathematical model that underlies the wave interpretation of ~ is Lebesgue measure on quantum mechanics, when X is Euclidean space and J..l X. L 2(J.l) 2(~) is a special case of a more general class of spaces called the Hilbert spaces. Other Hilbert spaces also play important roles in quantum mechanics, in fact, they are the mathematical models behind all quantum phenomenon. We will need some results about Hilbert spaces for our development of uniform differentiation in the next chapter. Hilbert spaces derive their name from David Hilbert who published a series of six papers between 1904 and 1910 titled Grundzuge einer allgemeinen Theorie der linearen Integralgleichungen I - VI in Nachr. Akad. Wiss. Gottingen Math.- Phys. Kl. that involved these spaces. They were republished in book form by Teubner Verlagsgesellschaft, Leipzig in 1912 and reprinted by Chelsea Publishing Co., New York in 1952. We now define these spaces.
A complex vector space H is called an inner product space if for each pair of vectors u, v E H there is a complex number (u, v) called the inner product or sometimes the scalar product or dot product that satisfies the following axioms:
L2()l) 11.2 The Space L 2(Jl) and Hilbert Spaces
(1) (2) (3) (4) (5)
327
(u, v) = (v, u)* (the * representing complex conjugation). = (u, w) + (v, w) for u, v, w E H. (u + v, w) = (cu, v) = c(u, v) for u, v E Hand c E C. (u, u) ~ 0 for each u E H. (u, u) = 0 if and only if u is the zero vector in H.
°
°
°
There are a number of observations we can make about these axioms. First, (3) Hand implies (0, x) = 0 for each x E H and (1) (I) and (3) together imply (x, cy) = c*(x, y) for each pair x, y E H and cC E C. Next we observe that (2) and (3) together = (x, y) for each x E His imply that for each y E H, the mapping defined by A(X) = a linear functional on H. (1) and (2) can be combined to show that (x, y + z) = (x, y) + (x, z) for x, y, z E H. Finally, by (4) we can define a nann norm Ix I for each x Ix I = (x, X)1I2 x)1!2 so that Ix Ix 1 122 = (x, x). E H by Ix PROPOSITION 11.5 (Schwarz Inequality) For each x, y E H, I(x, y) I :s; II y II where the norm on the left is the modulus of the complex number (x,y). ~ IIx II Proof: Ifx=Oory=Othen l(x,y)1 ~ Ixllyl soassumex:;tO:;ty. l(x,y)l:s; Ixllyl so assumext:O t:y. Let abean arbitrary complex number. Then (x + ay, x + ay) ~ 0 and (x + ay, x + ay) = 2 = 2Re(a(y, x» so IIx 112 + I I al a 1 22 11 y 12 /2 + a(y, x) + a*(x, y) and a(y, x) + a*(x, y) =
°
2 Ix1 I x 1 2 + IlaI2IyI2+2Re(a(y,x»~0. a /2 I y 1 2 + 2Re(a(y, x» ~ O.
°
Now each complex number a can be represented by a = re itil for some real number r ~ 0 and some complex number e itil for some real number t. Recall that it I = 1. Similarly, (Y, (y, x) = II(Y, (y, x) IIe isis for some real number s. Hence II eeill it it Re(a(y,x» (Y, x) leis) = Re(rl (Y, (Y, x) IRe(ei(s+t) Re(a(y,x» = Re(re II(y, (y, x) II ei(s+t) ei(s+I) = rl (y, IRe(ei(s+I) and (S+I) :s; Re(e i (s+t) ~ 1. Consequently,
°
2 1yl2 + 2rl(y, Ixl22 + Irl Irl21yl2 0. Put r = 2rl(Y, x)1 ~ 0 for each real number r ~ O. so Ixl -I x I/I I/ Iy I. I. Then substituting this value for r in the previous inequality yields 1.Ix II y I :s;~ 1 I(x, y) I.·
An immediate result of the Schwarz inequality is the so called triangle y I1 :s; I. It follows from the observation that IIx + Y y1 122 = ~ I1x I1 + I1y I. inequality: I1x + Y 2 y,x + y) = (X,x) (x, x) + (x,y) y» + lyl2 lyl2:s; Ixl22 (x + y,X (x, y) + (y,x) + (y,y):s; (Y, y) ~ Ixl /x1 2 + 2Re«x, y» ~ Ixl 2 + 21(x, y)/ + lyl2 ~ Ixl 21xllyl + ly/2 = (Ixl + lyl)2. +21(x,y)1 lyl2:s; Ixl 2 + +2lxllyl lyI2=(lxl lyl)2. Consequently, if we define the distance d(u, v) between two vectors u, v E H to be I1 u - vvi, I, it is easily shown that d satisfies the axioms of a metric space. In particular, if u, v, w E H we see that d(u, v) :s; dew, v) follows from I1 u - vi W ~ d(u, w) + d(w, v I :s; ~ I1 u - wi + I1 ww - v I.I. If the metric space (H, d) is complete, we call H a Hilbert space. We f.l is a positive measure, then L 2(J..1) Now observe that if J.l 2(J.l) is a Hilbert space.
328
11. Spaces of Functions
For this we define an inner product on L 2(J.!) 2(Il) by (j, (f, g) = fxfg*dJ.l. fxfg*dll. Since g 2(Il) implies g* EE L 2(J.!) 2(Il) and the exponents p = = 2 == q are conjugate L 2(J.!) exponents, Proposition 11.2 implies fg* ELI (J.!) so (j, ELI (Il) (f, g) is well defined on H. Ifl22 = = (j, (f, f) If II = f)l!2 Now we observe that if we define Ifl /) or equivalently, If = (f, if, /)112 then we have E E
2 2 L2-norm (Il) is equivalent to the inner product norm so the L -nonn II 12 1 2 on L 2 (J.!) nonn II II on 2 (Il). Since J.! Il was assumed to be positive, by Theorem 11.2 we know L 22(Il) L 2(J.!). (f.!) is nonn , so L 22(f.!) complete with respect to the L 22--norm, (Il) is complete with respect to II II and therefore is a Hilbert space.
PROPOSITION 11.6 JJ.6 For a given y E H, H. the mappings defined by f(x) g(x) = = (y, (y. x), h(x) = = IIxl x I for each x E X X are uniformly continuous functions on H.
= (x,y),
uniformly continuous let E > 0 and put 8 0= Proof: To show f is unifonnly Ix II - x 2211 < 80 we have by Proposition 11.5 that:
I I.
E/ I y I.
Then if
uniformly continuous on H. A similar argument shows that g is Therefore, f is unifonnly uniformly continuous on H. To show h is unifonnly uniformly continuous, let E > 0 also unifonnly 0= = E. By the triangle inequality, if IIx I -- x 21 < 8, 0, then IIx 1I II :0:; ~ IIx I -- x 21 and put 8 = IIX2 = + IIX21, x 21, so IIXII x I I -- IIX21:o:; x 21 ~ IIXI x 1 --x21. x 21. Similarly, IIX21 x 21 - IIXII xII = x 2 --xII X1I = Xl-x 2 1,so IIXI IIXI-x21,so 11 ~ IXI -x 1. Butthen IlxI -lx -lx211:o:; -x21. 2 2
so that h is also unifonnly uniformly continuous on H. •Recall that a subset S of a vector space V is called a subspace of V if S is a vector space with respect to the addition and scalar multiplication operations defined on V. A necessary and sufficient condition for S c V to be a subspace is E S and c EE C. If H is an inner product that u + v E E S and cu E E S for each u, v E space, a closed subspace of H is a subspace that is closed with respect to the metric topology generated by the inner product nonn. norm. A set E in a vector space is said to be convex if for each u, VEE ~x ~ 1, the vector (1 - x)u + xv is vEE and 0 :0:; x:O:; contained in E. One can visualize this property of convexity by imagining a segment" being traced out between u and v as x goes from 0 to 1, and that "line segment" all the vectors on this' 'line segment" this "line segment" are contained in E. If (u, v) = = 0 for some u, v EE H H we say u is orthogonal to v and denote this u.l1- be the collection of v EE H by u .l ..L v. Let u H which are orthogonal to u. If v, W EE u1= (u, v) + (u, w) = = 0 and if cC E u.l then (u, v + w) = E C then (u, cv) = = c(u, v) ==0 so v
11.2 The Space L2(~) L 2(1l) and Hilbert Spaces
329
W E E u.1 u1. and cv EE u.1. u1.. Hence u.1 u1. is a subspace of H. Now u.1 u1. is the set of H where the continuous function g(x) = (u, x) = 0, so u.1 u1. is a closed vectors x EE H set in H. If S is any subspace of H, let S.1 E H that S1. denote the collection of all v E are orthogonal to every Uu EE S. Clearly S.1 { u.1l U E S}. S1. = n n{u1.1 uE SI. Since each u.1 u1. is a closed subspace of H, so is S.1. S1..
+
PROPOSITION 11.7 I1.7 Each nonempty, closed, convex set in a Hilbert space has a smallest element with respect to the inner product norm. Proof: Let E be a nonempty, closed, convex subset of the Hilbert space H. Put b Then there exists a sequence {vnn IcE } c E such that { IIV vnn I} II Ix + yl2 + Ix - yl2 = (x + y, x + y) + converges to b. For any pair x, y EE H, Ix 2 (x-y,.x-y) = = 21 21xl2 21y12. (x-y,x-y) X 1 + 21 Y 1 2 • If we apply this identity to x/2 and y/2 we get 2 (1/4)lx-yI2= IxI 2/2 /2+ lyI2/2-I(x+y)/212. SinceEisconvex,(x+y)/2E (1/4)lx-yI2 = Ix1 + ly12/2 - I(x + y)/21 2. Since E is convex, (x + y)/2 E E which implies II(x + y)/21 ~ b, so
= = inf{ IIv II II vEE}. VEE I.
(11.2)
for each pair x, y EE H. If we replace x and y in this inequality by VVmm and VVnn we see that as m, n ~ 00, the right side of (11.2) tends to zero. Hence IVVmm -- VVnn I ~ o as m, n ~ 00 which implies {v n I} is a Cauchy sequence in E c H. Since H is complete, {v n I} converges to some v E H, and since E is closed, VEE. vEE. Also, norm function h(x) = = Ix II is continuous on H since the nonn H by Proposition 11.6, we have IIv II = limn IVVnn II = b. Consequently, there exists a vEE of smallest norm.
°
It remains to show that v is unique. If u is another member of H such that b = Ilvi, v I, then the inequality (11.2) implies Iluu - VvI2 12 ::;~ 0 so u = v. Therefore, E has a smallest element with respect to the inner product norm. •-
Ilui uI =
°
E H. Then the Let S be a closed subspace of the Hilbert space H and let u E set u + S = {u + v IIv EE S} S I is closed and convex. To see that u + S is closed, let w be a limit point of u + S. Then there exists a sequence {u + Vn V n}I in u + S that converges to w, so {u + vVnn }I is Cauchy. If E > 0, there exists a positive integer N such that if m, n > N then IIu + VVmm -- U - Vn Vn I V m -- V I < E which implies IVm Vnn II < E, so {v n I} is Cauchy in S. Since S is a closed subspace of the complete space H, S is complete which implies {v n I} converges to some v EE S. Then if E > 0, there exists a positive integer N such that if n > N then IIv - Vn vn II < E which implies I < e, E, so {u + v Vnn I} converges to u + VVEE U + S. But since limits II(u+v) - (u + vvn) n) I = W so W E are unique in metric spaces, u + v = E U + S. Hence u + S is closed.
To see u + S is convex, let u + v and u + W EE U + S and let 0 ::; ~ 'A::; A ~ 1. Then (1 - 'A)(u U + (1 - 'A)v E U A)(U + v) + 'A(u A(U + w) = U A)V + 'Aw AW E U + S. Therefore, u + S is convex. Consequently, we can apply Proposition 11.7 to u + S and get a Ps.L(u) in u + S with respect to the inner product norm. Next, smallest element Ps.l(u) putps(u) = = u - Ps.l(u). Ps.L(u). Thenps andps.l andps.L are functions on H. Since Ps.l(u) Ps.L(u) EE u + +
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S, 5, Ps(u) EE S. 5. The function Ps is called the orthogonal projection of H H onto S. 5. The function p s.l is called the orthogonal projection of H onto S1. 5.1.. For this later definition to make sense, we need to show that p s.l (u) E E S 1. . thatps.l(u) 5.1. For this let x = Ps.l(u). Then by the definition of the orthogonal projection onto 5.1 we have, for each yES y E 5 with Iy I = = 1, S1. 2 2 2 Ixl2 = (x (x-cy,x-cy) = IIxI IX 1 ~ IIx-cyl2 x - cy 1 2 = - cy, x - cy) = X 1 2 --c(y,x)-c*(x,y)+ c(y, x) - c* (x, y) + Icl I C 12
for each scalar c. If we substitute c = = (x, y) into this inequality, we get 0 ~ II y 1I = = 1. But then (x, y) = = 1 o for any yES. Therefore, x = Ps.l(u) E S1.. OforanYYE 5. Therefore,x=ps.l(u)E 5.1.
= 0 for each yES y E 5 with -I (x,y) 122 which implies (x, y) =
5 is a closed subspace of the Hilbert space H, then THEOREM 11.6 If S H onto Sand 5 and S1. 5.1 respectively have the the orthogonal projections Ps and Ps.l of H following properties: ps(u) + ps.l(u)for Ps.l(u)for each u E H. (1) u = Ps(u) (2) The orthogonal projections are unique. (3) The orthogonal projections are linear. (4)lfu 5 then thenps(u) andps.l(u) o. (4) ffu E S Ps(u) = u and Ps.l(u) = O. (5)lfu 5.1 thenps(u) = 0 andps.l(u) = u. (5) ffu E S1. = inf{ Iu - v II v EE S} 5j for each u EE H. (6) Iu - Ps(u) I = 2 2 Ips.l(u) 12 (7) II u 12 1 = Ips(u) 1122 + Ips.l(U) 1 for each u E E H.
=
I
I
I
=
=
II
Proof: (1) follows immediately from the definition of Ps(u). To show (2), first 5n5.1 = {OJ. 5n5.1 then (x, x) = 0 which {O}. This is because if x EE SnS1. note that SnS1. w where v E 5 and w wE 5.1. Thenps(u) + Ps.l(u) implies x = O. Next let u = v + w E Sand E S1..· = =U= v+w w which implies Ps(u) - v = w - Ps.l(u). Since the left side of this
=
=
=
=
equation is in S 5 while the right side is in S1., 5.1, we conclude that both sides are the w = Ps.l(u), so the orthogonal projections zero vector. Therefore, v = Ps(u) and w are unique.
(l),ps(cu + dv) + pS.l(cu Ps.l(cu + To show (3), let u, v E Hand c, dEC. Then by (1),ps(cu + = cfps(u) c[ps(u) + Ps.l(u)] + dv) = = cu + dv = + d[Ps(v) + Ps.l(v)] so Ps(cu + + dv) - cps(u) cps 1. (u) + dpsl.(v) dps.l(v) - Ps.l(cu + dv). Again, the left side of this equation dps(v) = CPs.l(u) 5 while the right side is in S1. 5.1 so we conclude both sides are the zero is in S vector. Therefore, Ps(cu + dv) = = cps(u) + dps(v) and Ps.l(cu Psl.(cu + dv) = = Cps.l(u) + dps 1. (v) so the orthogonal projections are linear.
5, then (1) implies u - Ps(u) = = Ps.l(u) To show (4) and (5), note that if u EE S, so the left hand side of this equation is in S 5 while the right hand side is in S1.. 5.1. and Ps.l(u) = O. Again, we conclude both sides are the zero vector, so Ps(u) = u andps.L(u) This proves (4). A similar argument proves (5). To show (6), note that by definition of Psl.(u) we have IIu - Ps(u) II = Ips.l(u)1I =in!{lu+vllvE =inf{lu+vllvE S} =inf{lu-vllvE Ipsl.(u) =in!{lu-vllvE S}. This proves (6). To show = (u, u) = = (Ps(u) + Psl.(u), Ps(u) + Psl.(u» Psl.(u» = Ips(u) 1122 + (7), observe that 1I u 1122 =
11.2 L 2(Jl) and Hilbert Spaces 1l.2 The Space L2(~)
331
2 (Ps(u), Ps-l(u» Psl.(u)) + (Ps-l(u), (Psl.(u), Ps(u» Ps(u)) + Ips-l(u) Ipsl.(u)12. Psl.(u)) = 0 = (Ps(U), 1 • Since (Ps(u), Ps-l(u» 2 2 (Psl.(u), Ps(u» Ps(u)) we have that 1 12 = Ips(u) 1 12 + Ips-l(u) Ipsl.(u) 1 122 which proves (7).(Ps-l(u), IUu 1 (7). •
COROLLARY 11.2 # H then there exists a U .1. S J J.2 If S :tu E H such that U u 1and u:t- O. andu#O. u = Ps-l(v). Psl.(v). Then v v:tu:tu 1Proof: Pick v E H - S. Put U # Ps(v) so U # O. But U .1. S since
UES.l.U E S.l.· In Proposition 11.6 we saw that the function f(x) = (x, y) for a fixed y E H is unifonnly continuous. Since (x, y) E C and since the definition of the inner product causes f to be linear, we see that f is a continuous linear functional. The Riesz Representation Theorem (Chapter 8) showed that positive linear functionals on CK(X) CK(X) could be represented as positive measures on X. It is therefore natural to ask if continuous linear functionals on a Hilbert space can be represented as inner product functions with respect to a given vector. The answer is affinnative as the next theorem shows. THEOREM 11.7 J J.7 If A is a continuous linear functional on a Hilbert space H, then there exists a unique v E H such that A(U) = = (u, v)for each Uu E H. Proof: If A(U) = 0 for each U u E H put v = O. Otherwise put K = {u I1 A(U) = O}. OJ. K is a subspace of H H and since A is continuous, K K is closed. Since A is linear, K {O j. Choose W wE Since A is not identically zero, Theorem 11.6 shows that K.l :t# {O}. 2 c = A(W)/ A(w)/II WW12 K.l such that w :t# O. Then w is not in K so A(W) :t# O. Put e 1 and let v 2 = c*W. c*w. Then v E K.l, vV:tc*c 1 cc*(w, w) = # 0 and A(V) = A(C*W) = C*A(W) = c*e I W 112 = ce*(w, = (c*W, (c*w, c*w) = (v, v). For any U u E H put x = U u - A(U)V/A(V). A(U)V/A(v). Then A(X) = A(U) A(U)A(V)/A(V) = 0 so X X E K which implies (x, v) = O. Now (u, v) = (X+A(U)V/A(V),V) [A(U)/A(V)](V, (X+A(U)V/A(V) ,v) = (x, v) + (A(U)V/A(V), v) = 0 + [A(U)/A(V)] (v, v) = [A(U)/A(V)]A(V) = A(U). Consequently, A(U) = (u, v) for each U U E H. w be another vector in H such that A(U) = To show that v is unique, let W = (u,w) for each U E H. Put z = v - w. Then (u, z) = (u, v - w) = (u, v) - (u, w) = A(U) - A(U) = 0 for each U E H. But then (z, z) = 0 which implies z = 0 so v = w. w.
•
Recall how a basis is defined in a vector space V. First we define a linear combination of vectors VI ... ..• V Vnn E V to be a vector sum of the fonn c 1 V v 1 + ... + en en Cn V Vnn for some e c 1 ••• .•• C VI'... .• V Vnn to be linearly n E C. We define the vectors VI independentifclvl + +",+cnvn=Oimpliesci=Oforanycj A independentifcIVl ... +cnvn=Oimpliesci=Oforanycl ... CnE C. A subset S of V is said to be linearly independent if every finite subset of S is linearly independent. The set V(S) of all linear combinations of finite subsets of S is clearly a vector space. In fact, it is easily seen to be the smallest subspace of V containing S. V(S) is called the span of S or the subspace spanned by S. If V(S) = V then S is called a spanning subset of V or we say S spans V. Finally,
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we define a basis of a vector space V to be a linearly independent subset that spans V.
PROPOSITION 11.8 A linearly independent subset of a vector space is a basis if if and only if if it is maximal. Proof: Let S be a maximal linearly independent subset of a vector space V. Assume V(S) :t '# V for otherwise S would be a basis for V. Then 0 E V(S) so there '# O. Let v I . . . Vn I v 1I + ... + exists a UU E V - V(S) with U U :t V n E S and suppose C 1 Cn V U = CiVJCn+I which Cn Vnn + Cn+l Cn+1 U = 0 for C I ••• Cn+I Cn+1 E C. If Cn+I Cn+1 :t '# 0 then U U= = Li=l Li'=1 CjVJCn+1 U E V(S) which is a contradiction. Therefore, Cn+l Cn+l = 0 which implies implies U = 0 so Ci Cj = = 0 for each i = = 1 ... n. But then Su{u} is linearly C I VI vI + ... + CnV cnv n = independent which implies S is not maximal which is a contradiction. Therefore, S is a basis for V. Conversely, assume {u a} is a basis for V and suppose {u i U a} a I is not '# 0 and {ua}u{u} {u a }u{ u} linearly maximal. Then there exists a U E H with U :t independent. Since {u a} spans V, there exists a finite subcollection UUI ... Un of {u aa }} and a finite collection CCI1 ••• C = CClUJ CnU nn•. Then cnn E C with U U = 1 U 1 + ... + CnU C lUI + ... . . . + Cn a }U { u} is not linearly CIUI CnUUnn + (-1) (-I)uU = 0 but (-1) :t '# 0, so {u {ua}u{u} independent after all. • A set of vectors {u a} in a Hilbert space H is said to be orthogonal if (ua,u~)=Oifa'#~. = 1 foreacha. (u a,u~) = 0 if a :t B. {u a} Ua I = for each a. If a } is said to be normalized if IIual normalized it is said to be orthonormal. Clearly, {u a} is both orthogonal and nonnalized {u a} is orthononnal orthonormal if and only if (u a, u~) = 1 if aex = B ~ and 0 otherwise. U 1I .... . . Un PROPOSITION 11.9 If U Un is an orthonormal set and v = C 1U I U1I + ... CnU nn,. then Ci = = (v, ui)for each i = = 1 ... n and I vl vl 22 = = Li=l Li'=I!c; ICi 12.2 . . " + CnU 1
= 1 ... n, (v, uJ u;) = = (c 1U I U1I + ... + CnU CnU nn ' u;) = Ci(Ui, Ui) Ui) = = Ci Ci since Proof: For each i = uJ = U 1 ... ••• Un v 1 22 = (c 1 U 1 + ... + CnU nn'' C 1lUI U1 UI Un is an orthononnal orthonormal set. Also, IIvl = (v, v) = = (CIUI CnU nn)) = C 1ICI ud CnC nn*(U *(u nn,, Un) un) = Li' =,1I CiCi 1122 •• .• + ... + CnU C 1 *(u" *(U 1, U 1) + ... + CnC Li=I
COROLLARY 11J.3 J.3 An orthonormal set is linearly independent. orthonormal set and let U! Un E {u a}. Proof: Let {u a} be an orthononnal u 1. ... • • • Un a }. Suppose C 1 U U1 + ... + CnU cnn E C. By Proposition 11.9, Ci CnU nn = = 0 for C 1I ..• ... C Ci = = (0, Ui) u;) = = 0 for each = 1 ... n. Consequently, {u a} is linearly independent. • i=
THEOREM 1JJ.8 J.8 Each Hilbert space has an orthonormal basis. u II = 1. Proof: If H is a non-trivial Hilbert space, then there exists a UU E H with IIU Then {u} is an orthononnal orthonormal subset of H. Let P be the collection of all orthononnal (u), partially ordered by set inclusion. orthonormal subsets of H containing {u}, Since {u} E P, P :t '# (0 0 so by the Hausdorff Maximal Principle (an equiValent
2(1!) and Hilbert Spaces 11.2 The Space L 2(Jl)
333
fonn form of the Axiom of Choice), there exists a maximal linearly ordered subcollection Q of P. Clearly {u} E Q, so Q :t =I- 0. Let R = = uQ. If v, W E R then wEB E Q for some A, BE v E A E A and WEB B E Q. Since Q is linearly ordered by inclusion, either v, W E A or v, wEB. Since both A and B are orthononnal orthonormal orthononnal. subsets of H, (v, w) = 1 if v = wand 0 otherwise. Therefore, R is orthonormal. Suppose R is not a maximal orthonormal orthononnal set. Then there exists an orthonormal "# 0. Now S orthononnal set S 5 containing R with S 5 - R =I5 is not in Q Q and S 5 contains each member of Q Q so we can adjoin S5 to Q Q and still have a linearly Q is not maximal which is a ordered set with respect to inclusion which implies Q orthononnal set. By Corollary 11.3, R contradiction. Therefore, R is a maximal orthonormal is linearly independent. Suppose R is not a maximal linearly independent set. Then there exists a linearly independent set T with ReT and T - R :t =I- 0. Let x V be the subspace of H spanned by R. Then x is not in V V which E T - R. Let V y = PV-l{x) Pv.i(x) :t =I- 0 by Theorem 11.7, so (y, v) = 0 for each v E R. Put z = implies Y y/ Iy I which implies IzzlI = y/I = 1 and (z, v) == 0 for each v E R. Therefore, z can be adjoined to R to obtain an orthonormal orthononnal set in H containing R as a proper subset which is a contradiction. Consequently, R is a maximal linearly independent set orthononnal, R is an orthonormal orthononnal basis. •so R is a basis. Since R is orthonormal,
E
LEMMA 11.1 If V is a closed subspace of the Hilbert space H and if if u {u) is closed. H - V, then the subspace W spanned by Vu {u}
V Proof: Suppose v is a limit point of W. Then v == limn(v n + cnu) where {v nn }} C V and {c n } C C. Consequently, there exists a b < 00 such that IVn Vn = = Cnu I < b for each positive integer n. Assume {c n} n } has no convergent subsequence. Since closed and bounded subsets in C are compact, this implies limn I Cn I = = 00. But then IV n + cnul/lcnl < b/lcnl for each n and limnb/lcnl limnbllcnl = = O. Therefore, vn!cnn + u I = limn(vn!c n)} = = -u which implies -u E V V since V V is closed limn Ivn/c = 0, so limn{vn/c which is a contradiction. 00
Hence we may assume {c n } has a convergent subsequence {c mn } that converges to some C E C. Now v = = limn(v mn + cmnu) cmn u) = = limnv limn vmn mn ++ cu which implies {v mn } converges to v - cu. Since {v mn } cC V and V V is closed, [v - cu] cul E V V cul E W. IS closed.· c1osed.and v = [v - cu + cu] W~ Therefore, W is THEOREM 11.9 If u 1 . Un is an orthonormal set of vectors in In the L;=! (v, Ui)Ui II :s; Li'=! CiUj CiUi II for all cc!J. .. ~ I1 v - Li=J. •• Hilbert space H and v E H, then I1 v - Li=l if and only if Cj Ci = (v, Ui) for each i = 1 ... n. The " Cn E C and equality holds if vector Li=l L;=! (v, Ui)Ui is the orthogonal projection of v onto the subspace W un} and if d is the distance from v to W then I1v 122 = d 22 + {u!1 .... generated by {u . . Un} 2 L;=! I (v, Ui) 12. . Li=l 0
0
1
1
H is obviously closed. By Lemma 11.1, the Proof: The subspace {O} to} of H subspace of H spanned by {u I} is closed since it is the subspace spanned by {u 1 }. Proceeding inductively, it is clear that W is closed. Then by {O} u {U
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Theorem 11.6, Pw(v) is an element of W such that IIv - Pw(v) II ~ IIv - w II for each W pw:c are unique, Pw(v) is unique. w E W and since the mappings Pw and Pw-L L?=1 CiUi II for each Consequently, Pw(v) pw(v) has the property that IIv - Pw(v) II ~ IIv - Li=l collection C 11 ... ••• Cn in C. .. an E C. By Proposition 11.9, ai = Let Pw(v) = Li=l L?=1 aiui aiUi for some a 11 ... = Ui) for each i = 1 ... n. Now v =Pw(v) + Ps.l(v) Ps.:c(v) so (v, Ui) = (Pw(v), uJ + (Pw(v), uJ (Pw.l(v), (Pw:c(v), Ui) = ai + 0 since PW-L(v) Pw:c(v) is orthogonal to all the Ui. Hence Pw(v) = = L?=1 (v, UJUi' L?=1 (v, Ui)Ui UJUi II ~ II v - Li=l L?=1 CiUi 1I for all C 1 Li=l Ui)Ui' Therefore, II v - Li=l 1 ... ••• cn E C Ci = = (v, uJ Ui) for each i = and equality holds if and only if Ci = 1 ... n. 0
Finally, the distance d from v to W is the minimum value of Iv - w I such Pw(v» = W. Therefore, d = Iv - Pw(v) I so d 22 = (v - Pw(v), v - Pw(v)) 22 (v-Pw(v),v) - (v - Pw(v), Pw(v)). = 0 so d = Pw(v». Now (v - Pw(v), Pw(v)) Pw(v» = = (v - Pw(v), = 1 ... n, (v - Pw(v), Ui) = (v - Li=l L?=1 (v, UJUi, UJUi' v). To see this, note that for each i = Ui) = (v, uJ Ui) - (v, UJ(Ui, Ui)(Ui, uJ = Pw(v» = uJ = = O. Therefore, (v - Pw(v), Pw(v)) = (v - Pw(v), L?=1 (V,Ui)UJ (V,Ui)Ui) = Li=l L?=1 (v, uJ(v Ui)(V - Pw(v), Ui) Li=l uJ = O. Consequently, d 22 = = (v - Pw(v), v) = 2 2 2 2 (V-L'?=I(V,Ui)Ui, L?=I(V, L?=II(v, UJI2 (V-L'i=l (V,UJUi, v) = Ivl IV1 (v, Ui)(Ui, v) = 1Ivl V 12 + + Li=l I (v, uJ 1 which -- Li=l 2 22 2 L?=1 =1 II(v, Ui) 12. 1 = d + Li 1 • •implies IIv 12 that
W E
orthonormal set in the Hilbert space H. For each vector v Let {U a} be an orthononnal 2 Va = = (v, uua) Lalval2 a ) for each a. By the symbol L a lv a l is meant the 2 form II VI 12 = supremum of the set of finite sums of the fonn 1 + ... + II Vn V n 12 /2 where Vi = {U a}. With this notation we can (v, Ui) for each finite collection U11 ... •.• Un E {u state and prove the following classical result: E
H put
2 2 va 12 THEOREM 11.10 (Bessel's Inequality) La IIVa 1 ~ II V 112. .
Proof: For any finite collection U11 .. . . Un E {{uUa}, a}, Theorem 11.9 gives 2 2 LI=1 1I(v,uJ (V,Ui) 12 = IIV Li=l /2 = V 112 - d 2 where d is the distance from v to the subspace W 2 spanned by the vectors U1 ... LI=1 II(v, Ui) ••• Un' Since d ~ 0, this means Li=l uJ 12 1 ~ IIV V 12 /2 2 sums is less than or equal to IIv 12. and hence the supremum of such finite sunlS 1 •• form LaC a where 0 ~ C a ~ 00 for each a and where the Sums of the fonn form C 11 + ... + Cnn summation is defined as the supremum of finite sums of the fonn where C 11 ... ••• Cn E {C a} are especially interesting in Hilbert spaces because they are used in the characterization of the structure of Hilbert spaces. We will now Il(E) = = 00 if E develop this characterization. Let X be a set. For each E E X put Jl(E) Il(E) = = card(E) if E is finite. It is easily seen that 11Jl is a measure is infinite and J.l(E) on the a-algebra of all subsets of X. The measure 11 Jl is called the counting If(x) I, where the measure on X. Let f:X ~ C. Then it is easily seen that LxeX I/(x) summation is the supremum of the finite sums of the fonn form I/(x If(x 1) II + ... · .. + E I/(x I where X l ' .. X X, is the Lebesgue integral of III with respect to the ) If(Xn) I Xl' .. Xn If I n n 2 L2-space counting measure on X. We use the notation [2(X) to denote the L -space L 22(11) (J.l) where 11 J.! is the counting measure on X. 00
2(fl) and Hilbert Spaces 11.2 The Space L 2(IJ)
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In particular, if {u a II a E A} is an orthononnal basis in H and for some v E ---+ C is defined by v'(a) = == Va = == (v, ua) for each aa EE A, then it is a H, v':A ---* consequence of Bessel's Inequality that v' E E [2(A) since:
The importance of the Riesz-Fischer Theorem that we will prove next is that if {{u U a II a E E A} is an orthononnal set of vectors in H then f E E [2 (A) implies that f ---+ C for some vector v E H. Before we prove the Rieszis of the fonn v':A ---* Fischer Theorem, we first observe that Bessel's Inequality implies an even ---+ C defined by v'(a) = == (v, ua) for each stronger statement about a function v':A ---* a E a E E A for which v'(a) "# # 0 is at E A. Bessel's Inequality implies the set of a most countable. To see this, suppose the set of a E A for which v'(a) "# # 0 is uncountable. For each positive integer n put En = == {a E E AI II v'(a) II > lIn}. lin}. Then for some positive integer m, Em must be uncountable. But then there exists a finite subset 2 S of Em with LaeS IIv'(a) 1122 > IIV V 112 which is a contradiction.
THEOREM 11.11 (Riesz-Fischer) If {Ua {u a Ia E A} is an orthonormal set Hand iffE in a Hilbert space H and iff E l2(A), [2(A), thenf= thenf == v' for some v E H. Proof: For each positive integer n put En = {a E A Ilf(a) I > lIn}. == {a lin}. Then each En must be finite, for otherwise there would exist a finite collection al al ... ... an E A such that Li'=1 Li=1 If(a;) 122 > If12' If12. For each positive integer n let VVnn = == LaeEn If(a) IIUa. Then each Vn V n is in H. For each n define v~:A ---* LaeEn ---+ C by v~(a) = == (vn,U ) for each aE A. Thenforagiven~E A,v~(~)=(LaeEnlf(a)lua,u~) (vn,ua)foreachaE ThenforagivenBE A,v~(B)==(LaeEnlf(a)lua,u~) a == LaeE LaeEnn If(a) I (u a , u~) = == If(~)1 If(B) I if ~BEEn == = E En and 0 otherwise. Therefore, v~ = If IIXEnn'• Consequently, IfIf - v~ II ~ Ifl Ifl22 which implies If - v~ II ~ If I. I. 1
Since Em C En if m < n, it is clear that limn v~ = == f, so by Theorem 8.13, limn If - v~ II = == 0 which implies limn lim" If - v~ II = == 0 since we can choose an N large 2 [LaeEnn If - v~ 12]112 LaeEnn If - v~ II and therefore 1 ]112 < LaeE enough so that n > N implies [LaeE 2 == [~aeEn [LaeEn If - V~ 112]112 LaeEn If - v~ II = == If - V~ v~ 11 II for n > N. If - V~ 1122 = ]112 < ~aeEn Then since limn lim" I!If - v~ 1 122 = == 0, {v~} is a Cauchy sequence in [2(A). Now for each n, VVnn = == ~aeEn LaeEn I!(a) If(a) 1 IUa and since En E" is finite we can apply Proposition 11.9 to get If(a) II = == (v (v", E". If m and n are positive integers with m n, ua) for each a EE En. < n, then Em C En· En. Put E(n, m) - En - Em. Then 2 Iv~ - v~ II = == [l:aeE(n,m) [LaeE(",m) 1I(vn, (v", Ua) - (v (vm, == [:EaeE(n.m) Ilf(a) I - 01 01 22]112 ]112 = == 1 ]112 = [~aeE(n,m) 11!(a) m, Ua) 12]1/2
1
(l:aeE(n,m) LaeE(n,m) I!(a) U a )1I2 )1/2 = (LaeE(n,m) If(a) 1 IUa' Ua, :EaeE(n,m) Ifla) IIU )1!2 = == (V (v n - Vvm, Vv"n - v m == IIVv"n - VVm m I, m )1I2
so 1I VVnn -- Vm V m II == 1I v~ 1 2 which implies {v n V~ -- v~ 12 H and therefore n }) is Cauchy in H converges to some v EE H.
=
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11. Spaces of Functions
= (v, uua) Then v':A ~ C defined by v'(a) = a ) for each a is the desired function. To see this, for a fixed B ~ E A, the function g(x) = = (x, u~) for each x E H is unifonnly uniformly continuous (Proposition 11.6). Therefore, since {v {v,,} n } converges to v, {g(v )} to g(v). Hence limn(v , u~) (v, u~). Then for each a {g(v,,)} converges lim"(v,,, = n n E A, v'(a) = , Ua) = limn v~(a) = f(a) since {v~} to f. = (v, Ua) Ua) = = limn(v lim"(v,,, U a) = lim" = converges n Consequently,f= Consequently,! = v'.· v'. PROPOSITION 111.10 J .10 Let {u a} be an orthonormal set in the Hilbert space H. Then {u a} is a basis for H if if and only if if the set S of finite linear combinations of members of {u a} is dense in H. Proof: Assume {u a} is a basis for H. By Proposition 11.8, {u a} is a maximal linear independent subset of H. Suppose {u a} is not a maximal orthononnal orthonormal set in H. Then there exists a U E H with I U I = = 1 and {u a} U {{u} orthonormal. By u} orthononnal. Corollary 11.3, {u a} U {u} { u} is linearly independent which is a contradiction. orthonormal set in H. Now assume S is not dense in Hence {u a} is a maximal orthononnal H. Then there exists a v E H - CI(S) which implies there exists ad> 0 such that Iv - U U I > d for each U U E CI(S). Now v = = Ps(v) + Ps-l(v) Ps~(v) and by the remarks Ps~(v)/IPs~(v)1 is in S.i. S-L. Since Iwl Iwl = 1, preceding Theorem 11.6, W w = Ps-l(v)/Ips-l(v)I {u w} is orthononnal { U a} U {w} orthonormal which is a contradiction. Therefore, S is dense in H. a }u{
Conversely, assume S is dense in H and suppose {u a} is not a basis for H. orthonormal set in H, so there exists a U E H with Then {u a} is not a maximal orthononnal {u,,} Cc S such that IUI = 1 and U 1.. Ua for each a. Now there exists a sequence {un} limnu lim"u"n = u. Clearly U 1.. Un u" for each positive integer n. Since the function f:H uniformly continuous, {f(u {f(u,,)} ~ C defined by f(v) = (u, v) for each v E H is unifonnly n)} must converge to feu) = (U, u) = = 1. But this is impossible since (u, un) = 0 for = (u, u,,) = each n. Hence {u a} is a basis for H. •The property that S is dense in H has some very interesting ramifications. In fact, it leads to a representation of all Hilbert spaces H as 12(A) [2(A) where card(A) = {Ii a}) where {u a} is a basis for H. To show this is the case, we = card( {II need the following lemma. LEMMA 11.2 JJ.2 Let {u a II a E A} be an orthonormal set in the Hilbert of finite linear combinations of members of space H and let S be the collection offinite {u Hletx',yJ'E = (x,ua)and {ua}. For each pair x, y E H let x', / E l2(A) [2(A) bedefinedbyx'(a) be defined by x'(a) = (x, ua) and a }. Foreachpairx,yE yJ'(a) = (y, U ira) ua)for a ) for each a E A. Then S is dense in H if and only if (x, y) = (x', y') for each pair x, y E H. Proof: First assume S is dense in H. Let v E H. Now choose e£ > O. Then there exists a finite finite. col~ec~ion UU1 . . . U" Un E {u a} ••• CnEe C" E C such tha,t thalt w = =c 1 UU 1 a } and c 1 ••. wlthm £e of v. By Theorem 11.9, Iv - L7;1 uJu; I ~ v - wi < e. + ... + C"U CnU nn IS wIthIn Li=l (v, Ui)Ui £. Putz=L7;I(V,U;)Uj. Iv-zl +ewhichinturn Put Z = Li=l (v, Ui)Ui. Then I v - z I <e < £ which implies IIvl v I < IIzl zI + £ which in turn a lv'(a)12 uI)12 + ... + II(v, La Iv'(a)12 by implies (Ivl - e)2 £)2 < Izl2 = I(v, ul)12 (v, U,,)12 Un )12 ~ L 2 = = Lav'(a)[v'(a)]* == (v', V'). Hence, for each v Proposition 11.9. But La II v'(a) 12 1
w'
0
0
•
11.2 The Space L L2(~) 2(Jl) and Hilbert Spaces
337
H, (v, v) = II VV 1122 ~ (v', v'). By Bessel's Inequality, (v', v') ~ (v, v) so (v, v) (v', v') for each v E H. E
=
Now let u, v E H. Then for each c E C we have (u + cv, U u + cv) = (u (u'' + cv', cv' , ' U' + cv') which implies (u, cv) + (cv, u) = (u', (u , cv') + (cv', (cv ' , U) u') or c*(u, v) + c(v, u) = c*(u', v') + c(v', uj. = u'). Since this holds for each cC E C, it holds for cC = = 1 and c = i. When c = 1 we have (u, v) + (v, u) = (u', v') + (v', u'). When c = i we have (u,v) - (v, u) = (u', v') - (v', uj. u'). Adding these two equations yields: 2(u, v) = = 2(u',v') which implies (u, v) = = (u', vj. v'). 2(u',vj
Conversely, assume (x, y) = = (x', y') for each pair x, Yy E H. Suppose S is not dense in H. Pick u E H - C/(S) CI(S) and let W = Ps.l(u). Clearly W =I- 0 and (w, u a ) = a = o for each a. Put x = WW = y. Then (x, y) = IIwI2 W 12 > a 0 but (x', yj y') = La(W,Ua)(w,u a )* = 0 so (x, y) =I- (x', y') which is a contradiction. Hence S must be dense in H.·
*a
a
*
Recall from Section 1.5 that two metric spaces X and M are isomorphic if uniform homeomorphism f:X ~ M that preserves distance. Two there exists a unifonn HI1 and H 2 are said to be isomorphic if there is an isomorphism Hilbert spaces H f:H 1 ~ H 2 that is also a linear transformation, i.e., one that preserves sums and scalar products. Such a mapping is called a Hilbert space isomorphism. {U a Ia E A} is an orthonormal basis for the THEOREM 11.12 If {u Hilbert space Hand iffor each x E H, x' is the element of 12(A) 12 (A) defined by x'(a) 2 = (x, u a ) for each a, then the mapping A:H ~ 12(A) = 1 (A) defined by A(X) = x' for 2 each x E H is a Hilbert space isomorphism of H onto 1[2(A). (A). Proof: Let u, v E Hand E H and c, dEC. Then A(CU + dv) = = (cu + dv)'. Then for each a E A, (cu + dv)'(a) = (cu + dv, u a ) = c(u, u a ) + d(v, Ua ) = cu'(a) + dv'(a) = = 2 12(A). CA(V) + dA(V) so A is a linear transformation from H into 1 (A). That A is onto follows from the Riesz-Fischer Theorem.
*
= v'. Then for each a E A, (u, u a ) == (v, u a ). Now Suppose u =I- v but u' = 0 for each a so (u-v)' E 12(A) 12 (A) and (u - v)'(a) = (u - v, u a) a ) = (u, u a) a ) - (v, u a) a) =a (u-v), is the zero element of 12(A). 0 which implies W = (u-v)' 12 (A). Since u =I- v, Iu - vi vI > a = (u-v)j 1.. U u aa for each a, so {ua}u{w} {u a}U {w} is (u-v)/IIu - vviI is a unit vector in H such that w w.1 an orthonormal set in H. But then {u a} is not maximal which is a contradiction. Hence u - v = a 0 so u = v. Therefore, A is one-to-one.
*
Since (u, u) = = (u', u') for each Uu E H, Apreserves inner products and hence 2 [2(A). distance. Therefore, A is a metric space isomorphism between Hand 1 (A). Since A Ais a linear transfonnation, transformation, it is a Hilbert space isomorphism. •
EXERCISES 1. Show that the vector space C*(X) of all real valued continuous functions on
11. Spaces of Functions
338
Jxfg*dx (where dx x = [0, 1] is an inner product space with respect to (j, if, g) = Jxlg*dx denotes integration with respect to Lebesgue measure) but not a Hilbert space. 2. Show that if S is a closed subspace of the Hilbert space H then (S-L)-L (S.l)1-
= S.
3. Show that a Hilbert space is separable if and only if it contains a countable orthonormal basis. {un) be an orthonormal sequence in the Hilbert space H. Let S be the set 4. Let {un} of all v E E H with v = 1:;;'=1 L;;'=l cnu CnU n Cn lin. Then S is isomomorphic with n I ~ l/n. n where I c the Hilbert cube and is an example of a closed and bounded subset of H that is not compact.
5. Show that no Hilbert space containing an orthonormal sequence is locally compact.
6. Show that for each pair of Hilbert spaces, one of them is isomorphic to a subspace of the other. 7. Let U be a member of the Hilbert space H and let S be a closed linear WE subspace of H. Show that mint min { IUu -- vvii II vv E S} S) = =max{ maxI I(u, w) II W E S.l S -L and Iwi = 1). I}.
THE TRIGONOMETRIC SYSTEM Let T be the unit circle in the complex plane. If F:T ~ C is any function f(x) = F(e iX) ix) for each x E R is a periodic on T then the function If defined by I(x) i.e., f(x + 2n) = = I(x) f(x) for each x E R. Conversely, if function of R of period 2n, i.e.,/(x f:R ~ C is periodic with period 2n, it is easily seen that there is a function F:T I:R ix ~ C such that I(x) ) for each x E R. Therefore, we can identify the f(x) = F(e ix complex valued functions on T with the complex valued 2n periodic functions LP (n, on R. Define U (D, where 1 ~ p ~ 00, to be the class of all complex valued, Lebesgue measurable, measurable. 2n periodic functions on R for which
Iflpp III
=
[1/2nfrr-Jr.Jr) If(x) IIpdx] lip < [1/2nJ -lt,ltl I/(x) p dX] lip
00. 00,
In other words, U LP (n LP (f.l) f.l is Lebesgue measure on [-n, n) divided (D = U (IA) where IA l/2n. The factor 1/2n is not essential here but it simplifies the following by 1/2n. LP -nonn development. For instance, with this definition, the U -norm I Ip of the constant function 1 is 1.
f, ggEL E L2(D (j, g) = l/2nJ1tJ(x)g(x)*dx for each x EE 8. For each pair I, 2(n define if, = 1/2nJ~J(x)g(x)*dx [-n,n). Show that (j, g) defines an inner product on L 2(n and that IfI f) L2(D If I~ = if, (j,!) for each f E L 22(D. foreachjE (n.
2(11) and Hilbert Spaces 11.2 The Space L 2(Jl)
339
9. A trigonometric polynomial is a finite sum of the form = a0 + fonn p(x) = L~=l [ancosnx + bnsinnx] for each x E R where the an's and bn's are complex = cosx + numbers. As is well known from calculus, for each x E R we have e ix = isinx. Show this implies that p(x) can be written as p(x) = L~~NCneinx for some = L~=-NCneinx . .. N. suitable Ccn's = -N ... n 's E C for n =
10. For each integer n (not necessarily positive) put un(x) = = e inx for each x E R. Then for each n, Un Un E E L 2(n. 2 (n. Show that for each pair of integers m, n that n and 0 otherwise. Consequently, {un IInn E Z} is an (um,u n) = 1 if m = n orthonormal orthononnal set in L 2(n. II. Show that for each positive integer n it is possible to choose Cn such that if 11. cosxt for each x E R then the sequence of functions {Pn} = 2-nn cn(1 + cosx)n converges uniformly unifonnly on [-n, -e]u[E, -e]u[e, n) for each eE > O.
Pn(x)
12. Show that {Pn} is a sequence of trigonometric polynomials such that (1) Pn(x) ~ 0 for each x E R. (2) 1/2nJ~1tPn(x)dx l/2nJ~llPn(x)dx = 1. -e]u[e, n)} then lim (3) If 8 bn(E) SUp {Pn(x) Ix E [-n, -E]U[E, limnbn(e) n8n(e) = 0 n(e) = SUp for each e> E > O.
°
°
13. Let C(n denote the class of all continuous, complex valued, 2n-periodic E C(n we can functions on R with norm nonn 1/100 III = = Supx I/(x) I. For each I E associate a sequence of functions {P n} defined by
=
l/2nfrr-lt,lt)l(Y - x)Pn(x)dx. Pn(y) = 1/21tf -lt,lt)/(Y -x)Pn(x)dx.
Show that {P n } is a sequence of trigonometric polynomials such that for each positive integer n,
unifonnly continuous on [-1t, [-n, n], if eE > 0, there 14. Let I E C(n. Since I is uniformly exists a 8b > 0 such that Vex) l(y) I < e whenever Ix - y I < 8. I/(x) - 1(Y) b. By (2) of Exercise 12, it follows that l(y) = = 1/2nf 1(Y)]Pn(x)dx. Pn(y) - !(Y) l/21tfrr-lt,lt)[f(Y - x) - !(Y)]Pn(x)dx.
°
limn-'>= IPn - II Show that Umn~oo 1100= = 0, and hence for each IE C(n and Ee > 0 there exists a trigonometric polynomial P such that I/(x) - P(x) II < eE for each x E R. In 15. Show that {un In a basis for L 2l{n. (D.
E
inx for each x Un is defined by un(x) un{x) = = e inx Z} where Un
E
R is
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11. Spaces of Functions
11.3 The Spaces U LP (11) (Jl) and Banach Spaces Hilbert space norms are fairly specialized, being based on the concept of an inner product. This is what accounts for there being so few of them, essentially one for each cardinal. The LP -spaces are a special case of a more general class of spaces called the Banach spaces. These spaces derive their name from Stefan Banach who published a series of papers between 1922 and 1932 in several journals that culminated in the now famous Theorie des Operations lineaires (Monografje Matematyczne, Volume 1, Warsaw, 1932). In the special (11) is both a Hilbert space and a Banach space. As we shall see case p = 2, L 2 (Jl) from the definition of Banach spaces, this is because all Hilbert spaces are Banach spaces.
(i.e., A normed linear space X is a vector space over the complex field C (Le., an abelian group in which multiplication of group members by complex numbers, called scalar multiplication, is defined that satisfies the distributive laws) in which a non-negative real number IIxl x I (called the norm of x) is associated with each x E X that satisfies the following properties: Bl. Ixl = 0 if and only if x = o. O. B2. Ix + yl ~ Ixl + Iyl X. yl :0:; Iyl for each pair x,y E EX. B3.laxl = lallxl foreachxE XandaE C. Here, IIa II denotes the modulus of the complex number a. A metric can be defined in a normed linear space in the following way. Define d:X x X ~ [0,00) by d(x,y) = = Ix - y I for each pair x, y E X. That d is actually a metric is left as an exercise. If the metric space (X, d) is complete, X is said to be a Banach space. The simplest Banach space is merely C itself, normed by absolute value, i.e., IIx II is simply the absolute value (modulus) of x for each x E C. One can also discuss real Banach spaces by restricting the field of scalars to R. The topology induced on X by d is called the norm topology and the set xl Ixl :0:;~ I} is the closed unit sphere in X. A mapping T of a normed linear space X into a normed linear space Y is said to be a linear transformation if T(x + y) = T(x) + T(y) and T(ax) = aT(x) for each pair x,y E X and a E C (linear transformations are sometimes called vector space X' homomorphisms - the definition does not depend on the norm, only on the vector spaces X and Y). Linear transformations are also commonly referred to as linear operators. In the special case the space Y is the Banach space C, Tis referred to as a linear functional. The kernel of a linear operator is the set of Ker(1) = {x E all elements in X that get mapped onto the zero element of Y, Le., Ker(n X IT(x) == O}. The proof of the following proposition is left as an exercise. S(O,I) = = {x E
PROPOSITION 11.11 The kernel of a linear operator T:X ~ Y from the linear space X into a linear space Y is a linear subspace ofX.
11.3 The Spaces LP(~) LP(fl.) and Banach Spaces
341
A linear transfonnation T is said to be bounded if there exists a real number a such that IT(x) II ~ a IIx II for each x E X. The smallest a with this property is called the norm of T and is denoted by ITI. It is easily seen that ITI = sup { IT(x)I/lxl # O}. Since I T(ax) I = I aT(x) I = IT(x)l/lxl I x E X and x ::f:. IIa IIII T(x) II for each x E X and a E C, we could restrict ourselves to unit vectors I) in the definition of the nonn of T. In this case, we (i.e., x E X such that IIx II = 1) would have ITI = sup{ IT(x)l/lxl I x E X and Ixl = I}. A bounded linear transfonnation T maps the closed unit sphere in X X into the closed sphere 5(0,ITI)= {YE lYE ylly-Ol ~ ITI}. Toseethis,letxE To see this,letx E S(O,I)inX. S(O,l)inX. Then Ixl ~ S(O,ITI)= 1. Since IT(x) I ~ ITllxl, we have IT(x)l/lxl ~ ITI. Since Ixl ~ 1I this implies IIT(x) II ~ IITI TI ). TI which in turn implies IIT(x) - 0 II ~ IITI so T(x) E S(O, 5(0, IITI). THEOREM 11.13 J 1.13 Let T:X ~ Y be a linear operator from a normed linear space X into a normed linear space Y. Then the following statements are equivalent: ( 1) T is bounded. (1) uniformly continuous. (2) T is uniformLy x. (3) T is continuous at some point of X. Proof: If T is bounded, IIT(x) - T(y) II = IIT(x - y) II ~ IITil x - y II for each pair x,y EX. Thenif£>Oand Ix-yl O. To show (3) ~ (I), Eo Then Then there exists a 0 > 0 such that IIx - z I < 0 implies IT(x) - T(z) II < £.Then IIxx II ~ 0 implies IIz + x - xxlI ~ 0 which in tum implies IIT(z + x) - T(z) II < £.Eo But IIT(z + x) - T(z) I = IIT(x) I so IIx I < 0 which implies IT(x) II < £. Hence IT(x) 1/0 I/o £/0. > Ix I/£ which implies IIT(x) III IIx I > 0/£ which in tum implies IIT(x) II Ix II < £/b. Therefore, Ix II < 0 implies IIT(x) III IIx I O. Since {Tn} is Cauchy, there exists a positive integer N such that m, n > N implies II T Tm E/lxo I. I. Now ITm(xo) m - Tn I < e/lxo T m - Tn II and therefore conver- Tn(xo) II ~ IITm II Xo II < E£ so {Tn(xo)} {Tn(xO)} is Cauchy in Y Yand
342
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ges to some xo' E Y. Define T:X ~ Y by T(x) = x~ x' for each x converges to T(x) for each x E X.
E
X. Then {Tn(x)} {Tn(x) I
To show T E L(X, Y), we must show T is linear and bounded. To show Tis linear, let x, y E X and a E C. Then T(x + y) = = (x + y)' and T(ax) = (ax)'. Let £E (x) I converges to T(x) for each x E X, there is a positive integer > O. Since {Tn (x)} Tn(x) E/2 and IITn(y) - y'l < £/2. E/2. Now N such that n > N implies IITn (x) - x'i x'i < £/2 <E. IITn(x+y)-(x'+y')l Tn (x+y) - (x' + y') I = IITn(x)-x'+Tn(y)-y'l::; Tn (x) - x' + Tn(y) - y'l ~ IITn(x)-x'l Tn (x) - x'i + IITn(y)-y'1 Tn(y) - y'l < £. {Tn(x y) I converges to Therefore, n > N implies I Tn(x Tn (x + y) - (x' + y') I < E £ so {Tn (x + y)} y') so T(x + y) = T(x) + T(y). Also, there exists a (x' + y'). Hence (x + y)' = (x' + YJ positive integer M such that n > M implies ITn (x) - x'i < £/ E/lal I a I which in tum E. Therefore, {Tn(ax)} {Tn(ax) I converges to ax' so (ax)' = = ax' implies ITn(ax) - ax' I < £. which implies T(ax) = aT(x). Consequently, T is linear. To show that T is bounded, let E £ > O. Since {Tn} {Tn I is Cauchy, there exists a positive integer M such that m, n > M implies I Tm E. Pick m> m > M. Then M M Tm - Tn I < £. ITml ITI ITI and IT'(w TI for each n. We conclude that IT'I 111 = ITI. •-
The complex numbers C, nonned normed by their absolute value fonn form a Banach space. L(X, C) is called the dual space of X and is denoted by X*. L(X, C) is a Banach space by Theorem 11.14. The interplay between X and its dual space X* is the basis of much of that field of mathematics known as functional x* analysis. To explore this interplay, one needs the Hahn-Banach Theorem. The Hahn-Banach Theorem essentially says that if Y = C, then we can drop the assumption that Z be dense in X. Linear transfonnations transformations from Z into C can then be extended to X in such a way that Theorem 11.15 still holds. By a real linear functional, we mean a linear operator from a real vector space (vector space over the real field). Letfbe a complex linear functional on a linear space X. Then for each x E X, f(x) = = u(x) + iv(x) for some real valued functions u and v on X. Since X is a vector space over the complex field, it is clearly a vector space over the real field as well. It is easily seen that the linearity of f implies the linearity of u and v, i.e., Le., u and v are real linear functionals. PROPOSITION 11.12 If X is a linear space andfis and f is a linear functional on X X then: (1) If thenf(x) = u(x) - iu(ix)for each x EX. E X. Ifuu is the real part off offthenj(x) X andf(x) = u(x) - iu(x)for each (2) Ifu is a real linear functional on X x E X, then f is a (complex) linear functional on X. (3)1fX (3) If X is a normed linear space andf(x) = u(x) - iu(ix)for each x E X, then IfI If I = = Iu I.
= a + i~, then the real part of iy Proof: If a, ~ E R and 'yY = i'Y is -~. Therefore, y =f(x). To show (2), it is Re(y) - iRe(iy) for each y E C. Then (1) follows with Y= clear that I(x f(x + Y) = f(x) + f(y) and that f(ax) f(a.x) = af(x) for each a E R. But we = must also show this second equation for a E C. It will suffice to show it for aex = i. For this note thatf(ix) that f(ix) = u(ix) - iu(-x) iu( -x) = u(ix) + iu(x) = if(x). To show (3) note that Iu(x) I ~ If(x) I for each x E X X which implies Iu I ~ If I. Let x E x. Put ~ = f(x)/ If(x) I. Then I ~ I = 1 and ~ If(x) I = f(x). Put aex = f(x)/lf(x) = ~-l. Then Iex alI == 1 and af(x) = = If(x) I. Hence If(x) I = =f(ax) f( ax) which implies ~-1.
11. Spaces of Functions
344
f(ax) isreal sof(ax) = u(ax)-S: lullaxl. Then If II =sup{lf(x)lllxl =1}~ =1}-S: f(ax)isrealsof(ax)=u(ax)~ lui. Iu I. Iu I. Therefore, If II = lul.One of the most important theorems in the theory of Banach spaces is the Hahn-Banach Theorem. It allows us to extend bounded linear functionals on subspaces of a normed linear space in such a way that the norm is preserved. 1I.16 (S. Banach Banach, H Hahn, 1932) If ffY THEOREM 11.16 Y is a subspace ofa of a normed linear space X and iff is a bounded linear functional on Y, thenf then f can be extended to a bounded linear functional F on X such that IF I == If I. g
Proof: We first prove the theorem assuming the field of scalars to be real, i.e., I.e., we assume X is a real nonned normed linear space and f is a real bounded linear functional on Y eX. If If II = 0 then clearly, the desired extension is F = O. If of. 1, If I > 0, we may assume, without loss of generality, that If I = 1 since if If I "# there exists an a E R with I allfl = 1 which implies I afl = 1. We can then prove the theorem for af and simply divide the extension F by IIa I. I. Assuming If II = 1, pick z E X - Y and let N be the subspace of X spanned by z and Y (i.e., N = {x + azlx az I x E Yand Y and a E R}). Define fN:N -7 = f(x) + aA for DefinefN:N ~ R by fN(X + az) =f(x) any fixed A E R. It is left as an exercise to show that fN is a linear functional on N such that fN restricted to Y is f. We next show that it is possible to pick A such that IIfN I, that Ifl If I -s: fN II = 1. For this, first note that by the definition of II I, ~ IIfN fN I. Also,
IfNI
= sup{
'{(X)+1 x+az
1
IXE YanduE R)
-s: Ix + azl for each x E Y and a E R, then IfNI ~ -s: 1. so that if If(x) + al..I aAI ~ al..II + IIxx + az azlI for each x E Yand Hence If(x) + aA Y and a E R which implies IfN I = 1. Therefore, it suffices to show that A can be chosen such that If(x) + aA II -s: ~ Ix+az I for each x E YYand and a E R. This can be shown if we choose Asuch that If (x)+aA II < Ia I --s:
II Haz x+az I ~
Ia I
h
Y
d
R
for each x E Y and lor eac an a E R..
But this is equivalent to showing that we can find a A such that If(x/a) + A AII ~ -s: ~ Ix/a + zl which in tum is equivalent to finding aWE Y such that If(w) - AI -s: )'(w) = few) IWw - z II in view of the substitution w = -x/a. For each Ww E Y put ')'(w) P(w) =f(w) + Iw-zl. If),(w)-S:A-s:P(w),then If(w)-AI-s: Iw-zl and B(w) If')'(w)~A~P(w),then I!(W)-AI ~ Iw-zl. Hence IfN I = 1 if )l(w) )'(w) ~ -s: A ~ -s: ~(w) pew) for each ·w W E Y" Y
I(w) be the interval [')'(w) [)'(w),, pew)] for each w E To show this, let lew) E Y. Then IfN I = 1 if n{ lew) I(w) Iw WEE Y} y} 7: of. 0, or equivalently if ')'(w) )'(w) ~ P(v) for each pair W,v I!N Now)'(w)-P(w)=f(w-v)-Iw-zl-Iz-vl. Since!(w-v)~ Sincef(w-v)~ If(w-v)1 If(w-v)l:5 E Y. Now')'(w)-P(w)=!(w-v)-Iw-zl-Iz-vl. $ 1!llw-vl If I Iw - v I = IIw-vl w - v I ~ Iw-zl Iw - z I + Iz-vl,')'(w)-p(v)~Oso')'(w)~p(v)foreach Iz - v I, )'(w) - P(v) ~ 0 so )'(w) ~ ~(v) for each
11.3 The Spaces LP(Jl) LP(Il) and Banach Spaces
345
pairw,vE pair w,v E Y. Hencen{I(w)lwE Hence n{I(w) IWE Y}#0. Y} # 0. PickAE Pick A E n{I(w)lwE n{I(w) IWE Y}. Then)'(w) Then y(w) : EE X*. Then cI>Ccu+dv)(q» (cu+dv)((v) (q». Hence cI>Ccu+dv) d(v) = ccI>u(q» cu(v(q» dv(u [cu + dcI>v] dv](u cu + dv = = cn(u) + dn(v). But then n(cu + dv) = cn(u) + dn(v) so n is linear. n I= n(x) II Ix II x EE To show n is bounded (continuous) recall that IQ =sup { IQ(x) X and Ixl y EE X with Iyl Ix I = I}. Let Y Iy I = 1. Then IIn(y) II = Iy(q»I/Iq>IIq> Let 'V EE x* = 1. Then sup {y(1 11 = = I}. Let", X* such that I'VI 1",1 = IIcI>y('V) 'V II /",) II = II'V(y) ",(Y) II ~ II", II y II = IIy I. I. Therefore, IIcI>y y II ~ IIy II = 1 which implies n(x) II ~ 1 for each x with Ixl Ix I = 1. Hencen Hence n is abounded a bounded (continuous) linear II(l(x) operator and IIn II ~ 1. Now if x E E X with IIx II = 0, then by the discussion above, IQ(x) O(x) I = O. If x X X with IIx II > 0 then by Corollary 11.5 there exists a q> E EX X with IIq> II = 1 and (x) x((x) = IIxl. x I. But then I x( 0 such that IIxl':s; £Ixl for each x EX. E x. elxl eachx I
Proof: Let X' denote the Banach space X with the nonn I II. Then the identity --7 X is clearly linear. Since Ix I :S; 81Ix I for each x E X, i is a ~ B operator i:X' ~ bounded linear operator. Since i is a one-to-one, onto, continuous linear i-I:x --7 ~ X' is also a operator it is a homeomorphism by Corollary 11.6 and i-I:X :S; E e Ix I bounded linear operator. Therefore, there exists an £e > 0 such that Ix II ~ for each x E X. •I.
I
I
ylf(x) = y} of an operator LEMMA 11.5 The graph r = {(x, y) E X x Ylf(x) f:X --7 ~ Y from a Banach space X into a Banach space Y is a closed subset of X x X such Y if and only if for any sequence {x lln }} Cc X that converges to some x E X havef(x) that {f(x nll )) )} converges to y E Y, we must have f(x) = y.
Proof: Assume r is closed and suppose {x {xlll n } is a sequence in X converging to a point x. Further, suppose {f(x n E Y. Then {x {xlll ll ))}I converges to y E n } is Cauchy in X and {f(x n {(XIl,f(X ll )} is Cauchy in Y. Hence the product sequence {(x n, f(x nIl ))} cC r is Cauchy in X x Y by Exercise 3 of Section 5.3. Since X and Yare complete, so is ))} converges to a point X x Y (Exercise 4, Section 5.3). Therefore, {(x n ll ,, f(x n ll »} (a,b) E r since r is closed. But then x = = a and y == b (Exercise 2, Section 5.3). 7: y. Then (x, y) is not in r which implies there exists open sets U Suppose f(x) "# c X and V c Y such that (x, y) E U X V and U x Vnr = 0. But {(xn,f(x {(XIl,f(X n c r Il ))} C converges to (x, y) which is a contradiction. Hence f(x) = y. Conversely, assume that for each sequence {x lln }} cC X that converges to Y, we have f(x) = y. some x E X such that {f(x n ) I converges to some y E ll )} Suppose r is not closed. Then there exists a point (x, y) E X X Y and a sequence {(x {(XIl,f(X {(XIl,f(X n Il ))} converges to (x, y) but (x, y) is not in r. n, f(x Iln))} c r such that {(xn,f(x Since {(x n , f(x ))} to (x, y), {x nll I} must converge to x and {f(x nll ))}} must , » I converges ll nll converge to y by Exercise 2 of Section 5.3. But then f(x) = y which implies (x,y) E r which is a contradiction. Hence r is closed. •-
THEOREM 11.21 (Closed Graph Theorem) The graph r of a linear operator f:X ~ --7 Y from a Banach space X X into a Banach space Y is closed if and iffis only iff is continuous. Proof: Assume r is closed. Define Ix I = = Ix I + If(x) I for each x E X. To show X first note that if x = 0 then Ix I = = 0 since the linearity of that I I is a nonn in X I = 0 then Ix I = -1/(x) -If(x) I which f implies f(O) = O. On the other hand, if Ix I' If(x) I since Ix I and I/(x) If(x) I are both nonnegative. But Butthen = I/(x) then x implies Ix I = 0 = I
I
I
I
351
11.3 The Spaces LP(fl) U(f.!) and Banach Spaces
0 since II II is a nonn on X. Therefore, II I' I' satisfies B 1. That I I' satisfies B2 =a and B3 is an easy exercise. Consequently, X is a nonned linear space with I'. Let X' denote the linear space X X equipped with the nonn I I'. I'. respect to II I'. To show that X' is complete let {x /x nn } C X be a Cauchy sequence with respect to II I'. I'. Then for a given ££ > a 0 there exists a positive integer N such that f(x n ) II < £. But then IX for m, n > N, IX Xmm - XXnn I + If(x If(xm) Xmm - X Xnn II < ££ and I!(x If(xm) m) - f(xn) m) !(x f(xn) I Xnn I} is Cauchy in X X and {!(x (f(xn)} n ) I < £ for each pair m, n > N. Hence {x n )} is X and {!(x I!(x nn)} ) I converges Cauchy in Y which implies {x n n }} converges to some x E X 11.5, f(x) = = y. to some y EE Y. Yo By Lemma 11.5,f(x)
{xn} I', let ££ > O. Then there To show that {x n } converges to x with respect to I I', Xnn -- x I < £/2 and I!(x If(xn) exists a positive integer N such that for each n > N, I X n) !(X) f(x) I < £/2 + £/2 = £. ) f(x) I < £/2. Therefore, IX IXnn -- xl' = IX IXnn -- xl + I!(x If(xn) I n Hence {xn} {x n } converges to x with respect to I I'. Therefore, XI' X' is complete. I' for each x EE X, by Corollary 11.7, there exists a () > a 0 such Since Ix II $~ 11 x I' Ixl'$()lxl foreachxE X. But then If(x) I $Ixi If(x) I = Ixl'$()lxl ~ Ixl + I!(x) Ixl'~()lxl that Ixl'~()lxl which implies! is a bounded linear operator and therefore continuous. impliesfis Conversely, if !f is continuous, then if {x {xn} X that n } is any sequence in X· {f(xn)} converges to some x E X, {f(x n )} converges to f(x) E Y by Proposition 1.12. But then by Lemma 11.5, r is closed. •-
LEMMA JJ.6 (Uniform Boundedness Principle) I! If X is a metric (pseudo-metric) space and F c C(X) such that for each x E X there exists a real If(x) I $~ bxx for eachf E F, then there exists a nonempty open set number bxx with I!(x) U c X and a real number b such that If(x) I $~ b for eachf E F and x E U.
F,r
Proof: For each positive integer nnand and each f E F, f- 1l [-n, nl n] is a closed subset of eachfE X since f is continuous. Therefore, the intersection En = = "JEFr) (If EFf- 1 [-n, n] nl is a X. By hypothesis, for each x E X there exists a positive integer closed subset of x. nnxx such that If(x) II $~ nnxx for each f E F. Therefore, x E En Enxx '• But then X = uEnn•. Em, is By Theorem 11.19, X is of the second category, so one of the En's, say Em' not nowhere dense in X. By Proposition 11.13, the open set U = Int(CI(E Int(CI(Em» i= m» i:0. Since Em is closed, U == Int(E ). Consequently, If(x) I ~ m for each f E F Int(Em). $ m and x E U. Put b = = m. Then b is the bound we were seeking. THEOREM 11.22 J J.22 (Uniform Boundedness Theorem) If F is a family of continuous linear operatorsfrom a Banach space X into a Banach space Y such that for each x E X there exists a real number bxx with If(x) I ~ f E F, $ b xx for each eachf $ b for eachf E F. then there exists a real number b such that If I ~
Proof: For each g EE F define a function ng:x ng:x --t R by ng(x) ng (x) = = Ig(x) I for each x E X {ng IIg EE -:}. -:}. Then G c C(X) and Ing(x) II $~ bxx for each x E X E X. Put G = {Og ngg EE G. By the prevIOus and n prevIous lemma, there exists a nonempty open set U eX c X
352
11. Spaces of Functions
0 > 0 such that IIOg(x) Q.g(x) II ~ 8 0 for each Og Q.g E G and every x E U. Let U u E U. and a 8 CI(S(u, E»~ e)) Cc U. Since U is open there exists an £e > 0 such that Cl(S(u, u.
Let b = = (8 (0 + b bu)/e. u )/£. To show If I ~ b for each f E F let f E F and x E X with Ixl = 1. Then I(u I (ex + u) - ui = lui lexl = elxl £Ixl = £. Hence (ex + u) E U. Therefore, If(x) II = If(eIf(£-11u) £-11 If(U) If(u + u - u) I ~ c £-11 (If(ex+u) II ex) II = eIf(ex) I = £-1 e- i If(ex 1 I) ~ £-1 e-- (8 (0 + bu Illx = I} ~ b for + If(u) I) u )) = b. Consequently, If II = sup{ If(x) Illx eachfE F.· COROLLARY JJ.8 J J.8 (Banach-Steinhaus Theorem) If {fn} is a sequence of continuous linear operators from a Banach space X into a normed linear space Y that converges pointwise to an operator f:X -7 --7 Y, then f is also a continuous linear operator.
Proof: Since limits are unique in Y, it is easily seen that f is a linear operator. I} < Let x E X. Since {fn(x)} converges to f(x), it follows that b x = sup { Ifn(x) I} 00. By the Unifonn Uniform Boundedness Theorem, there exists abE R such that Ifn iI ~ If(x) I = limn Ifn(x) I ~ blxl since the b for each n. Hence, for each x E X, I/(x) distance function d(x, y) = = Ix - y I is continuous in Y. But then If is bounded and therefore continuous. •
EXERCISES 1. Prove Proposition 11.11. 2. Prove Lemma 11.3.
3. Let II = [0, 1] and let f E £P LP (I) for p > 1. Show that there exists agE L q (I), If IIp = = IIg IIq and f(
.00
0
fonn Va = = {{f a) E V} IIv' V Hence coverings of the form { {f E F If( If(a) TtA uniformity on F. and V E V fonn a subbase for the TtA
E
•
V l where a E A A V}
To prove (2) first assume {fp} ~ I is Cauchy in F with respect to TtA' Tt;>.. Then for each covering of the fonn Va = {{f E F If( a) E V} V I IV E V} V I there exists a residual set R such that {fp R} C {fE {f~ IB I ~ E RI (fE Flf(a) E V} VI for some V E V. But then {fp(a) IB E RI R} c V so {fp(a)} {f~(a)l~ {f~(a)1 is Cauchy in (Y, v). Hence {fp(x)} {f~(x)1 is Cauchy in (Y, v) for each x E A. Conversely, assume {fp(x)} (f~(x) I is Cauchy in TtA' For some V E V (Y,v) for each xx E A. Let Va be a subbasic member of TtA' (f ~ (a) I ~. ~ E RIc (a)l is there exists a residual set R such that {f R} c V since {f {I p~ (a)} v)o But then {f131 (f~ I ~ E R} RI Cc {fE (f E Flf(a) E V} VI so tl13} tf~ i is Cauchy Cauchy in (Y, v). Tt;>. . with respect to 1tA
rn. III I (En). (En)· Then each En has a partition {E {Ennmm } such that 'f,m rn' Since {E Lnrn ~ {Ennmm In, m are positive integers} is a partition of E, we have 'f,nrn 'f,n.m {'f,nrnn I rrnn < IIII (En)for n) = 'f,n III I (En) Ln,m IIIl(E Jl(E nm ) I ~ IIIlI f.! I (E). Then sup {Lnr JlII (En) for each n} L n IJ.l1 sUp{'f,nrnn I rrnn < IJlI IIlI (En) for each n} ~ 'f,n,m and sup{Lnr Ln,m IIIl(E f.!(E nm ) I so 'f,n L n IIIlI JlI (En) ~ IIIlI JlI (E). To prove the opposite inequality, let {Am} {Am) be another partition of E. Then for a fixed n, {EnnA m }) is a partition of En and for a fixed m, {EnnA m } is a 'f,m 'f,m partItIOn of Am. partition Therefore, L Jl(A m) I L m I'f,nll(EnnA LnJl(EnnA m) I ~ m IIIl(Am) LmLn IIIl(EnnA Jl(EnnA m) I ~ 'f,n L n IIII f.! I (En). Since this holds for each partition {Am} 'f,m'f,n {Am) of E, III I (En)· (En). Consequently, 'f,n we have IIII JlII (E) ~ 'f,n L n IJ.l1 L n IIII JlII (En) == IIIIJlII (E) so IlJl is countably additive. That IIII countablyadditive. JlII satisfies the other properties of a positive measure are easily demonstrated. That IIII JlII is the smallest positive measure that J..l. follows from the fact that if A is another positive measure that dominates J.l dominates j..I., = 'f,nA(En) j..I.(En) {En) of E, so Jl, then A(E) = LnA(En) ~ 'f,n L n I Jl(E n) I for each partition {En} j..I.l (E) for each E A(E) ~ I1JlI E E M.·
LEMMA 12.1 If C 1 ESC I I ~ 1/6Li=1 1/6'f,;=1 II C that II'f,1 LiESCi! CiI I.
. . . Cn E
C there exists a subset of {1 C, {I ... n} such 9
Proof: Put r = 'f,i=1 Li=1 I1Ci CI I. C is the union of four "diagonal quadrants" quadrants" bounded by the lines y = x and y = -x. Let Q Q JJ. denote the one defined by II y I1 ~ x for each z= = (x, y) in Q I.1. Now one of these quadrants, say Q, has the property that the CI I 's for which Ci E Q Q is at least r/4. Assume first that Q Q = Q Q 1. I. For sum of the ICi 112 CEQ 1, Re(c) ~ I C 1- 1/2 . If S is the set of all i such that Ci Ci E Q Q 1, I, it then follows that:
Q =Q Q I. Q is one of the other quadrants, a similar This proves the lemma for Q 1. If Q
12. Uniform Differentiation
372
112 with the argument will work by replacing the fonnula ~ II cC 11-112 formula Re(c) ::::0: formula for that quadrant.· quadrant.appropriate fonnula
IfJl is a compLex complex measure on X JlI (X) < 00. PROPOSITION 12.2 If~ X then II~I (x).
Proof: Assume IIJlI ~I (X) = 00. Then for each r> 0 there exists a partition {X {Xn} n } of X such that L Lnn I~(Xn) Jl(X n) I > r. Now f..l(X) Jl(X) = c for some c E C so IJl(X) IJl(X) II < 00. Put s = L;;'=l I Jl(E n) I > s = 6( IJl(X) I + 1). Then there exists a partition {En} of X with L~=l Jl(E i ) for i = 1 ... m for some positive integer m. By Lemma 12.1, if we put Ci = = J.l(EJ LnESJl(En) ::::0: l/6L;;'=1 I. Put H there exists a subset S of {I ... m} with IILnesJ.l(E 1/6L~=1 IIJl(E n) I. n) II ~ = unesE = X-H. unesEnn and K = X - H. Then HnK = 0, X = HuK and IIJl(H) II = Jl(unesEn) = IILnEsJl(E LnesJl(En) = IIJ.l(X) Jl(X) II + 1 ~ ::::0: 1, so IIJl(unesE ~ 1/6L;;'=IIJl(E 1/6L~=1 IJ.l(E n))1I > s/6 = n) II ::::0: n) II = IJl(H) I ::::0:~ 1. l. (X).
(X).
For any pair a, b E C, it is easily seen that la Ia - bl bI ~ ::::0: Ihl Ib I -- lal. Ia I. Hence
Jl(K) I = IJ.l(X) Jl(X) - Jl(H) I ::::0: Jl(X) I > s/6 - IJ.l(X) Jl(X) I = 1. Since IJlI IJ.l(K) ~ IJl(H) I - IJ.l(X) ~ I(X) == IIJlI(H) J.l1 (H) + IIJlI(K) J.l1 (K) by Proposition 12.1, we have IIJlI(H) J.l1 (H) = or IIJlI(K) JlI (K) = = 00 (X)
00. (X).
Consequently, X is the union of two disjoint sets H and K such that for one of the sets, say H, IJlI (H) = = 00 and for the other, IJl(K) I ~::::0: 1. Put HI H1 = = H and K 1I = K. A similar argument can be used to show that H 1I can be partitioned into two disjoint sets H 2 and K 2 such that IIJlI JlI (H 2) = = 00 and IJl(K 2) I ::::0:~ 1. Continuing {Kn} inductively, we obtain a sequence {K n } of subsets of X such that KmnKnn = 0 if of. n and I Jl(K n) I ::::0: = uKn. Then Jl(Y) == Ln~(Kn). LnJl(Kn). But m 7= ~ 1 for each n. Put Y = Jl(K n) I ~ ::::0: 1 for each n, which is a LnJl(Kn) cannot converge absolutely since IJl(Kn) contradiction. Therefore, IIJlI (X) < 00. •Proposition 2.12 shows that complex measures are bounded in the sense that their range lies in a sphere of finite radius. To see this let Jl ~ be a complex JlI (E) :s; measure on a a-algebra M and E E M. Then IIJl(E) ~(E) II :s; ~ IIJ.l1 ~ IIJlI (X). This fact has many interesting and useful consequences, among which is the the CE following: if Jl ~ and A are complex measures on the same a-algebra M and if C C, we define J.l [J.1 + Al(E) A](E) = Jl + A and cJl CJl on M as follows: for each E E M put [Jl = Jl(E) + A(E) and [cJl](E) [cJll(E) = cJl(E). Clearly Jl + A and cJl are also complex measures on M. Therefore, the collection of complex measures on M is a vector space. In fact, if we put IJlI = = IJlI J.1/ (X), it is easily verified that the collection of normed linear space. complex measures on M is a nonned Now assume that Jl is a positive measure on M and A is either a complex measure or a positive measure on M. A is said to be absolutely continuous with = 0 for each E E M Jl(E) = = o. O. respect to Jl, denoted by A « Jl, if A(E) = M for which ~(E) If there exists aYE M such that Jl(E) = Jl(EnY) Jl(En Y) for each E EM, E M, we say J.l Jl is concentrated on Y. In other words, Jl is concentrated on Y if and only if J.l(E) Jl(E) = o whenever En YY == 0. If JlJ.1 and Af... are any measures on M such that there exists A concentrated on H and J.l disjoint sets H and K with f... Jl concentrated on K, K. then J.l Jl singular. denoted J.l.l Jl.l A. f.... and f... A are said to be orthogonal or to be mutually singular,
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PROPOSITION 12.3 If A, J.l a are measures on a a-aLgebra i2.3 ifA. f.! and 0' a-algebra M and If 0' a is positive, if positive. then: (J) Iff.! is concentrated on E E M, M. then so is 1IJ.l1. f.! I. (1) IfJI (2)/ff.!.lAthen (2) lf~ 1- Athen 1f.!ll.IAI. IJ.l1 1. I AI. (3)lff.!«athen 1f.!1 «a. «0'. (3)IfJ.l«athen 1J.l1 .1 a, 0'. then J.l f.! + A 1.1 0'. (4) iff.!.l IfJ.l1- 0' a and A 1. a. Iff.! « «0' 0'. then f.! (5) IfJ.l a and A« a, J.l + A« 0'. a. Iff.!« .1 a, 0'. then J.l1. f.! .1 A. (6) IfJ.l « 0'a and A 1. Iff.! « «0' f.! .1 0'. = O. (7) IfJ.l a and J.l1. a, then f.! J.l = The proof of this proposition is left as an exercise.
EXERCISES 1. Prove Proposition 12.3. f.! be a real valued measure on a a-algebra M and define Jl+(E) f.!+(E) = = 2. Let Jl 1/2[ IJ.l1 JlI (E) - J.l(E)] If.!1 (E) + J.l(E)] f.!(E)] and J.l-(E) WeE) = 1/2[ l/2[ I f.!1 f.!(E)] for each E in M. Show that both m+ and J.lare bounded positive measures on M. ware 3. Show that f.! J.l = = f.!+ J.l+ - J.l= J.l+ J.l+ - J.lW and that IJ.l1 If.! 1 = f.!+ + J.l-. W· Expressing J.l f.! as f.!+ W is known as the Jordan decomposition of f.!. J.l. Also, fJ,+ f.!+ and JlW are said to be the f.! respectively. positive and negative variations of fJ, J.l is positive and A is 4. Let A and J.l f.! be measures on a a-algebra M such that f.! complex. Show that A is absolutely continuous with respect to fJ, f.! if and only if 8 > 0 such that I1A(E) I1 < £ for each E E M with fJ,(E) f.!(E) for each £ > 0 there exists a b
. Since {XAJ} {xAnf} converges to If in LP (J.l) and A is continuous, limnA(XAnf). U(~) {A(XAnf>} converges to A(f). Therefore, A(f) = Ixfgd~. Jx!gdJ.l. That IIA II = IIg IIq {A(XAnf)} = g and IIh n IIq :5; follows from the fact that limnh n = ~ IIA II for each n. •-
EXERCISES 1. Let A A. denote Lebesgue measure on (0, 1) and let Jl ~ be the counting measure on the a-algebra of Lebesgue measurable sets in (0, 1). Show that A is absolutely continuous with respect to Jl ~ but there does not exist an If ELI ELI (Jl) (~)
380
12. Uniform Differentiation
= fEfd~ such that A(E) = fEfdJ.! for each Lebesgue measurable set E. In other words, the ~ is not a-finite. Radon-Nikodym derivative may not exist if Jl ~ is not a-finite, 2. Show that if 1I < P < 00, then Theorem 12.2 still holds even if J.l (~) is still the dual space of U (~) i.e., if q is an exponent conjugate to p then L q (J.!) LP (Jl) ~ is not a-finite. even if J.l
~ be two measures on M with 1.« ~ such that the Radon-Nikodym 3. Let A, J.l A « J.l derivative g exists. Show that for each E E M and measurable function f:X ~ fEfdA = fdg*. fEfgdJ.l. R with respect to J.l, ~, fddA
4. Show that the Radon-Nikodym Theorem (12.1) can be extended to the case ~ and Aare A are positive a-finite measures. where both J.l
12.3 Decompositions of Measures and Complex Integration We have already encountered a decomposition of a real valued measure J.l, ~, namely, the Jordan decomposition of J.l ~ defined in Exercise 2 and 3 of Section 12.1. This decomposition was helpful in proving the existence of the RadonNikodym derivative in the case where A was a complex measure, absolutely continuous with respect to J.l ~ and J.l ~ was positive and a-finite. In this section we introduce other useful decompositions, called the Lebesgue decomposition and the polar decomposition and also prove an important theorem about the Jordan decomposition called the Hahn Decomposition Theorem. All the major results in this section depend on the existence of the Radon-Nikodym derivative. The Lebesgue decomposition is introduced by means of the following theorem.
THEOREM 12.3 Let J.l ~ be a positive a-finite measure on a a-algebra A be a positive bounded measure or a complex measure on M that is absolutely continuous with respect to J.l. Then there exists a unique to~. J.l, ~ pair of measures Al and ~ on M such that A = Al + ~ and such that Al 1..l~, « -.L~. < < J.l ~ and Al ..1 ~. M M in a space X and let
Proof: We first prove the theorem for the case where J.l ~ and A are positive bounded measures on M. We can use the first part of the proof of the RadonNikodym Theorem to obtain a unique function gEL 2(J.l) 2(~) with g(x) E [0, 1] for each x such that equations (12.2) and (12.3) hold. Then, as in the proof of the = I} Radon-Nikodym Theorem, let H = {x E X I g(x) < I} and K = {x E X Ig(x) = AI and ~ and let Al (E) = A(KnE) and ~(E) = A(HnE) for each E E M. Then Al are both positive measures on M.
fKgd~ which implies JKdA If we take f = XK in (12.3) we get iJK(1 - g)dA = JKgdJl fKdA == JKdJ.! fKd~ so Jl(K) ~(K) = AI is JKdA = O. Therefore, J.!~ is concentrated on H. But clearly Al concentrated on K and HnK = J.!. As shown in the proof of the = 0. Hence Al AI 1..l~.
Radon-Nikodym Theorem, since g is bounded, (12.3) holds if we replace f by
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(l+g + ... + gn)XE for each positive integer n and each E E M, we get (12.4). At each point of K, g(x) = (x) = = 1 so 1 - gn+I gn+l(x) = O. On the other hand, at each point x of H, {gn+l (x)} converges to zero monotonically. Consequently, the left hand side of (12.4) converges to A(HnE) = = ~(E). Also, as shown in the proof of the ~ Radon-Nikodym Theorem, the right side of (12.4) converges to JEfdJl fddJ.! and n ---+ = JEfdJl fddJ.! for each E E M. 00 where f is the Radon-Nikodym derivative, so ~(E) = Therefore, f.!(E) J.!(E) = 0 implies ~(E) ~ (E) = 0 so ~ «J..l. «J.!. It remains to show that Al A) and ~ are unique since clearly Al ...L~. A) and ~ is easily seen. If A~ and A; are measures But the uniqueness of Al A') + A~ A; and such that A~ A') 1.. ..1 Jl J.! and A; J.!, then A') Ai = A~ A; - ~ A~ « f.1, A~ - Al such that A = A'~ are measures such that A'l A~ - Al A) l.. ..1 J.l J.! and A; - ~ «Jl. «J.!. But then it is easily shown that A~ - Al = 0 and A; - ~ = O. J.! is a positive a-finite measure but A Next we prove the case where J.l remains positive and bounded. Then, as shown in the proof of the RadonNikodym Theorem there is a disjoint sequence {En} c M with X = uE uEnn such J.!(E n) < 00 for each n. Since A(X) < 00 we have A(E A(En) ) < 00 for each n, so the that J.l(E n above result holds for each n. Therefore, there exists sequences {A~} and {A~ {A~}} of measures such that for each n, A~ and A~ are positive measures concentrated J.!, A~ «Jl «J.! = A~(EnnA) + A~(EnnA) for each A E M, A~ 1....1 J.l, on En with A(EnnA) = ..1 A~. Put Al A) = LnA~ and ~ = and A~ 1= LnA~. Then for each A E M, Al (A) + ~(A) = LnA~(AnEn) + LnA~(AnEn) = LnA(AnE LnA(AnEn) J.!(A) = 0 then n) = A(A). If fJ-(A) A~(A) = 0 for each nn so ~(A) = LnA~(A) = O. Therefore, ~ «fJ-. «J.!. Since A~ 1.. ..1 J.l J.! for each n and fJJ.! is concentrated on H, then A~ is concentrated on K for each n which implies Al A) is concentrated on K so Al 1.. ..1 Jl. J.!. Finally, since A~ is concentrated on K for each n, A~ is concentrated on H for each n which implies A) 1..~. ..l~. This proves the case where fJJ.! is a positive ~ is concentrated on H so Al a-finite measure but A remains positive and bounded since the uniqueness proof is the same as for the previous case.
To complete the proof, assume A is complex and Jl J.! is positive and a-finite. Then A = vY + ia for some real valued measures yY and a. y+ - y0'. Moreover, v Y = y+ Yand 0' a = = 0'+ a+ - ay+, y-, Y-, 0'+ a+ and a0'- (see Exercises 2 and 3 of Section 12.1) where y+, 0'are all positive bounded measures. Then applying the above result to these measures, we obtain the desired decomposition of A into a unique pair of +~, A) 1..~, ..l~, Al 1..1 Jl J.! and ~ «Jl. «J.!. measures Al and ~ on M such that A = Al + ~,AI The pair Al and ~ in Theorem 12.3 is known as the Lebesgue decomposition of A. Another useful decomposition of a complex measure is the so called polar decomposition. Like the Lebesgue decomposition we introduce it by means of a theorem. J.! is a complex valued measure on a a-algebra M THEOREM 12.4 If J.l in X, X. then there exists a complex measurable function h on X X with Ih(x) hex) I = 1J for each x E X X and such that J.1(E) J.!(E) = = JEhdl J.l1 J.!I for each E E M.
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382
Proof: Clearly J..L1 « IIJ..L11I so there exists a Radon-Nikodym derivative f E Ll(IJ..LI) ofJ..L thatJ..L(E) foreachEE L 1 (I J..11) of J..1 with respect to IIJ..LI J..11 such that J..1(E) = = fddlJ..L1 JEfd IJ..11 for each E E M. We would like to show that If(x) I = 1 for each x E X. Let a E R and put A = = {x E X Ilf(x) II < a). a}. Then IIJ..11 = SUP{Ln SUp{Ln IIJ..L(En) J..1(En) II J..LI (A) = II {En} {En) is a partition of A}. A). Let {An) be some partition of A. Then Ln J..L II ~ Lna IJ..L11I(An) = = {An} L n IJ..L(An) J..1(A n) I = Ln L n Ikfd JAnfd IJ..111 J..L I(A) so IJ..11 J..L I(A) ~ a IJ..11 J..L I(A). If a < 1 then IJ..11 J..L I(A) = = 0 so If(x) I ~ 1 almost a IJ..11
everywhere on X. M with 1J..11(E) IJ..LI(E) > 0 then II[1/1J..L1(E)lfddlJ..L1 Now if E E M [1/1J..11(E)]JEfdlJ..11 II = I[1/ [IIIIJ..11 II J..11 J..L I(E)]J..1(E) (E)]J..L(E) I = IJ..1(E) J..L(E) I1/1 J..L I(E) ~ 1. Then by Exercise 5 of Section 8.7, we have that If(x) II ~ 1 almost everywhere on X. But then If(x) II = 1 almost h(x) = 1 everywhere on X. If we define h:X ~ C by h(x) hex) = f(x) if If(x) II = 1 and hex) 1I (IJ..1I), otherwisethenJ fdlJ..11 =J hdlJ..11 foreachEE M,sohE L I h(x) I= 1 otherwise then fddlJ..L1 = fEhdlJ..L1 for each E EM, so h E L (1J..L1), Ih(x)1 E E for each x E X and J..1(E) JEhd IIJ..L11I for each E E M. J..L(E) = fEhd J..L I are said to be The function h of Theorem 12.4 and the total variation IJ..11
the polar decomposition of J..1. J..L. This is because when we define integration with J..L(E) = fEdJ..L JEdJ..1 respect to a complex measure, which we will do next, we will have J..1(E) for each E E M JEdJ..1 = fEhd JEhd IIJ..11. M so fEdJ..L J..L I. We use the notation dJ..1 dJ..L = hd IIJ..11 J..L I to denote fEdJ..L J..L I for each E E M. It is this notation that gives intuitive meaning JEdJ..1 = fEhd JEhd IIJ..11 to the term tenn polar decomposition. Notice that the polar decomposition is different in nature than the Jordan and Lebesgue decompositions where A = Al ~. We cannot write J..L1 = hi hi J..11 + Az. J..L I meaningfully. It is only meaningful in the sense that dJ..1 hdlIJ..11. dJ..L = hd J..L I. We defined integration with respect to a positive measure in Section 8.6 as aj with measures the supremum of finite sums of products of complex numbers ai LiaiXAj ~ f. That definition of integration was extended of sets E ij such that 0 ~ LjajMi to complex valued functions in Exercise 1 of Section 8.6. Since then we have been able to avoid the notion of integration with respect to a complex valued measure. But we now have a mechanism for defining integration with respect to a complex measure and a need for this idea in the next section. If J..L1 and Aare complex measures then so is J..L1 + A. Using our intuition about what integration with respect to a complex measure should mean and our experience with integration with respect to positive measures we would expect the following identities to hold for each E E M: (12.7)
/-!(E) = fEd/-! J..L(E) = fEdJ..L
(12.8)
feld(/-! + A) = feld/-! + fddA feldA fdd(J..L = fddJ..L
for each complex measurable function fEe oo(X). ~(X). That (12.8) holds for the special case where J..L1 and A are positive measures is an easy exercise (Exercise 1).
By Theorem 12.4, there exists a complex measurable function h with Ih I = 1 such that ).1(E) JEhd II ).11. J..L(E) = fEhd J..L I. Since we want (12.7) to hold, we expect our
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383
definition of integration with respect to a complex measure to yield 1l(E) J.!(E) == 1dll = = JfEEIhd I1111. = 1 (the unit function) then JEfdJ.! fddll = = JEfhd fEihd I1111. JfEEIdJ.! J.! I. If f = J.! I. Using J.! by this formula for motivation, we define the integral of g with respect to 11
(12.9) for each complex measurable function gEe ~(X) (X) and for each E E E M. Let us now see if the definition (12.9) implies (12.7) and (12.8). That (12.9) implies fEhd I1111 J.!(E) = JEhd J.! I so by (12.9), Il(E) J.!(E) = (12.7) is immediate since by Theorem 12.4, 1l(E) JEdJ.!. fEdll. That (12.9) implies (12.8) takes more work. 00
fxXEdll = = To show (12.9) implies (12.8), first note that by (12.9), JxXEdJ.! fXXEhd IJ.l 11111 == JEhd fEhd I1111 A are complex measures and JxXEhd J.! I = = 1l(E). J.!(E). Therefore, if 11 J.! and 'A (j J.! + 'A, A, then for each E E M we have a = 11
fxXEdcr f..l(E) + A(E) fxXEdA fxXEda = = cr(E) a(E) = = 1l(E) 'A(E) = = fxXEdf..l fXXEdll + fxXEd'A so (12.8) holds if f = XE for some E E M. Next suppose f is a nonnegative bounded measurable function. By Theorem 8.7 there exists a sequence {sn} of ~ f for each positive integer n and Sn ~ simple measurable functions with 0 ~ Sn ~ffor Sm if n < m such that {sn(x)} converges to f(x) for each x E X. For each Sn' Sn = Li=l L7'=l a(XAj aiM; for some positive integer m and some nonnegative real numbers a 1 •• am and characteristic functions M XA 1 • • • XA XAmm for measurable sets AI AI'. .. Am. Let da = kd I0a I. Then k be a complex measurable function with Ik I = 1 such that do
L7'=laiM;kdlal fxsn dcr == fxSnkdlal fxSnkdlcrl == fX L7':la fxsnda L7'=laJxXA;kdlal = iMj kdlcrl == L7':laJxMikdlcrl
L7':laJxMi dcr = L7':laJxXAj df..l + L7':JxMi dA. = fxL7':laiMi df..l + fxL7':la iMi dA. = fxsndf..l + fXSn dA . So (12.8) holds for E = = X and f a simple measurable function. Now let p be a complex measurable function with Ip I = = 1 such that dA d'A = = pd I'A A I. Then if E E Mwehave
Since Sn is a simple measurable function, XESn XEsn is also a simple measurable function. Therefore,
da = dA fXXESn dcr = ixXESndll fXXESnd'A JxXESn fXXESndf..l + JxXESn
= JEsnhdlf..l1 fESnhd 1111 =
+
fESnPdl).,1 fESndf..l + fESndA.. JESnPd I'A I = JESndj..t Consequently, (12.8) holds for the simple measurable functions. Define A A on
12.. Uniform Differentiation 12
384
~(X) by A(g) = = fxgdcr. Jxgda. By (12.9) A(g) = = fxgkd JxgkdlIcr a I so A is clearly a linear C oo(X) functional on C oo(X). ~(X). Since IIg 100 I~ < 00 and IIacr II(X) < 00 we have
IIA(g)1 A(g) I ~ fxlgkldlal fxlgkldlcrl
= fxlg1dlcrl fxlgldlal
~ IgIJcrl(X) Igl~lal(X)
so A is bounded and hence continuous by Theorem 11.13. Similarly define the continuous linear functionals .Q n and on C 00 (X) by n(g) ~(X) .Q(g) = fJxgd/-! = x gdJ.1 and (g) = fxgdA (X). Since {sn} JxgdA for each g E C 00 ~(X). (sn I converges to f we have {n(sn)} (.Q(sn) I converges to .Q(j), n(f), ((sn) {(sn)}I converges to (f) (j) and {A(sn)} (A(sn)) converges to A(f). A(j). Now for each positive integer n, A(sn) = s dJ.1 fxsndA {A(sn)} = fxsndo fxsnda = = ffxsnd/-! + JXSndA so (A(sn) I= x n {f s dJ.1 + fxsndA} = {n(sn) + (sn)}. {n(sn) + (sn)} converges to !fxsnd/-! JXSndA I (.Q(sn) (sn) I· Clearly, (.Q(sn) (sn) I x n n(f) fdJ.1 and {A(sn)} converges to fxldo. Since are .Q(j) + (f) (j) = fJxfd/-! + JxfdA (A(sn) I Jxfda. limits x Jxfda = = JJxfd/-! unique, fxfdo xfdJ.1 + JxfdA. Therefore, (12.8) holds whenfis a nonnegative bounded measurable function. It is a straightforward exercise (Exercise 2) to show that (12.8) holds for complex bounded measurable functions. Now if J.1 /-! is a complex measure then IJ. /-! /-!l + iJ.12 i/-!2 for some pair J.11 /-!l and J.12 /-!2 of real measures. If E E M then iJ.12(E) i/-!2 (E) = = = J.11 iJE iJ.!2(E) = d/-!2' Consequently, fJEd/-! = J.1(E) /-!(E) = = J.11 /-!l (E) + i/-!2(E) = JJEd/-!1 iJE d/-!2 so E dJ.11 + if E dJ.12 E dJ.1 = E dJ.12.
fEd(f.11 + if.12)
(12.10)
=
tf
fE df.11 + Edf.12
for each E E M. This concludes our discussion of integration with respect to complex measures. In the next section we will use these results to significantly generalize the Riesz Representation Theorem. 1(J.1), (/-!), where J.1 /-! is a positive measure on a PROPOSITION 12.4 If g E L 1 a-algebra M in X, and ifA(E) = fJEgd/-!for E M, then IIAI(E) AI (E) = JElgld/JJE Ig IdJ.1 MinX, E gdJ.1for each E EM, for each E E M.
*
= {x (x E Xlg(x) X Ig(x) = = O} 01 and B = = (x Ig(x) oF O}. 01. Then A(A) = =0= = Proof: Let A = {x E X Xlg(x) fA Ig IdJ.1. Since A(A) = 0 implies IIAI(A)=Owehave A I(A) = 0 we have IIAI(A)=JAlgld/-!. A I(A) = fA Ig IdJ.1. JAlgld/J-. =Oimplies = 0 implies A(EnB) = = 0 so fJEr>Bgd/-! = O. Next let E E M. Then IIAI (EnB) = EraB gdJ.1 = It can be shown that fJEr>Bgd/-! = 0 implies J.1(EnB) /-!(EnB) = = O. To see this note that EnB gdJ.1 = df.1 - dEnBg2 fEnBg df.1 == fEr>Bgid/-! fEnBgj df.1 - fEr>Bgjd/-! fEnBgi df.1 + dEnBgi fEr>Bgd/JJEr>Bg2:d/JJEr>Bg2 df.1 J.l
forsomenonnegativefunctionsgi,gj,g2: LetC Xlgi(x»OJ, for some nonnegative functions gi, gi, gi andg and g2. C 1I = {XE {x E X Igi(x) > O}, 2. Let = (x Xlgj(x) > OJ, = {x E xlgi(x) Xlg2(x) > 01 = {x (x E xlg2(x) Xlg2(x) > 01. C2 = {x E Xlgi(x) A}, C 3 = O} and C 4 = O}. . = C 1 uC UC 2UC UC 4. J.l(EnB) O. J.l(EnC for i ) Clearly B = uC 3 uC Suppose J.1(EnB) > Then > 0 4 gj(x) = 0 for each x E EnC 1 so some i = 1 ... 4. Assume J.l(EnC 1) > O. Then gl(X) = fC1nEgidJ.l JclrEgidJ.l > O. But EnC 1 c EnB so 0 = = A(EnC 1) = IEflC1gdJl, JEr.CIgdJ.l, and fJC1 g IdJ.l = oF J g 1 fJEr.C1 flC 1 gdJ.l 0 since IE nC 1 1 dJ.l > 0 which is a contradiction. Hence J.l(EnC 1) = ErCI E
o.O.
*
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385
Similarly f.!(EnC ll(EnC 2) = 0 = f.!(EnC ll(EnC 3) = f.!(EnC ll(EnC 4). But then f.!(BnE) Il(BnE) = O. Therefore, IIAI (EnB) = 0 which implies Il(EnB) f.!(EnB) = 0 so 11« J.l« IIAI on B. Hence dJ..LId IIAII exists on B since IIAII(X) < 00 by the Radon-Nikodym derivative dfJId (J.1) with IIhi = 1 such Proposition 12.3. By Theorem 12.4 there exists an h ELI (J.!) JEhd IIAII for each E E M M and by hypothesis, A(E) = fEgdJ.! JEgdll so fEhd JEhd IIAII that A(E) = fEhd = fEgdf.! JEgdll for each E E M. But then JEnBhdl JEnBhdlAI fEnBg[dJ..LIdIAI]dIAI = AI = JEnBg[dfJIdl AI ]dl AI for each M by Exercise 3 of Section 12.2. Thus h = gdfJId gdJ..LId IIAII almost everywhere EE M JE nB Ihid hid IAI with respect to IAI on B by Exercise 2 of Section 8.7. Therefore, IE = fJEnBlgl[dJ..LIdIAlldIAI J.l and III.I AI are both positive, so IIdfJId AIII = dJ..LIdlIAI EnB Ig I [dJlld IA I]d IAI since 11 dJlldiAI. dJ..LIdIAI. But then IAI(EnB) = fEnBdlAI JEnBdlAI = fEnBlgldJ.l JEnBlgldll by Exercise 3 of h I = 1. Consequently, IIA II(E) = JE Ig IIdJ.! dJ.1 for Section 12.2 and the fact that IIhi each E EEE M. • We conclude this section with the so called Hahn Decomposition Theorem. This theorem is a statement about the Jordan decomposition of a real valued measure Jl J.1 on a a-algebra M in a space X. It says that X can be decomposed into two disjoint sets A and B such that J.!+ J.1+ is concentrated on A and JlW is concentrated on B.
IfJ.1 is a real valued THEOREM 12.5 (Hahn Decomposition Theorem) IfJl M on a space X, there exists disjoint sets A and B with measure on a a-algebra M J.1+(E) = = J.l(EnA) J.1(EnA) and J.l-(E) J.1-(E) = = -J.l(EnB)for -J.1(EnB)for each E EE M. AuB = X such that Jl+(E) Proof: By Theorem 12.4, there exists a complex measurable function ff on X If I = 1 and J.l(E) J.1(E) = fEfd Jdd IIJ.l1 J.11 for each E E M. Since Jl J.1 is real ff must be real with IfI almost everywhere. Define h on X by hex) = f(x) if f(x) is real and hex) = 1
J.1(E) = otherwise. Then h is a real measurable function with IIh I = 1 and J.l(E)
JEhdlJ.11 foreachEE M. PutH={XE Xlh(x)=I}andK={xE Xlh(x)=l}andK={xE Xlh(x)=-I}. Xlh(x)=-I). fEhdlJ.l1 Clearly HuK = X and HnK = 0. Now J..l+ J.1+ = (I (I J.l1 J.11 + J.l)/2 J.1)/2 and (1 + h)j2 h)/2 = h on H and zero on K so for each E E M we have: J.l+(E) J.l1 +fE(1/2)dJ.1 + fE(l/2)dJ.l = fE[(1 + h)/2]dl J.l1 = J.1+(E) = fE(l/2)dl fE(1/2)dlJ.11 fd(1+h)/2]dlJ.11 = fEnH[(I+h)/2]dlJ.l1 fEnHhdlJ.l1 = fEnH[(1+h)/2]dlJ.11 + fEnK[(I+h)/2]dlJ.l1 fEnd(1+h)/2]dlJ.11 = = fEnHhdlJ.11 = J.l(EnH). J.1(EnH). Hence J.l+(E) J.1+(E) = J.1(EnH). J.1(E) = J.l(EnH) J.1(EnH) + J.1(EnK) and J.l(E) J.1(E) = 1l+(E) Jl(EnH). Now J..l(E) + J.l(EnK) J.l+(E) + J.1(EnK) Jl(EnK) = J.l+(E) J.l(EnH) we have J.l-(E) WeE) so J.!(EnH) J.1(EnH) + J.1+(E) - J.l-(E). WeE). Since J.l+(E) J.1+(E) = J.1(EnH) fJ,-(E) WeE) = -Jl(EnK). -J.1(EnK). •
EXERCISES 1. Show that the identity (12.8) holds when J.l J.1 and A are positive measures, without reference to complex measures.
12. Uniform Differentiation
386
Il and A ').., are complex measures, 2. Show that the identity (12.8) holds when J.l based on the fact that (12.8) holds whenfis when f is a nonnegative bounded measurable function. 3. Use a similar technique to the one used to prove the identity (12.8) to show that if c EE C and J.l E C oo(X), Il is a complex measure, then for each ff E ~(X), JEfd(cJ.l) JEfd(cll) = = cJEfdll for each measurable set E. cJEfdJ.l
12.4 The Riesz Representation Theorem One of the more important applications of the Radon-Nikodym Theorem is to generalize the version of the Riesz Representation Theorem presented as a series of exercises in Section 8.8, to include bounded linear functionals on (X) as opposed to the positive linear functionals on CK(X) as given in C ~(X) Section 8.8. For this we will find the following lemma useful. 00
LEMMA 12.2 Let Q be a bounded linear functional on C ~(X). (X). Then I) ~ If II~ there exists a positive linear functional A on CK(X) with IIQ(f) n(f) II ~ A( If I) CK(X). for eachfE CdX). 00
00
Proof: We first define A on the subset C"K(X) C1(X) consisting of all nonnegative real II g EE CK(X) and IIg II valued members of CK(X). For this put A(j) Aif) = = sup{ sup { IIQ(g) II ~f} foreachfEC1(X). ThenA(j)~OforeachfE C1(X) and IIn(j) Q(j) II ~ ~A(lfl) ~ f} for each f E C"K(X). Then Aif) ~ 0 for each f E C"K(X) A( If I) C1(X) withf~ withf~ g then A(j) ~ Ifl~. If I Moreover, if f, g E E C"K(X) Aif) ~ A(g) and if a > 0 then Aif) + A(af) = ClAif). aA(j). We want to show that if f, g EE C"K(X) C1(X) that A(f + g) = A(j) A(g). 00.
C1(X) and let e£ > O. Then there exists h, k E For this letf, g EE C"K(X) E CK(X) with
IIh II ~ f, IIk II ~ g and A(j) Aif) ~ IIQ(h) II + £e and A(g) ~ IIQ(k) II + £. e. Moreover, there are complex numbers a and ~ with IIa II = 1 = II ~ II such that a.il(h) a.D.(h) = IIn(h) Q(h) II and ~Q(k) = IIQ(k) n(k) I. Aif) + A(g) ~ IIn(h) n(k) II + 2£ I. Then A(j) Q(h) II + IIQ(k) 2e = a.il(h) a.D.(h) + ~Q(k) + 2£ 2e = n(ah Q(ah + ~k) + 2£. 2e. Clearly Q(ah = ~n(k) n(ah + ~k) ~ 0 so Q(ah n(ah + ~k) = IIn(ah+~k) Q(ah+~k) I. I. Therefore, Aif) A(j) + A(g) ~ IIQ(ah + ~k) II + 2£ 2e ~ A( IIah II + II~k I) I) + 2e. Since this inequality holds for each £e > 0 we have A(j) ~ A(f + g) + 2£. A(f) + A(g) ~ A(f + g).
2e 2£
To show A(f + g) ~ A(j) Aif) + A(g), let h EE CK(X) with Ih I ~ f + g. Put U = = Xlf(x) = X - U. Define ss and t on X by s(x) = {x E E X If(x) + g(x) > O} and F = f(x)h(x)f[f+g](x) and t(x) = g(x)h(x)/[f g(x)h(x)f[f + g](x) for each x E E U and s(x) = 0 = t(x) f(x)h(x)![f+g](x) for each x EE F. Clearly s and t are continuous on U and h = = s + t. Moreover, IIs(x) II ~ IIh(x) II and IIt(x) II ~ IIh(x) II for each x EE X so ss and t are also continuous on F since h(F) = 0 and h is continuous on X. Therefore, s, t E CK(X). Since IQ(h) I == In(s) IQ(s) + Q(t) ~A(f) lsi ~fand It II ~gwehave: In(h)1 n(t) II ~ I1 O(s) I1 + 10(/)1 In(t) I S:A(f) A(J + g) + A(g). Since this holds for each h E CK(X) with Ih I ~f + g we have A(f ~ A(f) + A(g). Hence A(f + g) = A(f) + A(g) for each pair f, g E Ck(X).
12.4 The Riesz Representation Theorem
387
To extend A to the real valued members of CK(X) note that if I is a real valued member of CK(X) that = (III (III + /)/2 and f1- = (If (IIII -- /)/2 so that both and f- are in C"K(X). Ae.r) - A(J-). CK(X) let andr CKeX). Define A(j) = A(f) A(J). To extend A to CK(X) g EE CK(X). Then g = h + ik for some real valued members h and k of CK(X). Define A(g) = A(h) + iA(k). Now A is defined for all g EE CK(X).
r
rr
It remains to show that A is linear on CK(X) I) ~ CK(X) and that IIO(j) II ~ A( If III) If I EE CK(X). We leave this as an exercise (Exercise 1). • IIII~ for each eachl 00
A complex Borel (Baire) measure Il J.l on X is said to be regular or almost regular if IIII J.l1I is regular or almost regular, respectively. The linear operator A fEldll is clearly bounded as shown in the last defined on C ~(X) (X) by A(j) = JEfdJl section and therefore continuous. Moreover, IIA II ~ IIII J.l1I (X) since IA IA II = sup{ IIfxldllll/E JxfdJlllf E C (X) and I/I~ If I = (X) and I/I~ If I C~(X) = I} ~ sup{J sup{fxl/ldlllll/E C~(X) x If Id IJ.llif E C = fxd IJlI III I = = IIII = I} == Jxd J.l1I(X). Our generalization of the Riesz Representation 00
00
00
00
00
Theorem shows that all bounded linear functionals on C ~ (X) are formed this way with respect to regular complex measures. 00
THEOREM 12.6 (Riesz Representation Theorem) Let 0 be a bounded linear lunctional functional on C ~(X) (X) where X is locally compact and Hausdorff Hausdorff. Then there exists a unique complex regular Borel measure Jl on X Il X such that O(f) Om = fxfdllior eachl each I EE C ~(X) 101. Jx.fdJ.llor (X) and such that IIII J.l1I(X) = 101. 00
00
Prool: Since CK(X) is a dense subspace of C ~(X) Proof: (X) with respect to the supremum 00
norm (Theorem 11.15) and each bounded linear functional A in CK(X) has a nonn unique extension to a bounded linear functional on C ~(X) (X) with the same nann CK(X). norm (Theorem 11.6), it suffices to prove the Theorem for CK(X). 00
(j) = = 0(/)/ O(j)/ I0 I. Then 1 I I = = 1 and is a linear For each I EE CK(X) put (/) CdX). By Lemma 12.2, there exists a positive linear functional functional on CK(X). ~A(I/I)~ IflooforeachfE 1/1~foreach/E CK(X). By Theorem 8.14 AonCK(X)with 1(j)1 1 (/) 1 ~A(lfl)~ 'A. such that A(/) A(j) = JxldA fxld'A. for there exists a positive almost regular Borel measure A I, Section 8.8) each II E CK(X). From the proof of Theorem 8.14 (see Exercise 1, 'A.(X) = sup sup{A(j) sup{A(j)I/E A(X) {A(/) II If < X} = sup {A(/) II E CK(X) and/(X) and I(X) c [0, I]}. Since for each I E CK(X) with If IIII~ ~ 1 we have IIA(/) A(j) II ~ 1 it is clear that A(X) 'A.(X) ~ 1. 'A. is regular. Moreover, II(/) (j) II ~ A( III) III) = J fxx If IIIIdA d'A. = If IIII Hence A I1 for each fIEE CK(X) where II 11 II denotes the L 1I (A) ('A.) norm. 00
II = sup{ II(j) III III E CK(X) with If IIII I1 = I} ~ 1 so the norm of Now II 11 ('A.) norm on CdX). is at most 1 with respect to the L 1I (A) CK(X). By Theorem 11.4, CK(X) CK(X) ('A.) and by Theorem 11.6 there exists a norm preserving extension is dense in L 1I (A) ('A.). By Theorem 12.2 there exists a unique Borel to a linear functional ' on L 1I (A). measurable function g with II g II~ = II' II such that 00
(12.11)
c'J>'(j) '(j) = ixfgd'A Jxlgd'A.
for each fI ELI E CK(X) ELI (A). ('A.). Consequently, IIg II~ ~ 1 so IIg II ~ 1. Let flEe K(X) with 00
388
12. Uniform Differentiation
Ifl~ = 10 Then I/(x) If(x) I $1 X soJxlgldA~JxlfgldA~ IIfx!gdAI JxfgdAI for 1/100 ~ 1 foreachxE XsoJ~lgldA~IxllgldA~ I~ = 1. Therefore, Ix IIg IdA ~ sup {{ IIIxfgdA JxfgdAllf CK(X) with IIf 1100 II I E CK(X) CK(X) each If E CK(X) If I~ = I} = sup sup{{ I(1) I(f) III lifE If I~ = I} = I I. Since I II = 1 with 1/100= E CK(X) with 1/100= I~ nann norm on CK(X» we have Ix Jx Ig IdA ~ 1. But since A(X) (with respect to the I1 100 $~ 1 and IIg I $~ 1, this can only happen if A(X) = 1 and IIg II = 1 almost everywhere with respect to A. Redefine g so that IIg(x) II = 1 for each x E X. Define~' Clearly we can do this without disturbing the results so far. Define J.!' on X by ~'(E) = IEgd JEgd I AI1 for each Borel set E c X. Then Then~' J.l'(E) J.l' is a complex measure and by ~'I (E) = IE JE iIg I1d IA I = IEdA JEdA for each Borel set E. But then Theorem 12.4, IJ.l'1 ~'I (X) = A(X) = 1 = I I. Now (12.11) and (12.9) yield IJ.l'1 (f) = = JxfgdA fxfgdA = = JxfgdlAI fxfgdl AI = = fxfd~' (j) JxfdJ.!'
(12.12)
n(f) = (1) (f)lnl Jxflnld~'. ~(E) = 10.1 Inl~'(E) for eachfE each IE CK(X). Then 0.(1) I0. I = Ix/l 0. IdJ.l'. Put J.l(E) J.l'(E) for each Borel set E c X. By Exercise 3 of Section 12.3 we have (12.13)
Q(j) Q I(j) = Q IJXfdJ.!' = Q IJxfd( IQ IJ.!) = fxfdJ.! n(f) = = IInl(f) = IInlfxf*' = IInlfxfd(lnl~') = fxfd~
for each IE f E CK(X), so the bounded linear functional 0. 0. on CK(X) is represented ~ in the sense that 0(1) n(f) = fxldJ.l Jxf* for each f E CK(X). CdX). by the complex measure J.! I~I(X) = SUp{L;=IIJ.!(E SUp{L;;'=ll~(En)11 Moreover, 1J.lI(X) n)11 {En} is a partition of X} = L;;'=l I J.l'(E ~'(En) II {En} is a partition of Xl Inll~'1 sup{ 10.1 10.1 L;=1 X} = 10.11 J.l'1 (X) = 10.1. 10.1. Hence n) II 10.1 = = IJ.!I(X). I~I(X). 10.1 ~ is regular, Since the measure A was regular, it is not difficult to show that J.! but we leave that as an exercise (Exercise 2). It remains to show the uniqueness of J.l. For this let Jl of~. ~ and v be two Borel measures on X such that for each f E CK(X), IxfdJ.l Jxfd~ = n(f) = Ixldv. Jxfdv. Then Ix/da Jxfda = 0 for each/E each f E CK(X) CK(X) where aa = f.1~CK(X), = v. Let h be a Borel measurable function on X with Ih I = 1 such that da = hdlal. E Ll(lal). Ll(lal) hd Ia I. Then h* ELI ( Ia I). Since CK(X) CK(X) is dense in L 1( Ia I) we can find a {fn} cC CK(X) such that Ih* - In fn I converges to 0 as n ~ 00. Now for sequence {In} each positive integer n, --fxfnda Ixlnda = 0 so --Jxfnhdl Ixfnhd IaalI = O. Therefore,
lal(X) = Jx fxlhldlal = Jxh*hd fxh*hdlal fxfnhdlal = fx(h* fx(h*-fn)hdlal. IaI (X) = Ihid IaI = IaI - JXfnhdl aI = - fn)hdl al.
Consequently, IaI Ia I(X) = II II aI a I(X) II :os; $ fx IIh* - fn Ida = IIh* - fn 1\. 11' Since IIh* fn 11 converges to 0 as n ~ 00 we see that Ia I (X) = o. In O. But then a(X) = 0 so ~(E) = veE) for each Borel set E a(E) = 0 for each Borel set E c X. Therefore, f.1(E) so~=v.sOf.1=v.-
EXERCISES 1. Show that the functional A constructed in Lemma 12.2 is linear and that IO(f)1 1[100 foreach[E In(f)1 ~A(I[I)~ $A(lfl)$lfl~ foreachfE CK(X).
12.5 Uniform Derivatives of Measures
389
2. Show that the measure constructed in Theorem 12.6 is regular.
12.5 Uniform Derivatives of Measures On the real line R there is a natural way to think about the derivative of a complex valued Borel measure f..l at a point x EE R. Let m denote Lebesgue J.l« -00, x» for each x EE R. Then it is measure on R. Define f:R ~ -7 C by f(x) = = f..l« possible to prove the following:
THEOREM 12.7 f is differentiable at x with derivative DEC if and only iffor each E > 0 there exists a b 0>> 0 with IJ.l(S)/m(S) f..l(S)lm(S) - D I < E whenever S is an open interval containing x with mrS) < b. o. This theorem, whose proof we leave as an exercise, suggests defining the Jl on more general spaces, on which derivative of a complex Borel measure J.t J.l(U)/m(U) for some form f..l(U)/m(U) Haar measure m exists, as a limit of quotients of the fonn suitably restricted class of open sets U, that contain some given point x, as these sets shrink unifonnly uniformly to the point poim x. In this more general setting, Haar measure replaces Lebesgue measure used in Theorem 12.7. Constructing such a generalization will be the program of this section.
uniform space with For this let (X, A) be a locally compact, isogeneous unifonn f..l be a complex Borel measure on X and let m denote isometric basis v. Let J.l Haar measure on X. If p E X and DEC such that whenever E > 0, there exists abE v with (12.14)
J.l(S(p,a» I f..l(S(p,a» m (S (p,a» m(S(p,a»
- DI
O. Then then fddm JEfdm = JEgdm. (12.17)
IIfE[Il(S(x,a» fE[ J..l(S (x,a» m(S(x,a»
_g(x)]dml -g(x)]dml
~
I
fE J..l(S (x,a» fEIIl(S(x,a» m(S(x,a»
- g(x) Idm < -g(x)ldm
0 so E [Il(S(x, a»/m(S(x, a»]dm - fEgdml (12.16) holds. Next we show that for m(E) < 00 the right hand side of (12.16) is equal to 1l(E), J.!(E), i.e., Le., 00
(12.18) dm < 00, and since Is classical theorem from analysis known as Fubini's Theorem, the order of integration can be changed in this case, i.e., (12.19) holds. Fubini's Theorem is developed in a series of exercises at the end of this section. Now the right hand f..l(E) since JEldm fddm = fJ-(E) f..l(E) (which can then be moved side of (12.19) is merely fJ-(E) outside the remaining integral sign) and fs JS(x,a) (x,a) [1/m(S(x, a»]dm a))]dm = 1 since for a given a, l/m(S(x, a» a)) is a constant. But then the limit of the right hand side of (12.19), with respect to a EE v, is simply f.1(E) f..l(E) so by (12.16) and (12.19), (12.18) holds. Therefore, for Borel sets E c X = f.1(E) X with m(E) < 00 we have JEldm fEldm = f..l(E) = = fEgdm. JEgdm. In particular, if K is a compact set then m(K) < 00 (since Haar measure as developed in Chapter 9 was derived from the Riesz Representation Theorem). {Kn} Since X is a-compact, m is regular and there exists a sequence {K n } of compact sets such that limnm(K limnm(Kn) n) = m(X). For each positive integer n, Jkgdm Kll gdm = fXXKn = fXXKJdm. {XK = {XKJ} JX XKll gdm and JfKJdm ldm = J XK ldm. Clearly limn {XK g} = g and limn {xK =fI· X Kll ll ll /} = lln Also, IIXK XKll l1 ~ III XKng I ~ IIg II and IIxKJI III so by Theorem 8.13 we have: ll g I (12.20)
limJKngdm = limJxXKngdm = JXgdm fxgdm and
Consequently, Jxldm fxldm = lim limnIKJdm = lim limnfKngdm nJKll ldm = nJKll gdm everywhere with respect to m. m,
= Jxgdm fxgdm
so
I =g 1=
almost
It remains to prove the theorem for the case where f.1 f..l is a complex measure. f..l is complex then f.1 f..l = f..ll if..l2 for some pair f..ll But this is now easy since if f.1 f.11 + if.12 f.11 and f.12 f..l2 of real measures. Since f.1 f..l is absolutely continuous with respect to m, so f..l2 so by Theorem 12.1 there exists unique functions functions/l,h E Ll(m) are f..ll f.11 and f.12 11 ,/2 ELI (m) = JEll fd 1 dm and f.12 f..l2 (E) = such that for each Borel measurable set E c X, f..ll f.11 (E) = fEhdm. a))/m(S(x, a»} a))} JE12dm. Also, there exists agE L 1 (m) such that {f..l(S(x, {f.1(S(x, a»/m(S(x, converges uniformly to g. Now g = g 1 + ig 22 for some real valued functions g 1 , 1 Ll(m) a))/m(S(x, a»} a))} = = {f..ll(S(X, a))/m(S(x, a» a)) + g2 E L (m) and {f..l(S(x, {f.1(S(x, a»/m(S(x, {f.11(S(X, a»/m(S(x, if..l2(S(x,a))/m(S(x, a»} a))} and the only way for this net to converge uniformly to if.12(S(x,a»/m(S(x, gl {Jll(S(X, a))/m(S(x, a»} a))} to converge uniformly to gland gl and g 1 + ig 2 is for {Jll (S(x, a»/m(S(x, {f.12(S(x,a»/m(S(x, {Jl2(S(x,a))/m(S(x, a»} a))} to converge uniformly to g2. g2' But then by what has already been shown,! shown, 111 = g 1I almost everywhere and 12 = g 22 almost everywhere with respect to m. Consequently,!= Consequently,! = g almost everywhere with respect to m.· m. -
ELI JE!dm for each measurable E c X, it is consistent If !IE L 1 (m) and J.1(E) Jl(E) = JEldm with the terminology of calculus to refer to Jl I. J.1 as the indefinite integral of f. Using this terminology, we can say that Theorem 12.8 states that the uniform derivative of 01 an indefinite integral is the integrand almost everywhere. We can 01 its uniform derivative. also say that Jl is the indefinite integral of
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12. Uniform Differentiation
EXERCISES 1. Prove Theorem 12.7. FUBINI'S THEOREM TIlEOREM Let (X, M) and (V, (Y, N) be measurable spaces. By M x N we mean the smallest a-algebra in X x Y that contains each set of the fonn A x B where A EE M and BEN. If E c X x Y and x EE X and y EE Y, we define the x-section, Ex and the y-section, E Y as follows:
Ex
= {yE lYE
VI(X,y)E YI(X,y)E E} and EY
= {XE
XI(X,y)E E}. XI(X,Y)E
2. Show that if E E M E N for each x E X y E Y. M x N, then Ex E M and EY EYE X and Y 3. Show that M x N is the smallest monotone class that contains all sets E that E M and BEN. are finite unions of sets of the fonn A x B where A E For each function Ion X x Y and x EE X, we associate the function Ix on Y defined by Ix(y) fx(y) =I(x, f(x, y). Similarly, for each Y yE E Y we defined the function p IY on X IY(x) = f(x, y). X by P(x) = I(x, 4. Show that if I is an M x N-measurable function on X x Y then for each x E E fx is an N-measurable function and for each y EE Y, P fY is an M-measurable X, Ix function.
Jl) and (Y, N, v) be a-finite measure spaces, and let E E xN. E M x N. If 5. Let (X, M, IJ,) f(x) = v(EJ v(Ex ) for each x EE X and g(y) = Jl(EY) IJ,(EY) for each y EE Y, show that I is I(x) M-measurable, N-measurable JxfdlJ, = fx Jxgdv. M -measurable, g is N -measurable and fxldJl gdv. Let (X, M, IJ,) E M x Jl) and (Y, N, v) be a-finite measure spaces. For each E E N define [IJ, [Jl x v](E) = = fyJl(EY)dv. Jl x v is said to be the = fxv(Ex)dJl Jxv(Ex)dlJ, = JylJ,(EY)dv. Then IJ, product of Jl IJ, and v, or the product measure on M x N.
Jl) and (Y, N, v) are a-finite measure spaces, and 6. [Fubini] Show that if (X, M, IJ,) if fI is M x N -measurable on X x Y, then:
m)
<j>(x) = = frfxdv ~ I(x) (x) JYfxdv and y(y) = fxpdJ.! JxlY dlJ, for (a) If 0