Elements of Operator Theory

Operator

Birkha" user

Carlos S. Kubrusly

Elements of Operator Theory

Birkhauser Boston Basel Berlin

Carlos S. Kubrusly Catholic University R. Marques de S. Vicente 225 22453-900, Rio de Janeiro, Brazil

e-mail: carloa@ele. puc-rio. br

Ubrary of Congress Cataloging-In-Publication Data Kubrusly, Carlos S., 1947-

Elements of operator theory / Carlos S. Kubrusly p. cm. Includes bibliographical references and index. ISBN 0-8176-4174-2 (acid-free paper) - ISBN 3-7643-4174-2 (acid-free paper) 1. Operator theory. I Title. QA329.K79 2001 515'.724-dc2I 2001018439

AMS Subject Classifications: 47-XX, 47-01, 47A05, 47A10, 47A12, 47AI5, 47A75, 47610, 47B15, 47B20, 47637, 47005, 47L05. 46-XX, 46-01 .46A22. 46A30, 46B 10, 461315, 46820.46B45, 46B50. 46C05,46C07, 46015, 54-XX, 54-01, 54A20, 54B05, 54B10, 54B15, 54005. 54020, 54E35. 54E45, 54850, 54E52, 15-XX, 15-01, 15A03, 15A04. 03Exx. 03E10. 03820

Printed on acid-free paper.

02001 Birkhiuser Boston

Birkhiiuser

rt) L11 1),

All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (BirkhBuser Boston, c/o 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.

ISBN 0-8176-4174-2

SPIN 10754156

ISBN 3-7643-4174-2

Reformatted from author's files in LATEX2e by TEXniques, Inc., Cambridge, MA Printed and bound by Hamilton Pnnting, Rensselaer, NY Printed in the United States of Amenca

987654321

To the memory of my father

The truth, he thought, has never been of any real value to any human being - it is a symbol for mathematicians and philosophers to pursue. In human relations kindness and lies are worth a thousand truths. He involved himself in what he always knew was a vain struggle to retain the lies. Graham Greene

Preface

"Elements" in the title of this book has its standard meaning, namely, basic principles and elementary theory. The main focus is operator theory, and the topics range from

sets to the spectral theorem. Chapter 1 (Set-Theoretic Structures) introduces the reader to ordering, lattices and cardinality. Linear spaces are presented in Chapter 2 (Algebraic Structures). Metric (and topological) spaces in Chapter 3 (Topological Structures). The purpose of Chapter 4 (Banach Spaces) is to put algebra and topology to work together. Continuity plays a central role in the theory of topological spaces,

and linear transformation plays a central role in the theory of linear spaces. When algebraic and topological structures are compatibly laid on the same underlying set, leading to the notion of topological vector spaces, then we may consider the concept of continuous linear transformations. By an operator we mean a continuous linear transformation of a normed space into itself. Chapter 5 (Hilbert Spaces) is central. There a geometric structure is properly added to the algebraic and topological structures. The spectral theorem is a cornerstone in the theory of operators on Hilbert spaces. It gives a full statement on the nature and structure of normal operators, and is considered in Chapter 6 (The Spectral Theorem). The book is addressed to graduate students, both in mathematics or in one of the sciences, and also to working mathematicians getting into operator theory and scientists willing to apply operator theory to their own subject. In the former case it actually is a first course. In the latter case it may serve as a basic reference on the so-called elementary theory of single operator theory. Its primary intention is to introduce operator theory to a new generation of students and provide the necessary background for it. Technically, the prerequisite for this book is some mathematical maturity that a first-year graduate student in mathematics, engineering or in one of

viii

Preface

the formal sciences is supposed to have already acquired. The book is largely selfcontained. Of course, a formal introduction to analysis will be helpful, as well as an introductory course on functions of a complex variable. Measure and integration are not required up to the very last section of the last chapter. Each section of each chapter has a short and concise (sometimes a compound) title. They were selected in such a way that, when put together in the contents, give a brief outline of the book to the right audience. The focus of this book is on concepts and ideas as an alternative to the computational approach. The proofs avoid computation whenever possible or convenient. Instead, I try to unfold the structural properties behind the statements of theorems, stressing mathematical ideas rather than long calculations. Tedious and ugly (all right, "ugly" is subjective) calculations were avoided where a more conceptual way to explain the stream of ideas was possible. Clearly, this is not new. In any event, every single proof in this book was specially tailored to meet this requirement but they (at least the majority of them) are standard proofs, perhaps with a touch of what may reflect some of the author's minor idiosyncrasies. In writing this book I kept my mind focused on the reader. Sometimes I am talking to my students and sometimes to my colleagues (they surely will identify in each case to whom I am talking). For my students the objective is to teach mathematics (ideas, structures and problems). There are 300 problems throughout the book, many of them with multiple parts. These problems, at the end of each chapter, comprise complements and extensions of the theory, further examples and counterexamples, or auxiliary results that may be useful in the sequel. They are an integral part of the main text, which makes them different from traditional classroom exercises. Many of these problems are accompanied by hints, which may be a single word or a sketch, sometimes long, of a proof. The idea behind providing these long and detailed hints is that just talking to students is not enough. One has to motivate them too. In my view, motivation (in this context) is to reveal the beauty of pure mathematics, and to challenge students with a real chance to reconstruct a proof for a theorem that is "new" to them. Such a real chance can be offered by a suitable, sometimes rather detailed, hint. At the end of each chapter, just before the problems, the reader will find a list of suggested readings that contains only books. Some of them had a strong influence in preparing this book, and many of them are suggested as a second or third reading. The reference section comprises a list of all those books and just a few research papers (82 books and 11 papers), all of them quoted in the text. Research papers are only mentioned to complement occasional historical remarks so that the few articles cited there are, in fact, classical breakthroughs. For a glance at current research in operator theory the reader is referred to recent research monographs suggested in Chapters 5 and 6. I started writing this book after lecturing on its subject at Catholic University of Rio de Janeiro for over 20 years. In general, the material is covered in two onesemester beginning graduate courses, where the audience comprises mathematics, engineering, economics and physics students. Quite often senior undergraduate students joined the courses. The dividing line between these two one-semester courses

Preface

ix

depends a bit on the pace of lectures but is usually somewhere at the beginning of Chapter 5. Questions asked by generations of students and colleagues have been collected. When the collection was big enough some former students, as well as current students, insisted upon a new book but urged that it should not be a mere collection of lecture notes and exercises bound together. I hope not to disappoint them so much. At this point, where a preface is coming to an end, one has the duty and pleasure to acknowledge the participation of those people who somehow effectively contributed in connection with writing the book. Certainly, the students in those

courses were a big help and a source of motivation. Some friends among students and colleagues have collaborated by discussing the subject of this book for a long time many on occasions. They are: Gilberto O. Correa, Oswaldo L. V. Costa, Giselle M. S. Ferreira, Marcelo D. Fragoso, Ricardo S. Kubrusly, Abilio P. Lucena, Helios Malebranche, Carlos E. Pedreira, Denise O. Pinto, MarcosA. da Silveira, and Paulo Cr sar M. Vieira. Special thanks are due to my friend and colleague Augusto C. Gadelha Vieira who read part of the manuscript and made many valuable suggestions. I am also grateful to Ruth F. Curtain who, back in the early seventies, introduced me to functional analysis. I wish to thank Catholic University of Rio de Janeiro for providing the release time that made this project possible. Let me also thank the staff of Birkhauser Boston and Elizabeth Loew of TeXniques for their ever efficient and friendly partnership. Finally, it is just fair to mention that this project was supported in part by CNPq (Brazilian National Research Council) and FAPERJ (Rio de Janeiro State Research Council). Carlos S. Kubrusly Rio de Janeiro November, 2000

Contents

Preface 1

Vii

Set-Theoretic Structures

1

Background 1.2 Sets and Relations . . 1.3 Functions . . . . . . 1.4 Equivalence Relations 1.5 Ordering . . . . . . . . 1.6 Lattices . . . . . . 1.7 Indexing . 1.8 Cardinality . . . . . 1.9 Remarks . . . . . . . . . Problems . . . . . 1.1

.

.

.

.. .. ..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

. .

.

.

.

.

.

.

.

.

. .

..

.

.

.

.

. .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

1

.

.

.

3

.. .. .. . ... . .. .. .. .. . ...... . .. . ... .. .. .. .

.

.

.

.

.

. . . .

.

.

. .

.

.

.

. .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

5

7

.

.

.

.

. .

.

.

.

. .

.

.

8 10 12 14 21

.

.

.

.

24

.

.

.

.

.

.

.

.

.

.

.

.

.

.. .. .. . ... .... . .. . ... . . ... . .. .. . ... . .. .. .. .. ..

.

.

.

.

.

.

.

.

.

2 Algebraic Structures 2.1 Linear Spaces . . . . 2.2 Linear Manifolds . . 2.3 Linear Independence . . . 2.4 Hamel Basis . . . . . . . 2.5 Linear Transformations .. . 2.6 Isomorphisms .. . . 2.7 Isomorphic Equivalence . 2.8 Direct Sum . . . . . . . .

.. .. .. .

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.. .. .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

37 .

.. .. .

.. .... ... .

.. .. .. .. .. .. .. .. . ... .. .. .. .. .. . ... . .. .

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

. ... .

.

.

.

. .

.

.

37 43

46 48 55 58 64 66

xii

Contents

2.9 Projections Problems . . . .

3

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Topological Structures

.

.

.

.

.

.

.

..

.. ... .

3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Convergence and Continuity ... ..

.

.

.

.

.

.

.

3.11 Sequential Compactness . Problems . . . . . .

.. ..

Banach Spaces 4.1

.

.

.

.

.

.

.

.

.

.

.

.. .. ..

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.. .

.

. .

.. ... .

.

70

.

75

.. .. .. ... . ... .. .. . .

.

.

.

.

. .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

. .

.

.

. .

.

. .

.

.

.

.

.

. . . .

.

.

.

.

.

..

.

.

. .

. .

. . . Open Sets and Topology . . Equivalent Metrics and Homeomt.rphisms . . . . Closed Sets and Closure . . . . Dense Sets and Separable Spaces . . . . . . Complete Spaces . . . . . . . . . . . . . . Continuous Extension and Completion . . . The Baire Category Theorem . . . . . . . . .

.

.

Metric Spaces

.

.. ..

.. .. . .... ... . .. . .... .

3.1

3.10 Compact Sets ..

4

.

.

.

.. .... . .. .. . .

. .... . .

.

.

. .

.

. .

.

.

. .

.. .... . .. .. .. .. ... . .. .. .

. .

.

.

. .... . .. .. .

.. ... . .. .. Examples . ... ... . .. .. .. .. . ... . Subspaces and Quotient Spaces . . ... .. .. ... . ... .. Normed Spaces ..

..

.

.

.

.

.

.

.

.

.

.

.

. . . . 4.2 . . . . . 4.3 4.4 Bounded Linear Transformations . 4.5 The Open Mapping Theorem and Continuous Inverses . . 4.6 Equivalence and Finite-Dimensional Spaces .. . . . . 4.7 Continuous Linear Extension and Completion . . 4.8 The Banach-Steinhaus Theorem and Operator Convergence 4.9 Compact Operators . . . 4.10 The Hahn-Banach Theorem and Dual Spaces . . . . . . Problems . . . .

.

. .

.. .. .... ... . .. .. .

.

. . .

.. .. . ... . ..

.

.

.

.

.

.

.

. ... . . ... . . ... .. .. .. .. .. . . ... .. .. . ... ..... .. .. . ... ..

5

Hilbert Spaces 5.1

5.2 5.3

Inner Product Spaces Examples Orthogonality

.

. .

.

.

..

.

. . .

.

.. . ..... .

.

.

100 106 113

119 127 134 142 148 155 163

197 197

202 209 215 223 230 237 242 250 258 269 311

.

. .

.

.

.

. .

. .

.

. ... ....... . .. . ..... . . ... .. ..... .... .. . ... ..... . .

85 85 93

.

.

.

311

318 323 328 335 339 343 352 359 367 376 387

.. .. .. .. . ..... .. .. .. . ... . .. .. .. . ... .... . .. . .... . .. .. .. .. .. .. .. .. .... . ... .. .. . ... .. .. .. .. .. .. .. .. .. 5.13 Self-Adjoint Operators ....................... 396 . 5.4 Orthogonal Complement .. ... .... ... . 5.5 Orthogonal Structure . . . . . . . . . . . . . . . . . . . . . . . 5.6 Unitary Equivalence . . . . . 5.7 Summability . . . . . . . . . . 5.8 Orthonormal Basis . . . . 5.9 The Fourier Series Theorem . . . . . 5.10 Orthogonal Projection . . . 5.11 The Riesz Representation Theorem and Weak Convergence . . . . 5.12 The Adjoint Operator . . . . . . . . . .

.

.

.

.

.

. . .

.

.

.

.

.

.

.

.

.

.

.

.

xiii

Contents

. .... ... . .. ..

.................................

5.14 Square Root and Polar Decomposition Problems 6

The Spectral Theorem 6.1 Normal Operators .

.

.

.

.

.

.

.

. .

.

.

..

. .

. .

.

.

. .

.

..

. ... ..

.. ..

.

.. . ... ... . . .... . . ... .. .. . .... . . ... .. .. . ... .

The Spectrum of an Operator . . . Spectral Radius . . . . . . . . . . Numerical Radius . . . . . . . . . . . . . . . . . . . . . . . Examples of Spectra . . . . . . . . . The Spectrum of a Compact Operator . . The Spectral Theorem for Compact Normal Operators . . . . A Glimpse at the Spectral Theorem for Normal Operators .. Problems . . . . . . . . . . 6.2 6.3 6.4 6.5 6.6 6.7 6.8

.

.

.

.. .. .. ....... . ..

. .

. . .

.

.

.

.

. .... . .

.

.

.

.

.

.

..

401

408 441 441

449 457 462 466 474 480 489 496

References

509

Index

517

1

Set-Theoretic Structures

The purpose of this chapter is to present a very brief review of some basic settheoretic concepts that will be needed in the sequel. By basic concepts we mean standard notation and terminology, and a few essential results that will be required in later chapters. We assume the reader is familiar with the notion of set and elements (or members, or points) of a set, as well as with the basic set operations. It is convenient to reserve certain symbols for certain sets, especially for the basic number systems.

The set of all nonnegative integers will be denoted by No, the set of all positive integers (i.e., the set of all natural numbers) by N, and the set of all integers by Z. The set of all rational numbers will be denoted by Q, the set of all real numbers (or the real line) by R, and the set of all complex numbers by C.

1.1

Background

We shall also assume that the reader is familiar with the basic rules of elementary (classical) logic, but acquaintance with formal logic is not necessary. The foundations of mathematics will not be reviewed in this book. However, before starting our brief review of set-theoretic concepts, we shall introduce some preliminary notation, terminology and logical principles as a background for our discourse. If a predicate P( ) is meaningful for a subject x, then P(x) (or simply P) will denote a proposition. The terms statement and assertion will be used as synonyms for proposition. A statement on statements is sometimes called a formula (or a secondary proposition). Statements may be true or false (not true). A tautology is a

2

1. Set-Theoretic Structures

formula that is true regardless of the truth of the statements in it. A contradiction is a formula that is false regardless of the truth of the statements in it. The symbol Q (whose logical definition is "either P is denotes implies and the formula P false or Q is true") means "the statement P implies the statement Q". That is, "if P is true, then Q is true", or "P is a sufficient condition for Q". We shall also use the symbol * for the denial of ==>, so that ¢i denotes does not imply and the formula

P

. Q means "the statement P does not imply the statement Q". Accordingly.

let 6 stand for the denial of P (read: not P). If P is a statement, then t is its contradictory. Let us first recall one of the basic rules of deduction called modus ponens: "if

a statement P is true and if P implies Q, then the statement Q is true" - "anything implied by a true statement is true". Symbolically, (P true and P Q) (Q true). A direct proof is essentially a chain of modus ponens. For instance, if P R ensures that R is true. Indeed, is true, then the string of implications P Q if we can establish that P holds, and also that P implies Q, then (modus ponens) Q holds. Moreover, if we can also establish that Q implies R, then (modus ponens again) R holds. However, modus ponens alone is not enough to ensure that such a reasoning may be extended to an arbitrary (endless) string of implications. In certain cases the Principle of Mathematical Induction provides an alternative reasoning. Let N be the set of all natural numbers. A set S of natural numbers is called inductive if n + 1 is an element of S whenever n is. The Principle of Mathematical Induction states that "if 1 is an element of an inductive set S, then S = N". This leads to a second scheme of proof, called proof by induction. For instance, for each natural number n let P be a proposition. If P1 holds true and if P,, P for every natural number n. The scheme of proof by induction works for N replaced with No. There is nothing magical about the number 1 as far as a proof by induction is concerned. All that is needed is a "beginning" and the notion of "induction". Example: Let i be an arbitrary integer and let Z; be the set made up of all integers greater than or equal to i. For each integer k in Zi let Pk be Pk+t for each k, then Pk holds true for a proposition. If Pi holds true and if Pk every integer k in Z; (particular cases: Zo = No and Zt = N). "If a statement leads to a contradiction, then this statement is false." This is the rule of a proof by contradiction - reductio ad absurdum. It relies on the Principle of Contradiction, which states that "P and ff are impossible". In other words, the Principle of Contradiction says that the formula "P and f " is a contradiction. But this alone does not ensure that any of P or t must hold. The Law of the Excluded

Middle (or Law of the Excluded Third - tertium non datur) does: "either P or i¢ hold". That is, the Law of the Excluded Middle simply says that the formula "P or r " is a tautology. Therefore, the formula 0 f means "P holds only if Q holds", or "Q is a necessary condition for P". If P = Q and Q = P, then we shall write P . Q which means "P if and only if Q", or "P is a necessary and sufficient condition for Q", or "P and Q are equivalent" (and vice versa). Indeed, t }. Such the formulas P Q and 0 = f are equivalent: (P = Q) t= (0

1.2 Sets and Relations

3

an equivalence is the basic idea behind a contrapositive proof "to verify that a proposition P implies a proposition Q prove, instead, that the denial of Q implies the denial of P". We conclude this introductory section by pointing out another usual but slightly different meaning for the term "proposition."We shall often say "prove the following proposition" instead of "prove that the following proposition holds true". Here the term proposition is being used as a synonym for theorem (a true statement for which we demand a proof of its truth), and not as a synonym for an assertion or statement (that may be either true or false). A conjecture is a statement that has not been proved yet - it may turn out to be either true or false once a proof of its truth or falsehood is supplied. If a conjecture is proved to be true, then it becomes a theorem. Note that

there is no "false theorem" - if it is false, it is not a theorem. Another synonym for theorem is lemma. There is no logical difference among the terms "theorem", "lemma" and "proposition" but it is usual to endow them with a psychological hierarchy. Generally, a theorem is supposed to bear a greater importance (which is subjective) and a lemma is often viewed as an intermediate theorem (which may be very important indeed) that will be applied to prove a further theorem. Propositions are sometimes placed a step below, either as an isolated theorem or as an auxiliary

result. A corollary is, of course, a theorem that comes out as a consequence of a previously proved theorem (i.e., whose proof is mainly based on an application of that previous theorem). Unlike "conjecture", "proposition", "lemma", "theorem" and "corollary", the term axiom (or postulate) is applied to a fundamental statement (or assumption, or hypothesis) upon which a theory (i.e., a set of theorems) is built. Clearly, a set of axioms (or, more appropriately, a system of axioms) should be consistent (i.e., they should not lead to a contradiction), and they are said to be independent if none of them is a theorem (i.e., if none of them can be proved by the remaining axioms).

1.2

Sets and Relations

If x is an element of a set X, then we shall write x E X (meaning that x belongs to X, or x is contained in X). Otherwise (i.e., if x is not an element of X), x 0 X. We also write A C B to mean that a set A is a subset of a set B (A C_ B {x E A X E B)). In such a case A is said to be included in B. The empty set, which is a subset of every set, will be denoted by 0. Two sets A and B are equal (notation: A = B) if A C B and B C A. If A is a subset of B but not equal to B, then we say that A is a proper subset of B and write A C B. In such a case A is said to be properly included in B. A nontrivial subset of a set X is a nonempty proper subset of it. If P( ) is a predicate that is meaningful for every element x of a set X (so that P(x) is a proposition for each x in X), then (x E X : P(x)) will denote the subset of X consisting of all those elements x of X for which the proposition P(x) is true. The complement of a subset A of a set X. denoted by X\A, is the

4

1. Set-Theoretic Structures

subset {x E X : x V A). If A and B are sets, the difference between A and B, or the relative complement of B in A, is the set

A\B = {x E A: x 4 B}. We shall also use the standard notations U and fl for union and intersection,

respectively (xEAUBb{xEAor xEB}and xeAf1BIX EAand

x E B}). The sets A and B are disjoint if A fl B = 0 (i.e., if they have an empty intersection). The symmetric difference (or Boolean sum) of two sets A and B is the set

AvB = (A\B) U (B\A) = (A U B)\(A fl B). The terms class, family and collection (as their related terms prefixed with "sub") will be used as synonyms for set (usually applied for sets of sets, but not necessarily) without imposing any hierarchy among them. If X is a collection of subsets of a given set X, then U X will denote the union of all sets in X. Similarly. n x will denote the intersection of all sets in X (alternative notation: UAEXA and nAEXA) An important statement about complements that exhibits the duality between union and intersection is the so-called De Morgan laws:

X\( U A) = n (X\A) AEX

and

X\(n A) =

AEX

AEX

U (X\A). AEX

The power set of any set X, denoted by P(X), is the collection of all subsets of X.

Note that UP(X) = X E P(X) and f P(X) = 0 E P(X). A singleton in a set X is a subset of X containing one and only one point of X (notation: (x) c X is a singleton on x E X). A pair (or a doubleton) is a set containing just two points, say {x, y), where x is an element of a set X and y is an element of a set Y. A pair of points x E X and y E Y is an ordered pair, denoted by (x, y), if x is regarded as the first member of the pair and y is regarded as the second. The Cartesian product of two sets X and Y, denoted by X x Y, is the set of all ordered pairs (x, y) where x E X and y E Y. A relation R between two sets X and Y is any subset of the Cartesian product X x Y. If R is a relation between X and Y, and if (x, y) is a pair in R C X x Y, then we say that x is related toy under R (or x and y are related by R), and write x Ry (instead of (x, y) E R). Tautologically, for any ordered pair (x, y) E X x Y, either (x, y) E R or (x, y) 0 R (i.e., either xRy or x$ y). A relation between a set X and itself is called a relation on X. If X and Y are sets and if R is a relation between X and Y, then the graph of the relation R is the subset of X x Y

GR = {(x, y) E X XY: xRy}. A relation R clearly coincides with its graph GR.

1.3 Functions

1.3

5

Functions

Let x be an arbitrary element of a set X and let y and z be arbitrary elements of a set Y. A relation F between the sets X and Y is a -function if xFy and xFz imply y = z. In other words, a relation F between a set X and a set Y is called a function from

X to Y (or a mapping of X into Y) if for each x E X there exists a unique y E Y such that x Fy. The terms map and transformation are often used as synonyms for function and mapping. (Sometimes the terms correspondence and operator are also used but we shall keep them for special kinds of functions.) It is usual to write F : X - Y to indicate that F is a mapping of X into Y, and y = F(x) (or y = Fx) instead of xFy. If y = F(x), we say that F maps x to y, so that F(x) E Y is the value of the function F at x E X. Equivalently, F(x), which is a point in Y, is the image of the point x in X under F. It is also customary to use the abbreviation "the function X -+ Y defined by x -> F(x)" for a function from X to Y that assigns to each x in X the value F(x) in Y. A Y-valued function on X is precisely a function from X to Y. If Y is a subset of the set C, IR or Z, then complex-valued function, realvaluedfunction or integer-valued function, respectively, are usual terminologies. An

X-valued function on X (i.e., a function F: X -+ X from X to itself) is referred to as a function on X. The collection of all functions from a set X to a set Y will be denoted by YX. Indeed, YX C P(X x Y). Consider a function F : X - Y. The set X is called the domain of F and the set Y is called the codomain of F. If A is a subset of X, then the image of A under F. denoted by F(A), is the subset of Y consisting of all points y of Y such that y = F(x) for some x E A:

F(A) = {yEY: y=F(x)forsome xEACX}. On the other hand, if B is a subset of Y, then the inverse image of B under F (or the pre-image of B under F), denoted by F (B), is the subset of X made up of all points x in X such that F(x) lies in B:

F-t(B)= IXEX: F(x)EBCY). The range of F, denoted by R (F), is the image of X under F. Thus

?Z(F) = F(X) = {y E Y: y = F(x)forsomex E X}. If R(F) is a singleton, then F is said to be a constant function. If the range of F coincides with the codomain (i.e., if F(X) = Y), then F is a surjective function. In this case F is said to map X onto Y. The function F is injective (or F is a one-toone mapping) if its domain X does not contain two elements with the same image. In other words, let x and x' be arbitrary elements of X. A function F: X -+ Y is injective if F(x) = F(x') implies x = x'. A one-to-one correspondence between a set X and a set Y is a one-to-one mapping of X onto Y. That is, a surjective and injective function (also called a bijective function).

6

1. Set-Theoretic Structures

If A is an arbitrary subset of X and F is a mapping of X into Y, then the function

G: A -+ Y such that G(x) = F(x) for each x E A is the restriction of F to A. Conversely, if G: A -> Y is the restriction of F: X -* Y to some subset A of X, then F is an extension of G over X. It is usual to write G = FIA. Note that R(FIA) = F(A). Let A be a subset of X and consider a function F: A -* X. An element x of A is a feed point of F (or F leaves x fixed) if F(x) = x. The function J : A -+ X defined by J (x) = x for every x E A is the inclusion map (or the embedding, or the injection) of A into X. In other words, the inclusion map of A into X is the function

J : A --). X that leaves each point of A fixed. The inclusion map of X into X is called the identity map on X and denoted by 1, or by Ix when necessary (i.e., the identity on X is the function 1: X - X such that 1(x) = x for every x E X). Thus the inclusion map of a subset of X is the restriction to that subset of the identity map on X. Now consider a function on X; that is, a mapping F: X -* X of X into itself. A subset of X, say A. is invariant for F (or invariant under F, or FX, has its invariant) if F(A) c A. In this case the restriction of F to A, FIA: A range included in A: R(FI A) = F(A) C A C_ X. Therefore we shall often think of the restriction of F : X -> X to an invariant subset A C X as a mapping of A into itself: FIA: A --+ A. It is in this sense that the inclusion map of a subset of X can be thought of as the identity map on that subset: they differ only in that one has a larger codomain than the other.

Let F : X -- Y be a function from a set X to a set Y, and let G : Y

Z be a

function from the set Y to a set Z. Since the range of F is included in the domain of G, R(F) C Y. consider the restriction of G to the range of F. G I R(F): R(F) -+ Z. The composition of G and F, denoted by G o F (or simply by GF), is the function

from X to Z defined by (G o F)(x) = GIR(F) (F(x)) = G(F(x)) for every x E X. It is usual to say that the diagram X

Fi

Y

H\

IG

Z

commutes if H = G o F. Although the above diagram is said to be commutative whenever H is the composition of G and F. the composition itself is not a commutative operation even when such a commutation makes sense. For instance, if X = Y = Z and F is a constant function on X, say F(x) = a E X for every x E X. then G o F and F o G are constant functions on X as well: (G o F)(x) = G(a) and (F o G)(x) = a for every x E X. However G o F and F o G need not be the same (unless a is a fixed point of G). Composition may not be commutative but it is always

associative. If F maps X into Y. G maps Y into Z and K maps Z into W, then we can consider the compositions K o (G o F): X -+ W and (K o G) o F: X --> W. It is readily verified that K o (G o F) = (K o G) o F. For this reason we may and

1.4 Equivalence Relations

7

shall drop the parentheses. In other words, the diagram

XF4

Y

H\ IG Z

K

+W

commutes (i.e., H= G o F, L= K o G, and K o H= L o F). If F is a function on set X, then the composition of F : X -+ X with itself, F o F, is denoted by F2. Likewise, for any positive integer n E N, F" denotes the composition of F with itself n times. F o ... o F : X -+ X, which is called the nth power of F. A function

F : X -+ X is idempotent if F2 = F (and hence F" = F for every n E N). It is easy to show that the range of an idempotent function is precisely the set of all its fixed points. In fact. F = F2 if and only if 7Z(F) = {x E X : F(x) = x}. Suppose F : X -+ Y is an injective function. Thus, for an arbitrary element of R(F), say y, there exists a unique element of X, say xy, such that y = F(xy). This

defines a function from 7Z(F) to X, F-1: R(F) -+ X. such that xy = F-1 (y). Hence Y = F(F-t (y)). On the other hand, if x is an arbitrary element of X. then F(x) lies in R(F) so that F(x) = F(F-t (F(x))). Since F is injective, x = F-t (F(x)). Conclusion: For every injective function F: X - Y there exists a (unique) function F-t : R(F) -+ X such that F-t F: X - X is the identity on X (and FF-t : R(F) --> R(F) is the identity on R(F)). F-t is called the inverse of F on R(F): an injective function has an inverse on its range. If F is also surjective, then F-t : Y --> X is called the inverse of F. Thus an injective and surjective function is also called an invertible function (in addition to its other names).

1.4

Equivalence Relations

Let x, y and z be arbitrary elements of a set X. A relation R on X is

reflexive if xRx for every x E X, transitive if xRy and yRz imply x Rz . and symmetric if xRy implies yRx. An equivalence relation on a set X is a relation - on X that is reflexive, transitive and symmetric. If - is an equivalence relation on a set X, then the equivalence class of an arbitrary element x of X (with respect to -) is the set

[x] _ (X' E X : x'. X ). Given an equivalence relation - on a set X, the quotient space of X modulo -, denoted by X/- , is the collection

X/- =

I[X]CX: xEX}

8

1. Set-Theoretic Structures

of the equivalence classes (with respect to --) of every x E X. For each x in X set

tr(x) = [x] in X/-. This defines a surjective map r : X -+ X/- which is called the natural mapping of X onto X/-. Let X be any collection of nonempty subsets of a set X. X covers X (or X is a covering of X) if X = U X (i.e., if every point in X belongs to some set in X). X is disjoint if the sets in X are pairwise disjoint (i.e., A fl B = 0 for every pair of distinct sets A and B in X). A partition of a set X is a disjoint covering of X Let -_ be a an equivalence relation on a set X, and let X/::z be the quotient space of X modulo --. It is clear that

X/.-- is a partition of X. Conversely, let X be any partition of a set X and define a relation - IX on X as follows: for every x, x' in X, x is related to x' under - /X (i.e., x - IX x') if x and x' belong to the same set in X. In fact,

-/X is an equivalence relation on X. which is called the equivalence relation induced by a partition X. It is readily verified that the quotient space of X modulo the equivalence relation induced by the partition X coincides with X itself, just as the equivalence relation induced by the quotient space of X modulo the equivalence relation on X coincides with . Symbolically,

X/(- /X) = X

and

-/(X/--) = --.

on X induces a partition X/.: of X, which in turn induces back an equivalence relation - /(X/) on X that coincides with..:. On the other hand, a partition X of X induces an equivalence relation - /X on X, which in turn induces back a partition X l(- /X) of X that coincides with X. Conclusion: The collection of all equivalence relations on a set X is in a one-to-one correspondence with the collection of all partitions of X.

Thus an equivalence relation

1.5

Ordering

Let x and y be arbitrary elements of a set X. A relation R on X is

antisymmetric if xRy and yRx imply x = y. A relation < on a nonempty set X is a partial ordering of X if it is reflexive, transitive

and antisymmetric. If < is a partial ordering on a set X, the notation x < y means x < y and x 0 y. Moreover, y > x and y > x are just another way to write x < y and x < y, respectively. Thus a partially ordered set is a pair (X, <) where X is a nonempty set and < is a partial ordering of X (i.e., a nonempty set equipped with a partial ordering on it). Warning: It may happen that x j y and y _ x for some

1.5 Ordering

9

(x, y) E X x X. Let (X, <) be a partially ordered set, and let A be a subset of X. Note that (A, <) is a partially ordered set as well. An element x E X is an upper bound for A if y < x for every y E A. Similarly, an element x E X is a lower bound for A if x < y for every y E A. A subset A of X is bounded above in X if it has an upper bound in X. and bounded below in X if it has a lower bound in X. A bounded subset of X is one that is bounded both above and below. If a subset A of a partially ordered set X is bounded above in X and if some upper bound of A belongs to A, then this (unique) element of A is the maximum of A (or the greatest or biggest element of A), denoted by max A. Similarly, if A is bounded below in X and if some lower bound of A belongs to A, then this (unique) element of A is the minimum of A (or the least or smallest element of A), denoted by min A.

An element x E A is maximal in A if there is no element y E A such that x < y (equivalently, if x >E y for every y E A). Similarly, an element x E A is minimal in A if there is no element y E A such that y < x (equivalently, if y tE x for every y E A). Note that x >E y (or y y x) does not mean that y < x (or x < y) so that the concepts of a maximal (or a minimal) element in A and that of the maximum (or the minimum) element of A do not coincide. Example 1A. A collection of many (e.g., two) pairwise disjoint nonempty subsets of a nonempty set, equipped with the partial ordering defined by the inclusion relation C , has no maximum, no minimum, and every element in it is both maximal and minimal. On the other hand, the collection of all infinite subsets of an infinite set, whose complements are also infinite, has no maximal element in the inclusion

ordering C. (The notion of infinite sets will be introduced later in Section 1.8 for instance, the set all even natural numbers is an infinite subset of N which has an infinite complement). Let A be a subset of a partially ordered set X. Let UA C X be the set of all upper bounds of A. and let VA C X be the set of all lower bounds of A. If UA is nonempty and has a minimum element, say u = min UA, then u E UA is called the supremum (or the least upper bound) of A (notation: u = sup A). Similarly. if VA is nonempty and has a maximum, say v = max VA. then v E VA is called the infimum (or the

greatest lower bound) of A (notation: v = inf A). A bounded set may not have a supremum or an infimum. However, if a set A has a maximum (or a minimum), then

sup A = max A (or inf A = min A). Moreover, if a set A has a supremum (or an infimum) in A, then sup A = max A (or inf A = min A). If a pair {x, y) of elements of a partially ordered set X has a supremum or an infimum in X, then we shall use the following notation: x v y = sup{x, y} and x A y = inf {x, y). Let F : X -- Y be a function from a set X to a partially ordered set Y. Thus the range of F, F(X) C Y, is a partially ordered set. An upper bound for F is an upper bound for F(X), and F is bounded above if it has an upper bound. Similarly, a lower bound for F is a lower bound for F(X), and F is bounded below if it has lower bound. If a function F is bounded both above and below, then it is said to be bounded.

The supremum of F, sup.EX F(x), and the infimum of F. infXExF(x), are defined by supXExF(x) = sup F(X) and infXEXF(x) = inf F(X). Now suppose X also is

1. Set-Theoretic Structures

10

partially ordered and take an arbitrary pair of points X1, x2 in X. F is an increasing

function if xi < x2 in X implies F(xi) < F(x2) in Y, and strictly increasing if x1 < x2 in X implies F(xj) < F(x2) in Y. (For notational simplicity we are using the same symbol < to denote both the partial ordering of X and the partial ordering of Y). Note that F is strictly increasing if and only if it is increasing and injective. In a similar way we can define decreasing and strictly decreasing functions between partially ordered sets. If a function is either decreasing or increasing, then it is said to be monotone.

1.6

Lattices

Let X be a partially ordered set. If every pair {x, y} of elements of X is bounded above, then X is a directed set (or the set X is said to be directed upward). If every pair {x, y} is bounded below, then X is said to be directed downward. X is a lattice if every pair of elements of X has a supremum and an infimum in X (i.e., if there exits

a unique u E X and a unique v E X such that u = x v y and v= x A y for every pair x E X and y E X). A nonempty subset A of a lattice X that contains x v y and x A y for every x and y in A is a sublattice of X (and hence a lattice itself). Every lattice is directed both upward and downward. If every bounded subset of X has a supremum and an infimum, then X is a boundedly complete lattice. If every subset of X has a supremum and an infimum, then X is a complete lattice. The following chain of implications complete lattice

boundedly complete lattice

lattice

directed set

is clear enough, and none of them can be reversed. If X is a complete lattice, then X has a supremum and an infimum in X, which actually are the maximum and the minimum of X, respectively. Since min X E X and max X E X. this shows that a complete lattice in fact is nonempty (even if this had not been assumed when we defined a partially ordered set). Likewise, the empty set 0 of a complete lattice X has a supremum and an infimum. Since every element of X is both an upper and

a lower bound for 0, it follows that UQ = Vo = X. Hence sup 0 = min X and inf 0 = max X. Example 1B. The power set P (X) of a set X is a complete lattice in the inclusion

ordering c, where AVB=AUBandAAB=AflBforevery pair (A. B) of subsets of X. In this case, sup 0 = min P(X) = 0 and inf 0 = max P(X) = X. Example 1C. The real line llt with its natural ordering < is a boundedly complete lattice but not a complete lattice (and so is its sublattice Z of all integers). The set A = (.x E R : 0 < x < 1), as a sublattice of (R, <), is a complete lattice where sup 0 = min A = 0 and inf 0 = max A = 1. The set of all rational numbers Q is a sublattice of E (in the natural ordering < ) but not a boundedly complete lattice -

1.6 Lattices

1l

e.g., the set (x E Q : x2 < 2) is bounded in Q but has no infimum and no supremum in Q.

Example 1D. The notion of connectedness needs topology and we shall define it in due course. However, if the reader is already familiar with it. then he can appreciate now a rather simple example of a directed set that is not a lattice (connectedness will be defined in Chapter 3). The subcollection of P(R2) made up of all connected subsets of the Euclidean plane R2 is a directed set in the inclusion ordering a (both upward and downward) but not a lattice.

Lemma 1.1. (Banach-Tarski). An increasing function on a complete lattice has a fired point.

Proof. Let (X, <) be a partially ordered set, consider a function F: X -- X, and set A = (x E X : F(x) < x ). Suppose X is a complete lattice. Then X has a supremum in X (sup X = max X). Since Max X E X, it follows that F(max X) E X so that F(max X) < max X. Conclusion: A is nonempty. Take x E A arbitrary and let a be the infimum of A (a = inf A E X). If F is increasing, then F(a) < F(x) < x because a < x and x E A. Therefore F(a) is a lower bound for A, and hence F(a) < a. Thus a E A. On the other hand, since F(x) < x and since F is increasing, F(F(x)) < F(x).Thus F(x) E A so that F(A) C A, and hence F(a) E A (for a E A), which implies that a = inf A < F(a). Therefore a < F(a) < a. Thus (antisymmetry) F(a) = a. The next theorem is an extremely important result that plays a central role in Section 1.8. Its proof is based on the previous lemma.

Theorem 1.2. (Cantor-Bernstein). If there exist an injective mapping of X into Y and an injective mapping of Y into X, then there exists a one-to-one correspondence between the sets X and Y. Proof. First note that the theorem statement can be translated into the following

problem. Given an injective function from X to Y and also an injective function from Y to X. construct a bijective function from X to Y. Thus consider two functions

F : X --+ Y and G : Y --* X. Let P (X) be the power set of X. For each A E P (X) set

4(A) = X\G(Y\F(A)). It is readily verified that 4): P(X) --> P(X) is an increasing function with respect to the inclusion ordering of P(X). Therefore, by the Banach-Tarski Lemma, it has a fixed point in the complete lattice &'(X). That is, there exists Ao E P(X) such that O(Ao) = Ao. Hence Ao = X\G(Y\F(Ao)) so that

X\Ao = G(Y\F(Ao)).

1. Set-Theoretic Structures

12

Thus X \A0 is included in the range of G. If F : X -+ Y and G : Y - X are both injective, then it is easy to show that the function H : X -> Y, defined by

H(x) is injective and surjective.

J F(x),

G-t (x),

X E AoA,

x E X \ Ao, Cl

If X is a partially ordered set such that for every pair x, y of elements of X either

x < y or y < x, then X is simply ordered (synonyms: linearly ordered, totally ordered). A simply ordered set is also called a chain. Note that, in this particular case, the concepts of maximal element and maximum element (as well as minimal element and minimum element) coincide. It is clear that every simply ordered set is a lattice. For instance, any subset of the real line R (e.g., lit itself or Z) is simply ordered.

Example IE. Let < be a simple ordering on a set X and recall that x < y means x < y and x # y. This defines a transitive relation < on X that satisfies the trichotomy law: for every pair {x, y) in X exactly one of the three statements x < y,

x = y, or y < x is true. Conversely, if < is a transitive relation on a set X that satisfies the trichotomy law, and if a relation < on X is defined by setting x < y whenever either x < y or x = y, then < is a simple ordering on X. Thus, according to the above notation, < is a simple ordering on a set X if and only if < is a transitive relation on X that satisfies the trichotomy law. If X is a partially ordered set such that every nonempty subset of it has a minimum, then X is said to be well-ordered. Every well-ordered set is simply ordered. Example: Any subset of the set No of all nonnegative integers is well-ordered.

1.7

Indexing

Let F be a function from a set X to a set Y. Another way to look at the range of F is: for each x E X set yx = F(x) E Y and note that F(X) = { yx E Y : x E X }, which can also be written as (yx }xEx. Thus the domain X can be thought of as an index set, the range (yx )xEx as a family of elements of Y indexed by an index set X (an indexed

family), and the function F: X -+ Y as an indexing. An indexed family {yx}xEx may contain elements ya and yb, for a and b in X. such that ya = yb. If (y.'),Ex has the property that ya * yb whenever a # b, then it is said to be an indexed family of distinct elements. Observe that {yx}xEx is a family of distinct elements if and only if the function F : X -+ Y (i.e., the indexing process) is injective. The identity mapping on an arbitrary set X can be viewed as an indexing of X, the self-indexing of X. Thus any set X can be thought of as an indexed family (the range of the self-indexing of itself).

1.7 Indexing

13

A mapping of the set N (or No, but not Z) into a set Y is called a sequence (or an infinite sequence). Notations: (y")"EN, {y"}">t, (y,,)', or simply (y"). Thus a Y-valued sequence (or a sequence of elements in Y, or even a sequence in Y) is precisely a function from N to Y, which is commonly thought of as an indexed family (indexed by N) where the indexing process (i.e., the function itself) is often omitted.

The elements y" of {y"} are sometimes referred to as the entries of the sequence [y"). If Y is a subset of the set C, R or Z, then complex-valued sequence, real-valued sequence or integer-valued sequence, respectively, are usual terminologies. Let {Xy)yer be an indexed family of sets. The Cartesian product of (Xy)YEr, denoted by flyer Xy, is the set consisting of all indexed families (xy )yEr such that xy E X y for every y E I'. In particular, if Xy = X for ally E T, where X is a fixed set, then nyErXy is precisely the collection of all functions from r to X. That is,

fl X = Xr.

yer

Recall: Xr denotes the collection of all functions from a set r to a set X. Suppose r = H", where D" = (i E N: i < n) for some n E N (II" is called an initial segment of N). The Cartesian product of {X; );Ej. (or (Xi )" 1), denoted by n;EiX; or f[; X; , is the set X 1 x .. x X. of all ordered n-tuples (xl, ... , x") with x; E Xi for every i E ll". Moreover, if Xi = X for all i E d", then IIjEj.X is the Cartesian product of n copies of X which is denoted by X" (instead of X' n). The n-tuples

(xI, ... , x") in X" are also called finite sequences (as functions from an initial segment of N into X). Accordingly, n"eNX is referred to as the Cartesian product of countably infinite copies of X, which coincides with XN: the set of all X-valued (infinite) sequences. A remarkably useful way of defining an infinite sequence is given by the Principle of Recursive Definition which says that, if F is a function from a nonempty set

X into itself and if x is an arbitrary element of X, then there exists a unique X-valued sequence (x")"EN such that xj = x and x"+l = F(x") for every n E N. The existence of such a unique sequence is intuitively clear, and it can be easily proved by induction (i.e., by using the Principle of Mathematical Induction). A slight generalization reads as follows. For each n E N let G" be a mapping of X" into X. and let x be an arbitrary element of X. Then there exists a unique X-valued sequence (xn)AEN such that xl = x and x"+i = G"(xr,... , x")for every n E N. Since sequences are functions of N (or of No) to a set X, the terms associated with the notion of boundedness clearly apply to sequences in a partially ordered set X. In particular, if X is a partially ordered set, and if (x") is an X-valued sequence, then sup"x" and inf"x" are defined as the supremum and infimum, respectively, of the partially ordered indexed family (x" }. Since N and No (with their natural ordering) are partially ordered sets (well-ordered, really), the terms associated with the property of being monotone also apply to sequences in a partially ordered set X. Let (ZI ),EN be a sequence in a set Z, and let {nk }kEN be a strictly increasing sequence of positive integers (i.e., a strictly increasing sequence in N). If we think of

14

1. Set-Theoretic Structures

{nk } and (Zn) as functions, then the range of the former is a subset of the domain of the latter (i.e., the indexed family Ink }kEN is a subset of N). Thus we may consider

the composition of (zn ) with Ink), say {Z-J' which is again a function of N to Z; that is, (Znk } is a sequence in Z. Moreover, since Ink) is strictly increasing, to each element of the indexed family {Znt )kEN there corresponds a unique element of the indexed family {Zn }nEN. In this case the Z-valued sequence (Z-k} } is called a subsequence of {Zn).

A sequence is a function whose domain is either N or No, but a similar concept could be likewise defined for a function on a well-ordered domain. Even in this case, a function with domain Z (equipped with its natural ordering) would not be a sequence. Now recall the string of (nonreversible) implications well-ordered

simply ordered

lattice

directed set.

This might suggest an extension of the concept of sequence by allowing functions whose domains are directed sets. Anet in a set X is a family of elements of X indexed by a directed set 1'. In other words, if r is a directed set and X is an arbitrary set, then an indexed family (xy )yEr of elements of X indexed by r is called a net in X indexed by r. Examples: Every X -valued sequence (xn) is a net in X. In fact, sequences are prototypes of nets. Every X-valued function on Z (notations: (xk }kEZ (xk )k° _OO or {xk ; k = 0, ± 1, ±2, ... }) is a net (sometimes called double sequences or bisequences, although these nets are not sequences themselves).

1.8

Cardinality

Two sets, say X and Y, are said to be equivalent (denoted by X H Y) if there exists a one-to-one correspondence between them. Clearly (see Problems 1.8 and 1.9), X H X (reflexivity), X H Y if and only if Y H X (symmetry), and X H Z whenever X .-+ Y and Y <- Z for some set Z (transitivity). Thus, if there exists a set upon which H is a relation, then it is an equivalence relation. For instance, if the notion of equivalent sets is restricted to subsets of a given set X. then H is an equivalence relation on the power set P(X). If C = {xy }yEr is an indexed family of distinct elements of a set X indexed by a set r (so that xa # xp for every a A # in I'), then c ea r (the very indexing process sets a one-to-one correspondence between P and C). Let N be the set of all natural numbers and, for each n E N. consider the initial segment Iln = {i E Nl: i < n).

A set X is finite if it is either empty or equivalent to Iln for some n E N. A set is infinite if it is not finite. If X is finite and Y is equivalent to X. then Y is finite. Therefore, if X is infinite and Y is equivalent to X, then Y is infinite. It is easy to show by induction that, for each n E N, Iln has no proper subset equivalent to it. Thus (see Problem 1.12), every finite set has no proper subset equivalent to it. That

1.8 Canlinali y

15

is, if a set has a proper equivalent subset, then it is infinite. Moreover, such a subset must be infinite too (since it is equivalent to an infinite set).

Example IF. N is infinite. Indeed, it is easy to show that No is equivalent to N (the

function F: No -> N such that F(n) = n + 1 for every n E No will do the job). Thus No is infinite, because N is a proper subset of No which is equivalent to it, and so is N. To verify the converse (i.e., to show that every infinite set has a proper equivalent subset) we apply the Axiom of Choice.

Axiom of Choice. If (Xy)yEr is an indexed family of nonempty sets indexed by a nonempty index set r, then there exists an indexed family (xy)ycr such that

x,EXyforeach yEr. Theorem 13. A set is infinite if and only if it has a proper equivalent subset.

Proof. We have already seen that every set with a proper equivalent subset is infinite. To prove the converse take an arbitrary element xo from an infinite set Xo, and an arbitrary k from No. The Principle of Mathematical Induction allows us to construct, for each k E No, a finite family (Xn)Ri of infinite sets as follows. Set be recursively Xt = Xo\{xo} and, for every nonnegative integer n < k, let defined by the formula

Xn+t = Xn\{xn}, where {xn }n.o is a finite set of pairwise distinct elements, each xn being an arbitrary element taken from each Xn. However, if we consider the (infinite) indexed family (Xn }nENp = UkeN0 { Xn )kn, then we need the Axiom of Choice to ensure the existence of the indexed family {xn (nENp = UkENO{xn}n where each xn is arbitrarily taken from each Xn. Now set AO = {xn }nENo C Xo, A = {xn }fEN C Ao, and

X = A U (Xo\Ao) C Ao U (Xo\Ao) = Xo Note that AO *- No and A E- N (since the elements of Ao are distinct). Thus Ao .-> A (because No H N), and hence Xo H X (see Problem 1.20). Conclusion: Any infinite set Xo has a proper equivalent subset (i.e., there exists a proper subset o X of Xo such that Xo + X). If X is a finite set, so that it is equivalent to an initial segment 1. for some natural number n, then we say that its cardinality (or its cardinal number) is n. Thus the cardinality of a finite set X is just the number of elements of X (where, in this case. "numbering" means "indexing" as a finite set may be naturally indexed by an index set In). We shall use the symbol # for cardinality. Thus #In = n, and so #X = n whenever X 4+ B. For infinite sets the concept of cardinal number is a bit more complicated. We shall not define a cardinal number for an infinite set as we did for finite sets (which "number" should it be?) but define the following concept instead.

1. Set-Theoretic Structures

16

Two sets X and Y are said to have the same cardinality if they are equivalent. Thus, to each set X we shall assign a symbol #X, called the cardinal number of X (or the X F-s Y - two cardinality of X) according to the following rule: #X = #Y sets have the same cardinality if and only if they are equivalent; otherwise (i.e., if they are not equivalent) we shall write #X 54 #Y. We say that the cardinality of a set X is less than or equal to the cardinality of a set Y (notation: #X < #Y) if there exists an injective mapping of X into Y (i.e., if there exists a subset Y' of Y such that

#X = #Y'). Equivalently, #X < #Y if there exists a surjective mapping of Y onto X (see Problem 1.6). If #X < #Y and #X i4 #Y, then we shall write #X < #Y.

Theorem 1.4. (Cantor). #X < #9 (X) for every set X.

Proof Consider the function F: X

9(X) defined by F(x) = (x) for every

x E X, which is clearly injective. Thus #X < #9(X). Hence #X < #9(X) if and only if #X 96 #9(X). Suppose #X = #9(X) so that there exists a surjective function G: X -+ 9(X). Consider the set A = {x E X : x ¢ G(x)) in 9(X) and take a E X such that G(a) = A (recall: G is surjective). If a E A. then a 0 G(a) and hence a ¢ A, which is a contradiction. Conclusion 1: a 0 A. On the other hand, if a 0 A, then a E G(a) and hence a E A, which is another contradiction. Conclusion 2: a E A. Therefore (a A and a E A), which is impossible (i.e., which also is a contradiction). Final conclusion: #X 0 #9(X). 11 Let A be a subset of a set X. The characteristic function of the set A is the map XA : X -- (0, 1 } such that 1,

0,

xEA, XEX\A.

It is clear that the correspondence between the subsets of X and their characteristic functions is one-to-one. Hence #9 (X) coincides with the cardinality of the collection of all characteristic functions of subsets of X. More generally, let 2X denote the collection of al I maps of a set X into the set (0, 1) (i.e., set 2X = {0, 1 IX).

Theorem 1.5. #9(X) = #2X for every set X. Proof. Let F be a function from the collection 2X to the power set P (X) that assigns to each map tp : X -+ {0, 1) the inverse image of the singleton (1) under W. That is, consider the function F : 2X -i 9(X) defined by F(cp) = (p-' ({ 1 }) for every p in 2X.

Claim 1. F is surjective. Proof. If A is a subset of X, then the characteristic function of A, XA in 2X. is such

that XA '({1}) = A. Thus F(XA) = A for every A E 9(X). Therefore, 9(X) _ UAE9(X) F(XA) C U,E2x F(p) = ?Z(F). o Claim

2. F is injective.

1.8 Caniinaliry

17

Take cp, }b E 2X. If co # ii, then V (x) 96 *(x) for some x E X. Thus cp(x) = 0 and *(x) = 1 (or vice versa), so that x if 9p-1({1)) and x E Hence'p-t({1}) # so that F((p) 0 F(*). E3 Proof.

Conclusion: F is a one-to-one correspondence between 2X and P(X).

o

Although the next theorem may come as no surprise, it is all but trivial. This actually is a rewriting, in terms of the concept of cardinality, of the rather important Theorem 1.2

Theorem 1.6. (Cantor-Bernstein). Let X and Y be any sets. If #X < #Y and #Y < #X, then #X = #Y. The Cantor-Bernstein Theorem exhibits an antisymmetry property. Note that reflexivity and transitivity are readily verified (see Problem 1.22) which, together with Theorem 1.6, lead to a partial ordering property. Behind the antisymmetry property of Theorem 1.6 there is in fact a simple ordering property. To establish it (Theorem 1.7 below) we shall rely on the following axiom. Zorn's Lemma. Let X be a partially ordered set. If every simply ordered subset of X (i.e., if each chain in X) has an upper bound in X, then X has a maximal element.

The label "Zorn's Lemma" is inappropriate but has already been consecrated in the literature. It should read "Zorn's Axiom" instead, for it really is an axiom equivalent to the Axiom of Choice.

Theorem 1.7. For any two sets X and Y, either #X < #Y or #Y < #X. Proof. Consider two sets X and Y. Let Z be the collection of all injective functions from subsets of X to Y. That is,

T={ F E yA : A C X and F is injective). Recall that (see Problem 1.17), as a subset of F = UAEP(X) YA, I is a partially ordered set in the extension ordering, and every simply ordered subset of it has a supremum in Z. Thus, according to Zorn's Lemma, I contains a maximal function. Let Fo : Ao -> Y be a maximal function of Z, where Ao g X and #Fo(Ao) = #AO (since Fo is injective). Suppose Ao 34 X and Fo(Ao) # Y. Take xo E X\Ao and yO E Y\Fo(Ao), and consider the function F1 : Ao U {xo} -s Y defined by J Fo(x) E Fo(Ao),

Fi (x) = lyoE Y\Fo(Ao),

X E AO,

x=xoE X\AO,

which is injective (because FO is injective and yo 0 Fo(Ao)). Since F1 E I and FO = F1IAO, it follows that FO < F1, which contradicts the fact that FO is a maximal of I (for Fo 0 F1). Hence, either AO = X or FO(AO) = Y. If AO = X, then

18

1. Set-Theoretic Structures

Fo: X -* Y is injective and so #X < #Y. If Fo(Ao) = Y, then #Y = #Fo(Ao) = #Ao < #X (for Ao c X - see Problem 1.21(a)). o We have already seen that N H No. Thus N and No have the same cardinality. It is usual to assign a special symbol (aleph naught) to such a cardinal number: #1V = #No = No. We have also seen (cf. proof of Theorem 1.3) that, if X is an infinite set, then there exists a subset of it, say A. which is equivalent to N. Thus #Nl = #A < #X, and hence tto is the smallest infinite cardinal number in the sense that No < #X for every infinite set X (see Problems 1.21(a) and 1.22). A set X such that #X = to is said to be countably infinite (or denumerable). Therefore, every infinite set has a countably infinite subset. A set that is either finite or countably infinite (i.e., a set X such that #X < so) is said to be countable; otherwise it is said to be uncountable (or uncountably infinite, or nondenumerable). Proposition 1.8. #(X x X) = #X for every countably infinite set X.

Proof. Suppose #X = #N. According to Problems 1.26, 1.23(b) and 1.25(a) we get #X _< #(X x X) < #(N x N) = #N = #X. Hence the identity #X = #(X x X) follows by the Cantor-Bernstein Theorem.

17

Note that #X < #(X x X) for any set X (see Problem 1.26). Moreover, it is easy to show that #X < #(X x X) whenever X is a finite nonempty set. Thus, if a nonempty set X is such that #X = #(X x X), then it is an infinite set. The previous proposition ensured the converse for countably infinite sets. The next theorem (which is another application of Zorn's Lemma) ensures the converse for every infinite set.

Therefore, the identity #X = #(X x X) actually characterizes the infinite sets (of any cardinality) among the nonempty sets. Theorem 1.9. If X is an infinite set, then #X = #(X x X). Proof. First we verify the following auxiliary result.

Claim 0. Let C, D and E be nonempty sets. If #(E x E) = #E, then #(C U D) < #E whenever #C < #E and #D < #E. Proof. The claimed result is a straightforward application of Problems 1.26, 1.23(b)

and 1.22: #(C U D) <#(CxD) <#(ExE)=#E.a Now, back to the theorem statement. Let X be a set and let .7 be the collection of all injective functions from subsets of X to X x X such that the range of each function in J coincides with the Cartesian product of its domain with itself. That is,

,7 = I F E (X X X)" : A C X, F is injective and F(A) = A xA I. Note that ,7 is nonempty (at least the empty function is there). From now on suppose X is infinite. Thus X has a countably infinite subset, and so Proposition 1.8 ensures

the existence of a function in j with infinite (at least countably infinite) domain.

1.8 Cardinaliry

19

Recall that J is a partially ordered set in the extension ordering, and that every chain in J (i.e., every simply ordered subset of 9) has an injective supremum (see Problem 1.17).

Claim 1. Such a supremum in fact lies in J.

Proof. Let (Fy) be an arbitrary chain in J. and let D(Fy) and 1Z(Fy) denote the domain and range of Fy, respectively. Thus each Fy is an injective function from

Ay to XxX, with Ay = D(Fy) c X and Fy(Ay) = AyxAy. Now let VyFy: U, Ay -, X x X be the supremum of (Fy ), and note that (see Problem 1.17)

(V,Fy)(UyAy) = R(VyFy) = U,R(Fy) = U,(AyxAy). Clearly, U,(AyxAy) c (UyAy)x(UyAy). On the other hand, if (Fx, Fu} is an arbitrary pair from (Fy ), then Fx < F,u (or vice versa), so that Ax x A,,, c A, x A4. c UY(AyxAy). Therefore (for Ax c Aµ). Thus (UyAy)x(UyAy)

(VyFy)(UyAy) = (UyAy)x(UyAy),andhence V,Fy E 3.13 Conclusion: Every chain in 3 has an upper bound (a supremum, actually) in J. Thus, according to Zorn's Lemma, 3 contains a maximal element. Let FO: A0 -> X x X be a maximal function of 3, where 0 A Ao C X and Fo(Ao) _ Ao x Ao. Since Fo is injective,

#(AoxAo) = #Ao, which implies that the nonempty set Ao is an infinite set.

Claim 2. If #Ao < #X, then #Ao < #(X\Ao). Proof.

If #(X\Ao) < #Ao, then (cf. Problems 1.26 and 1.23(b)) #X =

#(Ao U X\Ao) < #(Aox(X\Ao)) < #(A0 xA0) = #Ao. Conclusion: #(X\Ao) < #A0 implies #X < #Ao (see Problem 1.22). Equivalently, #Ao < #X implies #A0 < #(X\Ao) by Theorem 1.7. 0

Note that #A0 < #X (for A0 c X). Suppose #Ao < #X. In this case #A0 < #(X\Ao) by Claim 2. Thus there exists a proper subset of X\Ao, say At, such that #A0 = #A 1. Therefore (Problem 1.23(b)) #(A; x Aj) < #(A0 x AO) = #A0 for all combinations of i, j in (0, 1), and hence #[(Ao x A 1) U (A I X AO) U (A i x A 1)] < #Ao according to Claim 0. Since the reverse inequality is trivially verified, it follows by Theorem 1.6 that #[(Ao x A 1) U (A 1 x AO) U (A 1 x A 1)] = #A0. Set

A = AOUAi and observe that (AxA)\(AoxA0) _ (AoxA1) U (A1 xAo) U (A1 xA1) because Ao and At are disjoint. Thus

#[(AxA)\(AoxA0)] = #Ao = #A1,

20

1. Set-Theoretic Structures

which ensures the existence of an injective function F1: A t -> X x X such that F1 (A 1) = (A x A)\ (A0 x A0). Now consider a function F from A to X x X defined as follows.

F(x) _

FO(X) E Ao x Ao,

X E A0,

Fi(x) E (A x A) \ (Ao x A0),

x E At = A \ Ao.

F: A - X x X is injective (because Fo and F1 are injective functions with disjointranges) and F(A) = AxA (for F0(A0) U Ft(A1) = AxA). Since F E .land F0 = FIAO, it follows that Fo < F, which contradicts the fact that F0 is a maximal (for F0 * F). Therefore

of

#Ao = #X, and hence (cf. Problem 1.23(b)) #(Ao x Ao) = #(X x X). Conclusion: #X = #Ao = #(A0 x A0) = #(X x X). E3 Theorem 1.9 is a natural extension of Proposition 1.8 which in turn generalizes Problem 1.25(a). Another important and particularly useful result in this line is given by the next theorem and its corollary.

Theorem 1.10. Let X be a set and consider an indexed family of sets (X y )yEr. If

#X y < #X for all y E r, then

#(U Xy) < #(rxX). yEr

Proof. Take an indexed family of sets (Xy)yEr and suppose there exists a set X such that #Xy < #X for all y E r. Thus for each y E r there exists a surjective mapping of X onto X,,, say, Fy : X -+ X,. Now consider a function G : r x x -+ UyErXydefined by G(y,x) = Fy(x) forevery Y E randeveryx E X.Takeanarbitrary y E Uy e r X y so that y E X y for some y E r. Since Fy : X -> Xy is surjective, there exists x E X such that y = Fy (x). Thus y = G(y, x); that is, y E R(G). < Hence G : r x X --). UyEr Xy is surjective, and so #(UyerXy)

#(r x x).

E3

Corollary 1.11. A countable union of countable sets is countable.

Proof. Consider a countable family of countable sets {Xn)nEN, where #N < #N and #Xn < #N for all n E N. Theorem 1.10 ensures that #(UnENXf) < #(NxN). However, #(N x N) < #(N x N) = #N. Thus

#( U Xn) < #N. nEN

(See Problems 1.23(b), 1.25(a) and 1.22.)

1.9 Remarks

1.9

21

Remarks

We assume the reader is familiar with the definition of an interval of the real line R. An interval of R is nondegenerate if it does not collapse to a singleton. It is easy to show that the cardinality of the real line R is the same as the cardinality of any nondegenerate interval of R. A typical example: The function F: [0, 1] -+ [-1, 1] given by F(x) = 2x - I for every x E [0, 11 is injective and surjective, and so is the

function G : (-1, 1) --> R defined by G(x) _ (I - jxl)-lx for every x E (-1, 1). Thus #[0, 11 _ #[-1, 1] and #(-1, 1) = #R. Since (-1, 1) C [-1, 1] C R, it follows that #R = #(- 1, 1) < #[-1, 11 < #R, and hence (Cantor-Bernstein Theorem) #[ -1, 11 _ #R. Therefore,

#[0, 1] = #(-1, 1) = #[-1, 1] = #R. Moreover, we can also prove that #R = #2N. Indeed, consider the function F: 2N -- [0, 1] that assigns to each sequence {a") in 2N (i.e., to each sequence with values either 0 or 1) a real number in [0, 1], in ternary expansion, as follows.

F({a"}) = 0.(2a1)(2a2)

...

for every {a" } E 2N. It can be shown that F is injective.

Reason: Every real number x E [0, 11 can be written as En°_1 fl p-" for a given positive integer p greater than one (i.e., for a given base p). In this case 0.f1 $2 ... is a p-ary expansion of x, where (P")"EN is a sequence of nonnegative integers ranging from 0 to p - 1. That is, {8" )nEN is a sequence of digits with respect to the base p- e.g., if p = 2, 3, or 10, then 0.fi1 #2 ... is a binary, ternary or denary (i.e., decimal) expansion of x, respectively. A p-ary expansion (with respect to a base p)

is not unique - e.g., 0.499 ... and 0.500... are decimal expansions for x =. However, if two p-ary expansions of x differ, then the absolute difference between the first digits in which they differ is equal to 1. Thus, if we take a p-ary expansion whose digits are either 0 or 2 (as we did), then it is unique. This can only be done for a p-ary expansion with respect to a base p > 3, so that a ternary expansion is enough.

Thus #2N < #[0, 11. On the other hand, let G : 2N -- [0, 11 be the function that assigns to each sequence {a" } in 2N a real number in [0, 11, in binary expansion, as follows.

G({an}) = 0.a1a2

...

for every {a,,) E P. It can also be shown that G is surjective.

Reason: Every real number x E [0, 11 can be written asF01a"2' so that0.a1a2 is a binary expansion of it for some sequence {a") E 2N.

...

Thus #[0, 11 < #2N. Therefore #[0, 11 = #2N by the Cantor-Bernstein Theorem. Hence

#R=#2N

22

1. Set-Theoretic Structures

(for #[0, 1

#R). Using Theorems 1.4 and 1.5 we may conclude that

#N < #R. Such a fundamental result can also be derived by the celebrated Cantor's diagonal procedure as follows. Clearly, #N < #R (since N C R). Suppose #N = #R. This implies that #N = #[0, 1] (for #[O, 1] = #R). Thus the interval [0, 11 can be indexed Write each x in decimal expansion: by N so that [0, 1] = Xn =

where each a,, (k E N) is a nonnegative integer between 0 and 9. Now consider the point x E [0, 11 with the following decimal expansion.

x = 0.ata2 ... where, again, each a (n E N) is a nonnegative integer between 0 and 9 but al 96 a 11,

a2 * a221 and so on. That is, a,, 0 an for each n E N (e.g., take at. diametrically opposite to a,,,, so that, for each n E N. a = a,,,, + 5 if 0 < a,,. < 4 or a,, = a,,,, - 5 if 5 < a,,,, < 9). Thus x 0 x for every n E N. Hence x f [0, 1], which is a contradiction. Therefore #N 96 #R. Equivalently (since #N < #R), #N < #R. We have denoted #N by No. Let us now denote #2N by 2° so that

#N = Ko < 2K0 = #R. Cantor conjectured in 1878 that there is no cardinal number between No and 2K0. The conjecture is called the Continuum Hypothesis.

Continuum Hypothesis. There is no set whose cardinality is greater than #N and smaller than #R. The Generalized Continuum Hypothesis is the conjecture that naturally generalizes the Continuum Hypothesis.

Generalized Continuum Hypothesis. For any infinite set X, there is no cardinal number between #X and #2x. There are several different axiomatic set theories, each based on a somewhat different axiom system. The most popular is probably the axiom system ZFC. It comprises the axiom system ZF ("Z" for Zermelo and "F" for Fraenkel) plus the Axiom of Choice. The Axiom of Choice actually is a genuine axiom to be added to ZF. Indeed, Godel proved in 1939 that the Axiom of Choice is consistent with ZF, and Cohen proved in 1963 that the Axiom of Choice is independent of ZF. The situation of the Continuum Hypothesis with respect to ZFCis somewhat similar to that of the Axiom of Choice with respect to ZF, although the Continuum Hypothesis itself is not as primitive as the Axiom of Choice (even if the Axiom of Choice might be regarded

1.9 Remarks

23

as not primitive enough). Gbdel proved in 1939 that the Generalized Continuum Hypothesis is consistent with ZFC, and Cohen proved in 1963 that the denial of the Continuum Hypothesis also is consistent with ZFC. Thus both the Continuum Hypothesis and the Generalized Continuum Hypothesis are consistent with ZFC and also independent of ZFC: neither of them can be proved or disproved on the basis of ZFC alone (i.e., they are undecidable statements in ZFC). The Generalized Continuum Hypothesis in fact is stronger than the Axiom of Choice: Sierpinski showed in 1947 that the Generalized Continuum Hypothesis implies the Axiom of Choice. We have already observed that the Axiom of Choice and Zorn's Lemma are equivalent. There is a myriad of axioms equivalent to the Axiom of Choice. Let us mention just two of them.

Hausdorff Maximal Principle. Every partially ordered set contains a maximal chain (i.e., a maximal simply ordered subset).

Zermelo Well-Ordering Principle. Every set may be well-ordered. In particular, the set R of all reals may be well-ordered. This is a pure existence result, not exhibiting (or constructing or even defining) a well-ordering of R. In fact, Feferman showed in 1965 that no defined partial ordering can be proved in ZFC to well-order the set R. If X and Y are any sets, properly well-ordered, and if there exists a one-to-one

order-preserving correspondence between them (i.e., an injective and surjective

mapping 4>: X -+ Y such that xt < x2 in (X, <) if and only if 4)(xt) < 0(x2) in (Y, <)), then X and Y are said to have the same ordinal number. If two wellordered sets have the same ordinal number, then they have the same cardinal number. However, unlike the notion of cardinal number, the notion of ordinal number depends on the well-orderings that well-order the sets.

Proposition 1.12. There is an uncountable set X, well-ordered by a relation < on it, with the following properties. X has a greatest element S2 and the set {x E X : x < z) is countable for every z in X \ (9). Proof. Let Y be an uncountable set. By the Well-Ordering Principle, there exists a well-ordering of Y. Take C not in Y, set Z = Y U and extend the well-ordering of Y to Z by setting y < C for every y E Y. Consider the set A = lot E Z : {z E Z : z < Of } is an uncountable set). A is nonempty ( E A because Y = (Z E Z : z < (' ) is uncountable), and hence it has a minimum (for 0 # A C Z and Z is well-ordered). Set 0 = min A so that X = {z E Z : z < S2} is the required set. 17 Moreover, it can be shown that such a well-ordered set X is unique in the sense that, if Y is any well-ordered set with the same properties of X, then there exists a one-to-one order-preserving correspondence between X and Y (i.e., then X and Y have the same ordinal number). The greatest (or the last) element S2 in X is called the least or first uncountable ordinal, and the elements x of X such that x < 92

24

1. Set-Theoretic Structures

are called countable ordinals. The greatest elements of the finite subsets of X are called finite ordinals. If w is the first infinite ordinal (i.e., the least nonfinite ordinal), then the set (x E X : x < w) of all finite ordinals and the set of all natural numbers N (equipped with its natural ordering) have the same ordinal number. It is usual to assign the symbol w as the ordinal number of any well-ordered set that is in a one-to-one order-preserving correspondence with N.

Suggested Reading Binmore [1] Brown and Pearcy [2]

Kolmogorov and Fomin [1] Moore [1] Royden [1] Simmons [1] Suppes [I ]

Crossley et al. [ 1 ]

Dugundji [1] Fraenkel, Bar-Hillel and Levy [I ] Halmos [3] Kelley [1)

Vaught [ 1]

Wilder [1]

Problems Problem 1.1. Let A, B and C be arbitrary sets. Prove that (a) (A\B) U (B\A) = (A U B)\(A n B)

(symmetric difference);

(b) (AVB)U(BVC)=(AUBUC)\(AnBnC); (c) X\(A U B) = (X\A) n (X\B) and X\(A n B) = (X\A) U (X\B) whenever A and B are subsets of X

(De Morgan laws).

Problem 1.2. Consider a function F : X Y from a set X to a set Y. Let A, A 1 and A2 be arbitrary subsets of X, and let B. B1 and B2 be arbitrary subsets of Y. Verify the following propositions.

(a) F(X\A) = F(X)\F(A).

(b) F-(Y\B) = X\F"1(B) (c) A t C A 2

c B2

F(A ) c F(A 2 ) . 1

F-' ( B I) c F-

(d)

B,

(e)

F(A1 U A2) = F(A1) U F(A2).

(f)

F(A1 n A2) c F(A1) n F(A2).

( B 2).

Problems (g)

25

F-' (Bt U B2) = F-' (Bt) O F-' (B2).

(h) F- (BI n B2) = F-'(Bt) n F- I (B2). (i)

A c F-' (F(A)).

(j)

F(F-'(B)) C B.

Problem 1.3. Consider the setup of Problem 1.2. Show that

(a) F is injective if and only if the inverse image under F of each singleton in 'R(F) is a singleton in X;

(b) F is injective if and only if F(A1 n A2) = F(A1) n F(A2) for every At, A2 C X; (c) F is injective if and only if the images of disjoint sets in X are disjoint in Y;

(d) F is injective if and only if A = F-' (F(A)) for every A C- X; (e) F is surjective if and only if the inverse image under F of each nonempty subset of Y is a nonempty subset of X ; (f)

F is surjective if and only if F(F-' (B)) = B for every B C Y.

Problem 1.4. Verify that a function F: X -- X is idempotent if and only if the range of F coincides with the set of all fixed points of F.

Problem 1.5. A function L : Y -+ X is said to be a left inverse of a function F: X -+ Y if LF = Ix, the identity on X. F is injective if and only if it has a left inverse. The restriction of a left inverse of F to the range of F is unique and coincides

with the inverse of F on R(F). (Recall: an injective function has an inverse on its range.) Prove.

Problem 1.6. A function R : Y -> X is said to be a right inverse of a function F: X - Y if FR = ly, the identity on Y. Show that F is surjective if and only if it has a right inverse. Note that any right inverse of F is injective (for it has a left inverse). Similarly, any left inverse of F is surjective (for it has a right inverse). Conclusion: There exists an injective mapping of X into Y if and only if there exists a surjective mapping of Y onto X. Problem 1.7. A function F : X --* Y is injective and surjective if and only if there exists a function G: Y --+ X such that GF = IX (the identity on X) and FG = ly (the identity on Y). Prove the above proposition and show that, in this case, such a function G is unique and coincides with the inverse of F.

26

1. Set-Theoretic Structures

Problem 1.8. If F: X -+ Y is an invertible function, then so is its inverse F't : Y - X and (F-1)-' = F. Moreover, if A C X, then F I A : A -+ F(A) is invertible

and (FIB)-' = F-tlF(A) Prove. Problem 1.9. Verify the following propositions. (a) The composition of two injective functions is an injective function. (b) The composition of two surjective functions is a surjective function. (c) The composition of two invertible functions is an invertible function.

Note: When we speak of the composition G o F of two functions F and G it is assumed that the domain of G includes the range of F.

Problem 1.10. Let F : X - Y and G : Y - Z be invertible mappings. Show that

(Go F)-1 =F-1oG-'. Problem 1.11. A function F : X --* X is an involution if F2 = 1. Verify that an involution is precisely an invertible function on X that coincides with its inverse: F = F-1. Show that the composition of two involutions is again an involution if and only if they commute.

Problem 1.12. Let F : X - Y be a one-to-one mapping of a set X onto a set Y. Let G : X --j- A be a one-to-one mapping of X onto a subset A of X. Prove: If A is a proper subset of X, then F(A) is a proper subset of Y. Hint: Consider the commutative diagram

X t- Y i

GI

1"

A FIA

F(A).

Problem 1.13. Let X be a set with more than one element, and consider the following relations R1, R2, R3. R4 and R5 on the power set P(X) of X. For every pair (A, B) of subsets of X

A RI B if A V B= 0

(i.e., if A= B),

A R2 B if AV B is finite,

A R3 B if ACB or B C A A R4 B if ACB

(i.e., if A\B = 0 or B\A = 0),

(i.e., if A\B = 0).

Problems

27

Show that the table below properly classifies these relations according to reflexivity, transitivity, symmetry and antisymmetry.

Reflexive Rt

Transitive

Symmetric

Antisymmetric

,/

R2 R3 R4 R5

Hint: To verify that R2 is transitive use Problem 1.1(b) and recall that the union of two sets is finite if and only if each of them is finite.

Problem 1.14. Consider the functions 4) : P (X) -> P(X) and H : X -s Y defined in the proof of Theorem 1.2. Show that (D is an increasing function with respect to the inclusion ordering of P(X), and that H is injective and surjective. Problem 1.15. Let Yx denote the collection of all functions from a set X to a set Y. Suppose Y is partially ordered and let < be a partial ordering of Y. Now consider a relation on Yx, also denoted by < and defined as follows. For any pair IF, G) of functions in Yx,

F
if

F(x) < G(x) for every x E X.

(a) Show that < is a partial ordering of Yx. (b) Prove: If (Y, <) is a lattice, then (Yx, <) is a lattice.

Hint: Suppose (Y, <) is a lattice. Take F and G from Yx and let U and V be functions in Yx defined as follows. U(x) = F(x) v G(x) and V(x) _ F(x) A G(x) for every X E X. Show that FVG = Uand FAG = V. Problem 1.16. Set Y = (0, 1) and let XA : X -+ (0, 1) be the characteristic function of an arbitrary subset A of a set X. Thus, for every A E 9(X), XA E Yx = 2x Let

A and B be subsets of X, and consider the partial ordering of 2x introduced in Problem 1.15. Prove the following propositions. XA < XB

ACB.

XA V XB = XAUB-

XA A XB = XAnB = XAXB -

28

1. Set-Theoretic Structures

(d) A n B = 0

XAU8 = XA + X8

XAnB = 0

Problem 1.17. Let.F be the collection of all functions from subsets of a set X to a set Y. That is,

F= {FEYA: ACX) = U YA. AeP(X)

The unique function whose domain is the empty set 0 is called the empty function in F. Consider the following relation < on Y. For any pair IF, G) of functions in F. F < G if F is a restriction of G (equivalently, if G is an extension of F). That is, F G if and only if F: A Y, G: B - Y, A c_ B C X. and F = GIA (i.e.,

F(x) = G(x) for every x E A).

(a) Show that the relation < on F is a partial ordering (called the extension ordering).

A function V : C -+ Y in.F is a lower bound for a pair of functions F : A -+ Y and

CCB,CcAand .F is an upper bound for the pair IF, G I if A c D, B C D. U IA = F and U I B = G.

(b) Show that every pair of functions F and G in ,F has an infimum F A G on .F. (In particular, if the domain A of F and the domain B of G are disjoint, then F A G is the empty function - which function is F A G if A and B are not disjoint but F(A) and G(B) are disjoint?) Let (Fy) be an indexed family of functions in Y. For each Fy let D(Fy) and R(Fy) denote the domain and range of Fy, respectively. Prove the following propositions. (c) If (Fy) is simply ordered (i.e., if (FyI is a chain in ,F), then it has a supre-

mum Vy Fy in.F, whose domain and range are D(VY F),) = UYD(Fy) and R(Vy Fy) = UY R(Fy ). Moreover, if each Fy is injective, then so is Vr Fy.

(d) If the domains (D(Fy)} are pairwise disjoint, then the family (Fy) has a supremum VVFy: UYD(Fy) -+ UYR(Fy) in F. Moreover, if the ranges (7.(Fy)) are also pairwise disjoint, and if each F. is injective, then so is VY Fy.

Problem 1.18. Let {X }FEN be a sequence of sets. Set YI = X I and n+1

n

Yn+1 = (UXk)\(UXk) k=1

k=1

for each n E N. Show by induction that n

n

UYk=UXk k=1

k=1

Problems

29

for every n E N. (Hint: A U (B\A) = A U B.) Verify that Y. c Xn for each n E N, and Yn, n Yn = 0 for every pair of distinct natural numbers m and n. Moreover, show that 00

00

U Yn = U X1. n=1

n=I

Such a disjoint sequence (Yn}nEN is referred to as the disjointification of (Xn)nEN.

Problem 1.19. Let &'(X) be the power set of a set X and consider the inclusion ordering of P(X). Let (Xn}O° be an arbitrary 6'(X)-valued sequence (i.e., a sequence of subsets of a given set X). Recall that, for each n E N, (Xk) , is an indexed family of subsets of X, and hence a subset of the complete lattice P(X). 1

Let infn
n

and sup(Xk )k° n, respectively; and

00

Yn=inf Xk=nXk n
n
k=n

so that (Yn)O_1 and (Zn)O

(a) Verify that (Yn) O sequence.

00

Zn=SUpXk=UXk

and

1

1

k=n

are P(X)-valued sequences as well.

is an increasing sequence and (Zn)O° 1 is a decreasing

The union UO° Y,, and the intersection n0° I Zn, which are elements of usually denoted by 1

00

are

00

and n Zn = lim sup Xn,

U Yn = lim inf Xn n=1

n=1

n

n

and called limit inferior and limit superior of (Xn)n'

1,

respectively.

(b) Show that lim inf X. c lim sup Xn. n

n

If lim infnXn = lim sup,, Xn, then we say that the sequence (X.)O 1 converges to the limit lim Xn = lim inf X,, = lim sup Xn. n

n

n

Prove the following propositions. (c) If (Xn)O°1 is an increasing sequence, then Yn = Xn for each n E N and

Zn = UO° 1 Xn, = supn X. for every n E N. so that

lim inf Xn = lim sup Xn = sup X, . n

n

n

30

1. Set-Theoretic Structures

(d) If { Xn }O01 is itself decreasing, then Y. = nm t X. = inf m Xm for every n e N and Zn = Xn for each n E N, so that

lim inf X,, = lim sup X. = inf Xn. n

n

n

Therefore, an increasing sequence of sets converges to its union (sups X. = Un X.) and, dually, a decreasing sequence of sets converges to its intersection (inf, X,, = nAX,); so that every monotone sequence of sets converges. Thus, since (YY)°O and (Z,,) 1 are always increasing and decreasing, respectively, they do converge: { Yn }O0 converges to its union Y,, = Uz t n°__n Xk, and {Z,,1 converges to its intersection nn° 1 Zn = Uk .n Xk. 1

1

1

nn0=1

(e) Verify the following identities.

lim inf X,, = lim ( inf Xk) n

lim sup Xn = lira ( sup Xk).

and

n
n

n

n
n

(f) Now show that

X \1im sup Xn = lim inf (X \X.) and X \Iim inf X. = lim sup(X\X,). n

n

n

n

Thus a sequence (Xn )n_1 converges if and only if the sequence of its complements {X\Xn }O0 , converges and, in this case,

lim(X\Xn) = X\ Iim Xn. n

n

Problem 1.20. Let A and B be subsets of the sets X and Y, respectively. Show that

A H B and X\A .-> Y\B

imply

X H Y.

(Warning: X H Y and A H B do not imply X\A H Y\B). Problem 1.21. Let A and B be any sets. Prove that

#A < #B

(a)

ACB

(b)

A c B, B is finite, and #A = #B

implies

(hint: inclusion map), imply

A = B;

and show that assertion (b) does not hold if B is infinite.

Problem 1.22. For any sets A, B and C, verify that

#A < #B < #C

implies

#A < #C.

Problems

31

Moreover, if #A < #B < #C or #A < #B < #C, then #A < #C. (Hint: CantorBernstein Theorem.)

Problem 1.23. Suppose the sets A, B, C and D are such that

#A < #C

#B < #D.

and

Prove the following propositions.

(a) #(A U B) < #(C U D) whenever c n D = 0. (b) #(A x B) < #(C x D). Moreover, if #A = #C and #B = #D, then the cardinalities in (b) coincide too.

Problem 1.24. Let YX denote the collection of all mappings of a set X into a set Y. Show that, if Y has more than one element, then

#P(X) < #YX for every set X. (In fact, #P(X) = #2x = #YX if X is infinite and 2 < #Y < #X). Problem 1.25. Let N, Z and Q have their standard meanings and set No = #N as usual. Verify the following identities.

(a) #(N x N) = Mo. Hint: Consider the function F : N x N -s N defined by the formula F(m, n) _ (m+n-1 m+n-2 + m for every m, n E N. F is injective and surjective. The array

n? 5

11

4

7 12

3

4

2

2

1

1

8 5 3

13

9 14

6 10 15

m

12 3 4 5 -+ may be suggestive.

(b) #Z = No. Hint: Let Ne denote the set of all nonnegative even integers (including zero), and let No denote the set of all positive odd integers. Recall that #No = #N, Ne and note that #No = #Ne and #N = #N,,. (Reason: the functions F : No

and G : N - No, given by F(n) = 2n for every n E No and G(n) = 2n - 1 for every n E N, are injective and surjective). Set N. = {k E Z: - k E N) so that #N_ = #N,,. Use Problem 1.23(a) to show that #Z < #N.

32

1. Set-Theoretic Structures

(c) #Q = No. Hint: The function F: ZxN -+ Q defined by F(k, n) = A for every k E Z and every n E N is surjective. Use Problem 1.23(b). Problem 1.26. The purpose of this problem is to prove that, if X and Y are nonempty sets, then

#(X U Y) < #(X xY). (a) First verify that the above assertion holds whenever X and Y are both (nonempty) finite sets. Consider the relations ^-x and -y on the Cartesian product X x Y defined as follows. For every pair of pairs (x1, yl) and (x2, Y2) in X x Y, (xI , yi) ^-x (xZ, y2)

xi = x2,

(xl, yl) -.y (x2, y2)

yl = ),-2.

(b) Show that -x and -y are both equivalence relations on X x Y. Now take x E X and y E Y arbitrary and consider the equivalence classes, [x j c X x Y

and [y] c X x Y, of the point (x, y) E X x Y with respect to ^-x and --y, respectively.

(c) Show that [x] H Y and [y] H X. Hint: Y

[x]

y

[y]

X x

Next suppose one of the sets X or Y, say X, is infinite and consider the singleton {(x, y)) on (x, y) E X X Y.

(d) Verify that #X = #([y)\((x, y))) and #Y = #[x]. (e) Finally, apply the results of Problems 1.21(a), 1.22 and 1.23(a) to conclude that #(X U Y) < #(X x Y). Problem 1.27. Prove the following propositions.

Problems

33

(a) The union of two sets is countable if and only if each of them is countable. Hint: Problems 1.22, 1.23, 1.25 and 1.26. (b) Let X be an infinite set, and let A and B be arbitrary subsets of X. The relation

- on P(X) defined by A-B

AVB is countable

if

is an equivalence relation - compare with Problem 1.13.

Problem 1.28. (a) Let E be an infinite set. Suppose A and B are sets such that

#A < #E

and

#B < #E.

The following propositions, which are naturally linked to Problems 1.23 and 1.26, are in fact corollaries of Theorem 1.9. Prove them.

#(A U B) < #E

and

#(A x B) < #E.

(b) Now use Problem 1.23(b) and Theorem 1.9 to show by induction that, if X is an infinite set, then

#X" = #X for every n E N. Problem 1.29. Let I be the collection of all injective functions from subsets of a set X to itself. That is, set

Z= IF E XA : A C X and F is injective?. We know from Problem 1.17 that, as a subset of UAEP(X)XA, T is partially ordered

in the extension ordering. Let ,7 be the collection of all those functions F in I for which the range R(F) is disjoint with the domain D(F):

.7 = IF E Z: R(F) c X\D(F)}. Problem 1.17 also tells us that every chain { Fy ) in ,7 has a supremum Vy Fy in I.

and also that D('s/>Fy) = UYD(Fy) and R(VYFy) = UYR(Fy). (a) Show that V. Fy in fact lies in J.

Hint: Take Fa and F,L arbitrary from (Fy } c J so that Fa < F. (or vice

versa). Note that R(F,) fl D(F,) c R(Fµ) fl D(F,) = 0, and conclude that UYR(Fy) fl UYD(Fy) is empty too. Therefore, every chain in .7 has an upper bound (a supremum, actually) in J. Thus, according to Zorn's Lemma, ,7 contains a maximal function. Let FO be a maximal function of ,7 and let Ao be the domain of Fo, so that Fo(Ao) c X\Ao. Suppose

#Ao < #(X\Ao) and show that, if X is an infinite set, then

34

1. Set-Theoretic Structures

(b) there exist two distinct points, say xo and xt, in (X\Ao)\Fo(Ao) Hint: Under the above assumption #Fo(Ao) < #(X \Ao) (why?) and X\Ao is infinite - recall: the union of finite sets is finite.

Now set A i = Ao U {xo} and consider the function Ft : A t - X defined by the formula

Ft (x) =

X E Ao,

Fo(x) E Fo(Ao), X1 E X \ Fo(Ao),

x = xo E X \ Ao.

(c) Show that F1 E J. Since F0 = F, I& it follows that F0 < F1, which contradicts the fact that Fo is a maximal of 9 (for F0 A Fi ). Therefore (by Theorem 1.7),

#(X\Ao) < #Ao. Next verify that #A0 = #Fo(Ao) < #(X\Ao) and conclude (by the Cantor-Bernstein Theorem) that

#Ao = #(X\Ao). Finally, by using Problem 1.28(a), show that

#Ao = #X. Outcome: If X is an infinite set, then there exists a subset of it, say Ao, such that

#Ao = #X = #(X\Ao). Problem 1.30. Let A be an arbitrary set and write A' = A x { A).

(a) Show that #A'= #A. Hint: The function that assigns to each a E A the ordered pair (a, A) in the Cartesian product A x (A) is a one-to-one mapping of A onto A'.

Clearly, A n A' = 0. Moreover, if B is any set such that B i6 A. then A' n B' = 0 where B' = B x { B }. Conclusion 1: If A and B are any sets, then there exist sets C and D such that

#A=#C, #B=#D

and

CnD=O.

Now suppose C1, C2, DI and D2 are sets with the following properties. Ci n Dt = 0,

C2 n D2 = 0, #Ct = #C2 and #Dt = #D2. (b) Show that #(Ci U Di) = #(C2 U D2).

Problems

35

Conclusion 2: #(C U D) is independent of the particular pair of sets {C, D) employed in Conclusion 1. We are now prepared to define the sum of cardinal numbers. If A and B are sets, then

#A + #B = #(CUD) for any pair of sets {C, D) such that

#A=#C, #B=#D

and

CnD=O.

In particular, ifAnB=0, then #A+#B= #(AUB). (c) Use Problem 1.29 to show that #X + #X = #X for every infinite set X. The definition of product of cardinal numbers is much simpler:

#A #B = #(A x B) for any sets A and B. According to Theorem 1.9. #X #X = #X for every infinite set X.

(d) Prove: If X and Y are two sets, at least one of which is infinite, then

#X + #Y = #X #Y = max(#X, #Y).

Hint: If #Y < #X, then #X < #X + #Y < #X + #X and #X < #X #Y <

#X #X.

2 Algebraic Structures

The main algebraic structure that will be involved with the subject of this book is that of a "linear space" (or "vector space"). A linear space is a set endowed with an extra structure in addition to its set-theoretic structure (i.e., an extra structure that goes beyond the notions of functions and ordering, for instance). Roughly speaking, linear spaces are sets where two operations, called "addition" and "scalar multiplication", are properly defined so that we can refer to the "sum" of two points in a linear space, as well as to the "product" of a point in it by a "scalar". Although the reader is supposed to have already had some contact with linear algebra and, in particular, with "finite-dimensional vector spaces", we shall proceed from the very beginning. Our approach avoids the parochially "finite-dimensional" constructions (whenever this is possible), and focuses either on general results that do not depend on the "dimensionality" of the linear space, or on abstract "infinite-dimensional" linear spaces.

2.1

Linear Spaces

A binary operation on a set X is a mapping of X x X into X. If F is a function from X x X to X, then we generally write z = F(x, y) to indicate that z in X is the value of F at the point (x, y) in X x X. However, to emphasize the rule of the binary operation (the outcome of a binary operation on two points of X is again a point of X). it is convenient (and usual) to adopt a different notation. Moreover, in order to emphasize the abstract character of a binary operation, it is also common to use

38

2. Algebraic Structures

a noncommittal symbol to denote it. Thus, if * is a binary operation on X (so that * : X x X -+ X), then we shall write z = x * y instead of z = *(x, y) to indicate that z in X is the value of * at the point (x, y) in X x X. If a binary operation * on X has the property that

x*(y*z) = (x*y)*z

for every x, y and z in X, then it is said to be associative. In this case we shall drop the parentheses and write x * y * z. If there exists an element e in X such that

x*e = e*x = x for every x E X, then e is said to be the neutral element (or the identity element) with respect to the binary operation * on X. It is easy to show that, if a binary operation * has a neutral element e, then e is unique. If an associative binary operation * on X

has a neutral element e in X, and if for some x E X there exists x-t E X such that

x *x - = x-t *x = e, I

then x-1 is called the inverse of x with respect to *. It is also easy to show that. if the inverse of x exists with respect to an associative binary operation *, then it is unique. A group is a nonempty set X on which is defined a binary operation * such that (a) * is associative,

(b) * has a neutral element in X, and (c) every x in X has an inverse in X with respect to *.

If a binary operation * on X has the property that

x*y = y*x for every x and y in X, then it is said to be commutative. If X is a group with respect to a binary operation *, and if (d) * is commutative,

then X is said to be an Abelian (or commutative) group.

Example 2A. The collection of all injective mappings of a nonempty set X onto itself (i.e., the collection of all invertible mappings on X) is a non-Abelian group with respect to the composition operation o. The neutral element (or the identity element) of such a group is, of course, the identity map on X. An additive Abelian group is an Abelian group X for which the underlying binary operation is interpreted as an addition and denoted by + (instead of *). In this case

the element x + y (which lies in X for every x and y in X) is called the sum of

2.1 Linear Spaces

39

x and y. The (unique) neutral element with respect to addition is denoted by 0 (instead of e) and called zero. The (unique) inverse of x under addition is denoted by -x (instead of x-t) and called the negative of x. Thus x + 0 = 0 + x = x and x + (-x) = (-x) + x = 0 for every x E X. Moreover, the operation of subtraction is defined by x - y = x + (-y) and x - y is called the difference between x and Y. If o : X x X -> X is another binary operation on X, and if

xo(y*z)=(xoy)*(xoz)

and

(y*z)ox=(yox)*(zox)

for every x, y and z in X, then o is said to be distributive with respect to *. The above properties are the so-called distributive laws. A ring is an additive Abelian group X with a second binary operation on X, called multiplication and denoted by , such that

(e) the multiplication operation is associative and (f) distributive with respect to the addition operation.

In this case the element x y (which lies in X for every x and y in X) is called the product of x and y (alternative notation: x y instead of x y). A commutative ring is a ring for which (g) the multiplication operation is commutative. A ring with identity is a ring X such that

(h) the multiplication operation has a neutral element in X. In this case such a (unique) neutral element in X with respect to the multiplication operation is denoted by I (so that x 1 = 1 x = x for every x E X) and called the identity.

Example 2B. The power set P (X) of a nonempty set X is a commutative ring with identity if addition is interpreted as symmetric difference (or Boolean sum) and multiplication as intersection (i.e., A + B = AV B and A A. B = A fl B for all subsets A and B of X). Here the neutral element under addition (i.e., the zero) is the empty set 0, and the neutral element under multiplication (i.e., the identity) is X itself.

A ring with identity is nontrivial if it has another element besides the identity. (The set {0} with the operations 0 + 0 = 0. 0 = 0 is the trivial ring whose only element is the identity). If a ring with identity is nontrivial, then the neutral element

under addition and the neutral element under multiplication never coincide (i.e.,

0 # 1). In fact, x 0 = 0 x = 0 for every x in X whenever X is a ring (with or without identity). Incidentally (or not) this also shows that, in a nontrivial ring with identity, zero has no inverse with respect to the multiplication operation (i.e., there

is no x in X such that 0 x = x 0 = 1). A ring X with identity is called a division ring if

40

2. Algebraic Structures

(i) each nonzero x in X has an inverse in X with respect to the multiplication operation.

That is, if x 34 0 in X. then there exists a (unique) x-t E X such that x x-t =

x-1.x=1.

Example 2C. Let the addition and multiplication operations have their ordinary ("numerical") meanings. The set of all natural numbers N is not a group under addition; neither is the set of all nonnegative integers No. However, the set of all integers Z is a commutative ring with identity, but not a division ring. The sets Q. R and C (of all rational, real and complex numbers, respectively), when equipped with their respective operations of addition and multiplication, are all commutative division rings. These are infinite commutative division rings but there

are finite commutative division rings (e.g., if we declare that 1 + 1 = 0 + 0 = 0,

0+1 = 1+0= 1.0.0=0.1 = 1.0=0.and1.1 = 1, then (0, 1) is a commutative division ring).

Roughly speaking, commutative division rings are the number systems of mathematics, and so they deserve a name of their own. Afield is a nontrivial commutative division ring. The elements of a field are usually called scalars. We shall be particularly concerned with the fields R and C (the real field and the complex field). An arbitrary field will be denoted by F. Summing up: A field IF is a set with more than one element (at least 0 and 1 are distinct elements of it) equipped with two binary operations, called addition and multiplication, that satisfy all the properties (a) to (i) - clearly, with * replaced by + in properties (a) to (d).

Definition 2.1. A linear space (or vector space) over a field F is a nonempty set X (whose elements are called vectors) satisfying the following axioms. Vector Addition. X is an additive Abelian group under a binary operation + called vector addition. Scalar Multiplication. There is given a mapping of IFx X into X that assigns to each scalar a in IF and each vector x in X a vector ax in X. Such a mapping defines an operation, called scalar multiplication, with the following properties. For all scalars

a and 6 in F, and all vectors x and y in X.

lx a(flx)

= x,

= (a fl)x, a(x+y) = ax+ay. (a+f)x = ax +fix. Some remarks on notation and terminology. The underlying set of a linear space is the nonempty set upon which the linear space is built. We shall use the same notation

X for both the linear space and its underlying set, even though the underlying set

2.1 Linear Spaces

41

alone has no algebraic structure of its own. A set X needs a binary operation on it, a field, and another operation involving such a field with X to acquire the necessary algebraic structure that will grant it the status of a linear space. The scalar 1 in the above definition stands, of course, for the identity in the field F with respect to the multiplication in F, and + denotes the addition in F. Observe that + (addition

in the field F) and + (addition in the group X) are different binary operations. However, once the difference has been pointed out, we shall from now on use the same symbol + to denote both of them. Moreover, we shall drop the dot also from the multiplication notation in F. and write a# instead of a P. The neutral element under the vector addition in X (i.e., the vector zero) is referred to as the origin of the linear space X. Again, we shall use the same symbol 0 to denote both the origin in X and the scalar zero in F. A linear space over R is called a real linear space and a linear space over C is called a complex linear space. Example 2D. R itself is a linear space over R. That is, the plain set R when equipped with the ordinary binary operations of addition and multiplication becomes a field, also denoted by R. If vector addition is identified with scalar addition, then it be-

comes a real linear space, denoted again by R. More generally. for each n E N, let F" denote the Cartesian product of n copies of a field F (i.e., the set of all ordered n-tuples of scalars in F). Now define vector addition and scalar multiplication coordinatewise, as usual: and

ax=(att,...,at;")

f o r everyx = ( 1 , ... , f") and Y = (v1, ... , v") in F" and everya in F. This makes F" into a linear space over F. In particular, R" (the Cartesian product of n copies of R) and C" (the Cartesian product of n copies of C) become real and complex linear spaces, respectively, whenever vector addition and scalar multiplication are defined coordinatewise. However, if we restrict scalar multiplication to real multiplication only, then C" can also be made into a real linear space.

Example 2E. Let S be a nonempty set and let F be an arbitrary field. Consider the set X = F$

of all functions from S to F (i.e., the set of all scalar-valued functions on S, where "scalar-valued" stands for "F-valued"). Let vector addition and scalar multiplication be defined pointwise. That is, if x and y are functions in X and a is a scalar in F, then x + y and ax are functions in X defined by

(x + y)(s) = x(s) + y(s)

and

(ax)(s) = a(x(s))

for every s E S. Now it is easy to show that X, when equipped with these two operations, in fact is a linear space over IF. Particular cases: FN (the set of all scalarvalued sequences) and FIO' 1) (the set of all scalar-valued functions on the interval

42

2. Algebraic Structures

[0. 1]) are linear spaces over F, whenever vector addition and scalar multiplication are defined pointwise. Note that the linear space F" in the previous example also is a particular case of the present example, where the coordinatewise operations are identified with the pointwise operations (recall: F" = FL and fi = (i E N : i < n)).

Example 2F. What was the role played by the field F in the previous example? Answer: Vector addition and scalar multiplication in FS were defined pointwise by using addition and multiplication in F. This suggests the following generalization of Example 2E. Let S be a nonempty set, and let Y be an arbitrary linear space (over a field F). Consider the set x = YS

of all functions from S to y (i.e., the set of all y-valued functions on S). Let vector addition and scalar multiplication in X be defined pointwise by using vector addition and scalar multiplication in Y. That is, if f and g are functions in X (so that f (s)

and g(s) are elements of y for each s E S) and a is a scalar in F, then f + g and of are functions in X defined by

(f + g)(s) = f (s) + g(s)

and

(af)(s) = a(f (s))

for every s E S. As before, it is easily verified that X, when equipped with these operations, becomes a linear space over the same field F. The origin of X is the null function 0: S -+ Y (which is defined by 0(s) = 0 for all s E S). Examples 2D and 2E can be thought of as particular cases of this one.

Example 2G. Let X be a linear space over F, and let x. x', y and y' be arbitrary vectors in X. An equivalence relation - on X is compatible with vector addition if

x'- x and y'-. y

imply

x' + y'- x + y.

Similarly, it is said to be compatible with scalar multiplication if, for x and x' in X

ands in F,

x'- x

implies

ax'- ax.

If an equivalence relation - on a linear space X is compatible with both vector addition and scalar multiplication, then we shall say that - is a linear equivalence relation. Now consider X/-., the quotient space of X modulo - (i.e., the collection of all equivalence classes [x] with respect to ), and suppose the equivalence relation - on X is linear. In this case a binary operation + on X/- can be defined by setting

[xJ + [y] = [x + y] for every [x] and [y] in X/-r. Indeed, since - is compatible with vector addition, it follows that [x + y] does not depend on which particular members x and y of the equivalence classes [x] and [y) were taken. Thus the operation + actually is

2.2 Linear Manifolds

43

a function from (X/- x X/-) to X/-. . This defines vector addition in X/-- . Scalar multiplication in X/-. can be similarly defined by setting

a[x] = [ax] for every [x] in X/-. and every a in F. Therefore, if - is a linear equivalence relation on a linear space X over a field F, then X/- becomes a linear space over F whenever vector addition and scalar multiplication in X/- are defined this way. It is clear by the definition of a linear space X that x + y + z is a well-defined vector in X whenever x, y and z are vectors in X. Similarly, if {x; }"_i is a finite set of vectors in X, then the sum xj + - + x,,, denoted by x;, is again a vector in X. (The notion of infinite sums needs topology and we shall consider them in Chapters 4 and 5.)

2.2

Linear Manifolds

A linear manifold of a linear space X over F is a nonempty subset M of X with the following properties.

x+yEM and axEM for every pair of vectors x, y in M and every scalar a in F. It is readily verified that a linear manifold M of a linear space X over a field F is itself a linear space over the same field F. The origin 0 of X is the origin of every linear manifold M of X. The zero linear manifold is (0), consisting of the single vector 0. If a linear manifold M is a proper subset of X, then it is said to be a proper linear manifold. A nontrivial

linear manifold M of a linear space X is a nonzero proper linear manifold of it ({0) -A M A X). Example 2H. Let M be a linear manifold of a linear space X and consider a relation

- on X defined as follows. If x and x' are vectors in X, then

X'- X

if

x' - x E M.

That is, x'- x if x' is congruent to x modulo M - notation: x' = x (mod M). Since M is a linear manifold of X, the relation - in fact is an equivalence relation

on X (reason: 0 E M - reflexivity, x' - x" = (x' - x) + (x - x") E M whenever x' - x and x - x" lie in M - transitivity, and x - x' E M whenever x' - x E M - symmetry). The equivalence class (with respect to -)

[x] = {x'EX: x'-x{ = {x'EX: x'=x+zforsomeZEM} of a vector x in X is called the coset of x modulo M - notation: [x] = x + M. The set of all cosets [x] modulo M for every x E X (i.e.. the collection of all equivalence

44

2. Algebraic Structures

classes [x] with respect to the equivalence relation - for every x in X) is precisely the quotient space X/- of X modulo -. Following the terminology introduced in Example 2G, - is a linear equivalence relation on the linear space X. Indeed, if

x'-x E Mandy'-y E M, then (x'+y')-(x+y)=(x'-x)+(y'-y) E M and ax' - ax = a(x' - x) E M for every scalar a. so that x'- x and y'-' y imply x' + y'-' x + y and ax'- ax. Therefore, with vector addition and scalar multiplication defined by

[x] + [y] = [x + y]

and

a[x] = [ax],

X/- is made into a linear space. This is usually denoted by X/M (instead of XI-) and called the quotient space of X modulo M. The origin of X/M is. of course, [0] = M. If M and N are linear manifolds of a linear space X, then the sum of M and N, denoted by M + N, is the subset of X made up of all sums x + y where x is a vector in M and y is a vector in N:

M+N = {ZEX: Z=x+y, xEMand yEN}. It is trivially verified that M + Al' is a linear manifold of X. If (M i )" t is a finite family of linear manifolds of a linear space X, then the sum > 1Mj is the linear manifold MI + + Mn of X consisting of all sums E" txj where each vector x; lies in Mi. More generally, if {My}yer is an arbitrary indexed family of linear

manifolds of a linear space X, then the sum EyerMy is defined as the set of all sums with xy E My for each index y and xy = 0 except for some finite set of indices (i.e., EyErMy is the set made up of all finite sums with each summand being a vector in one of the linear manifolds My). Clearly, EYErMy is

itself a linear manifold of X, and Ma c EyerMy for every M. E (My)yer A linear manifold of a linear space X is never empty: the origin of X is always there. Note that the intersection m fl N of two linear manifolds M and M of a linear space X is itself a linear manifold of X. In fact, if (My )yEr is an arbitrary collection

of linear manifolds of a linear space X, then the intersection nyerMy is again a

linear manifold of X. Moreover, fyErMy c M. for every M. E (My)yEr. Now consider the collection Cat(X) of all linear manifolds of a linear space X.

Since £at(X) is a subcollection of the power set P(X), it follows that Cat(X) is partially ordered in the inclusion ordering. If (My)yEr is any subcollection of Cat (X), then F-yErMy in Gat (X) is an upper bound for {My }yEr and (l yErMy in Gat(X) is a lower bound for {My}yEr. If U in Cat(X) is an upper bound for

(My )yEr (i.e., if My c U for all y E I'), then EyErMy C- U. Therefore

E My = sup(My)yEr yEr

2.2 Linear Manifolds

45

Similarly, if V in Cat(X) is a lower bound for {My}yEr (i.e., if V c My for all Thus y E I'), then V c nyErMy.

n my = inf{My}yer yEr

Conclusion: Cat (X) is a complete lattice. The collection of all linear manifolds of a linear space is a complete lattice in the inclusion ordering. If {M, N} is a pair of

elements of Cat(X), then M v A(= M + N and M AN = M fl N. Let A be an arbitrary subset of a linear space X, and consider the subcollection (a sublattice. actually) GA of the complete lattice Cat(X),

CA = {MECat(X): AcM}, consisting of all linear manifolds of X that include A. Set

span A = inf CA = n CA, which is called the (linear) span of A. Since A c_ n CA (for A C_ M for every M E CA ), it follows that inf GA = min CA so that span A E CA. Thus span A is the smallest linear manifold of X that includes A, which coincides with the intersection of all linear manifolds of X that includes A. It is readily verified that span 0 = {0},

span M = M for every M E Cat (X), and A e_ span A = span (span A) for every A E &'(X). Moreover, if A and B are subsets of X, then

ACB

implies

span A C span B.

If M and N are linear manifolds of a linear space X, then it is clear that M U N c

M + N. Moreover, if K is a linear manifold of X such that M U N C K, then

x+ y E K for every x E M and every y E N, and hence M + N g K. Thus M + N is the smallest linear manifold of X that includes M U N, which means that

M +A( = span (M U AO. More generally, let {My}yEr be an arbitrary subcollection of Cat(X), and suppose K E Cat(X) is such that UyErMy c K. Then every (finite) sum Y-yErxy

with each xy in My is a vector in K. Thus >yErMy c K. Since UyerMy c EyerMy,it follows that EyErMy is the smallest elementofLat (X) that includes UyErMy.

Equivalently,

E My = span ( U My). yEr

yEr

46

2. Algebraic Structures

2.3

Linear Independence

Let A be a nonempty subset of a linear space X. A vector x E X is a linear combination of vectors in A if there exist a finite set (xi }" of vectors in A and a finite family of scalars (a,}.1 such that 1

n

X = Eaixi. i=1

Warning: A linear combination is, by definition, finite. That is, a linear combination of vectors in a set A is a weighted sum of a finite subset of vectors in A, weighted by a finite family of scalars, no matter whether A is a finite or an infinite set. Since X is a linear space, any linear combination of vectors in A is a vector in X.

Proposition 2.2. The set of all linear combinations of vectors in a nonempty subset A of a linear space X is a linear manifold of X that coincides with span A.

Proof. Let A be an arbitrary subset of a linear space X. consider the collection LA of all linear manifolds of X that include A. and recall that

span A = min LA. Suppose A is nonempty and let (A) denote the set of all linear combinations of vectors in A. It is clear that A c (A) (every vector in A is a trivial linear combination

of vectors in A), and that (A) is a linear manifold of X (if x, y E (A). then x + y and ax lie in (A)). Thus (A) E LA.

Moreover, if M is an arbitrary linear manifold of X, and if x E X is a linear combination of vectors in M, then x E M (because M is itself a linear space).

Thus (M) c M. Since M c (M), it follows that (M) = M for every linear manifold M of X. Furthermore, if M E LA, then A C_ M and hence (A) c (M) (reason: (A) C (B) whenever A and B are nonempty subsets of X such that A C_ B). Therefore, M E LA implies (A) C M.

Conclusion: (A) is the smallest element of LA. That is,

(A) = span A.

O

Following the notation introduced in the proof of Proposition 2.2, (A) = span A

whenever A 36 0. Set (0) = span 0 so that (0) = (0), and hence (A) is welldefined for every subset A of X. We shall use one and the same notation, viz. span A, for both of them: the set of all linear combinations of vectors in A and the (linear) span of A. For this reason span A is also referred to as the linear manifold

2.3 Linear Independence

47

generated (or spanned) by A. If a linear manifold M of X (which may be X itself) is such that span A = M for some subset A of X. then we say that A spans M. A subset A of a linear space X is said to be linearly independent if each vector x in A is not a linear combination of vectors in A\(x }. Equivalently, A is linearly independent if x ¢ span (A\(x)) for every x E A. If a set A is not linearly independent, then it is said to be linearly dependent. Note that the empty set 0 of a linear space X is linearly independent (there is no vector in 0 that is a linear combination of vectors in 0). Any singleton (x) of X such that x 0 0 is linearly independent. Indeed,

span ((x)\(x)) = span 0 = {0} so that x f span ((x)\(x)) if x A 0. However, 0 E span ({O}\{0}) = (0), and hence the singleton (0) is not linearly independent. In fact, every subset of X that contains the origin of X is not linearly independent (reason: if 0 E A C X and A has another vector besides the origin, say x A 0. then 0 = Ox). Thus, if a vector x is an element of a linearly independent subset of a linear space X, then x A 0.

Proposition 2.3. Let A be a nonempty subset of a linear space X. The following assertions are pairwise equivalent.

(a) A is linearly independent. (b) Each nonzero vector in span A has a unique representation as a linear combination of vectors in A. (c) Every finite subset of A is linearly independent. (d) There is no proper subset of A whose span coincides with span A. Proof. The statement (b) can be rewritten as follows.

(ti) For every nonzero vector x E span A there exist a unique finite family of scalars {ai }" t and a unique finite subset (ai )°_ t of A such that x = E"_ t ai ai .

Proof of (a)=>(b). Suppose A A 0 is linearly independent and take an arbitrary nonzero x E span A. Consider two representations of x as a linear combination of vectors in A:

n

x=

m

8ibi = >Yici,

where each bi and each ci are vectors in A (and hence nonzero because A is linearly independent). Since x # 0 we may assume that the scalars £i and yi are all nonzero. Set B = {bi }"_t and C = {ci )r t, both finite nonempty subsets of A. Take an arbitrary b E B and note that b is a linear combination of vectors in (B\{b)) U C. However, since b E A and A is linearly independent, it follows that b is not a linear combination of any subset of A\{b). Thus b E C. Similarly, take an arbitrary c E C and conclude that c E B by using the same argument. Hence

48

2. Algebraic Structures

B C C C B. That is, B = C. Therefore x = 1,6i b; = >"_t y; b; , which implies that Frs i (f; - y, )b; = 0. Since each b; is not a linear combination of vectors in B\(b; }, it follows that fl; = y; for every i. Summing up: The two representations of x coincide. Proof of (b)=(a). If A is nonempty and every nonzero vector x in span A has a unique representation as a linear combination of vectors in A, then the unique representation of an arbitrary a in A as a linear combination of vectors in A is a itself (recall: A C_ span A). Therefore, every a E A is not a linear combination of vectors in A\(a), which means that A is linearly independent.

Proof of (a). (c). If A is linearly independent, then every subset of it clearly is linearly independent. If A is not linearly independent, then either A = (0) or there exists x E A that is a linear combination of vectors, say (x; )"=t for some n E N, in

A\(x) # 0. In the former case A is itself a finite subset of A that is not linearly independent. In the latter case (x; )" 1 U (x) is a finite subset of A that is not linearly independent. Conclusion: If every finite subset of A is linearly independent, then A is itself linearly independent.

Proof of Recalling that B C_ A implies span B e_ span A. the statement (d) can be rewritten as follows.

(d) B C A implies

span B C span A.

Suppose A is nonempty and linearly independent. Let B be an arbitrary proper subset

of A. If B = 0, then (d) holds trivially (0 96 A 96 (0) so that span 0 C span A). Thus suppose B A 0 and take any x E A\B. If X E span B, then x is a linear combination of vectors in B. This implies that B U (x} is a subset of A that is not linearly independent, and hence A itself is not linearly independent, which is a contradiction. Therefore, x 0 span B for every x E A\B whenever 0 0 B C A. Since X E span A (because x E A), and since span B C span A (for B C A), it follows that span B C span A so that (d) holds true.

Proof of (d)=(a). If A is not linearly independent, then either A = (0) or there exists x E A that is a linear combination of vectors in A\(x). In the former case the only proper subset of A = (0) is B = 0, and span B = (0) = span A. In the latter case B = A\(x) is a proper subset of A such that span B = span A (reason: span B C_ span A because B C A. and span A C span B because every vector in A is a linear combination of vectors in A\ (x)). Therefore, (d) implies (a).

2.4

D

Hamel Basis

A linearly independent subset of a linear space X that spans X is called a Hamel basis (or a linear basis) for X. In other words, a subset B of a linear space X is a

2.4 Hamel Basis

49

Hamel basis for X if (i)

B is linearly independent, and

(ii)

span B = X.

Let B = (xy )yEr be an indexed Hamel basis for a linear space X. If x is a nonzero vector in X, then Proposition 2.3 ensures the existence of a unique (similarly in(which may depend on x) such that ay = 0 for all dexed) family of scalars {ay

but a finite set of indices y, and x = EYErayxy. The weighted sum EyErayxy (i.e., the unique representation of x as a linear combination of vectors in B, or the unique (linear) representation of x in terms of B) is called the expansion of x on B. and the coefficients of it (i.e., the unique indexed family of scalars (ay )yEr) are called the coordinates of x with respect to the indexed basis B. If x = 0, then its unique expansion on B is the trivial one whose coefficients are all null. Since 0 is linearly independent, and since span 0 = (0), it follows that the empty set 0 is a Hamel basis for the zero linear space (0). Now suppose X is a nonzero linear space. Every singleton {x) in X such that x # 0 is linearly independent. Thus every nonzero linear space has many linearly independent subsets. If a linearly independent subset A of X is not already a Hamel basis for X, then we can construct a larger linearly independent subset of X.

Proposition 2.4. If A is a linearly independent subset of a linear space X, and if there exists x E X\span A, then A U (x} is a linearly independent subset of X.

Proof. Suppose there exists a vector x in X\span A. Note that x * 0, and hence X 96 (0). If A = 0, then the result is trivially verified ({x) = 0 U (x) is linearly independent). Thus suppose A is nonempty and set C = A U (x) C X. Since x 0 span A, it follows that x 0 span (C\(x}). Take an arbitrary a E A. Suppose a E A is a linear combination of vectors in C\(a}. Clearly, a 96 ax for every scalar a (for x 0 span A and a 76 0 because A is linearly independent) so that n

a = apx+Eaiai, i=l

where each ai is a vector in A\{a} and each ai is a nonzero scalar (recall: 0 96 a 96 F"_I ai ai for A is linearly independent). Thus x is a linear combination of vectors in A, which contradicts the assumption that x l span A. Therefore, every a E A is not a linear combination of vectors in C\{a}. Conclusion: Every C E C is not a linear combination of vectors in C\(c), which means that C is linearly independent.

0 Can we proceed this way, enlarging linearly independent subsets of X in order to form a chain of linearly independent subsets, so that an "ultimate" linearly independent subset becomes a Hamel basis for X? Yes, we can; and it seems reasonable that the Axiom of Choice (or any statement equivalent to it as, for instance, Zorn's

50

2. Algebraic Structures

Lemma) might be called into play. In fact, every linearly independent subset of any

linear space X is included in some Hamel basis for X, so that every linear space has a large supply of Hamel bases. Theorem 2.5. If A is a linearly independent subset of a linear space X, then there exists a Hamel basis B for X such that A C- B.

Proof. Let X be a linear space and suppose A is a linearly independent subset of X. Set

TA = {B E P(X): B is linearly independent and A C B), the collection of all linearly independent subsets of X that include A. Recall that, as a nonempty subcollection (A E TA) of the power set 6a(X), TA is partially ordered in the inclusion ordering.

Claim 1. TA has a maximal element.

Proof. If X = {0}, then A = 0 and TA = A) = (0) 0 0, so that the claimed result is trivially verified. Thus suppose X # (0). In this case, the nonempty collection TA contains a nonempty set (e.g., if A = 0, then every nonzero singleton in X belongs to IA; if A # 0, then A E TA). Now consider an arbitrary chain C in TA containing a nonempty set. Recall that U C denotes the union of all sets in C. Take an arbitrary

finite nonempty subset of U C, say, a set D C U C such that #D = n for some n E N. Each element of D belongs to a set in C (for D C U C). Since C is a chain, we can arrange the elements of D as follows. D = (x;) t such that xi E C; E C for C_ C . Thus D C_ C. Since C is linearly independent each index i, where Cl g (because C E C C TA), it follows that D is linearly independent. Conclusion: Every finite subset of U C is linearly independent. Therefore U C is linearly independent by Proposition 2.3. Moreover, since A C C for all C E C (for C e TA), it also follows that A C_ U C. Hence U C E TA. Since U C clearly is an upper bound for C. we may conclude: Every chain in TA has an upper bound in TA. Thus TA has a maximal element by Zorn's Lemma. a Claim 2. Take B E TA. B is maximal in TA if and only if B is a Hamel basis for X.

Proof. Again, if X = (0), then B = A = 0 is the only (and so a maximal) element in TA and span B = X, so that the claimed result holds trivially. Thus suppose X # (0), which implies that TA contains nonempty sets, and take an arbitrary B in TA. If span B# X (i.e., if span B C X), then take x E X\span B so that B U {x) E TA (i.e., B U {x} is linearly independent by Proposition 2.4, and A C B U (x} because A C_ B). Hence B is not maximal in TA. Therefore, if B is maximal in ZA, then

span B = X. On the other hand, if span B = X, then B 96 0 (for X A (0)) and every vector in X is a linear combination of vectors in B. Thus B U {x) is not linearly independent for every x E X\ B. This implies that there is no B' E TA such that B C B', which means that B is maximal in TA. Conclusion: If B E ZA' then B

2.4 Hamel Basis

51

is maximal in ZA if and only if span B = X. According to the definition of Hamel basis, B in ZA is such that span B = X if and only if B is a Hamel basis for X. o Claims 1 and 2 ensure that, for each linearly independent subset A of X, there exists

a Hamel basis B for X such that A c B. Since the empty set is a subset of every set (in particular. of every linear space), and since the empty set is linearly independent, it follows that the preceding theorem holds for A = 0. In this case Zg is simply the collection of all linearly independent subsets of the linear space X. and the theorem statement just says that every linear

space has a Hamel basis. Moreover, Claim 2 says that a Hamel basis for a linear space is precisely a maximal linearly independent subset of it (i.e., a Hamel basis is a maximal element of Zm ). The idea behind the previous theorem was that of enlarging a linearly independent

subset of X to get a Hamel basis for X. Another way of facing the same problem (i.e., another way to obtain a Hamel basis for linear space X) is to begin with a set that spans X and then to weed out from it a linearly independent subset that also spans X. Theorem 2.6. if a subset A of a linear space X spans X. then there exists a Hamel basis B for X such that B C A.

Proof. Let A be a subset of a linear space X such that span A = X, and consider the collection 2A of all linearly independent subsets of A:

2A = {B E P(X): B is linearly independent and B C A). If X = {0}, then either A = 0 or A = {0}. In any case ZA = (0) trivially has a maximal element. If X 0 (0), then A has a nonzero vector (for span A = X) and every nonzero singleton {x) c A is an element of I . Thus, proceeding exactly as in the proof of Theorem 2.5 (Claim 1). we can show that ZA has a maximal element. Let Ao be a maximal element of IA. If A is linearly independent, then we

are done (i.e., A is itself a Hamel basis for X since span A = X). Thus suppose A is not linearly independent so that Ao is a proper subset of A. Take an arbitrary a E A\Ao and consider the set A0 U (a) c A, which is not linearly independent because Ao is maximal in 2A. Since Ao is linearly independent, it follows that a is a linear combination of vectors in Ao. Thus A\Ao c_ span A0. and hence A = Ao U (A\Ao) c span Ao. Therefore span A c span (span A0) = span A0 c span A, which implies that span Ao = span A = X. Conclusion: A0 is a Hamel basis for X. Since X trivially spans X, the above theorem holds for A = X. In this case T. is precisely the collection of all linearly independent subsets of X (i.e., I = Z0), and the theorem statement again says that every linear space has a Hamel basis. An ever-present purpose in mathematics is a quest for hidden invariants. The concept of Hamel basis supplies a fundamental invariant for a linear space, namely. the cardinality of all Hamel bases for X.

52

2. Algebraic Structures

Theorem 2.7. Every Hamel basis for a given linear space X has the same cardinality.

Proof. If X = (0), then the result holds trivially. Suppose X -A (0) and let B and C be arbitrary Hamel bases for X (so that they are nonempty and do not contain the origin). Proposition 2.3 ensures that for every nonzero vector x E X there exists a unique finite subset of the Hamel basis C, say C, such that x is a linear combination

of all vectors in Cz C C. Now take an arbitrary c E C and consider the unique representation of it as a linear combination of vectors in the Hamel basis B. Thus c is a linear combination of all vectors in (b) U B' for some (nonzero) b E B and some finite subset B' of B. Hence c = fib + d, where f is a nonzero scalar and d is a vector in X different from c (for fib 96 0). If d = 0, then c = fib so that Cb = (c), and hence c E Cb trivially. Suppose d 96 0. Recalling again that C also is a Hamel basis for X, consider the unique representation of the nonzero vector d as a linear

combination of vectors in C, so that fib = c - d 96 0 is a linear combination of vectors in C. Thus b is itself a linear combination of all vectors in (c) U C' for some subset C' of C. Since such a representation is unique, it follows that (c) U C' = Cb. Thereforec E Cb. Summing up: For every c E C there exists b E B such that c E Cb. Thus

CCUCb. bE B

Now we shall split the proof into two parts. One dealing with the case of finite Hamel bases, and the other with infinite Hamel bases.

Claim 0. Let X be a linear space. If a subset E of X with exactly n elements spans X, then every subset of X with more than n elements is not linearly independent.

Proof. Assume the linear space X is nonzero (i.e., X # (0)) to avoid trivialities. Take an integer n E N and let E = (e; )" be a subset (with n distinct elements) of X such that span E = X. Now take an arbitrary subset of X with n + 1 elements, say D = {d; }7I . Suppose D is linearly independent. Next consider the set S1 = {dt } U E

which clearly spans X (because E already does it). Since span E = X, it follows that di is a linear combination of vectors in E. Moreover, d, i4 0 because D is linearly independent. Thus di = y,"_ 1a; e; where at least one, say ak, of the scalars (a; }°_, is nonzero. Therefore, if we delete ek from S1, then the resulting set

S', = S,\(ek) = {d,}UE\(ek) still spans X. That is, in forming this new set S, that spans X we have traded off one vector in D for one vector in E. Rename the elements of S' by setting s; = e; for each i 54 k and Sk = d1, so that S, = (s; ]" . Since D has at least two elements. set

S2 = (d2)USi = (di,d2)UE\{ek)

2.4 Hamel Basis

53

which again spans X (for Si spans X). Since span Sj = X, it follows that d2 is a linear combination of vectors in S. say d2 = E" if;si for some family of scalars {M"_,. Moreover, 0 # d2 54 fksk = fkdi because D is linearly independent. Thus there exists at least one nonzero scalar in (f;)"_t different from ,8k, say $j. Therefore, if we delete sj from S2 (recall: sj = ej 96 ek), then the resulting set

S2 = S2\(ei) = (di, d2) U E\(ek, ej) still spans X. Continuing this way we eventually get down to the set

Sn = {d1 }n=1 U E\ (e1 )° = D\{dn+l } which once again spans X. Thus is a linear combination of vectors in D\{dn+1 }, which contradicts the assumption that D is linearly independent. Conclusion: Every

subset of X with n + 1 elements is not linearly independent. Recalling that every subset of a linearly independent set is again linearly independent, it follows that every subset of X with more than n elements is not linearly independent. 17

Claim 1. If B is finite, then #C = #B.

Proof. Recall that Cb is finite for every b in B. If B is finite, then UbEBCb is a finite union of finite sets. Hence any subset of it is finite. In particular. C is finite. Since C is linearly independent, it follows by Claim 0 that #C < #B. Dually (swap the Hamel bases B and C), #B < #C. Hence #C = #B. o Claim 2. If B is infinite, then #C = #B.

Proof. Since B is infinite, and since Cb is finite for every b in B, it follows that #Cb < #B for all b in B. Thus, according to Theorems 1.10 and 1.9,

#(U Cb) < #(BxB) = #B beB

because B is infinite. Therefore #C < #B (recall that C C UbEBCb and use Problems 1.21(a) and 1.22). Moreover, Claim 1 says that B is finite whenever C is finite. Thus C must be infinite because B is infinite. Since C is infinite we may reverse the argument (swapping again the Hamel bases B and C) and get #B < #C. Hence #C = #B by the Cantor-Bernstein Theorem (Theorem 1.6). a Claims 1 and 2 ensure that, if B and C are Hamel bases for a linear Space X, then o B and C have the same cardinal number. Such an invariant (i.e., the cardinality of any Hamel basis) is called the dimension (or the linear dimension) of the linear space X, denoted by dim X. Thus dim X = #B

for any Hamel basis B for X. If the dimension of X is finite (equivalently, if any Hamel basis for X is a finite set) then we say that X is a finite-dimensional linear space. Otherwise (i.e., if any Hamel basis for X is an infinite set) we say that X is an infinite-dimensional linear space.

54

2. Algebraic Structures

Example 21. The Kronecker delta (or Kronecker function) is the mapping in 2ZXZ (i.e., the function from Z x Z to (0, 1)) defined by 1,

i = j,

0,

i A j,

for all integers i, j. Now consider the linear space F" (for an arbitrary positive integer n, over an arbitrary field IF - see Example 2D). The subset B of F" consisting of the n-tuples ei = (Si 1, ... , Si" ), with 1 at the ith position and zeros elsewhere, constitute a Hamel basis for IF". This is called the canonical basis (or the natural basis) for F". Thus dim F" = n. As we shall see later, F" in fact is a prototype for every finite-dimensional linear space (of dimension n) over a field F. Example 2J. Let FN be the linear space (over a field F) of all scalar-valued sequences

(see Example 2E), and let X be the subset of FN defined as follows. x = (tk )kEN belongs to X if and only if k = 0 except for some finite set of indices k in N. That is, X is the set consisting of all F-valued sequences with a finite number of nonzero entries, which clearly is a linear manifold of FN, and hence a linear space itself over F. For each integer i E N let ei be an F-valued sequence with just one nonzero entry (equal to 1) at the ith position; that is, e; = (Sik )keN E X for every i E N. Now set B = (ei C X. It is readily verified that B is linearly independent and that span B = X (every vector in X is a linear combination of vectors in B). Thus B is a Hamel basis for X. Since B is countably infinite. X is an infinitedimensional linear space with dim X = No. Therefore (see Problem 2.6(b)), FN is an infinite-dimensional linear space. Note that B is not a Hamel basis for FN (reason: span B = X and X is properly included in FN). The next example shows that No < dim FN whenever F = Q, F = R, or F = C.

Example 2K. Let CN be the complex linear space of all complex-valued sequences. For each real number t E (0, 1) consider the real sequence xt = (tk-t }kEN = {tk}kENO = (1, t, 12, ..) E CN whose entries are the nonnegative powers oft. Set A = (x, 1) C CN. We claim that A is linearly independent. A bit of elementary real analysis (rather than pure algebra) supplies a very simple proof as follows. Suppose A is not linearly independent. Then there exists s E (0, 1) such that x, is a Iinear combination of vectors in A\(s). That is, x, = E°_taix,i for some n E N. where Jai )"= t is a family of nonzero complex numbers and (x,; )" t is

a (finite) subset of A such that x,, # x, for every i = I.... , n. Hence n > I (reason: if n = 1, then z, = aIx,, so that sk = attk for every k E No, which implies that x, = x,, ). As the set (ti }"_ t consists of distinct points from (0, 1), suppose it is decreasingly ordered (reorder it if necessary) so that ti < it for each i = 2.... , n. Since sk a; tk, it follows that (s/ti )k = at + E -..tai (ri /tt )k for every k E No. However limk "_,ai (ti /tt )k = 0, because each ti/ti lies in (0. 1), and

hence limk(s f tt )k = at. Thus at = 0 (recall: x, 0 x,, so that s 0 tt) which is a contradiction. Conclusion: A is linearly independent. Therefore, according to

2.5 Linear Transformations

55

Theorem 2.5, there exists a Hamel basis B for CN including A. Since A C B and

#A = #(0, 1) = 2K0, it follows that 2K0 < #B. However, #C = #R = 2K0 < #B = dim CN, so that #CN = dim CN (see Problem 2.8). Conclusion: CN is an infinite-dimensional linear space such that

2K0 < dim CN = #CN. Note that the whole argument applies for C replaced by R, so that

2k < dim RN = #RN; but it does not apply to the rational field Q (the interval (0, 1) is not a subset of Q, and hence the set A is not included in QN). However, the final conclusion does hold for the linear space QN. Indeed, if F is an arbitrary infinite field, then 2K0 = #2N < #IFN = max(#IF, dim FN) according to Problems 1.24 and 2.8. Therefore, since #Q = No < 2K0 (Problem 1.25(c)), it follows that

2"0 < dim QN = #QN.

2.5

Linear Transformations

A mapping L : X --)- Y of a linear space X over a field IF into a linear space y over the same field F is homogeneous if

L(ax) = aLx for every vector x E X and every scalar of E F. The scalar multiplication on the left-hand side is an operation on X and that on the right-hand side is an operation on ) (so that the linear spaces X and y must indeed be defined over the same field F). L is additive if

L(xj +x2) = L(xt)+L(x2) for all vectors xt, x2 in X. Again, the vector addition on the left-hand side is an operation on X while the one on the right-hand side is an operation on Y. If X and Y are linear spaces over the same scalar field, and if L is a homogeneous and additive

mapping of X into y, then L is a linear transformation: a linear transformation is a homogeneous and additive mapping between linear spaces over the same scalar field. When we say that L: X -i, Y is a linear transformation, it is implicitly assumed that X and Y are linear spaces over the same field F. If X = y and L : X -). X is a linear transformation, then we refer to L as a linear transformation on X. Trivial example: The identity 1: X -+ X (such that 1(x) = x for every x E X) is a linear transformation on X. Recall that a field F can be made into a linear space over F itself (see Example 2D). If X is a linear space over F, then a linear transformation f : X -+ F is called a linear functional: a linear functional is a scalar-valued linear transformation (i.e., a linear transformation of a linear space X into its scalar field).

56

2. Algebraic Structures

If Y E Y is the value of a linear transformation L : X -+ Y at x E X, then we shall often write y = Lx (instead of y = L(x)). Since y is a linear space, it has an origin. The null space (or kernel) of a linear transformation L : X -- Y is the subset

N(L) = {x E X : Lx = 01 = L-'(10)) of X consisting of all vectors in X mapped into the origin of Y by L. Since X also is a linear space, it has an origin too. The origin of X is always in N(L) (i.e., LO = 0

for every linear transformation L). The null transformation (denoted by 0) is the mapping O : X -+ Y such that Ox = 0 for every x E X, which certainly is a linear transformation. In fact, if L : X -+ ,y is a linear transformation, then L = 0 if and only if N(L) = X. Equivalently, L = 0 if and only if R(L) = (0). The null space, N(L) = L-' ((O)), of any linear transformation L : X --+ Y is a linear manifold of X, and the range of L, R(L) = L(X), is a linear manifold of Y (see Problem 2.10). These are indeed particular cases of Problem 2.11: The linear image of a linear manifold is a linear manifold, and the inverse image of a linear manifold under a linear transformation is again a linear manifold. The theorem below supplies an elegant and useful, although very simple, necessary and sufficient condition that a linear transformation be injective.

Theorem 2.8. A linear transformation L is injective if and only if N(L) = (0). Proof. let X and Y be linear spaces over the same scalar field, and consider a

linear transformation L : X -+ Y. If L is injective, then L-' (L((0))) = (0) (see Problem 1.3(d)). But L({0}) = (0) (for LO = 0) so that L-1((0)) = (0), which means N(L) = (0}. On the other hand, suppose N(L) = (0). Take xt and x2 arbitrary in X, and note that Lx1 - Lx2 = L(xi - x2) since L is linear. Thus, if Lx1 = Lx2, then L(xt - x2) = 0 and hence xt = x2 (i.e., xt - x2 = 0 because N(L) = {0)). Therefore L is injective. 0 The collection yX of all mappings of a set X into a linear space Y over a field IF is itself a linear space over F (see Example 2F). Now suppose X is a linear space (over the same field IF), and let C(X, y] denote the collection of all linear transformations

of X into Y. Since C[X, Y] is a linear manifold of YX (Problem 2.13), it follows that L[X, y] is a linear space over the same field F. Set LIX] = C[X, X] for short, so that L[X] C XX is the linear space of all linear transformations on X. The linear space C[X,F] of all linear functionals defined on a linear space X, which is a linear manifold of the linear space FX (see Example 2E), is called the algebraic dual (or algebraic conjugate) of X and denoted by X' (Dual spaces will be considered in Chapter 4.) Let X and Y be linear spaces over the same scalar field, and let LAM: M Y be the restriction of a linear transformation L : X - y to a linear manifold M of X. Since M is a linear space, it is readily verified that LAM is a linear transformation. Briefly: The restriction of a linear transformation to a linear manifold is again a linear transformation (Problem 2.14). The next result ensures the converse: If L E

2.5 linear Transformations

57

£(M, Y] and M is a linear manifold of X, then there exists T E L[X, y] such that L = TIM, which is called a linear extension of L over X.

Theorem 2.9. Let X and y be linear spaces over the same field F, and let M be a linear manifold of X. If L : M -> y is a linear transformation, then there exists a linear extension T : X -+ Y of L defined on the whole space X.

Proof. Set

IC = 1 K E C[N, Y] : N E Cat(X), M c N and L= KIM}, the collection of all linear transformations from linear manifolds of X to y that are extensions of L. Note that K is nonempty (at least L is there). Moreover, as a subcollection of .T = UAEP(X)yA, K is partially ordered in the extension ordering (see Problem 1.17). Problem 1.17 also tells us that every chain (K.) in X

has a supremum V1Ky in Y with domain D(V1Ky) = U1D(Ky) and range R(V1Ky) = U1R(Ky). Since D(K1) E Lat(X) (each K. is a linear transformation defined on a linear manifold of X), and since Lat(X) is a complete lattice,

it follows that D(V1Ky) is a linear manifold of X (i.e., U1D(Ky) E Lat(X)). Similarly, R(V1 Ky) is a linear manifold of Y. Claim. The supremum Vy Ky lies in K.

Proof. Take u and v arbitrary in D(V1K1). so that u E D(KA) for some Kx in {Ky} and V E D(K,,) for some K. in (Ky). Since (Ky) is a chain, it follows that Kx < K. (or vice versa), so that D(Kx) a D(KN,). Thus au + fJv E D(K0) and hence K,,(au + f v) = f K,,v for every a, f E F (recall: each K. is linear). However (V1Ky)ID(K,.) = Kµ, which implies that (V), Ky)(au + fv) = a(\/y Ky)u + f(Vy Ky)v. That is, Vy Ky : D(V1 Ky) -+ y is linear. Moreover, since each K. is such that K. I M = L, and since (Ky ) is a chain, it follows that (V), Ky) I m = L. Conclusion: V1 Ky E K. 0 Therefore, every chain in K has a supremum (and so an upper bound) in K. Thus, according to Zorn's Lemma, K contains a maximal element, say Ko : No -> Y. We shall show that No = X, and hence Ko is a linear extension of L over X. The proof goes by contradiction. Suppose No # X. Take xt E X\No (so that x1 34 0 because No is a linear manifold of X) and consider the sum of No and the one-dimensional linear manifold of X spanned by (xt ),

N1 = No + span {xt ), which is a linear manifold of X properly including M (because M C No C Nt ). Since No fl span (xt) = (0), it follows that every x in N, has a unique representation as a sum of a vector in No and a vector in span (xi ). That is, for each x E Nt there exists a unique pair (xo, a) in No xlF such that x = xo + axt. (Indeed, if

x = xo + axt = xo + a'xi, then xo - xo = (a' - a)xi E No n span (xt) = (0)

58

2. Algebraic Structures

so that xo = xo and a' = a - recall: xt 96 0.) Take an arbitrary y in y (for instance, y = 0) and consider the mapping Kt : Nt -+ Y defined by K t x= Kox0 + a y

for every x E N. Observe that K1 is linear (it inherits the linearity of Ko) and

Ko = KiINo (so that Ko < K1). Since M C No C Nt, it follows that L = KoIM = Ki I.&4. Thus KI E K, which contradicts the fact that K0 is maximal in K (for Ko y6 Ki ). Therefore N0 = X. o

Let X and y be nonzero linear spaces over the same field. Take x 0 0 in X and y * 0 in Y, set M = span [x} in Gat(X), and let L: M -+ Y be defined by Lu = a y for every u = ax E M. Clearly, L is linear and L * O. Thus Theorem 2.9 ensures that, if X and y are nonzero linear spaces over the same field, then there exist many T # 0 in £[X, yJ (at least as many as one-dimensional linear manifolds in Cat(X)).

2.6

Isomorphisms

Two exemplars of a mathematical structure are indistinguishable, in the context of the theory in which that structure is embedded, if there exists a one-to-one correspondence between them that preserves such a structure. This is a central concept in mathematics. From the point of view of the linear space theory, two linear spaces are essentially the same if there exists a one-to-one correspondence between them that preserves all the linear relations - they may differ on the set-theoretic nature of their elements but, as far as the linear space (algebraic) structure is concerned.

they are indistinguishable. In other words, two linear spaces X and y over the same scalar field are regarded as essentially the same linear space if there exists a one-to-one correspondence between them that preserves vector addition and scalar multiplication. That is, if there exists at least one invertible linear transformation from X toy whose inverse from y to X also is linear. The theorem below shows that the inverse of an invertible linear transformation is always linear.

Theorem 2.10. Let X and y be linear spaces over F. If L : X -+ y is an invertible linear transformation, then its inverse L-' : y -> X is a linear transformation. Proof. Recall that, by definition, a function is invertible if it is injective and surjective. Take yi and y2 arbitrary in y so that there exist xt and x2 in X such that yj = LxI and Y2 = Lx2 (for y = 1Z(L) - i.e., L is surjective). Since L is injective (i.e., L-' L is the identity on X - see Problems 1.5 and 1.7) and additive, it follows that

L-'(y, +y2) = L-'(Lx, +Lx2) = L-'L(xt +x2) = x, +x2 = L-' Lx1 + L-' Lx2 = L-' yl + L-' y2.

2.61somorphisms

59

and hence L-1 is additive. Similarly, since L is injective and homogeneous,

L-1(ay) = L-t(aLx) = L-'L(ax) = ax = aL-'Lx = aL-'y for everyy E Y = R(L) and everya E F, which implies that L-t is homogeneous. Thus L-t is a linear transformation. a An isomorphism between linear spaces (over the same scalar field) is an injective and surjective linear transformation. Equivalently, an invertible linear transformation. Two linear spaces X and Y over the same field F are isomorphic if there exists an isomorphism (i.e., a linear one-to-one correspondence) of X onto Y. Thus, according to Theorem 2.8, a linear transformation L: X -+ Y of a linear space X into a linear space Y is an isomorphism if and only if N(L) = (0) and R(L) = Y.

In particular, if N(L) = (0), then X and the range of L (R(L) = L(X)) are isomorphic linear spaces. We noticed in Example 21 that F" is a "prototype" for every n-dimensional linear space over F. What this really means is that every n-dimensional linear space over a field F is isomorphic to F", and hence two n-dimensional linear spaces over the same scalar field are isomorphic. In fact, such an isomorphism between linear spaces with the same dimension holds in general, either for finite or infinite dimensional linear spaces. We shall prove this below (Theorem 2.12) but first we need the following auxiliary result.

Proposition 2.11. Let X and Y be linear spaces over the same field, and let B be a Hamel basis for X. For each mapping F : B -+ Y there exists a unique linear transformation T: X -+ Y such that T I e= F.

Proof. If B = (xy)yer is a Hamel basis for X, indexed by an index set r (recall: any set can be thought of as an indexed set), then every vector x in X has a unique expansion on B, viz.,

x = E ayxy, yer where (ay}yEr is a similarly indexed family of scalars with ay = 0 for all but a finite set of indices y (the coordinates of x with respect to the indexed basis B). Now set

Tx = JayF(xy) yer

for every x E X. This defines a mapping T : X -+ Y of X into Y which is homogeneous, additive, and equals F when restricted to B. That is, T is a linear transformation such that T 18 = F. Moreover, if L : X -+ Y is a linear transformation of X into y such that LAB = F, then L = T. Indeed, for every x E X,

Lx = L(>ayxy) = EayF(xy) = T(Eayxy) = Tx. yer

yer

yEr

t7

60

2. Algebraic Structures

Theorem 2.12. Let X and ,y be linear spaces over the same scalar field. X and Y are isomorphic if and only if dim X = dim Y. Proof. (a) Let L : X -+ y be an isomorphism of X onto Y. and let BX be a Hamel basis for X. Set By = L(BX), a subset of Y. Claim 1. By is linearly independent.

Proof. Recall that L is an injective and surjective linear transformation. If By is not linearly independent, then there exists y E By, which is a linear combination of vectors in By\(y), say y = F"=ia;y; where each y; is a vector in By\{y). Thus x = L-t y in BX = L-i (By) is a linear combination of vectors in BX\{x). (Indeed,

x = F"=iaix; where each x; = L-t y; is a vector in BX = L-t (By) different from x = L-t y - recall: each y; is a vector in By different from y, and L is injective.) But this contradicts the fact that BX is linearly independent. Conclusion: By is linearly independent. El

Claim 2. By spans Y. Proof. Take Y E Y arbitrary so that y = Lx for some x E X (because L is surjective). Since span Bx = X, it follows that x is a linear combination of vectors in B. Hence y = Lx is a linear combination of vectors in By = L(Bx) (since L is linear) so that span By = Y. 0

Therefore, By is a Hamel basis for Y. Moreover, #By = #BX because L sets a one-to-one correspondence between BX and By. (In fact, the restriction L I Bx : BX -, By is injective and surjective, since L is injective and By = L(BX) by definition.) Thus dim X = dim Y. (b) Let BX and By be Hamel bases for X and y, respectively. If dim X = dim Y, then #By = #BX, which means that there exists a one-to-one mapping F: Bx By of BX onto By. Let T : X -> Y be the unique linear transformation such that T IBx = F (see Proposition 2.11), and hence T(BX) = F(BX) = By. Claim 3. T is injective.

Proof. If X = (0), then the result holds trivially. Thus suppose X # (0). Take any nonzero vector x in X and consider its (unique) representation as a linear combination of vectors in BX. Therefore, Tx has a representation as a linear combination of vectors in By = T(BX) because T is linear. Since By is linearly independent, it follows that Tx 74- 0. That is, N(T) = (0) which means, by Theorem 2.8, that T is injective. 17

Claim 4. T is surjective. Proof. Take any vector y E y and consider its expansion on By, say y =

rn_ i a, v,

ia;T(x,) with each x; in BX because By = with each y; in By. Thus y a;x, is a T(BX). Since T is linear, it follows that y = T(>"_ia;x;), where vector in X (since X is a linear space). Hence y E R(T). a

2.6 Isomorphisms

61

a

Therefore, T : X - y is an isomorphism of X onto Y.

Example 2L. Let X and Y be finite-dimensional linear spaces over the same field and By = (y) m be Hamel ;t, with dim X = n and dim Y = m. Let Bx = (x j bases for X and y, respectively. Take an arbitrary vector x in X and consider its unique expansion on Bx: 1

M

X = 1: jXj, j=1

consists of the coordinates of x with respect to B. Now let A : X --o- y be any linear transformation so that where the family of scalars (l; j

n

e 1: jAxj.

Ax

j=1

Each Axj is a vector in Y. Consider its unique expansion on By:

Axj =

aijyi i=1

where, for each j, (a}1 is a family of scalars - the coordinates of each Ax j with respect to By. Set y = Ax in Y and consider the unique expansion of y on By: M

y = Eviyi i=1

Again. (vi)" is a family of scalars consisting of the coordinates of y with respect to By. Thus the identity y = Ax can be written as 1

m

Eviyi

=

m

n

i=1

j=1

E(Djaij)yj.

i=1

Since the expansion of y on By is unique, it follows that

vi =

aij j j=1

for every i = i, ... , m. This gives an expression for each coordinate of Ax as a function of the coordinates of x. In terms of standard matrix notation, and according to the ordinary matrix operations, the matrix equation

(ui

all

...

aOn

t1

vm

aml

...

amn

to

62

2. Algebraic Structures

represents the identity y = Ax (the vector y is the value of the linear transformation A at the point x), and the in xn array of scalars

all [A]

...

Olin

.

amn

= (am,

I

is the matrix that represents the linear transformation A : X -+ Y with respect to the bases BX and By. The matrix [A] of a linear transformation A depends on the bases BX and By. If we change the bases, then the matrix that represents the linear transformation may change as well. Different matrices representing the same linear transformation are simply different representations of it with respect to different bases. However, if we fix the bases BX and By, then the representation [A] of A is unique. But uniqueness is not all. It is easy to show that (a) the set Fm xn of all m x n matrices with entries in F is a linear space over IF when equipped with the ordinary (entrywise) operations of matrix addition and scalar multiplication. Moreover, for fixed bases BX and By.

(b) Fmxn is isomorphic to £[X, Y]. If we fix the bases BX and By, then the relation between £[X, Y] and Fmxn defined by "[A l represents A with respect to BX and By" in fact is a function from £[X, Y] to Fmxn. It is readily verified that such a function, say 4': £[X, Y] - Fmxn, is homogeneous, additive, injective and surjective. In other words, 0 is an isomorphism.

For this reason we may and shall identify a linear transformation A E £[F', F'n] with its matrix [A] E Fmxn relative to the canonical bases for Fn and Fm (which were introduced in Example 21).

Example 2M. Let F denote either the real field or the complex field. Take an arbitrary nonnegative integer n, and let Pn [0, 1] be the collection of all polynomials in the variable t E [0, 1] with coefficients in F of degree no greater than n. That is,

{p E F10'1: p(t) = E"_oa;t'.t E [0, 1], with each ai in F)

Recall: The degree of a nonzero polynomial p is m if p(t) _ Via; t' with am 96 0 (e.g., the degree of a constant polynomial is zero), and the degree of the zero polynomial is undefined (thus not greater than any n E No). It is readily verified that P. [0, 1 ] is a linear manifold of the linear space Flo 11 (see Example 2E), and hence a linear space over F. Now consider the mapping L: F1+1 --- P. [0, 1] defined as follows. For each x = ( 4 o . . . . . kn) E Fn+I let p = Lx in P. [0, 1] be given by n

P(t) _ Dit' i=o

2.6Isomorphisnu

63

for every t E [0, 1]. It is easy to show that L is a linear transformation. More-

over, N(L) = (0) (i.e., if p(t) = E t...0 t' = 0 for every t E [0, 1], then x = = 0 - a nonzero polynomial has only a finite number of zeros) so that L is injective (see Theorem 2.8). Furthermore, every polynomial pin P. [0, 1] is of

the form p(t) = E"_oti t' for some x = (to,... , t") in F"+', which means that P, [0, 1] C R(L). Hence P"[0, 11 = R(L); that is, L is also surjective. Therefore, the linear transformation L is an isomorphism between the linear spaces F"+' and P. [0, 1). Thus, since dim F"+' = n + 1 (see Example 2I), it follows by Theorem 2.12 that

dim P"[0, 1] = n+ 1. Next consider the collection P[0, 1] of all polynomials in the variable I E [0, 1] with coefficients in F of any degree:

P[0, 11 = U P"[0,1]. neNO

Note that P[0, 11 contains the zero polynomial together with every polynomial of finite degree. It is again readily verified that, as a linear manifold of Flo,tl, P[0, 11 is itself a linear space over F. The functions pi : [0, 1] -> F, defined by pi (t) = t' for every t E [0, 1), clearly belong to P[0, 1 ] for each i E No. Consider the set B = (p; };ENO C P[0, 1]. Since any polynomial in P[0, 1] is, by definition, a (finite) linear combination of vectors in B. it follows that P[0, 1] e_ span B. Hence B spans P[0, 1] (i.e., span B = P[0, 1]). We claim that B is also linearly independent. Indeed, suppose B is not linearly independent. Then there exists in 1aj pi, for B a linear combination Pk of vectors in B\{ pk }. That is, pk some integer m E N, where (cry } , is a family of nonzero scalars and (pij IT t is a finite subset of B such that p;j * pk (i.e., i j 76 k) for every j = t, ... , m. Thus

p = Pk - F taj pij is the origin of P[0. I], which means that

P(t) = tk - E j=1

for all t E [0, 11. But this is a contradiction because p is a polynomial of degree equal

to max Ilk) U (i)1) > 1. Conclusion: B is linearly independent. Therefore, the set B = { pi };ENO is a Hamel basis for P[0, 11, and hence

dim P[0, 1] = NO

(for #B = #No = No). Thus P[0, 1] is isomorphic to the linear space X of all F-valued sequences with a finite number of nonzero entries (which was introduced in Example 21).

64

2.7

2. Algebraic Structures

Isomorphic Equivalence

Two linear spaces over the same scalar field are regarded as essentially the same linear space if they are isomorphic. Let X, Y and Z be linear spaces over the same field F. It is clear that X is isomorphic to itself (reflexivity), and Y is isomorphic to X whenever X is isomorphic to y (symmetry). Moreover, since the composition of two isomorphisms is again an isomorphism (see Problems 1.9(c) and 2.15) it follows that, if X is isomorphic to y and y is isomorphic to 2, then X is isomorphic to 2 (transitivity). Thus, if the notion of isomorphic linear spaces is restricted to a given set (for instance, to the collection of all linear manifolds Lat(X) of a linear space X), then it is an equivalence relation on that set. We shall now define an equivalence between linear transformations. Recall that GF: X -* 2 denotes the composition

of a mapping G : y - 2 and a mapping F : X -* Y. Definition 2.13. Let X, X, y and y be linear spaces over the same scalar field, where X is isomorphic to X and y is isomorphic to Y. Two linear transformations

T : X -> y and L : X - y are isomorphically equivalent if there exist isomorphisms X : X -i X and Y : y -- y such that

YT = LX. That is, T = Y-' L X (or equivalently, L = YT X -') which means that the diagram

x XI

T

y TY

X ` y commutes. Warning: If X is isomorphic to X and y is isomorphic to y, then there exists an uncountable supply of isomorphisms between X and X and between y

and y. If we take arbitrary linear transformations T : X - y and L : X -, y, it may happen that the above diagram does not commute (i.e., it may happen that YT * LX) for all isomorphisms of X onto X and all isomorphisms of y onto y. In this case T and L are not isomorphically equivalent. However, if there exists at least one pair of isomorphisms X and Y for which YT = LX, then T and L are isomorphically equivalent. It is readily verified that isomorphic equivalence deserves its name. In fact, every T E C[X, Y) is isomorphically equivalent to itself (reflexivity), and L E L[X, y] is isomorphically equivalent to T E L[X, yJ whenever T is isomorphically equivalent to L (symmetry). Moreover, if T E C[X, y] is isomorphically equivalent to L E L[X, yJ and L is isomorphically equivalent to K E L[X, yJ (so that X. X and X are isomorphic linear spaces, as well as Y, y and y), then it is easy to show that T is isomorphically equivalent to K (transitivity). Indeed, if X = x and y = y, and if we restrict the concept of isomorphic equivalence to the set LIX, yJ of all linear

2.7 Isomorphic Equivalence

65

transformations of X into Y, then isomorphic equivalence actually is an equivalence relation on C[X, Y).

An important particular case is obtained when X = y and X = y so that T lies in £[X] and L lies in £[X]. Let X and X be isomorphic linear spaces. Two linear transformations T : X -> X and L : X -> X are similar if there exists an isomorphism W : X --> X such that

WT = LW. Equivalently, if there exists an isomorphism W such the diagram

X

r>

wI

X 1w

XLX

commutes. It should be noticed now that the concept of similarity will be redefined later in Chapter 4 where the linear spaces are endowed with an additional (topological) structure. Such a redefinition will assume that all linear transformations involved in the definition of similarity are "continuous" (including the inverse of W).

Example 2N. Consider the setup of Example 21, where X and y are finitedimensional linear spaces over the same field F. Let X : X -> F" and Y : y -+ F' be two mappings defined by

Xx=(ti,...,i;n)

and

Yy=(vt, ..,vm)

for every x E X and every y E y, where (tj) f=t and {v; }"' t consist of the coordinates of x and y with respect to the bases BX and By, respectively. It is readily verified that X and Y are both isomorphisms (for fixed bases BX and By). Let F,,. t

denote the linear space (over the field F) of all n x 1 matrices (or, if you like, the linear space of all "column n-vectors" with entries in F - Example 2L). Now consider the map W,,: F" -> F,,,,1 that assigns to each n-tuple (fit , ... , k,,) in F" the n x 1 matrix i

"(fit, ")

Sn

in F. x t whose entries are the (similarly ordered) coordinates of the ordered n-tuple with respect to the canonical basis for F". It is easy to show that W. is an isomorphism between F" and F.. 1. This is called the natural isomorphism of F" onto F.. 1. Note that any mxn matrix (with entries in F) can be viewed as a linear transformation from F" 1 to F.. t : the action of an m x n matrix [a;i ] E F,,, xn on an n x 1 matrix

66

2. Algebraic Structures

e Fn

I

is simply the matrix product a] I

. .

.

aln amn

am l

1

to

which is an m x 1 matrix in lFm x 1. According to Example 2L let [A] E

1Fm xn be the unique matrix representing the linear transformation A E £[X, y] with respect to the bases BX and By. Now, if this matrix is viewed as a linear transformation of

Fn x I into Fm x 1, then the diagram

X

A)

y 1Y-'

[AI

commutes. This shows that the linear transformation A : X y is isomorphically equivalent to its matrix [A] with respect to the bases Bx and By when this matrix is viewed as a linear transformation [A]: IFnx1 lFmxl . That is,

(WmY)A = [A](WWX)

2.8 Let (Xi)

Direct Sum 1

be a finite indexed family of linear spaces over the same field F (but not

necessarily linear manifolds of the same linear space). The direct sum of (Xi " 1 denoted by (DR= 1 Xi , is the set of all ordered n-tuples (x 1, ... , xn ), with each x, in X, , where vector addition and scalar multiplication are defined as follows.

(x1.... ,xn)®(y1,... ,yn) = (xl +Y1,... ,xn+yn),

a(xl..... xn) = (axl, .... ax,,) for every (x l , .... xn) and (yl, ... , yn) in ®"..1 Xi and every a in F. It is easy to verify that the direct sum ®"-1Xi of the linear spaces {X1)°=1 is a linear space over F when vector addition (denoted by ®) and scalar multiplication are defined as above. The underlying set of the linear space ®"_1X, is the Cartesian product 1 1n=I X, of the underlying sets of each linear space X,. The origin of ®"=1 X, is the ordered n-tuple (01. .. - , On) consisting of the origins of each X,.

2.8 Direct Sum

67

If M and N are linear manifolds of a linear space X, then we may consider both their ordinary sum M + N (defined as in Section 2.2) and their direct sum M ® N. These are different linear spaces over the same field. There is however a natural

mapping 0: M ® N --> M + N, defined by

4'((x1,x2)) = xl +x2, which assigns to each pair (xl, x2) in M ® A (their sum in M + N C X. It is readily verified that 4) is a surjective linear transformation of the linear space M ®N onto the linear space M + N, but 4' is not always injective. We shall establish below a necessary and sufficient condition that 4> be injective, viz., M n N = (0). In such a case the mapping 4' is an isomorphism (called the natural isomorphism) of M ® N onto M + N, so that the direct sum M ® N and the ordinary sum M + N become isomorphic linear spaces.

Theorem 2.14. Let M and A (be linearmanifolds ofa linear space X. Thefollowing assertions are pairwise equivalent.

(a) M n N = (0). (b) For each x in M + N there exists a unique u in M and a unique v in N such

that x = u + v. (c) The natural mapping 0: M ® N -p M + N is an isomorphism. Proof. Take an arbitrary x in M +A(. If x = u t + v1 = u2 + v2, with u 1, u2 in - U2 = v1 -v2EMnN(forut-u2 E M and M and VI, V2 in N, then vi - v2 E N). Thus M n N = (0} implies that ui = u2 and vt = v2, and hence (a) =:>(b). On the other hand, if M n N # (0), then there exists a nonzero vector w in M n N. Take any nonzero vector x in M + N so that x = u + v with u in

M and vinN.Thusx = (u+w)+(v-w),whereu+wisinM andv-w is in N. Since w i 0, it follows that u + w 34 u. and hence the representation of x as a sum u + v with u in M and v in N is not unique. Therefore, if (a) does not hold, then (b) does not hold. Equivalently, (b)=(a). Finally, recall that the natural mapping 4> is linear and surjective. Since 45 is injective if and only if (b) holds (by its very definition), it follows that (b)q(c). 0

Two linear manifolds M and N of a linear space X are said to be disjoint (or algebraically disjoint) if M n N = (0). (Note that, as linear manifolds of a linear

space X, M and M can never be "disjoint" in the set-theoretical sense - the origin of X always belongs to both of them.) Therefore, if M and N are disjoint linear manifolds of a linear space X, then we may and shall identify their ordinary sum M + N with their direct sum M ® N. Such an identification is carried out

by the natural isomorphism 4> : M ®N M + N (Theorem 2.14). When we identify M 0 A (with M + N, which is a linear manifold of X, we are automatically

68

2. Algebraic Structures

identifying the pairs (u, 0) and (0, v) in M ® N with u in M and with v in N respectively. More generally, we shall be identifying the direct sums M ® (0) and

(0) ®NwithMandAf,respectively. Forinstance, ifx E M®NandMnN = (0), then Theorem 2.14 ensures that there exists a unique u in M and a unique v in

N such that x = (u, v). Hence x = (u, 0) ®(0, v) where (u, 0) E M ®(0) and (0, v) E {0) ® N (recall: M e (0) and (0) ® N are both linear manifolds of M ®N). Now identify (u, 0) with 0)) = u and (0, v) with 0((0, v)) = v, and write x = u ® v where u E M and V E N (instead of x = (u, 0) 6) (0, v) = 4 -' (u) ® 4>-'(v)). Outcome: If M and N are disjoint linear manifolds of a linear space X, then every x in M ®N has a unique decomposition with respect to M and N, denoted by x = u ® v, which is referred to as the direct sum of u in M and v in N. It should be noticed that u ® v is just another notation for (u, v) that reminds us of the algebraic structure of the linear space M ® N. What really is being added in M ®N is (u, 0) a (0, v). If M andN are disjoint linear manifolds of a linear space X, and if their (ordinary) sum is X, then we say that M and N are algebraic complements of each other. In other words, two linear manifolds M and N of a linear space X form a pair of algebraic complements in X if

X=M+N and MnN=(0). Accordingly, this can be written as X =Jet ® N and m n N = (0)

once we have identified the direct sum Me N with its isomorphic image 4 (M ® N) = M +A( = X through the natural isomorphism 0. Proposition 2.15. Let M and N be linear manifolds of a linear space X, and let BM and BN be Hamel bases for M and N, respectively.

(a) .M n N = {0) if and only if BM n BN = 0 and BM U BN is linearly independent.

(b) M + N = X and BM U BN is linearly independent if and only if BM U BN is a Hamel basis for X. In particular, if BM U BN C B. where B is a Hamel basis for X, then

(a) M n N = {0} if and only if BM n BN = 0, (8) M + N = X if and only if BM U BN = B. Proof. (a) Recall that

(0) C span (BM n BN) C span (.M n N) = M n N. Thus M nN= (0) implies span (BM n BN) = (0), which implies BM n BN = 0 (for 0 0 BM U BN). Moreover, if M n N = (0), then the union of the linearly

2.8 Direct Sum

69

independent sets BM and BN is again linearly independent (see Problem 2.3). On the other hand, recall that

(0) c 11.1 fl N = span BM f1 span BN = span (BM fl BN) when (BM U BN) is linearly independent (see Problem 2.4). Thus BM fl BN = 0 implies span (BM fl BN) = {0}, and hence M nN = (0).

(b) Next recall that

span (BM U BN) = span (M U N) = M + N c X whenever BM and BN are Hamel bases for M and N, respectively. If BM U BN is a Hamel basis for X, then BM U BN is linearly independent and X = span (BM U BN)

sothatM+N = X. On the other hand,ifM +N = X, then span (BMUBN) = X. Thus, according to Theorem 2.6, there exists a Hamel basis B' for X such that B' c BM U BN. If BM U BN is linearly independent, then Theorem 2.5 ensures that there exists a Hamel basis B for X such that BM U BN a B. Therefore B' c B. But a Hamel basis is maximal (see Claim 2 in the proof of Theorem 2.5) so that B' = B. Hence BM U BN = B. 0

Theorem 2.16. Every linear manifold has an algebraic complement.

Proof. Let M be a linear manifold of a linear space X, let BM be a Hamel basis for M. and let B be a Hamel basis for X such that BM C B (see Theorem 2.5). Set BN = B\BM (which, as a subset of a linearly independent set B, is linearly independent itself) and N = span BN (a linear manifold of X). Thus BM and BN are Hamel bases for M and N. respectively, both included in the Hamel basis B for X. Since BM fl BN = 0 and BM U BN = B. it follows by Proposition 2.15 that N is an algebraic complement of M. 17

Lemma 2.17. Let M be a linear manifold of a linear space X. Every algebraic complement of M is isomorphic to the quotient space X/M. Proof. Let M be a linear manifold of a linear space X over a field IF, and let X/M be the quotient space of X modulo M, which is again a linear space over F (see Example 2H). Let >r be the natural mapping of X onto X/M defined in Section 1.4. That is, for each x E X set 7r (x) = [x] = x + M E X/M. Example 2H shows that n : X -> X/M in fact is a linear transformation. Let )C be a linearmanifold of X and consider the restriction of n to K, rr IX: K -). X IM, which is again a linear transformation. Claim. If K is an algebraic complement of M, then n1K is invertible.

Proof. Take an arbitrary [x] in X/M so that [x] = x + M for some x in X. Since X = M + K, it follows that x = u + v with u in M and v in K. Thus [x] = [v]

(reason: [x] = [u + v] = [u] + [v] = [v] because [u] = u + M = M = [0] for

70

2. Algebraic Structures

every u in M). Hence [x] = 7rIZ(v) and so [x] E R(7rlx). Conclusion: 7riK is surjective. Moreover, if 7r f K(v) = [0] for some v E 1C, then v + M = [v] = [0] = M and hence v E M. Since m n; = (0), it follows that v = 0. That is, the null space of 7r Ix is the singleton (0),N(7rI,) = {v E IC: 7rIK(v) = [01) = (0), which means that 7r Ix is injective (see Theorem 2.8). L]

Therefore, 7r IX is an isomorphism of K onto X/M whenever 1C is an algebraic complement of M. D

Theorem 2.18. Let M be a linear manifold of a linear space X. Every algebraic complement of M has the same dimension. Proof. According to Theorem 2.12 the above statement can be rewritten as follows. If N and IC are algebraic complements of M, then IC and N are isomorphic. But

this is a straightforward consequence of the previous lemma: N and X are both isomorphic to X IM, and hence isomorphic to each other. 17 The dimension of an algebraic complement of M is therefore a property of M (i.e., it is an invariant for M). We refer to this invariant as the codimension of M: the codimension of a linear manifold M, denoted by codim M. is the (constant) dimension of any algebraic complement of M.

2.9

Projections

Aprojection is an idempotent linear transformation of a linear space into itself. Thus,

if X is a linear space, then P E £[X] is a projection if and only if P = P2. Briefly, projections are the idempotent elements of C[XI. Clearly, the null transformation 0 and the identity 1, both in C[X], are projections. A nontrivial projection in C[X] is a projection P such that 0 y6 P A 1. It is easy to verify that, if P is a projection, then so is I - P. Moreover, the null spaces and ranges of P and I - P are related as follows (see Problem 1.4).

R(P) = N(! - P)

and

N(P) = R(1 - P).

Projections are singularly useful linear transformations. One of their main properties is that the range and the null space of a projection form a pair of algebraic complements.

Theorem 2.19. If P E L[X] is a projection, then R(P) and N(P) are algebraic complements of each other.

Proof. Let X be a linear space, and let P: X -+ X be a projection. Recall that both the range 1Z(P) and the null space N(P) are linear manifolds of X (since P is linear). Since P is idempotent, it follows that

R(P) _ {x E X: Px = x}

2.9 Projections

71

(the range of an idempotent mapping is the set of all its fixed points - Problem 1.4). If x E R(P) n N(P), then x = Px = 0, and hence

R(P) n N(P) = (01. Moreover, for an arbitrary vector x in X. write x = Px + (x - Px). Since P(x - Px) = Px - P2x = 0 (recall: P is linear and idempotent), it follows that (x - Px) E N(P). Hence x = u + v with u = Px in R(P) and v = (x - Px) in N(P). Therefore

X = R(P) +N(P).

0

On the other hand, for any pair of algebraic complements there exists a unique projection whose range and null space coincide with them.

Theorem 2.20. Let M and N be linear manifolds of a linear space X. If M and N are algebraic complements of each other then there exists a unique projection

P : X -- X such that R(P) = M and N(P) = N. Proof. Suppose M and N are algebraic complements in a linear space X so that

M+N=X and MnN=(o1. According to Theorem 2.14, for each x E X there exists a unique u E M and a unique v E N such that x = u + v. Let P : X -* X be the function that assigns to each x in X its unique summand u in M (i.e., Px = u). It is easy to verify that P is linear. Moreover, for each vector x in X, P2x = P(Px) = Pu = u = Px (reason: u is itself its unique summand in M), so that P is idempotent. By the very definition

of P we get R(P) = M and N(P) = Al'. Conclusion: P: X -- X is a projection with R(P) = M and N(P) = N. Now let P': X -> X be any projection with IZ(P') = M and N(P') = N. Take an arbitrary x E X and consider again its unique representation as x = u + v with u E M = R(P') and V E N = N(P'). Since P' is linear and idempotent, it follows that P'x = P'u + P'v = u = Px. Therefore P = P. o Remark: An immediate corollary of Theorems 2.16 and 2.20 says that any linear manifold of a linear space is the range of some projection. That is, if M is a linear manifold of a linear space X, then there exists a projection P : X -* X such that

R(P) = M. If M and N are algebraic complements in a linear space X, then the unique projection P in L[X] with range R(P) = M and null space N(P) = N is called the projection on M along N. If P is the projection on M along H. then the projection onN along M is precisely the projection E = I - P in G[X]. Note that

EP=PE=0.

72

2. Algebraic Structures

Proposition 2.21. Let M and N be linear manifolds of a linear space X. If M and N are algebraic complements of each other then the unique decomposition of each x in X = M ® H as a direct sum

of u in M and v in N is such that

u = Px

and

v = (I - P)x

where P : X -> X is the unique projection on M along M. Proof. Take an arbitrary x in X and consider its unique decomposition x = u ® v in X = M ®M. Note that the identification of M ® M with M+M = X is implicitly assumed in the proposition statement. Now write x = (u, v) and set Px = (u, 0). The very same argument used in the proof of Theorem 2.20 can be applied here to

verify that this actually defines a unique projection P : M ® M -+ M ® N such that R(P) = M ® {0} and N(P) = (0) H. Finally, identify M ® (0) and (0) ®N with M and N (and hence (u. 0) and (0, v) with a and v), respectively. o

According to Theorem 2.16, every linear space X can be represented as the

sum X = M + M of a pair (M, N) of algebraic complements in X. If M N is identified with M + M, then this means that every linear space X has a decomposition X = M ® M as a direct sum of disjoint linear manifolds of X. Proposition 2.22. Let X be a linear space and consider its decomposition

X=M®N as a direct sum of disjoint linear manifolds M andN of X. Let P : X -+ X be the projection on M along M, and let E = I - P be the projection on N along M. Every linear transformation L : X - X can be written as a 2 x 2 matrix with linear transformation entries

L=A

B D

,

where A=PLIM:M-+M, B= PLIN:M-+M, C=ELIM:M-+Nand D = ELI,: M-*N. Proof. Let M and N be linear manifolds of a linear space X. Suppose M andN are algebraic complements of each other and consider the decomposition X = M ®N. Let L be a linear transformation on M ®M so that L E L(XJ. Take an arbitrary x E X and consider its unique decomposition x = u ® v in X = M ® M with u in

M and v in M. Now write x = (u, v) so that Lx = L(u, v) = L((u. 0) a (0, v)) = L(u. 0) ® L(0. v) = LAM(Dtol(u, 0) ® LIloli(0, v). Identifying M ® (0) and

2.9 Projections

73

(0) ® N with M and N (and hence (u, 0) and (0, v) with u and v), respectively, it follows that

Lx = LIMu ®LINv, where L IM u and L IN v lie in X = M ® N. Proposition 2.21 says that we may write

LIMu = PLIMu®ELIMu, LIArv = PLINv®ELINv, where P is the unique projection on M along N and E = I - P. Therefore

Lx = (PLIMu + PLINv) ®(ELIMu + ELINv), where PL IM u + PL IN u is in M and E L I M u + EL I N v is in N. Since the ranges

of PLIM and PLIN are included in R(P) = M, we may think of them as linear transformations into M. Similarly, ELIM and ELIN can be thought of as linear transformations into N. Thus set A = PLIM EC[M], B = PLIN E G[N, MI,

C= ELIM EL[M,N]and D=ELINEL[N]so that Lx = (Au + Bv, Cu + Dv) E M ® N for every x = (u, v) E M N. In terms of standard matrix notation, the vector Lx in M ® N can be viewed as a 2 x 1 matrix with the first entry in M and the other in N, namely (Au+Ov This is precisely the action of the 2x2 matrix with linear transformation entries, (C D), on the 2x 1 matrix with entries in M and N representing x, namely (uv). Thus Lx = (A BD)( "), and hence we write L = (c n).

0 Example 20. Consider the setup of Proposition 2.22. Note that the projection on M along .,V can be written as

P=(1O

0) 0/J

with respect to the decomposition X = M ® N, where 1 denotes the identity on M.

Thus L P = (A o) and P L P = (o o ), so that L P = PL P if and only if C = 0. Now note that M is L-invariant (i.e., L(M) c M) if and only if PLIM = LIM or, equivalently, if and only if ELIM = 0 (recall: E = I - P). Therefore,

L(M)cM

A = LIM

C=0

Conclusion 1: The following assertions are pairwise equivalent.

(a) M is L-invariant.

(b) L=(`' 0

DB)

LP = PLP.

74

2. Algebraic Structures

(c) LP =PLP. Similarly, if we apply the same argument to N, then

L(N)cN b D=LIN e== B=0

PL=PLP.

Conclusion 2: The following assertions are pairwise equivalent as well.

(a) M and N are both L-invariant.

(b) L =

(L 001M

°)

L 1q/

(c) L and P commute (i.e., PL = LP). Let M and N be algebraic complements in a linear space X. If a linear transformation L in £[X] is represented as L = (v v) in terms of the decomposition X = M ® A ((as in (a) above), where A E L[M] and D E C[N], then it is usual to write L = A (D D. For instance, the projection on M along N, which is represented

with respect to the same decomposition X = M ®N, is usually written as Po) = 19 0. These are examples of the following concept. Let (Xi)"=1 be a finite indexed family of linear spaces over the same scalar field and consider their direct sum ®"=1Xi. Now let (L1)"=1 be a (similarly indexed) family of linear transformations such that Li E G[Xi] for every index i. The direct

as P = (o

sum of (Li 1,1=1, denoted by ®in_ I Li, is the mapping of ®" I Xi into itself defined by n

®Li(xt,...

, xn) = (L1x1.... , Lnxn)

i=1

for every (x1, ... , xn) in ®"=1 Xi. It is readily verified that ®" I Li is linear (i.e.,

®" I Li E G[®in_I Xi J) and also that, for every index i, n

(DL)I x, =

Li.

i=1

Observe that the above identity actually is a short notation for the following assertion:

"if Oi is the origin of each Xi and Oi is the unique (linear) transformation of (0j) onto itself, then each linear manifold (01) ® - ®{0i _ I) ®Xi ® (Oi+I) ®... ® (On) of ®"=1 Xi is invariant for ®"_I Li and the restriction of ®°=1 Li to that invariant linear manifold is the direct sum 01 ® . . . ® 0j-1 ® Li ® 0i+1 ® . . . ® On". Of course, we shall always use the short notation. Conversely, if L E G[®" 1Xi] is such that LIxi E G[X1] for every index i, then L is the direct sum of (LIx1)" I. That is, if each Xi in ®"=I Xi is invariant for L E G[®"1 Xi ], then

Llx.

L= ,=1

Problems

75

Summing up: Set X = ®"_ i X; and consider linear transformations L; in L[X; ] for each i and L in L[X]. n

L=®L;

if and only if

L;=LIX;

-t

for every index i (so that each X;, viewed as a linear manifold of the linear space (D'"=Mi. is invariant for L). The linear transformations (Li ) are referred to as the direct summands of L. In particular, if we consider the decomposition X = M ® N of a linear space X into the direct sum of a pair of algebraic complements M and N in X, and if we take linear transformations L E G[X], A E G[M] and D E L[N], then

L = (A D) = A ® D

if and only if

A = LIM and D = LIN

(so that M and N are both L-invariant), where A and D are the direct summands of L with respect to the decomposition X = M ® N.

Suggested Reading Brown and Pearcy [2] Halmos [2] Herstein [1] MacLane and Birkhoff [1]

Naylor and Sell [1] Roman [I ] Simmons [ 1 ]

Taylor and Lay [ I ]

Problems Problem 2.1. Let X be a linear space over a field F. Take arbitrary a and 6 in F and arbitrary x, y and z in X. Verify the following propositions. (a)

(-a)x = -(ax).

(b) Ox =0=a0.

(c) ax = 0 ;

a = 0 or x = 0.

(d) x+y=x+z : y=z. (e) ax = ay = x=y if aA0. (f) ax =fix

a=0 if x54 0.

76

2. Algebraic Structures

Problem 2.2. Let X be a real or complex linear space. A subset C of X is convex if ax + (1 - a) Y E C whenever x, y E C and O< a< 1. A vector x E X is a convex linear combination of vectors in X if there exists a finite set (x;) t of vectors in X and a finite family of nonnegative scalars {a;}" t such that x = E"_ta;x; and E"_ t ai = 1. If A is a subset of X, then the intersection of all convex sets containing A is called the convex hull of A, denoted by co(A). (a) Show that the intersection of an arbitrary nonempty collection of convex sets is convex.

(b) Show that co(A) is the smallest (in the inclusion ordering) convex set that contains A.

(c) Show that C is convex if and only if every convex linear combination of vectors in C belongs to C.

Hint: To verify that every convex linear combination of vectors in a convex set C belongs to C proceed as follows. Note that the italicized result holds for any convex linear combination of two vectors in C (by the definition of a convex set). Suppose it holds for every convex linear combination of n vectors in C, for some n E N. This implies that aE"_ta-t a; xi lies in C

whenever {x;}"+ C C anda; = 1 with0 < (reason:

a

="_tai

a- a; x; E C). Now conclude the proof by induction.

(d) Show that co(A) coincides with the set of all convex linear combinations of vectors in A. Hint: Let clc(A) denote the set of all convex linear combinations of vectors in A. Verify that clc(A) is a convex set. Now use (b) and (c) to show that co(A) C clc(A) C clc(co(A)) = co(A).

Problem 2.3. Let M and N be linear manifolds of a linear space X, and let A and B be linearly independent subsets of M and N. respectively. If M n.h = (0). then A U B is linearly independent. (Hint: If a E A is a linear combination of vectors in A U B, then a = b + a' for some a' E M and some b E N.) Problem 2.4. Let A be a linearly independent subset of a linear space X. If B and C are subsets of A, then

span (B f1 C) = span B fl span C. Hint: Show that span B n span C c span (B U C) by Proposition 2.3.

Problem 2.5. If a subset A of a linear space X spans X, then the cardinality of every linearly independent subset of X is less than or equal to the cardinality of A.

Problems

77

Hint. Suppose A is a subset of a linear space X that spans X. Let B be an arbitrary Hamel basis for X, and let C be an arbitrary linearly independent subset of X. Show

that #C < #B < #A. (Apply Theorems 2.5, 2.6 and 2.7 - see Problems 1.21(a) and 1.22). Note that this generalizes Claim 0 in the proof of Theorem 2.7 for subsets of arbitrary cardinality.

Problem 2.6. Let X be a linear space, and let M be a linear manifold of X. Verify the following propositions.

(a) dim M = 0 if and only if M = (0). (b) dim M < dim X.

(Hint: Problem 2.5.)

Problem 2.7. If M is a proper linear manifold of afinite-dimensional linear space X, then dim M < dim X. Prove the above statement and show that it does not hold for infinite-dimensional linear spaces (e.g., show that dim X = dim Xp where X is the linear space of Example 2J and Xa = (x = (l;], l;Z, h, ... ,) E X: l;'t = 0}). Problem 2.8. Let X be a nonzero linear space over an infinite field, and let B be a Hamel basis for X. Recall that every nonzero vector x in X has a unique representation in terms of B. That is. for each x # 0 in X there exists a unique nonempty finite subset Bx of B and a unique finite family of nonzero scalars (ab}bEB, C iF such that

x = E abb. be B,

For each positive integer n E N, let X" be the set of all nonzero vectors in X whose representations as a (finite) linear combination of vectors in B have exactly n (nonzero) summands. That is, for each n E N, set

X" = {x E X: #Bs = n}. (a) Prove that #X" = #(IF x B) for all n E N. Hint: Show that #X, = #(IF" x B) and recall: if F is an infinite set, then #F" _ #IF (Problems 1.23 and 1.28).

(b) Apply Theorem 1.10 to show that #(U"ENXf) < #(Fx B). (c) Verify that {X"}IEN is a partition of X\{0}. Thus conclude from (b) and (c) (see Problem 1.28(a)) that

#X = #(IFx B) = max{#IF, dim X}.

78

2. Algebraic Structures

Problem 2.9. Prove the following proposition, which is known as the Principle of Superposition. A mapping L : X - Y. where X and Y are linear spaces over the same scalar field, is a linear transformation if and only if n

L(>aixi) i=1

n

= Ea1Lxi i=1

for all finite sets (xi)"_ 1 of vectors in X and all finite families of scalars (a i }" 1.

Problem 2.10. Let L : X -+ Y be a linear transformation. Show that the null space N(L) and the range R(L) of L are linear manifolds (of the linear spaces X and Y. respectively).

Problem 2.11. Let L : X --> y be a linear transformation of a linear space X into a linear space Y. Prove the following propositions. (a) If M is a linear manifold of X. then L(M) is a linear manifold of Y (i.e.. the linear image of a linear manifold is a linear manifold).

(b) If N is a linear manifold of y, then L-1 (N) is a linear manifold of X (i.e., the inverse image of a linear manifold under a linear transformation is again a linear manifold).

Problem 2.12. Let X and Y be linear spaces, and let L : X -+ Y be a linear transformation. Show that the following assertions are equivalent.

(a) A C X is a linear manifold whenever L(A) C y is a linear manifold.

(b) N(L) = (0). Hint: Give a direct proof for (b) =*(a) by using Problems 1.3(d) and 2.11(b). Give a recall: if x is a nonzero vector in X, then (x) contrapositive proof for is not a linear manifold of X.

Problem 2.13. Show that the set £[X, y] of all linear transformations of a linear space X into a linear space Y is itself a linear space (over the same common field of X and y) when vector addition and scalar multiplication in G[X, Y] are defined pointwise as in Example 2F.

Problem 2.14. The restriction L I M : M -- y of a linear transformation L : X -* Y to a linear manifold M of X is itself a linear transformation. Problem 2.15. Prove that a composition of two linear transformations is again a linear transformation.

Problems

79

Problem 2.16. Let L : X -s Y be a linear transformation. It is trivially verified that, if L is surjective. then dim R(L) = dim Y. Now verify that, if L is injective, then dim R(L) = dim X. Problem 2.17. Let L : X -+ Y be a linear transformation of a linear space X into a linear space Y. The dimension of the range of L is the rank of L, and the dimension of the null space of L is the nullity of L. Show that rank and nullity are related as follows.

dim N(L) + dim R(L) = dim X. (Hint: Addition of cardinal numbers was defined in Problem 1.30.)

Problem 2.18. If dim R(L) is finite, then L is called a finite-dimensional (or a finite-rank) linear transformation. Clearly, if)' is a finite-dimensional linear space, then every L E C[X, YJ is finite-dimensional. Verify that, if X is a finite-dimensional linear space, then every L E £[X. yJ is finite-dimensional. Moreover, if L : X -+ )' is a finite-dimensional linear transformation (so that R(L) is a finitedimensional linear manifold of y), then show that (a)

L is injective if and only if dim R(L) = dim X,

(b) L is surjective if and only if dim R(L) = dim Y. Problem 2.19. let X be a linear space over a field F. and let XNO be the linear space (over the same field F) of all X-valued sequences (x JnEN0 Suppose A is a linear transformation of X into itself. Take an arbitrary sequence u = (un )nENo in XNO and consider the (unique) sequence x = (xn )nENo in XNO which is recursively

defined as follows. Set xo = uo and, for each n E No, let

xn+l = Axn + un+1 Prove by induction that n

xn = E A"-'u1 i=o

for every n E No, where AO is (by definition) the identity I in LIX]. Let L: XNO - XNo be the map that assigns to each sequence u in XNO this unique sequence x in XNO, so that

x = Lu. Show that L is a linear transformation of XNo into itself. The recursive equation (or the difference equation) xn+l = Ax" + un+ 1 is called a discrete linear dynamical

system because L is linear. Its unique solution is given by x = Lu (i.e., xn = ui for every n E fro).

80

2. Algebraic Structures

Problem 2.20. Let F denote either the real or complex field, and let X and Y be linear spaces over F. For any polynomial p (in one variable in F, with coefficients in IF, and of any finite order n) set n

p(L) = >2aiL' i=o

in £[X] for every L E G[X], where the coefficients (ai )"O lie in IF (note: LO = 1).

Take L E £[X], K E C[y] and M E £[X,YJ arbitrary, and prove the following implication.

(a) If ML = KM, then Mp(L) = p(K)M for any polynomial p. Thus conclude: p(L) is similar to p(K) whenever L is similar to K. A linear transformation L in G[X J is called nilpotent if L" = 0 for some integer n E N, and algebraic if p(L) = 0 for some polynomial p. It is clear that every nilpotent linear transformation is algebraic. Prove the following propositions.

(b) A linear transformation is similar to an algebraic (nilpotent) linear transformation if and only if it is itself algebraic (nilpotent). (c) Sum and composition of nilpotent linear transformations are not necessarily nilpotent.

Hint: The matrices T = (o o) and L = (o

in £(C2] are both nilpotent. o) L + T is an involution. L T and TL are idempotent. Problem 2.21. Let IF denote either the real or complex field, and let X be a linear

space over F. A subset K of X is a cone (with vertex at the origin) if ax E K whenever x E K and a > 0. Recall the definition of a convex set in Problem 2.2 and verify the following assertions.

(a) Every linear manifold is a convex cone. (b) The union of nonzero disjoint linear manifolds is a nonconvex cone. Let S be a nonempty set and consider the linear space IFS. Show that (c) {x E IFS : x(s) > 0 for all s E S} is a convex cone in FS.

Problem 2.22. Show that the implication (a)=(b) in Theorem 2.14 does not generalize to three linear manifolds, say M, Al and TL, if we simply assume that they are pairwise disjoint. (Hint: R3.)

Problem 2.23. Let {M; }"_, be a finite collection of linear manifolds of a linear space X. Show that the following assertions are equivalent.

Problems

81

(a) M i n r_nj=i joiMj= (0) for every i = i , .... n. (b) For each x in F,,,= I Mi there exists a unique n-tuple (XI , ... ,

in ]l" 1 Mi

such that x = E"=Ixi. Hint. (a)=(b) for n = 2 by Theorem 2.14. Let n > 2 and suppose (a)=(b) for every 2 < m < n. Show that, if (a) holds true form + 1, then (b) holds true form + 1. Now conclude the proof of (a)=(b) by induction inn. Next show that (b)=(a) by Theorem 2.14.

Problem 2.24. Let {M, }"_, be a finite collection of linear manifolds of a linear

space X. and let Bi be a Hamel basis for each Mi. If Mi n E"_, i#iMi = (0) for every i = i, ... , n, then U"=i Bi is a Hamel basis for F_,"_1 Mi. Prove. Hint: The result holds for n = 2 by Proposition 2.15. Use the hint of the previous problem.

Problem 2.25. Let M and Al be linear manifolds of a linear space. (a) If M and N are disjoint, then

dim(M ®N) = dim(M + N) = dim M + dimN. Hint: Problem 1.30, Theorem 2.14 and Proposition 2.15. (b) If M and N are finite-dimensional, then

dim(M + N) = dim M + dim N - dim(M n N). Problem 2.26. Let M be a proper linear manifold of a linear space X so that M E Cat(X)\(X). Consider the inclusion ordering of Cat(X) and show that

M is maximal in ,Car(X)\(X}

codimM = 1.

Problem 2.27. Let rp be a nonzero linear functional on a linear space X (i.e., let (P be a nonzero element of X', the algebraic dual of X). Prove that

(a) N(cp) is maximal in Lat(X)\(X). That is, the null space of any nonzero linear functional in X' is a maximal proper lin-

ear manifold of X. Conversely, if M is a maximal linear manifold in Gat(X)\{X), then there exists a nonzero cp in X' such that M = N((p). In other words, prove the following assertion.

(b) Every maximal element of Lat(X)\(X} is the null space of some nonzero q in X'.

82

2. Algebraic Structures

Problem 2.28. Let X be a linear space over a field F. The set

H,,,,, = {x E X: q,(x) = a}, determined by a nonzero tp in X' and a scalar a in F. is called a hyperplane in X. It is clear that H,,,o coincides with N((p), but H., is not a linear manifold of X if Cr is a nonzero scalar. A linear variety is a translation of a proper linear manifold. That is, a linear variety V is a subset of X that coincides with the cosec of x modulo M.

V = M+x = (yEX: y=z+x forsome zEMI, for some x E X and some M E Lat(X)\(X). If M is maximal in Gat(X)\(X), then M + x is called a maximal linear variety. Show that a hyperplane is precisely a maximal linear variety. Problem 2.29. Let X be a linear space over a field IF, and let P and E be projections

in £[X]. Suppose E j4 0, and let a be an arbitrary nonzero scalar in F. Prove the following proposition.

(a) P + aE is a projection if and only if PE + EP = (1 - a)E. Moreover, if P + aE is a projection, then show that

(b) E and P commute (i.e., EP = PE) and EP is a projection,

(c) EP = O if and only if a = 1 and EP # O if and only if a = -1. Therefore, (d) if P + a E is a projection, then a = 1 or a = -1.

Finally conclude that

P + E is a projection if and only if EP = PE = 0, P - E is a projection if and only if EP = PE = E. Problem 2.30. An algebra (or a linear algebra) is a linear space A that is also a ring with respect to a second binary operation on A called product (notation: xy E A is the product of x E A and y E A). The product is related to scalar multiplication by the property

a(xy) = (ax)y = x(ay) for every x, y E A and every scalar a. We shall refer to a real or complex algebra if A is a real or complex linear space. Recall that this new binary operation on A (i.e., the product in the ring A) is associative,

x(yz) = (xy)z,

Problems

83

and distributive with respect to vector addition, x(y + z) = xy + xz

and

(y + z)x = yx + zx,

for every x, y and z in A. If A possesses a neutral element 1 under the product operation (i.e., if there exists I E A such that x I = lx = x for every x E A), then A is said to be an algebra with identity (or a unital algebra). Such a neutral element 1 is called the identity (or unit) of A. If A is an algebra with identity, and if x E A has an inverse (denoted by x I) with respect to the product operation (i.e., if there exists z' I E A such that xx - t = x - I x = 1), then x is an invertible element of A. Recall that the identity is unique if it exists, and so is the inverse of an invertible element of A. If the product operation is commutative, then A is said to be a commutative algebra.

(a) Let X be a linear space of dimension greater than one. Show that C[X] is a noncommutative algebra with identity when the product in L[X] is interpreted

as composition (i.e., LT = L o T for every L, T E C[X]). The identity I in £[X] is precisely the neutral element under the product operation. L is an invertible of C[X] if and only if L is injective and surjective. Asubalgebra of A is a linear manifold M of A (when A is viewed as a linear space) which is an algebra in its own right with respect to the product operation of A (i.e.,

uv E M whenever u E M and V E M). A subalgebra M of A is a left ideal of A if ux E M whenever u E M and X E A. A right ideal of A is a subalgebra M of A such that xu E M whenever x E A and u E M. An ideal (or a two-sided ideal or a bilateral ideal) of A is a subalgebra I of A that is both a left ideal and a right ideal.

(b) Let X be an infinite-dimensional linear space. Show that the set of all finitedimensional linear transformations in £[X] is a proper left ideal of G[X] with no identity. (Hint: Problem 2.25(b).)

(c) Show that, if A is an algebra and I is a proper ideal of A, then the quotient space All of A modulo I is an algebra. This is called the quotient algebra of A with respect to Z. If A has an identity 1, then the cosec 1 + I is the identity

of A/I. Hint: Recall that vector addition and scalar multiplication in the linear space

All are defined by

(x+I)+(y+I) = (x+y)+I, a(x + I) = ax + I, for every x, y E A and every scalar a (see Example 2H). Now show that the product of cosets in A/I can be likewise defined by

(x+I)(y+I) = xy+I

84

2. Algebraic Structures

forevery x,yEA(i.e.,ifx'=x+uandy'=y+v, withx,y E Aand U. V E Z, then there exists z E I such that x'y' + w = xy + z for any w E Z, whenever I is a two-sided ideal of A).

3 Topological Structures

The basic concept behind the subject of point-set topology is the notion of "closeness" between two points in a set X. In order to get a numerical gauge of how close together two points in X may be, we shall provide an extra structure to X, viz., a topological structure, that again goes beyond its purely set-theoretic structure. For most of our purposes the notion of closeness associated with a metric will be sufficient, and this leads to the concept of "metric space": a set upon which a "metric" is defined. The metric-space structure that a set acquires when a metric is defined on it is a special kind of topological structure. Metric spaces comprise the kernel of this chapter but general topological spaces are also introduced.

3.1

Metric Spaces

A metric (or metric function, or distance function) is a real-valued function on the Cartesian product of an arbitrary set with itself that has the following four properties, called the metric axioms.

Definition 3.1. Let X be an arbitrary set. A real-valued function d on the Cartesian product X x X,

d:XxX-->II8.

86

3. Topological Structures

is a metric on X if the following conditions are satisfied for all x, y and z in X. (i) d(x, y) > 0 and d(x, x) = 0 (ii) d(x, y) = 0 only if x = y (iii) d(x, y) = d(y, x) (iv) d (x, y) < d (x, z) + d (z, y)

(nonnegativeness), (positiveness), (symmetry),

(triangle inequality).

A set X equipped with a metric on it is a metric space. A word on notation and terminology. The value of the metric don a pair of points of X is called the distance between those points. According to the above definition a metric space actually is an ordered pair (X, d) where X is an arbitrary set, called

the underlying set of the metric space (X, d), and d is a metric function defined on it. We shall often refer to a metric space in several ways. Sometimes we shall speak of X itself as a metric space when the metric d is either clear in the context or is immaterial. In this case we shall simply say "X is a metric space". On the other hand, in order to avoid confusion among different metric spaces, we may occasionally insert a subscript on the metrics. For instance, (X, dX) and (Y, dy) will stand for metric spaces where X and Y are the respective underlying sets, dX denotes the metric on X, and dy the metric on Y. Moreover, if a set X can be equipped with more than one metric, say di and d2, then (X, d1) and (X, d2) will represent different metric spaces with the same underlying set X. In brief, a metric space is an arbitrary set with an additional structure defined by means of a metric d. Such an additional structure is the topological structure induced by the metric d. If (X, d) is a metric space, and if A is a subset of X, then it is easy to show that the restriction d I A x A: A x A-+ R of the metric d to A x A is a metric on A - the so-called relative metric. Equipped with the relative metric, A is a subspace of X. We shall drop the subscript A x A from dIA xA and say that (A, d) is a subspace of

(X, d). Thus a subspace of a metric space (X, d) is a subset A of the underlying set X equipped with the relative metric, which is itself a metric space. Roughly speaking, A inherits the metric of (X, d). If (A. d) is a subspace of (X, d) and A is a proper subset of X, then (A, d) is said to be a proper subspace of the metric space (X, d). Example 3A. The function d : R x R -+ R defined by

d(a, 0) = Ia - 0I for every a. fi E R is a metric on R. That is, it satisfies all the metric axioms in Definition 3.1, where la I stands for the absolute value of a E R: la I = (a2) 2. This is the usual metric on R. The real line R equipped with its usual metric is the most important concrete metric space. If we refer to R as a metric space without specifying a metric on it, then it is understood that R has been equipped with its

usual metric. Similarly, the function d: CxC - R given by d(, v) = l!; - vl for every , v E C is a metric on C. (Again, Ii; I stands for the absolute value (or

3.1 Metric Spaces

87

modulus) of a complex number r; : Il; I = (11;) i with the upper bar denoting complex

conjugate). This is the usual metric on C. More generally, let F denote either the real field R or the complex field C, and let F" be the set of all ordered n-tuples of scalars in F. For each real number p > 1, consider the function dP : F" xF" R defined by I41 - v1iP) r,

dp(x, y) i=1

and also the function doo: F" x F" -i R given by doo (x, y) = max

i
v; I,

f o r every x = (i;1, ... , ln) and y = (v1, ... , vn) in F". These are metrics on F". Indeed, all the metric axioms up to the triangle inequality are trivially verified. The triangle inequality follows from the Minkowski inequality (see Problem 3.4(a)). Note that (Q", dP) is a subspace of (R", dP) and (Q", dam) is a subspace of (R", doo).

The special (very special, really) metric space (R", d2) is called n-dimensional Euclidean space and d2 is the Euclidean metric on W. The metric space (C", d2) is called n-dimensional unitary space. The singular role played by the metric d2 will become clear in due course. Recall that the notion of a bounded subset was defined for partially ordered sets in Section 1.5. In particular, boundedness is well-defined for subsets of the simply ordered set (R, <): the set of all real numbers R equipped with its natural ordering < (see Section 1.6). Let us introduce a suitable and common notation for a subset of R that is bounded above. Since the simply ordered set R is a boundedly complete lattice (Example IC), it follows that a subset R of R is bounded above if and only if it has a supremum, sup R, in R. In such a case we shall write sup R < oo. Thus the notation sup R < oo simply means that R is a subset of R which is bounded above. Otherwise (i.e., if R C_ R is not bounded above) we write sup R = oo. With this in mind we shall extend the notion of boundedness from (R, <) to a metric space (X, d) as follows. A nonempty subset A of X is a bounded set in the metric space

(X, d) if sup d(x, y) < oo. x.yEA

That is, A is bounded in (X, d) if (d(x, y) E R: x, y E A} is a bounded subset of (R, <). Equivalently, if {d(x, y) E R: x, y E A} is bounded above in R, since 0 < d (x, y) for every x, y E X. An unbounded set is, of course, a set A that is not bounded in (X, d). The diameter of a nonempty bounded subset A of X (notation: diam(A)) is defined by

diam(A) = sup d(x, y) x,yEA

so that diam(A) < oo whenever a nonempty set A is bounded in (X, d). By convention the empty set 0 is bounded and diam(O) = 0. If A is unbounded we write

88

3. Topological Structures

diam(A) = oo. Let F be a function of a set S to a metric space (Y, d). F is a bounded

function if its range, R(F) = F(S), is a bounded subset in (Y. d); that is, if

sup d(F(s), F(t)) < oo. SEES

Note that R is bounded as a subset of the metric space R equipped with its usual metric if and only if R is bounded as a subset of the simply ordered set R equipped with its natural ordering. Thus the notion of a bounded subset of R and the notion of a bounded real-valued function on an arbitrary set S are both unambiguously defined.

Proposition 3.2. Let S be set and let F denote either the real field R or the complex field C. Equip F with its usual metric. A junction w E Fs (i.e., (p: S -)- F) is bounded if and only if sup Ik(s)I < 00. sES

Proof. Consider a function tp from a set S to the field F. Take s and t arbitrary in S, and let d be the usual metric on F (see Example 3A). Since sp(s), (p(t) E F. it follows by Problem 3.1(a) that Iw(S)I - Iw(t)I

-<

<

1Iw(s)I - Iw(t)Ij = I d (w(s), 0) - d(O. (p(t))I Iw(s)I + Iw(t)I. d(w(s), (p(t)) = Iw(s) - w(t)I

If sup5Eslw(s)I < oo (i.e., if (IQ(s)I E R: S E S} is bounded above) then

d(w(s), (p(t)) -< 2sup Iw(s)I, ZES

so that the function w is bounded. and hence sup,,,es d(w(s), (p(t )) < 2 On the other hand, if sups.rES d(w(s), (p(t)) < oo, then Iw(s)I <- sup d(w(s). w(t)) + Iw(t)I,

JjEs

and the real number sup,.IEs d(w(s), (p(t)) + Iw(t)I is an upper bound for (Iw(s)I E t] R: S E S) for each t E S. Thus Sup,ESlw(s)I < 00.

Example 3B. For each real number p > 1, let l+, denote the set of all scalarvalued (real or complex) infinite sequences M 1,N in CN (or in CNO) such that Fk° I Ilk I° < oo. We shall refer to this condition by saying that the elements of f+ are p-summable sequences. Notation: F_' t l k 1 = SuPnENEk-I 141P. Thus, according to Proposition 3.2, Fk°_ l Itk I° < co means that the nonnegative sequence {F_,=, IkI°},EN is bounded as a real-valued function on N. Note that, if (tk}kEN and (vk }kEN are arbitrary sequences in f+ P, then the Minkowski inequality (Problem

3.1 Metric Spaces

89

vk l p < oo. Hence we may consider the function

3.4(b)) ensures that Ekw 1

dp: tpxt+ -* Rgivenby 00

,

dp(x, y) = E I k - ukIp

F

k=1

for every x = (ik}kEN and y = (vk)kEN in l+. We claim that dp is a metric on tp.. Indeed, as it happened in Example 3A, all the metric axioms up to the triangle inequality are readily verified; and the triangle inequality follows from the Minkowski inequality. Therefore (t{p. , d p) is a metric space for each p > 1, and the metric d p is referred to as the usual metric on t.. Now let t+° denote the set of all scalar-valued bounded sequences; that is, the set of all real or complex-valued sequences (k }kEN such that supkENI4 I < oo. Again, the Minkowski inequality (Problem 3.4(b)) ensures that SupkEN Itk - uk I < co whenever ( k }kEN and (Uk )kEN lie in l+°, and hence

we may consider the function d.: t+° x t+ -+ R defined by

do0(x, y) = sup Ilk - Ukl kEN

for every x = ( k)kEN and y = (Uk)kEN in t.OO. Proceeding as before (using the Minkowski inequality to verify the triangle inequality) it follows that (t+, do") is a metric space, and the metric d00 is referred to as the usual metric on t+°. These metric spaces are the natural generalizations (for infinite sequences) of the metric spaces considered in Example 3A, and again the metric space (t+2, d2) will play a central role in the forthcoming chapters. There are counterparts of tp} and t+° for nets in CZ. In fact, for each p > 1 let tp denote the set of all scalar-valued (real or complex) nets {k }kEZ such that E0k°___00 Il;k Ip < oo (i.e., such that the nonnegative is bounded), and let £O0 denote the set of all bounded sequence (Ek=_,, Ilk I p

nets in CZ (i.e., the set of all scalar-valued nets (k)kEZ such that supkEZItkI < 00).

The functions d: tp x t P -* R and dd : tO0 x t0O -- R, given by 00

dp(x, y) = ( E I k=-0o

for every x =

k - U k I lr

and y = (vk)kEZ in V. and d0o(x, y) = sup Ilk - Uk I kEZ

for every x = (k}kEZ and y =

in tOO, are metrics on tp (for each p > 1)

and on £00, respectively.

Let (X, d) be a metric space. If x is an arbitrary point in X, and A is an arbitrary nonempty subset of X, then the distance from x to A is the real number

d(x. A) = inf d(x, a). aEA

90

3. Topological Structures

If A and B are nonempty subsets of X, then the distance between A and B is the real number d(A, B) = inf d(a,b). a be a metric space. Let B[S, Y] denote the subset of Ys consisting of all bounded mappings of S into (Y, d). According to Problem 3.6,

supd(f(s),g(s)) < diam(R(f))+diam(R(g))+d(R.(f),R(g)) .TES

so that supsES d (f (s), g(s)) E R for every f, g E B[S, Y]. Thus we may consider the function doo : B[S, Y] x B[S, Y] -+ R defined by

doo(f, g) = supd(f (s), g(s)) sES

for each pair of mappings f, g E B[S, Y]. This is a metric on B[S, Y]. Indeed, doo clearly satisfies conditions (i), (ii), and (iii) in Definition 3.1. To verify the triangle inequality (condition (iv)) proceed as follows. Take an arbitrary s E S and note that, if f, g, and h are mappings in B[S, Y], then (by the triangle inequality in (Y, d))

d(f(s),g(s)) < d(f(s),h(s))+d(h(s),g(s)) < doo(f,h)+doo(h,g). Hence dooff, g) < dooff, h) + dooff, g) and therefore (B[S, Y], doo) is a metric space. The metric d°o is referred to as the sup-metric on B[S, Y]. Note that the metric spaces (t?°, doo) and (t70O, dw) of the previous example are particular cases of (B[S, Y], doo). Indeed, t+°° = B[N, C] and £ = B[Z, C].

Example 3D. The general concept of a continuous mapping between metric spaces will be defined in the next section. However, assuming that the reader is familiar with the particular notion of a real-valued continuous function of a real variable, we shall consider now the following example. Let C[0, 1] denote the set of all scalarvalued (real or complex) continuous functions defined on the interval [0, 1]. For every x, y E C[0, 1] set 1

dp(x, y) _ (J Ix(t) - y(t)I°dt 0

where p is a real number such that p > 1, and

doo = sup Ix(t) - y(t)I. IE[0,11

These are metrics on the set C [O, 1]. That is, d: C[0, 1 ] x C [0, 11 -+ R and dO° : C[0, 1 ] x C[0, 1 ] - R are well-defined functions that satisfy all the conditions in Definition 3.1. Indeed, nonnegativeness and symmetry are trivially verified,

3.1 Metric Spaces

91

positiveness for d p is ensured by the continuity of the elements in C[0, 1], and the triangle inequality comes by the Minkowski inequality (Problem 3.4(c)): for every x, y, z E C[O, 1],

e

dp(x, y)

I

cI

<J

r I

Ix(t) - z(t) + z(t) - y(t)I p dt)

Ix(t)-z(t)Ipdt)A+(J Iz(t)-y(t)Ipdt) 0

0

dp(x, z) + dp(z, y),

doo(x,y) =

sup Ix(t)-z(t)+z(t)-y(t)I

te[o,I[

sup Ix(t) - z(t)I + sup WO - Y(01 1E[o,II

tE[0,1]

doo(x, z) + doo(z, y).

Let B [O, 1] denote the set B[S, Y] of Example 3C when S = [0, 11 and Y = IF (with

F standing either for the real field R or for the complex field C). Since C[0, 1] is a subset of B[0, 1) (reason: every scalar-valued continuous function defined on the interval [0, 11 is bounded), it follows that (C[0, 1], dam) is a subspace of the metric space (B[0, 1], dam). The metric doo is called the sup-metric on C[0, 1] and, as we shall see later, the "sup" in its definition in fact is a "max".

Let X be an arbitrary set. A real-valued function d on X x X, d : X x X -+ R, is a pseudometric on X if it satisfies the axioms (i), (iii) and (iv) of Definition 3.1. A pseudometric space (X, d) is a set X equipped with a pseudometric d. The difference between a metric space and a pseudometric space is that a pseudometric does not necessarily satisfy the axiom (ii) in Definition 3.1 (i.e., it is possible for a pseudometric to vanish at a pair (x, y) even though x 9& y). However, given a pseudometric space (X, d), there exists a natural way to obtain a metric space (X, d) associated with (X, d), where d is a metric on X associated with the pseudometric

d on X. Indeed, as we shall see next, a pseudometric d induces an equivalence relation - on X, and k is precisely the quotient space X/- (i.e., the collection of all equivalence classes [x] with respect to - for every x in X). Proposition 3.3. Let d be a pseudometric on a set X and consider the relation -' on X defined as follows. If x and x' are elements of X, then

x', x if

d(x', x) = 0.

The relation - is an equivalence relation on X with the following property. For every x, x', y and y' in X,

x'- x and y'-- y

imply

d(x', y') = d(x, y).

Let X/- be the quotient space of X modulo

X/-. xX/- set d([x], [y]) = d(x, y)

For each pair ([x], [y]) in

92

3. Topological Structures

for an arbitrary pair (x, y) in [x] x [y]. This defines a function d:

X/-r x X/s -+ R

which is a metric on the quotient space X l- . Proof. It is clear that the relation - on X is reflexive and symmetric because a

pseudometric is nonnegative and symmetric. Transitivity comes from the triangle

inequality: 0 < d(x, x") < d(x, x') + d(x', x") for every x, x', x" E X. Thus is an equivalence relation on X. Moreover, if x'- x and y'-' y (i.e., if x' E [x] and y' E [y]), then the triangle inequality in the pseudometric space (X, d) ensures that

d(x, y) < d(x, x') +d(x', y') +d(y', y) = d(x', y') and, similarly, d(x', y') < d(x, y). Therefore

d(x', y') = d(x, y)

whenever

x'- x and y'-. y.

That is, given a pair of equivalence classes [x] e X and [y] a X. the restriction of d to [x] x [y] c X x X, d I[xlxty] : [x] x [y] - R, is a constant function. Thus, for

each pair ([x], [y]) in X/- x X/-', set d([x], [y]) = dl[x]x[y](x y) = d(x, y) for any x E [x] and y E [y]. This defines a function d : X/-' x R which is nonnegative, symmetric, and satisfies the triangle inequality (along with d). The reason for defining equivalence classes is to ensure positiveness for d from the nonnegativeness of the pseudometric d: if d([x], [y]) = 0, then d(x, y) = 0 so that x -r y, and hence [x] = [y]. U Example 3E. The previous example exhibited different metric spaces with the same underlying set of all scalar-valued continuous functions on the interval [0, 1]. Here we shall allow discontinuous functions as well. Let S be a nondegenerate interval of the real line R (typical examples: S = [0, 11 or S = R). For each real number p > 1 , let rP (S) denote the set of all scalar-valued (real or complex) p-integrable functions on S. In this context, "p-integrable" means that a scalar-valued function x on S is

Riemann integrable and fslx(s)IPds < oo (i.e., the Riemann integral fslx(s)IPds exists as a number in R). Consider the function 8P: rP(S)xrP(S) -+ R given by

8P(x,Y) = tfs Ix(s)-y(s)IPds)

for every x, y E re(S). The Minkowski inequality (see Problem 3.4(c)) ensures that the function 8P is well-defined, and also that it satisfies the triangle inequality. Moreover, nonnegativeness and symmetry are readily verified but positiveness fails. For instance, if 0 denotes the null function on S = [0, 11 (i.e., 0(s) = 0 for all s E S),

3.2 Convergence and Continuity

93

and if x (s) = 1 for s = i and zero elsewhere, then SP (x, 0) = 0 although x 96 0 (for x(71) 0 0(2')). Thus 8P actually is a pseudometric on re(S) rather than a metric,

so that (rP(S), SP) is a pseudometric space. However, if we "redefine" re(S) by endowing it with a new notion of equality, different from the usual pointwise equality

for functions, then perhaps we might make SP a metric on such a "redefinition" of rn(S). This in fact is the idea behind Proposition 3.3. Consider the equivalence

relation - on re(S) defined as in Proposition 3.3: if x and x' are functions in re(S), then x'-' x if SP(x', x) = 0. Now set RP(S) = rP(S)/-, the collection of all equivalence classes [x] = (x' E rP(S) : SP(x', x) = 0) for every x E rP(S). Thus, according to Proposition 3.3, (RP(S), d,,) is a metric space where the metric

dP: RP(S)xRP(S) -> Risdefinedbyd,([x], [y]) =8P(x, y)forarbitraryx E [x] and y E [y] for every [x], [y] E RP(S). Note that equality in Re(S) is interpreted in the following way: if [x) and [y] are equivalence classes in RP(S), and if x and y are arbitrary functions in [x] and [y], respectively, then [x] = [y] if and only if 8P(x, y) = 0. If x is any element of [x] then, in this context, it is usual to write x for [x] and hence dp(x, y) for d,([x], [y]). Thus, following the common usage, we shall write x E RP(S) instead of [x] E RP(S), and also

dp(x, y) _

(1s Ix(s) - y(s)I P ds) A

for every x, y E RP(S) to represent the metric dP on RP(S). This is referred to as the usual metric on RP(S). Note that, according to this convention, x = y in RP(S) if and only if d,(x, y) = 0.

3.2

Convergence and Continuity

The notion of convergence, together with the notion of continuity, plays a central role in the theory of metric spaces.

Definition 3.4. Let (X, d) be a metric space. An X-valued sequence (xo} (or a sequence in X indexed by N or by No) converges to a point x in X if for each real number e > 0 there exists a positive integer ne such that

n > of

implies

d(x,,, x) < e.

If (x,,) converges to x E X, then (xo ) is said to be a convergent sequence and x is

said to be the limit of (xo ) (notations: lim x = x, limo x = x,

xn = X.

x,-*x,or xo->xas n--oo). As defined above, convergence depends on the metric d that equip the metric space (X, d). To emphasize the rule played by the metric d, it is usual to refer to an X-valued convergent sequence {xn } by saying that (xo ) converges in (X, d). If an Xvalued sequence (xn ) does not converge in (X, d) to the point x E X, then we shall

94

3. Topological Structures

either converges in (X, d) write x -4 x. Clearly, if x,, -# x, then the sequence to another point different from x or does not converge in (X, d) to any x in X. The notion of convergence in a metric space (X, d) is a natural extension of the ordinary notion of convergence in the real line R (equipped with its usual metric). Indeed, Let x be an let (X, d) be a metric space, and consider an X-valued sequence x)). According to arbitrary point in X and consider the real-valued sequence Definition 3.4.

x -s x

if and only if

d(x,,,x) -+ 0.

This shows at once that a convergent sequence in a metric space has a unique limit (as we had anticipated in Definition 3.4 by referring to the limit of a convergent +d (x , b) sequence). In fact, if a and b are points in X, then 0 < d (a, b) < d (a,

for every n. Thus, if x -- a and x -- b (i.e., d(a,

0 and d(x,,, b) - 0),

then d(a, b) = 0 (see Problem 3.10(c) and show that the sum of two convergent real-valued sequences (a.) and { fi } is a convergent real-valued sequence with lim a + lim &). Hence a = b. lim(a + Example 3F. Let C[0, 1] denote the set of all scalar-valued continuous functions on the interval [0, 1], and let be a C[0, 1]-valued sequence such that, for each integer n > 1, x : [0, 1] -* R is defined by

xn(t) =

1-nt, to,

t E [0, A], tE 1].

Consider the metric spaces (C[0, fl, dp) for p > 1 and (C[0, 1], do,) which were introduced in Example 3D. It is readily verified that the sequence Ix. 1 converges in (C[0, 1],dp) to the null function 0 E C[0, 11 for every p> 1. Indeed, take an arbitrary p > 1 and note that

f

i

dp(xn,0) _ (J Ixn(t)Ipdt)I < (,)II 0

for each n > 1. Since the sequence of real numbers ((A)F } converges to zero (when the real line R is equipped with its usual metric - apply Definition 3.4), it follows that dp(x,,, 0) -- 0 as n --r oo (Problem 3.10(c)). That is,

x _+ 0 in (C[0, 1],dp). However, (x,,) does not converge in the metric space (CIO, 11, dam). Indeed, if there

existsx E C(0, 1]suchthatd... (x, x) -- O, then it is easy to showthatx(0) = land x (s) = O for all e E (0, 1). Hence x ¢ C(0, 11, which is a contradiction. Conclusion: There is no X E C[0, 1] such that x -i x in (C[0, 1], dam). Equivalently,

(x,,) does not converges in (C[O, 1], dam).

3.2 Convergence and Continuity

95

Example 3G. Consider the metric space (B[S, Y], do.) introduced in Example 3C, where B[S, Y] denotes the set of all bounded functions of a set S into a metric space (Y, d), and dw is the sup-metric. Let (f.) be a B[S, Y]-valued sequence (i.e., a sequence of functions in B[S, Y]), and let f be an arbitrary function in B[S, Y]. Since

0 < d(fn(s), f(s)) < supd(fn(s), f(s)) = doo(fn, SES

f)

for each index n and all s E S, it follows by Problem 3.10(c) that

fn -+ f

in (B[S, Y], dam)

for every s E S. If fn

implies

f. (s) -- f (s) in (Y, d)

fin (B[S, Y]. dd), then we say that the sequence if,) of

functions in B[S, Y] converges uniformly to the function f in B[S, Y]. If f (s) -i f (s) in (Y, d) for every s E S, then we say that (f.) converges pointwise to f. Thus uniform convergence implies pointwise convergence (to the same

limit), but the converse fails. For instance, set S = [0, 1], Y = IF (either the real field R or the complex field C equipped with their usual metric d), and set B[0, 1] = B[[0, 11, F]. Recall that the metric space (C[0, 1], dam) of Example 3D is a subspace of (B[0, 1], dm). (Indeed, every scalar-valued continuous function defined on a bounded closed interval is a bounded function - we shall consider a generalized version of this well-known result later in this chapter). If {gn} is a sequence of functions in C[0, 1] given by s2

gn(s)

s2 + (I - ns)2

for each integer n > 1 and every s E [0, 1], then it is easy to show (Definition 3.4) that

g,, (s) -> 0 in

(R, d)

for every s E [0, 11, so that the sequence {gn } of functions in C[0, 11 converges pointwise to the null function 0 E C[0, I]. On the other hand, note that 0 < gn (s) < I for all s E [0, 1], and gn (.) = 1 for each n > 1. Therefore

doo(gn, 0) = SUP Ign(s)I = I sE(O.II

for every integer n > 1. Thus (gn } does not converge uniformly to the null function, and hence it does not converge uniformly to any limit (for, if it converges uniformly,

then it converges pointwise to the same limit). Conclusion: The C[0, 1]-valued sequence (g,,) does not converge in the metric space (C[0, I], dam). Briefly, (g,,) does not converge in (C[0, I], dm).

However, it does converge to the null function 0 E C[0, 1] in the metric spaces (C[0. 1]. d,,) of Example 3D. That is,

gn - 0

in

(C[0, 1], dp)

96

3. Topological Structures

for each 0 and 0 < t+tnt for every p ? 1. Indeed, since s E (0, l ), it follows that 0 < g (s) < for each n > 1 and every s E [0, 1 ]. )p, and hence

Therefore j

0 < d p ( an, 0) I + n2

for each n > 1 and all p > 1. Since the sequence of positive numbers verges to zero in (IR, d), it follows by Problem 3.10(c) that

dp (g , 0) -+ 0 as n -- oo for every

{-) con-

p > 1.

Proposition 3.5. An X -valued sequence (x,,) converges in a metric space (X, d) to a limit x e X if and only if every subsequence of it converges in (X, d) to x.

If every subsequence converges to a fixed limit, then, in particular, the x in sequence itself converges to the same limit. On the other hand, suppose x (X, d). That is, for every e > 0 there exists a positive integer n6 such that n > n£ implies x) < e. Take an arbitrary subsequence Since }kEN of Proof.

k < nk (reason: {nk }keN is a strictly increasing subsequence of the sequence {n}nEN

- see Section 1.7), it follows that k > n6 implies nk > n6 which in turn implies x) < E. Therefore x in (X, d) as k - oo. o As we saw in Section 1.7, nets constitute a natural generalization of (infinite) sequences. Thus it comes as no surprise that the concept of convergence can be generalized from sequences to nets in a metric space (X, d). Indeed, an X-valued net (xy )yEr (or a net in X) indexed by a directed set r converges to a point x in X if for each real numbers > 0 there exists an index Ye in i' such that y > Ye

implies

d(xy, x) < S.

If {xy }yEr converges to x, then it is said to be a convergent net and x is said to be the limit of {xy }yEr (notations: lira xy = x, limy xy = x, or xy -+ x). Just as in the particular case of sequences, a convergent net in a metric space has a unique limit.

The notion of a real-valued continuous function on R is essential in classical analysis. One of the main reasons for investigating metric spaces is the generalization

of the idea of continuity for maps between abstract metric spaces: a map between metric spaces is continuous if it preserves closeness.

Definition 3.6. Let F : X -* Y be a function from a set X to a set Y. Equip X and Y with metrics dx and dy, respectively, so that (X, dx) and (Y, dy) are metric

spaces. F: (X, dx) -* (Y, dy) is continuous at the point xp in X if for each real number e > 0 there exists a real number S > 0 (which certainly depends one and may depend on xo as well) such that

dx(x, xo) < 3

implies

dy(F(x), F(xo)) < E.

3.2 Convergence and Continuity

97

F is continuous (or continuous on X) if it is continuous at every point of X ; and uniformly continuous (on X) if for each real number e > 0 there exists a real number

6 > 0 such that

dx(x, x') < S

dy(F(x), F(x')) < e

implies

for all x and x' in X.

It is clear that a uniformly continuous mapping is continuous, but the converse fails. The difference between continuity and uniform continuity is that if

F is uniformly continuous, then for each s > 0 it is possible to take S > 0 (which depends only on e) so as to ensure that the implication (dx(x, xo) < dy(F(x), F(xo)) < e} holds for all points xo of X. We say that a mapping F: (X, dx) -+ (Y, dy) is Lipschitzian if there exists a real number y > 0 S

(called Lipschitz constant) such that

dy(F(x), F(x')) < y dx(x. x') for all x, x' E X (which is referred to as the Lipschitz condition). It is readily verified that every Lipschitzian mapping is uniformly continuous but, again, the converse fails (see Problem 3.16). A contraction is a Lipschitzian mapping F : (X, dx) -+ (Y, dy) with a Lipschitz constant y < 1. That is, F is a contraction if dy(F(x), F(x')) < dx(x, x') for all x, x' E X or, equivalently, if

F(x')) < 1. dx(x,x')

sup

dy(F(x),

X0x'

F is said to be a strict contraction if it is Lipschitzian with a Lipschitz constant y < 1, which means that su X Ox,

dy(F(x), F(x')) < 1. dx(x,x')

Note that, if dy(F(x), F(x')) < dx(x, x') for all x, x' E X. then F is a contraction but not necessarily a strict contraction. Consider a function F from a metric space (X, dx) to a metric space (Y, dy). If F is continuous at a point xo E X, then xo is said to be a point of continuity of F. Otherwise, if F is not continuous at a point xo E X, then xo is said to be a point of discontinuity of F, and F is said to be discontinuous at xo. F is not continuous if there exists at least one point xp E X such that F is discontinuous at xo. According to Definition 3.6 a function F is discontinuous at xo E X if and only if the following assertion holds true: there exists s > 0 such that for every S > 0 there exists xa E X with the property that

dx(x,, xo) < 3

and

dy(F(xs), F(xo)) > s.

98

3. Topological Structures

Example 3H. (a) Consider the set R2(R) defined in Example 3E. Put Y = R2(R) and let X be the subset of Y made up of all functions x in R2(R) for which the formula

y(t) = I r x(s) ds for each t E R J -00 defines a function in R2(R). Briefly,

X = )x E IY: f racx(s)dsldt f.2 < oo). Recall that a "function" in Y is, in fact, an equivalence class of functions as discussed in Example 3E. Thus consider the mapping F : X -* Y that assigns to each function x in X the function y = F(x) in Y defined by the above formula. Now equip R2(R) with its usual metric d2 (cf. Example 3E) so that (X, d2) is a subspace of the metric

space (Y, d2). We claim that F: (X, d2) -> (Y, d2) is nowhere continuous; that is, the mapping F is discontinuous at every xo E X (see Problem 3.17(a)). (b) Now let S be a (nondegenerate) closed and bounded interval of the real line R (typical example: S = [0, 1)), and consider the set R2(S) defined in Example 3E. If x is a function in R2(S) (so that it is Riemann integrable), then set

y(t) = f , x(s)ds for each t E S.

J

s

According to the Holder inequality in Problem 3.3(c) it follows that fslx(s)Ids <

(fsds) I (fslx(s)I2ds)i for every x E R2(S). Therefore Iy(t)I2 =I fox(s)dsI2 < (fslx(s)Ids)2 < diam(S) fslx(s)I2ds for each t E S, and hence

f ly(t)I2dt < diam(S)2fS Ix(s)I2ds < 00

Is

for everyx E R2(S). Thus the above formula defines a function y in R2(S). Let F be a mapping of R2(S) into itself that assigns to each function x in R2(S) this function y in R2(S), so that y = F(x). Equip R2(S) with its usual metric d2 (Example 3E). It is easy to show that F: (R2(S), d2) -> (R2(S), d2) is uniformly continuous. As a matter of fact, the mapping F is Lipschitzian (see Problem 3.17(b)). Comparing the example in item (a) with the present one we observe how different the metric spaces R2(R) and R2(S), both equipped with the usual metric d2, can be: the "same" integral transformation F that is nowhere continuous when defined on an appropriate subspace of (R2(R), d2) becomes Lipschitzian when defined on (R2(S), d2). The concepts of convergence and continuity are tightly intertwined. A particularly important result on the connection of these central concepts says that a function is continuous if and only if it preserves convergence. This supplies a useful necessary and sufficient condition for continuity in terms of convergence.

3.2 Convergence and Continuity

99

Theorem 3.7. Consider a mapping F: (X, dx) - (Y, dy) of a metric space (X,dx) into a metric space (Y, dy) and let xo be a point in X. The following assertions are equivalent. (a) F is continuous at xo.

(b) The Y-valued sequence (F(xn)} converges in (Y, dy) to F(xo) E Y whenever is an X-valued sequence that converges in (X, dx) to xo E X.

Proof If

xo in (X, dx) for some xo in is an X -valued sequence such that x X. then (Definition 3.4) for every S > 0 there exists a positive integer ns such that

n > na

implies

dx(x,,, xo) < S.

If F: (X, dx) --> (Y, dy) is continuous at xo, then (Definition 3.6) for each e > 0 there exists S > 0 such that dx(xn, xo) < S

implies

dy(F(x), F(xo)) < E.

Therefore, if x -+ xo and F is continuous at xo, then for each E > 0 there exists a positive integer nE (e.g., nE = na) such that n > nE

implies

dy(F(x,,), F(xo)) < E.

which means that (a)=*-(b). On the other hand, if F is not continuous at xo, then there exists s > 0 such that for every S > 0 there exists xa E X with the property that

dx(xa, xo) < S

and

dy(F(x3), F(xo)) ? E.

In particular, for each positive integer n there exists x E X such that dx(x,,, xo) < A

and

F(xo)) > E.

Thus x -- xo in (X, dx) (for dx(x,,, xo) - 0) and F(xo)) -4 0). That is, the Y-valued sequence (

F(xo) in (Y, dy) (for does not converge

to F(xo) (it may not converge in (Y, dy) or, if it converges in (Y, dy), then it does not converge to F(xo)). Therefore, the denial of (a) implies the denial of (b). o Equivalently,

Note that the proof of can be rewritten in terms of nets so that, if the mapping F: (X, dx) - (Y, dy) is continuous at xo E X, and if (xy}yEr is an Xvalued net that converges to xo, then (F(xy))yEr is a Y-valued net that converges to F(xo). That is, if (6) is the statement obtained from (b) by changing "sequence" (b) is a to "net" in (b), then particular case of (6)), it also follows that (b)=(a) (because (b) =*(a)).

We shall say that a mapping F: (X, dx) -+ (Y, dy) of a metric space (X, dx) into a metric space (Y, dy) preserves convergence if the Y-valued sequence (F(x))

100

3. Topological Structures

converges in (X, dX).

converges in (Y, dy) whenever the X-valued sequence and lim

Corollary 3.8. A map between metric spaces is continuous if and only if it preserves convergence.

Proof. Combine the above definition and the definition of a continuous function o with Theorem 3.7.

-

Example 31. Let C[0, 11 denote the set of all scalar-valued (real or complex) continuous functions defined on the interval [0, 1 ]. Consider the map qp : C[0, 1] C defined by I

p(x) = J x(t)dt 0

for every x in C[0, 1]. Equip C[0, 1] with the sup-metric d(,. and C with its usual metric d. Take an arbitrary convergent sequence in (C[0, 1], dam) (i.e., an arbitrary C[0, 1 )-valued sequence that converges in the metric space (C[0, I], dam)) and set xo = lim x E C[0, I]. Note that for each positive integer n

0<

(p(xO)) = I I

1 (x, (t) 0

J0

I

- x0(t))dtl

xn(t)dt <

f

J0 I

xo(t)dt

i

Ix, (t) - xo(t)I dt

0

t

<

sup te(O.II

-x0(t)11 dt = 0

(by the Holder inequality: Problem 3.3(c)). Since doo(x,,, x0) --> 0, it follows that W(xo)) --+ 0. Therefore V(xo) in (C, d) whenever x -+ xo in (C[0, 1 ], da,), so that tp : (C[0, I], dam) - (C, d) is continuous by Corollary 3.8. (In fact, (p is a contraction, thus Lipschitzian, and hence uniformly continuous.)

3.3

Open Sets and Topology

Let (X, d) be a metric space. For each point xo in X and each nonnegative real number p the set Bo(xo) = {x E X : d(x,xo) < p) is the open ball with center xo and radius p (or the open ball centered at xo with radius p). If p = 0, then Bo(xo) is empty; otherwise Bo(xo) always contains at least its center. The set Bp[x0] = {X E X : d(x, xo) < pI

3.3 Open Sets and Topology

101

is the closed ball with center x0 and radius p. It is clear that Bp(xo) a Bp[xo}. If p = 0, then Bo[xo} = (x0) for every xo E X. Definition 3.9. A subset U of a metric space X is an open set in X if it includes a nonempty open ball centered at each one of its points. That is, U is an open set in X if and only if for every u in U there exists a positive number p such that Bp(xo) C U. Equivalently, U C X is open in the metric space (X. d) if and only if for every u E U there exist p > 0 such that

d(x, u) < p

implies

x E U.

Thus, according to Definition 3.9, a subset A of a metric space (X, d) is not open if and only if there exists at least one point a in A such that every open ball with positive radius p centered at a contains a point of X not in A. In other words, A C X is not open in the metric space (X, d) if and only if there exists at least one point a E A with the following property: for every p > 0 there exists x E X such that

d(x, a) < p

and

x E X \A.

This shows at once that the empty set 0 is open in X (reason: if a set is not open then it has at least one point); and also that the underlying set X is always open in the metric space (X, d) (reason: there is no point in X\X). Proposition 3.10. An open ball is an open set. Proof. Let Bp (x0) bean open ball in a metric space (X, d) with center at an arbitrary

x0 E X and with an arbitrary radius p ? 0. Suppose p > 0 so that Bp(x0) i6 0 (otherwise Bp (x0) is empty and hence trivially open). Take an arbitrary u E Bp (xo),

which means that u E X and d (u, x0) < p. Set fl = p - d (u, x0) so that 0 < f < p, and let x be a point in X. If d(x, u) < fl, then the triangle inequality ensures that

d(x,x0) < d(x, u) + d(u, x0) < B+d(u,x0) = p. and hence x E Bp(xo). Conclusion: For every u E Bp(xo) there exists P > 0 such that

d(x, u) < fi

implies

x E Bp(x0).

That is, Bp(xo) is an open set.

o

An open neighborhood of a point x in a metric space is an open set containing x. In particular (see Proposition 3.10), every open ball with positive radius centered at a point x in a metric space is an open neighborhood of x. A neighborhood of a point x in a metric space X is any subset of X that includes an open neighborhood of x. Clearly, every open neighborhood of x is a neighborhood of x. Open sets can be used to give an alternative definition of continuity and convergence.

Lemma 3.11. Consider a mapping F : X -> Y of a metric space X into a metric space Y and let xo be a point in X. The following assertions are equivalent.

102

3. Topological Structures

(a) F is continuous at xo. (b) The inverse image of every neighborhood of F(xo) is a neighborhood of xo.

Proof Consider the image F(xo) E Y of xo E X. Take an arbitrary neighborhood N C Y of F(xo). Since N includes an open neighborhood U of F(xo), it follows that there exists an open ball BE(F(xo)) g U C Al with center at F(xo) and radius e > 0. If the mapping F : X - Y is continuous at xo (cf. Definition 3.6), then there exists 8 > 0 such that

dy(F(x), F(xo)) < e

whenever

dx(x, xo) < 8,

where dx and dy are the metrics on X and Y, respectively. In other words, there exists 8 > 0 such that x E Bs(xo)

implies

F(x) E BE(F(xo)).

Thus Ba(xo) C_ F-' (BE(F(xo))) C_ F-' (U) C_ F-' (N). Since the open ball BS (xo) is an open neighborhood of xo, and since Ba (xo) C F-' (N), it follows that F-' (N) is a neighborhood of xo. Hence (a) implies (b). Now suppose (b) holds true. Then, in particular, the inverse image F-' (BE (F(xo))) of each open ball BE (F(xo))

with center F(xo) and radius s > 0 includes a neighborhood N C X of xo. This neighborhood N includes an open neighborhood U of xo which in turn includes an open ball B8(xo) with center at xo and with a positive radius S (cf. Definition 3.9). Therefore, for each e > 0 there exists 8 > 0 such that

B3(xo) S U C N c F-' (BE(F(xo))) Hence (see Problems 1.2(c j))

F(B6(xo)) c Be(F(xo)).

Equivalently, if x E Bs(xo), then F(x) E BE(F(xo)). Thus for each e > 0 there exists 8 > 0 such that dx(x, xo) < 8

implies

dy(F(x), F(xo)) < e,

where dx and dy denote the metrics on X and Y, respectively. That is, (a) holds true (Definition 3.6). 13 Theorem 3.12. A map between metric spaces is continuous if and only if the inverse image of each open set is an open set.

Proof. Let F : X - Y be a mapping of a metric space X into a metric space Y.

(a) Take any neighborhood N C Y of F(x) E Y (for an arbitrary x c- X). Since N includes an open neighborhood of F(x), say U. it follows that F(x) E U C N which implies

x E F-'(U) c F-'(N).

3.3 Open Sets and Topology

103

If the inverse image (under F) of each open set in Y is an open set in X. then F-1 (U) is open in X, and hence F-' (U) is an open neighborhood of x. Therefore, the inverse image F- I (N) is a neighborhood of x. Conclusion: The inverse image of every neighborhood of F(x) (for any x E X) is a neighborhood of x. Thus F is continuous by Lemma 3.11.

(b) Take an arbitrary open subset U of Y. Suppose 1(F) fl u i4 0 and take x E F-1(U) C X arbitrary. Thus F(x) E U so that U is an open neighborhood of F(x). If F is continuous, then it is continuous at x. Therefore, according to Lemma 3.11, F-1(U) is a neighborhood of x, and hence it includes a nonempty open ball B3(x) centered at x. Thus B8 (x) C F-1(U) so that F-1(U) is open in X (reason: it includes a nonempty open ball of an arbitrary point of it). If R(F) fl U = 0, then F-1 (U) = 0 which is open. Conclusion: F-1(U) is open in X for every open subset U of Y. 0 Corollary 3.13. The composition of two continuous functions is again a continuous function.

Proof. Let X. Y and Z be metric spaces, and let F : X - Y and G : Y -* Z be continuous functions. Take an arbitrary open set U in Z. According to Theorem 3.12

the set G-1(U) is open in Y so that (GF)-1(U) = F-'(G-(U)) is open in X. Thus, using Theorem 3.12 again, we conclude that G F : X -)- Z is continuous. 13

An X-valued sequence is said to be eventually in a subset A of X if there exists a positive integer no such that

n > no

implies

x E A.

Theorem 3.14. Let {x } be a sequence in a metric space X. and let x be a point in X. The following assertions are equivalent.

(a) x - x in X. (b) {x } is eventually in every neighborhood of x. Proof.

If x -+ x, then (definition of convergence) {x } is eventually in every

nonempty open ball centered at x. Hence it is eventually in every neighborhood of x (cf. definitions of neighborhood and of open set). Conversely, if {x } is eventually in every neighborhood of x then, in particular, it is eventually in every nonempty open ball centered at x, which means that x,, - x. o

Observe that the above theorem is naturally extended from sequences to nets. A net {x),}yEr in a metric space X converges to x E X if and only if for every neighborhood N of x there exists an index yb E 1' such that xy E N for every

Y?Y)

104

3. Topological Structures

Given a metric space X, the collection of all open sets in X is of paramount importance. The fundamental properties of it are stated in the next theorem.

Theorem 3.15. If X is a metric space, then (a) the whole set X and the empty set 0 are open,

(b) the intersection of a finite collection of open sets is open,

(c) the union of an arbitrary collection of open sets is open.

Proof. We have already verified that assertion (a) holds true. Let (Un) be a finite collection of open subsets of X. Suppose nnUn 34 0 (otherwise nnUn is an open set). Take an arbitrary u E nn Un so that u E Un for every index n. As each Un is an open subset of X, there exist open balls Ban(u) c Un (with center at u and radius an > 0) for each index n. Consider the set {an) consisting of the radius of each Baju). Since {an} is a finite set of positive numbers, it follows that it has a

positive minimum. Set a = min{a) > 0 so that Ba(u) c nnUn. Thus nnUn is open in X (i.e., for each u E nnUn there exists an open ball Ba(u) c nnUn), which concludes the proof of (b). The proof of (c) goes as follows. Let U be an arbitrary collection of open subsets of X. Suppose U is nonempty (otherwise it is open by (a)) and take an arbitrary u E U U so that u E U for some U E U. As U is an open subset of X, there exists a nonempty open ball Bp(u) c U C_ UU, which means that U U is open in X. o

Corollary 3.16. A subset of a metric space is open if and only if it is a union of open balls. Proof. The union of open balls in a metric space X is an open set in X because open balls are open sets (cf. Proposition 3.10 and Theorem 3.15). On the other hand, let U be an open set in a metric space X. If U is empty, then it coincides with the empty open ball. If U is a nonempty open subset of X, then each u E U is the center of an open ball, say BpM(u), included in U. Thus U = UUEU{u) c UUEuBPM(u) E_ U,

and hence U =

UUEUBpM(u).

0

The collection T of all open sets in a metric space X (which is a subcollection of the power set P(X)) is called the topology (or the metric topology) on X. As the elements of T are the open sets in the metric space (X, d), and since the definition of an open set in X depends on the particular metric d that equips the metric space (X, d), the collection T is also referred to as the topology induced (or generated, or determined) by the metric d. Our starting point in this chapter was the definition of a metric space. A metric has been defined on a set X as a real-valued function on X x X that satisfies the metric axioms of Definition 3.1. A possible and different approach is to define axiomatically an abstract notion of open sets (instead of an abstract notion of distance as we did in

3.3 Open Sets and Topology

105

Definition 3.1), and then to build up a theory based on it. Such a "new" beginning goes as follows.

Definition 3.17. A subcollection T of the power set P(X) of a set X is a topology on X if it satisfies three axioms, viz., (i) (ii) (iii)

The whole set X and the empty set 0 belong to T. The intersection of a finite collection of sets in T belongs to T. The union of an arbitrary collection of sets in T belongs to T.

A set X equipped with a topology T is referred to as a topological space (denoted by (X, T) or simply by X), and the elements of T are called the open subsets of X with respect to T. Thus a topology T on an underlying set X is always identified with the collection of all open subsets of X : U is open in X with respect to T if and only if U E T. It is clear (see Theorem 3.15) that every metric space (X, d) is a topological space, where the topology T (the metric topology, that is) is that induced by the metric. This topology T induced by the metric d, and the topological space (X, T) obtained by equipping X with T, are said to be metrized by d. If (X, T) is a topological space, and if there exists a metric don X that metrizes T, the topological space (X, T) and the topology T are called metrizable. The notion of topological space is broader than the notion of metric space. Although every metric space is a topological space, the converse fails. There are topological spaces that are not metrizable.

Example 3J. Let X be an arbitrary set and define a function d : X x X - R by

d(x, y) =

0, 1,

x=y,

xy,

for every x and y in X. It is readily verified that d is a metric on X, the so-called discrete metric on X. A set X equipped with the discrete metric is called a discrete space. In a discrete space every open ball with radius p E (0, 1) is a singleton in X: Bp (xo) = (x0} for every xo E X and every p E (0, 1). Thus. according to Definition 3.9, every subset of X is an open set in the metric space (X, d) equipped with the discrete metric d. That is, the metric topology coincides with the power set of X. Conversely, if X is an arbitrary set, then the collection T = P(X) is a topology on X (since T trivially satisfies the above three axioms), called the discrete topology, which is the largest topology on X (any other topology on X is a subcollection of the discrete topology). Summing up: The discrete topology

7 = P(X) is metrizable by the discrete metric. On the other extreme lies the topology T = (0, X), called the indiscrete topology, which is the smallest topology on X (it is a subcollection of any other topology on X). If X has more than one point, then the indiscrete topology

T = (0, X) is not metrizable.

106

3. Topological Structures

Indeed, suppose there exists a metric d on X that induces the indiscrete topology. Take u in X arbitrary and consider the set X\{u). Since 0 96 X\{u} i4 X, it follows that this set is not open (with respect to the indiscrete topology). Thus there exists V E X \(u) with the following property: for every p > 0 there exists x E X such that d(x, v) < p and x E X\(X\{u)) = (u).

Hence x = u so that d (u, v) < p for every p > 0. Therefore u = v (i.e., d(u, v) _ 0), which is a contradiction (for v E X\{u}). Conclusion: There is no metric on X that induces the indiscrete topology. Continuity and convergence in a topological space can be defined as follows. A mapping F: X -* Y of a topological space (X, TX) into a topological space (Y, TO is continuous if F'1(U) E Tx for every U E Ty. An X-valued sequence converges in a topological space (X, T) to a limit x E X if it is eventually in every U E T that contains x. Carefully note that, for the particular case of metric spaces (or of metrizable topological spaces), the above definitions of continuity and convergence agree with Definitions 3.6 and 3.4, respectively, when the topological spaces are equipped with their metric topology. Indeed, these definitions are the topological-space versions of Theorems 3.12 and 3.14. Many (but not all) of the theorems in the next sections hold for general topological spaces (metrizable or not), and we shall prove them by using a topological-space style (based on open sets rather than on open balls) whenever this is possible and convenient. However, as we had anticipated at the introduction of this chapter, our attention will focus mainly on metric spaces.

3.4

Equivalent Metrics and Homeomorphisms

Let (X, d1) and (X, d2) be two metric spaces with the same underlying set X. The metrics dl and d2 are said to be equivalent (or dt and d2 are equivalent metrics

on X - notation: dt-d2) if they induce the same topology (i.e., a subset of X is open in (X, d1) if and only if it is open in (X, d2)). This notion of equivalence in fact is an equivalence relation on the collection of all metrics defined on a given set X. If Ti and T2 are the metric topologies on X induced by the metrics dt and d2, respectively, then

d1'-d2

if and only if

7 = T2.

If T c T2 (i.e., if every open set in (X, d i) is open in (X, d2)), then T2 is said to be stronger than T1. In this case we also say that T is weaker than 72. The terms finer and coarser are also used as synonyms for "stronger" and "weaker", respectively. If either 7 C_ T2 or T2 C_ T1, then 71 and T2 are said to be commensurable. Otherwise (i.e., if neither 7 C T2 nor T2 C Tt ), the topologies are said to be incommensurable. As we shall see below, if 71, is stronger than 71, then continuity with respect to 7 implies continuity with respect to T2. On the other hand, if T2 is stronger than Tt . then

3.4 Equivalent Metrics and Homeomorphisms

107

convergence with respect to T2 implies convergence with respect to T1. Briefly and roughly: "Strong convergence" implies "weak convergence" but "weak continuity" implies "strong continuity".

Theorem 3.18. Let dl and d2 be metrics on a set X, and consider the topologies Tj and T2 induced by dt and d2. respectively. The following assertions are pairwise equivalent. (a) T2 is stronger than T

(i.e., T C T2).

(b) Every mapping F : X -* Y that is continuous at xo E X as a mapping of (X, d j) into the metric space (Y, d) is continuous at xo as a mapping of (X, d2) into (Y, d). (c) Every X -valued sequence that converges in (X, d2) to a limit s E X converges in (X, dt) to the same limit x. (d) The identity map of (X, d2) onto (X, dt) is continuous.

Proof. Consider the topologies T and T2 on X induced by the metrics dl and d2 on X, respectively. Let T denote the topology on a set Y induced by a metric d on Y.

Proof of

If F : (X, di) -+ (Y, d) is continuous at xo E X, then (Lemma

3.11) for every U E T that contains F(xo) there exists U' E 7 containing xo

F_t (U). If T e 7-2, then U' E T2: the inverse image (under F) of such that U' C every open neighborhood of F(xo) in T includes an open neighborhood of xo in T2, which clearly implies that the inverse image (under F) of every neighborhood of F(xo) in T is a neighborhood of xo in T2. Thus, applying Lemma 3.11 again, F: (X, d2) -+ (Y, d) is continuous at xo.

Let {x,,) be an X-valued sequence. If x - x E X in (X, d2), Proof of then (Theorem 3.14) (x,,) is eventually in every open neighborhood of x in T2. If is eventually in every neighborhood of x in T. Tt g T2 then, in particular, Therefore, applying Theorem 3.14 again, x -+ x in (X, dj). Proof of (b)=(d). The identity map 1: (X, d1) - (X, dt) of a metric space onto itself is trivially continuous. Thus, by setting (Y, d) = (X, d1) in (b), it follows that (b) implies (d). Proof of (c)=*(d). Corollary 3.8 ensures that (c) implies (d).

(d) implies (a) (i.e., if the identity Proof of 1: (X, d2) -- (X, d1) is continuous, then U = 1-1 (U) is open in T2 whenever U 13 is open in TI, and hence 71 C T2). As the discrete topology is the strongest topology on X, the above theorem ensures that any function F : X --)- Y that is continuous in some topology on X is continuous

in the discrete topology. Actually, since every subset of X is open in the discrete

108

3. Topological Structures

topology, it follows that the inverse image of every subset of Y - no matter which topology equips the set Y - is an open subset of X when X is equipped with the discrete topology. Therefore, everyfunction defined on a discrete topological space is continuous. On the other hand, if an X-valued (infinite) sequence converges in the discrete topology, then it is eventually constant (i.e., it has only a finite number of entries not equal to its limit), and hence it converges in any topology on X. Corollary 3.19. Let (X, d j) and (X, d2) be metric spaces with the same underlying set X. The following assertions are pairwise equivalent. (a) d2 and dl are equivalent metrics on X.

(b) A mapping of X into a set Y is continuous at xo E X as a mapping of (X, d t ) into the metric space (Y. d) if and only if it is continuous at xo as a mapping of (X, d2) into (Y, d). (c) An X -valued sequence con verges in (X, d j) t o X E X if and only if it converges

in (X, d2) to x. (d) The identity map of (X, d i) onto (X, d2) and its inverse (i.e., the identity map of (X, d2) onto (X, dt)) are both continuous.

Proof. Recall that, by definition, two metrics dt and d2 on a set X are equivalent if the topologies T and T2 on X, induced by dt and d2 respectively, coincide (i.e., if T = T2). Now apply Theorem 3.18. o

A one-to-one mapping G of a metric space X onto a metric space Y is a homeomorphism if both G: X -- Y and G-t : Y -+ X are continuous. Equivalently, a homeomorphism between metric spaces is an invertible (i.e., injective and surjective) mapping that is continuous and has a continuous inverse. Obviously, G is a homeomorphism from X to Y if and only if G-t is a homeomorphism from Y to X. Two metric spaces are homeomorphic if there exists a homeomorphism between them. A function F : X -+ Y of a metric space X into a metric space Y is an open

map (or an open mapping) if the image of each open set in X is open in Y (i.e., F(U) is open in Y whenever U is open in X). Theorem 3.20. Let X and Y be metric spaces. If G : X -> Y is invertible, then

(a) G is open if and only if G-1 is continuous, (b) G is continuous if and only if G-1 is open, (c) G is a homeomorphism if and only if G and G

are both open.

Proof. If G is invertible, then it is trivially verified that the inverse image of B (B C Y) under G coincides with the image of B under the inverse of G (tautologically:

3.4 Equivalent Metrics and Homeomorphisms

109

G-I (B) = G-t (B)). Applying the same argument to the inverse G-t of G (which is clearly invertible), (G-t)-t(A) = G(A) for each A C_ X. Thus the theorem is a straightforward combination of the definitions of open map and homeomorphism by using the alternative definition of continuity in Theorem 3.12. o Thus a homeomorphism provides simultaneously a one-to-one correspondence between the underlying sets X and Y (so that X -> Y, for a homeomorphism is invective and surjective) and between their topologies (so that Tx as Ty, for a homeomorphism puts the open sets of Tx into a one-to-one correspondence with the open sets of Ty). Indeed, if Tx and Ty are the topologies on X and Y. respectively,

then a homeomorphism G: X Y induces a map Q: Tx - Ty, defined by G(U) = G(U) for every U E TX, which is injective and surjective according to Theorem 3.20. Therefore, any property of a metric space X expressed entirely in terms of set operations and open sets is also possessed by each metric space homeomorphic to X. We call a property of a metric space a topological property or a topological invariant if whenever it is true for one metric space, say X. it is true for every metric space homeomorphic to X (trivial examples: the cardinality of the underlying set and the cardinality of the topology). A map F : X -> Y of a metric space X into a metric space Y is a topological embedding of X into Y if it establishes a homeomorphism of X onto its range R(F) (i.e., F: X -+ Y is a topological embedding of X into Y if it is such that F : X -+ F(X) is a homeomorphism of X onto the subspace F(X) of Y).

Example 3K. Suppose G : X -> Y is a homeomorphism of a metric space X onto a metric space Y. Let A be a subspace of X and consider the subspace G(A) of Y. According to Problem 3.30 the restriction GIA: A - G(A) of G to A onto G(A) is continuous. Similarly, the restriction G-t IG(A): G(A) -- A of the inverse of G to G(A) onto G-1 (G(A)) = A is continuous as well. Since G-t IG(A) = (GIA)-t (Problem 1.8), it follows that

GIA: A --> G(A) is a homeomorphism, and hence A and G(A) are homeomorphic metric spaces (as subspaces of X and Y, respectively). Thus, as we might expect, the restriction GIA: A -> Y of a home-

omorphism G : X --). Y to any subset A of X is a topological embedding of A into Y.

The notions of homeomorphism, open map, topological invariant and topological embedding are germane to topological spaces in general (and to metric spaces in particular). For instance, both Theorem 3.20 and Example 3K can be likewise stated (and proved) in a topological-space setting. In other words, the metric has played no role in the above paragraph, and "metric space" can be replaced with "topological space" there. Next we shall consider a couple of concepts that only make sense in a metric space. A homeomorphism G of a metric space (X, dx) onto a metric space (Y, dy) is a uniform homeomorphism if both G and G-1 are uniformly continuous.

110

3. Topological Structures

Two metric spaces are uniformly homeomorphic if there exists a uniform homeomorphism mapping one of them onto the other. An isometry between metric spaces is a map that preserves distance. Precisely, a mapping J : (X, dx) -* (Y, dy) of a metric space (X, dx) into a metric space (Y, dy) is an isometry if

dy(J(x), J(x')) = dx(x,x') for every pair of points x, x' in X. It is clear that every isometry is an injective contraction, and hence an injective and uniformly continuous mapping. Thus every surjective isometry is a uniform homeomorphism (the inverse of a surjective isometry is again a surjective isometry - trivial example: the identity mapping of a metric space into itself is a surjective isometry on that space). Two metric spaces are isometric (or isometrically equivalent) if there exists a surjective isometry between them, so that two isometrically equivalent metric spaces are uniformly homeomorphic. It is trivially verified that a composition of surjective isometries is a surjective

isometry (transitivity), and this shows that the notion of isometrically equivalent metric spaces deserves its name: it is indeed an equivalence relation on any collection of metric spaces. If two metric spaces are isometrically equivalent, then they can be thought of as being essentially the same metric space - they may differ on the set-theoretic nature of their points but, as far as the metric space (topological) structure is concerned, they are indistinguishable. A surjective isometry not only preserves open sets (for it is a homeomorphism) but also distance. Now consider two metric spaces (X, d1) and (X, d2) with the same underlying set X. According to Corollary 3.19 the metrics dl and d2 are equivalent if and only if the identity map of (X, d1) onto (X, d2) is a homeomorphism (i.e., if and only if 1: (X, dj) -+ (X, d2) and its inverse 1-1: (X, d2) --- (X, d1) are both continuous). We say that the metrics di and d2 are uniformly equivalent if the identity map of (X, d1) onto (X, d2) is a uniform homeomorphism (i.e., 1: (X, d1) --> (X, d2) and its inverse /-1: (X, d2) --* (X, d1) are both uniformly continuous). For instance, if I and /-1 are both Lipschitzian, which means that there exist real numbers a > 0 and fi > 0 such that

adi(x,x') < d2(x,x') < 0dt(x,x') for every x, x' in X, then the metrics di and d2 are uniformly equivalent, and hence equivalent. Thus, if di and d2 are equivalent metrics on X, then (X, dj) and (X, d2) are homeomorphic metric spaces. However, the converse fails: there exist uniformly homeomorphic metric spaces with the same underlying set for which the identity is not a homeomorphism.

Example X. Take two metric spaces (X, dj) and (X, d2) with the same underlying set X and consider the product spaces (X x X, d) and (X x X, d'), where

d((x, y), (u, v)) d'((x, y), (u, v))

= dj(x, u) +d2(y. v), = d2(x, u) + di (y, v),

3.4 Equivalent M e t r i c s a n d H o m e o m o r p h i s m s

Ill

for all ordered pairs (x, y), (u, v) in X x X. That is, (X x X, d) = (X, dt) x (X, d2) and (X x X, d') = (X, d2) x (X, dl) - see Problem 3.9. Suppose the metrics dl

and d2 on X are not equivalent so that either the identity map of (X,dt) onto (X, d2) or the identity map of (X, d2) onto (X, dt) (or both) is not continuous. Let 1: (X, dt) --+ (X, d2) be the one that is not continuous. The identity map

1: (XxX,d) -- (XxX,d') is not continuous. Indeed, if it is continuous, then the restriction of it to any subspace of X x X, d is continuous (Problem 3.30). In particular, the restriction of it to (X, d i) - viewed as a subspace of (X x X. d) = (X, dt) x (X, d2) - is continuous. But such a restriction is clearly identified with the identity map of (X, d1) onto (X, d2), which is not continuous. Thus 1: (X x X, d) -+ (X x X, d') is not continuous, and hence the metrics d and d' on X x X are not equivalent. Now let J: X x X -- X x X be the involution on X x X defined by J((x, y)) = (y, x) for every (x, y) E X X X (Problem 1.11). It is readily verified that J: (X x X, d) -+ (X x X, d') is a surjective isometry so that

J: (XxX,d) -+ (XxX,d') is a uniform homeomorphism. Summing up: The metric spaces (X x X, d) and (X x X, d'), with the same underlying set X x X, are uniformly homeomorphic (more than that, they are isometrically equivalent) but the metrics d and d' on X x X are not equivalent.

Since two metric spaces with the same underlying set may be homeomorphic even if the identity between them is not a homeomorphism, it follows that a weaker version of Corollary 3.19 is obtained if we replace the homeomorphic identity with an arbitrary homeomorphism. This in fact can be formulated for arbitrary metric spaces (not necessarily with the same underlying set).

Theorem 3.21. Let X and Y be metric spaces and let G be an invertible mapping of X onto Y. The following assertions are pairwise equivalent.

(a) G is a homeomorphism.

(b) A mapping F of X into a metric space Z is continuous if and only if the composition FG-t : Y -* Z is continuous. (c) An X -valued sequence {xn) converges in X to a limit x E X if and only if the Y-valued sequence converges in Y to G(x).

Proof. Let G : X - Y be an invertible mapping of a metric space X onto a metric space Y.

112

3. Topological Structures

Proof of (a)=(b). Let F: X --* Z be a mapping of X into a metric space Z, and consider the commutative diagram

Z

so that H = FG't : Y - Z. Suppose (a) holds true, and consider the following assertions.

(b l) F : X -+ Z is continuous. (b2) F-' (U) is an open set in X whenever U is an open set in Z (b3) G(F-1 (U)) is an open set in Ywhenever U is an open set in Z

(b4) (FG-t)-t (U) is an open set in Ywhenever U is an open set in Z

(b5) H = FG-t : Y -> Z is continuous. Theorem 3.12 says that (b1) and (b2) are equivalent. But (b2) holds true if and only if (b3) holds true by Theorem 3.20 (the homeomorphism G : X -- Y puts the open sets of X into a one-to-one correspondence with the open sets of Y). Now note that, as G is invertible,

G(F-'(A)) = (G(x) E Y: F(x) E Al =

(y E Y: F(G-'(y)) E A}

= (FG-t)-'(A)

for every subset A of Z. Thus (b3) is equivalent to (b4), which in turn is equivalent to (b5) (cf. Theorem 3.12 again). Conclusion: (bt).e(b5) whenever (a) holds true.

Proof of (b) =>(a). If (b) holds, then it holds in particular for Z = X and for Z = Y. Thus (b) ensures that the following assertions hold true. (b)

(b)

If a mapping F of X into itself is continuous, then the mapping H = FG-t : Y -+ X is continuous. A mapping F of X into Y is continuous whenever the mapping H = FG-t : Y - Y is continuous.

Since the identity of X onto itself is a continuous mapping, (1) implies that G-t :

Y -> X is continuous. By setting F = G in (b") it follows that G: X -> Y is continuous (because the identity I = GG-' : Y --> Y is continuous). Summing up: (b) implies that both G and G-1 are continuous, which means that (a) holds true.

Proof of (a)q(c). According to Corollary 3.8 an invertible mapping G between metric spaces is continuous and has a continuous inverse if and only if both G and G-t preserve convergence. 13

3.5 Closed Sets and Closure

3.5

113

Closed Sets and Closure

A subset V of a metric space X is closed in X if its complement X\ V is an open set in X.

Theorem 3.22. If X is a metric space, then (a) the whole set X and the empty set 0 are closed, (b) the union of a finite collection of closed sets is closed,

(c) the intersection of an arbitrary collection of closed sets is closed. Proof. Apply the De Morgan laws to each item of Theorem 3.15.

0

Thus the concepts "closed" and "open" are dual of each other (U is open in X if and only if its complement X\U is closed in X, and V is closed in X if and only if its complement X\ V is open in X); but they are neither exclusive (a set in a metric space may be both open and closed) nor exhaustive (a set in a metric space may be neither open nor closed). Theorem 3.23. A map between metric spaces is continuous if and only if the inverse image of each closed set is a closed set.

Proof. Let F : X -+ Y be a mapping of a metric space X into a metric space Y. Recall that F'1(Y\B) = X\F-1(B) for every subset B of Y (Problem 1.2(b)). Suppose F is continuous and take an arbitrary closed set V in Y. Since Y\V is open

in Y. it follows by Theorem 3.12 that F-1(Y\V) is open in X. Thus F-t (V) = X\F-t (Y\ V) is closed in X. Therefore, the inverse image under F of an arbitrary closed set V in Y is closed in X. Conversely, suppose the inverse image under F of each closed set in Y is a closed set in X and take an arbitrary open set U in Y. Thus F-t(Y\U) is closed in X (since Y\U is closed in Y) so that F-1(U) = X\F-t (Y\U) is open in X. Conclusion: The inverse image under F of an arbitrary D open set U in Y is open in X. Therefore F is continuous by Theorem 3.12. Y of a metric space X into a metric space Y is a closed map A function F : X (or a closed mapping) if the image of each closed set in X is closed in Y (i.e., F(V )

is closed in Y whenever V is closed in X). In general, a map F : X -+ Y may possess any combination of the attributes "continuous", "open" and "closed" (i.e., the are independent concepts). However, if F : X -> Y is invertible (i.e., injective and surjective), then it is a closed map if and only if it is an open map.

Theorem 3.24. Let X and Y be metric spaces. If a map G : X -+ Y is invertible, then

(a) G is closed if and only if G- 1 is continuous,

1 14

3. Topological Structures

(b) G is continuous if and only if G-t is closed, (c) G is a homeomorphism if and only if G and G-t are both closed. Proof. Replace "open map" with "closed map" in the proof of Theorem 3.20 and use Theorem 3.23 instead of Theorem 3.12. Let A be a set in a metric space X and let VA be the collection of all closed subsets of X that include A:

VA = IVEP(X): V is closed in X and Ac V1. The whole set X always belongs to VA so that VA is never empty. The intersection of all sets in VA is called the closure of A in X, denoted by A- (i.e., A - = n VA). According to Theorem 3.22(c) it follows that A- is closed in X

and

ACA-.

If V E VA, then A- = n VA c V. Thus, with respect to the inclusion ordering of

P(X). A - is the smallest closed subset of X that includes A,

and hence (since A- is closed in X) A is closed in X

if and only if

A = A-.

From the above displayed results it is readily verified that

0- = 0,

(A-)- = A-

X - = X,

and, if B also is a set in X,

ACB

A- C B-.

implies

Moreover, since both A and B are subsets of A U B, it follows that A- C (A U

and B- c (A U B)- so that A- U B- C_ (A U B)-. On the other hand, since (A U BY is the smallest closed subset of X that includes A U B, and since A - U B-

is closed (Theorem 3.22(b)) and includes A U B (for A C A- and B C B' so that A U B C A' U B-), it follows that (A U B)- c A' U B-. Therefore, if A and B are subsets of X, then

(AUB)- = A- U B-. It is easy to show by induction that the above identity holds for any finite collection of subsets of X. That is, the closure of the union of a finite collection of subsets of X coincides with the union of their closures. In general (i.e., by allowing infinite collections as well) one has inclusion rather than equality. Indeed, if (Ay )yEr is an arbitrary indexed family of subsets of X. then

UAY C (UAY) Y

y

3.5 Closed Sets and Closure

115

because A. C U,, Ay and hence Aa c (UyAy)- for each index a E T. Similarly,

IAy) c n Ay

I

y

Y

because ny Ay c fly Ay and ny Ay is closed in X by Theorem 3.22(c). However, the above two inclusions are not reversible in general so that equality does not hold. Example 3M. Set X = R with its usual metric and consider the following subsets of R. An = [0, 1-1r ], which is closed in R for each positive integer n, and A = [0, 1), which is not closed in It Since 00

UAn

n=1

= A,

it follows that the union ofan infinite collection of closed sets is not necessarily closed

(see Theorem 3.22(b)). In particular, as An = An for each n and A _ [0, 1], 00

00

[0, 1) = U An C (U An) _ [0, 1], n=1

n=1

which is a proper inclusion. If B = [1, 2] (so that B- = B), then

0 = (A fl B)- c A- f1 B- = [1], so that the closure of any (even finite) intersection of sets may be a proper subset of the intersection of their closures. A point x in X is adherent to A (or an adherent point of A, or a point of adherence

of A) if it belongs to the closure A- of A. It is clear that every point of A is an adherent point of A (i.e., A c A-). Proposition 3.25. Let A be a subset of a metric space X and let x be a point in X. The following assertions are pairwise equivalent. (a) x is a point of adherence of A.

(b) Every open set U in X that contains x meets A (i.e., if U is open in X and

xEU,then AflU#0). (c) Every neighborhood N of x contains at least one point of A (which may be x itself). Proof. Suppose there exists an open set U in X containing x for which A n u = 0.

Then A C X\U, the set X\U is closed in X, and x tl X\U. Since A- is the

116

3. Topological Structures

smallest closed subset of X that includes A, it follows that A- C_ X\U so that x f A-. Thus the denial of (b) implies the denial of (a), which means that (a) implies (b). Conversely, if x f A-, then x lies in the open set X\A- which does not meet A- (A- fl X\A' = 0). Therefore, the denial of (a) implies the denial of (b); that is, (b) implies (a). Finally note that (b) is equivalent to (c) as an obvious

0

consequence of the definition of neighborhood.

A point x in X is a point of accumulation (or an accumulation point, or a cluster point) of A if it is a point of adherence of A\{x ). The set of all accumulation points of A is the derived set of A, denoted by A*. Thus X E A* if and only if x E (A\{x})-.

It is clear that every point of accumulation of A is also a point of adherence of A; that is, A* C A- (for A\{x} e A implies (A\{x})- C A-). Actually,

A- = A U A*. Indeed, since A* c A- and A e A-, it follows that A U A* C A-. On the other hand, if x A U A*, then (A\{x})- = A (because A\(x) = A whenever x V A), and hence x V A- (for x V A* so that x (A\{x})-). Therefore, X E A- implies x E AUA*, which means A- C AUA*. Hence A- = A U A*. Thus

A = A-

if and only if

A* C A.

That is, A is closed in X if and only if it contains all its accumulation points. It is trivially verified that A C B implies A* C B* whenever A and B are subsets of X. Also note that A - = 0 if and only if A = 0 (for

0- = 0 and 0 C A C A-), and A* = 0 whenever A = 0 (because A* C A-), but the converse fails (e.g., the derived set of a singleton is empty).

Proposition 3.26. Let A be a subset of a metric space X and let x be a point in X. The following assertions are pairwise equivalent. (a) x is a point of accumulation of A.

(b) Every open set U in X that contains x also contains at least one point of A other than x. (c) Every neighborhood N of x contains at least one point of A distinct from x.

Proof. Since X E X is a point of accumulation of A if and only if it is a point of adherence of A\(x), it follows by Proposition 3.25 that the assertions (a), (b) and (c) are pairwise equivalent (replace A with A\(x) in Proposition 3.25). 0 Everything that has been written so far in this section pertains to the realm of topological spaces (metrizable or not). However, the next results are typical of metric spaces.

Proposition 3.27. Let A be a subset of a metric space (X, d) and let x be a point in X. The following assertions are pairwise equivalent.

3.5 Closed Sets and Closure

117

(a) x is a point of adherence of A.

(b) Every nonempty open ball centered at x meets A.

(c) A # 0 and d(x, A) = 0. (d) There exists an A-valued sequence that converges to x in (X, d).

Proof. The equivalence (a)..(b) follows by Proposition 3.25 (recall: a nonempty open ball centered at x is a neighborhood of x and, conversely, every neighborhood of x includes a nonempty open ball centered at x, so that every nonempty open ball

centered at x meets A if and only if every neighborhood of x meets A). Clearly (b).'(c) (i.e., for each e > 0 there exists a E A such that d (x, a) < c if and only if A # 0 and infaEA d(x, a) = 0). Theorem 3.14 ensures that (d)=(b). On the other hand, if (b) holds true, then for each positive integer n the open ball Bi (x) meets A

(i.e., B i (x) fl A 76 0). Take X,, E B* (x) fl A so that X. E A and 0 < d(x,,, x) < W is an A-valued sequence such that d(x,,, x) - 0. Therefore for each n. Thus

(b) *(d).

o

Proposition 3.28. Let A be a subset of a metric space (X, d) and let x be a point in X. The following assertions are pairwise equivalent. (a) x is a point of accumulation of A.

(b) Every nonempty open ball centered at x contains a point of A distinct from x. (c) Every nonempty open ball centered at x contains infinitely many points of A.

(d) There exists an A\(x)-valued sequence of pairwise distinct points that converges to x in (X, d).

Proof. Note that (d)=*(c) by Theorem 3.14. (c)=(b) trivially, and (d) (a) (b) by the previous proposition. To complete the proof it remains to show that (b)=(d). Let BE(x) be open ball centered at x E X with radius e > 0. We shall say that an A-valued sequence (xk)kEN has Property P,,, for some integer n E N, if xk E (Bi (x)\(x)) for each k = i, ... , n+I and d(xk+t, x) < d(xk, x) for every

k = t, .. , n. Claim. If assertion (b) holds true, then there exists an A-valued sequence that has Property P. for every n E N.

Proof. Suppose (b) holds true so that (BE(x)\{x)) fl A A 0 for every e > 0. Take an arbitrary x1 E (BI (x)\(x)) fl A and an arbitrary x2 E (B£2(x)\(x)) fl A where e2 = min( I', d(xj, x)}. Every A-valued sequence whose first two entries coincide with x I and x2 has Property P1. Suppose there exists an A-valued sequence that has Property P. for some integer n E N. Take any point from (B,.12 (x)\(x )) fl A where

118

3. Topological Structures

x)}, and replace the (n+2)th entry of that sequence with an+2 = min{ nom, d this point. The resulting sequence has Property P.+1. Thus there exists an A-valued

whenever there exists one that has Property P,,, sequence that has Property and this concludes the proof by induction. 0 However, an A-valued sequence (xk )kEN that has Property P. for every n c- N in fact

is an A\{x}-valued sequence of pairwise distinct points such that 0 < d(xk, x) < I for every k E N. Therefore (b)=(d). 0 Recall that "point of adherence" and "point of accumulation" are concepts defined for sets, while "limit of a convergent sequence" is, of course, a concept defined for sequences. But the range of a sequence is a set, and it can have (many) accumulation be an X-valued sequence. A point x points. Let (X, d) be a metric space and let in X is a cluster point of the sequence {x) if some subsequence of converges to x. The cluster points of a sequence are precisely the accumulation points of its range (Proposition 3.28). If a sequence is convergent, then (Proposition 3.5) its range has only one point of accumulation which coincides with the unique limit of the sequence.

Corollary 3.29. The derived set A' of every subset A of a metric space (X, d) is closed in (X, d).

Proof. Let A be an arbitrary subset of a metric space (X, d). We want to show that (A*)- = A' (i.e., A' is closed) or, equivalently, (A')- a A* (recall: every set is included in its closure). If A is empty, then the result is trivially verified

(0* = 0 = 0-). Thus suppose A is nonempty. Take an arbitrary x- E (A*)and an arbitrary e > 0. Proposition 3.27 ensures that BE(x-) fl A* 96 0. Take

x' E BE(x-) fl A* and set S = e - d(x', x-). Note that 0 < S < e (for 0 < d(x', x-) < e). Since x' E A', it follows by Proposition 3.28 that Ba(x') fl A contains infinitely many points. Take x E Ba (x*) fl A distinct from x- and from x'.

Thus 0 < d(x, x-) < d(x, x') + d(x*, x-) < S + d(x*. x-) = e by the triangle inequality. Therefore X E BE(x-) and x A x-. Conclusion: Every nonempty ball BE(x-) centered at x- contains a point x of A other than x-. Thus X_ E A' by 13 Proposition 3.28, and therefore (A*)- C A'. The above corollary does not hold in a general topological space. Indeed, if a set X containing more than one point is equipped with the indiscrete topology (where the only open sets are 0 and X), then the derived set (x)* of a singleton (x) is X\(x) which is not closed in that topology.

Theorem 3.30. (The Closed Set Theorem). A subset A of a set X is closed in the metric space (X, d) if and only if every A -valued sequence that converges in (X, d) has its limit in A. Proof. (a) Take an arbitrary A-valued sequence

that converges to x E X in

(X, d). By Theorem 3.14 (x,,) is eventually in every neighborhood of x, and hence every neighborhood of x contains a point of A. Thus x is a point of adherence of

3.6 Dense Sets and Separable Spaces

119

A (Proposition 3.25); that is, x E A-. If A = A- (equivalently, if A is closed in (X, d)), then x E A.

(b) Take an arbitrary point x E A- (i.e., an arbitrary point of adherence of A). According to Proposition 3.27, there exists an A-valued sequence that converges to x in (X, d). If every A-valued sequence that converges in (X, d) has its limit in A,

then x E A. Thus A- C A, and hence A = A- (for A c A- for every set A). That o is, A is closed in (X, d). This is a particularly useful result that will be often applied throughout this book.

Note that part (a) of the proof holds for general topological spaces but not part (b). The counterpart of the above theorem for general (not necessarily metrizable) topological spaces is stated in terms of nets (instead of sequences).

Example 3N. Consider the set B[X, Y] of all bounded mappings of a metric space (X, dx) into a metric space (Y, dy), and let BC[X, Y] denote the subset of B[X, Y] consisting of all bounded continuous mappings of (X, dx) into (Y, dy). Equip B[X, Y] with the sup-metric doo as in Example 3C. We shall use the Closed Set Theorem to show that

BC[X, Y] is closed in (B[X, Y], dam). Take an arbitrary BC[X, Y]-valued sequence {fn} that converges in (B[X, Y), dam) to a mapping f E B[X, Y]. The triangle inequality in (Y, dy) ensures that

dy(f(u), f(v)) _< dy(f(u), fn(u))+dy(fn(u),.fn(v))+dy(fn(v), f(v)) for each integer n and every u, v E X. Take an arbitrary real number E > 0. Since fn -+ f in (B[X, Y], dx), it follows that there exists a positive integer nE such

that doo(fn, f) = sup

f(x)) < E, and hence

f(x)) <'

for all x E X, whenever n > ne (uniform convergence - see Example 3G). Since each fn is continuous, it follows that there exists a real number SE > 0 (which may

depend on u and v) such that dy(fn, (u), fn, (v)) < 3 whenever dx(u, v) < S. Therefore dy(f (u), f (v)) < E whenever dx(u, v) < SE, so that f is continuous. That is, f E BC[X, Y]. Thus, according to Theorem 3.30, BC[X, Y] is a closed subset of the metric space (B[X, Y]. dm). Particular case (see Examples 3D and 3G): C[0, 11 is closed in (B[0, 11, dm).

3.6

Dense Sets and Separable Spaces

Let A be a set in a metric space X and let UA be the collection of all open subsets of X included in A:

UA = { U E P (X) : U is open in X and U CA).

120

3. Topological Structures

The empty set 0 of X always belongs to UA so that UA is never empty. The union of all sets in UA is called the interior of A in X, denoted by A° (i.e., A° = U UA). According to Theorem 3.15(c) it follows that A° is open in X

A° C A.

and

If U E UA, then U e U UA = A°. Thus, with respect to the inclusion ordering of

P(X). A° is the largest open subset of X that is included in A,

and hence (since A° is open in X)

if and only if

A is open in X

A° = A.

From the above displayed results it is readily verified that

0° = 0,

X° = X,

(A°)° = A°

and, if B also is a set in X,

AcB

A° a B°.

implies

Moreover, since A n B is a subset of both A and B, it follows that (A n B)° C A° n B°. On the other hand, since (A n B)° is the largest open subset of X that is included in A n B, and since A° n B° is open (Theorem 3.15(b)) and is included in A n B (for A° c_ A and B° C B so that A° n B° c A n B), it follows that A° n B° c (A n B)°. Therefore, if A and B are subsets of X, then

(A n B)° = A° n B°. It is easy to show by induction that the above identity holds for any finite collection

of subsets of X. That is, the interior of the intersection of a finite collection of subsets of X coincides with the intersection of their interiors. In general (i.e., by allowing infinite collections as well) one has inclusion rather than equality. Indeed, if {Ay is an arbitrary indexed family of subsets of X, then

(I

I

Ay)° C I

Y

I

Ay

y

because f y Ay g A° and hence (f y Ay )° a A,,, for each index a E r. Similarly,

UAY C (UAY Y

y

because UY A°y C UY Ay and UyA°y is open in X by Theorem 3.15(c). However. the above two inclusions are not reversible in general so that equality does not hold.

3.6 Dense Sets and Separable Spaces

121

Example 30. This is the dual of Example 3M. Consider the setup of Example 3M and set Cn = X\An, which is open in R for each positive integer n, and C = X\A, which is not open in R. Since 00

00

00

n Cn = n(X\An) = X` U An = X\A = C, n=1

n=1

n=1

it follows that the intersection of an infinite collection of open sets is not necessarily open (see Theorem 3.15(b)). In particular, as C° = Cn, 00

00

(X\A)° = C° = (n Cn)° c n C° = C = X\A n=1

n=1

which is a proper inclusion. Now set D = X\B = (-oo, 1) U (2, oo) (so that D° = D). Thus C° U D° is a proper subset of (C U D)°:

R\(I) = CO U D° c (C U D)° = R. Remark: The duality between "interior" and "closure" is clear:

(X\A)` = X\A°

and

(X\A)° = X\A-

for every subset A of X. Indeed, U E UA if and only if X \U E VX\A (i.e., U is open

in X and U C A if and only if X\U is closed in X and X\A C X\U) and, dually, V E Vx\A if and only ifX\V E UA.ThusA° = X\nUEUA(X\U) = X\nvEVI\AV = X\(X\A)- and so X\A° = (X\A)-; which implies (swap A and X\A) that X \(X \A)° = A- and hence (X \A)° = X\A-. This confirms the above identities and also their equivalent forms:

A° = X\(X\A)-

and

A- = X\(X\A)°.

A point x E X is an interior point of A if it belongs to the interior A° of A. It is clear that every interior point of A is a point of A (i.e., A° C A), and it is readily verified that x E A is an interior point of A if and only if there exists a neighborhood of x included in A (reason: A° is the largest open neighborhood of every interior point of A that is included in A). A subset A of a metric space X is called dense in X (or dense everywhere) if its closure A- coincides with X (i.e., if A- = X). More generally, suppose A and B are subsets of a metric space X such that A C B. A is dense in B if B C A or, equivalently, if A- = B-. (Why?) Clearly, if A C_ B and A- = X. then B- = X. Note that the only closed set dense in X is X itself. Proposition 3.31. Let A be a subset of a metric space X. The following assertions are pairwise equivalent.

122

3. Topological Structures

(a) A- = X

(i.e., A is dense in X).

(b) Every nonempty open subset of X meets A. (c) VA = {X}.

(d) (X\A)° = 0

(i.e., the complement of A has empty interior).

Proof. Take any nonempty open subset U of X, and take an arbitrary u E U C X. If (a) holds true, then every point of X is adherent to A. In particular, u is adherent to A. Thus Proposition 3.25 ensures that U meets A. Conclusion: (a)=(b). Now take an arbitrary proper closed subset V of X, so that 0 96 X\V is open in X. If (b) holds true, then (X\V) fl A 96 0. Thus V does not include A. and hence V f VA. Since A- surely belongs to VA, it follows that (c)=>(a). The Therefore o equivalence (a). (d) is obvious from the identity A- = X\(X\A)°.

The reader has probably observed that the concepts and results so far in this section apply to topological spaces in general. From now on the metric will play its role. Note that a point in a subset A of a metric space X is an interior point of A if and only if it is the center of a nonempty open ball included in A (reason: every nonempty open ball is a neighborhood and every neighborhood includes a nonempty open ball). We shall say that (A, d) is a dense subspace of a metric space (X, d) if the subset A of X is dense in (X, d). Proposition 3.32. Let (X, d) be a metric space and let A and B be subsets of X such that 0 96 A C B C X. The following assertions are pairwise equivalent.

(a) A-= B-

(i.e., A is dense in B).

(b) Every nonempty open ball centered at any point b of B meets A. (c) infOEA d(b, a) = 0 for every b E B.

(d) For every point bin B there exists an A-valued sequence (a.) that converges in (X, d) to b.

Proof. Recall that A- = B- if and only if B C_ A-. Let b be an arbitrary point in B. Thus assertion (a) can be rewritten as (a) every point b in B is a point of adherence of A. Now notice that assertions (a), (a). (b) and (c) are pairwise equivalent by Proposition 3.27. o

Corollary 3.33. Let F and G be continuous mappings of a metric space X into a metric space Y. If F and G coincide on a dense subset of X, then they coincide on the whole space X.

3.6 Dense Sets and Separable Spaces

123

Proof. Suppose X is nonempty, to avoid trivialities, and let A be a nonempty dense subset of X. Take an arbitrary x E X and let (a"} be an A-valued sequence that converges in X to x (whose existence is ensured by Proposition 3.32). If F: X --0, Y and G : X -- Y are continuous mappings such that FIA = GIA, then

F(x) = F(lima") = limF(a") = limG(a") = G(lima") = G(x) (Corollary 3.8). Thus F(x) = G(x) for every x E X; that is, F = G.

El

A metric space (X, d) is separable if there exists a countable dense set in X. The density criteria in Proposition 3.32 (with B = X) are particularly useful to check separability.

Example 3P. Take an arbitrary integer n > 1, an arbitrary real p > 1, and consider the metric space (R", dp) of Example 3A. Since the set of all rational numbers Q is dense in the real line R equipped with its usual metric, it follows that Q" (the

set of all rational n-tuples) is dense in (R", dp). Indeed, Q- = R implies that

inf ,QI - vI = 0 for every E R, which in turn implies that n

inf d p(x, y) = YEQ"

inf

/y

Y=(UI .....Vn)E Q"

Iii - v; Ip) r = 0

i=I

for every x = (I, ... , k") E R". Hence

R", according to Proposition 3.32. Moreover, since #Q" = #Q = No (Problems 1.25(c) and 2.8), it follows that Q" is countably infinite. Therefore Q" is a countable dense subset of (R", dp), and hence (R", dp) is a separable metric space. Now consider the metric space (e+, d p) for any p > 1 as in Example 3B, where £. is the set of all real-valued p-summable infinite sequences. Let t. be the subset of £. made up of all real-valued infinite sequences with a finite number of nonzero entries, and let X be the subset of eO consisting of all rational-valued infinite sequences with a finite number of nonzero entries. The set eO is dense in (e+, d p) - Problem 3.44(c).

Since Q- = R, it follows that X is dense in (8+, dp) - the proof is essentially the same as the proof that Q" is dense in (R", dp). Thus X- = (e+)- = f+ p, and hence X is dense in (P+,dp). Next we show that X is countably infinite. In fact, X is a linear space over the rational field Q for which dim X = No (see Example 2J). Thus, according to Problem 2.8, #X = max{#Q, dim X} = No. Conclusion: X is a countable dense subset of (e+, d p), and so (e+ p, d p) is a separable metric space.

The same argument is readily extended to complex spaces so that (C", dp) also is separable, as well as (e+, d p) when e+ is made up of all complex-valued p-summable infinite sequences. Finally we show that

(C[0, 1], dd) is a separable metric space

124

3. Topological Structures

(see Example 3D). Actually, the set P[0, 11 of all polynomials on [0, 11 is dense in (C[0, 1], dam). This is the well-known Weierstrass Theorem, which says that every continuous function in C[0, 1) is the uniform limit of a sequence of polynomials

in P[0, 1] (i.e., for every x E C[0, 1] there exists a P[0, 1]-valued sequence (p.1 such that doo (p., x) -+ 0). Moreover, it is easy to show that the set X of all polynomials on [0, 1 ] with rational coefficients is dense in (P[0, I], dam), and hence X is dense in (C[0, 1 ], dam). Since X is a linear space over the rational field Q. and since dim X = No (essentially the same proof as that in Example 2M), it follows by Problem 2.8 that X is countable. Thus X is a countable dense subset of (C[0, 1 ], doo).

A collection B of open subsets of a metric space X is a base (or a topological base) for X if every open set in X is the union of some subcollection of B. For instance, the collection of all open balls in a metric space (including the empty ball) is a base for X (cf. Corollary 3.16). Note that the above definition of base forces the empty set 0 of X to be a member of any base for X.

Proposition 3.34. Let X be a metric space and let B be a collection of open subsets of X that contains the empty set. The following assertions are pairwise equivalent. (a) B is a base for X.

(b) For every nonempty open subset U of X and every point x in U, there exists a set B in B such that x E B C U.

(c) For every x in X and every neighborhood N of x, there exists a set B in B such that x E B C N. Proof. Take an arbitrary open subset U of the metric space X and set BU = (B E B: B C U). If B is a base for X, then U = U Bu by definition of base. Thus, if x E U, then x E B for some B E BU so that x E B C_ U. That is, (a) implies (b). On the other hand, if (b) holds, then any open subset U of X clearly coincides with U BU, which shows that (a) holds true. Finally note that (b) and (c) are trivially equivalent:

every neighborhood of x includes an open set containing X, and every open set containing x is a neighborhood of x. D Theorem 3.35. A metric space (X, d) is separable if and only if it has a countable base. Proof. Suppose B is a countable base for X. For each nonempty set B in B take an arbitrary point b in B. Now consider the set (bn) consisting of one point of each nonempty set B in B. The set {b,) is countable and every nonempty open subset U of X meets (since U is the union of some subcollection of so that U n (b,,) 36 0). Therefore (cf. Proposition 3.31) is a countable dense subset of X, and hence X is separable. On the other hand, suppose X is separable. Then there exists a countable subset A of X that is dense in X. Consider the collection

3.6 Dense Sets and Separable Spaces

125

of nonempty open balls B = (Bi(a): n E N and a E A). Observe that B = UaEAIBi(a))nEN, a countable union of countable collections, so that B is itself a countable collection (Corollary 1.11).

Claim. For every x E X and every neighborhood N of x there exists a ball in B containing x and included in N. Proof. Take an arbitrary x E X and an arbitrary neighborhood N of x. Let BE(x) be an open ball of radius e > 0, centered at x, and included in N. Now take a positive integer n such that 1 < , and a point a E A such that a E B i (x) (recall:

since A- = X, it follows by Proposition 3.32 that there exists a E A such that a E B,, (x) for every x E X and every p > 0). Obviously, X E Bi (a). Moreover, Bi (a) a BE(x). Indeed, if y E BR (a), then d(x, y) < d(x, a) + d(a, y) < so that y E Be (x). Thus X E Bi (a) c B.(x) c N.

<E

Therefore the countable collection B U (0) of open balls is a base for X by Proposition 3.34.

Corollary 3.36. Every subspace of a separable metric space is itself separable. Proof. Let S be a subspace of a separable metric space X and, according to Theorem 3.35, let B be a countable base for X. Set Bs = (Sfl B : B E B), which is a countable collection of subsets of S. Since the sets in B are open subsets of X, it follows that the sets in Bs are open relative to S (see Problem 3.38(c)). Take an arbitrary nonempty

relatively open subset A of S so that A = S fl U for some open subset U of X (Problem 3.38(c)). Since U = U B' for some subcollection B' of B, it follows that

A = S fl UBEB,B = UBEBfS n B = UBs, where BS = IS fl B: B E B') is a subcollection of Bs. Thus Bs is a base for S. Therefore the subspace S has a countable base, which means by the previous theorem that S is separable.

Let A be a subset of a metric space. An isolated point of A is a point in A that is not an accumulation point of A. That is, a point x is an isolated point of A if x E A\A`. Proposition 3.37. Let A be a subset of a metric space X and let x be a point in A. The following assertions are pairwise equivalent.

(a) x is an isolated point of A. (b) There exists an open set U in X such that A U U = {x}.

(c) There exists a neighborhood N of x such that A U N = (x). (d) There exists a nonempty open ball B,,(x) centered at x such that A U B,,(x) _ IX).

126

3. Topological Structures

Proof. Assertion (a) is equivalent to assertion (b) by Proposition 3.26. Assertions (b), (c) and (d) are trivially pairwise equivalent.

0

A subset A of X consisting entirely of isolated points is a discrete subset of X. This means that in the subspace A every set is open, and hence the subspace A is homeomorphic to a discrete space (i.e., to a metric space equipped with the discrete metric). According to Theorem 3.35 and Corollary 3.36, a discrete subset of a separable metric space is countable. Thus, if a metric space has an uncountable discrete subset, then it is not separable.

Example 3Q. Let S be a set, let (Y, d) be a metric space, and consider the metric space (B[S, Y], dd) of all bounded mappings of S into (Y, d) equipped with the sup-metric doo (see Example 3C). Suppose Y has more than one point, and let yo and y1 be two distinct points in Y. As usual, let 2s denote the set of all mappings on S with values either yo or yi (i.e., the set of all mappings of S into (yo, yt } so that 2S = (yo. y1 )S C B[S, Y]). If f, g E 2s and f 96 g (i.e., if f and g are two distinct mappings on S with values either yo or yt ), then

dm(f g) = supd(f (s), g(s)) = d(yo, yt) A 0. ses

Therefore, any open ball Bp (g) = (f E 2s: doo (f, g) < p) centered at an arbitrary point g of 2s with radius p = d (yo, Y1)/2 is such that 2s n B,.(g) = (g}. This means that every point of 2S is an isolated point of it, and hence

2S is a discrete set in (B[S, Y], dam). If S is an infinite set, then 2S is an uncountable subset of B[S, YJ (recall: if S is infinite, then No < #S < #2S according to Theorems 1.4 and 1.5). Thus (B[S, Y], dd) is not separable whenever 2 < #Y and No < #S. Concrete example:

(P°+°, do,) is not a separable metric space.

Indeed, set S = N and Y = C (or Y = R) with its usual metric d, so that (B[S, Y], doo) = (e{°, doo): the set of all scalar-valued bounded sequences equipped with the sup-metric, as introduced in Example 3B. The set 2N, consisting of all sequences with values either 0 or 1, is an uncountable discrete subset of (& 0, dd).

In a discrete subset every point is isolated. The opposite notion is that of a set where every point is not isolated. A subset A of a metric space X is dense in itself if A has no isolated point or, equivalently, if every point in A is an accumulation point

of A. That is, if A C_ A*. Since A- = A U A' for every subset A of X. it follows that a set A is dense in itself if and only if A' = A-. A subset A of X that is both closed in X and dense in itself (i.e., such that A' = A) is a perfect set: a closed set without isolated points. For instance, Q n [0, 11 is a countable perfect subset of the metric space Q, but it is not perfect in the metric space R (since it is not closed in R). As a matter of fact, every nonempty perfect subset of R is uncountable because R is a "complete" metric space, a concept that we shall define next.

3.7 Complete Spaces

3.7

127

Complete Spaces

Consider the metric space (R, d), where d denotes the usual metric on the real line ;.R, and let (A, d) be the subspace of (R, d) with A = (0, 11. Let be the A-valued sequence such that an = >s for each n E N. Does (an } converge in the metric space (A, d)? It is clear that {an } converges to 0 in (R, d), and hence we might at a first glance think that it also converges in (A, d). But the point 0 simply does not exist in A so that it is nonsense to say that "an --> 0 in (A, d)". In fact, (a } does not converge in the metric space (A, d). However, the sequence (an} seems to possess a "special property" that makes it apparently convergent in spite of the particular underlying set A, and the metric space (A, d) in turn seems to bear a "peculiar characteristic" that makes such a sequence fail to converge in it. The "special property" of the sequence (an ) is that it is a Cauchy sequence in (A, d) and the "peculiar characteristic" of the metric space (A, d) is that it is not complete. Definition 3.38. Let (X, d) be a metric space. An X -valued sequence (xe) (indexed either by N or N0) is a Cauchy sequence in (X, d) (or satisfies the Cauchy criterion) if for each real number e > 0 there exists a positive integer ne such that

n, in > n£

implies

d (xm, xn) < E.

A usual notation for the Cauchy criterion is limm,n d(xm, xn) = 0. Equivalently, an

X-valued sequence (xn) is a Cauchy sequence if diam({xk}n
Proposition 3.39. Let (X, d) be a metric space. (a) Every convergent sequence in (X, d) is a Cauchy sequence. (b) Every Cauchy sequence in (X, d) is bounded.

(c) If a Cauchy sequence in (X, d) has a subsequence that converges in (X, d), then it converges itself in (X, d) and its limit coincides with the limit of that convergent subsequence. Proof. (a) Take an arbitrary e > 0. If an X -valued sequence (x,,) converges to a point x E X, then there exists an integer n£ > 1 such that d(xn, x) < ' whenever n > ne. Since d(xm, xn) < d(xm, x) + d(x. for every pair of indices in. n (triangle inequality), it follows that d(xm, e whenever in. n > nf. (b) If {xn) is a Cauchy sequence, then there exists an integer n i > 1 such that

d(xm, xn) < 1 whenever m, n > n1. Note that the set {d(xm, xn) E R: in, n < n1)

128

3. Topological Structures

has a maximum in R, say $, because it is finite. Thus d(xm, d (x,,, , 2 max( 1, P) for every pair of indices m, n.

d(xm, xn,) +

(c) Suppose (x,,,) is a subsequence of an X-valued Cauchy sequence (xn) that converges to a point x E X (i.e., x,,, -> x as k --> oo). Take an arbitrary e > 0. Since (x,,) is a Cauchy sequence, it follows that there exists a positive integer of such that d (xm ,

whenever m, n > n f. Since {x,,, } converges to x, it follows

that there exists a positive integer kE such that d(x,,,, x) < ' whenever k > kE. Thus, if j is any integer with the property that j > kE and nj > nE (for instance, if j = max(nF, k5}), then d(x,,, x) < d(x,,, x) < e for every n > nE, and therefore (x,,) converges to X.

El

Although a convergent sequence always is a Cauchy sequence, the converse may fail. For instance, the (0, 1]-valued sequence is a Cauchy sequence in the metric space ((0, 1 ], d), where d is the usual metric on R, which does not converge in ((0, 1], d). There are however many metric spaces with the notable property that Cauchy sequences in it are convergent. Metric spaces possessing this property are so important that we give them a name. A metric space X is complete if every Cauchy sequence in X is a convergent sequence in X.

Theorem 3.40. Let A be a subset of a metric space X. (a) If the subspace A is complete, then A is closed in X.

(b) If X is complete and if A is closed in X, then the subspace A is complete. Proof. (a) Take an arbitrary A-valued sequence (an) that converges in X. Since

every convergent sequence is a Cauchy sequence, it follows that (an) is a Cauchy sequence in X, and therefore a Cauchy sequence in the subspace A. If the subspace A is complete, then {a } converges in A. Conclusion: If A is complete as a subspace of X, then every A-valued sequence that converges in X has its limit in A. Thus, according to the Closed Set Theorem (Theorem 3.30), A is closed in X. (b) Take an arbitrary A-valued Cauchy sequence If X is complete, then converges in X to a point a E X. If A is closed in X. then Theorem 3.30 (the Closed Set Theorem again) ensures that a E A, and hence {a } converges in the subspace A. Conclusion: If X is complete and A is closed in X, then every Cauchy sequence in the subspace A converges in A. That is, A is complete as a subspace of X. El An important, although immediate, corollary of the above theorem says that "inside" a complete metric space the properties of being closed and complete coincide.

Corollary 3.41. Let X be a complete metric space. A subset A of X is closed in X if and only if the subspace A is complete. Example 3R. (a) A basic property of the real number system is that every bounded sequence of real numbers has a convergent subsequence. This and Proposition 3.39

3.7 Complete Spaces

129

ensure that the metric space R (equipped with its usual metric) is complete; and so is the metric space C of all complex numbers equipped with its usual metric (reason: if (ak) is a Cauchy sequence in C, then (Re ak } and (Im ak) are both Cauchy sequences

in R so that they converge in R, and hence (ak } converges in Q. Since the set Q of all rational numbers is not closed in R (recall: Q- = R), it follows by Corollary 3.41 that the metric Q is not complete More generally (but similarly), for every positive integer n. R" and C" are complete metric spaces

when equipped with any of their metrics d p for p > 1 or dO (as in Example 3A); while

Q" is not a complete metric space.

(b) Now let F denote either the real field R or the complex field C equipped with their usual metrics. As we have just seen, F is a complete metric space. For each real number p > 1 let (P+, dp) be the metric space of all F-valued p-summable sequences equipped with its usual metric dp as in Example 3B. Take an arbitrary Cauchy sequence in (£+, dp), say (Xn)nEN. Recall that this is a sequence of sequences; that is, x,, = (in(k))kEN is a sequence in e+ for each integer n E N. The Cauchy criterion says: for every e > 0 there exists an integer ne > I such that d p(xm. xn) < e whenever m, n > nE. Thus 00

i

Im(i)-+;n(i)Ip)II = dp(xm,Xn) < E i=1

for every k E N whenever m, n > n, . Therefore, for each k E N the scalar-valued sequence (rn(k)}nEN is a Cauchy sequence in F, and hence it converges in F (since F is complete) to, say, r; (k) E F. Consider the scalar-valued sequence x = (r; (k))kEN consisting of those limits i; (k) E F for every k E N. First we show that x E e+. Since

(x"}nEN is a Cauchy sequence in (e+, dp), it follows by Proposition 3.39 that it is bounded (i.e., supm,ndp(xm, xn) < oo), and hence supmdp(xm, 0) < oo where 0 denotes the null sequence in a+. (Indeed, for every m E N the triangle inequality ensures that dp(xm, 0) < supn,ndp(xm, xn) + dp(xn, 0) for an arbitrary n E N). Therefore, n(k)I")

It r

<

dp(Xn,0) < supdp(Xm,0)

k=1

k=1

M

for every n E N and each integer j ? 1. Since ln(k) -+ !; (k) in F as n -> oo for each k E N, it follows that

I

j

I

supdp(xm.0)

lirn k=1

k=I

M

130

3. Topological Structures

for every j E N. Thus 00

EI

1

(k)IP)

k=1

=sup( J

l

!

E It(k)It')' < supdp(xm, 0), m

k=1

which means that x = (i; (k))kEN E P+. Next we show that xn -> x in (e+, d p). Again, as {xn }fEN is a Cauchy sequence in (f,°p, dp), for any E > 0 there exists an integer nE > 1 such that dp(xm, xn) < E whenever m, n > na. Thus 00

In(k) - m(kllt' < EP

I Sn(k) - lm(k)I p < k=1

k=1

for every integer j > 1 whenever m, n > ne . Since limn tm (k) = (k) for each k E N, it follows that rk_1 Iln(k) - 1; (k)Ip < e°, and hence 00

1

i

k=1

I

In(k) - (k)Ip)' < E,

Itn(k) - (k)Ip) F = Sup

dp(xn, x)

k=1

whenever n > nE; which means that x(n) -> x in (e+, d p). Therefore

(t-P, d p) is a complete metric space for every p > 1. Similarly (see Example 3B), for each p > 1,

(1p, dp) is a complete metric space. Example 3S. Let S be a nonempty set, let (Y, d) be a metric space, and consider the metric space (B[S, Y], d00) of all bounded mappings of S into (Y, d) equipped with the sup-metric do0 (Example 3C). We claim that

(B[S, Y], d00) is complete if and only if (Y, d) is complete. (a) Indeed, if (fn } is a Cauchy sequence in (B[S, Y], d00), then { fn (s)) is a Cauchy

sequence in(Y,d)foreverys E S(ford(fm(s), fn(s)) d00 (fn, fn) for each pair of integers m, n and every s E S), and hence (f. (s)) converges in (Y, d) for every s E S whenever (Y, d) is a complete metric space. For

each s E S set f(s) = limn f, (s) (i.e., fn (s) -* f(s) in (Y,d)), which defines a function f of S into Y. We shall show that f E B[S, Y] and that fn --

f

in (B[S, Y], d00), thus proving that (B[S, Y], d00) is complete whenever (Y, d) is complete. First note that, by the triangle inequality,

d(f(s), f(t)) < d(f(s), f. (s)) +d(fn(s), fn(t))+d(fn(t), f(t))

3.7 Complete Spaces

131

for each positive integer n and every pair of points s, t in S. Now take an arbitrary

e > 0. Since (fn) is a Cauchy sequence in (B[S, Y], dam), it follows that there exists a positive integer nE such that dd(fm, fn) = sup:Esd(fm(s), fn(s)) < s, and hence

d(fm(s), fn(s)) < -for all S E S, whenever m, n > ne. Moreover, since fm(s) -+ f (s) in (Y, d) for y): Y -> R is a every s E S, and since the metric is continuous (that is, continuous function from the metric space Y to the metric space R for every y E Y),

it also followsthatd(f(s), fn (s)) =d(limm f. (s), fn(s)) = limmd(fm(s), fn(s)) for each positive integer n and every s E S (see Problem 3.14 or 3.34 and Corollary 3.8). Thus

d(f(s), fn (s)) < e for all s E S whenever n > ne. Furthermore, as each fn lies in B[S, Y], there exists a real number yn, such that sup d(fn,(s). fn, (t)) < yn,.

S.IEs

Summing up: For an arbitrary real number e > 0 there exists a positive integer nE such that

d(f(s), f(t)) < 2e+yn, for all s, t E S, so that f E B[S, Y]; and

doo(f fn) = supd(f(s), fn (s)) < e SES

whenever n > ne, so that fn -* fin (B[S, Y], do.). (b) Conversely, take an arbitrary Y-valued sequence (y,,). Suppose S is nonempty and set fn (s) = yn for each integer n and all s E S. This defines a sequence [f.) of constant mappings of S into Y which clearly lie in B[S, Y] (a constant mapping is obviously bounded). Note that doo(fm, f,,) = supsEsd(fm(s), fn(s)) = d(ym, yn) for every pair of integers m, n. Thus (f.) is a Cauchy sequence in (B[S, Y], dam) if and only if { yn } is a Cauchy sequence in (Y, d). Moreover, (fn } converges in (B[S, Y], dam) if and only if (yn) converges in (Y, d) (reason: if d(yn, y) - 0 for

some y E Y. then doo(fn, f) --> 0 where f E B[S. Y] is the constant mapping

f (s) = y for all s E S; and if doo(f,,, f) --> 0 for some f E B[S, Y], then d(yn, f (s)) = d(fn(s), f (s)) for each n and every s, so that d(yn, f (s)) -* 0 for all S E S - and hence f must be a constant mapping). Now suppose (Y, d) is not complete, which implies that there exists a Cauchy sequence in (Y, d), say (y.), that fails to converge in (Y, d). Thus the sequence (f.) of constant mappings f" (s) = yn for each integer n and all s E S is a Cauchy sequence in (B[S, Y], dam) that fails to converge in (B[S, Y], dam), and hence (B[S, Y], dd) is not complete. Conclusion: If (B[S, Y], doo) is complete, then (Y, d) is complete.

132

3. Topological Structures

(c) Concrete example: Set S = N or S = Z and Y = IF (either the real field R or the complex field C equipped with their usual metric). Then (e.°, d0,) and (100, dam) are complete metric spaces.

Example 3T. Consider the set B[X, Y] of all bounded mappings of a nonempty metric space (X, dX) into a metric space (Y, dy) and equip it with the sup-metric dO0 as in the previous example. Let BC[X, Y] be the set of all continuous mappings from B[X, Y] (Example 3N), so that (BC[X, Y], dam) is the subspace of (B[X, Y], doo) made up of all bounded continuous mappings of (X, dX) into (Y, dy).

If (Y, dy) is complete, then (B[X, Y], doo) is complete according to Example 3S. Since BC[X, Y] is closed in (B[X, Y], doo) - Example 3N - it follows by Theorem 3.40 that (BC[X, Y], doo) is complete. On the other hand, the very same construction used in item (b) of the previous example shows that (BC[X, Y], dam) is not complete unless (Y, dy) is. Conclusion:

(BC[X, Y], dam) is complete if and only if (Y, dX) is complete. In particular (see Examples 3D, 3G and 3N),

(C[0, fl, d..) is a complete metric space because