Schaum’s Outline of General Topology (Schaum’s Outlines)

SCHAUM'S OUTLINE OF

THEORY AND PROBLEMS OF

GENERAL TOPOLOGY •

BY

SEYMOUR LIPSCHUTZ, Ph.D. Associate Professor of Mathematics Temple University

•

SCHAUM'S OUTLINE SERIES McGRAW-HILL BOOK COMPANY New York, St. Louis, San Francisco, Toronto, Sydney

Copyright © 1965 by McGraw-Hill, Inc. All Rights Reserved. Printed in the United States of America.

No part of this publication may be reproduced,

stored in a retrieval system, or transmitted,

in any form or by any means,

electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. 37988 2 3 4 5 6 7 8 9 0 SHSH 7 2 1 0 6 9 8

Preface General topology, also called point set topology, has recently become an essential part of the mathematical background of both graduate and undergraduate students. This book is designed to be used either as a textbook for a formal course in topology or as a supplement to all current standard texts. It should also be of considerable value as a source and reference book for those who require a comprehensive and rigorous introduction to the subject. Each chapter begins with clear statements of pertinent definitions, principles and theorems together with illustrative and other descriptive material. This is followed by graded sets of solved and supplementary problems. The solved problems serve t6 illustrate and amplify the theory, bring into sharp focus those fine points without which the student continually feels himself on unsafe ground, and provide the repetition of basic principles so vital to effective learning. Numerous proofs of theorems are included among the solved problems. The supplementary problems serve as a complete review of the material of each chapter. Topics covered include the basic properties of topological, metric and normed spaces, the separation axioms, compactness, the product topology, and connectedness. Theorems proven include Urysohn's lemma and metrization theorem, Tychonoff's product theorem and Baire's category theorem. The last chapter, on function spaces, investigates the topologies of pointwise, uniform and compact convergence. In addition, the first three chapters present the required concepts of set theory, the fourth chapter treats of the topology of the line and plane, and the appendix gives the basic principles of the real numbers. More material is included here than can be covered in most first courses. This has been done to make the book more flexible, to provide a more useful book of reference, and to stimulate further interest in the subject. I wish to thank many of my friends and colleagues, especially Dr. Joan Landman, for invaluable suggestions and critical review of the manuscript. I also wish to express my gratitude to the staff of the Schaum Publishing Company, particularly to Jeffrey Albert and Alan Hopenwasser, for their helpful cooperation. SEYMOUR LIPSCHUTZ Temple University May, 1965

CONTENTS Page Chapter 1

SETS AND RELATIONS Sets. Subsets. Set operations. Product sets. Composition of relations.

Chapter

2

1 Relations. Equivalence relations.

FUNCTIONS

17

Functions. Indexed sets. Cartesian products. Generalized operations. Associated set functions. Algebra of real-valued functions.

Chapter

3

CARDINALITY, ORDER

32

Equivalent sets. Denumerable and countable sets. Continuum. SchroederBernstein theorem. Cardinality. Partially ordered sets. Subsets of ordered sets. First and last elements. Maximal and minimal elements. Upper and lower bounds. Zorn's lemma.

Chapter

4

TOPOLOGY OF THE LINE AND PLANE

47

Real line. Open sets. Accumulation points. Bolzano-Weierstrass theorem. Closed sets. Heine-Borel theorem. Sequences. Convergent sequences. Subsequences. Cauchy sequences. Completeness. Continuous functions. Topology of the plane.

Chapter

5

TOPOLOGICAL SPACES: DEFINITIONS

66

Topological spaces. Accumulation points. Closed sets. Closure of a set. Interior, exterior, boundary. Neighborhoods and neighborhood systems. Convergent sequences. Coarser and finer topologies. Subspaces, relative topologies. Equivalent definitions of topologies.

Chapter

6

BASES AND SUBBASES

87

Base for a topology. Subbases. Topologies generated by classes of sets. Local bases.

Chapter

7

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

97

Continuous functions. Continuous functions and arbitrary closeness. Continuity at a point. Sequential continuity at a point. Open and closed functions. Homeomorphic spaces. Topological properties. Topologies induced by functions.

Chapter

8

METRIC AND NORMED SPACES Metrics. Distance between sets, diameters. Open spheres. Metric topologies, metric spaces. Properties of metric topologies. Equivalent metrics. Metrization problem. Isometric metric spaces. Euclidean m-space. Hilbert space. Convergence and continuity in metric spaces. Normed spaces.

111

CONTENTS

Page Chapter

9

COUNTABILITY

131

First countable spaces. Second countable spaces. Linde leirs theorems. Separable spaces. Hereditary properties.

Chapter

10

SEPARATION AXIOMS

139

T 1 -spaces. Hausdorff spaces. Regular spaces. Normal spaces. Urysohn's lemma and metrization theorem. Functions that separate points. Completely regular spaces.

Chapter

II

COMPACTNESS

151

Covers. Compact sets. Subsets of compact spaces. Finite intersection property. Compactness and Hausdorff spaces. Sequentially compact sets. Countably compact sets. Locally compact spaces. Compactification. Compactness in metric spaces. Totally bounded sets. Lebesgue numbers for covers.

Chapter

12

PRODUCT SPACES

167

Product topology. Base for a finite product topology. Defining subbase and defining base for the product topology. Tychonoff product theorem. Metric product spaces. Cantor set.

Chapter

13

CONNECTEDNESS

180

Separated sets. Connected sets. Connected spaces. Connectedness on the real line. Components. Locally connected spaces. Paths. Arcwise connected sets. Homotopic paths. Simply connected spaces.

Chapter

14

COMPLETE METRIC SPACES

195

Cauchy sequences. Complete metric spaces. Principle of nested closed sets. Baire's category Completeness and contracting mappings. Completions. theorem. Completeness and compactness.

Chapter

15

FUNCTION SPACES

207

Uniform Point open topology. Pointwise convergence. Function spaces. convergence. Function space C[0,1]. Uniform boundedness. Equicontinuity. Ascoli's theorem. Compact open topology. Topology of compact convergence. Functionals on normed spaces.

Appendix

PROPERTIES OF THE REAL NUMBERS

225

Subsets of R. Positive numbers. Order. Absolute value. Least upper bound axiom. Nested interval property.

Field axioms.

Real line.

INDEX INDEX OF SYMBOLS

235 239

Chapter

1

Sets and Relations SETS, ELEMENTS The concept set appears in all branches of mathematics. Intuitively, a set is any welldefined list or collection of objects, and will be denoted by capital letters A,B,X,Y, The objects comprising the set are called its elements or members and will be denoted by lower case letters a, b, x, y, . . . . The statement "p is an element of A" or, equivalently, "p belongs to A" is written p E A The negation of p E A is written p a A. There are essentially two ways to specify a particular set. One way, if it is possible, is by actually listing its members. For example, A =

la, e, o, ul

denotes the set A whose elements are the letters a, e, j, o and u. Note that the elements are separated by commas and enclosed in braces { }. The other way is by stating those properties which characterize the elements in the set, For example,

B = {x: x is an integer, x > 0} which reads "B is the set of x such that x is an integer and x is greater than zero," denotes the set B whose elements are the positive integers. A letter, usually x, is used to denote an arbitrary member of the set; the colon is read as 'such that' and the comma as 'and'.

a B,

Example 1.1:

The set B above can also be written as B = {1,2,3, ...}. and 77- ig B.

Example 1.2:

Intervals on the real line, defined below, appear very often in mathematics. Here a and b are real numbers with a < b.

Note that —6

Open interval from a to b

= (a, b) = {x : a < x < b}

Closed interval from a to b

= [a, bl =

3G B

b1

Open-closed interval from a to b = (a, b] = {x:a<x-b} Closed-open interval from a to b

= [a, b) = {x : a -- x < b}

The open-closed and closed-open intervals are also called half-open intervals.

Two sets A and B are equal, written A = B, if they consist of the same elements, i.e. if each member of A belongs to B and each member of B belongs to A. The negation of A =B is written A B. Example 1.3:

Let E -= {x : x 2 — 3x + 2 = 0), F = {2,1} and G = {1, 2, 2, 1.}. Then E = F =G. Observe that a set does not depend on the way in which its elements are displayed. A set remains the same if its elements are repeated or rearranged.

Sets can be finite or infinite. A set is finite if it consists of n different elements, where n is some positive integer; otherwise a set is infinite. In particular, a set which consists of exactly one element is called a singleton set.

1

2

SETS AND RELATIONS

[CHAP. 1

SUBSETS, SUPERSETS A set A is a subset of a set B or, equivalently, B is a superset of A, written A c B or B D A

iff each element in A also belongs to B; that is, if x E A implies x E B. We also say that A is contained in B or B contains A. The negation of A cB is written AltB or Bfi A and states that there is an x E A such that x B. Example 2.1:

Consider the sets

A

= {1, 3, 5, 7,

B = {5, 10, 15, 20, . .

C = 1x : x is prime, x > 21 = {3, 5,7,11, .. .} Then C c A since every prime number greater than 2 is odd. On the other hand, B (;1 A since 10 G B but 10 A. Example 2.2:

We will let N denote the set of positive integers, Z denote the set of integers, Q denote the set of rational numbers and R denote the set of real numbers. Accordingly,

NcZcQcR

Observe that A cB does not exclude the possibility that A =B. to restate the definition of equality of sets as follows:

Definition:

I

In

fact, we are able

Two sets A and B are equal if and only if A cB and B c A.

In the case that A c B but A B , we say that A is a proper subset of B or B contains A properly. The reader should be warned that some authors use the symbol C for a subset and the symbol c only for a proper subset. Our first theorem follows from the preceding definitions.

Theorem 1.1:

Let A, B and C be any sets. Then (i) A c A; (ii) if A cB and B cA then A = B; and (iii) if A cB and B cC then A C C.

UNIVERSAL AND NULL SETS In any application of the theory of sets, all sets under investigation are subsets of a fixed set. We call this set the universal set or universe of discourse and denote it in this chapter by U. It is also convenient to introduce the concept of the empty or null set, that is, a set which contains no elements. This set, denoted by 9), is considered finite and a subset of every other set. Thus, for any set A, 9)c A c U. Example 3.1:

In plane geometry, the universal set consists of all the points in the plane.

Example 3.2:

Let A

Example 3.3:

Let B = (0).

= (x : x2 = 4, x is odd). Then A is empty, i.e. A = Then B 0 (6 for B contains one element.

CLASSES, COLLECTIONS, FAMILIES AND SPACES Frequently, the members of a set are sets themselves. For of lines is a set of points. To help clarify these situations, "collection" and "family" synonymously with set. Usually we and collection or family for a set of classes. The words subfamily have meanings analogous to subset. Example 4.1:

example, each line in a set we use the words "class", use class for a set of sets, subclass, subcollection and

The members of the class C2, 3), {2 } , 15,61} are the sets 12,31, {2 } and { 5,6).

CHAP. 1]

Example 4.2:

SETS AND RELATIONS

3

Consider any set A. The power set of A, denoted by ‘NA) or 2A, is the class of all subsets of A. In particular, if A = {a, b, c},_then

p(A) =

{A, {a, b } , {a, e}, {b, c}, {a}, {b}, {c}, 01

In general, if A is finite, say A has n elements, then P(A) will have 24 elements.

The word space shall mean a non-empty set which possesses some type of mathematical structure, e.g. vector space, metric space or topological space. In such a situation, we will call the elements in a space points.

SET OPERATIONS The union of two sets A and B, denoted by A uB, is the set of all elements which belong to A or B, i.e., AUB = {x:xEA or x E B} Here "or" is used in the sense of "and/or". The intersection of two sets A and B, denoted by A belong to both A and B, i.e., AnB

=

(x:xGA

nB, is the set of elements which

and x E B)

If A n B = Q, that is, if A and B do not have any elements in common, then A and B are said to be disjoint or non-intersecting. A class exi of sets is called a disjoint class of sets if each pair of distinct sets in cA is disjoint. The relative complement of a set B with respect to a set A or, simply the difference of A and B, denoted by A \B, is the set of elements which belong to A but which do not belong to B. In other words, A\B = fx: xEA, x(ZBI Observe that A \B and B are disjoint, i.e. (A \B)

n B = Q.

The absolute complement or, simply, complement of a set A, denoted by Ac, is the set of elements which do not belong to A, i.e., Ac

= {X :

X

E U,

X

In other words, Ac is the difference of the universal set U and A. Example 5.1:

The following diagrams, called Venn diagrams, illustrate the above set operations. Here sets are represented by simple plane areas and U, the universal set, by the area in the entire rectangle.

AuB is shaded

A\ B is shaded

Ac is shaded

4

SETS AND RELATIONS

[CHAP. 1

Sets under the above operations satisfy various laws or identities which are listed in the table below (Table 1). In fact, we state

Sets satisfy the laws in Table 1.

Theorem 1.2:

LAWS OF THE ALGEBRA OF SETS Idempotent Laws la.

A uA = A

lb.

AnA = A

Associative Laws 2a.

(AuB)uC = Au(BuC)

3a.

AuB = BuA

2b.

(AnB)nC = An(BnC)

Commutative Laws 3b.

AnS = BnA

Distributive Laws

Au(BnC) = (AuB)n(AuC)

4a.

4b.

An(BuC) = (AnB)u(AnC)

Identity Laws 5a.

A u0 = A

5b.

6a.

AuU = U

6b.

AnU = A An0 = 0

Complement Laws 7a.

AuAc = U

7b.

AnAc = 0

8a.

(ile)c = A

8b.

Uc =

0, Oc = U

De Morgan's Laws a..

(AuB)c = AcnBc

9b.

(AnB)c = AcuBc

Table 1

Remark:

Each of the above laws follows from an analogous logical law.

AnB =

fx:xE.A

and x E B1 =

fx:xEB

For example,

and xeAl = BnA

Here we use the fact that the composite statement "p and q", written p A q, is logically equivalent to the composite statement "q and p", i.e. q A p.

The relationship between set inclusion and the above set operations follows. Theorem 1.3:

Each of the following conditions is equivalent to AcB:

(i) AnB = A

(iii) .BccAc

(ii) AuB = B

(iv) AnBe = Os

(y) Bu Ac = U

PRODUCT SETS Let A and B be two sets. The product set of A and B, written A ordered pairs (a, b) where a E A and b G B, i.e., A x B =

{(a, b.):

The product of a set with itself, say A x A,

X

B, consists of all

a e A, b E B)

will be denoted by A 2.

Example 6.1:

The reader is familiar with the Cartesian plane R2 = R X R (Fig. 1-1 below). Here each point P represents an ordered pair (a, b) of real numbers and vice versa.

Example 6.2:

Let A = {1, 2,3} and B = {a, b}.

A xB=

Then

{(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}

CHAP. 1]

5

SETS AND RELATIONS

a

2

3

A

Fig. 1-2

Fig. 1-1

Since A and B do not contain many elements, it is possible to represent A X B by a coordinate diagram as shown in Fig. 1-2 above. Here the vertical lines through the points of A and the horizontal lines through the points of B meet in 6 points which represent A X B in the obvious way. The point P is the ordered pair (2, b). In general, if a set A has s elements and a set B has t elements, then A x B has s times t elements.

Remark:

The notion "ordered pair" (a, b) is defined rigorously by (a, b) From this definition, the "order" property may be proven: (a, b = (c, (1)

Ha), ta, b».

implies a= c and b

The concept of product set can be extended to any finite number of sets in a natural way. The product set of the sets Al, ..., A., denoted by A,

X A2 X ' ' X

Am or

11-1-1L consists of all m-tuples (a1, a2, ..., a.) where a, e Ai for each i. RELATIONS A binary relation (or relation) R from a set A to a set B assigns to each pair (a, b) in A X B exactly one of the following statements: (i) "a is related to b", written a R b

(ii) "a is not related to b", written a $ b A relation from a set A to the same set A is called a relation in A. Example 7.1:

Set inclusion is a relation in any class of sets. For, given any pair of sets A and B, either A cB or A(03.

Observe that any relation R from a set A to a set B uniquely defines a subset R* of A x B as follows: R* {(a,b): a R b} On the other hand, any subset R* of A a Rb

X

B defines a relation R from A to B as follows: iff (a, b) E R*

In view of the correspondence between relations R from A to B and subsets of A redefine a relation by

X

B, we

Definition: I A relation R from A to B is a subset of A x B. The domain of a relation R from A to B is the set of first coordinates of the pairs in R and its range is the set of second coordinates, i.e., domain of R = { a: (a, b G R 1,

range of R

b : (a, b E R 1

The inverse of R, denoted by R -1, is the relation from B to A defined by

R-

(b, a: (a, b) E R )

6

[CHAP. /

SETS AND RELATIONS

Note that R -1 can be obtained by reversing the pairs in R. Example 7.2:

Consider the relation

R = { (1, 2), (1, 3), (2, 3) } in

A = (1, 2, 3).

Then the domain of R = ( 1,2 ) , the range of R = {2,3), and

R -1 = { (2, 1), (3, 1), (3, 2) ) Observe that R and R -1 are identical, respectively, to the relations < and > in A, i.e., (a, b) E R -1 if a> b (a, b) E R if a < b and

The identity relation in any set A, denoted by A or AA, is the set of all pairs in A X A with equal coordinates, i.e., AA = { (a, a): aeA} The identity relation is also called the diagonal by virtue of its position in a coordinate diagram of A X A.

EQUIVALENCE RELATIONS A relation R in a set A, i.e. a subset R of A X A, is termed an equivalence relation . iff it satisfies the following axioms:

[Ed

For every a E A, (a, a) E R.

[E2]

If (a, b) E R, then (b, a) e R.

[E3]

If (a, b.) e R and (b, 0 e R, then (a, 0 G R.

In general, a relation is said to be reflexive if it satisfies [E1], symmetric if it satisfies [E21 and transitive if it satisfies [E3]. Accordingly, a relation R is an equivalence relation if it is reflexive, symmetric and transitive. Example 8.1:

Consider the relation c, i.e. set inclusion. Recall, by Theorem 1.1, that A CA for every set A, and if A cB and B c C then ACC Hence c is both reflexive and transitive. On the other hand,

A cB and A

B

implies BçLA

Accordingly, c is not symmetric and hence is not an equivalence relation. Example 8.2:

In Euclidian geometry, similarity of triangles is an equivalence relation. For if and y are any triangles then: (1) a is similar to itself; (ii) if a is similar to 13, then 13 is similar to a; and (iii) if a is similar to [I and 13 is similar to y then a is similar to y. a, ft

If R is an equivalence relation in A, then the equivalence class of any element a G A, denoted by [a], is the set of elements to which a is related: [a]

= tx : (a, x) E R}

The collection of equivalence classes of A, denoted by AIR, is called the quotient of A by R: AIR = { [a] : a e AI The quotient set All? possesses the following properties:

Theorem 1.4:

Let R be an equivalence relation in A and let [a] be the equivalence class of a E A. Then: (i) For every a E A, a E [a]. (ii) [a] = [b] if and only if (a, b.) e R. (iii) If [a]

[b], then [a] and [b] are disjoint.

7

SETS AND RELATIONS

CHAP. 1]

A class eA of non-empty subsets of A is called a partition of A if (1) each a GA belongs to some member of eA and (2) the members of e4 are pair-wise disjoint. Accordingly, the previous theorem implies the following fundamental theorem of equivalence relations:

Theorem 1.5:

Let R be an equivalence relation in A. partition of A.

Example 8.3:

Then the quotient set AIR is a

Let R5 be the relation in Z, the set of integers, defined by

x

(mod 5)

y

which reads "x is congruent to y modulo 5" and which means "x - y is divisible by 5". Then R5 is an equivalence relation in Z. There are exactly five distinct equivalence classes in Z/R5 :

E0

..., -10, -5, 0, 5, 10, ...) =

[-10] = [-5] = [0] = [5]

{..., -9, -4, 1, 6, 11, ...}

=

= [-9]

, -8, -3, 2, 7, 12, ...}

=

E3 = {

, -7, -2,3,8,13, ... }

=•

[ - 8] [-7]

= [ - 3] = [2] = [7] = [-2] = [3] = [8]

E4 = {

, -6, -1,4,9, 14, ...}

= •

= [ -6]

= [ -1 ] = [ 4 ] = [9]

E2

.

= [-4] = [1] = [6] = .-

Observe that each integer x, which is uniquely expressible in the form x 5q r where 0 r < 5, is a member of the equivalence class Er where r is the remainder. Note that the equivalence classes are pairwise disjoint and that

Z = E0 uE 1 uE2 uE3 uE4-

COMPOSITION OF RELATIONS Let U be a relation from A to B and let V be a relation from B to C, i.e. UcAxB and V c B x C. Then the relation from A to C which consists of all ordered pairs (a, c) e A xC such that, for some b GB, (a, b) E U

and (b,

E V

is called the composition of U and V and is denoted by Vo U. warned that some authors denote this relation by Uo V.)

(The reader should be

It is convenient to introduce some more symbols: 3, there exists We may then write: VoU Example 9.1:

s.t., such that

{ (x, : x Let

G A, y

e C;

V, for all

3b E B s.t.

implies U, (b, E V }

A = {1, 2, 3, 4 } , B = {x, y, z, tv} and C = {5, 6, 7, 8 } , and let

U = { (1, x), (1, y), (2, x), (3, iv), (4, iv) }

and

V = { (y, 5), (y, 6), (z, 8), (w, 7)

That is, U is a relation from A to B and V is a relation from B to C. illustrate U and V as follows:

A 1

Accordingly,

y E B and (1, y) E U,

(y, 5) E V

(1,5) E VoU

since

(1,6) E V 01/

(y, 6) E V (3, tv) E U, (w, 7) E V E B and since w since w E B and (4, tv) E U, (tv, 7) E V

(3, 7) E Vo U (4, 7) E Vo U

since

y E B and (1, y) E U,

We may

8

[CHAP. 1

SETS AND RELATIONS

No other ordered pairs belong to V 0 U, that is, Vo U =-- { (1,5), (1, 6), (3, 7), (4, 7) } Observe that V 0 U consists precisely of those pairs (x, y) for which there exists, in the above diagram, a "path" from x E A to y E C composed of two arrows, one following the other. Example 9.2:

Let U and V be the relations in R defined by

U = { (x, y) : x2 + y2 = 1}

V = {(y, z) : 2y 3z = 4}

and

Then the relation V 0 U, the composition of U and V, can be found by eliminating y from the two equations x2 + y2 = 1 and 2y + 3z = 4. In other words, Vo U = f(x, z) : 4x2 9z2 — 24z + 12 = 01 Example 9.3:

Let N denote the set of positive integers, and let R denote the relation < in N, i.e. (a, b) E R if a < b. Hence (a, b) E R -1 iff a> b. Then

R 0 R -1 = { (x, y) : x,y E N; 3 b E N s.t. (x, b) E R -1 , (b, y) E R = {(x, y) : x,y E N; 3b E N s.t. b < x, b < y } (N\{1}) X (N\{1}) =

{(x,y): x,y E N; x,y 1)

and

R -1 OR = { (x, y)

x,y E N; 3b E N s.t. (x,

{ (x, y) : x,y E N;

E R, (b,y) E R -1 }

3b E N s.t. b > x, b> y}

NXN Note that R 0 R -1 R -1 o R.

Solved Problems SETS, ELEMENTS, SUBSETS 1. Let A = {x: 3x = 61. Does A= 2? Solution: A is the set which consists of the single element 2, i.e. A = {2}. The number 2 belongs to A; it does not equal A. There is a basic difference between an element p and the singleton set { p).

2.

Determine which of the following sets are equal: 0, {01, {Q}. Solution: Each is different from the other. The set {0} contains one element, the number zero. The set 0 contains no elements; it is the null set. The set {0} also contains one element, the null set.

3. Determine whether or not each of the following sets is the null set: (i) X = {x: x 2 =9, 2x= 41, (ii) Y = fx : xx1, (iii) Z= {x: x +8 = 81. Solution:

(1)

There is no number which satisfies both x2 = 9 and 2x = 4; hence X =

(11)

We assume that any object is itself, so Y is empty.

Q.

In fact, some texts define the null set by

—= {x : x x } . (iii)

The number zero satisfies x -I- 8 = 8; hence Z = {0}.

4. Prove that A

{2,3,4,5 } is not a subset of B

Accordingly, Z

Q.

fx:x is even).

Solution: It is necessary to show that at least one member of A does not belong to B. 3 B, A is not a subset of B.

Since 3 E A and

CHAP. 1]

5.

9

SETS AND RELATIONS

Prove Theorem 1.1 (iii):

If AcB and BcC then ACC.

Solution: We must show that each element in A also belongs to C. Let x E A. Now A cB implies x E B. But BcC, so x E C. We have therefore shown that x E A implies x E C, or ACC.

6.

Prove: If A is a subset of the null set Q, then A= Q. Solution: The null set 0 is a subset of every set: in particular, (/) cA. by Definition 1.1, A=0.

7.

Find the power set P(S) of the set S

But, by hypothesis, A 46; hence,

{1,2,31.

Solution: Recall that the power set 41)(S) of S is the class of all subsets of S. {1, 2, 3 } , {1, 2), {1, 3 } , {2, 3), 111, {2 } , {3} and the empty set V). Hence

P(S)

The subsets of S are

IS, {1,31, {2, 3), {1, 2), {1}, {2 } , {3},

=

Note that there are 23 =--- 8 subsets of S.

8.

Find the power set P(S) of S = (3, (1,41). Solution: Note first that S contains two elements, 3 and the set (1, 4). Therefore `1)(S) contains 22 = 4 elements: S itself, the empty set (6, the singleton set {3} containing 3 and the singleton set {{1, 4)) containing the set {1, 4 } . In other words,

=

'KS)

KS, 01, (0,411, 0)

SET OPERATIONS

9.

Let U = {1, 2, . ., 8, 91, A = {1, 2, 3, 4 } , B = {2, 4, 6, 8 } and C = {3, 4, 5, 6 } . Find: (i) Ac, (ii) (A n C )c , (iii) B \C , (iv) (Au By . Solution: Ac consists of the elements in U that are not in A; hence Ac = {5, 6, 7, 8, 9). (I)

(ii)

A

nC

consists of the elements in both A and C; hence

nc =

A

{3,4}

and

(A

n c)c =

{1, 2, 5, 6, 7, 8, 9 }

(iii)

B\C consists of the elements in B which are not in C; hence B\C = {2, 8).

(iv)

A UB consists of the elements in A or B (or both); hence AUB = {1, 2, 3, 4, 6, 8 }

10. Prove-.

and

(AuB) =

9}

(A \B) n B =

Solution:

{x: x E B, x E A\B}

(A \B)nB

{x: x E B, x E A, x B} since there is no element x satisfying x E B and x

11. Prove De Morgan's Law:

12. Prove: Solution:

(A u B)c = Ac n Bc. {x e Au B}

(AuB)c

Solution:

B.

=

{x : x

=

{x: x E Ac, x E Bc}

A, x

B} AenBc

B \ A = B n Ac. B \A

{x x E B, x A} = {x x

E B, x E Ac) =

BnAc

10

[CHAP. 1

SETS AND RELATIONS

13. Prove the Distributive Law: A n (B UC) = (An B) U (A n C). : x E A; xEBuC} x E A; x E B or x E C} {x x E A, xE B; or x E A, x E C}

An(BuC)

Solution:

{x xEAnB

or xEAnC}

(A nB) u (AnC) Observe that in the third step above we used the analogous logical law p

A

(q v r) = (p

q)v (p

A

r)

where A reads "and" and y reads "or".

14. Prove: For any sets A and B, A nBcA cA uB. Solution: Let x E AnB; then xEA and xEB. In particular, xEA. Accordingly, AnBcA. Hence A cA u B. In other words, AnBcA c A u B. then x E A or x E B, i.e. x E AuB.

If xEA,

15. Prove Theorem 1.3 (i): A cB if and only if A nB = A. Solution: Suppose A cB. Let x E A; then by hypothesis, x E B. Hence x E A and x E B, i.e. Accordingly, A cA nB. But by the previous problem, A nB cA. Hence AnB = A. On the other hand, suppose A nB = A. Then in particular, A cA n B. problem, A nB cB. Hence, by Theorem 1.1, A c B.

x E A nB.

But, by the previous

PRODUCT SETS, RELATIONS, COMPOSITION OF RELATIONS 16. Let A= {a, b}, B= (2,31 and C = {3,4 }. Find: (1) A x (B U C), (ii) (A x B)U (A x Solution: First compute BUG = {2, 3, 4 } . (i)

A x (B u C) (ii)

First find A x B

C).

Then

=

(a, 2), (a, 3), (a, 4), (b, 2), (b, 3), (b, 4)1

and A x C:

A x B = { (a, 2), (a, 3), (b, 2), (b, 3) 1,

A X C = { (a, 3), (a, 4), (b, 3), (b, 4) 1

Then compute the union of the two sets:

{ (a, 2), (a, 3), (b, 2), (b, 3), (a, 4), (b, 4) 1

(A X B)u (A X C)

Observe, from (i) and (ii), that A x (B u C) = (A X B) u (A X C).

17. Prove: Solution:

A x (B nC) = (A X B) n (A x C). A x (B n C) =

{(x,y): x EA, yEBnC}

•

{(x, y) : xEA,yEB,yEC1

•

{ (x,

•

: (x, y) E Ax B, (x, y) EA xC} (A x B) n (A x C)

18. Let R be the relation < from A = {1,2,3,4} to B = {1, 3, 5}, i.e., (a, e R iff a < b. (i)

Write R as a set of ordered pairs.

(ii)

Plot R on a coordinate diagram of A x B.

(iii) Find domain of R, range of R and R -1 . (iv) Find R 0R -1 .

5

•

•

3

•

•

•

•

A 2

4

CHAP. 1]

11

SETS AND RELATIONS

Solution:

R consists of those ordered pairs (a, b) E A XB such that a < b; hence R (1,5), (2, 3), (2, 5), (3, 5), (4, 5) 1 (ii)

R is displayed on the coordinate diagram of A x B as shown above.

(iii)

The domain of R is the set of first coordinates of the pairs in R; hence domain of R = {1, 2, 3, 41. The range of R is the set of second coordinates of the pairs in R; hence range of R = 0,51. R ' can be obtained from R by reversing the pairs in R; hence

R -1 (iv)

=

{ (3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5,4) }

To find R0R -1 , construct diagrams of R -1 and R as shown below. second factor in the product R 0R 1 , is constructed first. Then

R R -1

=

Observe that R -1 , the

{ (3, 3), ( 3, 5 ), (5 , 3), (5, 5 ) }

19. Let T be the relation in the set of real numbers R defined by xT y if both x rn,n +11 and y G [n, n +11 for some integer n Graph the relation T. Solution: T consists of the shaded squares below. 3 2-

-

3

-2 -

20. Let T be the relation in the set of real numbers R defined by xTy if 0 (i) Express T and T -1 as subsets of R x R and graph.' (ii) Show that T 0T -1 = {(x, z) Solution: 1} (i) T = {(x,y): x,y E R, 0 x - y T-1 = ( (x, y) : (y, E T } = ((x, y) : x,y E R, 0

y - x 1}

The relations T and T -1 are graphed below, -2

I

I

4,4.

1

2

--2

Graph of T

Graph of T - 1

x-y

1.

SETS AND' RELATIONS

12 (ii)

[CHAP. 1

By definition of composition of relations,

T T -1 = { (x, z) : 3y E R s.t. (x, y) E T -1 , ( y, z) E T (x, z) : 3y E R s.t. (y, x), (y, z) E {(x,z): 3yER s.t. 0=y - x=1, 0=y-z=1} Let

S = { (x, z) : lx - = 1).

Let (x, z) belong to T o T -1 .

We want to show that T T -1

Then 3y s.t. 0 y- x, y - z 1.

0 =yœx, y-z=1

Also,

•

y-z=1 y-z

•

x- z=1

y - x, y - z = 1

0

= S.

y-x=1 •

• 0 = y - x, y - z = 1 Accordingly, (x, z) E S, i.e. To T -1 C S.

y-x=1-1-y- z -1 -= x - z

-1 = x - z = 1

In other words,

But

if x - zi -‹ 1

Now let (x, z) belong to S; then Ix - zi -= 1. Let y = max (x, z); then

0=y-x ,<1 and 0 y - z 1.

Thus (x, z) also belongs to To T -1 , i.e. Sc To T -1 . Hence TOT - ' =- S.

21. Prove: For any two relations R cXx Y and ScYx Z, (S R) --1 = R -1 0S -- '. Solution:

=

(SoR) -1

{(z,x): (x,z)E SoR) (z, x): 3y E Y s.t. (x, y) E R, (y, z) E S)

=

(z,

: 3y E Y s.t. (z, y) E S -1 , (y, x) E R -1 } = R-1 0 S -1

22. Prove: For any three relations RcWxX, To (S o R). Solution:

(T 0 S) OR

ScXxY and Tc YxZ,

(ToS)0.1? =

s.t. (w, x) E R, (x, z)E TOS} 3x E X, 3y E Y s.t. x) E R, (x, y) E S, (y, z) E T } : ay E Y s.t. (w, y) E S o R, (y, E T1

{ (w, : 3x E X

•

{ (tv,

•

(w, To(SoR)

REFLEXIVE, SYMMETRIC, TRANSITIVE AND EQUIVALENCE RELATIONS 23. Prove:

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

R R R R (v.) R (vi) R

Let R be a relation in A, i.e. RcAx A. Then: is reflexive if AA C R; is symmetric if R -= R -1 ; is transitive iff R oR c R; reflexive implies R oR D R and R 0 R is reflexive; symmetric implies R 0 R -1 = R -1 R; transitive implies R 0 R is transitive.

Solution: (i) Recall that the diagonal (a, a) E R if AA C R.

AA

= {(a, :

a E A}.

Now R is reflexive if, for every a E A,

(ii)

Follows directly from the definition of R -1 and symmetric.

(iii)

Let (a, e) E R 0 R; then 3 b E A such that (a, b) E R and (b, e) E R. (a, b), (b, E R implies (a, OE R. Consequently, RoR c R. On the other hand, suppose other words, R is transitive.

RoRcR.

If (a, b),(b,c)ER,

then

But, by transitivity, (a, e) ER0R c R.

In

CHAP. 11

(iv)

13

SETS AND RELATIONS

Let (a, b) E R.

R 0R

Now,

= { (a, c) : 3b E A

s.t. (a, b) E R, (b, e) E R). (a, b)ER0R,

But (a, b)ER and, since R is reflexive, (b,b)ER. Thus Furthermore, A A CR c R oR implies R o R is also reflexive.

(y)

R R -1

{ (a, c) : 3b E A

s.t. (a, b) E R -1 , (b,

{ (a, : 3b E A

s.t. (a, b) E R, (b, e) E R -1 }

i.e.

R cRoR.

E R)

= R -1 0 R (vi)

Let (a,b), (b,c) E R 0 R. transitive.

By (iii), R oR c R; hence

So (a, c) E R 0 R, i.e. R 0 R is

(a,b), (b,c) E R.

24. Consider the relation R = (1,1), (2,3), (3,2)} in X = {1,2,31. Determine whether or not R is (1) reflexive, (ii) symmetric, (iii) transitive. Solution:

(i)

R is not reflexive since

2 E X but (2, 2) E R.

(ii)

R is symmetric since R -1 = R.

(iii)

R is not transitive since (3, 2) E R and (2, 3) E R but (3, 3) E R.

25. Consider the set N x N, i.e. the set of ordered pairs of positive integers. Let R be the relation = in N x N which is defined by (a, b) (c, d) if ad = be Prove that R is an equivalence relation. Solution: Note that, for every Suppose (a, h)

(c, d).

(a, b) E N X N, Then ad

(a, b)

(a, b) since ab = ba;

be, which implies cb

hence R is reflexive.

da. Hence (c, d)

(a, b)

and, therefore

R is symmetric. Now suppose (a, h)

(c, d) and

(c, d)_^,z (e, f).

Then ad = be and cf--=--- de.

Thus

(ad)(cf) = (bc)(de) and, by cancelling from both sides, af =- be. Accordingly, (a, h)

(e, f)

and R is transitive.

Since R is reflexive, symmetric and transitive, R is an equivalence relation. a then the above relation R is, Observe that if the ordered pair (a, b) is written as a fraction

a

usual definition of the equality of two fractions, i.e. -b- =

c

if infact,he ad = be.

26. Prove Theorem 1.4: Let R be an equivalence relation in A and let [a] be the equivalence class of a e A. Then: (1) For every a E A, a E raj. (ii) [a] = [b] if and only if (a, b) E R. (iii) If [a] [b], then [a] and [b] are disjoint. Solution: ,

Proof of (i).

Since R is reflexive, (a, a) E R for every a E A

and therefore

a E [a].

[b]. Let x E [b]; then Proof of (ii). Suppose (a, b) E R. We want to show that [a] (b, x) E R. But by hypothesis, (a, b) E R; hence by transitivity, (a, x) E R. Accordingly, x E [a], i.e. [b] c [a]. To prove that [a] c [b], we observe that (a, b) E R implies, by symmetry, that (b, a) E R. Then by a similar argument, we get [a] c [b]. So [a] = [b]. On the other hand, if [a]

[b], then by reflexivity, b E [h] = [a], i.e. (a, b) E R.

Proof of (iii). We prove the equivalent contrapositive statement, i.e. if [a] n [b] 0, then [a] = [b]. If Fain [b] 0, there exists an element x E A with x E [a] n [b]. Hence (a, x) E By symmetry, (x, b) E R and, by transitivity, (a, h) E R. Consequently by (ii), and (b, x) E R. [a] = [b].

14

[CHAP. 1

SETS AND RELATIONS

Supplementary Problems SETS, ELEMENTS, SUBSETS

27. Determine which of the following sets is the empty set: (i) {x:1<x< 2, xER} (iii) {x : x E (ii) { x : 1 < x < 2, x E N } (iv) {x : x 2 < x, x E 28. Let A = {1, 2, ..., 8, 9 } , B = {2, 4, 6, 81, C = {1, 3, 5, 7, 9 } , D = {3, 4, 5} and E = {3, 5 } . Which of these sets can equal X if we are given the following information? (iii) XcA and X4C, (i) X and B are disjoint, (ii) XcD and X03, (iv) X c C and 2CA. 29. State whether each of the following statements is true or false. (i) Every subset of a finite set is finite. (ii) Every subset of an infinite set is infinite. 30. Discuss all inclusions and membership relations among the following three sets: (6, {0}, {0, {0}}. 31. Prove that the closed interval [a, b] is not a subset of the open interval (a, b). 32. Find the power set "P(U) of U =---

21 and the power set P(V) of V = {0, { 1 , 2 } 1-

33. State whether each of the following is true or false. Here S is any non-empty set and 2s is the power set of S. (iii) {S} E 2s (iv) {S} c 2s (i) S E 2s (ii) S c 2s SET OPERATIONS Find: A uB, A nB, A \ B, B \A.

34. Let A = {1, 2, 3, {1, 2, 3 } ), B = {1,2, {1, 2 } ).

(i) A n (B u C), (ii) C\ (A nB).

35. In each of the Venn diagrams below shade:

(b)

(a) 36.

Prove and show by Venn diagrams: /lc \Be = B A.

37. (i) (ii)

Prove

A n (B\C) = (A n B)\ (A n C).

Give an example to show that A u (B \C)

2A UB.

38.

prove: 24c 213

39.

Each of the following conditions is equivalent to A cB: (i) A nB = A, (ii) Au B = B, (iii) Be cile, (iv) A nBe =Ø, (y) B uAc (Note. A nB = A was already proven equivalent to A cB in Problem 15.)

40.

2A n B ; 2A

(A u B)\ (A u C).

U2B C2AUB.

Give an example to show that 2A u 2B

Prove Theorem 1.3:

Prove that A cB iff (B n C)u A = B n (Cu A)

U

for any C.

PRODUCT SETS, RELATIONS, COMPOSITION OF RELATIONS

A x (BuC) = (A x B)u (A X C).

41.

Prove:

42.

Using the definition of ordered pair, i.e. (a, b) = {{a}, {a, b1), a= e and b= d.

prove that (a, b) = (e, d)

if

43. Determine the number of distinct relations from a set with m elements to a set with n elements, where m and n are positive integers.

CHAP. 11

15

SETS AND RELATIONS

44. Let R be the relation in the positive integers N defined by R = { (x,

(i) Write R as a set of ordered pairs. (iv) Find 1? -1 0 R.

: x,y E N, x

2y = 12 }

(ii) Find domain of R, range of R and R -1 . (iii) Find RoR.

1 (4, 5), (1, 4), (4, 6), (7, 6), (3, 7) 45. Consider the relation R (i) Find domain of R, range of R and R -1 . (ii) Find R 0 R.

46. Let U and V be the relations in R defined by (i) Find Vo U. (ii) Find U o V. 47. Consider the relations < and

in N.

(iii) Find R -1 0R.

U = {(x, y) : x 2 + 2y = 5)

in R. Show that < u A

and

V = {(x, y) : 2x — y = 3).

= where A is the diagonal.

EQUIVALENCE RELATIONS

48. State whether each of the following statements is true or false. Assume R and S are (non-empty) relations in a set A. (1)

If R is symmetric, then R -1 is symmetric.

(2)

If R is reflexive, then RnR -1

(3)

If R is symmetric, then RnR -1

(4)

If R and S are transitive, then RuS is transitive.

(5)

If R and S are transitive, then RnS is transitive.

(6)

If R and S are symmetric, then RuS is symmetric.

(7)

If R and S are symmetric, then RnS is symmetric.

(8)

If R and S are reflexive, then R nS is reflexive.

Ç.

49. Consider N X N, the set of ordered pairs of positive integers. Let (a, b)

(i) Prove = is an equivalence relation.

be the relation in N X N defined by

(c,d) if a + d = b c

(ii) Find the equivalence class of (2, 5), i.e. [(2,

50. Let — be the relation in R defined by x — y if x — y is an integer. Prove that — is an equivalence relation.

51. Let — be the relation in the Cartesian plane R2 defined by (x, y (w, z) if x = Prove that — is an equivalence relation and graph several equivalence classes.

W.

52. Let a and b be arbitrary real numbers. Furthermore, let — be the relation in R2 defined by (x, — (w z) if 3k E Z s.t. x — w = ka, y — z = kb ,

Prove that — is an equivalence relation and graph several equivalence classes.

Answers to Supplementary Problems 27. The sets in (ii) and (iii) are empty. 31.

a G [a, b]

but a 4 (a, b).

32. 'P(V) = (V, (0), {(1, 2)), 01 33.

(i) T, (ii) F, (iii) F, (iv) T

34. AuB = (1, 2, 3, {1, 2 } , { 1, 2, 3 } 1, AnB = { 1, 2 } , A\B = (3, {1, 2, 33, B\A = ((1, 2 ) ).

16

SETS AND RELATIONS

[CHAP. 1

35.

(i

)

37. (ii) C = 0, A= B

38.

Example:

A

= {1 } , B = ( 2)

43. 2inn

(i) (ii)

{ ( 1 0, 1), (8, 2), (6, 3), (4, 4), domain of R = {10, 8, 6, 4, 21,

R

R -'

(2,5) } range of

R = {1, 2, 3, 4, 5 } ,

{ (1, 10), (2, 8), (3, 6), (4, 4), (5, 2)

{ (8, 5), (4,4) } (iii) R oR (iv) R -1 0 R = (10, 10), (8, 8), (6„6), (4, 4), (2, 2) ) 45. (i)

domain of

R = {4, 1, 7, 3},

range of

R = {5, 4, 6, 7}, R -1 = { (5, 4), (4, 1), (6, 4), (6, 7), (7, 3) )

(ii) R 0 R = { (1, 5), (1, 6), (3, 6) 1 (iii) R -1 OR = { (4, 4), (1, 1), (4, 7), (7, 4), (7, 7), (3, 3) } 46. V 0 U = (x, y) : x2 + y = 2 1,

U o V = (x, y) : 4x 2 - 12x + 2y + 4 = 0 1

48.

(1) T, (2) T, (3) T, (4) F, (5) T, (6) T, (7) T, (8) T

49.

(ii)

[(2, 5)] =

(1, 4), (2, 5), (3, 6), (4, 7),

(n, n

3),

.}

51.

The equivalence classes are the vertical lines.

52.

•

•

•

• •

•

•

•

a

• •

•

The above gives a typical equivalence class. The distance between adjacent horizontal points is a and the distance between adjacent vertical points is b.

Chapter

2

Functions FUNCTIONS Suppose that to each element of a set A there is assigned a unique element of a set B; the collection, f, of such assignments is called a function (or mapping) from (or on) A into B and is written A -f--> B or f:A B The unique element in B assigned to a e A by f is denoted by f(a), and called the value of f at a or the image of a under f. The domain of f is A, the co-domain is B. To each function f: A -> B there corresponds the relation in A x B given by aE A

We call this set the graph of f. f[A] = {f(a): a E A}.

}

The range of f, denoted by f[A1, is the set of images, i.e.

Two functions f:A->B and g:A->B are defined to be equal, written f=g, iff f(a) = g(a) for every a E A, i.e. if they have the same graph. Accordingly, we do not distinguish between a function and its graph. A subset f of A x B, i.e. a relation from A to B, is a function if it possesses the following property:

[F]

Each a E A appears as the first coordinate in exactly one ordered pair (a, b) in f.

The negation of f g is written f Example 1.1:

g and is the statement:

3a e A for which f(a) g(a).

Let f :R It be the function which assigns to each real number its square, i.e. for each x E R, f(x) = x 2. Here f is a real-valued function. Its graph, { (x, x 2 ) : x E R}, is displayed in Fig. 2-1 below. The range of f is the set of non-negative real numbers, i.e. f[R] = {x : x E R, x 0}. -5 - 4 -3 - 2 —1

Fig. 2-1

Fig. 2-2

Example 1.2:

Then the diagram in Fig. 2-2 above Let A = {a, b, c, d} and B = fx, y, z, wl. Here f[A] = {x, y, w } . The graph of f is the B. function f from A into a defines relation { (a, y), (1), x), (c, y), (d, iv)}

Example 1.3:

13 is called a constant function if, for some 1,0 E B, f (a) = 1,0 A function f: A for all a E A. Hence the range f[A] of any constant function f is a singleton set, i.e. f[24] = {b0}.

17

18

[CHAP. 2

FUNCTIONS

Consider now functions f :A-›B and g:B->C, illustrated below: The function from A into C which maps the element a e A into the element g(f(a)) of C is called the composition or product of f and g and is denoted by g o f. Hence, by definition, (g o f)(a) = g (f (a))

We remark that, if we view f c A xB and gCBxC as relations, we have already defined a product g • f (Chapter 1). However, these two products are the same in that if f and g are functions then g • f is a function and g • f = got If f: X -> Y and A c X, then the restriction of f to A, denoted by f IA, is the function

from A into Y defined by

f A(a) f(a)

for all a E A

Equivalently, flA = f n (Ax Y). On the other hand, if f:X->Y is the restriction of some function g: X* -> Y where X c X*, then g is called an extension of f. ONE-ONE, ONTO, INVERSE AND IDENTITY FUNCTIONS A function f: A -> B is said to be one-to-one (or one-one, or 1-1) if distinct elements in A have distinct images, i.e. if a =- a' f (a) f (a') A function f: A -> B is said to be onto (or f is a function from A onto B, or f maps A onto B) if every b E B is the image of some a G A, i.e. if b eB

3a e A for which f(a)

b

Hence if f is onto, f[Al = B. In general, the inverse relation f -1 of a function f C A x/3 need not be a function. However, if f is both one-one and onto, then f - I is a function from B onto A and is called the inverse function. The diagonal AA c AxA is a function and called the identity function on A. It is also denoted by lA or 1. Here, 1A(a) = a for every a G A. Clearly, if f: A -> B, then

1 13 0 f = f = f lA

Furthermore, if f is one-one and onto, and so has an inverse function f1, then riot = 1 4 and fofl =11, The converse is also true:

Proposition 2.1:

Let f:A->B and g:B->A satisfy gof --= l A and

Then f' :B -› A exists and g Example 2.1:

fog =--

rt.

Let f :R —> R, g :R --> R and h:R --> It be defined by f(x) = ex , g(x) = x 3 — x

and

h(x) = x 2

The function f shown in Fig. 2-3(a) below is one-one; geometrically, this means that each horizontal line does not contain more than one point of f. The function g shown in Fig. 2-3(b) below is onto; geometrically this means that each horizontal line contains at least one point of g. The function h shown in Fig. 2-3(c) below is neither one-one nor onto, for h(2) = h( —2) = 4 and h[R1 is a proper subset of R, e.g. —16 h[R].

CHAP. 2]

19

FUNCTIONS

(a) f(x) -= ex

(e) h(x) = x 2 Fig. 2-3

INDEXED SETS, CARTESIAN PRODUCTS An indexed class of sets, denoted by

i E /), tA,),„ 1 or simply {A il assigns a set A, to each i E /, Le. is a function from / into a class of sets. The set I fs the index set, the sets Ai are called indexed sets, and each i e I is called an index. caled When the index set / is the set of positive integers, the indexed class {A1, A2, . . } is called a sequence (of sets). Example 3.1:

For each n E N, the positive integers, let .13„2 = { x : x EN, x is a multiple of n-} Then D 1 = {1, 2, 3, . } , D2 = {2, 4, 6, . } , D3 = {3, 6, 9, . . .1, ...

The Cartesian product of an indexed class of sets, eA = {A,: i G

n {A,:

i

I},

denoted by

e / } or rk A, or simply n A, ,

,

is the set of all functions p: I -> U1Ai such that p(i) = a, E A. We denote such an element of the Cartesian product by p = (a,: i G /). For each io E / there exists a function 77 20 into the ioth coordinate set called the ioth projection function, from the product set Aio defined by

nA,

i E /)) =

Example 3.2:

aio

Recall that R3 =RXRXR consists of all 3-tuples p = (a 1 , a2, a3) of real numbers. Now let R / , R2 and R3 denote copies of R. Then p can be viewed as a function on I = {1, 2, 31 where p(1) =-- al E R1 , p(2) = a2 E R2 and p(3) = a3 E R3. In other words, R3 = n fRi : i E I, Ri =

GENERALIZED OPERATIONS The notion of union and intersection, originally defined for two sets, may be generalized to any class e4 of subsets of a universal set U. The union of the sets in eA, denoted' by UfA : A e 0.,4), is the set of elements which belong to at least one set in 0.//: e/1- s.t. x G A) {X : X U, 321 U {A : A E ) The intersection of the sets in cA, denoted by ntA: A e e/i- }, is the set of elements which belong to every set in e/1:

n{A:A. E

cif}

=

{x:xEU, xeAforeveryAecA}

For an indexed class of subsets of U, say eA = {A c : i e /), we write

U{A,:

/}, U 11 A2 or

for the union of the sets in 0A, and

n{Ai: i e I),

n ie/ Ai or

UiAi

20

FUNCTIONS

for the intersection of the sets in e/i.

We will also write

U7-, Ai = A 1 UA2U • • •

[CHAP. 2

flr1A = A l UAzU • • • for the union and intersection, respectively, of a sequence ( A1, A2, . ) of subsets of U. Example 4.1:

{x: x E N, x is a multiple of n}

For each n E N, the positive integers, let D (see Example 3.1). Then

U {D i : i Example 4.2:

and

10} = {10,11,12, • • •

Let I = [0, 1 ] and, for each

and

i E I, let A i

Then

n i A i = {0}

u i A i = [0,1] and

The distributive laws and De Morgan's laws also hold for these generalized operations: Theorem 2.2:

For any class of sets c/I = {A,} and any set B,

(i) Theorem 2.3:

Bu(n i Ai) = n i (BuA,)

(ii)

Bn(uiAi) = u i (BnA,)

Let 0,4 -= {A i } be any class of subsets of U. Then:

(i) ( U Ai)' = n A

(ii)

(n, Ai)" =

The following theorem will be used frequently. Theorem 2.4: Remark:

Let A be any set and, for each p EA, let Gp be a subset of A such that p Gp c A. Then A= U{Gp : p A}.

In the case of an empty class 9) of subsets of a universal set U, it is convenient to define = 0 and MA: A E =U U{A: A E

0}

Hence

and

ÇZ

not:Jew = U

ASSOCIATED SET FUNCTIONS Let f: X -> Y. Then the image f[A] of any subset A of X is the set of images of points in A, and the inverse image ri [B] of any subset B of Y is the set of points in X whose images lie in B. That is, f[A] = {f(x): x Example 5.1:

e A}

and

f -1 [B] = {x : x E X, f(x) E

Let f : R --* It be defined by f(x) = x 2 .

f[{1, 3, 4, 7H Also,

Then

{1, 9, 16, 49 } , f[(1, 2) ] =-- (1, 4)

f - I [ { 4, 9} ] = { -3, —2,2, 3 } ,

f

[(1, 4) ] = (1,2) u (-2, —1)

Thus a function f:X-> Y induces a function, also denoted by f, from the power set P(X) of X into the power set '13 (Y) of Y, and a function f - t from T(Y) into P(X). The induced functions f and f --1 are called set functions since they are maps of classes (of sets) into classes. We remark that the associated set function f -1 is not in general the inverse of the associated set function f. For example, if f is the function in Example 5.1, then f -1 of { (1 ,2) ] = f

[(1, 4) ] = (1,2) U (-2, —1)

Observe that different brackets are used to distinguish between a function and its associated set functions, i.e. f (a) denotes a value of the original function, and f[A] and f [B] denote values of the associated set functions.

CHAP. 2]

21

FUNCTIONS

The associated set functions possess various properties.

Theorem 2.5:

In particular we state:

Let f: X -> Y. Then, for any subsets A and B of X, (i)

f[A U

(ii) f[A

f

(iii)

U f[B]

n B] c f[A] n f[B]

\

(iv)

D

f[A] \ r/31

A c B implies f [A] c f[B]

(.4,} of subsets of X, (ii') f[ n i c n i f[A,]

and, more generally, for any indexed class (i')

f[u i Ai] = U,f[A,]

The following example shows that the inclusions of replaced by equality. Example 5.2:

Consider the subsets A = [1,2] X [1,2 ]

B

and

(ii) and (iii) cannot in general be

[1,2] X [3,4]

of the plane R 2 and the projection 77: R2 R, into the first coordinate set, i.e. the x-axis. Observe that 77[11] [1,2] and 7[B] = [1,2], and that A n B implies 71-[A nB1 = Q. Hence 7T[A]

7[A n131 _-_-

n z[B] = [1,2]

A \B = A, so 77-[A \ B] = [1,2] Q = 77[A] \ 77[B]

Furthermore,

On the other hand, the inverse set function is much more "well-behaved" in the sense that equality holds in both cases. Namely, Theorem 2.6: Let f: X- Y. Then for any subsets A and B of Y, (i) (ii)

I [A U B] f

(iii) ri

[A

n B] \ 131

[A] U f - 1 [B] f -, [A] n f -1 [B] f- [Al f-

f-1 [A] c f - ' [B] and, more generally, for any indexed class fM of subsets of Y, (i') f [U 1 Ai] = u I f ÇAd (W) f --1 [n, = ntf - i Since f1 [Y]= X, we have, as a special case of (iii), = (f 1 [A])e. Corollary 2.7: Let f: X- Y and let A c Y. Then f 1 (iv)

A c B implies

Next follows an important relationship between the two set functions.

Theorem 2.8:

Let

f : X -> Y and let A cX and B

c Y. Then: (i) A c f -1 0 f [A] (ii) B D

As shown previously, the inclusion in

fof - I[B]

(i) cannot in general be replaced by equality.

ALGEBRA OF REAL-VALUED FUNCTIONS Let T(X,R) denote the collection of all - real-valued functions defined on some set X. Many operations are inherited by T(X, R) from corresponding operations in R. Specifically, let f:X->R and g:X->R and let k E R: then we define (f + g) : X -> R (k • f): X -> R (If) : X

by by

-> R by (f g): X -> R by

(f + g)(x) —= f(x) + g(x) (k • f)(x)

(1f1)(x) (f g)(x)

k(f (x)) (x)1 f (x) g(x)

22

[CHAP. 2

FUNCTIONS

It is also convenient to identify the real number k e R with the constant function f(x) = k for every x E R. Then (f+k):X-->it is the function f(x) + (f + le)(x) Observe that (fg) : X -> R is not the composition of f and g discussed previously. Example 6.1:

Consider the functions (a, 1), (b, 3) }

f= with domain X = {a, M.

and

g = { (a, 2), (b, —1) }

Then

(3f — 2g)(a) (3f —2g)(b)

3f(a) — 2g(a) = 3(1) — 2(2) = —1 3f(b) — 2g(b) - -= 3(3) — 2(-1) = 11 3f — 2g = { (a, —1), (b ,11)1

that is, Also, since

igl(x) Ig(x) 1 and (g + 3)(x) = {(a,2), (b, 1)1 and

g(x)+ 3, g+ 3 -= { (a, 5), (b, 2) }

The collection 7' (X,R) with the above operations possesses various properties of which some are included in the next theorem.

Theorem 2.9:

The collection .T(X, R) of all real-valued functions defined on a non-empty set X together with the above operations satisfies the following axioms of a real linear vector space:

[V1 ]

The operation of addition of functions f and g satisfies: (1) (2)

(3) (4)

(f + g) + h = f +(g+h) f+g:= g+f

30 E T(X, R), i.e. 0 : X -> R, such that f + 0 = f. R), 3 — f e F(X, R), i.e. —f : X --> R, such that For each f G f+( f) = 0. —

[V2]

The operation of scalar multiplication k • f of a function f by a real number k satisfies: (1) k • (k' • f) = (kk') • f (2)

[V3]

Example 6.2:

1. f =f

The operations of addition and scalar multiplication satisfy: (1) k•(f+g) k•f + kg (2) (k + lc') • f = k • f + k' • f Let X = {1, 2, . ,m } . ordered m-tuple (f(1),

Then each function f E T(X, R) Am)). Furthermore, if

f = (a1, ..., am) then

f

and

g = (b1 ,

may be written as an

bm)

g = (a 1 b1, a2 b2, ..., am bm)

k • f = (ka i , . . .,karn) and, for any k E R, In this case, the real linear (vector) space T(X, R) is called m-dimensional Euclidean space. Example 6.3:

A function f E T(X, R) is said to be bounded if aM E R such that If(x)1 M for every x E X Let p (X , R) denote the collection of all bounded functions in T(X,R). Then 13(X, R) possesses the following properties: (i) If f,g E p (X , R), then f + g E 13(X, R). (ii) If f E fl(X, R) and k ER, then k • f E fl(X, R). Any subset of T(X, R) satisfying (i) and (ii) is called a (linear) subspace of T(X, R).

CHAP. 2]

23

FUNCTIONS

Solved Problems FUNCTIONS 1. State whether or not each of the diagrams defines a function from A --= ( a, b, e) into B = {x, y, z }.

(ii)

(i) Solution:

2.

(i) (ii)

No. There is nothing assigned to the element b E A. No. Two elements, x and z, are assigned to c E A.

(iii)

Yes.

Let X = ( 1,2,3,4). State whether or not each of the following relations is a function from X into X.

(i) (ii) (iii)

f = t (2, 3), (1, 4), (2, 1), (3, 2), (4, 4) ) g = f (3, 1), (4, 2), (1, 1) } h =-- f (2, 1), (3, 4), (1, 4), (2, 1), (4, 4) )

Solution: Recall that a subset f of XXX is a function coordinate in exactly one ordered pair in f.

3.

f:X--2(

if each

xEX

appears as the first

(i)

No.

Two different ordered pairs (2, 3) and (2, 1) in f have the same first coordinate.

(ii)

No.

The element 2 E X does not appear as the first coordinate in any ordered pair in g.

(iii)

Yes. Although 2 E X appears as the first coordinate in two ordered pairs in h, these two ordered pairs are equal.

Consider the functions

f = ( (1,3), (2, 5), (3, 3), (4, 1), (5, 2) 1 g = f (1,4), (2, 1), (3, 1), (4, 2), (5, 3) )

from X = (1, 2, 3, 4, 5) into X. (i) Determine the range of f and of g. (ii) Find the composition functions g 0 f and fo g. Solution: Recall that the range of a function is the set of image values, i.e. the set of second coordinates. (i) Hence and range of g = {4, 1, 2, 3 } range of f = { 3, 5, 1, 2} (ii)

Use the definition of the composition function and compute: ( f 0 g) (1) -= (g 0 f)(1) = g(f (1)) = g(3) = 1

fig(1)) = f(4) =-- 1 -

(g 0 f)(2) = g(f (2)) = g(5) = 3

(f 0 g)(2) = f(g(2)) = f(1) = 3

(g 0 f)(3) = g(f(3)) = g(3) = 1

(f 0 g)(3)

= f (g (3)) = f(1) = 3

(g 0 f)(4)

g(f (4)) = g(1) = 4

(f 0 g)(4)

(g 0 f)(5) = g(f(5)) = g(2) = 1

(f 0 g) (5)

f(g(4)) = f(2) = 5 ye ( g (5)) -,2= f(3) = 3

In other words, Observe that fo g

g of.

-

-"

g0f

= { (1 , 1) ,

fo g

= { (1,1), (2, 3), (3, 3), (4, 5), (5, 3) )

(2, 3), (3, 1), (4, 4), (5, 1) )

-

4.

[CHAP. 2

FUNCTIONS

24

Let the functions f : R -> R and g : R -> R be defined by 1(x) = 2x -r 1, g (x) = X2 - 2 Find formulas defining the product functions go f and f o g. Solution:

Compute gof:R

It as follows:

(g 0 j)(x)

g(f(x)) = g(2x +1) = (2x +1)2 — 2 = 4x2 + 4x — 1

Observe that the same answer can be found by writing y = f(x) = 2x + 1,

g(y) = y2 — 2

z

and then eliminating y from the two equations:

z = y2 —2 = (2x+1)2 — 2 Now compute

fog:R R: Pg(x)) = f(x 2 2) -= 2(x2 — 2) + 1 -= 2x 2 — 3

(f ° g)(x)

5.

4x 2 + 4x —1

Prove the associative law for composition of functions, i.e. if ho (g o f). then (hog)of

f:A->B, g:B->C and

h: C- > D,

Solution: Since the associative law was proven for composition of relations in general, this result follows. We also give a direct proof:

Hence

((h 0g) 0 f)(a) = (h 0 g)(f(a)) -= h(g(f(a))),

Va E A

(h 0 (g o f))(a) = h((g 0 f)(a)) = h(g(f(a))),

Va E A

(hog)of _= ho(goo.

ONE-ONE AND ONTO FUNCTIONS 6. Let f: A -> B, g :B -> C. Prove: (i) If f and g are onto, then gof:A->C is onto. (ii) If f and g are one-one, then gof:A->C is one-one. Solution: (i)

Let c E C. Since g is onto, 3 b E B s.t. g(b) = c. Since f is onto, (g 0 f)(a) = g(f(a)) = c, i.e. g o f is also onto.

3a E A s.t. f (a) = b.

But

then (ii)

7.

Suppose (g o f)(a) = (g o f)(a'): i.e. g (f (a)) =- g(f(a')). So f(a) -= fiat) since g is one-one; hence a = a' since f is one-one. Accordingly, g of is also one-one.

Let A = 1-1,1) and let

f:A->A, g:A->A

f(x) =--

and h:A->A be defined by

sin x, g (x) = sin 7r x , h(x) = sin ix

State whether or not each of the functions is (i) one-one, (ii) onto, (iii) bijective (i.e. one-one and onto). Solution: The graphs of the functions are as follows:

—1

f(x)

g(x)

CHAP. 2)

The function f is one-one; each not onto since, for example, sin x line contains at least one point of f. function h is both one-one and onto;

8.

25

FUNCTIONS

horizontal line does not contain more than one point of f. It is 1 for any x E A. On the other hand, g is onto; each horizontal But g is not one-one since, for example, g(-1) = g(0) = 0. The each horizontal line contains exactly one point of h.

Prove: Let f:A->B and g:B->C be one-one and onto; then (gof) ---1 :C--->A exists C -> A. and equals f -1 Solution: Utilizing Proposition 2.1, we show that: 0

g1)

o

0

1A

(g 0f) 0 (f-1 o g -1)

and

Using the associative law for composition of functions, (f —

log - 1)0 (g o f)

(g - 1 (g of))

f =

0

((g - 100 Of)

„.= f -1 0 (10f)

of

= = since g --1 o g = 1 and lof=f=fal.

1 2,11

Similarly,

(g oho(f—lo g —i)

g ovou—to g —r» g

o g - 1)

of

go(log -1) —

9.

When will a projection function

7r :

gOgl

e I) ---> Aio, A io 9),

be an onto function?

Solution: A projection function is always onto, providing the Cartesian product i.e. provided no A i is the empty set.

II {A, : i E I) is non-empty,

INDEXED SETS, GENERALIZED OPERATIONS 10. Let An = {x: x is a multiple of n), where n E N, the positive integers, and let B i = [i, i +1], where i E Z, the integers. Find: (i) A3 n A5; (ii) t J{Ai : i e P), where P is the set of prime numbers; (iii) B3 n B4; (iv) U{Bi : j E Z } ; (v) (U fBi: j 7)) n A5. Solution: Those numbers which are multiples of both 3 and 5 are the multiples of 15; hence

(i) (ii)

A 3 n A 5 = A /5,

Every positive integer except 1 is a multiple of at least one prime number; hence U {A i : i E a, 4, ...} = N \ {1 } .

=

{2, (iii)

B3 n B4 = {x : 3

(iv)

Since every real number belongs to at least one interval of real numbers.

(v)

(

{Bi : i

4, 4

x

x

5} = {4 }

7 } ) n A5 = {x : x is a multiple of 5, x

E Z1

[i, j + 1], U

7} = A5\ {5}

{10, 15, 20, ...I.

11. Let Ds = (0, 1/n), where n E N, the positive integers. Find: (1) D3 U D7 (ii) D3

n D20

(v. )

(iv) Ds n D s

(vi) fl {Di : i e

Solution: (i)

Since (0, 1 1 7) c (0, 1/3),

(ii)

Since (0, 1/20) c (0, 1/3),

U {Di :iEAC N)

(iii) Ds U Dt

D3 U D7 = D3. D3 n D20 = D zo .

R, the set

26

[CHAP. 2

FUNCTIONS

(iii)

Let m = min {s, t}, i.e. the smaller of the two numbers s and t; then Dm equals D s or D t and contains the other. So D .s uD t = Dm .

(iv)

Let M =- max {s, t), Le. the larger of the two numbers. Then

(y)

Let a E A

be the smallest number in A.

(vi) If x E R, then

3i E N s.t. x

U{Di :iEA c

Then

(0, 1/i).

D,nD t -= DM.

n{D i : i E

Hence

D a.

--=

12. Prove (Distributive Law) Theorem 2.2 (ii): Bn(u i „A i) = Solution:

{x : x E B, x E u tet A t}

B n(u te.t A t)

{x: x E B,

aio E I s.t. x E

{x : 3i0 E I s.t.

x EBn 24. 40)

u iGr (BnA i)

13. Prove: Let {A,: i E I) be an indexed class of sets and let io E I. n ie/ A, c A40 c Uiei Ai

Then

Solution: Let x E m e/ A.,; Now let y E414 40.

then

xEA 1 for every iEl.

Since

In particular,

x EA 40.

Hence flI EJ A I C 21 10.

io E I, y E u ie/ A i. Hence A io c

14. Prove Theorem 2.4: Let A be any set and, for each p EA, let Gp be a subset of A such that p E G, c A. Then A = U{G, : p E A). Solution: Let x E U {Gp : p E Al. Then

3p0 E A

s.t. x E Gpo C A; hence x E A, so

U{Gp : p E A} c A.

(In other words, if each Gp is a subset of A, then the union of the Gp is also a subset of A.) Now let y E A. sets are equal.

Then y E Gy , so

y E U{Gp : p E A } .

Thus

Ac U{G p : p E A } and the two

ASSOCIATED SET FUNCTIONS

15. Let A = {1,2,3,4,51 and let f: A --> A be defined by the diagram:

Find

(i) f[{1, 3., 5 } ],

(ii)

[{2, 3, 4)],, (iii) f -1 [{3, 5)].

Solution: (i) f[{1, 3,5)] -= {f( 1), f( 3), f( 5)1 = { 4 } (ii) f 1 [{2, 3,4)] = {4, 1, 3, 5) (iii) f -1 [{3,51] = since no element of A has 3 or 5 as an image.

16. Consider the function f : R -> R defined by f(x) =-- x 2 . Find: [{-9)], (iii) f -1 [fx : x 0}], (iv) f - I[{x: 4 (i) f -1 [{25}], (ii)

x

25)].

Solution: f —11{25 } ] = {5, —5) since f(5) -= 25, f(-5) = 25 and since the square of no other number is 25. Ø since the square of no real number is —9. (ii) f ' [{-9)]

CHAP. 2]

(iii)

27

FUNCTIONS

f ' [{x : x

01] = {0 } since 1 (0) = 0

0 and since the square of every other real number is

greater than Q. (iv)

VI [{x

4= x = - 25 } ] consists of those numbers x such that 4 = x2 = 25. f — l[fx : 4 -= x 251] = [2,5] u [-5, —2]

17. Prove: Let f: X -> Y is also one-one.

be one-one.

Accordingly,

Then the associated set function f: ‘1) (X) -> 99 (Y)

Solution: If X = (1), then T(X) = {0}; hence f: P(X) T(Y) is one-one, for no two different members of P(X) can have the same image, as there are no two different members in T(X). If X (6, P(X) has at least two members. Let A , B E P(X), but A B. Then 3p E X s.t. p EA, p E B (or p E B, p S A). Thus f(p) E f[Al and, since f is one-one, f (p) f[Bi (or f(p) E f[Bi and f(p) E f[A]). Hence f[A] f[B], and so the induced set function is also one-one.

18. Prove (Theorem 2.5, (i) and (iii)): (a) f[A u B] = flA] U f[B], (b) f[A]\f[B] f[A\B]. Solution: c f [A] u f [B]. (a) We first show f[A u either x E A or x E B, but x E A

Let

x E B

or

y E f[A u B],

i.e.

ax EA uB s.t. f(x) = y.

implies

f(x) = y E f[A]

implies

f(x)

Then

y E f[B]

In either case, y E f[A] u f[B]. We now prove the reverse inclusion, i.e. f[A] u f[B] c f[A u 13]. Let y E f[A] or y E f[B], but y E f[A] implies 3x E A s.t. f(x) z= y y E f[B] In either case, y = f(x)

implies

with xEAuB,

y E f[A] u f[B].

Then

ax E B s.t. f(x) = y

i.e. yEf[AuB ] .

(b) Let y E f[A] \ f[B]. Then 3x E A s.t. f(x) x E B \ A. Accordingly, y E f[A \ B].

y, but y E {f (x) : x E B).

Hence

x B,

or

f -1 [A] n f -1 [B]. Let x E f -1 [A n B ] . Then f(x)EA n B x E f -1 [A] and x E f -1 [B]. Hence x E f —1 [A] n f -1 [B].

so

19. Prove (Theorem 2.6, (ii) and (iii)):

(a) f -1 [A n B] =

[A]

n f -- ' [B],

Solution: (a) We first show f —1 [A n B] f(x) E A and f(x) E B, or

(b) f -1 [A \B] = f -1 [A] \ f [B].

For the reverse inclusion, let x E f -1 [A] n f -1 [B ] . f(x) E A n B. Hence x E f-1 [A n B].

f(x) E B,

i.e.

(b) To show f —1 \B] c f1 [A] \ f l [B], assume x E f - 1 [A \B]. Then f(x) E A \B, f(x) E A and f(x) E B. Thus x E f ' [A] but x f — 1 [B], i.e. x E f -1 [A] \ f -1 [B].

i.e.

For the reverse inclusion, let x E f(x) E A\ B. Hence x E f -1 [A \ B].

[A] \ f -1 [13].

Then

Then

ALGEBRA OF REAL-VALUED FUNCTIONS 20. Let X = {a, b, e) and let f,g E f(X, R) be as follows: f = { (a, 1), (b, —2), (e, 3) I, g = (a, Find: (i) f + 2g, (ii) fg —2f, (iii) f + 4, (iv) f , (v) f2 . Solution: (i) Compute as follows:

(.f

2g)(a) —= f(a) 2g)(b) 2g)(c)

f(x) E A

f(x) E A

(b, 0), (e,

2g(a) = 1 — 4 = —3 f(b) 2g(b) = —2 + 0 = —2 f(c) 2g(c) = 3 + 2 = 5

and

but

}

f(x) B,

i.e.

28

[CHAP. 2

FUNCTIONS

In other words, (ii)

f + 2g =

Similarly,

(a, —3), (b, —2), (c, 5) }.

(fg —2f)(a) —= f(a) g(a) — 2f(a) = (fg —2f)(b)

f(b)g(b) — 2f (b) = (-2)(0) —

(fg —2f)(c) —= f(c)g(e) — 2f(c) =

That is,

fg —

f+

Since Ifi(x)

(iv)

if(x)I,

Since f 2(x) = (ff)(x) square. Thus

(3)(1) — 2(3) = —3

add 4 to each image value, i.e. to the second

4 = { (a, 5), (b, 2), (c, '7) }

replace the second coordinate of each pair in f by its absolute value. Thus

(fi = (v)

2(-2) = 4

(a, —4), (b, 4), (c, —3) }

2f =

Since, by definition, (f + 4)(x) f(x) + 4, coordinate in each pair in f. Thus

(iii)

(1)(-2) — 2(1) = —4

(a, I), (b, 2), (C, 3 ) }

f(x) f(x) = (f(x)) 2 , replace the second coordinate of each pair in f by its

f2 =

(a, 1), (1), 4), (c, 9) }

21. Let /6 E T(X, R) be defined by 'Os (x) = 0 for all x E X. Prove: For any f E l'(X, R), (1) f + = f and (ii) f =

.

Solution: (i)

for every x G X; hence f+ Ô = f. Ô satisfies the conditions of the 0 in the axiom [V1 ] of Theorem 2.9.

(ii)

(fÔ)(x)

(f + ) x

f(x) + (x) = f(x) + O = f(x)

f(x)Ô (x)

f(x) •

(0) = 0 = Ô(x)

Observe that

for all x E X; hence fÔ = 0.

22. Prove: T(X,R) satisfies the axiom [V3] of Theorem 2.9, i.e. if f,g e 7"(X, R) and k, k' E R, then: (i) k • (f + g) k•f +k•g, (ii) (k + kl•f --=k•f +ki•f. Solution: [k • (f + g)](x) = k[(f + g)(x)1 = k[f(x) + g(x)1 = k(f(x)) + k(g(x))

(i)

(k • f + k • g)(x) = (k • f)(x) + (k • g)(x) = k(f(x)) + k(g(x))

for all xE X; hence k • (f + g) = k • f k • g. Observe that we use the fact that k, f(x) and g(x) are real numbers and satisfy the distributive law. (ii)

((k + k') • f)(x) = (k k') f(x) = k(f(x)) + k'(f(x)) (k • f + k' • f)(x) = (k • f)(x) + (k' • f)(x) = k(f(x)) + k'(f(x))

for all xEX;

so

(k-l-k')•f = k•f + k'•f.

Supplementary Problems FUNCTIONS

23. Let f Find

IL and g :R -*R be defined by

f(x)

if x> 2 = { 2x — 5 x2 — 21xI if x = 2

g ( x)

(i) f(-2), (ii) g(-3), (iii) f(4), (iv) (g 0 f)(1), (y) (f ° g)(2), (vi) (f o f)(3).

It be defined by f(x) = x 2 -I- 3x + 1, g(x) = 2x— 3. f : R R and g Find formulas which define the composition functions (i) fog, (ii) go f, (iii) f o f.

24. Let

3x+1.

CHAP. 21

FUNCTIONS

29

25.

Let k:X-->X be a constant function. Prove that for any function f:X-4X, kof =k. be said about f k?

26.

Consider the function f(x) x functions is an extension of f.

27.

What can

x E R, x Q. State whether or not each of the following

where

(i)

g1 (x)

x

for all x E R

(iii)

g 3(x) = (x

(ii)

g2(x)

x where x E [-1, 1 ]

(iv)

1R R R

ixi)/2 for all x E R

Y. The inclusion function j from A into X, denoted by j: A c X, is Let A c X and let f: X defined by j(a) = a for all a E A. Show that f A, the restriction of f to A, equals the composition f j, i.e. f A =To j.

ONE-ONE, ONTO, INVERSE AND IDENTITY FUNCTIONS

28.

Prove: For any function f: A B, f 01 A

29.

Prove:

30.

Prove: If f:A--*B

31.

If f:A-->B

15°f.

is both one-one and onto, then fl of and

Prove Proposition 2.1: exists and

f -1 : B > A

f

g:B-4A

gof = 1A, then f is one-one and g is onto.

satisfy

f:A.-*B

Let

1 A and f 0 f -1 =- 1.

g:B--4/1

and

gof = 1 A and

satisfy

fog = 1 B .

Then

g= 7rio n {A,: i E

A 10 be one-to-one?

32.

Under what conditions will the projection

33.

Let f: (-1,1)-4 R be defined by f(x) = x1(1— oc!).

34.

Let R be an equivalence relation in a non-empty set A. The natural function i from A into the quotient set AIR is defined by n(a) [a], the equivalence class of a. Prove that n is an onto function.

35.

Let f: A -4 B. Let

7

Prove that f is both one-one and onto.

The relation R in A defined by a R a' iff f (a) =- f(a') is an equivalence relation. [al -> f(a). denote the correspondence from the quotient set AIR into the range AA] of f by

7:

7: AIR gill

is a function which is both one-one and onto.

(i)

Prove that

(ii)

Prove that f = j 070 77, where inclusion function.

AIR

77: A

is the natural function and

A 214 AIR 1-> f[A ]

j: f[A] c B

is the

B

INDEXED SETS AND GENERALIZED OPERATIONS

36. Let

A n = {x : x is a multiple of n}

{n, 2n, 3n, ...},

where nE N, the positive integers. Find:

(i) A 2 nA 7 ; (ii) A 6 nA 8 ; (iii) A 3 uA l2; (iv) A 3 nA l2; (y) A s uA s „ where where s,t E N. (vii) Prove: If J c N is infinite, then n E = Q. 37.

38.

Let B = (i, i+ 11, an open-closed interval, where i E Z, the integers. Find: (i)

Bg U B5

(ii)

B6

n B7

(iii) ur_ 4/3 i

(iv)

(v) ,

and

T

[0, 1/n) where n G N, the positive integers. Find:

n {D 0 : n E N}, (ii) 11{S„ : n E N}, (iii)

Prove DeMorgan's Laws:

(i) (u 1 A i )c =

ull o Bs+i

(vi) u jez B s+ ,

B s u B+1 uB +2 s E Z

Let D, = [0, 1/n], S = (0, 1/ni

(i) 39.

s, t E N; (vi) A s nA st,

n i A, (ii)

n E N}.

(n i A i)c = U 1 A.

30 40.

[CHAP. 2

FUNCTIONS

Let 0..4 = {A l : j E I} be an indexed class of sets and let

(i) U{A: E J} c U {A i : E K},

n {A i :

(ii)

Jc.KCI.

Prove:

D n{A i : E K}

E

ASSOCIATED SET FUNCTIONS 41.

42.

Let f: R R be defined by f(x) = x2 + 1. (iv) f -1 [(5, 1 0) ] , (v) f[R], (vi) f -1 [It].

(i) f[{ -1, OM], (ii) f

Find:

[ ( 10, 17) ] , (iii) f[( - 2,2)),

Prove: A function f: X -41 7 is one-one if and only if f[AnB] = f[A] n f[B], for all subsets

B of X.

43.

Prove: Let f: X -o Y.

44.

Prove: Let f: X -0 Y.

f -1

(a)

45.

c f[A] n f[B],

(b)

A cB

implies f[A] c f[B]

Then, for any subsets A and B of Y,

[A uB]

Prove Theorem 2.8:

and

Then, for any subsets A and B of X,

frAn

(a)

A

f -1 [A] u f -1 [B],

(b)

A c B implies f -1 [A] c f -1 [B ]

Let f : X -0 Y and let A c X and B c Y. Then

(i) 46.

Prove: Let f: X -o Y

47.

Prove: A function f: X subset A of X.

48.

Prove: A function f :X -o Y

be onto.

A c f --1 0

(ii) B

D fof -1 [B]

Then the associated set function

f: T(X) -o T(Y)

is also onto.

is both one-one and onto if and only if f[Ac] = (f[A])c

Y

f-lo f[A]

is one-one if and only if A

for every

for every subset A of X.

ALGEBRA OF REAL-VALUED FUNCTIONS 49.

Let X -= {a, b, c}

and let f and g be the following real valued functions on X:

f=

Let

(c, -1) },

A

(a, -2), (b, 0),

g =

(i) 3f, (ii) 2f- 5g, (iii) fg, (iv) f, (y) f 3, (vi) 13f

Find

50.

(a, 2), (b, -3),

(c, 1) }

fgi.

be any subset of a universal set U. Then the real-valued function

xA U -0 R defined by

if x E A 0 if x E A

{1

X A (X) is called the characteristic function of A. Prove: (i) X

51.

Prove:

(i)

52.

A flB

XA

XS

(ii) X AUB

(iii) x A \ B = X A X A n n •

T(X, R) satisfies the axiom [V2 ] of Theorem 2.9; i.e. if f E T(X,R)

k • (k' • f)

(kk') • f,

le

For each k E R, let ' E (i)

xA n

= xA

and k, k' E R,

then

(ii) 1 • f =- f

y(x, R)

denote the constant function

Show that the collection C of constant functions, i.e.

re (x)

C =

= k for all x E X. k E R1, is a linear subspace of

T(X, R). (ii)

Let a : C --0 R be defined by any k, k' E R,

k.

a(k +

Show that a is both one-one and onto and that, for

= a(k)

a(fe')

CHAP. 2 1

31

FUNCTIONS

Answers to Supplementary Problems 23. (i) 0, (ii) -8, (iii) 3, (iv) -2, (v) 9, (vi) -1 24. (i) (f g)(x) 25.

4x 2 - 6x +1,

(ii) (g f)(x)

2x2 + 6x -1,

(iii) (f f)(x) = x 4 + 6x3 + 14x 2 + 15x + 5

The function f o k is a constant function.

26. (i) yes, (ii) no, (iii) yes, (iv) yes 32. A i is a singleton set, say A i = {ai), for i jo. 36. (i) A 14, (ii) A2 4, (iii) A3, (iv) Al2, (y) 37. (i) (4, 6], (ii) 0, (iii) (4, 21],

As,

(iv) (s, s + 3],

41. (1) {1, 2 } , (ii) {3, -3, 4, -4), (iii) (1, 5), 3f = { (a, 6), (b, -9), (c, -3)

(ii)

2f - 5g = { (a, 14), (b,

(iii)

fg = { (a, -4), (b, 0), (e, -1)

(iv)

(N) (s, s + 16], (vi) R

0, (iii) {0 }

38. (i) {0 } , (ii)

49. (i)

(Vi) Ast

{

2

), (b, 3), (e, 1 ) }

(v)

f3 = { (a, 8), (b,

(vi)

13f

fg(

(e, -7) )

-

27), (e, -1) )

{ (a, 10), (b, 9), (e, 2)

(iv) (-3, -2), (2, 3),

(N) {x: x

1), (vi) R

Chapter

3

Cardinality, Order EQUIVALENT SETS A set A is called equivalent to a set B, written A- B, if there exists a function f:A---)B which is one-one and onto. The function f is then said to define a one-to-one correspondence between the sets A and B. A set is finite if it is empty or equivalent to {1,2, ..., n} for some n e N; otherwise it is said to be infinite. Clearly two finite sets are equivalent if they contain the same number of elements. Hence, for finite sets, equivalence corresponds to the usual meaning of two sets containing the same number of elements. Example 1.1:

Let N =-- {1,2, 3, . . .} and E = {2, 4, 6, ...}. The function f : N --0 E defined by f (x) ----- 2x is both one-one and onto; hence N is equivalent to E.

Example 1.2:

lx1) is both one-one and onto. The function f: (-1,1) --+ R defined by 1(x) = x1(1 Hence the open interval (-1,1) is equivalent to R, the set of real numbers. -

Observe that an infinite set can be equivalent to a proper subset of itself. This property is true of infinite sets generally. Proposition 3.1:

The relation in any collection of sets defined by A -B is an equivalence relation.

DENUMERABLE AND COUNTABLE SETS Let N be the set of positive integers {1, 2,3, ... }. A set X is called denumerable and is said to have cardinality t o (read: aleph-mull) if it is equivalent to N. A set is called countable if it is finite or denumerable. Example 2.1:

The set of terms in any infinite sequence al , a2 , a3 , ... of distinct terms is denumerable, for a sequence is essentially a function 1(n) = an, whose domain is N. So if the au are distinct, the function is one-one and onto. Accordingly, each of the following sets is denumerable: {1, 1, .1, . . . },

Example 2.2:

{1, -2, 3, -4, . . . },

{ (1, 1), (4, 8), (9, 27), . .., (n 2, n3), ... }

Consider the product set N X N as exhibited below. (1, 4) --.-

2)-0- (1, 3) 2) 2) 2)

32

/ /1 /

(2, 3) (3, 3) (4, 3)

/ (2, 4) / • / /

(3, 4) (4,4)

/

• •

CARDINALITY, ORDER

CHAP. 3]

33

The set N X N can be written in an infinite sequence of distinct elements as follows:

(1, 1), (2, 1), (1, 2), (1, 3), (2, 2), ... (Note that the sequence is determined by "following the arrows" in the above diagram.) Thus we see that N X N is denumerable. Example 2.3:

N u {O. Now each positive integer Let M = {0,1, 2, 3, . „} written uniquely in the form a = 2r(2s + 1) where r,s E M. f N III X M defined by f(a)

a E N can be The function

(r,

where r and s are as above, is one-one and onto. Note that N X N is a subset of M X M.

Hence M X M is denumerable.

The following theorems concern denumerable and countable sets.

Theorem 3.2:

Every infinite set contains a denumerable subset.

Theorem 3.3:

Every subset of a countable set is countable.

Lemma 3.4:

Let {A1 A2, ... } be a denumerable disjoint class of denumerable sets. Then ur_ l Ai is also denumerable.

Theorem 3.5:

Let A,: i E I) be a countable class of countable sets, i.e. / is countable and Ai is countable for each L EI. Then U{A i : L E /} is countable.

A set which is neither finite nor denumerable is said to be non-denumerable or noncountable.

THE CONTINUUM Not every infinite set is denumerable; in fact, the next theorem gives a specific and extremely important example. Theorem 3.6:

The unit interval [0,11 is non-denumerable.

A set X is said to have the power of the continuum or is said to have cardinality c if it is equivalent to the unit interval [0,1]. We show, in a solved problem, that every interval, open or closed, has cardinality c. By Example 1.2, the open interval (-1,1) is equivalent to R. Hence,

Proposition 3.7: R, the set of real numbers, has cardinality c. SCHROEDER-BERNSTEIN THEOREM B if A is equivalent to a subset of B, i.e.,

We write A

A

B

if 3 B* c B

such that

A — B*

We also write A < B if A < B but A 4 B, i.e. A is not equivalent to B. Example 3.1:

Since N is a subset of R, we may write N < R. On the other hand, by Proposition 3.7, R is not denumerable, i.e. R 11' N. Accordingly, N < R.

Given any pair of sets A and B, then at least one of the following must be true: (i) A — B, (ii) A < B or B A, (iii) A < B and B A, (iv) A B, A lc B and B i A The celebrated Schroeder-Bernstein Theorem states that, in Case (iii) above, A is equivalent to B. Namely,

Theorem (Schroeder-Bernstein) 3.8: If A B and B

A, then A — B.

The Schroeder-Bernstein Theorem can be restated as follows:

Theorem 3.8:

Let XD YD X1 and let X — X1.

Then X

Y.

CARDINALITY, ORDER

34

We remark that Case (iv) above is impossible.

Theorem (Law of Trichotomy) 3.9:

[CHAP. 3

That is,

Given any pair of sets A and B, either A
CONCEPT OF CARDINALITY If A is equivalent to B, i.e. A — B, then we say that A and B have the same cardinal number or cardinality. We write #(A) for "the cardinal number (or cardinality) of A". So #(A) = #(B) if A — B On the other hand, if A -GB then we say that A has cardinality less than B or B has cardinality greater than A. That is, #(A) < #(B) if A < B So #(A) #(B) if A < B. stated as follows:

Theorem 3.8:

If #(A)

Accordingly, the Schroeder-Bernstein Theorem can be re-

#(B) and #(B)

#(A), then #(A) = #(B).

The cardinal number of each of the sets 0, {0 } , 10, {0 }} , (0, {0 } , {0,

WM,

•••

is denoted by 0, 1, 2, 3, . . . , respectively, and is called a finite cardinal. The cardinal numbers of N and [0, 1] are denoted by tA0 = #(N), Accordingly, we may write

c =-- #({0,1])

0 < 1 <2 <3 < • • • < ti o < c

CANTOR'S THEOREM AND THE CONTINUUM HYPOTHESIS It is natural to ask if there are infinite cardinal numbers other than tA 0 and c. The answer is yes. In fact, Cantor's Theorem determines a set with cardinality greater than any given set. Namely,

Theorem (Cantor) 3.10:

The power set P(A) of any set A has cardinality greater than A.

It is also natural to ask if there exists a set whose cardinality lies between tA 0 and c. The conjecture that the answer to this question is negative is known as the Continuum Hypothesis. That is,

Continuum Hypothesis:

There does not exist a set A with the property that tA o < #(A) < c.

In 1963 it was shown that the Continuum Hypothesis is independent of our axioms of set theory in somewhat the same sense that Euclid's Fifth Postulate on parallel lines is independent of the other axioms of geometry.

PARTIALLY ORDERED SETS A relation in a set A is called a partial order (or order) on A if, for every a, b,c G A: (i) a-< a; (ii) a b and b < a implies a = b; and (iii) a b and b < c implies a c. The set A together with the partial order, i.e. the pair (A, ), is called a partially ordered set.

CHAP. 3]

35

CARDINALITY, ORDER

Recall that a relation is reflexive if it satisfies (1), and transitive if it satisfies (iii). A relation is said to be anti-symmetric if it satisfies (ii). In other words, a partial order is a reflexive, anti-symmetric, transitive relation. Example 4.1:

Set inclusion is a partial order in any class of sets since: (i) AcA for any set A; (ii) A cB and B c A implies A =B; and (iii) A c B and B cC implies ACC.

Example 4.2:

Let A be any set of real numbers. Then the relation in A defined by x partial order and is called the natural order in A.

Example 4.3:

Let X = {a, b,c,d, e) .

y is a

Then the diagram

ct‘‘N c

b

11

d

e

defines a partial order in X as follows: x•< y if x = y or if one can go from x to y in the diagram, always moving in the indicated direction, i.e. upward.

If a b in an ordered set, then we say that a precedes or is smaller than b and that b follows or dominates or is larger than a. Furthermore, we write a < b if a < b but a b. A partially ordered set A is said to be totally (or linearly) ordered if, for every a, b EA, either a b or b a. R, the set of real numbers, with the natural order defined by xy is an example of a totally ordered set. Example 4.4:

Let A and B be totally ordered. Then the product set ordered as follows: (a, b) < (a', b')

if

a < a' or if a = a' and b

This order is called the lexicographical order of A X B way words are arranged in a dictionary.

Remark:

AXB

can be totally

b'

since it is similar to the

If a relation R in a set A defines a partial order, i.e. is reflexive, anti-symmetric and transitive, then the inverse relation R -1 is also a partial order; it is called the inverse order.

SUBSETS OF ORDERED SETS Let A be a subset of fa partially ordered set X. Then the order in X induces an order in A in the following natural way: If a, b E A, then a b as elements in A if a b as elements in X. More precisely, if R is a partial order in X, then the relation RA = Rn (A x A), called the restriction of R to A, is a partial order in A. The ordered set (A, RA) is called a (partially ordered) subset of the ordered set (X, R). Some subsets of a partially ordered set X may, in fact, be totally ordered. Clearly if X itself is totally ordered, every subset of X will also be totally ordered. Example 5.1:

Consider the partial order in

W = {a, b,c,d, el defined by the diagram a ‘41

c

d The sets {a, e, d} and {b, el are totally ordered subsets; the sets {a, b, el and fd, el are not totally ordered subsets.

FIRST AND LAST ELEMENTS Let X be an ordered set. An element ao e X is a first or smallest element of X if o < x for all x E X. Analogously, an element bo E X is a last or largest element of X a iff x bo for all x E X.

[CHAP. 3

CARDINALITY, ORDER

36 Example 6.1:

Let

X = {a, b, c, d, el be ordered by the diagram

a /v d

e

Then a is a last element since a follows every element. Note that X has no first element. The element d is not a first element since d does not precede e. Example 6.2:

The positive integers N with the natural order has 1 as a first element. integers Z with the natural order has no first element and no last element.

The

MAXIMAL AND MINIMAL ELEMENTS Let X be an ordered set. An element ao EX is maximal if ao-< x implies x = a, i.e. if no element follows ao except itself. Similarly, an element bo e X is minimal if x bo implies x 1) 0, i.e. if no element precedes bo except itself. Example 7.1:

Let X = {a, b, e, d, el be ordered by the diagram in Example 6.1. Then both d and e are minimal elements. The element a is a maximal element.

Example 7.2:

Although R with the natural order is totally ordered it has no minimal and no maximal elements.

Example 7.3:

Let A {al, a2, am } be a finite totally ordered set. Then A contains precisely one minimal element and precisely one maximal element, denoted respectively by am} min {a l , ..., am) max {a1 , and

UPPER AND LOWER BOUNDS Let A be a subset of a partially ordered set X. An element m e X is a lower bound of A if m x for all x EA, i.e. if m precedes every element in A. If some lower bound of A follows every other lower bound of A, then it is called the greatest lower bound (g.l.b.) or infimum of A and is denoted by inf (A). Similarly, an element M EX is an upper bound of A if x-< M for all x E A, i.e. if M follows every element in A. If some upper bound of A precedes every other upper bound of A, then it is called the least upper bound (1.u.b.) or supremum of A and is denoted by sup (A). A is said to be bounded above if it has an upper bound, and bounded below if it has a lower bound. If A has both an upper and lower bound, then it is said to be bounded. Example 8.1:

Let X = {a, b, e, d, e, f,

be ordered by the following diagram!

Let B = {e, d, el. Then a, 1 and c are upper bounds of B, and f is the only lower bound of B. Note that g is not a lower bound of B since g does not precede d. Furthermore, c = sup (B) belongs to B, while f = inf (B) does not belong to B. Example 8.2:

Let A be a bounded set of real numbers. Then a fundamental theorem about real numbers states that, under the natural order, inf (A) and sup (A) exist.

CHAP. 3]

CARDINALITY, ORDER

Example 8.3:

Let Q be the set of rational numbers.

37

Let

B =-- fx! xEQ, x> 0, 2 < x2 < 3} that is, B consists of those rational points which lie between '/ -1,5 on the real line. Then B has an infinite number of upper and lower bounds, but inf (B) and sup (B) do not exist. Note that the real numbers Ari and 15 do not belong to Q and cannot be considered as upper or lower bounds of B.

ZORN'S LEMMA Zorn's Lemma is one of the most important tools in mathematics; it asserts the existence of certain types of elements although no constructive process is given to find these elements.

Zorn's Lemma 3.11:

Remark.

Let X be a non-empty partially ordered set in which every totally ordered subset has an upper bound. Then X contains at least one maximal element.

Zorn's Lemma is equivalent to the classical Axiom of Choice and the Wellordering Principle. The proof of this fact, which uses the concept of ordinal numbers, is beyond the scope of this text.

Solved Problems EQUIVALENT SETS, DENUMERABLE SETS 1.

Consider the concentric circles Ci

= {(x, : x2 + y 2 - a2 j,

C2 =

{(x, y) : x2 + y2 = b2)

where, say 0
2.

Thus f defines a one-

Prove: The set of rational numbers is denumerable. Let Q+ be the set of positive rational numbers and let Q -- be the set of negative rational numbers. is the set of rational numbers. Then Q = Q - u { 0} u Q+ Let the function

f: Q+ N X N

be defined by

f(plq) = (p,q) where plq is any positive rational number expressed as the ratio of two positive integers. Note f is one-one; hence Q+ is equivalent to a subset of N X N. But N X N is denumerable (see Example 2.2); hence Q+ is also denumerable. Similarly Q - is denumerable. Accordingly, by Theorem 3.5, the union of Q - , {0} and Q+, i.e. the set of rational numbers, is also denumerable.

3.

Prove Proposition 3.1: The relation in any collection of sets defined by A —B is an equivalence relation. That is, (i) A A for any set A; (ii) if A —B then B—A; and (iii) if A — B and B— C then A — C. Solution: The identity function (i)

lA : A A

is one-one and onto; hence A - A.

CARDINALITY, ORDER

38 (ii)

If A—B, then there exists f:A--013 which is one-one and onto. But then f has an inverse f -1 :B--*A which is also one-one and onto. Hence A — B

(iii)

[CHAP. 3

implies

B—A

If A—B and B—C, then there exist functions f:A-->B and g:B—>C which are one-one and onto. Thus the composition function gof:A-->C is also one-one and onto. Hence A — B

and B — C

implies

A —C

4. Prove: The collection P of all polynomials p(x) =

ao +

with integral coefficients, i.e. where

ao,

ax + • • • + anam

al,

...,

am are integers, is denumerable.

Solution: For each pair of positive integers (n, m) E N x N, let P,,,,, denote the set of polynomials p(x) of degree m in which lac)! + lail + • • • -I- land — n Observe that Pm?, is finite. Accordingly,

P

= U{P,,,,,,: (n,m) E N x N}

is countable since it is a countable union of countable sets. In particular, since P is not finite, P is denumerable.

5.

A real number r is called an algebraic number if r is a solution to a polynomial equation p(x)

with integral coefficients.

ao + aix + • • • + am en

=

Prove that the set A of algebraic numbers is denumerable.

Solution: Note, by the preceding problem, that the set E of polynomial equations is denumerable:

E = {p i (x) = 0, p2 (x) = 0, p3(x) = 0, ...} Ai =

Let

{x : x is a solution of p i (x) = 0}

Since a polynomial of degree n can have at most n roots, each A i is finite. Hence A = U{A i : i E N) is denumerable.

6. Prove Theorem

3.2:

Every infinite set X contains a subset D which is denumerable.

Solution: Let f: T(X) -4 X be a choice function, i.e. for each non-empty subset A of X, f (A) E A. a function exists by virtue of the Axiom of Choice.) Consider the following sequence:

(Such

ai = f(X) a2 = f(X\ fail)

a3 =

f(X\ fai , a21)

a n = f(X\ {au . . • , an _ 1 1)

Since X is infinite, choice function,

X \ {a1 , ..., an _i}

Thus the an are distinct and

is not empty for every n EN. Furthermore, since f is a an ai for

i
D = {al , a2, ...} is a denumerable subset of X.

Essentially, the choice function f "chooses" an element ai E X, then chooses an element a2 from those elements which "remain" in X, etc. Since X is infinite, the set of elements which "remain" in X is non-empty.

CHAP. 31

39

CARDINALITY, ORDER

7. Prove: Let X be any set and let C(X) be the collection of characteristic functions on X, i.e. the collection of functions f: X { 1,0 } . Then the power set of X is equivalent to C(X), i.e. 619 (X) C(X). Solution: Let A be any subset of X, i.e. A E

P(X).

= xA

f (A )

Then

f

Hence

is one-one and onto.

Let

f :P(X) --0 C(X) { 0 if x 4 A 1 if x E A

be defined by

P(X) C(X).

8. Prove: A subset of a denumerable set is either finite or denumerable, i.e. is countable. Solution: Let X = {a 1 , a2 , ...} be any denumerable set and let A be a subset of X. If A = (6, then A is 0, then let n i be the least positive integer such that a ril E A; let n2 be the least positive finite. If A

n2 > n i and a n2 E A; etc. Then A = {a 1 , a 2 , ...}. In 1 ot2, . . } is bounded, then A is finite. Otherwise A is denumerable.

integer such that

If the set of integers

9. Prove Theorem 3.3: Every subset of a countable set is countable. Solution: If X is countable, then

X

In either case, its subsets are countable.

is either finite or denumerable.

10. Prove Lemma 3.4: Let {A1, A2, ...} be a denumerable disjoint class of denumerable sets. Then u7_, A, is denumerable. Solution: Since the sets A i are denumerable, we can write

A1 =

{a 11 , a12, a13,

A2 = {a21, Cr'22 ,

=

.}

a23 , • • • }

an2 , a3, . .

Then u7=1 Ai (i,j) E Nx N). The function f: u i7. 1 A i --4N X N defined by f(a) =(i, is clearly one-one and onto. Hence u7_ 1 A i is denumerable since N x N is denumerable.

j)

11. Prove: Let A be an infinite set, let B = {b1, b2, ...} be denumerable, and let A and B be disjoint. Then A uB - A. Solution: Since A is infinite, A contains a denumerable subset defined by the following diagram:

A u

B=

A

(A

=

u

\D) u (DUB) =

(A

\D) u

D

if

d27 _ 1 if

, is one-one and onto; hence A

Let

d2, d3, . . b 1 , b2, b3,

I■

f (x) = f

= {d1, d2 , ...}.

\ P 4 u {d 1 , d2, d3, d4, d5, d6, . • •

=

In other words,

Observe that

D

uB -

if

A.

xEA\D x = x=

f:

A

uB --+ A

be

40

CARDINALITY, ORDER

[CHAP. 3

CONTINUUM, CARDINALITY 12. Prove that the intervals (0, 1), [0, 1) and (0, 11 have cardinality c, i.e. is equivalent to {0, 1]. Solution: (i) Note that

[0, 1 ] = {0, 1, 1/2, 1/3, ...} u A,

(0, 1) = {1/2, 1/3, 1/4, ...} u A

A --= [0, 1] \ {0, 1, 1/2, 1/3, ...} = (0, 1 ) \ {1/2, 1/3, 1/4, ...}

where

Consider the function

f:

[0, 1 1 --> (0, 1) defined by the following diagram

A

In other words,

f(x) =

if x = 0 if x =

{ 1/2 1/(n + 2)

nEN 0, 1/n, n E N, i.e. if x E A

if x The function f is one-one and onto. (ii)

The function

f:

[0, 1]

Accordingly,

[0, 1] - (0, 1).

[0, 1 ) defined by f(x) =

1/(n + 1)

if x = 1/n, n E N if x 1/n, n E N

is one-one and onto. (It is similar to the function in Part (i)). (iii)

Let f: [0,1) -> (0, 11 be defined by f (x) = 1- x. [0, 1) - (0, 1] and, by transitivity, [0, 1] - (0, 1).

Then

Hence [0,1] - [0, 1).

f is one-one and onto.

Hence

In other words, (0, 1), [0, 1) and (0, 1] have cardinality c.

13. Prove: Each of the following intervals has the power of the continuum, i.e. has cardinality c: la , b] , (a, b), la, b) and (a, bl. Here a G b. Solution:

f(x) = a + (b

Let each of the following functions be defined by [0,1] ---1> [a, b]

[0,1) -4 [a, b)

(0,1) 4 (a, b)

a)x: (0,1) -4 (a, b]

Each function is one-one and onto. Hence by the preceding problem and Proposition 3.1, each interval is equivalent to [0, 1], i.e. has cardinality c.

14. Prove Theorem 3.6:

The unit interval A = [0, 1] is non-denumerable.

Solution:

Method 1.

Assume the contrary; then

A = {x i , x 2 , x 3, .

}

i.e. the elements of A can be written in a sequence. Each element in A can be written in the form of an infinite decimal as follows: xi

= 0. all a 12 a 13

a in

X2

=

a2n

x3

0. a2 1 a22 a23 31 a32 a330.a

x n -= 0. an1 an2 an3

ann

CHAP. 3]

CARDINALITY, ORDER

41

where a,i E {0,1, ..., 9} and where each decimal contains an infinite number of non-zero elements, i.e. for those numbers which can be written in the form of a decimal in two ways, e.g.,

1/2 = .5000... = .4999... we write the infinite decimal in which all except a finite set of digits are nines. Now construct the real number y

= O. b i b2 b3

which will belong to A, in the following way: and b2 0, etc.

choose b 1 so b 1 a ll and b 1 0, choose b2 so b2 a22

Observe that y x1 since b 1 a n , y2 x2 since b2 a22, etc., that is, y x n, for n EN. Hence y 4 A, which is impossible. Thus the assumption that A is denumerable has led to a contradiction. Consequently, A is non-denumerable. Method 2.

Assume the contrary. Then, as above,

A = {x 1 , x2 , x 3 , .

•}

Now construct a sequence of closed intervals as follows: Consider the following three closed sub-intervals of A = [0,1], [0, -1], [-A, 1], [I, 11 (I) each having length *. Now x1 cannot belong to all three intervals. Let I = [a 1 , 1) 1 ] be one of the intervals in (I) such that x / E I.

b 1 ],

Now consider the following three closed sub-intervals of I I [a l , ai each having length

al +

[ai

[ai +

4, an

(2)

Similarly, let 1 2 be one of the intervals in (2) such that x2 4 1 2 .

By continuing in this manner, we obtain a sequence of closed intervals

12 D /3 D • • •

(3)

such that x n E In for all n E N. By the Nested Interval Property (see Appendix A) of the real numbers, there exists a real number y E A = [0,1 ] such that y belongs to every interval in (3). But

y E A = {x i , x2, ... } implies

y x„,,

o

for some mo E N

Then by our construction y =-- x mo /,„„0, which contradicts the fact that y belongs to every interval in (3). Thus our assumption that A is denumerable has led to a contradiction. In other words, non-denumerable.

A is

15. Prove Theorem (Schroeder-Bernstein) 3.8: Let XD YD X 1 and let X — X1; then X

Y.

Solution: Since X — X1 , there exists a function f: X —> X 1 which is one-one and onto. But XD Y; hence the restriction of f to Y, which we shall also denote by f, is also one-one. So Y is equivalent to a subset of X 1 , i.e. Y Y 1 where

XDYDX 1 DY 1

and f Y

Y 1 is one-one and onto. But now YD X i ; hence, for similar reasons, X

—

X2 where

XDYDX1DY 1 DX2 and f: X 1 —> X2 is one-one and onto. Consequently, there exist equivalent sets such that equivalent sets Y1 , Y2, Y3, . . .

XD YDX 1 DY1 DX2 DY2 D•-• B =

Let Then

X

=

(X

XnYnX1 nY1 nX2 nY2 n•-• Y) u (Y \X1) u (X 1 \ Y 1 ) u • • u B

Y = (Y \ X i ) u (X i \ Yi ) u (Y 1 \ X2) u • • • u B Note further that Specifically, the function

(X \ Y) — (X 1 \ Y1 ) f: (X\ Y,,) (Xn +1\ Yn+i)

(

X2 \ Y2)

•••

is one-one and onto.

X1 , X2, X3,

.

.

and

CARDINALITY, ORDER

42 Consider the function g:X—>Y

[CHAP. 3

defined by the following diagram:

X

Y

In other words, g(x) =

{f (x)

if x E Xi \ Yi or xEX\Y if x E Y\ X, or x E B

Therefore X — Y.

Then g is one-one and onto.

16. Prove Theorem (Cantor) 3.10: The power set 9)(A) of any arbitrary set A has cardinality greater than A, i.e. A -‹ 613 (A) and hence #(A) < #( 6P(A)). Solution: which sends each element a E A The function g: A —> T(A) g (a) = {a}, is one-one; hence A -< P(A).

into the singleton set {a}, i.e.

If we show that A is not equivalent to P(A), then the theorem will follow. Suppose the contrary, i.e. let there exist a function f: A —> P(A) which is one-one and onto. Call aEA a "bad" element if a is not a member of the set which is its image, i.e. if a E f(a). Let B be the set of "bad" elements, i.e., B = {x : x E A, x f(x)} 61)(A) is onto, there exists Observe that B is a subset of A, that is, B E P(A). Since f: A an element b E A with the property that f(b) = B. Question: Is b "bad" or "good"? If b EB then, by definition of B, b f(b) = B which is a contradiction. Likewise, if bEB, then b E f(b) = B which is also a contradiction. Thus the original assumption, that A — TEA), has led to a contradiction. Accordingly A — T(A) is false, and so the theorem is true.

ORDERED SETS AND SUBSETS 17. Let N, the positive integers, be ordered as follows: each pair of elements a,a' C N can be written uniquely in the form a = 2r(2s + 1), a' = ar(as'+ 1) where r,r`,s,s' C {0, 1, 2, 3, . . . }. Let a < a' if r < r' or if r = r' but s <s' Insert the correct symbol, -< or >, between each of the following pairs of numbers. (Here x > y iff y -< x.)

14, (ii) 6

(i) 5

9, (iii) 3_20,

(iv) 14

21

Solution:

The elements in N can be written as follows:

2

0

1

2

1

3

2

6

4

12

3

4

5

6

7

7

9

11

13

15

10

14

18

22

26

20

28

36

44

52

30 60

Then a number in a higher row precedes a number in a lower row and, if two numbers are in the same row, the number to the left precedes the number to the right. Accordingly,

(i) 5 -< 14,

(ii) 6 > 9,

(iii) 3 -< 20,

(iv) 14> 21

CHAP.

3]

CARDINALITY, ORDER

43 a

18. Let A = {a, b, c) be ordered as in the diagram on the right. Let 04 be the collection of all non-empty totally ordered subsets of A, and let cA/ be partially ordered by set inclusion. Construct a diagram of the order of e.g. Solution: The totally ordered subsets of A are: {a}, {b}, {c}, {a, b}, {a, c } . inclusion, the order of cAr is the following:

Since c 4 is ordered by set

(a, c}

{a, b}

{ {b}

{a}

N\ {1 } , and let A be ordered by "x divides y". (i) Determine the minimal elements of A. (ii) Determine the maximal elements of A.

19. Let A = {2,3,4, Solution:

(i)

(ii)

If p E A is a prime number, then only p divides p (since 1 A); hence all prime numbers are minimal elements. Furthermore, if a E A is not prime, then there is a number b E A such that b divides a, i.e. b-< a; hence a is not minimal. In other words, the minimal elements are precisely the prime numbers. There are no maximal elements since, for every a E A, a divides 2a, for example.

20. Let B = {2, 3,4, 5, 6, 8, 9, 10) be ordered by "x is a multiple of elements of B. (ii) Find all minimal elements of B.

y".

(i) Find all maximal

Solution: Construct a diagram of the order of

B as follows:

3

2

5 t

9

6

19

(i) The maximal elements are 2, 3 and 5. (ii) The minimal elements are 6, 8, 9 and 10.

21. Let W = {1, 2, ..., 7, 8) be ordered as follows: 1 2 3 /4 \ / 6

\ /7 8

/

Consider the subset V = {4, 5, 6) of W. (1) Find the set of upper bounds of V. (ii) Find the set of lower bounds of V. (iii) Does sup (V) exist? (iv) Does inf (V) exist? Solution: Each of the elements in {1, 2, 3 } , and only these elements, follows every element in (i) is an upper bound. Only 6 and 8 precede every element in V; hence {6,8 } is the set of lower bounds. (ii) (iii) (iv)

V and hence

Since 3 is a first element in the set of upper bounds of V, sup (V) = 3. Note that 3 E V. Since 6 is a last element in the set of lower bounds of V, inf (V) = 6. Note that 6 E V.

CARDINALITY, ORDER

44

[CHAP. 3

22. Let e4 be a collection of sets partially ordered by set inclusion, and let 73 be a subcollection of eA1 . (i) Prove that if A E e4 is an upper bound of 6/3, then U{B : B E 93) C A. (ii) Is U : B E (13) an upper bound of 93? Solution: (i) Let x G LJIB : B GB); then 3B 0 st. x E B o . But A is an upper bound of B; so Bo c A c A. and hence x A. Accordingly, U{B : B G (ii) Even though B is a subcollection of C.4, it need not be true that the union of members of B, i.e. 1.11/3 : B E BI, is a member of &g. In other words, UIB: B E B1 is an upper bound of B if and only if it belongs to e-4.

APPLICATIONS OF ZORN'S LEMMA 23. Prove: Let X be a partially ordered set. Then there exists a totally ordered subset of X which is not a proper subset of any other totally ordered subset of X. Solution: Let &I be the class of all totally ordered subsets of X. Let &I be partially ordered by set inclusion. We want to show, by Zorn's Lemma, that &4 possesses a maximal element. So suppose B {B i : i G I} is a totally ordered subclass of A. Let A = i G II.

B, c X for all B. G B

Observe that

We next show that A is totally ordered.

aBj ,

Bk G B

implies

A c X

Let a,b G A; then such that

a G By b E Bk

But B is totally ordered by set inclusion; hence one of them, say Bp is a subset of the other. Consequently, a,b G Bk. Recall that Bk G B is a totally ordered subset of X; so either a b or b a. Then A is a totally ordered subset of X, and so A G c./1. But B,c A for all B,GB; hence A is an upper bound of B. Since every totally ordered subset of has an upper bound in e-4, by Zorn's Lemma, &4 has a maximal element, i.e. a totally ordered subset of X which is not a proper subset of any other totally ordered subset of X.

24. Prove: Let R be a relation from A to B, i.e. Rc A x B, and suppose the domain of R is A. Then there exists a subset f* of R such that f* is a function from A into B. Solution: Let &61 be the class of subsets of R such that each f G cAl is a function from a subset of A into B. Partially order &4 by set inclusion. Observe that if f : A 1 —> B is a subset of g : A2 B then

A 1 C A 2. {f,: A t B} icI is a totally ordered subset of &g. Then (see Problem 44) f U]1 is a function from u i A i into B. Furthermore, f cR. Hence f is an upper bound of B. By Zorn's Lemma, cA possesses a maximal element f* : A* —> B. If we show that A* = A, then the Now suppose

B

theorem is proven. Suppose A.* A. A. Then aa G A s.t. a By hypothesis, the domain of R is A; hence there exists an ordered pair (a,1» G R. Then f* u ((a, b} is a function from A* u {a} into B. But this contradicts the fact that f , is a maximal element in &I. So A = A, and the theorem is proven.

Supplementary Problems EQUIVALENT SETS, CARDINALITY

25. Prove:

Every infinite set is equivalent to a proper subset of itself.

26. Prove:

If A and B are denumerable, then A X B is denumerable.

27. Prove:

The set of points in the plane R2 with rational coordinates is denumerable.

CARDINALITY, ORDER

CHAP. 3]

45

28. A real number x is called transcendental if x is not algebraic, i.e. if x is not a solution to a polynomial equation

p(x) = ao ai x + • • • + ani xm = 0

with integral coefficients (see Problem 5).

29.

For example,

(ii)

Prove that T has the power of the continuum, i.e. has eardinality c.

An operation of multiplication is defined for cardinal numbers as follows:

(i)

Prove:

and #(B) = #(B') implies

A — A' and B B'

or, equivalently, (ii)

= #(A X B)

Show that the operation is well-defined, i.e., #(A) = #(A')

(a) N O N ° =

0,

(b) o e = c,

implies

#(A) #(B) = #(A') #(B')

(A X B)

(A' X B')

(c) c c = c.

An operation of addition is defined for cardinal numbers as follows:

#(A) #(B) = #(A X

{1 } u B X {2 } )

(i)

Show that if A nB = 0, then #(A) #(B) = #(Au B).

(ii)

Show that the operation is well-defined, i.e.,

#(A) 31.

and e are transcendental numbers.

Prove that the set T of transcendental numbers is non-denumerable.

#(A)#(B)

30.

7T

(i)

#(A') and #(B) = #(B')

implies

#(A)

#(B) = #(A')

#(B')

An operation of powers is defined for cardinal numbers as follows:

#(A) 4t(B) (i)

Show that if #(A) = m and

=

#({f : f:B— A})

#(B) = n are finite cardinals, then #(A)#(B) —

mIt

i.e. the operation of powers for cardinals corresponds, in the case of finite cardinals, to the usual operation of powers of positive integers.

(ii)

Show that the operation is well-defined, i.e., #(A) = #(A') and #(B) = #(B')

(iii)

Prove: For any set A,

implies

#(A)#(B ) = #(21')# ( 3 ')

#(cP(A))

32. Let — be the equivalence relation in R defined by x — y if x — y is rational. Determine the cardinality of the quotient set RI—.

33.

Prove: The cardinal number of the class of all functions from [0,1] into R is 2'.

34.

Prove that the following two statements of the Schroeder-Bernstein Theorem 3.8 are equivalent:

35.

(i)

If A -< B and B

(ii)

If XDYD X 1 and X

A, then A — B.

X 1 , then X

Y.

Prove Theorem 3.9: Given any pair of sets A and B, either A -
ORDERED SETS AND SUBSETS

36.

Let A = (N, the positive integers with the natural order; and let B = (N, =), the positive integers with the inverse order. Furthermore, let A X B denote the lexicographical ordering of N x N according to the order of A and then B. Insert the correct symbol, -‹ or >, between each pair of elements of N x N. (i) (3, 8)_ (1, 1), (ii) (2, 1

8), (iii) (3, 3)_ (3, 1),

(iv) (4, 9)_(7, 15).

46 37.

CARDINALITY, ORDER

Let X = {1,2,3,4,5,6} be ordered as in the adjacent diagram. Consider the subset A = {2, 3, 4 } of X. (i) Find the maximal elements of X. (ii) Find the minimal elements of X. (iii) Does X have a first element? (iv) Does X have a last element? (v) Find the set of upper bounds of A. (vi) Find the set of lower bounds of A. (vii) Does sup (A) exist? (viii) Does

[CHAP. 3

if 3 ............„/". ."`-............... 2 4

inf (A) exist?

5

7

38.

Consider Q, the set of rational numbers, with the natural order, and its subset A = {x : x E Q, x3 < 3 } . (i) Is A bounded above? (ii) Is A bounded below? (iii) Does sup (A) exist? (iv) Does inf (A) exist?

39.

Let N, the positive integers, be ordered by "x divides y", and let A c N. (ii) Does sup (A) exist?

40.

Prove:

41.

Give an example of an ordered set which has exactly one maximal element but does not have a last element.

42.

Prove: If R is a partial order on A, then R -1 is also a partial order on A.

(i) Does inf (A) exist?

Every finite partially ordered set has a maximal element.

ZORN'S LEMMA

43.

Consider the proof of the following statement: There exists a finite set of positive integers which is not a proper subset of any other finite set of positive integers. Proof. Let cAl be the class of all finite sets of positive integers. Partially order c../ by set inclusion. Now let B = {B i : i E I} be a totally ordered subclass of Consider the set A = u B. Observe that B, c A for every B l E B; hence A is an upper bound of B. Since every totally ordered subset of c./ has an upper bound, by Zorn's Lemma, c// has a maximal element, a finite set which is not a proper subset of another finite set.

Question:

Since the statement is clearly false, which step in the proof is incorrect?

44.

Prove the following fact which was assumed in the proof in Problem 24: Let {f A. B} be a class of functions which is totally ordered by set inclusion. Then u, f, is a function from u i A i into B.

45.

Prove that the following two statements are equivalent:

46.

The product

n {A i : i E I}

(i)

(Axiom of Choice.) non-empty.

of a non-empty class of non-empty sets is

(ii)

If eAr is a non-empty class of non-empty disjoint sets, then there exists a subset B c U {A : A E c./} such that the intersection of B and each set A E c./ consists of exactly one element.

Prove: If every totally ordered subset of an ordered set X has a lower bound in X, then X has a minimal element.

Answers to Supplementary Problems 32. c 36. (i) >, (ii) >, (iii) 37.

(iv) -‹

(i) {1 }; (ii) {5, 6}; (iii) No; (iv) Yes, 1;

(r) {1, 2}; (vi) {5, 6}; (vii) Yes, 2; (viii) No

38. (i) Yes, (ii) No, (iii) No, (iv) No 39.

(i) inf (A) exists if A

41.

a

(1).

(ii) sup (A) exists if A is finite.

1 —0- 2 —0-3 —0- 4 —0- • • • Here a is maximal but a is not a last element.

Chapter

4

Topology of the Line and Plane REAL LINE The set of real numbers, denoted by R, plays a dominant role in mathematics and, in particular, in analysis. In fact, many concepts in topology are abstractions of properties of sets of real numbers. The set R can be characterized by the statement that R is a complete, Archimedean ordered field. These notions are explained in the Appendix. Here we use the order relation in R to define the "usual topology" for R. We assume the reader is familiar with the geometric representation of R by means of the points on a straight line. As in Fig. 4-1, a point, called the origin, is chosen to represent 0 and another point, usually to the right of 0, to represent 1. Then there is a natural way to pair off the points on the line and the real numbers, i.e. each point will represent a unique real number and each real number will be represented by a unique point. For this reason we refer to the line as the real line or real axis. Furthermore, we will use the words point and number interchangeably. -2

0

-1

1

2

Fig. 4-1

OPEN SETS IN R Let A be a set of real numbers. A point p E A is an interior point of A if p belongs to some open interval Sp which is contained in A: p E S, c A

The set A is open (or 11-open) if each of its points is an interior point. of V in 71-open will appear in the next chapter.) Example 1.1: Example 1.2:

(The significance

An open interval A = (a, b) is an open set, for we may choose Sp = A for each p E A. The real line R, itself, is open since any open interval Sp must be a subset of R, i.e. E Sp c R.

p

Observe that a set is not open if there exists a point in the set that is not an interior point. Example 1.3:

The closed interval B = [a, b] is not an open set, for any open interval containing a or b must contain points outside of B. Hence the end points a and b are not interior points of B.

Example 1.4:

The empty set 0 is open since there is no point in 0 which is not an interior point.

Example 1.5:

The infinite open intervals, i.e. the subsets of R defined and denoted by

{x : x E R, x > a) = (a, co), {x : x E R, x < a} = (—cc , a), {x : x E RI = R are open sets. On the other hand, the infinite closed intervals, i.e. the subsets of R defined and denoted by

{x : x E R, x = a} = [a, co), {x : x E R, x a} = (---co, a] are not open sets, since a E R is not an interior point of either [a, co) or (—cc, al.

47

TOPOLOGY OF THE LINE AND PLANE

48

[CHAP. 4

We state two fundamental theorems about open sets.

Theorem 4.1:

The union of any number of open sets in R is open.

Theorem 4.2:

The intersection of any finite number of open sets in R is open.

The next example shows that the finiteness condition in the preceding theorem cannot be removed. Example 1.6:

Consider the class of open intervals and, hence, open sets = (-1/n, 1/n) : n G NI, i.e. {( -1 , 1 ), Observe that the intersection

n =1 A n

=

(-4, l), , ,

{0 }

of the open intervals consists of the single point 0 which is not an open set. other words, an arbitrary intersection of open sets need not be open.

In

ACCUMULATION POINTS Let A be a subset of R, i.e. a set of real numbers. A point p E R is an accumulation point or limit point of A if every open set G containing p contains a point of A different from p; i.e., G open, p E G implies A n (G {p}) The set of accumulation points of A, denoted by A', is called the derived set of A. Example 2.1:

Let A = {14, 4 , , ...}. The point 0 is an accumulation point of A since any open set G with 0 E G contains an open interval (—a l , a2) c G with —a, < 0 < a2 points of A. whicontas

4

—1_ 2

— et'

6, 2

I

I 4

Observe that the limit point 0 of A does not belong to A. Observe also that A does not contain any other limit points; hence the derived set of A is the singleton set {0 }, i.e. A' = {0}. Example 2.2:

Consider the set Q of rational numbers. Every real number p E R is a limit point of Q since every open set contains rational numbers, i.e. points of Q.

Example 2.3:

The set of integers Z = { , —2, —1, 0, 1, 2, . .1 does not have any points of accumulation. In other words, the derived set of Z is the empty set 0.

Remark:

The reader should not confuse the concept "limit point of a set" with the different, though related, concept "limit of a sequence". Some of the solved and supplementary problems will show the relationship between these two concepts.

BOLZANO-WEIERSTRASS THEOREM The existence or non-existence of accumulation points for various sets is an important question in topology. Not every set, even if it is infinite as in Example 2.3, has a limit point. There does exist, however, an important general case which gives a positive answer.

Theorem (Bolzano-Weierstrass) 4.3:

Let A be a bounded, infinite set of real numbers. Then A has at least one accumulation point.

CLOSED SETS A subset A of R, i.e. a set of real numbers, is a closed set if its complement Ac is an open set. A closed set can also be described in terms of its accumulation points.

Theorem 4.4:

A subset A of R is closed if and only if A contains each of its points of accumulation.

CHAP. 4]

TOPOLOGY OF THE LINE AND PLANE

49

(b,

Example 3.1:

The closed interval [a, h] is a closed set since its complement (--cc, a) u union of two open infinite intervals, is open.

Example 3.2:

The set A = {1, 12_, 1_, 1, ...} is not closed since, as seen in Example 2.1, 0 is a limit point of A but does not belong to A.

Example 3.3:

The empty set 0 and the entire line R are closed sets since their complements R and 0, respectively, are open sets.

),

the

Sets may be neither open nor closed as seen in the next example. Example 3.4: Consider the open-closed interval A = (a, b]. Note that A is not

open since b G A is not an interior point of A, and is not closed since a G A but is a limit point of A.

HEINE-BOREL THEOREM One of the most important properties of a closed and bounded interval is given in the next theorem. Here a class of sets, cA - = { Ai }, is said to cover a set A if A is contained in the union of the members of cA, i.e. A C u{ A,. Theorem (Heine-Borel) 4.5: Let A = [c, d] be a closed and bounded interval, and let Ç = {Gi: i c / } be a class of open intervals which covers A,. i.e. A C Ui G. Then g contains a finite subclass, say {Gil, • • • , which also covers A, i.e., A c Gil U G,2 U U Both conditions, closed and bounded, must be satisfied by A or else the theorem is not true. We show this by the next two examples. Example 4.1:

Consider the open, bounded unit interval A = (0,1). 1

G

Observe that the class

1

= {Gn =( n + 2' nj:nEN

of open intervals covers A, i.e.,

(1,1) u (i,-k) u —

A c (i-,1) u

?—='"? 4;,

0 But the union of no finite subclass of

Example 4.2:

Consider the closed infinite interval

G contains A.

A = [1, 00).

The class

G = {(4,2), (1,3), (2,4), • • •I of open intervals covers A, but no finite subclass does. (3,5) (1,3) = I

I

—2

—1

I

0

(2, 4) — ?

=9 1 !

1

4

ç 1

4 1

2

3

9

9

4

SEQUENCES A sequence, denoted by n EN or (sii) is a function whose domain is N = {1, 2, 3, .. }, i.e. a sequence assigns a point s. to each positive integer n E N. The image sn or s(n) of n E N is called the nth term of the sequence.

TOPOLOGY OF THE LINE AND PLANE

50 Example

5.1:

[CHAP.

4

The sequences

(sa) = (1, 3, 5, • • 0,

(4) =

—1{,

A, • • •),

(un) = (1, 0,1, 0, • • • )

can be defined, respectively, by the formulas

s(n) --= 2n — 1,

t(n) = (-1)fl/2n, u(n) = 1(1 + (-1)n+1

{

A sequence On : n Example

5.2:

1

if

0

if

)

n is odd n is even

e N) is said to be bounded if its range is,,: n E NI is a bounded set.

Consider the three sequences in Example 5.1. The range of (s,,) is {1,3, 5, ...}; so (sn) is not a bounded sequence. The range of (6,) is {-1, —i, ...} which is bounded; hence (tn) is a bounded sequence. The range of (u„) is the finite set {0,1 } ; so (u „) is also a bounded sequence.

Observe that (s.: n E N) denotes a sequence and is a function. N} denotes the range of the sequence and is a set.

On the other hand,

{s.: n e

CONVERGENT SEQUENCES The usual definition of a convergent sequence is stated as follows:

Definition:I The sequence (a1, a, ...) of real numbers converges to b G R or, equivalently, the real number b is the limit of the sequence (a.: n E N), denoted by lim an = b, lim an = b or an -> b if for every E >

n there exists a positive integer no such that

lan - bl <

n> no implies

E

Observe that cur,- b < E means that b - E
Definition:I The sequence (a,,: n E N) converges to b if every open set containing b contains almost all, i.e. all but a finite number, of the terms of the sequence. Example

6.1:

), such as (1, 1, 1, ... ) or (-3, —3, —3, ...), conA constant sequence (ao, a0, ao, verges to ao since each open set containing ao contains every term of the sequence.

Example

6.2:

Each of the sequences

(1 ,1, (1 , 0, 1, 0, 0, 0, ( 1 , — 1, A-, converges to 0 since any open interval containing 0 contains almost all of the terms of each of the sequences. Example

6.3:

Consider the sequence

11- II,

A,

...), i.e. the sequence

1

{

1

n is even

1

2(n+1)/2 if

n is odd

The points are displayed below:

a4

9

0

a2

ai

a3

1

1

3 4

I' 1

a5

11 9 4

2

1

Observe that any open interval containing either 0 or 1 contains an infinite number of the terms of the sequence. Neither 0 nor 1, however, is a limit of the sequence. Observe, though, that 0 and 1 are accumulation points of the range of the sequence, that is, of the set {1,1,1, 4,i, ...}.

CHAP. 4]

51

TOPOLOGY OF THE LINE AND PLANE

SUBSEQUENCES If (i,„) is a sequence of positive integers such that

Consider a sequence (ai, a2, a3, then

< i2 < • •

ai2 , a15 , . . .

is called a subsequence of (a.: n e Is1). Example 7.1:

Consider the sequence (a n) = (1, • • • ). Observe that (1, I, A-, ...) is a subsequence of (an), but that (4,1, if if tf if • • ) is not a subsequence of (an) since 1 appears before in the original sequence.

Example 7.2:

Although the sequence (4, ...) of Example 6.3 does not converge, it does have convergent subsequences such as A, •• ..) and (21,-2-, 1,, a • ..). On the other hand, the sequence (1, 3, 5, ... ) does not have any convergent subsequences.

+,

As seen in the preceding example, sequences may or may not have convergent subsequences. There does exist a very important general case which gives a positive answer.

Theorem 4.6:

Every bounded sequence of real numbers contains a convergent subsequence.

CAUCHY SEQUENCES A sequence (an : n G 1s ) of real numbers is a Cauchy sequence iff for every exists a positive integer n o such that

n,m >

no

implies Ian — (toil <

E>

0 there

E

In other words, a sequence is a Cauchy sequence if the terms of the sequence become arbitrarily close to each other as n gets large. Example 8.1:

Let (a, n E N) be a Cauchy sequence of integers, i.e. each term of the sequence belongs to Z = {..., —1,0,1, ...}. Then the sequence must be of the form

(a1 , a2,

ant), b, b, b,

p)

i.e. the sequence is constant after some noth term. For if we choose e =

an,a,„ E Z and lan — am l < Example 8.2:

implies a, = am

We show that every convergent sequence is a Cauchy sequence. e> 0. Then there exists n o E N sufficiently large such that

n > no implies l an — bl < le

and

then

in

Let an —> b and let

> no implies lam — bl <

Consequently, n,m > no implies la,— a m l

Ian — b b — and

lan — bl + lb — am

l < 1e +

e

2 —

Hence (an) is a Cauchy sequence.

COMPLETENESS A set A of real numbers is said to be complete if every Cauchy sequence (a. EA:ne 1% ) of points in A converges to a point in A. Example 9.1:

The set Z = { —2, —1, 0, 1, 2, .. .} of integers is complete. For, as seen in Example 8.1, a Cauchy sequence (a n : n E N) of points in Z is of the form

(a1 , a2,

ano, b,

bp

b, ...)

which converges to the point b E Z. Example 9.2:

The set Q of rational numbers is not complete.

For we can choose a sequence of rational numbers, such as (1, L4,1.41,1.412, ...) which converges to the real number 12-, which is not rational, i.e. which does not belong to Q.

52

[CHAP. 4

TOPOLOGY OF THE LINE AND PLANE

A fundamental property of the entire set R of real numbers is that R is complete. Namely, Theorem (Cauchy) 4.7:

Every Cauchy sequence of real numbers converges to a real number.

CONTINUOUS FUNCTIONS The usual

Definition:

I

E -

8 definition of a continuous function is stated as follows:

A function f : R ---> R is continuous at a point xo if for every a 8 > 0 such that — x0 < 8 implies 11(x)— f(x0)I< E

> 0 there exists

The function f is a continuous function if it is continuous at every point. Observe that Ix — xol < 8 means that x o — 8 < x < x 0 + 8, or equivalently that x belongs to the open interval (xo — 8, xo + 8). Similarly, 11(x) — f(x0)1 < E means that 1(x) belongs to the open interval (f(x o) — c, f (x 0) + f). Accordingly, the statement

— x 0( < 8

implies

(f(x)— f(x o)( <

is equivalent to the statement x E (xo

—

8, xo + 8)

f(x) E (1(x 0)—

implies

E,

f(X0)

which is equivalent to the statement f[(x o — 8, xo + 8)1

is contained in

(1(x0)—

E,

1(X0)

Hence we can restate the previous definition as follows.

Definition: I

A function f: R ---> R is continuous at a point p E R if for any open set Vf(p) containing f(p) there exists an open set Up containing p such that ff Up] C Vf(p). The function f is a continuous function if it is continuous at every point.

The Venn diagram below may be helpful in visualizing this definition.

A continuous function can be completely characterized in terms of open sets as follows:

Theorem 4.8:

A function is continuous if and only if the inverse image of every open set is open.

Observe that Theorem 4.8 also states that a function is not continuous if there exists an open set whose inverse image is not open. Example 10.1:

Consider the function f:R—>R

1(x) =

x -1 if x

defined by 3

if x > 31(x+5)

and illustrated in the adjacent diagram. Note that the inverse of the open interval (1, 3) is the open-closed interval (2, 31 which is not an open set. Hence the function f is not continuous.

CHAP. 4}

TOPOLOGY OF THE LINE AND PLANE

53

We now state one important property of continuous functions which we will refer to later in the text.

Theorem 4.9:

Let f R --> R be continuous on a closed interval [a, b]. assumes every value between f(a) and f(b).

In other words, if yo is a real number for which f(a) then

3x 0 E R

such that

a

yo

x o b and f(x 0 )

Then the function

f(b) or f(b)

yo 1(a),

yo

This theorem is known as the Weierstrass Intermediate Value Theorem.

Remark:

A function f: R --> R is said to be continuous on a subset D of R if it is continuous at each point in D.

TOPOLOGY OF THE PLANE An open disc D in the plane R2 is p (a i , a2) and radius 8 > 0, i.e.,

D

= ((x, : (x —

the

set of points inside a circle, say, with center

air + (y — a 2)2 < 8 2 1

{q E R2 : deD,

<

Here d(p, q) denotes the usual distance between two points p = (a l, a2) and q (b1, b2) in R2 :

d(p,q) = V(ai— bi) 2 + (a2 — b2) 2 The open disc plays a role in the topology of the plane R2 that is analogous to the role of the open interval in the topology of the line R. Let A be a subset of R2. A point p e A is an interior point of A iff p belongs to some open disc D, which is contained in A: p D, c The set A is open (or If-open) if each of its points is an interior point. Example 11.1:

Clearly an open disc, the entire plane R2 and the empty set 0 are open subsets of R2. We now show that the intersection of any two open discs, say

D1 =

{ q E R2 : d(pi , q) < 8 1 }

is also an open set.

For let po E D i

and

n

D2 =

{ q E R2 d(P2, q) < 62}

D2 SO

d(pi, Po) < S i and

d(P2, Po) < 8 2

Po

r = min {SI — d(Pi) PO) 8 2 — d(P2) Po

Set and let Then

D

)}

> 0

= Iq E R2 : d(po, q) <

po E D c D i n D2 or, po is an interior point of D I n D2-

A point p e R2 is an accumulation point or limit point of a subset A of R2 if every open set G containing p contains a point of A different from p, i.e., implies A n (G\ {p}) 9) G C R2 open, p E G

[CHAP. 4

TOPOLOGY OF THE LINE AND PLANE

54 Example 11.2:

Consider the following subset of R2:

= {(x, y) : y = sin -I- , x > 0}

A

The set A is illustrated in the adjacent diagram. Observe that the curve, going from right to left, fluctuates faster and faster, i.e. that the points where the curve crosses the x-axis become (0,4) closer and closer. The point p is a limit point of A since A will eventually pass through any open disc containing p. In fact, each point on the y-axis between —1 and 1, i.e. each point in the set

(

B = {(x, y) : x = 0, —1 y 1} -1 1

is a limit point of A.

A subset A of R2 is closed if its complement Ac is an open subset of R2 . A sequence (p1, p2, of points in R2 converges to the point g E R2 if every open set containing g contains almost all of the terms of the sequence. Convergence in the plane R2 can be characterized in terms of convergence in R as follows.

Proposition 4.10:

Consider the sequence (p 1 = the point g (a, b) E R2 . Then pn --> g

202= (a2,b4,

if and only if

of point in R2 and

an --> a and bn ---> b

A function f: R2 --> R2 is continuous at a point p E R2 if for any open set ing f(p) there exists an open set Up containing p such that f[Up] C Vf(p).

Vf(p)

contain-

We list theorems for the plane R2 which are analagous to theorems for the line R stated earlier in this chapter.

Theorem 4.1*:

The union of any number of open subsets of R2 is open.

Theorem 4.2*:

The intersection of any finite number of open subsets of R2 is open.

Theorem 4.4*:

A subset A of R2 is closed if and only if A contains each of its accumulation points.

Theorem 4.8*:

A function f: R2 ---> R2 is continuous if and only if the inverse image of every open set is open.

Solved Problems OPEN SETS, ACCUMULATION POINTS 1.

Determine the accumulation points of each set of real numbers: (i) N;

(ii) (a, bl; (iii) Qe, the set of irrational points.

Solution:

(i)

N, the set of positive integers, does not have any limit points. For if a is any real number, we can find a S > 0 so small that the open set (a — 8, a + 8) contains no point of N other than a.

(ii)

Every point p in the closed interval [a, b ] is a limit point of the open-closed interval (a, b}, since every open interval containing p E [a, b ] will contain points of (a, bl other than p.

(iii)

Every real number p E R is a limit point of Qe since every open interval containing p E R will contain points of Qe, i.e. irrational numbers, other than p.

CHAP. 4]

2.

Recall that A' denotes the derived set, i.e. set of limit points, of a set A. Find sets A such that (i) A and A' are disjoint, (ii) A is a proper subset of A', (iii) A' is a proper subset of A, (iv) A = A'. Solution: (i) The set A = ...} A and A' are disjoint. (ii) (iii)

(iv)

has 0 as its only point of accumulation.

Hence

Let A 7-- (a, h.], an open-closed interval. As seen in the preceding problem closed interval, and so A c A'. Let A = {0,1, A' = {0}

3.

55

TOPOLOGY OF THE LINE AND PLANE

...}.

A' n"--- {0}

and

A' = [a, b], the

Then 0, which belongs to A, is the only limit point of A.

Hence

and A' c A.

Let A ---- [a, b], a closed interval. Then each point in A is a limit point of A and they are the only limit points. So A A' [a, b ] .

Prove Theorem 4.1*:

The union of any number of open subsets of R2 is open.

Solution: Let c// be a class of open subsets of R2, let H = U{G: G G cc-4}, and let p E H. The theorem is proved if we show that p is an interior point of H, i.e. there exists an open disc D p containing p such that D, is contained in H.

p E H = L1 {G: G E cA1 }, aGo cA1 such that p E G o hence there exists But Go is an open set; an open disc D, containing p such that Since

p G D, C Go

Since G o is a subset of H -= U{G: G E 0AL Dp is also a subset of H.

4.

Prove:

Thus H is open.

Every open subset G of the plane R2 is the union of open discs.

Solution: Since G is open, for each point pEG there is an open disc D, such that G= U{Dp : p E G}.

5.

Prove Theorem 42*: is open. Solution:

pE D, c G.

Then

The intersection of any finite number of open subsets of R2

We prove the theorem in the case of two open subsets of R2. The theorem will then follow by induction. Let G and H be open subsets of R2 and let pEG n H; so peG and pEH. Hence there exist open discs D I and D2 such that p D i c G and p E D2 C H Then pED 1 n D2 C G n H. By Example 11.1, the intersection of any two open discs is open; so there exists an open disc D such that

pEDcD 1 nD2 cGnH Hence p is an interior point of G

6.

n

H and, so, G

n

H is open.

Prove: Let p e G, an open subset of R2. Then there exists an open disc D with center p such that p ED c G. Solution: By definition of an interior point, there exists an open disc

D I = {q G R2 : d(p t ,q) < 8}, with center p i and radius 8, such that p G D 1 c G. So d(p i ,p) < 8. Set r = 8 d(p i , p) > 0 —

and let

D

=

1g G R2 : d(p, q) < irl

Then, as indicated in the diagram, pED c D I c G where D is an open disc with center p.

7.

[CHAP. 4

TOPOLOGY OF THE LINE AND PLANE

56

Prove: Let p be an accumulation point of a subset A of the plane R2 . open set containing p contains an infinite number of points of A.

Then every

Solution: Suppose G is an open set containing p and containing only a finite number of points, say a t , ..., am , of A different from p. By the preceding problem, there exists an open disc D, with center p and, say, radius 8 such that p G D, c G. Choose r > O to be less than 8 and less than the distance from p to any of the points at , ..., am; and let D = {q G R2 : d(p, q) < --r} Then the open disc D containing p does not contain a t, ..., am and, since D c D, c G, does not contain any other points of A different from p. The last statement contradicts the fact that p is a limit point of A. containing p contains an infinite number of points of A.

Hence every open set

Remark: A similar statement is true for the real line R, i.e. if a G R is a limit point of A cR, then every open subset of R containing a contains an infinite number of points of A.

8.

Prove: Consider any open disc D, with center p E R2 and radius 8. Then there exists an open disc D such that (i) the center of D has rational coordinates, (ii) the radius of D is rational, and (iii) p ED c D. Solution: Suppose p c and d such that

(a, h.).

a < c < a +

Then there exist rational numbers and

b
Let q = c, d). Note that d(p,q) < *8. Now choose a rational number r such that i8 < r < 18; and let D be the open disc with center q, which has rational coordinates, and radius r which is rational. Then, as indicated in the diagram, pED c D.

9.

Prove: Every open subset G of the plane R2 is the union of a countable number of open discs. Solution: Since G is open, for each point pEG there exists an open disc Dp with center p such that pED, c G. But, by the preceding problem, for each disc Dp there exists an open disc Ep such that (i) the center of Ep has rational coordinates, (ii) the radius of Ep is rational, and (iii) p EEp c Dr,. SO

pG

Ep c D p c G

G = U{Ep : p G

Accordingly,

The theorem now follows from the fact that there are only a countable number of open discs whose center has rational coordinates and whose radius is rational.

10. Prove Theorem (Bolzano-Weierstrass) 4.3: Let A be a bounded infinite set of real numbers. Then A contains at least one accumulation point. Solution: Since A is bounded, A is a subset of a closed interval It = [al , b1 ] . that both of the closed subintervals of It , [a t ,

(cc t

b1)]

and

[4-(a1 +

Bisect I t at -1-(a1 + b 1 ). Note

b 1 ), b 1 ]

cannot contain a finite number of points of A since A is infinite. intervals in (1) which contains an infinite number of points of A.

Let

(1)

== [a2, b2]

be one of the

CHAP. 4]

Now bisect 12.

TOPOLOGY OF THE LINE AND PLANE

57

As before, one of the two closed intervals

[a2, -1(a 2 + b2)] must contain an infinite number of points of A.

[1(a2 b2 ), b 2 ]

and

Call that interval 1 3 .

Continuing this procedure we obtain a sequence of nested closed intervals D /2 D

13 D

such that each interval In contains an infinite number of points of A and

lim l ,

7-- 0

where /n i denotes the length of the interval In . By the Nested Interval Property of the real numbers (see Appendix A), there exists a point in each interval I. We show that p is a limit point of A and then the theorem will follow. Let Sp = (a, b)

be an open interval containing p.

3n0 G N

such that

Since lim i/n = 0,

1/„0 < min (p — a, b— p)

a subset of the open interval S p = (a, b)

Then the interval I

as indicated in the diagram below.

/no a Since / no contains an infinite number of points of A, so does the open interval S p . Thus each open interval containing p contains points of A other than p, i.e. p is a limit point of A.

CLOSED SETS 11. Prove: A set F is closed if and only if its complement Fe is open. Solution: Note that (Fey

F; so F is the complement of Fe. Thus, by definition, F is closed if Fe is open.

12. Prove: The union of a finite number of closed sets is closed. Solution: Let F1 , ..., Fm be closed sets and let F = F1 U - • - UFni .

By DeMorgan's Law,

Fe = (F1 U • • u Fm)e = Fet n F; n • So Fe is the intersection of a finite number of open sets F complement Fee F is closed.

,

and thus Fe is also open.

Hence its

13. Prove: The intersection of any number of closed sets is closed. Solution: Let {F,} be a class of closed sets and let F = Fe =

n

(n,F,)e

F.

By DeMorgan's Law,

uF

So Fe is the union of open sets and, hence, is open itself. Consequently, Fee = F is closed.

14. Prove Theorem 4.4*: accumulation points.

A subset of R2 is closed if and only if it contains each of its

Solution: Suppose p is a limit point of a closed set F. Then every open disc containing p contains points of F other than p. Hence there cannot be an open disc Dp containing p which is completely contained in the complement of F. In other words, p is not an interior point of Fe. But Fe is open since F is closed; so p does not belong to Fe, i.e. p G F. On the other hand, suppose a set A contains each of its limit points. We claim that A is closed or, equivalently, that its complement A C is open. Let p G AC. Since A contains each of its limit points, p is not a limit point of A. Hence there exists at least one open disc Dp containing p such that Dp does not contain any points of A. So Dp c Ac, and hence p is an interior point of Ac. Since each point p Ac is an interior point, Ac is open and so A is closed.

58

[CHAP. 4

TOPOLOGY OF THE LINE AND PLANE

15. Prove: The derived set A', i.e. set of accumulation points, of an arbitrary subset A of R2 is closed. Solution: Let p be a limit point of A'. By Theorem 4.4*, the theorem is proved if we show that p E A', that is, that p is also a limit point of A. Let G„, be an open set containing p. Since p is a limit point of A', Gp contains at least one point q E A' different from p. But G0 is an open set containing q G A'; hence G0 contains (infinitely many) points of A. So, 3a G A p, a q, and a E G0 such that a That is, each open set containing p contains points of A other than p; so p E A'.

16. Prove: Let A be a closed and bounded set of real numbers and let sup (A) = p. p E A.

Then

Solution: Suppose p G A. Let G be an open set containing p. Then G contains an open interval (b, c) containing p, i.e. such that b < p < c. Since sup (A) = p and p E A,

3a E A

such that

b
for otherwise b would be an upper bound for A. So a E (b, c) C G. Thus each open set containing p contains a point of A different from p; hence p is a limit point of A. But A is closed; hence, by Theorem 4.4*, p E A.

17. Prove Theorem (Heine-Borel) 4.5: Let /1 = [CI, d1] be covered by a class g = {(ai, b i): i E / } of open intervals. Then Ç contains a finite subclass which also covers Solution: Assume that no finite subclass of G covers I. We bisect I I =- [c i , d11 at +(el + d 1) and consider the two closed intervals (1) 1(c 1+ d1)1 and [1(c1+ dt), dti At least one of these two intervals cannot be covered by a finite subclass of G or else the whole interval I I will be covered by a finite subclass of G. Let 12 [c2, d2} be one of the two intervals in (1) which cannot be covered by a finite subclass of G. We now bisect 12 . As before, one of the two closed intervals [ 2 , 14 c2 + d2)

cannot be covered by a finite subclass of G.

and

[4(c2 +

d2) I

C1 2]

Call that interval 1 3 .

We continue this procedure and obtain a sequence of nested closed intervals I I 0 12 0 13 0 such that each interval In cannot be covered by a finite subclass of G and lim 141 = 0 where denotes the length of the interval I.

14 1

By the Nested Interval Property of the real numbers (see Appendix), there exists a point p in each interval In . In particular, p E I. Since G is a cover of 1 1 , there exists an open interval (a, b. ) in G which contains p. Hence a < p < b . Since lim 141 = 0,

lo

3no G N

io

such that

i/no i < min (p — ctio, b io

—

p)

Then, as indicated in the diagram below, the interval 40 is a subset of the one interval (ail), bio) in G. Ino

aio But this contradicts our choice of / . Thus the original assumption that no finite subclass of G no covers / 1 is false and the theorem is true.

CHAP. 4}

TOPOLOGY OF THE LINE AND PLANE

59

SEQUENCES 18. Write the first six terms of each of the following sequences:

(i)

s(n)

{ n - 1 if n is odd n2 if n is even

(ii)

1 2 t( - 1) + t(n - 2)

t(n) =

if n = 1 if n = 2 if n> 2

Solution: (i)

Two formulas are used to define this function. Substitute 1, 3 and 5 into s(n) = n -1 to get s 1 = 0, s3 ----- 2 and 3 5 = 4. Then substitute 2, 4 and 6 into s(n) = n2 to get 32 = 4, 3 4 = 16 and 3 6 = 36. Thus we have (0, 4, 2, 16, 4, 36, ... ).

(ii)

Here the function is defined recursively. Each term after the second is found by adding the two previous terms. Thus: t1 = 1

t4 = t3

t2 = 2

t5 =

ta Hence we have

t 2 + t i -= 2 + 1 =-- 3

t4

t2 t3

3 + 2 -= 5

- 5 + 3 -= 8

77

t6 = t5 + t4 = 8 + 5 = 13

(1, 2, 3, 5, 8, 13, . ..).

19. Consider the sequence (an = (-1)n -1 (2n - 1)): (1, -3, 5, - 7, 9, -11 , 1 3 , - 15, Determine whether or not each of the following sequences is a subsequence of (an). (i)

(bn) = (1, 5, -3, -7, 9, 13, -11, -15, . . .

(ii)

(cn) = (1, 3, 5, 7, 9, 11, 13, .. . )

(iii) (dn) ------ (-3, -7, -11, -15, -19, -23, . ) Solution: (i)

Note that 5 appears before -3 in (b,), but -3 appears before 5 in (an). subsequence of (an).

(ii)

The terms 3, 7 and 11 do not even appear in (an); hence (c a) is not a subsequence of (an).

(iii)

The sequence (du) is a subsequence of (an), for (in = 2n) = (2, 4, 6, . ) • ; so integers such that i t < i2 < i3 <

(a2 , a4 , a6 ,

...)

Hence (b i ) is not a

is a sequence of positive

,) = (-3, -7, -11, ...)

is a subsequence of (a„).

20. Determine the range of each sequence: 1,k, 1, k, (ii) (1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,

(i)

(1,

(iii) (2, 4, 6, 8, 10, . .

1,

)

Solution: The range of a sequence is the set of image points.

(i) {1,

+, I,

},

21. Prove: If the range of a sequence subsequence. Solution:

(ii) {1, 0, -11,

(an)

Hence the ranges of the sequences are

(iii) {2, 4, 6, 8, . • • }

is finite, then the sequence has a convergent

If the range {an } of (an) is finite, then one of the image points, say b, appears an infinite number of times in the sequence. Hence (b,b,b,b, ...) is a subsequence of (an) and it converges.

TOPOLOGY OF THE LINE AND PLANE

60

[CHAP.

4

b and lim a„ = c, then b = c.

22. Prove: If lim a

Solution: lb — ci > 0. Then the open intervals B = (b 0, Suppose that b and c are distinct. Let 8 b 1.8) and C =-- (c c 1.8), containing b and c respectively, are disjoint. Since (an) converges to b, B must contain all except a finite number of the terms of the sequence. Hence C can only contain a finite number of the terms of the sequence. But this contradicts the fact that (a„) converges to c. Accordingly, b and c are not distinct.

23. Prove: If the range {an} of a sequence (a n) contains an accumulation point b, then the sequence (a„) contains a subsequence (a,n) which converges to b. Solution: Since

b is a limit point of {a n }, each of the open intervals S2 = (b - 4, b ) , S3 Si = (b —1, b 1),

contains an infinite number of elements of the set {a n } and, hence, an infinite number of the terms of the sequence (an). We choose a sequence (a,) as follows: Choose a il to be a point in Si . Choose

a 12 to be a point in S2 such that i2 > j1 i.e. such that a,2 appears after a11 in the sequence (an).

Choose

a13 to be a point in

,

S3 such that

i3

> i2.

We continue in the same manner.

Observe that we are always able to choose the next term in the sequence (ao since there are an infinite number of the terms of the original sequence (an) in each interval S n . We claim that (ain) satisfies the conditions of the theorem. Recall that we choose the terms of the sequence (a, n) so that j 1 < i2 < is < • ; hence (aid is a subsequence of (an). We need to show that lirn a, , = b, Let G be an open set containing b. Then G contains an open interval (d i , d2) containing b; so d1 < b

d2. Let 8 = min (b — di , d2 — b) > 0; then no G N

Hence

.5,20 c (d i , d2) c G,

such that 1/n 0

< 8

and so

n > n o implies a 1 ES c Sno c Thus G contains almost all the terms of the sequence (a,n); that is,

d2) c G

lim a

b.

24. Prove Theorem 4.6: Every bounded sequence (a.) of real numbers contains a convergent subsequence. Solution: Consider the range {an } of the sequence (an). If the range is finite, then by Problem 21 the sequence contains a convergent subsequence. On the other hand, if the range is infinite, then, by the Bolzano-Weierstrass Theorem, the bounded infinite set {an } contains a limit point. But then, by the previous problem, the sequence in this case also contains a convergent subsequence.

25. Prove: Every Cauchy sequence (a.) of real numbers is bounded. Solution: Let e z-- 1. Then, by definition of a Cauchy sequence, alto G

In particular, Let

N

such that

ni no implies a

n,m

n o implies

ia n — am j < 1

jano a m l < 1, or, an() — 1 < am < ano + 1 = max (a i , a2,

ano , ano + 1)

=-- min (a i , a2,

ano, ano — 1)

Then a is an upper bound for the range {an } of the sequence (an) and [3 is a lower bound. Accordingly, (an) is a bounded sequence.

CHAP.

4]

61

TOPOLOGY OF THE LINE AND PLANE

26. Prove: Let (a,t) be a Cauchy sequence. If a subsequence (a,n) of (a„) converges to a point b, then the Cauchy sequence itself converges to b. Solution: Let e > O.

We need to find a positive integer no such that

n > n o implies

c:c.„ — b < e

Since (an ) is a Cauchy sequence,

3n0 E N

n,m > no implies

such that

la„ — am l <

Also, since the subsequence (a i ) converges to b,

GN Observe that we can choose

such that !a im — bl <

im so that im > n o . Accordingly,

n> no implies la,

— bI

— aim -I- a im — bl

=

— aim ! + a1 — b! le + an

< Hence (a n ) converges to b. Observe that we need im

> no in order to state that:

27. Prove Theorem (Cauchy) 4.7: to a real number.

n > no implies a„,— aH <

Every Cauchy sequence (a of real numbers converges

Solution : By Problem 25, the Cauchy sequence (a,,) is bounded. Hence, by Theorem 4.6, the bounded sequence (a„) contains a convergent subsequence (a,„ ). But, by the preceding problem, the Cauchy sequence (a„) converges to the same limit as its subsequence (a, n). In other words, the Cauchy sequence

(a,,,) converges to a real number.

28. Determine whether or not each of the following subsets of R is complete: (i) N, the set of positive integers; (ii) Qc, the set of irrational numbers. Solution: (i)

Let (a„) be a Cauchy sequence of positive integers. If

E

-I2 ' then

— am ! < E implies an 7-- am Therefore, the Cauchy sequence (a„) is of the form (a 1 , a2 , a no, b, b, b, ... to the positive integer b. Hence N is complete. (ii)

which converges

Observe that each of the open intervals ( - 1,

1), ( -4, 4), ( - 1, I), • • • contains irrational points. Hence there exists a sequence (a„) of irrational numbers such that an belongs to the open interval (-1/n, 1/n). The sequence (a„) will be a Cauchy sequence of points in Qt and it will converge to the rational number O. Hence Qc is not complete.

CONTINUITY 29. Prove: If the function f: R R is constant, say f(x) = a for every x E R, then

is continuous. Solution:

f

Method 1. The function f is continuous if the inverse f -1 [G] of any open set G is also open. f(x) = a for every x G R, G if a f [G] {(116' if a G G

Since

—

for any open set G. In either case,

f -1 [G] is open since both R and Ø are open sets.

Method 2. We show that f is continuous at any point xo using the > O. Then for any 8 > 0, say 8 = 1,

— xo l < 1 Hence f is continuous.

implies

If(x)

E

8 definition of continuity. Let

= — = O<e

62

[CHAP. 4

TOPOLOGY OF THE LINE AND PLANE

30. Prove: The identity function continuous. Solution:

that is, the function defined by 1(x) x, is

f: R -> R,

Method 1. Let G be any open set. Then f -1 [G] = G is also an open set. Accordingly, Method 2. We show that f is continuous at any point xo using the > O. Then choosing E 8,

E-

f is continuous.

8 definition of continuity. Let

e

Ix

—

4<8

1f(x)— f(x 0)1 = Ix — xo

implies

<

8

E

Accordingly, f is continuous.

31. Prove: Let the functions f : R -> R and g : R -> R be continuous. Then the composition function g of : R -> R is also continuous. Solution: We show that the inverse (g 0 -1 [G] of any open set G is also open. Since g is continuous, the inverse g -1 [G] is an open set. But since f is continuous, the inverse f -1 [g -1 [Gil of g -1 [G] is also open. Recall that (g f) - 1

(g 0 .0- 1 [G] =(f 0 g -1)[G] -

Hence is an open set.

f—1 [g -1

gof: R R

Thus the composition function

[Gil

is continuous.

32. Prove: Let f : R -> R be continuous and let f(q) = 0 for every rational number Then f(x) = 0 for every real number x G R.

E Q.

Solution: Suppose f(p) is not zero for some real number p G R, i.e. suppose

3p G R Choose

E

117 .

>0

f(p) = y,

such that

Since f is continuous,

38 -> O

such that

x — p1 < 8

implies

f(x) —

f(P)1 <

E

1-171

Now there are rational points in every open interval. In particular,

3q E Q which implies

such that

1f(q) — .f(1))1

q G {x : !x — p < )

1.f(P)1 = 171 <

1•17

an impossibility. Hence f(x) = O for every x G R.

33. Prove Theorem 4.8:

A function f: R2 -> R2 is continuous if and only if the inverse

image of every open set is open. Solution: Let f : R2 —> R2 be continuous and let V be an open subset of R2 . We want to show that f -1 [V] is also an open set. Let p E f 1 [V]. Then f(p) e V. By definition of continuity, there exists an open set Up containing p such that f[U] c V. Hence (as indicated in the diagram below)

C

f

[f[171,]1 C f — I [V]

We have shown that, for every point p G f 1 [V], there exists an open set Up such that

p E U c f -1 [V] Accordingly,

f - I [V]

= U {U

pef —1 [V] }

So f -1 [V] is the union of open sets and is, therefore, open itself.

63

TOPOLOGY OF THE LINE AND PLANE

CHAP. LI]

On the other hand, suppose the inverse of every open set is open. We want to show that f is continuous at any point p G R. Let V be an open set containing f(p), i.e. f(p) e V. Then f -1 -17] is an open set containing p with the property that f[f — I [V]] c V. Hence f is continuous at p.

34. Give an example of two functions f: R —> R and g: R R such that f and g are each discontinuous (not continuous) at every point and such that the sum f + g is continuous at every point in R. Solution: Consider the functions f and g defined by f(x)

J O if x is rational

if x is irrational

g(x) = {

1 0

if x is rational if x is irrational

The functions f and g are discontinuous at every point in R, but the sum f -I- g is the constant function (f g)(x) = 1 which is continuous.

35. Prove: Let the function f : R —> R be continuous at a point p E R. (i)

If f(p) is positive, i.e. f(p) > 0, then there exists an open interval S containing p such that f is positive at every point in S.

(ii)

If f(p) is negative, i.e. f(p) < 0, then there exists an open interval S containing p such that f is negative at every point in S. Solution: We prove (i). The proof of (ii) is similar and will be omitted. Suppose f(p) = E > 0. continuous at p, 38 > 0 such that — Pi < 8 implies If(x) — f(P)I < e or, equivalently, x G (p

8, p 8)

implies

f(x) G (f(p) — E, f(p)

Since f is

= (0, 2e)

Thus for every point x in the open interval (p — 8, p 8), f(x) is positive.

36. Prove: Let f: R R be continuous at every point in a closed interval [a, b], and let f(a) < 0 < f(b). Then there exists a point p E [a, b] such that f(p) = O. (In other words, the graph of a continuous function defined on a closed interval which lies both below and above the x-axis must cross the x-axis at at least one point, as indicated in the diagram.) Solution: Let A be the set of points in [a, b] at which f is negative, i.e., A

= {x : x

[a,b], f(x) < 0}

Observe that A is not empty since, for example, a G A. Let p = sup (A) be the least upper bound b. So p belongs to the interval for A. Since a GA, a p; and since b is an upper bound for A, p [a, b].

64

[CHAP. 4

TOPOLOGY OF THE LINE AND PLANE

(p

We claim that f (p) O. If f(p) < 0, then, by the preceding problem, there is an open interval 8, p + 8) in which f is negative, i.e.,

(p — 8, p

8) c A

So p cannot be an upper bound for A. On the other hand, if f(p) > 0, then there exists an interval (p — 8, p ± 8) in which f is positive; so

(p — , p + 8) which implies that p cannot be

Remark.

a

nA =

least upper bound for A.

Thus f (p) can only be zero, i.e. f (p)

O.

The theorem is also true and proved similarly in the case f(b) < 0 < f(a).

37. Prove Theorem (Weierstrass) 4.9: Let f: R > R be continuous on a closed interval la, bl. Then the function assumes every value between f(a) and f(b). Solution: —

Suppose f(a) < f(b) and let y o be a real number such that f(a) < y o < f(b). We want to prove that there is a point p such that f(p) y o . Consider the function g(x) /(x) y o which is also continuous. Observe that g(a) < O < g(b). By the preceding problem, there exists a point p such that g(p)

f(p) — y o

O. Hence f(P)

Yo•

The case when f(b) < f (a) is proved similarly.

Supplementary Problems OPEN SETS, CLOSED SETS, ACCUMULATION POINTS 38. Prove: If A is a finite subset of R, then the derived set A' of A is empty, i.e. A' = Ø. 39. Prove:

Every finite subset of R is closed.

40. Prove: If A cB, then A' cB' .

41, Prove: A subset B of R2 is closed if and only if inf {d(p, q) : q B}.

d(p, B)

0

implies

p G B,

where

d(p, B) =

42. Prove:

Au A' is closed for any set A.

43. Prove:

A UA' is the smallest closed set containing A, i.e. if F is closed and AcFcAuA'

then

44, Prove: The set of interior points of any set A, written int (A), is an open set. 45.

Prove:

The set of interior points of A is the largest open set contained in A, i.e. if G is open and

int (A) c G c A, then int (A) G. 46. Prove:

The only subsets of R which are both open and closed are 0 and R.

SEQUENCES

47. Prove: If the sequence (an) converges to b GR, then the sequence

bl ) converges to O.

48. Prove: If the sequence (a n) converges to 0, and the sequence (b n ) is bounded, then the sequence (ab) also converges to O. 49. Prove: If a n a and

b, then the sequence (a„ + bn) converges to a + b.

50, Prove: If art a and b ,, —> b, then the sequence (a,„b„„) converges to ab.

CHAP. 41

TOPOLOGY OF THE LINE AND PLANE

a and bn —> b where bn 0 and b

65

51.

Prove:

If a,

52.

Prove:

If the sequence (ai,) converges to b, then every subsequence (aiu) of (a„) also converges to b.

53.

Prove: If the sequence (an) converges to b, then either the range {an } of the sequence (an) is finite, or b is an accumulation point of the range fo it ).

54.

Prove: If the sequence (an) of distinct elements is bounded and the range {an } of (an) has exactly one limit point b, then the sequence (a n) converges to b.

(Remark: The sequence (1,1., 2, 3, 1, 4, , ,) removed from this theorem.)

0, then the sequence (an/bn) converges to alb.

shows that the condition of boundedness cannot be

CONTINUITY 55.

Prove: A function f R —> R is continuous at a E R if and only if for every sequence (an) converging to a, the sequence (f(a n)) converges to f(a).

56.

Prove: Let the function f:R—>R be continuous at p G R. Then there exists an open interval S containing p such that f is bounded on the open interval S.

57.

Give an example of a function f R —> R which is continuous at every point in the open interval but which is not bounded on the open interval S.

S = (0,1) 58.

Prove: Let f : R R be continuous at every point in a closed interval A = [a, bj, Then f is bounded on A. (Remark: By the preceding problem, this theorem is not true if A is not closed.)

59.

Prove: Let f:R—)R and g:RR where f g is defined by (f +g)(x)

be continuous. Then the sum

(f + g) :R—>R

is continuous,

f(x) g(x).

60.

Prove: Let f:R—>R be continuous, and let k be any real number. Then the function is continuous, where kf is defined by (k.f)(x) k(f(x))•

61.

Prove: Let f : R —> R

62.

Prove: The projection

63.

Consider the functions

f(x) =

and

g:RR

be continuous. Then

{x ER : f(x)=---- g(x)1

T x : R2 —> R is continuous where ,T-x is defined by f:RR

f sin (1/x)

10

and

g:R—>R

if x 0 if x = 0 '

(kf) : R —> R

is a closed set.

T-x ((a, b)) = a.

defined by

g(x) = f x

1°

sin (1/x)

0 if x = 0 if x

Prove g is continuous at 0 but f is not continuous at 0. 64.

Recall that every rational number qGQ can be written uniquely in the form q = alb where a b e N, and a and b are relatively prime. Consider the function f R —> R defined by

fo 1 1/b

f(x)

if x is irrational if x is rational and x = alb as above

Prove that f is continuous at every irrational point, but f is discontinuous at every rational point.

Answers to Supplementary Problems 57.

Consider the function

f(x) =

f —x

1/x

0 if x if x > 0

The function f is continuous at every point in R except at 0 as indicated in the adjacent graph of f. Hence f is continuous at every point in the open interval (0,1). But f is not bounded on

(0,1). Use the result stated in Problem 56 and the Heine-Borel Theorem.

65. Hint.

e z,

Chapter

5

Topological Spaces: Definitions TOPOLOGICAL SPACES Let X be a non-empty set. A class T of subsets of X is a topology on X if T satisfies the following axioms. [0 1 1

X and 0 belong to T.

[2]

The union of any number of sets in T belongs to T.

[3]

The intersection of any two sets in T belongs to T.

The members of T are then called T-open sets, or simply open sets, and X together with T, i.e. the pair (X, T) is called a topological space. Example 1.1:

Let 1,1 denote the class of all open sets of real numbers discussed in Chapter 4. Then 'V is a topology on R; it is called the usual topology on R. Similarly, the class I( of all open sets in the plane R2 is a topology, and also called the usual topology, on R2 . We shall always assume the usual topology on R and R2 unless otherwise specified.

Example 1.2:

Consider the following classes of subsets of X = {a, b, c, d, e}. 0, lal, {c, d}, {a, c, d } , {b, c, d, ell

T1

=

{X,

T2

=

( X, 0,

T3

=

IX,

{a},

0, {a},

{c, d}, {a, c, d}, {b,c,d}} {c, d } , {a, e, d } , {a, b, d, ell

Observe that T 1 is a topology on X since it satisfies the necessary three axioms [0 1 ], [02 ] and [03 ]. But T2 is not a topology on X since the union {a, c, d} U {b, c, d} = {a, b, c, d } of two members of T2 does not belong to T2, i.e. T2 does not satisfy the axiom [0 2 ]. Also, T3 is not a topology on X since the intersection {a,c,d}

n

{a,b,d,e} = {a,d}

of two sets in T3 does not belong to T3, i.e. T3 does not satisfy the axiom [0 3 ]. Example 1.3:

Let,,,D denote the class of all subsets of X. Observe that D satisfies the axioms for a topology on X. This topology is called the discrete topology; and X together with its discrete topology, i.e. the pair (X, ,0), is called a discrete topological space or simply a discrete space.

Example 1.4:

As seen by axiom [O l t a topology on X must contain the sets X and 0. The class rq = {X, 0 }, consisting of X and 0 alone, is itself a topology on X. It is called the indiscrete topology; and X together with its indiscrete topology, i.e. (X, S), is called an indiscrete topological space or simply an indiscrete space.

Example 1.5:

Let T denote the class of all subsets of X whose complements are finite together with the empty set Q. This class T is also a topology on X. It is called the co/mite topology or the T 1 -topology on X. (The significance of the T 1 will appear in a later chapter.)

66

CHAP. 5]

TOPOLOGICAL SPACES: DEFINITIONS

Example 1.6:

67

The intersection T 1 n T2 of any two topologies T 1 and T2 on X is also a topology on X. For, by [01 1 X and 0 each belongs to both T 1 and T2 ; hence X and 0 each belongs to the intersection T1 n T2 , i.e. T1 n T2 satisfies [0 1 ]. Furthermore, if G,H E T 1 n T2 then, in particular, G,H G T 1 and G,H G T2. But since T1 and T2 are topologies, G n H C T1 and GnHE T2. Accordingly,

Gni ET 1 n T2 In other words

Similarly, T1 n T 2 satisfies [021.

T 1 n T2 satisfies [03 1.

The statement in the preceding example can, in fact, be generalized to any collection of topologies. Namely,

Theorem 5.1:

Let {Ti: i E I) be any collection of topologies on a set X. section fl T4 is also a topology on X.

Then the inter-

In our last example, we show that the union of topologies need not be a topology. Example 1.7:

Each of the classes

T 1 = {X, 0, {a}} is a topology on X = {a, b, c}.

and

T2 -= {X,

0, {WI

But the union

T 1 U T2 =

(X, 0, {a}, Ibll

is not a topology on X since it violates [02 ]. That is, {a} G T I U T2 , {b} G T1 U T2 but {a} u {b} = {a, b } does not belong to T1 U T2.

If G is an open set containing a point p EX, then G is called an open neighborhood of p. Also, G without p, i.e. G\ {p}, is called a deleted open neighborhood of p.

Remark:

The axioms [01], [02] and [03] are equivalent to the following two axioms:

[Orl

The union of any number of sets in T belongs to T.

1 0:1

The intersection of any finite number of sets in T belongs to T.

For [0 *,1 implies that 0 belongs to T since U{GET :GE9)) = 9)

i.e. the empty union of sets is the empty set. T since n {G

Furthermore, [Or] implies that X belongs to

ET:GE

9)}

= X

i.e. the empty intersection of subsets of X is X itself.

ACCUMULATION POINTS Let X be a topological space. A point p E X is an accumulation point or limit point (also called cluster point or derived point) of a subset A of X if every open set G containing p contains a point of A different from p, i.e., G open, p E G

implies

(G\ {p}) n A

The set of accumulation points of A, denoted by A', is called the derived set of A. Example 2.1:

The class

T

{X, 0, {a}, { c,

{a, e, d } , {b, e, d ell

defines a topology on X = {a, b, e, d, e}. Consider the subset A = {a, b, el of X. Observe that b e X is a limit point of A since the open sets containing b are {b, e, d, el and X, and each contains a point of A different from b, i.e. c. On the other hand, the point a G X is not a limit point of A since the open set {a}, which contains a, does not contain a point of A different from a. Similarly, the points d and e are limit points of A and the point c is not a limit point of A. So A' = {b, d, el is the derived set of A.

[CHAP. 5

TOPOLOGICAL SPACES: DEFINITIONS

68 Example 2.2:

Let X be an indiscrete topological space, i.e. X and 0 are the only open subsets of X. Then X is the only open set containing any point p G X. Hence p is an accumulation point of every subset of X except the empty set 0 and the set consisting of p alone, i.e. the singleton set {p}. Accordingly, the derived set A' of any subset A of X is as follows:

0 A'

=

{p } c = X\ {p} X

if A = 0 if A = {p} if A contains two or more points

Observe that, for the usual topology on the line R and the plane R2, the above definition of an accumulation point is the same as that given in Chapter 4. CLOSED SETS Let X be a topological space. an open set. Example 3.1:

The class

A subset A of X is a closed set if its complement Ac is T =

{X, 0, {a},

{c,

defines a topology on X = {a, b, e, d, e}. 0, X, {b,c,d, e} ,

d},

e,

{b,c,d,e}}

The closed subsets of X are

{a, b, e}, {b, e}, {a}

that is, the complements of the open subsets of X. Note that there are subsets of X, such as {b, e, d, e} , which are both open and closed, and there are subsets of X, such as {a, b}, which are neither open nor closed. Example 3.2:

Let X be a discrete topological space, i.e. every subset of X is open. Then every subset of X is also closed since its complement is always open. In other words, all subsets of X are both open and closed.

Recall that Acc = A, for any subset A of a space X. Hence Proposition 5.2: In a topological space X, a subset A of X is open if and only if its complement is closed. The axioms Theorem 5.3:

[Oa

[02] and [Os] of a topological space and DeMorgan's Laws give

Let X be a topological space. Then the class of closed subsets of X possesses the following properties: (i)

X and

0

are closed sets.

(ii) The intersection of any number of closed sets is closed. (iii) The union of any two closed sets is closed. Closed sets can also be characterized in terms of their limit points as follows: Theorem 5.4:

A subset A of a topological space X is closed if and only if A contains each of its accumulation points.

In other words, a set A is closed if and only if the derived set A' of A is a subset of A, i.e. A' c A. CLOSURE OF A SET Let A be a subset of a topological space X.

The closure of A, denoted by

or A is the intersection of all closed supersets of A. In other words, if {F, : i of all closed subsets of X containing A, then

A = nF

e I) is the class

CHAP. 5]

69

TOPOLOGICAL SPACES: DEFINITIONS

Observe first that A is a closed set since it is the intersection of closed sets. Furthermore, A is the smallest closed superset of A, that is, if F is a closed set containing A, then

AcÀcF According/y, a set A is closed if and only if A= A. We state these results formally: Proposition 5.5: Let A be the closure of a set A. Then: (i) A is closed; (ii) if F is a closed superset of A, then A CAC F; and (iii) A is closed if A = A. Example 4.1:

Consider the topology T on subsets of X are

X = {a, b, e, d, el of Example 3.1 where the closed

0, X, {b,c,d, e } , {a, b, e } , {b, el, {al

Accordingly, Example 4.2:

{b}

{b, e } ,

{a, cl = X,

{b,d}

{b,c,d,e}

Let X be a cofinite topological space, i.e. the complements of finite sets and 0 are the open sets. Then the closed sets are precisely the finite subsets of X together with X. Hence if A c X is finite, its closure A is A itself since A is closed. On the other hand, if A c X is infinite then X is the only closed superset of A; so À is X. More concisely, for any subset A of a cofinite space X, A

A

lx

if A is finite if A is infinite

The closure of a set can be completely described in terms of its limit points as follows: Theorem 5.6: Let A be a subset of a topological space X. Then the closure of A is the union of A and its set of accumulation points, i.e.,

= A u A point p EX is called a closure point or adherent point of A c X if p belongs to the closure of A, i.e. p EA. In view of the preceding theorem, p E X is a closure point of A c X if p e A or p is a limit point of A. Example 4.3:

Consider the set Q of rational numbers. As seen previously, in the usual topology for R, every real number a E R is a limit point of Q. Hence the closure of Q is the entire set R of real numbers, i.e. Ci --- R.

A subset A of a topological space X is said to be dense in BcX if B is contained in the closure of A, i.e. BcA. In particular, A is dense in X or is a dense subset of X if A=X. Example 4.4:

Observe in Example 4.1 that {a, where X is not.

Example 4.5:

= X

and

{b,

Ib, c, d,

{a, b, c, d, e}. Hence the set {a, el is a dense subset of X but the set {b, d}

As noted in Example 4.3, (1 = R. In other words, in the usual topology, the set Q of rational numbers is dense in R.

The operator "closure", assigning to each subset A of X its closure A eX satisfies the (four properties appearing in the proposition below, called the Kuratowski Closure Axioms. In fact, these axioms may be used to define a topology on X, as we shall prove subsequently.

Proposition 5.7: (i) Ç2 = 0; (ii) AcA;

(iii) A u B = AuR;

and (iv)

= A.

INTERIOR, EXTERIOR, BOUNDARY Let A be a subset of a topological space X. A point p e A is called an interior point of A if p belongs to an open set G contained in A:

pEGCA

where G is open

TOPOLOGICAL SPACES: DEFINITIONS

70

(CHAP. 5

The set of interior points of A, denoted by is called the interior of A.

int (A), A or A° The interior of A can also be characterized as follows:

The interior of a set A is the union of all open subsets of A. Furthermore: (i) A° is open; (ii) A° is the largest open subset of A, i.e. if G is an open subset of A then GC A° CA; and (iii) A is open iff A A°. The exterior of A, written ext (A), is the interior of the complement of A, i.e. int (Ac). The boundary of A, written b (A), is the set of points which do not belong to the interior or the exterior of A. Next follows an important relationship between interior, exterior and closure. Theorem 5.9: Let A be any subset of a topological space X. Then the closure of A is the union of the interior and boundary of A, i.e. A = A° U b (A).

Proposition 5.8:

Example 5.1:

Consider the four intervals [a, b], (a, b), (a, b] and [a, b) whose endpoints are a and b. The interior of each is the open interval (a, b) and the boundary of each is the set of endpoints, i.e. {a, 1)}.

Example 5.2:

Consider the topology T =

{X, 0, {a}, {c, d}, {a, c, d}, {b, c, cl, en

{a, b, c, d, e}, and the subset A =-- {b, c, c/} of X. each interior points of A since c,d E {c, d} c A on X

The points c and d are

where le, c/1 is an open set. The point b G A is not an interior point of A; so int (A) = {e, cl}. Only the point a e X is exterior to A, i.e. interior to the complement Ac = {a, el of A; hence int (Ac) = {a}. Accordingly the boundary of A consists of the points b and e, i.e. b (A) {b, e}. Example 5.3: Consider the set Q of rational numbers. Since every open subset of R contains both rational and irrational points, there are no interior or exterior points of Q; so int (Q) 0 and int (QC) = Ø. Hence the boundary of Q is the entire set of real numbers, i.e. b (Q) = R.

A subset A of a topological space X is said to be nowhere dense in X if the interior of the closure of A is empty, i.e. int (A) = Q. Example 5.4:

Consider the subset A ={1, 1,

=

one limit point, O. Hence A {0, 1, points; so A is nowhere dense in R.

Example 5.5:

of R.

4,

As noted previously, A has exactly

Observe that A has no interior

Let A consist of the rational points between 0 and 1, i.e. A = {x x E Q, 0 < x < 1}. Observe that the interior of A is empty, i.e. int (A) = Ø. But A is not nowhere dense in R; for the closure of A is [0, 1], and so

int (A) = int ([0, 1]) = (0,1) is not empty.

NEIGHBORHOODS AND NEIGHBORHOOD SYSTEMS Let p be a point in a topological space X. A subset N of X is a neighborhood of p iff N is a superset of an open set G containing p: p EGCN where G is an open set In other words, the relation "N is a neighborhood of a point p" is the inverse of the relation "p is an interior point of N". The class of all neighborhoods of p e X, denoted by N p, is called the neighborhood system of p. Example 6.1:

Let a be any real number, i.e. a G R. Then each closed interval [a — S, a+ S], with center a, is a neighborhood of a since it contains the open interval (a— 8, a+ S) containing a. Similarly, if p is a point in the plane R2 , then every closed disc {q E R2 : d(p, q) < 8 0}, with center p, is a neighborhood of p since it contains the open disc with center p.

TOPOLOGICAL SPACES:

CHAP. 5]

71

DEFINITIONS

The central facts about the neighborhood system Np of any point p e X are the four properties appearing in the proposition below, called the Neighborhood Axioms. In fact, these axioms may be used to define a topology on X, as we shall note subsequently. Proposition 5.10:

(i) Np is not empty

and p belongs to each member of Np.

Np belongs to Np. Every superset of a member of Np belongs to N. Each member N e Np is a superset of a member G E Np where G

(ii) The intersection of any two members of (iii) (iv)

is a neighborhood of each of its points, i.e. G E Ng for every g E G.

CONVERGENT SEQUENCES A sequence (ai, of points in a topological space X converges to a point b or b is the limit of the sequence (an), denoted by b lim a = b, lim an = b or an if for each open set G containing b there exists a positive integer n o

E

X,

e N such that

n > no implies an e G that is, if G contains almost all, i.e. all except a finite number, of the terms of the sequence. Example 7.1:

Let (a 1 , a 2 , ...) be a sequence of points in an indiscrete topological space (X, (q). Note that: (1) X is the only open set containing any point b G X; and (ii) X contains every term of the sequence (a n ). Accordingly, the sequence (0, 1 , a2, ...) converges to every point b G X.

Example 7.2:

Let (a 1 , a2 , ...) be a sequence of points in a discrete topological space (X, D) . Now for every point b G X, the singleton set {b} is an open set containing b. So, if an b, then the set {b} must contain almost all of the terms of the sequence. In other words, the sequence (a n ) converges to a point b G X if the sequence is of the form (a 1 , a 2 , a no , b, b, b,

Example 7.3:

Let T be the topology on an infinite set X which consists of 0 and the complements of countable sets (see Problem 56). We claim that a sequence (a 1 , a 2 , ...) in X ano , b, b, b, ...), converges to b G X if the sequence is also of the form (a 1 , a2 , b is finite. Now A is i.e. the set A consisting of the terms of (an) different from countable and so Ac is an open set containing b. Hence if an —> b then Ac contains all except a finite number of the terms of the sequence, and so A is finite.

COARSER AND FINER TOPOLOGIES Let T1 and T2 be topologies on a non-empty set X. Suppose that each Ti-open subset of X is also a T2-open subset of X. That is, suppose that T1 is a subclass of T2, i.e. T 1 C T2, Then we say that T1 is coarser or smaller (sometimes called weaker) than "r2 or that T2 is finer or larger than T1. Observe that the collection T {T,} of all topologies on X is partially ordered by class inclusion; so we shall also write Ti < 'T2 for

Ti c

and we shall say that two topologies on X are not comparable if neither is coarser than the other. Example 8.1:

Consider the discrete topology ,:1), the indiscrete topology g and any other topology T on any set X. Then T is coarser than e'D and T is finer than (5. That is,

T

Example 8.2:

Consider the cofinite topology T and the usual topology I( on the plane R2. Recall that every finite subset of R2 is a II-closed set; hence the complement of any finite subset of R2 , i.e. any member of T, is also a 1{-open set. In other words, T is coarser than It, i.e. T<

TOPOLOGICAL SPACES: DEFINITIONS

72

[CHAP. 5

SUBSPACES, RELATIVE TOPOLOGIES Let A be a non-empty subset of a topological space (X, T). The class T A of all intersections of A with T-open subsets of X is a topology on A; it is called the relative topology on A or the relativization of T to A, and the topological space (A, TA) is called a subspace of (X, T). In other words, a subset H of A is a TA -open set, i.e. open relative to A, if and only if there exists a T-open subset G of X such that H = GnA Example 9.1:

Consider the topology T =

{X, 0, {a}, {c, d} , {a, e, d}, {b, e, d, e}}

on X = {a, b, c, d, e}, and the subset A = {a, d, el of X.

X n A = A, 0 n A = 0,

{a} n A = {a},

{c, d} n A = {d},

Observe that

{a, c, d} n A = {a, d}

{b, e, d, el n A == {d, el

Hence the relativization of T to A is

TA Example 9.2:

=

{A, 0, {a}, {d}, {a, d } , {d, e}}

Consider the usual topology II on R and the relative topology TA on the closed interval A = [3,8]. Note that the closed-open interval [3,5) is open in the relative topology on A, i.e. is a TA-open set, since [3,5) = (2,5) n A where (2, 5) is a T-open subset of R. Thus we see that a set may be open relative to a subspace but be neither open nor closed in the entire space.

EQUIVALENT DEFINITIONS OF TOPOLOGIES Our definition of a topological space gave axioms for the open sets in the topological space, that is, we used the open set as the primitive notion for the topology. We now state two theorems which exhibit alternate methods of defining a topology on a set, using as primitives the notions of "neighborhood of a point" and "closure of a set". Theorem 5.11: Let X be a non-empty set and let there be assigned to each point pEX a class cil, of subsets of X satisfying the following axioms: [Ail c//, is not empty and p belongs to each member of [A2] The intersection of any two members of ey6i, belongs to elI p. [A3] Every superset of a member of c//, belongs to A,. [114 1

Each member N e A, is a superset of a member G e cA, such that G e eA, for every g E G.

Then there exists one and only one topology T on X such that T-neighborhood system of the point p G X.

0././,

is the

Theorem 5.12: Let X be a non-empty set and let k be an operation which assigns to each subset A of X the subset Ac of X, satisfying the following axioms, called the Kuratowski Closure Axioms: [Ki] 0" = 0 [K2]

A c Ak

[K3]

(A U B) k = A" U B k

[ li4] (Ak)k = A" Then there exists one and only one topology T on X such that A" will be the T-closure of the subset A of X.

CHAP. 5]

73

TOPOLOGICAL SPACES: DEFINITIONS

Solved Problems TOPOLOGIES, OPEN SETS 1.

Let X = {a, b, c, d, el. Determine whether or not each of the following classes of subsets of X is a topology on X.

(i)

T 1 = {X, Q {a}, {a, b} , {a, c}}

(ii)

T2

=

{X, 0, {a, b, c}, {a, b, d} , {a, b, c, d}}

(Hi)

T3

=

{X, 0, {a), {a, b), {a, c, d } , {a, b,c,d}}

,

Solution:

(i)

(ii)

T 1 is not a topology on X since {a, b}, {a, cl E T 1

2.

T1

{a, b, cl n {a, b, dl =-- {a, b }

but

T2

T3 is a topology on X since it satisfies the necessary axioms.

Let T be the class consisting of R, 0 and all infinite open intervals A, = (q, Do) with E Q, the rationals. Show that T is not a topology on R. Solution: Observe that

A

==

is the union of members of T, but A not a topology on R.

3.

{a, b,

{a, b } U {a, c}

T2 is not a topology on X since {a, b, cl, {a, b, dl G T2

(iii)

but

{A l : q E Q, q > ‘1. }

e T since

(V)

is irrational. Hence T violates [02 ] and is therefore

Let T be a topology on a set X consisting of four sets, i.e. T =

{X, 0, A, B}

where A and B are non-empty distinct proper subsets of X. A and B satisfy? Solution: Since A n B

Case!.

must also belong to T, there are two possibilities:

A n B :=

Then AU B cannot be A or B; hence AUB =-- X.

Case II.

What conditions must

A n B =-- A

or

Thus the class {A, B} is a partition of X.

A n B =_- B

In either case, one of the sets is a subset of the other, and the members of T are totally ordered by inclusion: 0c Ac Bc X or 0c BC A c X.

4.

List all topologies on X =- {a, b, c} which consist of exactly four members. Solution: Each topology T on X with four members is of the form correspond to Case I or Case II of the preceding problem.

T

{X, 0, A, B } where A and B

Case I. {A, B} is a partition of X. The topologies in this case are the following:

T1 Case II.

{X, 0, {a}, {b, e}l,

T2 = {X, 0, {b}, {a, cfl, T3 = {X, 0, {c}, {a, bll

The members of T are totally ordered by inclusion.

The topologies in this case are the following:

T4 = (X, 0, fal, {a, b1)

T7 = {X, 0, {b}, {a, b } l

T5 = IX, 0, {a}, {a, e}}

T8 •= {X, 0, {c}, {a, C}}

T6 = {X, 0, {b}, {b,

T9 = {X, 0,

tel, fb,

5.

(CHAP. 5

TOPOLOGICAL SPACES: DEFINITIONS

74

Y be a function from a non-empty set X into a topological space (Y,14). Let f: X Furthermore, let T be the class of inverses of open subsets of Y: [G] : G E 1,11 = Show that T is a topology on X. Solution: Since It is a topology, satisfies [0 1 ].

Y,0 E II.

But X

f -1 [Y]

and

0 = f -1 [0], so X,0 E T and T

Let {A i} be a class of sets in T. By definition, there exist Gi G IA foT which A i -= f -1 [Gd.

u, A i = Since IA is a topology, U G i E It, so

f-1

=

But

uf

u, A E T, and T satisfies [02 ].

Lastly, let A I ,A 2 E T. Then

3

G1,

G2 G '14

A 1 n A2 = f -1 [Ga]

But

A 1 = f -1 [Ga], A2

such that ii

f -1 [G 2]

f

1

[G2]

f --1 [G 1 n G 2]

and G1 n G2 E U. Thus A 1 n A2 G T and [03 ] is also satisfied.

6.

Consider the second axiom for a topology T on a set X: [02] The union of any number of sets in T belongs to T. Show that [02] can be replaced by the following weaker axiom: [0'2 ] The union of any number of sets in T\ {X, 0} belongs to T. In other words, show that the axioms [Oil, [02] and [03].

[Ol], [0'2]

and [03] are equivalent to the axioms

Solution: Let T be a class of subsets of X satisfying (0 1 ], [02'] and [03 and let cA be a subclass of T. We want to show that T also satisfies [02 ], i.e. that U{E E G cs1 } G T. ],

Case I. X E Then U{E : E G cA } --= X

Case II.

X

and therefore belongs to T by [0 1 ].

e U{E : E E

Then

U{E : E

\ {X}}

But the empty set 0 does not contribute any elements to a union of sets; hence U{E : E E cA}

=

E G cA'\{X}} = U {E : E G cA\{X,0}}

Since cA is a subclass of T, 0.4 \ {X, 0} is a subclass of T \ {X, 0}, belongs to T.

7.

(1)

so by [0'2 ] the union in (1)

Prove: Let A be a subset of a topological space X with the property that each point Then A is open.

p E A belongs to an open set G„ contained in A. Solution: For each point p G A, p E Gp c A. and, by [02], is open.

8.

Hence U {G p : p G Al = A and so A is a union of open sets

Let T be a class of subsets of X totally ordered by set inclusion. Show that T satisfies [Os], i.e. the intersection of any two members of T belongs to T. Solution : Let A,B E T. In either case

Since T is totally ordered by set inclusion, either AnB =A or AnB=B

AnBGT,

and so T satisfies [03].

CHAP. 5]

9.

TOPOLOGICAL SPACES: DEFINITIONS

75

Let T be the class of subsets of R consisting of R, p and all open infinite intervals E. = (a, co) with a E R. Show that T is a topology on R. Solution: Since R and Q belong to T, T satisfies [0 1 ]. Observe that T is totally ordered by set inclusion; hence T satisfies [0 3 ]. Now let a be a subclass of T \ {X, 0 }, that is a {E i : i E /} where I is some set of real numbers. We want to show that U E i belongs to T. If I is not bounded from below, i.e. if inf (I) then U = R. If I is bounded from below, say inf (I) io , then U, E, 7-- (i0 ,00)= E,0. In either case, U 1 E i G T, and T satisfies [OD. -, --

10. Let T be the class of subsets of N consisting of Q and all subsets of N of the form En = {n, n +1, n+2, ... } with n e N.

(i) Show that T is a topology on N. (ii) List the open sets containing the positive integer 6. Solution: (i)

Since 0 and E 1 = {1, 2, 3, ... } = N belong to T, T satisfies [O i l. totally ordered by set inclusion, T also satisfies [OA

Furthermore, since T is

{E n : n E /} where I is some set of Now let a be a subclass of T \ {N, 0 }, that is, a positive integers. Note that I contains a smallest positive integer n o and U {E„: n G which belongs to (ii)

{no , no + 1, no + 2, ...} = E no

T. Hence T satisfies [O], and so T is a topology on N.

Since the non-empty open sets are of the form

E„

{n, n + 1, n + 2, ...}

with n E N, the open sets containing the positive integer 6 are the following: E l = N = {1, 2, 3, ...}

E4 = {4, 5, 6, ...I

E2 = {2, 3, 4, ...}

E5 = {5, 6,7, ...}

E3 = {3, 4, 5, ...}

E6 = {6,7,8, ...1

ACCUMULATION POINTS, DERIVED SETS

11. Let T be the topology on N which consists of

=

p

and all subsets of N of the form In, n +1, n+2, ... } where n E N as in Problem 10.

(i) Find the accumulation points of the set A (ii) Determine those subsets E of N for which

= {4,13,28,37 } . E' = N.

Solution: (1)

i p. If no 36, Observe that the open sets containing any point p E N are the sets E . then every open set containing no also contains 37 e A which is different from n o ; hence no 36 is a limit point of A. On the other hand, if no > 36 then the open set Eno = {n o , no +1, no + 2, ...1 contains no point of A different from n o . So no > 36 is not a limit point of A. Accordingly, the derived set of A is A' = {1,2,3, ..., 34, 35, 36 }.

(ii)

If E is an infinite subset of N then E is not bounded from above. So every open set containing any point p E N will contain points of E other than p. Hence E' -= N. On the other hand, if E is finite then E is bounded from above, say, by no E N. Then the open set E 0 + 1 contains no point of E. Hence n o + 1 G N is not a limit point of E, and so E' N.

76

TOPOLOGICAL SPACES: DEFINITIONS

[CHAP. 5

12. Let A be a subset of a topological space (X, T). When will a point p E X not be a limit point of A? Solution: The point p E X is a limit point of A iff every open neighborhood of p contains a point of A other than p, i.e., p E G and G G T implies (G\ {p{) n A So p is not a limit point of A if there exists an open set G such that pEG or, equivalently,

pEG

and

and

GnA= 0 or GnA

pE G

or, equivalently,

(p})n A =

(G

GnAc {P}

and

13. Let A be any subset of a discrete topological space X. of A is empty.

Show that the derived set A'

Solution: Let p be any point in X. Recall that every subset of a discrete space is open. Hence, in particular, the singleton set G = {p) is an open subset of X. But pE G Hence, by the above problem, p

G n A = ({p} n A) c {p}

and

A' for every p E X, Le. A'

Ø.

14. Consider the topology IX, p, {a}, {a,b}, {a, e, cl}, {a, b, c, cl}, fa, b,e)} on X = (a, b, c, d, el.

Determine the derived sets of (i) A = tc, d, e} and (ii) B = (b.).

Solution: Note that {a, b} and {a, b, el are open subsets of X and that

(1)

a, b G {a, b} eC

{a, b, el

and {a, b} n A = and

{a, b, el n A = {e}

Hence a, b and e are not limit points of A. On the other hand, every other point in X is a limit point of A since every open set containing it also contains a point of A different from it. Accordingly, A' = { c, d } . (ii)

Note that {a}, {a, b} and {a, c, d } are open subsets of X and that a E {a } and

b E {a, b} G {a, c, d}

{a} n B

0

and {a, b} n B = { b } and

{a, c, d} n B = c,d

bl. But e is a limit point of B since the open sets Hence a, b, c and d are not limit points of B containing e are {a, b, el and X and each contains the point b GB different from e. Thus B' (e).

15. Prove: If A is a subset of B, then every limit point of A 'is also a limit point of B, i.e., A c B implies A' c B'. Solution: Recall that p G A' if (G\ {p}) n A

0 for every open set G containing p.

(G\ {p}) n B D (G\ {p}) So p E A'

implies p E B', i.e. A' c B'.

n A

But B D A; hence

CHAP. 51

77

TOPOLOGICAL SPACES: DEFINITIONS

16. Let T1 and T2 be topologies on X such that T1 C T2, i.e. every Ti-open subset of X is also a T2-open subset of X. Furthermore, let A be any subset of X.

Show that every T2-limit point of A is also a T i-limit point of A. (ii) Construct a space in which a T i-limit point is not a T2-limit point. Solution: 0 for every G E T2 such that p E G. But (i) Let p be a T2-limit point of A; i.e. (G\{p}) n A T i c T2 ; so, in particular, (G\ {p}) n A 0 for every GE T i such that p E G, i.e. p is a T 1 -limit point of A. (ii)

Consider the usual topology '1,1 and the discrete topology rz, on R. Note that I/ C since 1) contains every subset of R. By Problem 13, 0 is not a „/)-limit point of the set A = {1, A, ...} since A' is empty. But 0 is a limit point of A with respect to the usual topology on R.

17. Prove: Let A and B be subsets of a topological space (X, T). Solution: Utilizing Problem 15,

So

A' u B' c (AuB)',

AcAuB

implies

A'c(AuB)'

BcAuB

implies

B'c(AuB)'

Then (A U B)' = A' U B'.

and we need only show that

(AuB)' c A' u B' Assume

p E A' u B'; pE

3G,H E T such that G and GnAC {p} and p e H and HnBC {P} thus

But Gni-icT, 23EGnH and

(GnH)n(AuB) Thus p E (Au B)',

(GnHnA)u (GnHnB)

c (GnA)u (HnB)

c {p} u {p} = {p}

and so (AuB)' C (A'uB').

CLOSED SETS, CLOSURE OPERATION, DENSE SETS 18. Consider the following topology on X = {a, b, c, d, e}:

T = (X, 9), {a}, (a, b}, {a, c, d}, {a,b,c,d}, ta, b, e}} (i) List the closed subsets of X. (ii) Determine the closure of the sets {a}, {b} and {c, e}. (iii) Which sets in (ii) are dense in X? Solution: (i) A set is closed if its complement is open. Hence write the complement of each set in T: 0, X, {b, c, d, e}, {c, d, e}, {b, e}, {e}, {c, d} (ii)

The closure A of any set A is the intersection of all closed supersets of A. The only closed superset of {a} is X; the closed supersets of {b} are {b, e), {b, c, d, el and X; and the closed supersets of {c, e} are {c, d, e}, {b, c, d, el and X. Thus, = X,

(iii)

{-b-I = {b, e},

{c,

= {c, d, el

A set A is dense in X if A --= X; so {a} is the only dense set.

19. Let T be the topology on N which consists of 9) and all subsets of N of the form ET, = {n, n +1, n+2, ... } where n N as in Problem 10.

Determine the closed subsets of (N, T). (i) (ii) Determine the closure of the sets {7, 24, 47, 85} and {3, 6, 9, 12, . . . (iii) Determine those subsets of N which are dense in N. Solution: (i) A set is closed if its complement is open. Hence the closed subsets of N are as follows: , {1, 2, , m}, N, Q, {1}, {1, 2}, {1, 2, 31,

78

TOPOLOGICAL SPACES: DEFINITIONS

(ii)

The closure of a set is the smallest closed superset. So {7, 24, 47, 85 } = {1, 2, • . , 84, 85},

(iii)

[CHAP. 5

{3, 6, 9, 12, . .} = {1, 2, 3,

.} = N

If a subset A of N is infinite, or equivalently unbounded, then À = N, i.e. A is dense in N. If A is finite then its closure is not N, i.e. A is not dense in N.

20. Let T be the topology on R consisting of R, Ø and all open infinite intervals Ea = (a, where a e R. (i) Determine the closed subsets of (R, T). (ii) Determine the closure of the sets [3, 7), (7,24, 47, 85 } and (3, 6, 9, 12, .

co)

Solution: (i) A set is closed if its complement is open. Hence the closed subsets of (R, T) are 0, R and all closed infinite intervals E a' = (-00, a].

(ii)

The closure of a set is the smallest closed superset. Hence [3,7) = (-00,71,

{7,24,47,85} = (-09,851,

{3,6,9,12, ...} = (--r4,00)

R

21. Let X be a discrete topological space. (i) Determine the closure of any subset A of X. (ii) Determine the dense subsets of X. Solution: (i)

Recall that in a discrete space X any A c X is closed; hence A = A.

(ii)

A is dense in X if

A X.

But A = A, so X is the only dense subset of X.

22. Let X be an indiscrete space. (i) Determine the closed subsets of X. (ii) Determine the closure of any subset A of X. (iii) Determine the dense subsets of X. Solution: (i)

Recall that the only open subsets of an indiscrete space X are X and 0; hence the closed subsets of X are also X and Q.

(ii)

If A= 0, then A = Q. If A 0, then X is the only closed superset of A; so for any AcX, { 0 if A = A X if A

(iii)

AcX is dense in X if

A= X;

A= X.

That is,

hence every non-empty subset of X is dense in X.

23. Prove Theorem 5.4: A subset A of a topological space X is closed if and only if A contains each of its accumulation points, i.e. A' C A. Solution: Suppose A is closed, and let p E A, i.e. pe Ac. hence p A' for Ac is an open set such that p E Ae and Thus A' c A if A is closed.

But Ac, the complement of a closed set, is open;

Ac n A =

Now assume A'cA; we show that Ac is open. Let pEAe; then pEA.', so such that p EG and (G\ {p}) n A = But p

A; hence

So G c Ac.

3 an open set G

G n A = (G \ {P} ) n A =

Thus p is an interior point of Ac, and so Ac is open.

24. Prove: If F is a closed superset of any set A, then A' C F. Solution: By Problem 15, A c F implies A' c F'. A' c F' c F, which implies A' c F.

But F' c F, by Theorem 5.4, since F is closed. Thus

CHAP. 5]

TOPOLOGICAL SPACES: DEFINITIONS

25. Prove:

79

A U A' is a closed set.

Solution: Let p G (A u A')c.

Since p

a A',

3 an open set G such that

pE G

G n A = Q.

However, p E A; hence, in particular, We also claim that

G n A' = O.

For if g G G, then

g G where G is an open set.

So g E A'

G n A = 0 or {p}

and

and

and thus G

Gn A=

n A' = Q. Accordingly,

G n (Au A') = (Gn A) u (Gn A') = QuQ = and so G c (Au A')c, Hence Au A' is closed.

Thus p is an interior point of (A u A')c which is therefore an open set.

26. Prove Theorem 5.6: A= A

U

A'.

Solution: Since A c A and A is closed, A' c (A)' c containing A, so AcitcAuA'. Thus A

A

and hence A u A' ci A. But Au A' is a closed set

AuA'.

27. Prove: If A c B then A C Solution: If A cB, then by Problem 15, A' cB'.

28. Prove:

AUB

So A u A' c Bu B' or, by the preceding problem,

A c B.

AUB.

Solution: Utilizing the preceding problem, AcAuB and EcAuB; hence (A. u /3) c A u B. But (A u B) c (A u E), a closed set since it is the union of two closed sets. Then (Proposition 5.5) and therefore AuB = Au f3. (AuB)cAuBc(AuB)

29. Prove Proposition 5.7: (1)

= 9);

(ii) A c A; (iii) A UB=A U .6'; and (iv) (A - ) - = A - .

Solution: (i) and (iv): 0 and A are closed; hence they are equal to their closures. (iii) Preceding problem. (Problem 26).

(ii)

AcAuA'

INTERIOR, EXTERIOR, BOUNDARY

30. Consider the following topology on X = {a, b, c, d, el: = {X, 9), {a}, (a, b), (a, c, dl, {a, b, c, d}, {a, b, el} Find the interior points of the subset A = {a, b, cl of X. points of A. (iii) Find the boundary points of A. (i)

(ii) Find the exterior

Solution: (i) The points a and b are interior points of A since a,b G { a, b} c A

=

{a, b,c}

where {a, b} is an open set, i.e since each belongs to an open set contained in A. Note that c is not an interior point of A since c does not belong to any open set contained in A. Hence int (A) = {a, b} is the interior of A. (ii)

The complement of A is Ac = {d, e}. Neither d nor e are interior points of Ac since neither 0, i.e. there are no exterior belongs to any open subset of Ac = {d, e}. Hence int (Ac) points of A.

(iii)

The boundary b (A) of A consists of those points which are neither interior nor exterior to A. So b (A)--= lc, d, el .

80

TOPOLOGICAL SPACES: DEFINITIONS

[CHAP. 5

31. Prove Proposition 5.8: The interior of a set A is the union of all open subsets of A. Furthermore: (1) A° is open; (ii) A° is the largest open subset of A, i.e. if G is an open subset of A then GCA°c A; and (iii) A is open if A = A°. Solution: Let {G i l be the class of all open subsets of A. If x E A°, then x belongs to an open subset of A, i.e.,

3io such that

x G Gio

Hence x G U G, and sa A° C U G. On the other hand, if Thus y G A°, and u ,G i C A°. Accordingly, A° =

(1)

A° = u G, is open since it is the union of open sets.

(ii)

If G is an open subset of A then G E {G,};

(iii)

If A is open then A C A° c A

so

ye

Ll

G,, then y e Gio for some io .

G c uG1 = A° c A.

or A -= A°. If A

= A° then A is open since A° is open.

32. Let A be a non-empty proper subset of an indiscrete space X. exterior and boundary of A.

Find the interior,

Solution: X and 0 are the only open subsets of X. Since X A, 0 is the only open subset of A; hence int (A) = Q. Similarly, int (Ac) = 0, i.e. the exterior of A is empty. Thus b (A) = X.

33. Let T be the topology on R consisting of R, Ø and all open infinite intervals Ea = (a, 00) where a E R. Find the interior, exterior and boundary of the closed infinite interval A = 17 , co). Solution : Since the interior of A is the largest open subset of A, int (A) = (7, 00). Note that Ac = (—c,o, 7) contains no open set except 0; so int (Ac) = ext (A) --= Q. The boundary consists of those points which do not belong to int (A) or ext (A); hence b (A) = (— , 7].

A = int (A) U b (A)

34. Prove Theorem 5.9: Solution: Since

X = int (A) u b (A) U ext (A),

(int (A) U b (A))c =

ext (A)

and

it suffices to

show

(A)c = ext (A). Let p E ext (A);

then 3 an open G such that

pEGC Ac

GnA=O

which implies

So p is not a limit point of A, i.e. p G A',

and p V A.

p V A' u A = A.

Hence

In other words,

ext (A) c (A )c . Now assume

pG

VW'

= (Au Alc. p

But also p

A,

so

eG

Thus p and

A', so 3 an open set G such that

(G\ {p})

G n A = Q and pEGc Ac.

n

A =

Thus p

ext (A), and (A )° ext (A).

35. Show by a counterexample that the function f which assigns to each set its interior, i.e. f(A) = int (A), does not commute with the function g which assigns to each set its closure, i.e. g(A) = A. Solution: Consider Q, the set of rational numbers, as a subset of R with the usual topology. Recall (Example 5.3) that the interior of Q is empty; hence

°f)(Q) On the other hand,

=

g(f(Q)) = g(int (Q))

R and the interior of R is R itself. So

(f o g)(Q) = f(g(Q)) Thus gof

g(0) = Q =

fog, or f and g do not commute.

f()

f(R)

R

CHAP. 51

81

TOPOLOGICAL SPACES: DEFINITIONS

NEIGHBORHOODS, NEIGHBORHOOD SYSTEMS 36. Consider the following topology on X = {a, b, c, d, e}: T =

{X, 0,

(a), (a, b), la,c,d1, {a, b, c, d}, {a, b, el}

List the neighborhoods (i) of the point e,

( ii) of the point c.

Solution: (i) A neighborhood of e is any superset of an open set containing e. The open sets containing e are {a, b, el and X. The supersets of {a, b, el are {a, b, e}, {a, b, c, e}, {a, b, d, el and X; the only superset of X is X. Accordingly, the class of neighborhoods of e, i.e. the neighborhood system of e, is Ne = {la, b, el, {a, b, c, e}, {a, b, d, e}, X} (ii)

The open sets containing e are {a, c, d}, {a, b, e, d } and X. {{a, e,

{a, b, c,

Hence the neighborhood system of c is {a, c, d, e}, Xl

37. Determine the neighborhood system of a point p in an indiscrete space X. Solution: X and 0 are the only open subsets of X; hence X is the only open set containing p. X is the only superset of X. Hence N1,, = {X}.

38. Prove: The intersection N also a neighborhood of p.

nM

In addition,

of any two neighborhoods N and M of a point p is

Solution: N and M are neighborhoods of p, so 3 open sets G,H such that

pEG c N Hence p E GnH C NnM,

and

pEHcM

and GnH is open, or NnM is a neighborhood of p.

39. Prove: Any superset M of a neighborhood N of a point p is also a neighborhood of p. Solution: N is a neighborhood of p, so 3 an open set G such that

pEGcNcM

which implies

pEGCN.

By hypothesis, N C M, so

pEGcM

and hence M is a neighborhood of p.

40. Determine whether or not each of the following intervals is a neighborhood of 0 under (ii) (-1, 01, (iii) [0, 4 (iv) (0, 1]. the usual topology for the real line R. (i) (—i, Solution: (i) Note that 0E (-3,3) c

3]

and (-3, 3) is open; so (-3, -I] is a neighborhood of 0.

(ii)

Any Ifropen set G containing 0 contains an open interval (a, b) containing 0, i:e. and (iii) a < 0 < b; hence G contains points both greater and less than 0. So neither (-1, 0] nor [0, 3) is a neighborhood of 0.

(Iv)

The interval (0,1] does not even contain 0 and hence is not a neighborhood of 0.

41. Prove: A set G is open if and only if it is a neighborhood of each of its points. Solution: Suppose G is open; then each point p EG belongs to the open set G contained in G. neighborhood of each of its points.

Hence G is a

Conversely, suppose G is a neighborhood of each of its points. So, for each point p EG, 3 an open set G, such that p C G,, c G. Hence G U {G o : 23EG} and is open since it is a union of open sets.

82

TOPOLOGICAL SPACES: DEFINITIONS

[CHAP.

5

42. Prove Proposition 5.10: Let N, be the neighborhood system of a point p in a topological space X. Then: (i) N p is not empty and p belongs to each member of Np. (ii) The intersection of any two members of

N„

belongs to

(iii) Every superset of a member of N, belongs to

Np.

Np.

(iv) Each member N e N, is a superset of a member G e Nr where G is a neighborhood of each of its points. Solution: (i)

If N Np, then 3 an open set G such that an open set containing p; so Np

(ii)

Proven in Problem 38. (iii) Proven in Problem 39.

(iv)

If NE Np , then N is a neighborhood of p, so 3 an open set G such that pEGCAT. the preceding problem G G and G is a neighborhood of each of its points.

pEG n N; hence p E N.

Note X 7\L p since X is

But by

SUBSPACES, RELATIVE TOPOLOGIES

43. Consider the following topology on X = {a, b, c, d, e}: T = {X, 0, {a}, {a, bl, {a, c, d } , {a, b,c,d},

{a,

b, e»

List the members of the relative topology TA on A = {a, c, el. Solution: TA

=

{AnG:GET},

so the members of

TA

are:

A nX= A

A n {a} = {a}

A n {a, e, d } = {a, el

An0=0

A n {a, b} = {a}

A n {a, b, c, d} ,--- {a, cl

In other words, TA -= {A, 0, {a}, {a, a}, {a, e}). relatively open in A, i.e. is TA-open.

A n {a, b, el = {a, el

Observe that {a, cl is not open in X, but is

44. Consider the usual topology 11 on the real line R. Describe the relative topology on the set N of positive integers.

14N

Solution: Observe that, for each positive integer n o G N,

Inol =

N n (no

—

no +

and (n o — 1, no + 1) is a It-open set; so every singleton subset {n o} of N is open relative to N. Hence every subset of N is open relative to N since it is a union of singleton sets. In other words, TIN is the discrete topology on N.

45. Let A be a T-open subset of (X, T) and let AcYc X.

Show that A is also open on Y, i.e. A is a Ty-open relative to the relative topology subset of Y. Solution: Ty

=

{ YnG : G E T } .

But A c Y and

AET; so A= Yn A E T.

46. Consider the usual topology U on the real line R. Determine whether or not each of the following subsets of I = [0, 1] are open relative to I, i.e. Tropen: (i) 4,4 (1•, I), (iii) (0, j. Solution:

(1)

Note that

(ii)

Since (1, i) is open in R, i.e. (i-, i) G II, it is open relative I by the preceding problem.

(1 (iii)

= In

(1,1]

/ n (, 3)

and

(, 3) is open in R; hence (1,1] is open relative to I.

(I 1 ).

Since (0, 1] is not the intersection of / with any li-open subset of R, it is not Iii-open.

In fact,

CHAP. 5]

83

TOPOLOGICAL SPACES: DEFINITIONS

47. Let A be a subset of a topological space (X, Ty well-defined. In other words, show that TA

Show that the relative topology TA is (AnG:GETI is a topology on A.

Solution : Since T is a topology, X and 0 belong to T. Hence AnX=A and An0=-0 both belong to TA, which then satisfies [O i l. Now let {H i : i E II be a subclass of TA. By definition of TA, for each jE I ] a T-open set G such that Hi = A n G. By the distributive law of intersection over union,

u,Hi = u,(AnG,) = A n But

(LJ 1

G, E T as it is the union of T-open sets; hence u 1 H E

U

Now suppose H I ,H2 e TA. Then 3 G i , G2 E T But G 1 n G2 E T since T is a topology. Hence belongs to TA.

such that

G,)

TA.

Thus TA satisfies [02 ].

A

H1

n

G 1 and

H2 -= A n G2.

n H2 == (A n G I ) n (A n G2) = A n (GI n G2) Accordingly, TA satisfies [03 ] and is a topology on A.

48. Let (X, T) be a subspace of (Y, T"') and let (17 , T*) be a subspace of (Z, T**). that (X, T) is also a subspace of (Z, T**).

Show

Solution: Since

T,

T. Let G E T; now XcYc Z, (X, T) is a subspace of (Z, T') if and only if T;';'' But T* = T y", so such that so 3G" E T for which G -= X n G*. 3G** G T"

G* -= Y n G.

Thus

since X c Y; so G G T,

G = xnG* = XnYnG" = xnG** Accordingly, T c T.

Now assume GE T', i.e. T. Since

31-/ G T**

X n (YnH) E cr" we have

G G T. Accordingly,

such that

G

XnH

X n H.

But

YnHE Ty" = T*

so

G

c T and the theorem is proved.

MISCELLANEOUS PROBLEMS 49. Let 'P(X) be the power set, i.e. class of subsets, of a non-empty set X. Furthermore, let k:c1)(X).- P(X) be the identity mapping, i.e. for each A c X, k(A)= A. (i) Verify that k satisfies the Kuratowski Closure Axioms of Theorem 5.12. (ii) Determine the topology on X induced by k. Solution:

(i) (ii)

k(AuB) -= A u B = k(A) u k(B), so [K3 ] is satisfied. k(k(A))=-- k(A), so [K4 ] is satisfied. A subset Fc X is closed in the topology induced by k if and only if k(F) = F. But k(A) = A for every AcX, so every set is closed and k induces the discrete topology. k(0) = 0, so [K 1 ] is satisfied. k(A) = A A, so [K2 ] is satisfied.

50. Let T be the cofinite topology on the real line R, and let 04, a2, ...) be a sequence in R with distinct terms. Show that (an) converges to every real number p e R. Solution: Let G be any open set containing p E R. By definition of the cofinite topology, Ge is a finite set and hence can contain only a finite number of the terms of the sequence (an ) since the terms are distinct. Thus G contains almost all of the terms of (a„), and so (a n ) converges to p.

51. Let T be the collection of all topologies on a non-empty set X, partially ordered by class inclusion. Show that T is a complete lattice, i.e. if S is a non-empty subcollection of T then sup (S) and inf (S) exist. Solution: Let T1 = n {T T G S}. By Theorem 5.1, T1 is a topology; so

T1

E T and T1 =- inf (S).

Now let B be the collection of all upper bounds of S. Observe that B is non-empty since, for example, the discrete topology j) on X belongs to B. Let T2 n {T: T G B } . Again by Theorem 5.1, T2 is a topology on X and, furthermore, T2 = sup (S).

84

TOPOLOGICAL SPACES: DEFINITIONS

[CHAP. 5

52. Let X be a non-empty set and, for each point p E X, let cAp denote the class of subsets of X containing p. (i) Verify that cA, satisfies the Neighborhood Axioms of Theorem 5.11. (ii) Determine the induced topology on X. Solution: (i) Since p G X, X G crip and, so, 0.4 [A i ] is satisfied.

O. By hypothesis, p belongs to each member of ap.

Hence

If M,NEc4 p , then p E M and p E N, and so p E M n N. Hence M nN Ea p and so [A 2 1 is satisfied. If NE0.4 1, and N C M, i.e. if pENc M, then pCM. Hence M E 0,4, and so [A3] is satisfied. By definition of cA p , every A c X has the property that A G c.41., for every p G A. is satisfied.

Hence [A4]

A subset A C X is open in the induced topology if and only if A G 0.40 for every p E A. every subset of X has this property, the induced topology on X is the discrete topology.

(ii)

Since

Supplementary Problems TOPOLOGICAL SPACES

X

{a, b }.

53.

List all possible topologies on the set

54.

Prove Theorem 5.1; Let {T, : i C /} be any collection of topologies on a set X. fl Ti is also a topology on X.

55.

Let X be an infinite set and let T be a topology on X in which all infinite subsets of X are open. Show that T is the discrete topology on X.

56.

Then the intersection

Let X be an infinite set and let T consist of Q and all subsets of X whose complements are countable. Prove that (X, T) is a topological space.

(i)

If X is countable, describe the topology determined by T.

(ii) 57.

Let T = {R2 , 0} u {G k : k C RI be the class of subsets of the plane R2 where

{(x,y): x,y E R, x > y + k} (1)

Prove that T is a topology on R2 .

(ii)

Is T a topology on R2 if "k G R" is replaced by "k G N" ? by "k E Q" ?

58.

Prove that (R 2 , T) is a topological space where the elements of T are Q and the complements of finite sets of lines and points.

59.

R u {p} and Let {p} be an arbitrary singleton set such that p R; e.g. {R}. Furthermore, let R* let T be the class of subsets of 11 .' consisting of all II-open subsets of R and the complements (relative to IV) of all bounded 14-closed subsets of R. Prove that T is a topology on R*.

60.

Furthermore, let T R u {R}. Let {p} be an arbitrary singleton set such that p R; and let R* be the class of subsets of R" consisting of all subsets of R and the complements (relative to R*) of all finite subsets of R. Prove that T is a topology on R.

ACCUMULATION POINTS, DERIVED SETS 61.

Prove;

A' u B' = (Au Br ,

62.

Prove:

If p is a limit point of the set A, then p is also a limit point of A\ {p}.

63.

Prove:

Let X be a cofinite topological space.

64.

Consider the topological space (R, T) where T consists of R, Q and all open infinite intervals Ea =(a, .0), a G R. Find the derived set of: (i) the interval [4, 10); (ii) Z, the set of integers.

Then A' is closed for any subset A of X.

CHAP. 5]

65.

Let T be the topology on R* = R u {p}

defined in Problem 59.

Determine the accumulation points of the following sets: (1) open interval (a, b), a, b G R (2) infinite open interval (a,

(1) (ii) 66.

85

TOPOLOGICAL SPACES: DEFINITIONS

aER

(3) R.

Determine those subsets of R* which have p as a limit point.

Let T 1 and T, be topologies on a set X with T 1 coarser than T2, Le. T 1 C T2 . Show that every T2-accumulation point of a subset A of X is also a T1 -accumulation point. Construct an example in which the converse of (i) does not hold.

(i) (ii)

CLOSED SETS, CLOSURE OF A SET, DENSE SUBSETS 67.

Construct a non-discrete topological space in which the closed sets are identical to the open sets.

68.

Prove:

AnBcAnE.

69.

Prove:

A\Pc (A\ B).

70.

Prove:

If A is open, then AnBcAnB.

Prove;

Let A be a dense subset of (X, T), and let B be a non-empty open subset of X. Then 0.

71.

A n B 72.

Construct an example in which equality does not hold. Construct an example in which equality does not hold.

Let T 1 and T2 be topologies on X with T1 coarser than T2.

Show that the T2 -closure of any subset

A of X is contained in the T 1-closure of A. 73.

Show that every non-finite subset of an infinite cofinite space X is dense in X.

74.

Show that every non-empty open subset of an indiscrete space X is dense in X.

INTERIOR, EXTERIOR, BOUNDARY 75.

Let X be a discrete space and let Ac'X.

76.

Prove: (i) b (A) c A (ii) b (A) (iii) b (A)

(ii) ext (A),

Find (i) int (A),

and (iii) b (A).

if and only if A is closed.

nA =

0 if and only if A is open.

0 if and only if A is both open and closed.

0, then b (A u B)

AnE

b (A) u b (B).

77.

Prove: If

78.

Prove: (i) A° n B° = (A nB)°; (ii) A° u B° c (Au B)° . in (ii) does not hold.

79.

Prove: b (A°) C b (A).

80.

Show that int (A) u ext (A) need not be dense in a space X. (It is true if X = R.)

81.

Prove:

Construct an example in which equality

Construct an example in which equality does not hold.

(i)

Let T 1 and T2 be topologies on X with T 1 coarser than T 2, i.e. T 1 c T9, and let A c X. The Tr-interior of A is a subset of the T2-interior of A.

(ii)

The T2-boundary of A is a subset of the T 1 -boundary of A.

Then:

NEIGHBORHOODS, NEIGHBORHOOD SYSTEMS 82.

Let X be a cofinite topological space. Show that every neighborhood of a point p G X is an open set.

83.

Let X be an indiscrete space. Determine the neighborhood system N I, of any point p G X.

84.

Show that if

N,

is finite, then

n{N : N E Np

}

belongs to

SUBSPACES, RELATIVE TOPOLOGIES 85.

Show that every subspace of a discrete space is also discrete.

86.

Show that every subspace of an indiscrete space is indiscrete.

86 87.

TOPOLOGICAL SPACES: DEFINITIONS

[CHAP. 5

Let (Y, Ty) be a subspace of (X, T). Show that Ec Y is Ty-closed if and only if E=YnF, where

F is a T-closed subset of X. 88.

Let (A, T 4 ) be a subspace of (X, T). Prove that TA consists of the members of T contained in A, i.e. TA 1G : G c A, G G T1, if and only if A is a T-open subset of X.

89.

Let (Y, Ty) be a subspace of (X, T).

For any subset A of Y, let

A

and A° be the closure and interior

of A with respect to T and let (A)y and (A°) y be the closure and interior of A with respect to T. Prove (ii) A° = (A°) y n Y°. (i) (A)y = A n Y, 90.

Let A, B and C be subsets of a topological space X with CcAuB. If A, B and AuB are given the relative topologies, prove that C is open with respect to Au B if and only if C n A is open with respect to A and C n B is open with respect to B.

EQUIVALENT DEFINITIONS OF TOPOLOGIES 91. Prove Theorem 5.11: Let X be a non-empty set and let there be assigned to each point pEX a class „ of subsets of X satisfying the following axioms: [A ] ] c4 1, is not empty arid p belongs to each member of c-4 0. [A2 ] The intersection of any two members of oA belongs to eigv. [ A3 ] Every superset of a member of c-40 belongs to ,c40. [A4 ] Each member N C c.4 0 is a superset of a member G E / such that G E for every g E G. Then there exists one and only one topology T on X such that cA1 the point p E X.

is the T-neighborhood system of

Let X be a non-empty set and let k: 'P(X) P(X) Kuratowski Closure Axioms:

92 , Prove Theorem 5.12:

[K1 ] k(0) = 0,

[K2 ] A c k(A),

[K3 ] k(Au B) = k(A) u k(B),

satisfy the following

[K4 ] k(k(A)) = k(A)

Then there exists one and only one topology T on X such that k(A) will be the T-closure of AcX.

93.

Prove: Let X be a non-empty set and let

(i) i(X) = X,

(ii) i(A)

c A,

i: 'P(X) —> P(X) (iii) i(Au B) = i(A)

satisfy the following properties:

u i(B),

(iv) i(i(A)) = i(A)

Then there exists one and only one topology T on X such that i(A) will be the T-interior of A c X.

94.

be a class of subsets of X satisfying the following Prove: Let X be a non-empty set and let properties: (i) X and 0 belong to T. (ii) The intersection of any number of members of T belongs to T. (iii) The union of any two members of T belongs to T. Then there exists one and only one topology T on X such that the members of

T are precisely the

T-closed subsets of X. 95.

Let a neighborhood of a real number p E R be any set containing p and containing all the rational numbers of some open interval (a, b) where a < p < b. Show that these neighborhoods actually satisfy the neighborhood axioms and hence define a ( 1) topology on the real line R. (ii) Show that any set of irrational numbers does not contain any accumulation points. ( 7/2, 7/3, w!4, . . . ), does not converge. (iii) Show that any sequence of irrational numbers, such as

Answers to Supplementary Problems 53.

{X, 0 } , (X, {a}, 0), (X, {b} , 0) and (X, {a} , {b} ,

.

56. (ii) Discrete topology. 64. (i) (-00,10]

(ii) R

65. (1): (1) [a, I],

(2) [a, .0) u (p), (3) W.

67.

X = (a, b, c}, T = (X, 0, la, bl, Ic11

75. (i) A, (ii) Ac, 80.

(ii) Unbounded subsets of R.

Let X

(iii)

la, b} be an indiscrete space and let A = {a}.

6

Chapter Bases and Subbases BASE FOR A TOPOLOGY Let (X, T) be a topological space. A class 93 of open subsets of X, i.e. 93 c T, is for the topology T iff (i) every open set G E T is the union of members of 93.

a

base

Equivalently, 93 c T is a base for T ill (ii) for any point p belonging to an open set G, there exists 13 e 93 with pEBc. G. Example 1.1:

The open intervals form a base for the usual topology on the line R. For if G C R is open and p E G, then by definition, 3 an open interval (a, b) with p G (a, b) c G. Similarly, the open discs form a base for the usual topology on the plane R2 .

Example 1.2:

The open rectangles in the plane R2 , bounded by sides parallel to the x-axis and y-axis, also form a base 673 for the usual topology on R2 . For, let G C R2 be open and p G G. Hence there exists an open disc D, centered at p with p E Dp c G. Then any rectangle BECB whose vertices lie on the boundary of 1), satisfies

pEBcD p cG

or

pEB cG

as indicated in the diagram. In other words, '13 satisfies (ii) above, Example 1.3:

{{p} : p E X) of all Consider any discrete space (X, 0). Then the class `73 singleton subsets of X is a base for the discrete topology „D on X. For each singleton set {p} is 0-open, since every A c X is 0-open; furthermore, every set is the union of singleton sets. In fact any other class 673* of subsets of X is a base for 0 if and only if it is a superclass of 93, i.e. 93* D

We now ask the following question: Given a class /3 of subsets of a set X, when will the class 93 be a base for some topology on X? Clearly X = utB: B G 931 is necessary since X is open in every topology on X. The next example shows that other conditions are also needed. Example 1.4:

Let X -= {a, b, c}. We show that the class q3 consisting of {a, b} and {b, i.e. {{a, 1)}, c}}, cannot be a base for any topology on X. Fo'-then {a, b} and b,c} would themselves be open and therefore their intersection {a,b} n {b,c} = { b} would also be open; but {b} is not the union of members of '13,

The next theorem gives both necessary and sufficient conditions for a class of sets to be a base for some topology.

Theorem 6.1:

Let 93 be a class of subsets of a non-empty set X. Then 93 is a base for some topology on X if and only if it possesses the following two properties: (i) X = U{B : B E 93}. (I1) For any B,B* E93, B n B* is the union of members of B, or, equivalently, if pEBn B* then 3B, G 93 such that p GBpcBnB*. 87

BASES AND SUBBASES

88 Example L5:

[CHAP. 6

Let 93 be the class of open-closed intervals in the real line R: 673

= {(a, b] : a,b E R, a< b)

Clearly R is the union of members of '13 since every real number belongs to some open-closed intervals. In addition, the intersection (a, b] n (e, d] of any two openclosed intervals is either empty or another open-closed interval. For example, if a
then

(a, b] n (c,

(c, bl

as indicated in the diagram below.

•

411 d

a

Thus the class T consisting of unions of open-closed intervals is a topology on R, i.e.

'13 is a base for a topology T on R. This topology T is called the upper limit topology on R. Observe that T Similarly, the class of closed-open intervals,

93*

{[a, b) : a,b E R, a G b }

is a base for a topology T* on R called the lower limit topology on R.

SUBBASES Let (X, T) be a topological space. A class ef of open subsets of X, i.e. ef c T, is a subbase for the topology T on X if finite intersections of members of J form a base for T. Example 2.1:

Observe that every open interval (a, b) in the line R is the intersection of two infinite open intervals (a, 00) and (—ac, b): (a, b) ----- (a, n b). But the open intervals form a base for the usual topology on R; hence the class ef of all infinite open intervals is a subbase for R. 00)

Example 2.2:

The intersection of a vertical and a horizontal infinite open strip in the plane R 2 is an open rectangle as indicated in the diagram below.

But, as noted previously, the open rectangles form a base for the usual topology on R2. Accordingly, the class 4 of all infinite open strips is a subbase for R 2.

TOPOLOGIES GENERATED BY CLASSES OF SETS Let cri be any class of subsets of a non-empty set X. As seen previously, eyti may not be a base for a topology on X. However, ey'l always generates a topology on X in the following sense: Theorem 6.2: Any class a of subsets of a non-empty set X is the subbase for a unique topology T on X. That is, finite intersections of members of e}I form abase for the topology T on X. Example 3.1:

Consider the following class of subsets of X = {a, b, c, d}:

cA =

{{a, b}, {b, c}, {d}}

Finite intersections of members of c/I gives the class {{a, b), {b, cl, {d}, {b}, 0, X}

89

BASES AND SUBBASES

CHAP. 61

(Note X E '73 since, by definition, it is the empty intersection of members of eA.) Taking unions of members of eB gives the class T

{{a, b}, {b, c}, {d}, {b}, 0, X, {a, b, d}, {b, c, d}, {b,

{a, b, c}}

which is the topology on X generated by the class cA. Let (X, <) be any non-empty totally ordered set. The topology on X generated by the subsets of X of the form {xEX:p<x,pEX} {xEX : x
Example 3.2:

is called the order topology on X. Observe, by Example 2.1, that the usual topology on R is, in fact, identical to the (natural) order topology on R.

The topology generated by a class of sets can also be characterized as follows: Proposition 6.3: Let ell be a class of subsets of a non-empty set X. Then the topology T on X generated by e/1 is the intersection of all topologies on X which contain Q.T.

LOCAL BASES Let p be any arbitrary point in a topological space X. A class 93 p of open sets containing p is called a local base at p if for each open set G containing p, 3Gp G-93p with the property p G G, c G. Example 4.1: Consider the usual topology on the plane R2 and any point p E R2 . Then the class 6Bp of all open discs centered at p is a local base at p. For, as proven previously, any open set G containing p also contains an open disc Dp whose center is p.

I

p„

E)

T Dp

Similarly, the class of all open intervals (a — 8, a+ 6) in the line R with center a E R is a local base at the point a.

The following relationship between a base ("in the large") for a topology and a local base ("in the small") at a point clearly holds:

Proposition 6.4:

Let 93 be a base for a topology T on X and let p

G X. Then the members of the base 93 which contain p form a local base at the point p.

Some concepts previously defined in terms of the open sets containing a point p can also be defined merely in terms of the members of a local base at p. For example,

Proposition 6.5:

A point p in a topological space X is an accumulation point of A C X iff each member of some local base ‘B, at p contains a point of A different from p.

Proposition 6.6:

A sequence (a1, a2, ) of points in a topological space X converges to p E X if each member of some local base 93, at p contains almost all of the terms of the sequence.

The previous three propositions imply the following useful corollary.

Corollary 6.7:

Let 93 be a base for a topology T on X. Then: (i) p G X is an accumulation point of A c X if each open base set B E containing p contains a point of A different from p; (ii)

a sequence (ai, a2, ) of points in X converges to p G X if each open base set B E 93 containing p contains almost all of the terms of the sequence.

[CHAP. 6

BASES AND SUBBASES

90 Example 4.2:

Consider the lower limit topology T on the real line R which has as a base the class of closed-open intervals [a, b), and let A -= (0,1). Note that G = (1,2) is a T-open set containing 1 E R for which GnA -= 0; hence 1 is not a limit point of A. On the other hand, 0 C R is a limit point of A since any open base set [a, b) con0 < b, contains points of A other than 0. taining 0, i.e. for which a

Solved Problems BASES 1.

Show the equivalence of both definitions of a base for a topology, that is, if 113 is a subclass of T then the following statements are equivalent: (i)

Each G E T is the union of members of '1 3 .

(ii) For any point p belonging to an open set G, Solution: If G = u i B, where B i e 53, then each point B 10 in the union; so

3B, G 6/3 such that p

p eG

G

B,

C

G.

u f /3 1 belongs to at least one member

p E Bio c u i B i -= G

On the other hand, if for each p E G, 3B, E eB

such that p E B p cG, then

G = L.1{B p p e and G is the union of members of `73.

2.

Determine whether or not each of the following classes of subsets of the plane R2 is a base for the usual topology on R2 : (i) the class of open equilateral triangles; (ii) the class of open squares with horizontal and vertical sides. Solution: Both of the above classes are a base for the usual topology on R2. For let G be an open subset of R2 and let p C G. Then 3 an open disc D, centered at p such that p E Dp c G. Observ that either an equilateral triangle or a square can be inscribed in D, as indicated in the diagrams below.

Hence each class satisfies the second definition of a base for a topology.

3.

Let '73 be a base for a topology T on X and let '73 * be a class of open sets containing 673, i.e. 673 C c T. Show that 673* is also a base for T. Solution: Let G be an open subset of X. Since '73 is a base for (X, T), G is the union of members of 53, i.e. G = U 1 B1 where B ,e53. But 53 c 53*; hence each B i e ‘13 also belongs to 53*. So G is the union of members of 53*, and therefore 73 * is also a base for (X, T).

CHAP. 6]

4.

91

BASES AND SUBBASES

Let X be a discrete space and let B be the class of all singleton subsets of X, i.e. = {{p} : p e X). Show that any class B* of subsets of X is a base for X if and only if it is a superclass of B. Solution: Suppose is a base for X. Since any singleton set {p} is open in a discrete space, {p} must be a union of members of `73*. But a singleton set can only be the union of itself or itself with the empty set 0. Hence {p} must be a member of '73* , so '13 c '73* . On the other hand, since '73 is a base for the discrete space X (see Example 1.3), any superset of B is also a base for X.

5.

Prove Theorem 6.1: Let GB be a class of subsets of a non-empty set X. Then B is a base for some topology on X if and only if it satisfies the following two properties .

(i)

X = U{B B E

(ii)

For any 11, B* E B, B n B* is the union of members of B, or, equivalently, if p eBn B* then 3 1? „ E B such that p e B„ c B n 13*.

B).

Solution: Suppose '13 is a base for a topology T on X. Since X is open, X is the union of members of D, Hence X is the union of all the members of 173, i.e. X = U{B: B e D}. Furthermore, if B, Bt E 613 then, in particular, B and B* are open. Hence the intersection B n B* is also open and, since D is a base for T, it is the union of members of 'B. Thus (i) and (ii) are satisfied. Conversely, suppose D is a class of subsets of X which satisfy (i) and (ii) above. Let T be the class of all subsets of X which are unions of members of B. We claim that T is a topology on X. Observe that 673 c T will be a base for this topology. By (i), X

U{B: B E D } ; so X G T.

Note that 0 is the union of the empty subclass of D, i.e.

= U{B: BE0c93 }; hence 0 G T, and so T satisfies W i t Now let {G i } be a class of members of T. By definition of T, each G. is the union of members of D; hence the union Li G, is also the union of members of B and so belongs to T. Thus T satisfies [02 ]. Lastly, suppose G, H G T. We need to show that G n H also belongs to T. By definition of T, there exist two subclasses {B i : i G and {Bi: je J } of 1H such that G -= U,B1 and H .-= ui B i . Then, by the distributive laws, G n H

(u i B i ) n (u i B i)

U{B i n Bi : i EI, j

J}

But by (ii), B i n B i is the union of members of '73; hence G n H = U{B i n B i E I, j G J } is also the union of members of '73 and so belongs to T which therefore satisfies [03 ]. Hence T is a topology on X with base 'B.

6.

Let B and B* be bases, respectively, for topologies T and T* on a set X. Suppose that each B E 13 is the union of members of CB* . Show that T is coarser than T*, i.e.

T c T*. Solution: Let G be a T-open set. Then G is the union of members of D, i.e. G U,B i where B,G'73. But, by hypothesis, each B, G D is the union of members of D*, and so G U,B 1 is also the union of members of 73'. which are V. -open sets. Hence G is also a T*-open set, and so T c T*.

7.

Show that the usual topology on the real line R is coarser than the upper limit topology T on R which has as a base the class of open-closed intervals (a, N. Solution: Note first that any open interval is the union of open-closed interva ) s.

(a, b) = U {(a, b

For example,

1/n] : n G NI

Since the class of open intervals is a base for If, by the preceding problem, 14 c T, i.e. any CU-open set is also T-open.

BASES AND SUBBASES

92 8.

[CHAP. 6

Consider the upper limit topology T on the real line R which has as a base the class of open-closed intervals (a, b]. (i) Show that the open infinite interval (4, co) and the closed infinite interval (-00, 2] are T-open sets. (ii) Show that any open infinite interval (a, 00) and any closed infinite interval (-00, b] are T-open sets. (iii) Show that any open-closed interval (a, b] is both T-open and T-closed. Solution: (i) Observe that

(4, 00)

= (4,5] U (4, 6] U (4,7] U (4,8] U • • •

(—c*, 2]

==

(0,2] U (-1,2] u (-2,2] u • • •

Hence each is T-open since each is the union of members of the base for T. (ii)

(a, .0) -= (a, a + 1] U (a, a + 2] U (a, a + 3] U • • •

Similarly,

b]

(b — 1, b] U (b — 2, b] u (b —3, b] U (b

—

4, b] u . • •

Hence eaA is T-open. (iii)

(a, ble -= (b, Go), and the two intervals on the right are open, so their union is open and therefore (a, bl is closed. But (a, b] belongs to the base for T and so is also open.

SUBBASES, TOPOLOGIES GENERATED BY CLASSES OF SETS 9. Let X = a, b, c, d, e} and let c/T b, cl, (e, d), ( d, e)}. Find the topology -ein X generated by cA . Solution: First compute the class 6/3 of all finite intersections of sets in c.4

73

6

{X, {a, b, c}, {c,

{c}, {d}, 01

{d,

(Note that X '73, since by definition X is the empty intersection of members of cif.) Taking unions of members of cB gives the class T

{X, {a, b, c}, {c, d}, {d, c} , {c}, {d}, 0,

{a, b, c,

{c, d, e}}

which is the topology on X generated by e4.

10. Determine the topology T on the real line R generated by the class c4 of all closed intervals [a, a + 1] with length one.

Solution: Let p be any point in R. Note that the closed intervals as they have length one. Hence [p —1, p] n [p, p 1]

[p — 1, p]

and

[p, p + 1]

belong to ce/

{p}

belongs to the topology T, i.e. all singleton sets {p} are T-open, and so T is the discrete topology on X.

11. Let c./61 be the class of all open half-planes

H

H in the plane R2 of the form

{(x , v) : x < a, or x > a, or

y < a, or

y > a}

(See diagrams below.)

x < a

x > a

Find the topology on R2 generated by

y< a e.g.

y>a

CHAP. 6]

93

BASES AND SUBBASES

Solution: Observe that every open rectangle four half-planes

{(x,

H2 -=

B = {(x, y) : a < x < b, c < y < d} : a < x}

H3 H4 =

{(x,y) : x <

y)

is the intersection of the

c<

{(x, y) : y <

Since each H G cif is 71-open, and since the class of all open rectangles B is a base for the usual topology II on R2, the class cif is a subbase for U. That is, 0.-4 generates the usual topology on the plane R2.

12. Consider the discrete topology -D on X -= {a, b, c, d, e}. Find a subbase 5 for .D which does not contain any singleton sets. Solution: Recall that any class D of subsets of X is a base for the discrete topology I) on X if it contains gives all singleton subsets of X. Hence (-5 is a subbase for /i) if finite intersections of members of 1, {d} and {e}. So (4 = Oa, b} , {b, c}, {c, , {d, e}, {e, a}} is a subbase for ,j), {a}, {b},

13. Let H be a subbase for a topology T on X and let A be a subset of X. Show that the class (SA = {AnS : SECS} is a subbase for the relative topology T4 on A. Solution: Let H be a TA-open subset of A. hypothesis, d is a subbase for T; so

G H

Hence

= u (S

Then

H -= A n G

S12 n • n

Sin d

= AnG = A n [u i (S11 n =

where G is a T-open subset of X. where S.Pk

n

By

e (7,5

Sin)]

u i 1(A n S11 ) n • • • n (A n

Sin )]

Thus H is the union of finite intersections of members of JA and therefore ,T,JA is a subbase for TA.

14. Show that all intervals (a, 1] and [0, b), where 0 < a, b < 1, form a subbase for the relative usual topology on the unit interval I = [0, 1]. Solution: Recall that the infinite open intervals (a, oc) and (— co, b) form a subbase for the usual topology on the real line R. The intersection of these infinite open intervals with I = [0,1] are the sets 0,1, (a,1] and [0, b) which, by the preceding problem, form a subbase for I = [0,1]. But we can exclude the empty set 0 and the whole space I from any subbase; so the intervals (a,1) and [0, b) form a subbase for I.

15. Show that if cf is a subbase for topologies T and T* on X, then T = Solution: Suppose G G T. Since J is a subbase for T, G

4 •• n

But

S

= U (sii n • • • n Sind where Sik G J.

E Since T* is a topology, is also a subbase for T* and so cf c T*; hence each and hence G G T*. Thus T c V'. Similarly T* c T, and so T = T*. Si,, G

16. Prove Theorem 6.2: Any class ei/ of subsets of a non-empty set X is the subbase for a unique topology on X. That is, finite intersections of members of ei/ form a base for a topology T on X. Solution : We show that the class '73 of finite intersections of members of Theorem 6.1 for it to be a base for a topology on X:

(i) (ii)

c/I satisfies the two conditions in

X -= U{B : B B}. For any G,H e 93, G n H is the union of members of D.

94

ICHAP. 6

BASES AND SUBBASES

Note X E D, since X by definition is the empty intersection of members of e4; so

X ----- U{B B E 93} Furthermore, if G,H E D, then G and H are finite intersections of members of e/I. Hence H is also a finite intersection of members of cA and therefore belongs to D. Accordingly, D is a base for a topology T on X for which e4 is a subbase. The preceding problem shows that T is unique.

G

n

17. Prove Proposition 6.3: Let c..4 be a class of subsets of a non-empty set X. Then the topology T on X generated by eA is the intersection of all topologies on X which contain cif. Solution: Let {Ti } be the collection of topologies on X containing cA, and let T* n 1 Ti. Note that cA c T. We want to prove that T = T*. Since T is a topology containing cif, and T* is the intersection of all such topologies, we have T* c T. On tile other hand, suppose G E T. Then by the definition of T,

u

G But

M. C T*,

hence each

(sii n Si2 n • • n

Sik G T*.

Accordingly,

G = U, We have shown that

G E T implies

where

Sil n • • • n

n • • • n Sin

d

S T*

cA and so

G T*

G E T*; hence T c T*.

Consequently T T*.

LOCAL BASES 18. Prove Proposition 6.5: A point p in a topological space (X, T) is an accumulation point of A C X if each member of some local base '73p at p contains a point of A different from p. Solution: Recall p G X is an accumulation point of A if (G\{p}) n A 0 for all G But Dv c T, so in particular (B\{p}) n A 0 for all B E DI9

e

T such that

p E G.

Conversely, suppose (B\{p}) n A f12) for all B E 'B p , and let G be any open subset of X containing p. Then 3B 0 G `Bp for which p E B o c G. But then

(G\{p}) n A D (B 0 \ {p}) n A So (G\{p}) n A

0,

or p is an accumulation point of A.

19. Prove Proposition 6.6: A sequence (a1, a2, of points in a topological space (X, T) converges to p EX iff each member of some local base 930 at p contains almost all of the terms of the sequence. Solution: Recall that a„ --> p if every open set G G T containing p contains almost all the terms of the sequence. But 613 p c T, so in particular each B E V3 1, contains almost all the terms of the sequence. On the other hand, suppose every B G 93, contains almost all the terms of the sequence, and let G be any open set containing p. Then 3/30 E '13, for which p E Bo c G. Hence G also contains almost all the terms of the sequence, and so (a) converges to p.

20. Show that every point p in a discrete space X has a finite local base. Solution: Note that the singleton set {p} is open since every subset of a discrete space is open. Accordingly {{p}}, i.e. the class consisting of the singleton set {p}, is a local base at p since the class every open set G containing p must be a superset of {p}.

CHAP. 6]

95

BASES AND SUBBASES

21. Consider the upper limit topology T on the real line R which has as a base the class of open-closed intervals (a, b]. Determine whether or not each of the following sequences converges to 0:

(i) ( 1,

A-,

Solution: (i) No.

For the T-open set (-2, 0] containing 0 does not contain any term of the sequence.

(ii)

•••)

(ii) ( -1 , -4-,

.

)

Yes. For any open basic set (a, Id containing 0, i.e. for which a < —1/n 0 < O. Hence n > n0 implies —1/n G (a,

a
b, 3n0 G N such that

Supplementary Problems BASES FOR TOPLOLOGIES

22. Show that the class of closed intervals [a, bl, where a and b are rational and a < b, is not a base for a topology on the real line R. 23. Show that the class of closed intervals [a, b ] , where a is rational and b is irrational and a < b, is a base for a topology on the real line R. 24. Let 6/3 be a base for a topology T on X and let A c X. is a base for the relative topology TA on A.

Show that the class

TA

{AnG :GCB}

25. Let B be the class of half-open rectangles in the -plane R2 indicated in the diagram on the right, i.e. of the form {(x, y) a = x b, c y d}

,a

base for a topology T on R2.

(i)

Show that '73 is

(ii)

Show that the relative topology TA on the line

A

a

={xy) : x y = 0}

is the discrete topology on A. (iii)

Show that the relative topology Tg on the line B = {x, y) x yl is not the discrete topology on B.

26.

Let '73 be a class of subsets of a non-empty set X totally ordered by set inclusion. Show that '73 is a base for a topology on X provided that X = U {B : B E 6/3}.

27.

Show that a topology T-on X is finite if and only if T has a finite base.

SUBBASES 28. Let 29.

X = {a, b, e, d, .

Find the topology T on X generated by cif = {{a}, {a, b, c } , {c, d}} .

Determine the smallest subbase ef for the discrete topology T on any non-empty set X.

30. Let d be the class of all closed intervals [a, h] where a and b are rational, i.e. a, b E Q, and a G b. Show that J U {{p} : p C Q}, is a base for the topology T on the real line R generated by J.

x, 0}

31.

Show that if ef is a subbase for a topology T on X, then

32.

Let T and T* be the topologies on X generated respectively by cA and c/I*. c cif* c T implies (i) e4 c e.g. * implies T c T*; and (ii) Show that:

is also a subbase for T.

T = T*.

96

BASES AND SUBBASES

[CHAP. 6

33. Let j be a subbase for a topological space X and let G c X be an open set containing a point p E X. Show that there exists a finite number of members of cf, say S I , S2 , . , Sm , with the property that P E S 1 fl S2 n • • • n Sm c G. LOCAL BASES

34.

Let (X, T) be a topological space and let cA be a T-local base at p G X. Consider any subset A of X such that pEA c X, and consider the relative topology TA on A. Show that the following class of subsets of A is a TA-local base at p E A: cAA {AnG:GEcAl}.

35.

Let X be a topological space, let p E X, let Np be the neighborhood system at p and let Op be a local base at p. Show that every neighborhood of p contains a member of the local base at p; i.e. for every N E .7\z 3G E Bp for which G c N.

36.

Show that if a point p has a finite local base Bp then it also has a local base consisting of exactly one set.

37.

Consider the upper limit topology T on the real line R which has as a base the class of open-closed intervals (a, b]. Determine whether or not each of the following sequences converges:

(i) (1, 38.

39.

(ii) (-1,

...),

(iii) (-1,

...).

Let T be the topology on the real line R generated by the class j of all closed intervals la, I)] where a and b are rational (see Problem 30). (i)

Determine whether or not each of the following sequences converges: (a) (2 + 4, 2 + k 2 + ...), (b) + -V2- +*, Vi+ 1, ...).

(ii)

Determine the closure of each of the following subsets of R: (b) (c) (d) A = {1, ..}. (a) (2, 4),

(iii)

Show that any finite subset of R is T-closed.

Let of be a subbase for a topological space X and let p E X. es,

=

{SE ef:pES}

need not be a local base at p.

(i)

Show by a counterexample that the class

(ii)

Show that finite intersections of members of cf, do form a local base at p.

(iii)

Show that a sequence (a n) in X converges to p if and only if every S E finite number of the terms of the sequence.

4,

Answers to Supplementary Problems 28. T = {X, 0, {a} , {c} , {a, c} , {c,

37. (i) No 38. (i):

(ii) Yes

(a) No,

, {a, b, c}, {a, c,

, {a, b, c, d}}

(iii) No.

(b) Yes.

39. (ii) Hint. Use Problem 33.

(ii):

(a) (2, 4), (b)

(c) (-3, Id , (d) A.

contains all except a

Chapter

7

Continuity and Topological Equivalence CONTINUOUS FUNCTIONS Let (X, T) and (Y, T*) be topological spaces. A function f from X into Y is continuous relative to T and T*, or T-T* continuous, or simply continuous, if the inverse image [H] of every T*-open subset H of Y is a T-open subset of X, that is, iff H E T*

implies

E T

We shall write f: (X, T) -> (Y, T*) for a function from X into Y when it is convenient to indicate the topologies involved. Example 1.1:

Consider the following topologies on X =---• {a, b, e, d} and Y = {x, y, z, w} respectively: T = (X, 0, {a}, {a, b } , {a, b, c}}, T* = (Y, 0, {x}, {y}, {x, y}, {y,z,w}} Also consider the functions below:

f: X

Y

and

v. X —> Y defined by the diagrams

The function f is continuous since the inverse of each member of the topology T* on Y is a member of the topology T on X. The function g is not continuous since {y,z,w} G T*, i.e. is an open subset of Y, but its inverse image g -1 [{y,z,w}1 = {c, d} is not an open subset of X, i.e. does not belong to T.

Example 1.2: Consider any discrete space (X, 0) and any topological space (Y, T). Then every function f : X Y is 0-T continuous. For if H is any open subset of Y, its inverse f l [11] is an open subset of X since every subset of a discrete space is open. Example 1.3:

Let f:X . --.17 where X and Y are topological spaces, and let be a base for the topology on Y. Suppose for each member B G 93, f —1 [B] is an open subset of X; then f is a continuous function. For let H be an open subset of Y; then H = U a union of members of D. But

f

[111 = f Eu

= uff and each f -1 [B,] is open by hypothesis; hence f -1 [H] is the union of open sets and is therefore open. Accordingly, f is continuous.

We formally state the result of the preceding example. Proposition 7.1:

A function f:X->Y is continuous if the inverse of each member of a base B for Y is an open subset of X.

This proposition can in fact be strengthened as follows:

Theorem 7.2:

Let ef be a subbase for a topological space Y. Then a function f: X -› Y is continuous iff the inverse of each member of ef is an open subset of X.

97

[CHAP. 7

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

98 Example 1.4:

The projection mappings from the plane R2 into the line R are both continuous relative to the usual topologies. Consider, for example, the projection IT R 2 R y. Then the inverse of any open interval (a, b) is an infinite defined by 27-((x, y)) open strip as illustrated below:

fr -1

[(a, b)] is shaded

Hence by Proposition 7.1, the inverse of every open subset of R is open in R2, i.e.

17- is continuous.

Example 1.5:

loci The absolute value function f on R, i.e. 1(x) For if A = (a, b) is an open interval in R, then {0 f 1 [A]

=

(—b, b) (—b, a) U (a, b) —

as illustrated below.

for

x G R,

is continuous.

if a < b 0 if a < 0< b if 0--a< b

In each case f -1 [Al is open; hence f is continuous.

f' [A] = Ø

f 1 [A] = (—b , b)

f — ' [A] = (—b , — a)

u (a, b)

Continuous functions can be characterized by their behavior with respect to closed sets, as follows: Theorem 7.3: A function f: X -> Y is continuous if and only if the inverse image of every closed subset of Y is a closed subset of X. CONTINUOUS FUNCTIONS AND ARBITRARY CLOSENESS Let X be a topological space. A point p E X is said to be arbitrarily close to a set A c X if either (i) p E A or (ii) p is an accumulation point of A Recall that Az---- ALJA'; so the closure of A consists precisely of those points in X which are arbitrarily close to A. Recall also that A. = A° u b (A); hence p is arbitrarily close to A if p is either an interior or a boundary point of A. Continuous functions can also be characterized as those functions which preserve arbitrary closeness, namely, Theorem 7.4: A function f:X->Y is continuous if and only if, for any p E X and any A c X, p arbitrarily close to A f(p) arbitrarily close to f[A] or or

PG

f(p)E

A f[Al c

f[A]

A

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

CHAP. 71

99

CONTINUITY AT A POINT Continuity as we have defined it is a global property, that is, it restricts the way in which a function behaves on the entire set X. There also exists a corresponding local concept of continuity at a point. A function f: X 4 Y is continuous at p E X if the inverse image f --1 [I/1 of every open set H C Y containing f(p) is a superset of an open set G C X containing p or, equivalently, if the inverse image of every neighborhood of f(p) is a neighborhood of p, i.e., E Np N eN Notice that, with respect to the usual topology on the real line R, this definition coincides with the E — 8 definition of continuity at a point for functions f: R R. In fact, the relationship between local and global continuity for functions f: R -> R holds true in general; namely, Theorem 7.5: Let X and Y be topological spaces. Then a function f:X4Y is continuous if and only if it is continuous at every point of X.

SEQUENTIAL CONTINUITY AT A POINT A function f: X Y is sequentially continuous at a point p E X if for every sequence (an ) in X converging to p, the sequence (f(a)) in Y converges to f(p), i.e., -› p implies f (an) --> f (p)

Sequential continuity and continuity at a point are related as follows: Proposition 7.6: If a function f:X-4Y is continuous at p E X, then it is sequentially continuous at p. Remark:

The converse of the previous proposition is not true. Consider, for example, the topology T on the real line R consisting of V) and the complements of countable sets. Recall (see Example 7.3 of Chapter 5) that a sequence (an) converges to p if and only if it has the form (a 1 , a2, . • •

2

an02 PP

PY • • •

Then for any function f: (R, T) -› (X, T*), f(a), f(P), f(P), RP), . converges to f(p). In other words, every function on (R, T) is sequentially continuous. On the other hand, the function gR, T) (R, II) defined by f(x) ,--- x, i.e. the identity function, is not T-11 continuous since f -1 [(0,1)] =- (0,1) is not a T-open subset of R. (f(an)) = (f(ai),

OPEN AND CLOSED FUNCTIONS A continuous function has the property that the inverse image of every open set is open and the inverse image of every closed set is closed. It is natural then to ask about the following types of functions:

(1) A function f: X Y is called an open (or interior) function if the image of every open set is open. (2) A function g: X 4 Y is called a closed function if the image of every closed set

is closed. In general, functions which are open need not be closed and vice versa. In fact, the function in our first example is open and continuous but not closed.

[CHAP. 7

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

100 Example 2.1:

Consider the projection mapping 7T : R2 R of the plane R2 into the x-axis, i.e. =-- x. Observe that the projection 7[D] of any open disc D c R 2 is an open interval. Hence any point 77-(p) in the image 77-[G] of an open set G c R2 belongs to an open interval contained in 7T[G], or 7T[G1 is open. Accordingly, 7T is an open function. On the other hand, r is not a closed function, for the set A = {(x,y) : xy 1, x > 01 is closed, but its projection 77-[A] (0,00 is not closed. (See diagrams below.)

17 A] [

HOME OMORPHIC SPACES A topological space (X, T) is, as we have seen, a set X together with a distinguished class T of subsets of X, satisfying certain axioms. Between any two such spaces (X, T) Y. We choose to discuss continuous, or open, and (Y, T*) there are many functions f: X or closed functions rather than arbitrary functions since it is these functions which preserve some aspect of the structure of the spaces (X, T) and (Y, T*). Y. Then Now suppose there is some bijective (i.e. one-one and onto) mapping f: X f induces a bijective function f: P(X) P(Y) from the power set of X, i.e. the class of subsets of X, into the power set of Y. If this induced function also takes T onto T*, i.e. defines a one-to-one correspondence between the open sets in X and the open sets in Y, then the spaces (X, T) and (Y, T*) are identical from the topological point of view. Specifically: Definition: I Two topological spaces X and Y are called homeomorphic or topologically equivalent if there exists a bijective (i.e. one-one, onto) function f: X -› Y such that f and f ' are continuous. The function f is called a homeomorphism. A function f is called bicontinuous or topological if f is open and continuous. Thus f : X -› Y is a homeomorphism iff f is bicontinuous and bijective. Example 3.1:

Let X = - (-1,1). The function f: X —> R defined by f(x) = tan 4-- ,n-x is one-one, onto and continuous. Furthermore, the inverse function f --1 is also continuous. Hence the real line R and the open interval (-1,1) are homeomorphic.

Example 3.2:

Let X and Y be discrete spaces. Then, as seen in Example 1.2, all functions from one to the other are continuous. Hence X and Y are homeomorphic if theee exists a one-one, onto function from one to the other, i.e. if they are cardinally equivalent.

Proposition 7.7: The relation in any collection of topological spaces defined by "X is homeomorphic to Y" is an equivalence relation. Thus, by the Fundamental Theorem on Equivalence Relations, any collection of topological spaces can be partitioned into classes of topologically equivalent spaces.

TOPOLOGICAL PROPERTIES A property P of sets is called topological or a topological invariant if whenever a topological space (X, T) has P then every space homeornorphic to (X, T) also has P.

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

CHAP.?]

101

Example 4.1:

As seen in Example 3.1, the real line R is homeomorphic to the open interval X = (-1,1). Hence length is not a topological property since X and R have different lengths, and boundedness is not a topological property since X is bounded but R is not.

Example 4.2:

Let X be the set of positive real numbers, i.e. X = (0, co). The function f: X X defined by f(x) 1Ix is a homeomorphism from X onto X. Observe that the sequence (a.) ( 1 ,1, A-, corresponds, under the homeomorphism, to the sequence

(Pan» =

( 1 , 2 , 3, —)

The sequence (ad is a Cauchy sequence; the sequence (f(an )) is not. property of being a Cauchy sequence is not topological.

Hence the

Most of topology is an investigation of the consequences of certain topological properties as compactness and connectedness. In fact, formally topology is the study of topological invariants. In the n ext example, connectedness is defined and is shown to be a topological property. Example 4.3:

A topological space (X, T) is disconnected ill' X is the union of two open, non-empty, disjoint subsets, i.e.

X= GuH

where

G,HET,GnH=0 but

G,HO

If f : X Y is a homeomorphism then X = G U H if and only if Y and so Y is disconnected if and only if X is. The space (X, T) is connected

f[H)

it is not disconnected.

TOPOLOGIES INDUCED BY FUNCTIONS Let f (17 ., T,)} be any collection of topological spaces and for each Y, let there be given a function f : X Yi defined on some arbitrary non-empty set X. We want to investigate those topologies on X with respect to which all the functions f are continuous. Recall that f i is continuous relative to some topology on X provided the inverse image of each open subset of Yi is an open subset of X. Thus we consider the following class of subsets of X: .

=U

{f [H] : H G T,}

That is, cf' consists of the inverse image of each open subset of every space Y,. The topology T on X generated by J is called the topology induced (or generated) by the functions f The main properties of T are listed in the next theorem. Theorem 7.8: (i) (ii)

All the functions f are continuous relative to T. T is the intersection of all the topologies on X with respect to which the functions f are continuous.

(iii) T is the smallest, i.e. coarsest, topology on X with respect to which the functions f are continuous. (iv)

cf is a subbase for the topology T.

We shall call -4 the defining subbase for the topology induced by the functions f,, i.e. the coarsest topology on X with respect to which the functions f , are continuous. Example 5.1:

Let 77-1 and

72

be the projections of the plane R2 into R, i.e., 7r1((x, Y)) = x

and

72 ((x,y)) = y

Observe, as illustrated below, that the inverse image of an open interval (a, b) in R is an infinite open strip in W.

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

102

7,-T 1 [(a, b)]

[CHAP. 7

2

[(a ' b)]

Recall that these infinite open strips form a subbase for the usual topology on R2. Accordingly, the usual topology on R2 is the smallest topology on R2 with respect to which the projections 7 1 and 72 are continuous.

Solved Problems CONTINUOUS FUNCTIONS 1.

Prove: Let f: X --> Y be a constant function, say f(x) = p e Y, for every x Then f is continuous relative to any topology T on X and any topology T*

e X.

Solution: We need to show that the inverse image of any T*-open subset of Y is a T-open subset of X. Let H G T*. Now f(x) = p for all x É X, so X 0 if p G H f 1 [H] = if p H ■

{

a

0

In either case f -1 [H] is an open subset of X since X and

2.

Prove: Let f: X -› Y be any function. f: (X, 7)-4 (Y, S) is continuous for any T.

belong to every topology T on X.

If (Y, c9) is an indiscrete space, then

Solution: We want to show that the inverse image of every open subset of Y is an open subset of X. Since (Y, g) is an indiscrete space, Y and 0 are the only open subsets of Y. But

f -1 [Y] = X, f-1 [0] = 0 and X and 0 belong to any topology T on X. Hence f is continuous for any T.

3.

Let 14 be the usual topology on the real line R and let T be the upper limit topology on R which is generated by the open-closed intervals (a, b]. Furthermore, let f: R --> R be defined by f(x) =

x

if x -- 1

x+2

if x > 1

(See diagram on the right.) (i) Show that f is not 14-11 continuous. (ii) Show that f is T-T continuous. Solution: (i) Let A = (-3,2). continuous.

Then f --" 1 [A] = (-3, 1].

Now

A e 14 but f -1 [A] V It,

SO

f

is not

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

CHAP. 7]

Let A = (a, b] . Then:

(ii)

I (a, b] (a, 11 (a, b — 2]

f —1 [A]

0 (1,b-2] (a — 2, b — 2]

In each case, f —1 [A] is a T-open set.

4.

if a
Hence f is T-T continuous.

Suppose a function f: (X, T1) (Y, T2) is not T1-T2 continuous. Show that if T;' is a topology on X coarser than T1 and if T: is a topology on Y finer than T2, i.e. V C Ti and T2 C 71, then f is also not T,* -71 continuous. Solution: Since

f: (X, T 1 ) —> (Y,

T2)

is not continuous,

3G G

T2

for which

f [G E T 1

Now, T i c T 1 and T2 C T. Hence G G T2 implies G G 7";, f 1 [G] E T i . Thus f is not continuous with respect to T i and T2

5.

103

Show that the identity function j: (X, finer than T*, i.e. T* c T.

T)

(X,

T*)

and

f -1 [G] E T1 implies

is continuous if and only if T is

Solution: By definition, i is T-T* continuous if and only if

i 1 [G] CT

G

[G] = G, so i is T-T* continuous, if and only if

But

GE T

G ET*

that is, T* c T.

6.

Prove Theorem 7.2: Let f: (X, T) -› (Y, T'), and let cf be a subbase for the topology 7- * on Y. Then f is continuous if and only if the inverse of every member of the subbase cf is an open subset of X, i.e. f - l[S] E T for every S E J. Solution: Suppose f — / [S] G T for every S C J. We want to show that f is continuous, i.e. G E T* implies V [G] e T. Let G e T*. By definition of subbase,

G = U (S il n • • • n Sind Hence,

f —1 [G]

=

f1 [u (S i]

n••n

(f -1 [Si ] n • • •

n

S,k E

Sik G

ui f

Sin)]

[S

• • • n Sin i]

f —1 [Si, j)

implies f 1 [S,k ] E T. Hence f —1 [G] E T of open sets. Accordingly, f is continuous.

But

where

since it is the union of finite intersections

On the other hand, if f is continuous then the inverse of all open sets, including the members of cf are open.

7. Let f be a function from a topological space X into the unit interval [0,1]. Show that if f -1 [(a, 11] and f -1 [[0, b)] are open subsets of X for all 0 < a, b < 1, then f is continuous. Solution: Recall that the intervals (a, 1] and [0, b) form a subbase for the unit interval / = [0,1]. Hence f is continuous by the preceding problem, i.e. by Theorem 7.2.

8.

[CHAP. 7

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

104

Prove: Let the functions f: X -› Y and g: Y Z sition function gof:X.4Z is also continuous.

be continuous. Then the compo-

Solution.. Let G be an open subset of Z. Then g -1 [GI is open in Y since g is continuous. continuous, so V' [g -1 [G]] is open in X. Now (g o f) -1 [G]

But f is also

= f -1 [g -1 [G]]

Thus (g o1)1 [G] is open in X for every open subset G of Z, or, g of is continuous.

9.

Prove: Let {T,} be a collection of topologies on a set X. If a function f: X Y is continuous with respect to each T,, then f is continuous with respect to the intersection topology T = n, T. Solution : Let G be an open subset of Y. Then, by hypothesis, f -1 [G] belongs to each Ti . Hence f -1 [G] belongs to the intersection, i.e. f -1 [G] G n,Ti T, and so f is continuous with respect to T.

10. Prove Theorem 7.3: A function f: X Y is continuous image of every closed subset of Y is a closed subset of X. Solution: Suppose f: X Y is continuous, and let F be a closed subset f -1 [Fc] is open in X. But f -1 [Fe] = (f -1 [F])C; therefore f -1 [F] Conversely, assume F closed in Y implies f -1 [F] closed in X. Then Ge is closed in Y, and so f -1 [Ge] = (f-1 [G])e is closed in X. therefore f is continuous.

if and only if the inverse

of Y. Then Fe is open, and so is closed. Let G be an open subset of Y. Accordingly, f -1 [G] is open and

11. Prove Theorem 7.4: A function f: X Y is continuous if and only if, for every subset A c X, f[À] c f[A]. Solution: Suppose f: X -> Y is continuous. Now f[A] c f[A], so A c f -1 [f[A]] c f -1 [f[A]] But f[A] is closed, and so f -1 [f[A]] is also closed; hence A c A c f -1 [f[A]]

f[A] c f[A] =

and therefore

[f[A]]]

Conversely, assume f[A] c JUL] for any A c X, and let F be a closed subset of Y. Set A = f-1 [F]; we wish to show that A is also closed or, equivalently, that A = A. Now f[ii] = f[f -1 [F]] c f1f -1 [Fj] = F= F A C f 1 [f[A]] c f 1 [F]

Hence But A C A,

so A

A

A and f is continuous.

12. Prove: Let f: (X, T)--* (Y, T*) be continuous. Then tinuous, where A C X and f 4 is the restriction of f to A.

(A, TA

)

-->

(

Y, T*)

is con-

Solution: Observe that

.G 1 [G] = A

n f -1 [G] for any G c Y.

Then f 1 [G] E T, and so A n f -1 [G] C TA_ by definition of the induced topology. f1 [G] = f, 1 [G] E TA, SO fA is continuous. Thus A n Let G G

CHAP. 71

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

CONTINUITY AT A POINT 13. Under what conditions will a function f: X

105

Y not be continuous at a point p E X?

Solution: A function f:X->Y is continuous at p E X iff, for every open set HcY containing f(p), f [H] is a superset of an open set containing p. Hence f is not continuous at p e X if there exists at least one open set H c Y containing f (p) such that f -1 [H] does not contain an open set containing p.

Equivalently, f: X Y is not continuous at p E X if 3 a neighborhood N of f (p) such that f -1 [N] is not a neighborhood of p.

14. Consider the following topology defined on {a, b,c,d}: T = {X, 0, {a}, {b}, {a,b}, {b,c,d}}

X =

Let ing (i) (ii)

the function f:X->X be defined by the adjoindiagram. Show that f is not continuous at c. Show that f is continuous at d. Solution : (i) Observe that {a, b } is an open set containing f(c) = b and that J.-1 [{ a, bl] = {a, c } . Hence f is not continuous at c since there exists no open set containing e which is contained in e} (ii) The only open sets containing f(d) X and c are {b,c,d} and X. Note that f [{b,c, cl}] f -1 [X] = X. Hence f is continuous at cl since the inverse of each open set containing f (d) is an open set containing cl. 15. Suppose a singleton set {p} is an open subset of a topological space X. Show that for any topological space Y and any function f:X->Y, f is continuous at p e X. Solution: Let H c Y

be an open set containing f (p). f (p) EH

But

. pEf'[H]

{p} c f

[H]

Hence f is continuous at p.

16. Prove: If f:X->Y is continuous at peX, then the restriction of f to a subset containing p is also continuous at p. More precisely, let A be a subset of a topological space (X, T) such that peACX, and let f A : A --> Y denote the restriction of f: X. -> Y to A. Then if f is T-continuous at p, f A will be TA-continuous at p where TA is the relative topology on A. Solution: Let H c Y be an open set containing f (p). Since f is continuous at p, ]GET such that pE Gc f -1 [H] and so

p EA nGc A nf -1 [H]

But, by definition of the induced topology, AnG G

TA;

fA-1 [H]

hence fA is TA-continuous at p.

17. Prove Theorem 7.5: Let X and Y be topological spaces. Then a function f:X->Y is continuous if and only if it is continuous at every point p E X. Solution: Assume f is continuous, and let H c Y be an open set containing f (p). and f -1 [H] is open. Hence f is continuous at p.

But then

p G f -1 [H],

Now suppose f is continuous at every point p G X, and let H c Y be open. For every p E f - [H], there exists an open set Gp c X such that p E Gp c f -1 [H]. Hence f 1 [H] = U {Gp : p G f -1 [H]} a union of open sets. Accordingly, f -1 [H] is open and so f is continuous.

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

106

[CHAP. 7

18. Prove Proposition 7.6: If a function f: X-> Y is continuous at p e X, then it is sequentially continuous at p, i.e. an -> p f (a.) -> f (P). Solution : We need to show that any neighborhood N of f(p) contains almost all the terms of the sequence (.f(ai), f(a2), • .). Let N be a neighborhood of f (p). By hypothesis, f is continuous at p; hence M [N] is a neighborhood of p. If the sequence (an) converges to p, then M contains almost all the terms of the sequence (a 1 , a2, ...), i.e. a„ E M for almost all n G N. But a,E M

f(a) E

N

f [M] = f [f 1 [N]]

Hence f(a) E N for almost ail n E N, and so the sequence (f(a n)) converges to f(p). Accordingly, f is sequentially continuous at p.

OPEN AND CLOSED FUNCTIONS, HOMEOMORPHISMS 19. Give an example of a real function f : R R such that f is continuous and closed, but not open. Solution: Let f be a constant function, say f(x) = 1 for all x E R. Then f[A] = {1} for any A C R. Hence f is a closed function and is not an open function. Furthermore, f is continuous. 20. Let the real function f : R -> R be defined by f(x) = x 2 . Show that f is not open. Solution: Let A = (-1,1), an open set. Note that f[A] = [0,1), which is not open; hence f is not an open function. 21. Let g3 be a base for a topological space X. Show that if f: X -> Y has the property that f[B] is open for every B e B , then f is an open function. Solution: We want to show that the image of every open subset of X is open in Y. Let G c X be open. By definition of a base, G = u,B, where B. 'B. Now f[G] = f[u i B] = u f[B]. By hypothesis, each f [B i] is open in Y and so f[G], a union of open sets, is also open in Y; hence f is an open function. 22. Show that the closed interval A = [a, b] is homeomorphic to the closed unit interval = [0, 1.1. Solution: The linear function f:I >A defined by Hence f is a homeomorphism. —

f(x) =

(b

—

is one-one, onto and bicontinuous.

a)x + a

23. Show that area is not a topological property. Solution: {(r, e) : r < 1} with radius 1 is homeomorphic to the open disc The open disc D D" = {(r, e): r < 2} with radius 2. In fact, the function f: D > D" defined by f ((r, 0)) = (2r, e) is a homeomorphism. Here (r, e) denotes the polar coordinates of a point in the plane R2. —

24. Let f: (X, T) -> (Y, T*) be one-one and open, let A C X, and let f[A] = B. Show that the function fA : (A, T ) -> (B, 71) is also one-one and open. Here f, denotes the restriction of f to A, and TA and 7-;,' are the relative topologies. Solution: If f is one-one, then every restriction of f is also one-one; hence we need only show that f A is open.

CHAP. 7]

107

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

Let H c A be TA-open. Then by definition of the relative topology, H" , AnG Since f is one-one, f[A n G] = f[A] n f[G], and so

where

G

e T.

f[H] = f[AnG] = f[A1 n f[G] = B n f[G1

fA [H]

Since f is open and G G T, f[G] E T*. Thus

B n f[G] G T'B'

and so

f A : (A, TA ) —> (B, T) is open.

25. Let f: (X, T) -> (Y, T*) be a homeomorphism and let (A, T A) be any subspace of (X, T). Show that fA : (A,T A )->(B,T B*) is also a homeomorphism where f A is the restriction of f to A, frAl= B, and T the relative topology on B. Solution: Since f is one-one and onto, f A : A —> B, where B = f[A], is also one-one and onto. Hence we need only show that fA is bicontinuous, i.e. open and continuous. By the preceding problem fA is open. Furthermore, the restriction of any continuous function is also continuous; hence f A : (A, A ) —> (B,T)) is a homeomorphism.

26. Show that any interval A = (a, b) is connected as a subspace of the real line R. Example 4.3 for the definition of connectedness.) Solution: Suppose A is not connected. Then 3 open sets G,H C R such that A n G and A empty, disjoint and satisfy (AnG)u(AnH)= A. Define the function f;A—>R

f(x)

=--

(See

n H are nonby

1 if xGAnG 0 if xEAnH

Then f is continuous, for the inverse of any open set is either A But then the intermediate value theorem applies, so 3x 0 G A impossible, so A is connected.

n G, A n H, Ø or A and so is open. for which

f(x 0) = 1.

But this is

27. Show that the following subsets of the plane R2 are not homeomorphic, where the topologies are the relativized usual topologies:

X = {x : d(x,p0) = 1 or d(x,p i)= 1; po -= (0, -1), Y

=

pi =

(0, 1)}

x : d(x,p)=1, p= (0, 5))

Y Solution: {q}. Suppose there exists a homeomorphism f: X Y; let q = f(0), X* = X\ {0}, and Y* = Y Then f: X* —> Y* is also a homeomorphism with respect to the relative topologies (see Problem 25). We show that Y* is connected. g: (0, 27) —> Y*

For if q = (5 4- cos 00, sin 0 0), then the function defined by

g(e) = (5 4- cos (e 0 e), sin (o 0 e))

is a homeomorphism. But the interval (0, 27-r) is connected, so Y* is also connected. On the other hand, X* is not connected; for the sets G

{(x, y) : x > 0}

and

H = {(x, y)

x < 0}

are both open in R2 , so G* = X*nG and H* = x*nki are open subsets of X*. Furthermore, G* and H* are non-empty, disjoint and satisfy GuH* = X*. Since connectedness is a topological property, X* is not homeomorphic to Y* and therefore there can exist no such function f.

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

108

[CHAP. 7

TOPOLOGIES INDUCED BY FUNCTIONS 28. Let {fi : X -> (Yi, Ti)} be a collection of constant functions from an arbitrary set X into the topological spaces (Yi, Ti). Determine the coarsest topology on X with respect to which the functions f i are continuous. Solution:

Recall (see Problem 1) that a constant function f:X—>Y is continuous with respect to every topology on X. Hence all the constant functions fi are continuous with respect to the indiscrete topology {X, 0 } on X. Since the indiscrete topology {X, 0 } on X is the coarsest topology on X, it is also the coarsest topology on X with respect to which the constant functions are continuous.

29. Consider the following topology on Y = {a, b, c, d}: T = tY, 0, {c}, {a,b,c}, {c, d}} Let X = {1, 2, 3, 4} and let the functions f: X -> (Y, T) and g: X -> (Y, T) be defined by

Find the defining subbase ef for the topology T* on X induced by f and g, i.e. the coarsest topology with respect to which f and g are continuous. Solution : T} {g -1 [H] : H Recall that J = {f -1 [H]: HC that is, d consists of the inverses under f and g of the open subsets of Y. Hence

=

0, {1, 2, 4 } , {3 } , {2, 3 } , {1, 2, 3 } , {2, 3, 4 } 1

30. Let T be the topology on the real line R generated by the closed-open intervals [ a, b), and let T* be the topology on R induced by the collection of all linear functions f : R-> (R, T)

defined by

f(x) = ax + b,

a, b

eR

Show that T* is the discrete topology on R. Solution: We want to show that, for every p G R, the singleton set {p} is a T*-open set. Consider the T-open set A = [1,2) and the functions f: R —> (R, T) and g: R --> (R, T) defined by f(x)

and illustrated below.

x p + 1

and

g(x) =

CHAP. 71

109

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

Now A G T implies

f

[A]

rp, p

1)

and

g -1 [A1 = (p - 1, pl

belong to the defining subbase 4 for the topology T. Hence the intersection (lo - 1 ,

PI n FP) P + 1 )

belongs to T, and so T* is the discrete topology on R.

31. Prove Theorem 7.9: Let f : X -* (Y, T)) be a collection of functions defined on an arbitrary non-empty set X, let = LI, {f' [H] H e Ti ) and let T be the topology on X generated by ci. Then: (i) All the functions fi are continuous relative to T. (ii) If T 4' is the intersection of all topologies on X with respect to which the functions continuous, then T = f (iii) T is the coarsest topology on X with respect to which the functions f i are continuous. (iv) J is a subbase for T. :

.

Solution: (i) For any function f (X, T) (Y are continuous with respect to T.

T i), if H G T,

then

f [H] G g c T. Hence all the f

(ii)

By Problem 9, all the functions f1 are also continuous with respect to T*; hence ey c 7- * and, since T is the topology generated by 4, T C T. On the other hand, T is one of the topologies with respect to which the f, are continuous; hence T c T and so T = T*.

(iii)

Follows from (ii).

(iv)

Follows from the fact that any class of sets is a subbase of the topology it generates.

Supplementary Problems CONTINUOUS FUNCTIONS

32. Prove that f: X -> Y

is continuous if and only if f - ' [A°] c

33. Let X and Y be topological spaces with

X = EuF.

-1 [A])°

Let f:E->Y

for every A c X.

and

Y, with f=g on

EnF, be continuous with respect to the relative topologies. Note that h = fug is a function from X into Y. (i) Show, by an example, that h need not be continuous. (ii) Prove: If E and F are both open, then h is continuous. (iii) Prove: If E and F are both closed, then h is continuous. 34. Let f;X->Y

be continuous.

Show that f:X->f[X1

is also continuous where f[X] has the relative

topology.

35. Let X be a topological space and let xA : X R be the characteristic function for some subset A of X. Show that xA is continuous at p E X, if and only if p is not an element of the boundary of A. (Recall xA (x) = 1 if x G A. and xA (x) = 0 if x E Ac,) 36.

Consider the real line R with the usual topology. Show that if every function f: X then X is a discrete space.

OPEN AND CLOSED FUNCTIONS

37.

Let f: (X, T) -> (Y, T*).

Prove the following:

(i)

f is closed if and only if f[A] c f[A]

for every A c X;

(ii)

f is open if and only if f[A °] c (PAD°

for every A c X.

R is continuous,

CONTINUITY AND TOPOLOGICAL EQUIVALENCE

110

[CHAP. 7

38. Show that the function f: (0, cc) > [ 1, 1] defined by f(x) = sin (1/x) is continuous, but neither open nor closed, where (0, m) and [-1, 1] have the relativized usual topologies. -

39. Prove: Let f: (X, T) (Y, T') is a base for T*.

-

be open and onto, and let 93 be a base for T. Then {f[B] : B E CB}

40. Give an example of a function f X -> Y and a subset A c X such that f is open but fA , the restriction of f to A, is not open.

HOMEOMORPHISMS, TOPOLOGICAL PROPERTIES 41. Let f:X.1 7 and g:17 ->Z be continuous, Show that if gof:X->Z g one-one (or f onto) implies that f and g are homeomorphisms. 42. Prove that each of the following is a topological property: (i) (iii) boundary, (iv) density, and (v) neighborhood.

is a homeomorphism, then

accumulation point, (ii)

interior,

43. Prove: Let f: X -> Y be a homeomorphism and let A c X have the property that A n A' = Then f[A] n (f[A]y = Q. (A subset A c X having the property A n A' 0 is called isolated. The property of being isolated is thus a topological property.)

TOPOLOGIES INDUCED BY FUNCTIONS {a, b, c, d } : T 44. Consider the following topology on Y and let f:XY and g:X->Y be as follows:

f

{(1, a), (2, a), (3, b), (4, b), (5, d)},

{Y, 0, {a, b }, {c, d}1.

Let X = {1, 2, 3, 4, 5}

g = {(1, c), (2, b), (3, d), (4, a), (5, c)1

Find the defining subbase for the topology on X induced by f and g. 45. Let f: X -> (Y ,T*), Show that if g is the defining subbase for the topology T induced by the one function f, then 4 = T. 46. Prove: Let {fi : X (1 7,,T 1)} be a collection of functions defined on an arbitrary set X, and let 4, be a subbase for the topology T i on Y. Then the class g* U {f, -1 [S]: S E eh} has the following properties: (i) jc is a subclass of the defining subbase g of the topology T on X induced by the functions f,; (ii) ,--fro is also a subbase for T. ,

47. Show that the coarsest topology on the real line R with respect to which the linear functions f: R > (R, II) -

defined by

f(x)

ax b, a,b G R

are continuous is also the usual topology 11.

Answers to Supplementary Problems 33. (1) Let X = (0, 2) and let E = (0, 1) and tinuous, but h = fug is not continuous. 44.

{X, 0, {1, 2,3, 41, {5}, {2,4}, {1, 3, 5}1

45.

Hint.

Show that

4

is a topology.

F = [1, 2).

Then f(x) = 1 and g(x)= 2 are each con-

Chapter

8

Metric and Normed Spaces METRICS Let X be a non-empty set. A real-valued function d defined on X x X, i.e. ordered pairs of elements in X, is called a metric or distance function on X iff it satisfies, for every a, b, c e X, the following axioms: [Mil

d(a, h)

[M2]

(Symmetry)

[M3] [M41

(Triangle Inequality) d(a, c) If a b, then d(a,b) > O.

0 and d(a, a)

O.

d(a , b) = d(b, a). d(a, b) + d(b , c).

The real number d(a , h) is called the distance from a to b. Observe that [Mi] states that the distance from any point to another is never negative, and that the distance from a point to itself is zero. The axiom [M2 ] states that the distance from a point a to a point b is the same as the distance from b to a; hence we speak of the distance between a and b. [M3] is called the Triangle Inequality because if a, b and c are points in the plane R2 as illustrated on the right, then [M3] states that the length d(a, c) of one side of the triangle is less than or equal to the sum d(a , h) + d(b , c) of the lengths of the other two sides of the triangle. The last axiom [M4] states that the distance between two distinct points is positive. We now give some examples of metrics. That they actually satisfy the required axioms will be verified later. Example 1.1:

The function d defined by d(a, b) = a — bl, where a and b are real numbers, is a metric and called the usual metric on the real line R. Furthermore, the function d defined by

d(p, q) =

1/(a i — b i) 2

(a2 b 2) 2

(a 1 , a2 and q where p (b 1 , b 2 are points in the plane R2, is a metric and called the usual metric on R2. We shall assume these metrics on R and R2, respectively, unless otherwise specified. Example 1.2:

Let X be any non-empty set and let d be the function defined by

fo

d(a, b) Then d is a metric on X. metric on X. Example 1.3:

if a -= b if a b

This distance function d is usually called the trivial

Let C[0,11 denote the class of continuous functions on the closed unit interval 10,11. A metric is defined on the class C[0,1] as follows:

d(f, g) =

111

f o

(x) — g(x)1 dx

METRIC AND NORMED SPACES

112

[CHAP. 8

Here d(f, , g) is precisely the area of the region which lies between the functions as illustrated below.

d(f, , g) is shaded Example 1.4:

Again let C[0,11 denote the collection of continuous functions on [0,1]. Another metric is defined on C[0,1] as follows: d*(f, , g) -= sup {I f (x) g(x)1 : x E [0, 1]} Here d*(f, , g) is precisely the greatest vertical gap between the functions as illustrated above.

Example 1.5:

b 1 , b2) be arbitrary points in the plane R2, i.e. ordeted Let p a2) and q pairs of real numbers. The functions d 1 and d2 defined by b 2 1), d2 (p, q) la i — b i l di (p, q) -= max (la i — a2 — b 2 1 are distinct metrics on R.

A function p satisfying [M1], [M2] and [M3], i.e. not necessarily [M4], is called a pseudometric. Many of the results for metrics are also true for pseudometrics.

DISTANCE BETWEEN SETS, DIAMETERS Let d be a metric on a set X. The distance between a point p E X and a non-empty subset A of X is denoted and defined by d(p, A) = inf {d(p,a): a e Al i.e. the greatest lower bound of the distances from p to points of A. The distance between two non-empty subsets A and B of X is denoted and defined by inf {d(a, h) : a E A, b E B }

d(A, B) =

i.e. the greatest lower bound of the distances from points in A to points in B. The diameter of a non-empty subset A of X is denoted and defined by d(A) = sup {d(a, a') : a, a'

e A}

i.e. the least upper bound of the distances between points in A. If the diameter of A is finite, i.e. d(A) < o, then A is said to be bounded; if not, i.e. d(A) = co, then A is said to be unbounded. Example 2.1:

Let d be the trivial metric on a non-empty set X.

d(p, A) Example 2.2:

-=

1 O

if p if p

eA A

Then for p G X and Ad3 c X, 1

d(A, B)

Consider the following intervals on the real line R:

{

0

if A n B if AnB0

A = [0,1), B -= (1,2].

If d denotes the usual metric on R, then d(A, B)-= O. On the other hand, if (1* denotes the trivial metric on R, then d*(A, B) = 1 since A and B are disjoint.

The next proposition clearly follows from the above definitions:

Proposition 8.1:

Let A and B be non-empty subsets of X and let p e X. Then: (1) d(p, A), d(A, B) and d(A) are non-negative real numbers. (ii) If p E A, then d(p, A) = O.

113

METRIC AND NORMED SPACES

CHAP. 8]

(iv)

n B is non-empty, then d(A ,

= O. If A is finite, then d(A) < .0, i.e. A is bounded.

(iii) If A

Observe that the converses of (ii), (iii) and (iv) are not true. For the empty set 0, the following conventions are adopted: d(0) = d(p, 0) = c , d(A, 0) = d(Ø, A) = co ,

OPEN SPHERES Let d be a metric on a set X. For any point p e X and any real number 8> 0, we shall let Sd(p, 8) or simply S(p, 8) denote the set of points within a distance of 8 from p: S(p, 8) =

{

oc : d(p, x) < SI

call S(p, 8) the open sphere, or simply sphere, with center p and radius 8. called a spherical neighborhood or ball. We

Example 3.1:

Consider the point p = (0, 0) in the plane R2 , and the real number 6 --= 1. If d is the usual metric on R2, then Sd(P, 8) is the open unit disc illustrated on the right. On the other hand, if d1 and d2 are the metrics on R2 which are defined in Example 1.5, then Sdi(p, 6) and Sd, 2 (P, 8) are the subsets of R2 which are illustrated below.

Sd2(p, 8) is shaded

S cii (p , S) is shaded Example 3.2:

It is also

Let d denote the trivial metric on some set X, and let p G X. Recall that the distance between p and every other point in X is exactly 1. Hence X

if 6 > 1

{p}

If 61

Example 3.3:

Let d be the usual metric on the real line R, i.e. d(a, b) = a — b. sphere S(p, 6) is the open interval (p — 6, p 6).

Example 3.4:

Let d be the metric on the collection (10,1] of all continuous functions on [0,11 defined by d(f , g) = sup {If (x) — x E [0,1]}

Then the open

(see Example 1.4). Given 6 > 0 and a function fo E C[0,1], then the° open sphere S(fo, 6) consists of all continuous functions g which lie in the area bounded by to 6 and fo + 8, as indicated in the diagram below:

114

METRIC AND NORMED SPACES

[CHAP. 8

One important property of open spheres in metric spaces is given in the next lemma.

Lemma 8.2:

Let S be an open sphere with center p and radius 8. Then for every point q e S there exists an open sphere T centered at q such that T is contained in S. (See the adjacent Venn diagram.)

METRIC TOPOLOGIES, METRIC SPACES In general, the intersection of two open spheres need not be an open sphere. However, we will show that every point in the intersection of two open spheres does belong to an open sphere contained in the intersection. Namely, Lemma 8.3: Let S1 and S2 be open spheres and let p e Si n S2. Then there exists an open sphere S, with center p such that p G Sp c si n S2. Hence by virtue of Theorem 6.1 we have

Theorem 8.4: Definition:

The class of open spheres in a set X with metric d is a base for a topology on X. Let d be a metric on a non-empty set X. The topology T on X generated by the class of open spheres in X is called the metric topology (or, the topology induced by the metric d). Furthermore, the set X together with the topology T induced by the metric d is called a metric space and is denoted by (X, d).

Thus a metric space is a topological space in which the topology is induced by a metric. Accordingly, all concepts defined for topological spaces are also defined for metric spaces. For example, we can speak about open sets, closed sets, neighborhoods, accumulation points, closure, etc., for metric spaces. the usual metric on the real line R, i.e. d(a, b) = la — bl, then the open in R are precisely the .finite open intervals. Hence the usual metric on R the usual topology on R. Similarly, the usual metric on the plane R 2 the usual topology on R2 ,

Example 4.1:

If d is spheres induces induces

Example 4.2:

Let d be the trivial metric on some set X. Note that for any p G X, S(p, = {p}. Hence every singleton set is open and so every set is open, In other words, the trivial metric on X induces the discrete topology on X.

Example 4.3:

Let (X, d) be a metric space and let Y be a non-empty subset of X. The restriction of the function d to the points in the subset Y, also denoted by d, is a metric on Y. We call (Y, d) a metric subspace of (X, d). In fact, (Y, d) is a subspace of (X, d), i.e. has the relative topology.

Frequently the same symbol, say X, is used to denote both a metric space and the underlying set on which the metric is defined.

PROPERTIES OF METRIC TOPOLOGIES Since the topology of a metric space X is derived from a metric, one would correctly expect that the topological properties of X are related to the distance properties of X. For example, Theorem 8.5: Let p be a point in a metric space X. Then the countable class of open spheres, {S(p, 1), S(p, S(p,4), . .4 is a local base at p. Theorem 8.6:

The closure A of a subset A of a metric space X is the set of points whose distance from A is zero, Le. A = foe: d(x,A) = 01.

METRIC AND NORMED SPACES

CHAP. 8]

115

Observe that axiom [M4] implies that the only point with zero distance from a singleton set {p} is the point p itself, i.e.,

d(x, {p}) = 0

implies x = p Hence by the preceding theorem, singleton sets {p} in a metric space are closed. Accordingly, finite unions of singleton sets, i.e. finite sets, are also closed. We state this result formally: Corollary 8.7: In a metric space X all finite sets are closed. Thus we see that a metric space X possesses certain topological properties which do not hold for topological spaces in general. Next follows an important "separation" property of metric spaces.

Theorem 8.8 (Separation Axiom): Let A and B be closed disjoint subsets of a metric space X. Then there exist disjoint open sets G and H such that A c G and B c H. (See Venn diagram below.) ...." ---

0 /

.._

N.

/

(

-....,

0

.....

N

G

\ N. .---..--- ..."

/

I

I

i

1,

//

\

/

B \

\ 1

H1 / N. ..... --.... _ ....-- .0

One might suspect from the above theorem that the distance between two disjoint closed sets is greater than zero. The next example shows that this is not true. Example 5.1: Consider the following sets in the plane R2 which are illustrated below: A

= {(x, y) : xy —1, x <

B=

{(x,y) xy 1, x > 0}

Observe that A and B are both closed and disjoint. However, d(A, B) = O.

EQUIVALENT METRICS Two metrics d and d* on a set X are said to be equivalent if they induce the same topology on X, i.e. if the d-open spheres and the d*-open spheres in X are bases for the same topology on X. Example 6.1:

The usual metric d and the metrics d1 and d2, defined in Example 1.5, all induce the usual topology on the plane R2, since the class of open spheres of each metric (illustrated below) is a base for the usual topology on R2.

I

/.-7"' .-T , •D / ,....../

•

\ .

S d(P, 6 )

Hence the metrics are equivalent.

S(p, 6 )

Sd2 (p, 5)

116

METRIC AND NORMED SPACES

[CHAP. 8

Example 6.2: Consider the metric d on a non-empty set X defined by 2 if a b 0 if a Observe that s 1 (p,1) {p}; so singleton sets are open and d induces the discrete topology on X. Accordingly, d is equivalent to the trivial metric on X which also induces the discrete topology.

d(a, b)

The next proposition clearly follows from the above definition. Proposition 8.9: The relation "d is equivalent to d*" is an equivalence relation in any collection of metrics on a set X.

METRIZATION PROBLEM Given any topological space (X, T), it is natural to ask whether or not there exists a metric d on X which induces the topology T. The topological space (X, T) is said to be metrizable if such a metric exists. Example 7.1:

Every discrete space (X, c7)) is metrizable since the trivial metric on X induces the discrete topology :D.

Example 7.2:

Consider the topological space (R,14), the real line R with the usual topology U. Observe that (R,14) is metrizable since the usual metric on R induces the usual topology on R. Similarly, the plane R2 with the usual topology is metrizable.

Example 7.3:

An indiscrete space (X, ref) where X consists of more than one point is not metrizable. For X and 0 are the only closed sets in an indiscrete space (X,,9). But by Corollary 8.7 all finite sets in a metric space are closed. Hence X and 0 cannot be the only closed sets in a topology on X induced by a metric. Accordingly, (X, S) is not metrizable.

The metrization problem in topology consists of finding necessary and sufficient topological conditions for a topological space to be metrizable. An important partial solution to this problem was given in 1924 by Urysohn as a result of his celebrated Urysohn's Lemma. It was not until 1950 that a complete solution to this problem was given independently by a number of mathematicians. We will prove Urysohn's results later. The complete solution to the metrization problem is beyond the scope of this text and the reader is referred to the classical text of Kelley, General Topology.

ISOMETRIC METRIC SPACES A metric space (X, d) is isometric to a metric space (Y, e) if there exists a one-one, onto function f: X ---> Y which preserves distances, i.e. for all p, q e X,

d(P, q) (P), f (q)) Observe that the relation "(X, d) is isometric to (Y, e)" is an equivalence relation in any collection of metric spaces. Furthermore, Theorem 8.10: If the metric space (X, d) is isometric to (Y, e), then (X, d) is also homeomorphic to (Y, e). The next example shows that the converse of the above theorem is not true, i.e. two metric spaces can be homeomorphic but not isometric. Example 8.1:

Let d be the trivial metric on a set X and let e be the metric on a set Y defined by f2 if a b 10 if a= b Assume that X and Y have the same cardinality greater than one. Then (X, d) and (Y, e) are not isometric since distances between points in each space are different. But both d and e induce the discrete topology, and two discrete spaces with the same cardinality are homeomorphic; so (X, d) and (Y, e) are homeomorphic.

e(a, b)

CHAP. 8]

METRIC AND NORMED SPACES

117

EUCLIDEAN m-SPACE Recall that WI denotes the product set of m copies of the set R of real numbers, i.e. consists of all m-tuples (ai, a2, ...,a,n) of real numbers. The function d defined by m m d(p, q) V(ai - bi) 2 + (ai- b i) 2 a, - b1 2 + (a.- b..) 2 =

E

1=1

1=1

where p = (al, . , bm), is a metric, called the Euclidean metric on , am) and R'n. We assume this metric on R'n unless otherwise specified. The metric space R"' with the Euclidean metric is called Euclidean m-space and will also be denoted by Ent. Theorem 8.11: Euclidean m-space is a metric space. Observe that Euclidean 1-space is precisely the real line R with the usual metric, and Euclidean 2-space is the plane R2 with the usual metric.

HILBERT SPACE The class of all infinite real sequences a2, ...) such that

a2„ < 71=-

i.e. such that the series a', + Example 9.1:

c'0

1

converges, is denoted by Ir.

Consider the sequences p = (1, 1, 1, . ..)

q = (1, 1•,

and

Since 12 + 1 2 + • • • does not converge, p is not a point in Ir. On the other hand, the series 1 2 + (*) 2 (i) 2 + • • • does converge; hence q is a point in Ir.

Now let p = (an) and q = (bn) belong to fr. d(p q) =

The function d defined by

!an - b 2

is a metric and called the 12-metric on Ir. We assume this metric on 1r unless otherwise specified. The metric space consisting of fr with the (2-metric is called Hilbert space Or 12-space and will also be denoted by H. We formally state: Theorem 8.12: Hilbert space (or /9-space) is a metric space. Example 9.2:

Let H m denote the subspace of Hilbert space H consisting of all sequences of the form (a 1 , a9, . • . arpt IT a nt, 0 1 ° Y ° Y • • Observe that Hm is isometric and hence homeomorphic to Euclidean m-space by the natural identification ■

(a1 , . .

am , 0, 0, ...)

am) <---> (a l , .

Hilbert space exhibits two phenomena (not occurring in Euclidean m-space) described in the examples below: Example 9.3:

Consider the sequence (pd of points in Hilbert space where pf, = (a ik , a 2k, ...) is defined by ail, = (5, k ; i.e. ad, = 1 if i = k, and aik 0 if i k . Observe, a•s illustrated below, that the projection (.7 1 (p,)) of (N) into each coordinate space converges to zero: =- (1, 0, 0, 0, ...) (0, 1, 0, 0, . ..) 192 p3 = (0, 0, 1, 0, .) P4 = (0, 0, 0, 1, ...) . . . . .

.

.

.

11 11 0

= (0, 0, 0, 0, ...)

But the sequence (p„) does not converge to 0, since d(pk , 0) = 1 for every k G N; in fact, (IV has no convergent subsequence.

[CHAP. 8

METRIC AND NORMED SPACES

118 Example 9.4:

Let W denote the Proper subspace of H which consists of all points in H whose first coordinate is zero. Observe that the function f : H --> W.- defined by f((ai, az, • • -)) = (0, a l , a9, ...) is one-one, onto and preserves distances. Hence Hilbert space is isometric to a proper subspace of itself.

CONVERGENCE AND CONTINUITY IN METRIC SPACES The following definitions of convergence and continuity in metric spaces are frequently used. Observe their similarity to the usual E 8 definitions. Definition:

The sequence (a1, a2, b e if for every E >

) of points in a metric space (X, d) converges to there exists a positive integer no such that

x

n > no implies

Definition:

d(a„,b) < e

Let (X, d) and (Y, d*) be metric spaces. A function f from X into Y is continuous at p e X if for every E > 0 there exists a 8 > 0 such that d(p, x) < 8

implies

d* (f (P), f (x))

<E

The above definitions are equivalent to the definitions of convergence and continuity (in the metric topology) which were given for topological spaces in general. NORMED SPACES

Let V be a real linear vector space, that is, V under an operation of vector addition and of scalar multiplication by real numbers satisfies the axioms [VI], [V2] and [V3] of Chapter 2, Page 22. A function which assigns to each vector y E V the real number Ily is a norm on V if it satisfies, for all y, w E V and k G R, the following axioms: [N i ]

Iv

[N2I

v+w

0 and

lv = 0 iff v = O.

Iv! 1.

[N3]

A linear space V together with a norm is called a normed linear vector space or simply a normed space. The real number vl is called the norm of the vector v.

Theorem 8.13:

Let V be a normed space.

The function d defined by d(v , w) =

!Iv

w

where v,w G V, is a metric, called the induced metric on V. Thus every normed space with the induced metric is a metric space and hence is also a topological space. Example 10.1:

The product set Rm is a linear vector space with addition defined by

(a„ ..., a m) + (b„

bm) =

(a l + b i , ..., am +

bm)

and scalar multiplication defined by

k

..

am )

=(kai ,

k am )

The function on Rm defined by

(al , ..

am ) I

=

a 21

+ • + am2 = NIM a i2 =

2

is a norm and called the Euclidean norm on WI. Note that the Euclidean norm on Rm induces the Euclidean metric on R. If p = (a1 , a2 a3) is a point in R3 , then 11p11 corresponds precisely to the "length" of the arrow (or vector) from the origin to the point p as illustrated below. ,

METRIC AND NORMED SPACES

CHAP. 8]

Example 10.2:

119

The following two functions are also norms on the linear space Rm. :

ano

max (1a11, la21, •

and)

0,1 1 + a2 1+ • • • + laml

l I (al, • • •, am) I

Let T(X, R) be the collection of all real-valued functions on a non-empty set X. Recall (see Theorem 2.9) that T(X, R) is a linear space with vector addition and scalar multiplication defined as follows: f (x) + g (x) and (kf)(x) k f(x) (f + g)(x) We shall frequently want to study classes of functions with certain other properties such as boundedness, continuity, etc. We shall use the following result from linear algebra Proposition 8.14:

Let c4(X,R) be a non-empty subcollection of ''(X, R) satisfying the following two properties: (i)

If f,g e c,l(X, R.), then the sum f + g E

I (X , R).

(ii) If f e eA- (X,R) and k E R, then the scalar multiple kf E 0_4(X, R). Then 0.4(X, R) is, itself, a linear vector space. Example 10.3:

The class C[0, 1] of all continuous real functions on the interval / = [0,1] is a linear space since the sum and scalar multiples of continuous functions are continuous. The function on C[0,1] defined by

If(x)1 dx 13

is a norm which induces the metric on CEO, 1] defined in Example 1.3. Example 10.4:

The function on the linear space C[0,1] defined by

HfH

=

sup {1f(x)1 : x G [0, 1]}

is also a norm. This norm induces the metric on Example 10.5:

C1.0,11 defined in Example 1.4.

Let '73(X, R) denote the subcollection of T(X, R) consisting of all bounded functions : X R. Then 613(X, R) is a linear space since the sum and scalar multiples of bounded functions are also bounded. The function on '13(X, R) defined by sup { f(x)1 x

e

X}

is a norm. Example 10.6:

We show later that the class Ir of all real sequences is a linear space. The function on Rx' defined by

= is a norm and called the 12-norm on IV° . 12-metric in Hilbert space.

(an)

such that M la,1 2 <

..\/M la,1 2 Observe that this norm induces the

120

(CHAP. 8

METRIC AND NORM ED SPACES

Solved Problems METRICS 1. Show that in the definition of a metric the axiom 1M3 1 can be replaced by the following (weaker) axiom: [M1 If a, b,c EX are distinct then d(a, c) Solution: Suppose a = b.

Then

d(a , c) = d(b, c) -= d(b, b) If b

d(ab) + d(b, c).

d(b, c)

d(a, b)

d(b, c)

c, the argument is similar. Lastly, suppose a = c; then d(a, c)

0

d(a, b)

d(b, c)

Thus the Triangle Inequality follows from [M1 ] if the points a, b and c are not all distinct.

2. Show that the trivial metric on a set X is a metric, i.e. that the function d defined by

fl

d(ab)

o

if a b if a= b

satisfies [Mil, [M2], [M: ] and 1 M41. Solution : Let a, b G X. Then d(a, b) = 1 or d(a, b) O. In either case, then, by definition of cl, d(a, b) = O. Hence d satisfies [M 1 ].

O. Also, if a = b

d(a, b)

Let a, b E X. If a b, then b a. Hence d(a, b) = 1 and d(b, a) = 1. Accordingly, d(a, b) = d(b , a). Hence d satisfies On the other hand, if a = b then b = a and therefore d(a, b) = 0 1M 2 ].

d(b , a).

Now let a, b,c G X

be distinct points.

d(a, c) = 1

and d satisfies [*]. Lastly, let a, b G X

and a

b. Then

Then

d(a, c) = 1, d(a, b) = 1

1 4- 1 = d(a, b) d(a, b)

3. Let d be a metric on a non-empty set X.

1.

and

d(b, c)

1.

Hence

d(b , c)

Hence

d(a, b)

0,

and d satisfies Did.

Show that the function e defined by

e(a,b) = min (1, d(a,b)) where a,b E X, is also a metric on X. Solution: Let a, b G X. Since d is a metric, d(a, b) is non-negative. Hence e(a, b), which is either 1 or d (a, b), is also non-negative. Furthermore, if a = b then

e(a, b) = min (1, d(a, b)) -= min (1, 0) = 0 Hence e satisfies [Mil Now let a, b G X. By definition e(a, b) = d(a, b) or e(a, b) = 1. Suppose then d(a, b) < 1. Since d is a metric, d(b, a) = d(a, b) < 1. Consequently,

e(a, b) = d(a, b);

e(b, a) = d(b, a) = d(a, b) == e(a, b) On the other hand, suppose

e(a, b) = 1; then d(a, b) e(b, a)

1. Hence d(b, a) = d(a, b)

1.

Consequently,

1 = e(a, b)

In either case e satisfies [M2 ]. Now let a,b,c E X.

We want to prove the Triangle Inequality

e(a, c)

e(a, b)

e(b, c)

1 or e(b, c) =-- 1, the Triangle 1. Hence if e(a, b) Observe that e(a, c) = min (1, d(a, c)) e(a, b) d(a, b) and e(b, c) = d(b, c). 1, then Inequality holds. But if both e(a, b) < 1 and e(b, c) < Accordingly, d(a, c) d(a, b) d(b, c) = e(a, b) 4- e(b , c) e(a, c) =- min (1, d(a, e)) Thus in all cases the Triangle Inequality holds. Hence e satisfies [M3 ]. Finally, let a, b G X and a not zero. Thus e satisfies [M4 1.

b. Then

d(a, b)

O. Hence

e(a, b) = min (1, d(a, b))

is also

▪

121

METRIC AND NORMED SPACES

CHAP. 8]

4.

-

Let d be a metric on a non-empty set X.

Show that the function e defined by d(a,b)

e(a,b) =

1+d(a,b)

where a,b E X, is also a metric on X. Solution: Since d is a metric, e clearly satisfies [M1 ] , [M2 ] and [MA Hence we only need to show that e satisfies [M2 ], the Triangle Inequality. Let a,b,c e X; then

d(a, b) 1 + d(a, b) 4 d(b, c) -

and

d(b, c) d(a, b)

1 d(a, c)

Since d is a metric,

e(a, c)

d(a, b) -4- d(b, c).

d(a, c) 1 + d(a, c)

=-

d(b, c)

1

d(a, b) d(a, b)

e(a, b)

1

d(b, c) d(b, c)

e(b, e)

Hence

d(a, b) -4- d(b, c) d(b,c) d(a, b)

d(a, b) d(b, c) d(a, b)

1

1

d(b, c)

1 + d(a, b)

d(b, c)

e(a, b)

e(b, c)

Thus e is a metric.

OPEN SPHERES 5. Prove Lemma 8.2: Let S be an open sphere with center p and radius 8, i.e. S = S(p, 8). Then for every point q E S there exists an open sphere T centered at q such that T is contained in S. Solution: Now

d(q, p) < b

since e

q C S = S(p, 8).

S—

d(q, p) > 0

We claim that the open sphere and radius e, is a subset of S. Let x E T S(q, E). the Triangle Inequality,

d(x , p)

d(x, q)

Then

Hence

T = S(q, E),

with center q

d(x, q) <e = 8 — d(q, p).

d(q, p) < [8 — d(q, p)]

So, by

d(q, p) = 8

Thus x E S = S(p, 8) since its distance from p is less than S. So x G T implies x G S, i.e. T is a subset of S (as indicated in the adjacent Venn diagram).

6. Let SI and 8 2 be real numbers such that 0 < 61 is a subset of the open sphere S(p, 80. Solution: Let x E S(p, S i ). Then d(x, P) <

Hence

82.

Show that the open sphere S(p, 81)

x G S(P, 6 2)

and thus S(p, 51) c S(p,

62).

7. Show that if S and T are open spheres with the same center, then one of them is a subset of the other. Solution: S(p, 8 2), i.e. S and T have the same center p with radii S i and 8 2 Say S S(p,8 1) and T or 8 2 8 1 . Hence by the preceding problem either 5 C T or T C S. I 8 2 either S respectively. But

122

8.

METRIC AND NORMED SPACES

[CHAP. 8

Prove Lemma 8.3: Let S1 and S.) be open spheres and let p e Si exists an open sphere S, with center p such that p e S, C S1 n S2.

n

S2.

Then there

Solution : Since p E S 1 and S 1 is an open sphere, there exists by Lemma 8.2 an open sphere S';' with center p such that p E S C S I . an open sphere S2'. with center p such thatSimlarythexs E p C S9. Now S; and each has center p; so by Problem 7 one of them, say S, is contained in the other. Thus we have

s,

s.';

p e ,sr;. c

and

Accordingly, p G S c S i (See adjacent diagram.)

n S2.

pEScS cS2 Hence we may take Sp =S.

METRIC TOPOLOGIES 9.

Prove: Let X be a metric space, and let B T, denote the class of open spheres with center p e X. Then D, is a local base at p. Solution: Let G be an open subset of X containing p. Since the open spheres in X form a base for the metric topology, 3 an open sphere S such that pESc G. But by Lemma 8.2 3 an open sphere S„ E A i„ i.e. with center p, such that p G Sp c S C G. Hence 00 is a local base at p.

10. Prove Theorem 8.5:

Let X be a metric space. Then the countable class of open spheres IS(P, 1), S(P,

S(P,

D, • • •

with center p E X, is a local base at p. Solution: Let G be an open subset of X containing p. By the preceding problem, 3 an open sphere S(p, S) with center p such that p E S(p,S) c G. Since 5 > 0,

3n 0 E N Accordingly,

p G S(p, 1/n0) c S(p, S) c G

such that where

1/n0 < 8

S(p, 1/n0) e

7.

Hence

7

is a local base at p.

11. Prove Theorem 8.6: The closure A of a subset A of a metric space X is the set of 0} . points whose distance from A is zero: A = {x: d(x, A) Solution: Suppose d(p, A) = O. Then every open sphere with center p, and therefore every open set G containing p, also contains at least one point of A. Hence p G A or p is a limit point of A, and so p E A. On the other hand, suppose d(p, A) = e > O. Then the open sphere S(p, ie) with center p contains A. Accordingly, A .= {x d(x, A) = 0 } . no point of A. Hence p belongs to the exterior of A, and so p

12. Show that a subset F of a metric space X is closed if and only if

{x : d(x,F) = 01 C F.

Solution: This follows directly from Problem 11 and the fact that a set is closed ill* it is equal to its closure.

13. If F is a closed subset of a metric space X and p e X does not belong to F, i.e. p 0. then d(p,F) Solution: If d(p,

d(p, F)

O.

0 and

F

is closed, then by Problem 12, p G

F,

F. But by hypothesis p F; so

CHAP. 81

METRIC AND NORMED SPACES

123

14. Prove Theorem 8.8: Let A and B be closed disjoint subsets of a metric space X. there exist disjoint open sets G and H such that A C G and B C H.

Then

Solution: If either A or B is empty, say A 0, then 0 and X are open disjoint sets such that A C and B C X. Hence we may assume A and B are non-empty. Let a G A. Since A and B are disjoint, a B. But B is closed; hence by the preceding problem, d(a, B) S a > O. Similarly, if b E B, then d(b, A) = S b > O. Set Sa = S(a, i-S a) so a e Sa and gram.)

b E Sb.

and

Sb

S(b,k8b)

(See the adjacent Venn dia-

We claim that the sets

G =- U {S a G A } and

H = U {S b : b B}

satisfy the required conditions of the theorem. Now G and H are open since they are each the union of open spheres. Furthermore, et G Sa implies A C G, and b G Sb implies B c H. We must show that G n H Q. Suppose

Gn H

0, say pEGn H.

3 ao G A, b o E B Let d(ao ,

= e> O. Then d(ao,B)

Then

p E Sao, p E Sbo

such that

s ao e and d(bo, A) --= 600 e. But p G Sao and

d(ao,p) < k

a0

and

d(p,b 0) < kho

Therefore by the Triangle Inequality,

d(a0,b0) = e

d(ao,p)

d(p,b 0) < -A-S ao

pbo

tœt +

=

an impossibility. Hence G and H are disjoint and the theorem is true.

EQUIVALENT METRICS 15. Let d and e be metrics on a set X such that for each d-open sphere Sd with center p e X there exists an e-open sphere Se with center p such that Se c S. Show that the topology Td induced by d is coarser (smaller) than the topology Te induced by e, i.e. Td C Te. Solution: Let G E Td. We want to show that G is also an e-open set. Let p E G. Since G is d-open there exists a d-open sphere Sd with center p such that p G Sd C G. By hypothesis, there exists an e-open sphere S e(p) with center p such that p E S(p) C Sd C G. Accordingly, G -= U {S e (p) p E G } . Thus G is the union of e-open spheres, and so it is e-open. Hence Td C Te .

16. Let d and e be metrics on a set X such that for each d-open sphere Sd with center p e X there exists an e-open sphere S a with center p such that S e C Sd, and for each e-open sphere S: with center p E X there exists a d-open sphere S: such that S: C S. Show that d and e are equivalent metrics, i.e. that they induce the same topology on X. Solution : By Problem 15, the topology Td induced by d is coarser than the topology Te induced by e, i.e. Te. Td c T e . Also by Problem 15, Te c T. Therefore Td

METRIC AND NORMED SPACES

124

[CHAP. 8

17. Show that the usual metric d on the plane 112 is equivalent to the metrics d1 and d2 on R2 defined in Example 1.5. Solution: Observe that we can inscribe a square in any circle as shown in Fig. (a) below, and we can inscribe a circle in a square as shown in Fig. (b). Now the points inside a circle form a d-open sphere and the points inside a square form a di -open sphere, so the metrics d and di are equivalent by Problem 16. Furthermore, we can inscribe a "diamond" in any circle as shown in Fig. (c), and we can inscribe a circle in any diamond as shown in Fig. (d). Since the points inside a "diamond" form a d2 -open sphere, the metrics d and d2 are equivalent by Problem 16.

Fig. (h)

Fig. (a)

Fig. (e)

Fig. (d)

C10, 1] denote the collection of all real continuous functions defined on I Consider the metrics d and e on C[0, 1] defined by

18. Let

d(f,, g)

= sup

{V(x)

— g(x)1: x E I),

10, 1].

e(f,, g) = f f(x) — g(x) i dx 0

(see Example 1.3 and Example 1.4). Show that the topology T d induced by d is not coarser than the topology Te induced by e, i.e. Td Solution: Let p be the constant function P(x) =-- 2 and let E = 1. Then the sphere Sd(P,€) consists of all functions g for which g lies between the functions p — 1 and p 1, i.e. such that 1 < g(x) < 3 for all x 4

p+1

3 2

p -- 1 o

-1It is sufficient to show that Sd (p, E) contains no e-open sphere with center p; Le. for every 8 > 0, S e (p, S) S d (p, e). Let 8 > 0. Consider the function q consisting of the line segments between the points (0,4) and (p, 2) and between (1,3,2) and (1,2), i.e. defined by g(x)

=

f (-4x/S) + 4 t 2

if 0 if

x< c

1

(see diagram above). Observe that the "area" between p and q is y, i.e. e(p, q) = IA. Then > 0. Hence q G S e (p, 6). S d(p,E). Thus S e (p,S) S d(p, e) for any But d(1), 2 ; so q Td T.

19. Let C[a, b] denote the collection of all continuous functions on a closed interval X = [a, b]. Consider the metrics d and e on C[a, b] defined by

d(f,, g) = sup tif(x)

—

g(x): x

5 coarser than the topology

e X),

e(f, , g) =

f(x) — g(x) J dx

a

Show that the topology Te induced by e is TeC T de

Td

induced by d,

i.e.

CHAP. 8]

METRIC AND NORMED SPACES

125

Solution: Let S,(p, E) be any e-open sphere in Ca, bl with center p C CIa, b]. Let 8 --= E/( b — a). In view of Problem 15 it is sufficient to show that S d(p, S), the d-open sphere with center p and radius 3, is a subset of S e (p,e), i.e. S d (p, S) c S e (p,c). Let f G S d (p, S); Hence

then

f G S,(p, E.)

.b

b

.b

f sup flp(x) — f(x)11 dx < f

e(p, f) =-- f p(x) — f(x)1 dx = So

f(x)11 < 8 = €1(b — a)

sup {p(x)

Egb — a) dx

S d(p, 8) c S,(p, c).

and therefore

NORMED SPACES 20. Prove Theorem 8.13: The function d defined by d(v, w) are vectors in a normed space V, is a metric on V. Solution: Note that by [N I

= 1 v — wi

, where y and w

],

d(v , w)

v — wi= 0

d(v , v) = v v11 =

and

110N =

Hence d satisfies [Md. Also, by [N 3 ],

= 1-11

=

—

d(v, w) =

that is,

d(a, e)

Finally, if v

d(a, b)

d(b, c).

w then v — w

Accordingly if a, b, c

ilwil =

Ht/]-wH

!(a — b)

e V.

+ !7.,(HI for all v, w we have

y + wI Hence d satisfies [M2 ] . BY [N2 ] , b—e then substituting v = a— b and w

!a —CH

w —vI= d(w, v)

w — vjj

bil

e V,

11b-0

Hence d satisfies [M 3 ] .

0; hence by [N I ], d(v, w) = Iv — w11 > 0.

For any pair of points p = (a1, ..., am) and

21. Prove the Cauchy-Schwarz Inequality: = (b„ . . .,b,n) in R", bt

IP

Thus d satisfies [M 4 ].

af, 2

q

J±

\

i=

bi1 2

where HpIi is the Euclidean norm. Solution: If p = 0 or q 0, then the inequality reduces to 0 0 and is therefore true. consider the case in which p 0 and q 0, i.e. in which ili O and Now for any real numbers

x, y G R, 0

x2

(x — 02

2xy

2 --HP1 1 iqH

Pt

2 LPH 1 qT

i=1

(1 )

x = !ad 411

111)112

=1.

bi 2

!1PH 2

2

I 1 q!1

So,

(2

and

1170

IqH2.

)

So summing (2) with

tt1

cf,0

in (I).

MqH2

11,7,112 =1402

a bi

that is,

and y = IbMIqt

bi l2

Hpll2

But by definition of the Euclidean norm, 1,{ai! 2 = respect to i and using ab = la bf , we have

a b1

2xy y 2 or, equivalently,

x2 4- y2

Since x and y are arbitrary real numbers, we can let for any i,

Pt

So we need only

1

Multiplying both sides by Ip1I Hqii gives us the required inequality.

11,7 112 2 dq= 112

126

METRIC AND NORMED SPACES

For any pair of points p =

22. Prove Minkowski's Inequality: (b1, bm) i• n It, m

111/ Solution: If

1129 1

(CHAP. 8

-\/I 1a + bir

i.e.

11q11

,ai

, an,) and 2

+ A[1 Ibil 2

+ q!1 = 0, the inequality clearly holds. Hence we need only consider the case in which

+

O.

ai, bi G R, we have

Observe that, for real numbers

+ qH 2 =

■ b i l.

laf bij

kt i + bi l 2 =

Hence

!a• +

bi (kt i + =--

+ bi l

+ bil Icti1 +

But by the Cauchy-Schwarz Inequality,

e

and

1+ !P +

Then I lp +

Since we are considering the case inequality.

ÎqH

=

p -1- q

0, we can divide by

PI 1

+

p + qH., this yields the required

23. Prove that the Euclidean norm, P = (a1, ..., am) G satisfies the required axioms [NI], [N2] and [N3]. where

=

Solution: Now [N 1 ] follows from properties of the real numbers, and [N2 ] is Minkowski's Inequality which was proven in the preceding problem. Hence we only need to show that [N 3 ] holds. But for any vector p = ...,a„,) and any real number k G R,

. , a„,)

=

-\1m

Ikai{ 2

.. 2

2

Abk1 2 2

-01c la1 12

2

ka ni )H

IIPI

Hence [N 3 ] also holds.

24. Prove Theorem 8.11: Euclidean m-space is a metric space, i.e. the Euclidean metric on R satisfies the axioms [Mil to Mi. '

Solution: Use Problem 23 and the fact that the Euclidean metric on Wn is induced by the Euclidean norm on Ra'.

25. Let

(a1, a2, ...) be a convergent sequence of real numbers with the property that b. b for all n G N. Show that lim an

Solution: Suppose lima,,

=a > b and set 3n0 G N

Thus —a

n°

E

a— b >0. Since an -*a,

such that a — ano

< —b and therefore b < a

n' o

which contradicts the hypothesis. Accordingly,

26. Prove Minkowski's inequality for infinite sums:

bn)11

1 Ra,,) H+ If(bn)i

a no i < e 1-- a — b

i.e.

Solution: By Minkowski's inequality for finite sums,

If (a,,), (b.) G Rc%

+ b.r2

lin=1

lim a,, = - b.

then

la,d 2

1,b„)2

• •

CHAP. 8]

127

METRIC AND NORMED SPACES

Ian + bn1 2

an 2

1

+ •\

an 2

n=,

'NI

ktn 1 2

n=1

bri.1 2

Since the above is true for every m G N, by the preceding problem it is also true in the limit.

27. Show that the 12-norm on R", i.e. [N1], [N2 ] and [ N3].

Ian

I (a.)1 ,

2,

satisfies the required axioms

Solution: This is similar to the proof in Problem 23 that the Euclidean norm satisfies the axioms [N 1 ], [N2 ] and [N3 ].

28. Prove Theorem 8.12:

Hilbert space (or 12-space) is a metric space.

Solution: Use Problem 27 and the fact that the 12-metric on R is induced by the 12 -norm.

VA, for every c > O. Show

29. Let a and b be real numbers with the property that a that a b. Solution: Suppose a > b. Then a = b+ 8 where 3 > O. Set t E > O. But this contradicts the hypothesis; so a -= b.

30. Let I = [0,1].

IS.

Now a > b+As = b+e where

Show that the following is a norm on c[0,

= sup f(x)I}.

Solution: Recall that a real continuous function on a closed interval is bounded; so If is well-defined. Since If(x) 0 for every x G /, 0; also Hf = 0 if If(x)1 = 0 for every x E /, Le. if f = O. Thus [N i ] is satisfied. Let e> O.

Then 3x 0 E / such that sup {/(x)

+ g(x)1}

•

If(X0)1

Ig(x0) 1

•

sup { f (x) } + sup

If + q11

f +g

11f11 + l!g{1

1

E

+ {g(x)j} +

E

Ifi + g +

• Hence by Problem 29,

f(x 0) + g(xo) +

and [N 2] is satisfied.

Now let k E R. Then

I kfll =

sup { (kf)(x)J

=

sup {f(x)1}

=

sup {lk f(x)1} = sup ticHf(x)}

k

IlfU

and [N 3 ] is satisfied.

Supplementary Problems METRICS

31. Let 93(X, Y) be the collection of all bounded functions from an arbitrary set X into a metric space (Y, d). Show that the function e is a metric on (13(X, Y): e(f,, g) =

sup {d(f(x), g (x)) : x G X}

di , ..., dm be metrics on Xi , ..., X i,, metrics on the product set X = ni X

32. Let

d(p, q) = max {d1 (q 1 , b i ), . . Here,

P = (aii, • •

am), q = (bi, • •

respectively.

dm (am , b m )} ,

b) G X =

111 X.

Show that the following functions are

e(p, q) =

b i ) + • + dm (am , bm )

128

METRIC AND NORMED SPACES

[CHAP. 8

33.

Let IV = R u { , — .0} be the extended real line and let f: IV —> [-1, 1 be defined by f(s) = x/(1 + ;x) if x G It, .f() 1 and f(— ) = —1. Show that the following function is a metric on IV: d(x, y) = tf(x) — f(y)!.

34.

Let R+ denote the non-negative real numbers, and let f: R+ —> R+ be a continuous function such that (i) PO) = 0, (ii) f(x + f(x) + f(y), and (iii) x < y implies f(x) < f(y). Show that if d is a metric on any set X then the composition function [o d is also a metric on X.

35.

Let

p

be a pseudometric on some set X.

Let — be the relation in X defined by P(a, b) = o

a — b if (i)

Show that — is an equivalence relation in X.

(ii)

Show that the following function is a metric on the quotient set XI— d([al,b1) = p(a, b). Here ral denotes the equivalence class of a G X.

{[a] : a G

36. Let JO, 11 denote the collection of (Riemann) integrable functions on [0, 1). Show that the following function is a pseudometric on RjO, 11:

•1 1.f(x) — g(x) dx

P, (f g)

o

Also show by a counterexample that p is not a metric. 37.

Show that a function d is a metric on a set X if it satisfies the following two conditions: (i) d(a, b) = 0 iff a --,- b; (ii) d(a, c) d(a, b) + d(c, b).

DISTANCES BETWEEN SETS, DIAMETERS 38.

Give an example of two closed subsets A and B of the real line R such that d(A, B)

0

but

AnB ---=-0

39.

Let d be a metric on X. Show that for any subsets A,B c X: (i) d(AuB) d(A) + d(B) + d(A,B) and (ii) d(A) = d(A).

40.

Let d be a metric on X and let A be any arbitrary subset of X. defined by f(x) d(x, A) is continuous.

41.

Consider the function d: R2 —> R defined by d((a, I))) = — bl (i.e. the usual metric on R). that d is continuous with respect to the usual topologies on the line R and the plane R 2.

42.

Let A be any subset of a metric space X.

Show that the function

f:X—> R

Show

Show that d(A) = d(À).

METRIC TOPOLOGIES 43.

Let (A, d) be a metric subspace of (X, d). Show that (A, d) is also a topological subspace of (X, d), i.e. the restriction of d to A induces the relative topology on A.

44.

Prove: If the topological space (X, T) is homeomorphic to a metric space (Y, d), then (X, T) is metrizable.

45.

Prove Theorem 8.10:

46.

Give an example to show that the closure of an open sphere

If (X, d) is isometric to (Y, e), then (X, d) is also homeomorphic to (Y, e).

S(p, 8) = {x : d(p, x) < s} need not be the "closed sphere" {x : d(p, x) 47.

Show that a closed sphere

3(p, 8)

{x : d(p, x)

8}

8}

is closed.

CHAP. 81

METRIC AND NORMED SPACES

129

48.

Prove: The sequence (a l , a2 , ) converges to the point p in a metric space X if and only if the sequence of real numbers (d(a l , p), d(a2 , p), .) converges to 0 E R, i.e. lirn an =-- p if lim d(a„, p) = O.

49.

Prove: If Tim a

p and lirn b q in a metric space X, then the sequence of real numbers .) converges to d(p, q) E R, i.e. lim d(an , b ) = d(lim an , lim

b 1 ), d(a,2, b2 ),

EQUIVALENT METRICS 50.

Let d be a metric on X.

51.

Let d be a metric on X.

52.

Let d and e be metrics on X.

Show that the following metric is equivalent to d: e(a, b) = min {1, d(a, b)} .

Show that the following metric is equivalent to d:

Suppose

d(a, b)

e(a, b) =

1

d(a, b) d(a, b) .

3k,k' G R such that, for every a, b G X,

k e(a, b)

and

e(a,

b)

k' d(a, b)

Show that d and e are equivalent metrics.

EUCLIDEAN m-SPACE, HILBERT SPACE be points in Euclidean m-space. Show that Pi = (an, a12, • • • , alm), P2 (C121 , a22, • • • a2m), . . pn —> q = (1)1 , b2 , . . . , b m ) if and only if, for k 1, . . . , m, a2k , a 3k , ...) converges to b k ; i.e. the projection (rrk(Pn)) converges to r(q) in each coordinate space.

53.

Let

54.

Show that if G is an open subset of Hilbert Space H, then

55.

Let H* denote the proper subspace of Hilbert Space H which consists of all points in H whose first coordinate is zero. (1) Show that H* is closed. (ii) Show that H* is nowhere dense in H, i.e.

3p = (an)

G

G such that al * O.

int (H*) = Ø.

56.

Let Pi = (an, a12, • ), P2 = (a21, a22, • • ), . . be points in IC and suppose that the sequence of real numbers (rrk(Pn)) = (aik, ((sky a3k, ...) converge to bk G R for every k G N.

q = (111 , b 2,

) belongs to R.

(i)

Show that

(ii)

Show that the sequence (pi, p2 ,

) converges to q.

HILBERT CUBE

57.

The set I of all real sequences (a1 , a2 ,

)

such that

for every

0

nG

N,

is called the

Hilbert cube. (i)

Show that I is a subset of R.

(ii)

Show that I is a closed and bounded subset of R.

NORMED SPACES 58. Let '73(X, R) denote the class of all real bounded functions f: X —> R defined on some non-empty set X. Show that the following is a norm on B(X,R): f = sup { f(x) : x G

59.

i and • • • 2, on a linear space X are equivalent iff they induce equivalent metrics Two norms, is equivalent to H • • • H2 if and •• on X, i.e. iff they determine the same topology on X. Show that only if 3a1 , a2, b 1 , b 2 G R such that, for all x G X, a1

X1

< k°C 2 < b1

XIII

and

a2 X12 <

x

METRIC AND NORMED SPACES

130 60.

[CHAP. 8

Let H• • • H be the Euclidean norm and let d be the induced Euclidean metric on the plane R2. Consider the function c defined by if HAI HqH

e(p, q)

=

if

61.

(i)

Show that e is a metric on R2 .

(ii)

Describe an open sphere in the metric space (R2 , e).

1.11 -

Show that the following is a norm on C[0,1]:

62. Let X be a normed space. Show that the function

1f(x)1

dx•

le defined by f(x) = 11x11

f: X

is continuous.

Answers to Supplementary Problems 36.

The function

f: [0,1]

R defined by {1 0

f(x) =

if x = 0 if 0 < x 1

is (Riemann) integrable, i.e. belongs to 1[0, 1]. The zero function g: [0,1] R, i.e. g(x) = 0 for all But p(f, g) = 0 and f g. Hence p is not a metric as it does not satisfy [M4 ].

x G [0,1], also belongs to c7([0,11.

38.

Let A .7-- {2, 3, 4, 5, ... } and

441., ...}.

B

46. Let d be the trivial metric on a set X containing more than one point.

Then, for any p G X,

S(p,1) = {x d(p, x) < 1} = {p} g'(p, 1)

{x d(p, x)

11

X

But d induces the discrete topology on X, and so every subset of X is both open and closed. Thus

•S(P, 1) = {PI = {p} 58. Hint.

:S(P, 1 )

Proof is similar to that of Problem 30

then S(p, s) is an arc of the circle {x : Hxn =-- HAI. If IlpU < S, then S(p, 8) consists and the points on an arc of the circle /1:)11} of the points interior to the circle {x: 11xd = 8 { x: I*1 = 117)11}.

60. (ii) If Hpll

<

61.

+ 9 11

=

i f (x)

o f(x),

g(x)1 dx

If(x)1 dx

Ig(x) 1 dx

i g(x) 1 ) dx

= 11f11 + 11g11

S

Chapter 9 Countability FIRST COUNTABLE SPACES A topological space X is called a first countable space if it satisfies the following axiom, called the first axiom of countability. [CI]

For each point p e X there exists a countable class Bp of open sets containing p such that every open set G containing p also contains a member of Bp.

In other words, a topological space X is a first countable space if there exists a countable local base at every point p e X. Observe that [CI] is a local property of a topological space X, i.e. it depends only upon the properties of arbitrary neighborhoods of the point p e X. Example 1.1:

Let X be a metric space and let p E X. Recall that the countable class of open spheres {S(p,1), S(p, S(p, i), ...} with center p is a local base at p. Hence every metric space satisfies the first axiom of countability.

Example 1.2:

Let X be any discrete space. Now the singleton set {p} is open and is contained in every open set G containing p G X. Hence every discrete space satisfies [C 1 ].

First countable spaces possess the following property which was proven for the special case of the real line R. Theorem 9.1:

A function defined on a first countable space X is continuous at p and only if it is sequentially continuous at p.

e

X if

Y is continuous at p e X if for every In other words, if X satisfies [Cd, then f: X sequence (a„) converging to p in X, the sequence (f (a„)) converges to f (p) in Y, i.e., an. Remark:

p

implies

f (an)

f(p)

Let 93, be a countable local base at the point p E X. Then we can index the members of 'Bp by N, i.e. we can write Bp -= {131, B2, . }. (We permit repetitions in the case that Bp is finite.) If, in addition, B1 D B2 D B3 D .. ., then we call B„ a nested local base at p. We show, as a solved problem, that we can always construct a nested local base from a countable local base.

SECOND COUNTABLE SPACES A topological space (X, T) is called a second countable space if it satisfies the following axiom, called the second axiom of countability. [C2]

There exists a countable base 93 for the topology T.

Observe that second countability is a global rather than a local property of a topological space. Example 2.1:

The class 93 of open intervals (a, b) with rational endpoints, i.e. a, b E Q, is countable and is a base for the usual topology on the real line R. Thus R is a second countable space, i.e. R satisfies [C2 ].

131

132

[CHAP. 9

COUNTABILITY

Example 2.2:

Consider the discrete topology 0 on the real line R. Recall that a class (13 is a base for a discrete topology if and only if it contains all singleton sets. But R, and hence the class of singleton subsets (p) of R, are non-countable. Accordingly, (R, I)) does not satisfy the second axiom of countability.

Now if B is a countable base for a space X, and if Bp consists of the members of 93 which contain the point p e X, then '13, is a countable local base at p. In other words,

Proposition 9.2:

A second countable space is also first countable.

On the other hand, the real line R with the discrete topology does not satisfy [C21 by Example 2.2 but does satisfy [C I ] by Example 1.2. Thus we see that the converse of Proposition 9.2 is not true.

LINDELOF'S THEOREMS It is convenient to introduce some terminology. Let A subsets of X such that A c U{E : E E cA'}

C

X and let 0.4 be a class of

Then c..1 is called a cover (or, covering) of A, or cil is said to cover A. If each member of c/1/ is an open subset of X, then 0.1 is called an open cover of A. Furthermore, if c.1 contains a countable (finite) subclass which also is a cover of A, then 0,4 is said to be reducible to a countable (finite) cover, or c4 is said to contain a countable (finite) subcover. The central facts about second countable spaces are contained in the next two theorems, due to Lindel6f.

Theorem 9.3:

Let A be any subset of a second countable space X. cover of A is reducible to a countable cover.

Theorem 9.4:

Let X be a second countable space. Then every base '13 for X is reducible to a countable base for X.

Then every open

The preceding theorem motivates the definition of a Lindel6f space. A topological space X is called a Lindelat space if every open cover of X is reducible to a countable cover. Hence every second countable space is a Lindell:3f space.

SEPARABLE SPACES A topological space X is said to be separable if it satisfies the following axiom.

[S]

X contains a countable dense subset.

In other words, X is separable if there exists a finite or a denumerable subset A of X such that the closure of A is the entire space, i.e. A---- X. Example 3.1:

The real line R with the usual topology is a separable space since the set Q of R. rational numbers is denumerable and is dense in R, i.e.

(1. .

Example 3.2:

Consider the real line R with the discrete topology D. Recall that every subset of R is both 0-open and 0-closed; so the only 0-dense subset of R is R itself. But R is not a countable set; hence (R, .0) is not a separable space.

We will show that every second countable space is also separable. Namely,

Proposition 9.5:

If X satisfies the second axiom of countability, then X is separable.

CHAP. 9]

COUNTABILITY

133

The real line R with the topology generated by the closed-open intervals (a, b) is a classical example of a separable space which does not satisfy the second axiom of count. ability. So the converse of the previous proposition is not true in general. We do, though, have the following special case.

Theorem 9.6:

Every separable metric space is second countable.

Example 3.3:

Let C[0,11 denote the linear space of all continuous functions on the closed interval 1] with the norm defined by

1 .f

sup {lf(x)1 : 0

x

1}

By the Weierstrass Approximation Theorem, for any function f G C(0, 1 1 and any > 0, there exists a polynomial p with rational coefficients such that

fpH <e

(x) — p(x)I < c

i.e.

for all x

[0, t]

Hence the collection ci) of all such polynomials is dense in C[0,1 ] . But T is a countable set; so q0, 1] is separable and, by Theorem 9,6, second countable.

In our last example we show that a metric space need not be separable. Example 3.4:

Consider the metric e on the plane R2 defined by

e(p, q)

ilqll Pil ld(p, q)

if

if

lipil

i1q11

P 1 = 110

where !I. —11 is the Euclidean norm on R2 and d is the induced usual metric (see Problem 60, Chapter 8). (0,0) and 8< p I, Recall that if p e-open sphere S(p, 8) consists only the then of points on the circle

P

fx 114 = ilP11)

and so p cannot be an accumulation point of any A c R2 unless A contains points of the circle P. But there are an uncountable number of circles with center (0, 0), so A c R2 cannot be dense in R2 unless A is uncountable. Thus the metric space (R2 e) is not separable. ,

HEREDITARY PROPERTIES

A property P of a topological space X is said to be hereditary UT every subspace of X also possesses property P. We will show that every subspace of a second countable space In is second countable and every subspace of a first countable space is first countable. other words, the properties 1C11 and [C2I are both hereditary. On the other hand, we will show by a counterexample that a subspace of a separable space need not be separable, i.e. separability is not hereditary. We conclude with the following diagram which gives the only relationship between the three axioms in this chapter: separable

second countable

first countable

Here an arrow denotes implication as stated in Propositions 9.2 and 9.5.

134

[CHAP. 9

COUNTABILITY

Solved Problems FIRST COUNTABLE SPACES 1.

Show that any subspace (Y, TO of a first countable space (X, T) is also first countable. Solution: Let p E Y. Since Y c X, p E X. By hypothesis, (X, T) is a first countable space, so 3 a {B n : n E NI/ at p. By a previous problem, B o* countable T-local base 'B p IY n Br, : n E is a Ty-local base at p. Since "B; is countable, (Y, TO satisfies [C I ].

2.

Let Bp = Gi, G 2 ) • ) be a countable local base at p e X. Show that: (i) There exists a nested local base at p. (ii) If X satisfies [CT I then there exists a nested local base at every p E X. Solution: (i) Set

B 1 = GI ,

Then B 1 D B 2 D • • containing p, then

B2 = G1 n G2, . ,

B n = G1 n • • • n G„,

and each Bk is open and contains p.

]n0

E

N

such that

Furthermore, if G is an open set

B oo c G no c G

Accordingly, IB 1 , B, .) is a nested local base at P. (ii)

3.

If X satisfies [C 1 I and p G X, then 3 a countable local base at p by [C 1 ] and, by (i), there exists a nested local base at p.

Let Bp= 131, B2, . ) be a nested local base at p E X and let (a1, a2, . .) be a sequence such that al E /3 1 , a2 e 132 , .... Show that (TO converges to p. Solution: Let G be an open set containing p.

Since '1311 is a local base at p,

3n0 E N

such that

137, 0 c G

But '13 1, is nested; hence n > no implies ay, E B„ 0 c G, and so an —> p.

4.

Let T be the cofinite topology on the real line R, i.e. T contains Q and the complements of finite sets. Show that (R, T) does not satisfy the first axiom of countability. Solution: Suppose that

(R, T) does satisfy [C 1 ]. Then 1 E R possesses a countable open local base Since each B ,, is T-open, its complement .13 is T-closed and hence finite. Accordingly, A = U {B ca : n E NI is the countable union of finite sets and is therefore countable. But R is not countable; hence there exists a point p C R different from 1 which does not belong to A, i.e. p E Ac. E NI.

Now, by DeMorgan's Law we have p E Ac = ( U{B ric : n EN }) ('

n{Bn"

: n

=

n{B„ :

nE

Hence p E B„ for every n G N. On the other hand, {p } c is a T-open set since it is the complement of a finite set, and fpl( contains 1 since p is different from 1. Since 'B 1 is a local base, there exists a member B ,,0 G 93 1 such that B„0 c ;plc. Hence p B,, 1 . But this contradicts the statement that p G B i, for every n E N. Consequently, the original assumption that (R, T) satisfies the first axiom of countability is false.

5.

Prove Theorem 9.1: Let X satisfy the first axiom of countability. Then f:X-1 1 is continuous at p e X if and only if it is sequentially continuous at p. Solution: It suffices to show that if f is sequentially continuous at p then f is continuous at p, since the converse has been proven for an arbitrary topological space. We shall in fact prove the contrapositive statement: if f is not continuous at p then f is not sequentially continuous at p.

CHAP. 9]

135

COUNTABILITY

Let ‘B„ = {B 1 , B2 , . ..} be a nested local base at p and suppose f is not continuous at p. there exists an open subset H of Y such that

f(p) G H

but

B„ f -1 [111

for every

Then

nGN

Hence, for every n C N,

GB

an V f -1 [HI

that

which implies

f(a) V H

Now by a previous problem the sequence (a„) converges to p; but the sequence (flan)) does not converge to f(p), since the open set H containing f(p) does not contain any of the terms of the sequence. Accordingly, f is not sequentially continuous at p.

SECOND COUNTABLE SPACES 6.

Show that the plane R2 with the usual topology satisfies the second axiom of countability. Solution: Let '73 be the class of open discs in R2 with rational radii and centers whose coordinates are rational. Then '73 is a countable set and, furthermore, is a base for the usual topology on R2 . Hence R2 is a second countable space.

7.

Show that every subspace of a second countable space is second countable. Solution: Let '13 = {B„ : n E NI be a countable base for the second countable space X, and let Y be a {Y n B, : n E N) is a base for Y. Since (By is subspace of X. By a previous problem, 'B countable, Y satisfies [C2].

8.

Prove Theorem (Lindeldf) 9.3: Let A be any subset of a second countable space X. g is an open cover of A, then g is reducible to a countable cover. Solution: Let B be a countable base for X. Since A C U{G: G that p G G p. Since '13 is a base for X, for every p G A,

3B, G (73

Hence

A c U{B,: p G AI.

But

=

pe AI

for every p G A,

3G p G G such

p G B, c G,

such that

{B,: p G A) c B,

{Bp

e GI,

If

so it is countable; hence

{B„: n

NI

where N is a countable index set. For each n G N choose one set G, E G such that B,„ C G. Then

A

c U {B, : n E NI

c

U{G„: n G N)

and so {G,: n G N) is a countable subcover of G.

9.

Prove Theorem (Lindel6f) 9.4: Let g be a base for a second countable space X. is reducible to a countable base for X. Solution: Since X is second countable, X has a countable base for X, for each n G N,

B, = U{G : G G

(in}

B = {B„. : n G Nl. with

Then

Since G is also a base

gn c g

So G„ is an open cover of B T, and, by the preceding theorem, is reducible to a countable cover ci*, i.e., for each n G N, with VT':C G and G: countable = U{G : G E G: } But

G*

{G: G E G r*,, n G NI

is a base for X since '13 is. Furthermore, G* C

g

and G* is countable.

136

COUNT ABILITY

[CHAP. 9

SEPARABILITY

10. Let T be the cofinite topology on any set X. a countable dense subset.

Show that (X, 7-) is separable, i.e. contains

Solution: If X is itself countable, then clearly X is a countable dense subset of (X, T). On the other hand, suppose X is not countable. Then X contains a denumerable, i.e. non-finite countable, subset A. Recall that the only T-closed sets are the finite sets and X; hence the closure of the non-finite set A is the entire space X, i.e. A = X. But A is countable; hence (X, T) is separable.

11. Show that a discrete space X is separable if and only if X is countable. Solution: Recall that every subset of a discrete space X is both open and closed. Hence the only dense subset of X is X itself. Hence X contains a countable dense subset if X is countable, i.e. X is separable if X is countable.

12. Prove Proposition 9;5: separable.

If X satisfies the second axiom of countability, then X is

Solution: Since X satisfies [C2 ], X has a countable base 93 = {B n : n G N}. For each n G N, choose a point a. G B. Then the set A = {an : n G N} is also countable. We show that A = X or, equivalently, that each point p G Ac, the complement of A, iS an accumulation point of A.,

B no c G. Let G be an open set containing p. Then G contains at least one set B G B. Hence a no Now ano is different from p since p E Ac but ano G A. Accordingly, p is an accumulation point of A since every open se(G containing p also contains a point of A different from p.

13. Let T be the topology on the plane R2 generated by the half-open rectangles. x < b, c y < d) [a, b) x [c, d) = {(x, y) : a Show that (R2,-T) is separable. Solution: Now there are always rational numbers x o and y o such that a < xo < b and c
14. Show by a counterexample that a subspace of a separable space need not be separable, i.e. separability is not a hereditary property. Solution: Consider the separable topological-space (R2 , T) of the preceding problem. Recall (see Problem 25 of Chapter 6) that the relative topology Ty on the line Y = {(x, y) : x + y = 0} is the discrete topology since each singleton subset {p} of Y is Ty-open. But an uncountable discrete space is not separable. Thus the separability of (R2 , T) is not inherited by the subspace (Y, Ty).

15. Let S(p, E) be an open sphere in a metric space X, and let d(p, a) <. Show that if j'E < 8 < i-E, then p E S(a, 8) c S(p,€) Solution: Now d(p, a)
Accordingly, we

CHAP. 91

COUNTABILITY

Let x E S(a, 8).

137

Then d(a, x) < 8 and, by the Triangle Inequality,

d(p,x)

d(p, a) + d(a, x) < ke

<

E

E

Hence x G S(p, e), or S(a, 8) c S(p, e).

16. Prove Theorem 9.6: Let X be a separable metric space. Then X satisfies [C2], i.e. X contains a countable base.

Solution: Since X is separable, X contains a countable dense subset A. spheres with centers in A and with rational radii, i.e.,

Let '13 be the class of all open

'13 = {S(a,):aGA, 8G Q } Note that B is a countable set. We claim that '73 is a base for the topology on X, Le. for every open set G c X and every p G G, 3 S(a, 8) G (73 such that p E S(a, 8) c G Since p G G, in X,

3 an open sphere S(p,

E)

with center p such that p G S(p,

3 a0 G A

such that

.4E < 50 < le.

Let 8 0 be a rational number such that

E)C

G.

Since A is dense

d(P, a0) < Then, by the preceding problem,

p E S(a0, 8 0) c S(p, e) c G But S(ao, S o) G '13, and so 93 is a countable base for the topology on X.

Supplementary Problems FIRST COUNTABLE SPACES

17.

Show that the property of being a first countable space is a topological property.

18. Let 'B p = {B 1 , B 2 , . .}

be a nested local base at p E X.

Show that any subsequence

{B L 1,B 1 2,

.}

of 93, is also a nested local base at p.

19. Let T be the topology on the real line R generated by the closed-open intervals [a, b). Show that (R, T) satisfies [C 1 ] by exhibiting a countable local base at any point p G R. 20.

Let T be the topology on the plane R2 generated by the half-open rectangles [a, b) X [c, d)

=

{(x , y) :

x
y<

Show that (R2, T) satisfies [C I ] by exhibiting a countable local base at any point p G R2.

21. Let T and T ' be topologies on X with T coarser than T*, i.e. T c T*. (i) Show that if (X, T*) can be first countable, but (X, T) not. (ii) Show that (X, T) can be first countable, but (X, T*) not.

SECOND COUNTABLE SPACES 22. Show that the property of being a second countable space is a topological property. 23. Show that if X has a countable subbase then X satisfies [C2 ]. 24.

Exhibit a countable base for Euclidean m-space.

25.

Let c/1 be any collection of disjoint open subsets of a second countable space X. countable collection.

Show that c.,4 is a

138

[CHAP. 9

COUNTABILITY

26.

Let A be an uncountable subset of a second countable space X. Show that A has at least one point of accumulation.

27.

Let T be the topology on the real line R generated by the closed-open intervals a, b). (R, T) does not satisfy [C 2 ].

28.

Show that 1 2-space (Hilbert Space) is second countable.

Show that

SEPARABLE SPACES 29.

Show that the property of being a separable space is a topological property.

30.

Show that Euclidean m-space is separable.

31.

Show that 12-space (Hilbert Space) is separable.

32.

Let T be the topology on the real line R generated by the closed-open intervals a, b) . (R, T) is separable.

33.

Let T and T* be topologies on X with T coarser than T*, i.e. T c T*. (i) (ii)

34.

Show that if (X, T

Show that

is separable, then (X, T) is also separable.

Show by a counterexample that the converse of (i) is not true.

Let C[0,11 denote the class of continuous functions on [0,1] with norm .1

I f (x)I dx

If

Show that C[0, 1] is separable and therefore second countable.

LINDEL6F SPACES 35.

Show that a continuous image of a Lindeleif space is also a Lindel8f space.

36.

Let A be a closed subset of a Lindeltif space X. Lindeltif space.

37.

Show that a discrete space X is Lindelifif if and only if X is a countable set.

38.

Let T be the topology on the plane R2 generated by the half-open rectangles [a, b) X [c, d)

Show that A, with the relative topology, is also a

{(x, y): a

x < b, c

y<

Recall (see Problem 14) that T induces the discrete topology on the line Y = {(x , y) : x + y = 1}. Show that (R2 , T) is not Lindeltif and thus (R2 , T) is a separable first countable space which does not satisfy the second axiom of countability.

Chapter

10

Separation Axioms INTRODUCTION Many properties of a topological space X depend upon the distribution of the open sets in the space. Roughly speaking, a space is more likely to be separable, or first or second countable, if there are "few" open sets; on the other hand, an arbitrary function on X to some topological space is more likely to be continuous, or a sequence to have a unique limit, if the space has "many" open sets. The separation axioms of Alexandroff and Hopf, discussed in this chapter, postulate the existence of "enough" open sets. TI-SPACES A topological space X is a T1-space if it satisfies the following axiom: [TI ] Given any pair of distinct points a, b e X, each belongs to an open set which does not contain the other. In other words, there exist open sets G and H such that a E G, b G

and

b E H, a E H

The open sets G and H are not necessarily disjoint. Our next theorem gives a very simple characterization of 7'1-spaces. Theorem 10.1:

A topological space X is a TI-space if and only if every singleton subset {p} of X is closed.

Since finite unions of closed sets are closed, the above theorem implies: Corollary 10.2: (X, T) is a T1-space if and only if T contains the cofinite topology on X. Example 1.1:

Every metric space X is a 7' 1 -space, since we proved that finite subsets of X are closed.

Example 1.2:

Consider the topology T = {X, 0, {a}} on the set X = {a, b } . Observe that X is the only open set containing b, but it also contains a. Hence (X, T) does not' satisfy [T 1 1, i.e. (X, T) is not a 7' 1 -space. Note that the singleton set {a} is not closed since its complement {a}c {b} is not open.

Example 1.3:

The cofinite topology on X is the coarsest topology on X for which (X, T) is a T1 -space (Corollary 10.2). Hence the cofinite topology is also called the T 1 -topology.

HAUSDORFF SPACES A topological space X is a Hausdorff space or T2-space if it satisfies the following axiom: [T2]

Each pair of distinct points a, b

e X belong respectively to disjoint open sets. 139

140

SEPARATION AXIOMS

[CHAP. 10

In other words, there exist open sets G and H such that aeG, bEH

and

GnH=

Observe that a Hausdorff space is always a Ti-space. Example 2.1: We show that every metric space X is Hausdorff. Let a,b G X be distinct points; hence by IIVI 4 ] d(a, b) = > 0. Consider the open spheres G = S(a, zt3 e) and H = S(b, Tti c), centered at a and b respectively. We claim that G and H are disjoint. For if pGGn H, then d(a, p) < and d(p, b) < A-6.; hence by the Triangle Inequality,

d(a, b)

d(a, p) + d(p, b) <

+ &e. = ic

But this contradicts the fact that d(a, b) = e. Hence G and H are disjoint, i.e. a and b belong respectively to the disjoint open spheres G and H. Accordingly, X is Hausdorff.

We formally state the result in the preceding example, namely:

Theorem 10.3: Every metric space is a Hausdorff space. Example 2.2:

Let T be the cofinite topology, i.e. T rtopology, on the real line R. We show that (R, 2") is not Hausdorff. Let G and H be any non-empty T-open sets. Now G and H are infinite since they are complements of finite sets. If G n H = 0, then G, an infinite set, would be contained in the finite complement of H; hence G and H are not disjoint. Accordingly, no pair of distinct points in R belongs, respectively, to disjoint T-open sets. Thus T i -spaces need not be Hausdorff.

As noted previously, a sequence a 1 a2, ) of points in a topological space X could, in general, converge to more than one point in X. This cannot happen if X is Hausdorff: ,

Theorem 10.4:

If X is a Hausdorff space, then every convergent sequence in X has a unique limit.

The converse of the above theorem is not true unless we add additional conditions.

Theorem 10.5: Remark:

Let X be first countable. Then X is Hausdorff if and only if every convergent sequence has a unique limit.

The notion of a sequence has been generalized to that of a net (Moore-Smith sequence) and to that of a filter with the following results: Theorem 10.4A: X is a Hausdorff space if and only if every convergent net in X has a unique limit.

Theorem 10.4B:

X is a Hausdorff space if and only if every convergent filter in X has a unique limit.

The definitions of net and filter and the proofs of the above theorems lie beyond the scope of this text.

REGULAR SPACES A topological space X is regular if it satisfies the following axiom: [R] If F is a closed subset of X and p E X does not belong to F, then there exist disjoint open sets G and H such that F C G and p e H. A regular space need not be a Ti-space, as seen by the next example. Example 3.1:

Consider the topology T = {X, 0, {a}, {b, c}} on the set X = {a, b, e}. Observe that the closed subsets of X are also X, 0, {a} and {b, cl and that (X, T) does satisfy [RI. On the other hand, (X, T) is not a T I -space since there are finite sets, e.g. {b}, which are not closed.

CHAP. 10]

SEPARATION AXIOMS

141

A regular space X which also satisfies the separation axiom IT11, i.e. a regular T1-space, is called a T 3-space. Example 3.2:

Let X be a T3-space. Then X is also a Hausdorff space, i.e. a T2 -space. For let a,b E X be distinct points. Since X is a 7' 1 -space, cf,} is a closed set; and since a and b are distinct, b fal. Accordingly, by [R], there exist disjoint open sets G and H such that {a} c G and b G H. Hence a and b belong respectively to disjoint open sets G and H.

NORMAL SPACES A topological space X is normal if X satisfies the following axiom: [N] If F, and F2 are disjoint closed subsets of X, then there exist disjoint open sets G and H such that F 1 c G and F2 C H. A normal space can also be characterized as follows:

Theorem 10.6:

A topological space X is normal if and only if for every closed set F and open set H containing F there exists an open set G such that FcGc6'c H.

Example 4.1:

Every metric space is normal by virtue of the Separation Theorem 8.8.

Example 4.2:

- {a, b, c}. Consider the topology T = {X, 0, {a}, {b}, {a, b}} on the set X = Observe that the closed sets are X, 0, {b, c } , {a, cl and {O. If Fi and F2 are disjoint closed subsets of (X, T), then one of them, say F1 , must be the empty set Q. Hence 0 and X are disjoint open sets and F1 c 0 and F2 C X. In other words, (X, T) is a normal space. On the other hand, (X, T) is not a Ti-space since the singleton set {a} is not closed. Furthermore, (X, T) is not a regular space since a EZ {c}, and the only open superset of the closed set {c} is X which also contains a.

A normal space X which also satisfies -the separation axiom [Td, i.e. a normal T1-space, is called a T4-space. Example 4.3:

Let X be a T4-space. Then X is also a regular T 1 -space, i.e. T3-space. For suppose F is a closed subset of X and p G X does not belong to F. By [Ti t {p} is closed; and since F and {p} are disjoint, by [N], there exist disjoint open sets G and H such that F c G and p G {p} c H.

Now a metric space is both a normal space and a Ti-space, i.e. a 114-space. The following diagram illustrates the relationship between the spaces discussed in this chapter. Topological spaces T1 -spaces

T2-spaces (Hausdorff) T3 -spaces (regular Ti -spaces) T4-spaces (normal T1 -spaces) Metric spaces

[CHAP. 10

SEPARATION AXIOMS

142

URYSOHN'S LEMMA AND METRIZATION THEOREM Next cornes the classical result of Urysohn. Theorem (Urysohn's Lemma) 10.7: Let F1 and F.) be disjoint closed subsets of a normal space X. Then there exists a continuous function f: X- [0,1] such that f[Fi] = (01

and f[F2] = {11

One important consequence of Urysohn's Lemma gives a partial solution to the metrization problem as discussed in Chapter 8. Namely, Urysohn's Metrization Theorem 10.8: Every second countable normal Ti-space is metrizable. In fact, we will prove that every second countable normal Ti-space is homeomorphic to a subset of the Hilbert cube in R.

FUNCTIONS THAT SEPARATE POINTS Let cA = f i : i G /} be a class of functions from a set X into a set Y. functions is said to separate points if for any pair of distinct points a, b a function f in e4 such that f (a) f (b). Example 5.1:

The class cA of

e X there exists

Consider the class of real-valued functions

= {fi(x) = sin x, f2(x) = sin 2x, f3(x) = sin 3x, ...I defined on R. Observe that for every function fn O. Hence fn(w) fr,( 0) the class c4 does not separate points. Example 5.2:

Let C(X,R) denote the class of all real-valued continuous functions on a topological space X. We show that if C(X,R) separates points, then X is a Hausdorff space. Let a,b G X be distinct points. By hypothesis, there exists a continuous function f : X —> R such that f(a) f(b). But R is a Hausdorff space; hence there exist disjoint open subsets G and H of R containing f(a) and f(b) respectively. Accordingly, the inverses [G] and f —1 [H] are disjoint, open and contain a and b respectively. In other words, X is a Hausdorff space.

We formally state the result in the preceding example.

Proposition 10.9:

If the class C(X, R) of all real-valued continuous functions on a topological space X separates points, then X is a Hausdorff space.

COMPLETELY REGULAR SPACES A topological space X is completely regular if it satisfies the following axiom: [CRI

If F is a closed subset of X and p E X does not belong to F, then there exists a continuous function f: X- [0,1] such that f(p) = 0 and f[F] = (1 1.

We show later that

Proposition 10.10:

A completely regular space is also regular.

A completely regular space X which also satisfies IT1], i.e. a completely regular T,-space, is called a Tychonoff space. By virtue of Urysohn's Lemma, a T4-space is a Tychonoff space and, by Proposition 10.10, a Tychonoff space is a T 3-space. Hence a Tychonoff space, i.e. a completely regular 7'1-space, is sometimes called a 7151/2-space. One important property of Tychonoff spaces is the following:

Theorem 10.11:

The class C(X, R) of all real-valued continuous functions on a completely regular Ti-space X separates points.

CHAP. 10]

SEPARATION AXIOMS

143

Solved Problems T1-SPACES 1.

Prove Theorem 10.1: A topological space X is a 7'1-space if and only if every singleton subset of X is closed. Solution: Suppose X is a T1 -space and p G X. We show that {plc is open. Let x E {p}c. Then x p, and so by [Ti ] 3 an open set G, such that x G G, but p G, Hence x E G1 c {p}c, and hence {p}c = U{G,: x G ;plc). Accordingly {p}c, a union of open sets, is open and {p} is closed. Conversely, suppose {p} is closed for every p E X. Let a, b E X with a b. Now bE {a}e ; hence {a}c is an open set containing b but not containing a. Similarly {b}c is an open set containing a but not containing b. Accordingly, X is a 1' 1 -space.

ab

2.

Show that the property of being a Ti-space is hereditary, i.e. every subspace of a Ti -space is also a T 1 -space. Solution: Let (X, T) be a T 1 -space and let (Y, Ty) be a subspace of (X, T). We show that every singleton subset {p} of Y is a Ty-closed set or, equivalently, that Y \ {p} is T y-open. Since (X, T) is a T 1 -space, X \ {p} is T-open. But pGYcX

Yn(X\ {p}) = Y {p}

Hence by definition of subspace, Y \ {p} is a Ty -open set.

3.

Thus (Y, Ty) is also a 7' 1 -space.

Show that a finite subset of a T 1-space X has no accumulation points. Solution: Suppose A c X has n elements, say A = {a 1 , . . . , an } . Since A is finite it is closed and therefore contains all of its accumulation points. But {a2, ..., an } is also finite and hence closed. Accordingly, the complement {a2, a n }c of {a2, an } is open, contains a l , and contains no points of A different from a l . Hence a l is not an accumulation point of A. Similarly, no other point of A is an accumulation point of A and so A has no accumulation points.

4.

Show that every finite Ti-space X is a discrete space. Solution: Every subset of X is finite and therefore closed. Hence every subset of X is also open, i.e. X is a discrete space.

5.

Prove: Let X be a T1 -space. Then the following are equivalent: (i) p e X is an accumulation point of A. (ii)

Every open set containing p contains an infinite number of points of A.

Solution: By definition of an accumulation point of a set, (ii) (i) (ii).

(i);

hence we only need to prove that

Suppose G is an open set containing p and only containing a finite number of points of A different from p; say B = (G\ {p}) n A = {a 1 , a2 , ..., an } Now B, a finite subset of a T1 -space, is closed and so BC is open. Set H = G n BC. Then H is open, p E H and H contains no points of A different from p. Hence p is not an accumulation point of A and so (i) (ii).

[CHAP. 10

SE PARA TION -AXIOMS

144

'13p be a local base at p e X. from p, then some member of Bp does not contain q.

6. Let X be a Ti-space and let

Show that if

q

ex

is distinct

Solution: Since p q and X satisfies [T1 ], 3 an open set G c X containing p but not containing q. Bp is a local base at p, so G is a superset of some B G 'B y and B also does not contain q.

Now

'

7. Let X be a Ti-space which satisfies the first axiom of counta,bility. Show that if p E X is an accumulation point of A c X, then there , exists a sequence of distinct terms in A converging to p. Solution: Let ‘13=-- {B} be a nested local base at p. Set B11 = B 1 . Since p is a limit point of A, B i contains a point al E A different from p. By the preceding problem,

3 B12 G

al B12

such that

Similarly B12 contains a point a2 E A different from p and, since al /3 12 , different from a l . Again by the preceding problem,

3 B 13 G B

Furthermore,

such that 13 13

a2 G B1 2 , a2

Continuing in this manner we obtain

a2 131 2

a subsequence

(a1 , a2 , .) of distinct terms in A with al G B 1 , a2 E base at p; hence (an ) converges to p.

131 3 B13

{B 11 , B12, ...}

B12,.... But

{BO

of (73

and

a sequence

is also a nested local

HAUSDORFF SPACES 8. Show that the property of being a Hausdorff space is hereditary, i.e. every subspace of a Hausdorff space is also Hausdorff. Solution: Let (X, T) be a Hausdorff space and let (Y, Ty) be a subspace of (X, T). a, bEYc X with a b. By hypothesis, _(X, T) is Hausdorff; hence

3 G,HET

Furthermore, let

such that a E G, bEll and G n H =

By definition of a subspace, Y n G and Y

nH

are Ty-open -sets. Furthermore,

aGG,aGY • aGYnG bEH, bEY

• bGYnH

GnH

• (YnG) n (YnH) = Y n (GnH) = Y n =

(as indicated in the diagram below). Accordingly (Y, Ty) is also a- Hausdorff space.

9. Let T be the topology on the real line R generated by the open-closed intervals (a, b]. Show that (R, T) is Hausdorff. Solution: Let a, b G R with

a

b, say a < b.

G,HGT, Hence (X, T) is Hausdorff.

a G G,

Choose G = (a — 1, a]

G II

and

and H = (a, b].

Gni =

Then

CHAP. 10]

145

SEPARATION AXIOMS

10. Prove Theorem 10.4: Let X be a Hausdorff space. Then every convergent sequence in X has a unique limit. Solution: Suppose (a1 , a2, ...) converges to a and b, and suppose a b. Since X is Hausdorff, 3 open sets G and H such that aGG, b G H and GnH= 0 By hypothesis, (an ) converges to a; hence

3 no E N

n > no implies an E G

such that

i.e. G contains all except a finite number of the terms of the sequence. But G and H are disjoint; hence H can only contain those terms of the sequence which do not belong to G and there are only a finite number of these. Accordingly, (an) cannot converge to b. But this violates the hypothesis; hence a = b.

11. Prove Theorem 10.5: Let X be a first countable space. Then the following are equivalent: (i) X is Hausdorff. (ii) Every convergent sequence has a unique limit. Solution: By the preceding problem, (i) (ii); hence we need only show that (ii) (i). Suppose X is not Hausdorff. Then 3 a, b G X, a b, with the property that every open set containing a has a non-empty intersection with every open set containing b. Now let {G„} and {H} be nested local bases at a and b respectively. Then every n E N, and so

3 (a1 , 0 2, ...)

such that

Accordingly, (an ) converges to both a and b.

al G G1

n

111, a2 G

G2

(ii)

(i).

In other words,

n

G„

n H„

for

H21

NORMAL SPACES AND URYSOHN'S LEMMA 12. Prove Theorem 10.6: Let X be a topological space. Then the following conditions are equivalent: (i) X is normal. (ii) If H is an open superset of a closed set F, then there exists an open set G such that FCGCOC H. Solution:

F

(i) (ii). Let F c H, with F closed and H open. Then He is closed, and X is normal; hence 3 open sets G, G'''

GnG=Q

But

F c G, lit c G*

such that

GcG ,'('

HccG

and

and

n

Fit

= Q.

But

GnG = G.ecH

Furthermore, G'''' is closed; hence FCGc6"-cG,:( c H.

(ii)

(i).

Let F1 and

F2

be disjoint closed sets. Then F, c F , and PI is open.

3 an open set G But Furthermore,

OCF

6, is open. Thus

F 2 ce

such that and

By

F1 c Gc Gc 0 .

GnGe = 0

F 1 c G and F9 C Ôe with G, .6 , disjoint open sets; hence X is normal.

13. Let GB be a base for a normal Ti-space X. Show that for each G e p E Gi, there exists a member G ; G 6/3 such that p E G c G.

B

and any point

Solution: Since X is a T1 -space, {p} is closed; hence Gi is an open superset of the closed set {p}. Theorem 10.6,

By

146

SEPARATION AXIOMS

3 an open set G

[CHAP. 10

{p} cGcdc G i

such that

Since p G G, there is a member Gj of the base '73 such that 6 c G i ; hence p E d i c G.

p E G, c G;

SO

p E

c G.

But

14. Let D be the set of dyadic fractions (fractions whose denominators are powers of 2) in the unit interval [0, 11, i.e.,

f 1.,

D

if if

2af

15 11 6f • •

•7

16'

Show that D is dense in [0, 1]. Solution: To show that D

[0,1], it is sufficient to show that any open interval (a — 8, a + 8) centered 1 at any point a E [0,11 contains a point of D. Observe that lim — 0 =-- 0; hence there exists a n power q =-- 2no such that 0 < 1/q < 6. Consider the intervals 12

q- 2 q-1

q qj

qq

1,

Since [0,11 is the union of the above intervals, one of them, say

q

——

m+1 q

.

1 But — < 8;

rq -1 1

L

m

q

1

'

m+

contains a, i.e.

hence

a-3 <

In other words, the open interval (a— 8,

a < a+ 8

a, + 6)

contains the point m/q which belongs to D.

Thus

D is dense in [0,1].

15. Prove Theorem (Urysohn's Lemma) 10.7: Let F1 and F2 be disjoint closed subsets of a normal space X. Then there exists a continuous function f: X 3 [0, 1] such that f[F1 ] = {0} and f[F2] -

Solution: By hypothesis, F1 n F2 (6; open superset of the closed set F 1 .

hence F1 c F. In particular, since F2 is a closed set, P-'2 is an By Theorem 10.4, there exists an open set G1 12 such that C G112 C G112 C

Observe that G1/2 is an open superset of the closed set F1 , and F2' is an open superset of the closed set 61/2 . Hence, by Theorem 10.4, there exist open sets G114 and G3 14 such that C G112 C

C G 1/4 C

C G844 C

3 /,4n F2('

We continue in this manner and obtain for each t E D, where D is the set of dyadic fractions in [0,1 ] , an open set G, with the property that if t 1 , t2 G D and t, < t2 then d ti c G 12 . Define the function f on X as follows:

{ inf {t : x G G,}

if x

F2

if x G F2

1

1, i.e. f maps X into [0,11. Observe also that F 1 c G, f(x) Observe that, for every x G X, 0 for all t G D; hence f[F,I =: {0 } . Moreover, by definition, f[F2] = {1 } . Consequently, the only thing left for us to prove is that f is continuous. Now f is continuous if the inverses of the sets [0, a) and (b,11 are open subsets of X (see Problem 7, Chapter 7). We claim that f l [[0,a) ]

f

[An]

=

1.1 {G t : t < a}

= U{6`t' : t >

Then each is the union of open sets and is therefore open.

(I)

(2)

CHAP. 10]

147

SEPARATION AXIOMS

We first prove (1). Let xEf --1 [10, a) . Then f(x) E [0, a), i.e. 0 in 10, 11, there exists t r G D such that f(x) < t < a. In other words,

f(x)

=

f(x) < a.

Since D is dense

inf {t : x G G,} < t < 0, ,

Accordingly x G G, where tx < a. Hence x G U{G, t < al. We have just shown that every element in f -1 r[0, a)] also belongs to U {G, : t < a}, i.e., f• 1 HO, a)]

c U {G t : t < al

3 ty G D such that t y < a and y G G t .

On the other hand, suppose y G U{G, : t < al. Then Therefore f(y) = inf {t : y G G,} Hence y also belongs to f -1 r[0, a)].

t < a

In other words,

U {G, : t < a}

c f ' [[0, a)1

The above two results imply (1). We now prove (2) . Let x G 1(b,111. Then f(x) G (b, 11, i.e. b < f(x) in (0, 1, there exist t 1 , t.2 E D such that b < ti < t2 < f(x). In other words,

f(x)

=

1.

Since D is dense

inf {t : x E G,} > t2

Hence x G. Observe that t, < to implies Ô t ,

C G. Hence

ingly, x E dte, where t, > b; hence x E UCa te : t > bl.

f - 1-1(b,1‘

c

x does not belong to 6, 1 either. Accord-

Consequently,

U{G`i : t>1)

On the other hand, let y G U{d te e t > Then there exists ty G D such that tu > b and y E 6 c ; hence y does not belong to . But t < t, implies Gt c G t c ; hence y G Gt for every

t

Y

less than ti, .

Consequently,

Hence y E f ((b, 11).

f(y) =

inf {t : y E G t }

ty > b

In other words,

: t > b} c f The above two results imply (2).

[(b,

Hence f is continuous and Urysohn's Lemma is proven.

16. Prove Urysohn's Metrization Theorem 10.8: Every second countable normal Ti-space X is metrizable. (In fact, X is homeomorphic to a subset of the Hilbert cube I of R'.) Solution: If X is finite, then X is a discrete space and hence X is homeomorphic to any subset of H with an equivalent number of points. If X is infinite, then X contains a denumerable base '13 = {G1 , G-2 , where none of the members of 73 is X itself.

G 3„ . . }

By a previous problem, for each Gi in '13 there exists some G i in '73 such that qi c G i . The class of all such pairs KG » G), where 0.i c Gt, is denumerable; hence we can denote them by P I ,

P2, . . .

where of X. and

P, = (G,Jn, G i'n).

Observe that d

G in implies that din and GZ, are disjoint closed subsets

Hence by Urysohn's Lemma there exists a function f „[GZ,1 ----- In.

Now define a function

f:X—>I f(x) =

Observe that, for all n G N,

0

f: X —> [0,11

such that

f„[Gl = {0}

as follows: /fi(x) f 2 (x) f 3 (X)

f(x) -' 1

2 ' implies

22 ' 1 'n .nx,

)

23 '

.>

= 1;

hence f(x) is a point in the Hilbert

R, n E N, 0 ay -- 1/0, see Page 129.) We now show that f is one-to-one. Let x and y be distinct points in X. Since X is a T,-space, there exists a member G 1 of the base 93 such that x E G i but y V Gi. By a previous problem, there c G i . By definition, f m (x) = 0 since x E di, and exists a pair I'm = (G, G i) such that x G since they differ in the mth coordinate. Thus G i.e. y f(x) f(y) Hence G G1, since y G. (y) = 1 fm f is one-to-one. cube I. (Recall that I ,-- {Kan) : a

a,

148

[CHAP. 10

SEPARATION AXIOMS

We now prove that f is continuous. Let e > 0. Observe that f is continuous at p E X if there exists an open neighborhood G of p such that x G G implies If(x)— f(p)11 < f or, equivalently, _ f(p)H2 < Recall that 29, f „ (x) — f (P)I 2

Ilf(x) — f(P )I12

22n

n=1

=2

1/22n. Furthermore, since the values of f„ lie in [0,11, { f (x) f „ (P) 2)/22n Note that M T, 1/224 converges; hence there exists an no =' no (e), which is independent of x and p, such that

11(x

if(x) )

f(P)11 2

€2

fn(P)1 2 22n

+

2

is continuous; hence there exists an open neighborhood G.„ of p fn(P) 2 < e2/2no. Let G = G1 n • • • n G no . Since G is a finite

Now each function f„ : X —> [0,1] such that x E G„ implies

intersection of open neighborhoods of p, G is also an open neighborhood of p. then

!Y(x)

f(p)11 2

=

(PP =t

22n

<

Furthermore, if x G G

( E2

2no /

Hence f is continuous. Now let Y denote the range of f, i.e. Y f[X] c I. We want to prove that f -1 : Y —> X is also continuous. Observe that continuity in Y is equivalent to sequential continuity; hence f -1 is continuous at f(p) E Y if for every sequence (f(y„)) converging to f(p), the sequence (y„) converges to p. Suppose f 1 is not continuous, i.e. suppose (y„) does not converge to p. Then there exists an open neighborhood G of p such that G does not contain an infinite number of the terms of (y„). Hence we can choose a subsequence (x,) of (y„) such that all the terms of (x 7,) lie outside of G. Since p G G, there exists a member Gi in the base c13 such that p E G i c G. Furthermore, by a previous problem, there exists a pair Pm = (G1, G1) such that p G 0,1 c G1 c G. Observe that, for all n E N, x„ g G; hence oc„ G G. Accordingly, f„,(p) = 0 and f m (x.„) = 1. Then fm (x,,) — fm(P) 1 2 = 1 and If(x .)

f(p)12

fk (P)1 2

Ifk (xn) k=z1

22k

1 221tri

f(x„)— 1(p)1 > 1/2m. Therefore the sequence (f(x„)) does not In other words, for every n G N, converge to f(p). But this contradicts the fact that every subsequence of (((y„)) should also converge to f(p). Hence f -1 is continuous. Hence f is a homeomorphism and X is homeomorphic to a subset of the Hilbert cube. Accordingly, X is metrizable.

REGULAR AND COMPLETELY REGULAR SPACES 17. Prove Proposition 10.10: A completely regular space X is also regular. Solution: Let F be a closed subset of X and suppose p G X does not belong to F. By hypothesis, X is 0 and completely regular; hence there exists a continuous function f: X —> [0,1] such that f(p) J[F] = {1 } . But R and its subspace [0,1] are Hausdorff spaces; hence there are disjoint open sets G and H containing 0 and 1 respectively. Accordingly, their inverses f ' [G] and f -1 [1-1] are disjoint, open and contain p and F respectively. In other words, X is also regular.

18. Prove Theorem 10.11: The class C(X, R) of all real-valued continuous functions on a completely regular T 1 -space X separates points. Solution: Let a and b be distinct points in X. Since X is a T 1 -space, {b} is a closed set. Also, since a and b are distinct, a g {b}. By hypothesis, X is completely regular; hence there exists a real-valued con{11. Accordingly, f(a) f(b). tinuous function f on X such that f(a) = 0 and f[0, 1]

be a subspace of (X, T) and let p EY and A c Yc X. Show that if p does not belong to the Ty-closure of A, then p a A, the T-closure of A.

19. Let (Y,

Ty)

Solution: Now, by a property of subspaces (see Problem 89, Chapter 5),

CHAP. 101

SEPARATION AXIOMS

T v -closure of A But p E Y and p subset of Y and p

A.

Ty -closure of A; hence p

=

149

Yn A

(Observe that, in particular, if F is a Ty -closed

F, then p (7 P.)

20. Show that the property of being a regular space is hereditary, i.e. every subspace of a regular space is regular. Solution: Let (X, T) be a regular space and let (Y, Ty) be a subspace of (X, T). Furthermore, let p e Y and let F be a Ty-closed subset of Y such that p q F. Now by Problem 19, p P, the T.-closure of F. By hypothesis, (X, T) is regular; hence

3 G,HET But Y n G and

P c G, p G H

such that

and

GnH =

Y n H are Ty-open subsets of Y, and FEY, FEPEG • FcY nG

pe peH

• peYnH

GnH =

• (Y nG) n (YnH) = 0

Accordingly, (Y,Ty) is also regular.

Supplementary Problems T1 -SPACES 21.

Show that the property of being a 7' 1 -space is topological.

22.

Show, by a counterexample, that the image of a T1 -space under a continuous map need not be 7'1 .

23.

Let (X, T) be a T i -space and let T T*.

24.

Prove: X is a T i -space if and only if every p e X is the intersection of all open sets containing it, i.e. {p} = n{G: G open, p C G}.

25.

A topological space X is called a To-space if it satisfies the following axiom: [To ]

Show that (X, T*) is also a T1 -space.

For any pair of distinct points in X, there exists an open set containing one of the points but not the other.

(i)

Give an example of a To-space which is not a T i -space.

(ii)

Show that every T 1 -space is also a To-space.

26. Let X be a T1 -space containing at least two points. Show that if 93 is a base for X then 93

{X} is

also a base for X.

HAUSDORFF SPACES 27.

Show that the property of being a Hausdorff space is topological.

28. Let (X, T) be a Hausdorff space and let T < T*.

Show that (X, T*) is also a Hausdorff space.

29.

Show that if a i , ..., a. are distinct points in a Hausdorff space X, then there exists a disjoint class {G I , . . G.} of open subsets of X such that a t E G I , ..., an, G am .

30.

Prove: Let X be an infinite Hausdorff space. Then there exists an infinite disjoint class of open subsets of X.

150 31.

!CHAP. 10

SEPARATION AXIOMS

Prove: Let f: X —> Y and g: X —> Y be continuous functions from a topological space X into a Hausdorff space Y. Then A = {x : f(x) = g(x)) is a closed subset of X.

NORMAL SPACES 32.

Show that the property of being a normal space is topological.

33.

Let T be the topology on the real, line R generated by the closed-open intervals ra, b). (R, T) is a normal space.

34.

Let T be the topology on the plane R2 generated by the half-open rectangles, [a, b) x [c,

=

{(x,y) : a

Furthermore, let A consist of the points on the line are rational and let B = Y \A.

(1) (ii)

< b, c

Show that

y < d}

Y = {(x, y): x I- y = 1} c R2 whose coordinates

Show that A and B are closed subsets of (R2, T). Show that there exist no disjoint T-open subsets G and H of 1t2 such that A c G and B c H; and so (R2 , T) is not normal.

35.

Let A be a closed subset of a normal 7' 1 -space. Show that A with the relative topology is also a normal T1 -space.

36.

Let X be an ordered set and let T be the order topology on X, i.e. T is generated by the subsets of X of the form {x : x < a} and {x : x > a). Show that (X, T) is a normal space.

37.

Prove:

Let X be a normal space. Then X is regular if and only if X is completely regular.

URYSOHN'S LEMMA 38.

Prove: If for every two disjoint closed subsets F 1 and F2 of a topological space X, there exists a continuous function f: X —> [0,11 such that f[Fi ] = {0} and fi F2 ] = {1}, then X is a normal space. (Note that this is the converse of Urysohn's Lemma.)

39.

Prove the following generalization of Urysohn's Lemma: Let F, and F2 be disjoint closed subsets of a normal space X. Then there exists a continuous function f: X —> [a, h] such that f[F i] = {a} and

f [F21

{b}.

40.

Prove the Tietze Extension Theorem: Let F be a closed subset of a normal space X and let f: F —> [a, b] be a real continuous function. Then f has a continuous extension f* : X —> [a, b].

41.

Prove Urysohn's Lemma using the Tietze Extension Theorem.

REGULAR AND COMPLETELY REGULAR SPACES 42.

Show that the property of being a regular space is topological.

43.

Show that the property of being completely regular is topological.

44.

Show that the property of being a completely regular space is hereditary, that is, every subspace of a completely regular space is also completely regular.

45.

Prove: Let X be a regular Lindeled space. Then X is normal.

Answers to Supplementary Problems 25. (i)

Let

X = {a., b) and T = {X, {a,}, (6).

11

Chapter Compactness COVERS

Let 04 = G,} be a class of subsets of X such that A C U G I for some A C X. Recall that c4 is then called a cover of A, and an open cover if each Gi is open. Furthermore, if a finite subclass of 0,4 is also a cover of A, i.e. if

e A c Gi, U such that then 0.4 is said to be reducible to a finite cover, or contains a finite sub cover. 3

Example 1.1:

Example

1.2..

Consider the class a = {D : pG ZX Z}, where Dp is the open disc in the plane R2, with radius 1 and center p = (m, n), m and n integers. T1/en a is a cover of R2 , i.e. every point in R2 belongs to at least one member of a . On the other hand, the class of open discs '13 = {D I; : pG ZX Z}, where Dp has center p and radius is not a cover of R2. For example, the point (1-, A-) G R2 does not belong to any member of '1 3 , as shown in the figure.

'13 is displayed

Consider the classical Heine-Borel Theorem: Let A = [a, bl be a closed and bounded interval and let {GI} be a class of open sets such that A c u i G Then one can select a finite number of the open sets, say Gi , , so that

A c G, i u •

u Gim.

By virtue of the above terminology, the Heine-Borel Theorem can be restated as follows: Heine-Borel Theorem: Every open cover of a closed and bounded interval A = [a, bl is reducible to a finite cover.

COMPACT SETS

,

The-,concept of compactness is no doubt motivated by the property of a closed and bounded interval as stated in the classical Heine-Borel Theorem. Namely, Definition:11 A subset A of a topological space X is compact if every open cover of A is reducible to a finite cover.

In other words, if A is compact and AC U t where the Gt are open sets, then one can select a finite number of the open sets, say Gil , ..., Gino so that A c Gt, U • • U Glm . -

Example

2.1:

Example 2.2:

By the Heine-Borel Theorem, every closed and bounded interval line R is compact.

a, b] on the real

Let A be any finite subset of a topological space X, say A = {ct i , Then A is necessarily compact. For if g = {G I is an open cover of A, then each point }

in A belongs to one of the members of ingly, A c Gil u G u - u G.

151

G, say a t E Gt /,

G GI .

Accord-

152

[CHAP. 11

COMPACTNESS

Since a set A is compact if every open cover of A contains a finite subcover, we only have to exhibit one open cover of A with no finite subcover to prove that A is not compact. Example 2.3:

The open interval A = (0,1) on the real line R with the usual topology is not compact. Consider, for example, the class of open intervals

g

{(A,1),

Observe that A = of A. • ••

G6

I

1

0

i f

G2

1

i ,,I

o 1 1

I I

I

I

1

1

i I IG4' •••10-0 k_.) I! !I

ii

1 + 2 '

1

; hence G is an open cover

0

,

(DO , 1 ,

.• •}

G3 0

G5 i

i

A),

G„, where G, =

0-0

•••

(-k,

1 i i

be any finite subclass of G.

0 1 i

But G contains no finite subcover.

=

Gz

For let

{(al, 1)1), (a2, b2), • • •,

If

E -----

min (a l ,

(am, bm)}

am) then f >

and

(ai , 1) 1 ) u • • • U (am , bm ) c (E, 1) But (0, e] and (e, 1) are disjoint; hence G* is not a cover of A, and so A is not compact. Example 2.4:

We show that a continuous image of a compact set is also compact, i.e. if the function f:X->Y is continuous and A is a compact subset of X, then its image f[A] is a compact subset of Y. For suppose G = {G i} is an open cover of f[A], i.e. c u i Gi. Then

A c f - I Lf[Ali c

[u i Gi] = ui f -1 [Gil

Hence ,g{ = ff -1 [G i l} is a cover of A. Now f is continuous and each G I is an open set, so each f [Gi] is also open. In other words, ,qi is an open cover of A. But A is compact, so q-{ is reducible to a finite cover, say

A c f -1 [Gill u • • • u f -, [Gim ] Accordingly,

f[Al c ftf - i [G il ] u • • • u f - ILG im]] c G il

U•••U

G i.

Thus _CA} is compact.

We formally state the result in Example 2.4: Theorem 11.1: Continuous images of compact sets are compact. Compactness is an absolute property of a set. Namely, Theorem 11.2: Let A be a subset of a topological space (X, T). Then A is compaet with respect to T if and only if A is compact with respect to the relative topology TA on A. Accordingly, we can frequently limit our investigation of compactness to those topological spaces which are themselves compact, i.e. to compact spaces.

SUBSETS OF COMPACT SPACES A subset of a compact space need not be compact. For example, the closed unit interval [0, 1] is compact by the Heine-Borel Theorem, but the open interval (0, 1) is a subset of [0, 11 which, by Example 2.3 above, is not compact. We do, however, have the following Theorem 11.3: Let F be a closed subset of a compact space X. Then F is also compact.

CHAP. 111

153

COMPACTNESS

Let G = (G I 1be an open cover of F, i.e. F C Ui G. Then X = (U G) U Fe, Proof: that is, g* tG,; u tF(1 is a cover of X. But F is open since F is closed, so g* is an open cover of X. By hypothesis, X is compact; hence g* is reducible to a finite cover of X, say Gi k e X Gl i u • • • u G u Fc, ,

But F and Fc are disjoint; hence

F C G11 u • • • u Gi m ,

Gik G

We have just shown that any open cover G = {Gi) of F contains a finite subcover, i.e. F is compact.

FINITE INTERSECTION PROPERTY A class {A,1 of sets is said to have the finite intersection property if every finite subclass {A,,, A,„r } has a non-empty intersection, i.e. /1, 1 n • • n O. Example 3.1:

Consider the following class of open intervals:

e/61

=

{(0, 1), (0,

(0, ?,), (0, +),

}

Now eA has the finite intersection property, for

(0, a1 ) where b = min (a 1 , Example 3.2:

n

(0, a2 )

n • • • n

(0, a m )

= (0, b)

a m ) > O. Observe that cA itself has an empty intersection.

Consider the following class of closed infinite intervals:

=

— 2], (—'°, —1 1, (—°°, 0 1,

1 ], (—°°, 2], ...I

Note that '13 has an empty intersection, Le. n(B n : nE Z} But any finite subclass of '13 has a non-empty intersection. fies the finite intersection property.

0 where B(—,n]. In other words, '13 satis-

With the above terminology, we can now state the notion of compactness in terms of the closed subsets of a topological space. Theorem 11.4:

A topological space X is compact if and only if every class {F1) of closed subsets of X which satisfies the finite intersection property has, itself, a non-empty intersection.

COMPACTNESS AND HAUSDORFF SPACES Here we relate the concept of compactness to the separation property of Hausdorff spaces. Theorem 11.5:

Every compact subset of a Hausdorff space is closed.

The above theorem is not true in general; for example, finite sets are always compact and yet there exist topological spaces whose finite subsets are not all closed. Theorem 11.6:

Let A and B be disjoint compact subsets of a Hausdorff space X. Then there exist disjoint open sets G and H such that A c G and B CH.

In particular, suppose X is both Hausdorff and compact and F1 and F2 are disjoint closed subsets of X. By Theorem 11.3, F1 and F2 are compact and, by Theorem 11.6, F, and F2 are subsets, respectively, of disjoint open sets. In other words, Corollary 11.7: Every compact Hausdorff space is normal.

154

COMPACTNESS

[CHAP. 11

Thus metric spaces and compact Hausdorff spaces are both contained in the class of T4-spaces, i.e. normal T1-spaces.

compact Hausdorff spaces

metric spaces

T4 -spaces (normal T 1 -spaces)

The following theorem plays a very important role in geometry. Theorem 11.8: Let f be a one-one continuous function from a compact space X into a Hausdorff space Y. Then X and f[X] are homeomorphic. The next example- shows that the above theorem is not true in general. Example 4.1:

Let f be the function from the half-open interval X = [0, 1) into the plane R2 defined by f(t) = (cos 27t, sin 277-t). Observe that f maps X onto the unit circle and that f is one-one and continuous.

• 0

For But the half-open interval [0,1) is not homeomorphic to the circle. from X, X will not be connected; but if we example, if we delete the point t = delete any point from a circle, the circle is still connected. The reason that Theorem 11.8 does not apply in this case is that X is not compact. Example 4.2:

Let f be a one-one continuous function from the closed unit interval I = [0, 1] into Euclidean n-space R. Observe that I is compact by the Heine-Borel Theorem and that Rn is a metric space and therefore Hausdorff. By virtue of Theorem 11.8, I and f(li are homeomorphic.

SEQUENTIALLY COMPACT SETS

A subset A of a topological space X is sequentially compact if every sequence in A contains a subsequence which converges to a point in A. Example 5.1:

Let A be a finite subset of a topological space X. Then A is necessarily sequenis a sequence in A, then at least one of the tially compact. For if (s,, s2, elements in A, say cto, must appear an infinite number of times in the sequence. HenCe (a 0, ao, ao, ...) is a subsequence of (s„), it converges, and furthermore it converges to the point (to belonging to A.

Example 5.2:

The open interval A = (0,1) on the real line R with the usual topology is not It, ...) sequentially compact. Consider, for example, the sequence (sn) = in A. Observe that (sn) converges to 0 and therefore every subsequence also converges to 0. But 0 does not belong to A. In other words, the sequence (sd in A does not contain a subsequence which converges to a point in A, i.e. A is not sequentially compact.

In general, there exist compact sets which are not sequentially compact and vice versa, although in metric spaces, as we show later, they are equivalent. Remark:

Historically, the term bicompact was used to denote a compact set, and the term compact was used to denote a sequentially compact set.

CHAP. 11]

155

COMPACTNESS

COUNTABLY COMPACT SETS A subset A of a topological space X is countably compact if every infinite subset B of A has an accumulation point in A. This definition is no doubt motivated by the classical Bolzano-Weierstrass Theorem: Every bounded infinite set of real numbers has an accumulation point. Example 6.1:

Every bounded closed interval A [a, bl is countably compact. For if B is an infinite subset of A, then B is also bounded and, by the Bolzano-Weierstrass Theorem, B has an accumulation point p. Furthermore, since A is closed, the accumulation point p of B belongs to A, i.e. A is countably compact.

Example 6.2:

The open interval A = (0,1) is not countably compact. For consider the infinite subset B of A 2-- (0, 1). Observe that B has exactly one limit point which is 0 and that 0 does not belong to A. Hence A is not countably compact.

The general relationship between compact, sequentially compact and countably compact sets is given in the following diagram and theorem. compact Theorem 11.9:

--> countably compact

sequentially compact

Let A be a subset of a topological space X. If A is compact or sequentially compact, then A is also countably compact.

The next example shows that neither arrow in the above diagram can be reversed. Example 6.3:

Let T be the topology on N, the set of positive integers, generated by the following sets: {1, 2 } , {3, 4 } , {5, 61, ... Let A be a non-empty subset of N, say no E A. If no is odd, then no + 1 is a limit point of A; and if no is even, then n o — 1 is a limit point of A. In either case, A has an accumulation point. Accordingly, (N, T) is countably compact. On the other hand, (N, T) is not compact since

=

({1, 2} , {3,4 } , {5,61, ...)

is an open cover of N with no finite subcover. Furthermore, (N, T) is not sequentially compact, since the sequence (1, 2, 3, ...) contains no convergent subsequence.

LOCALLY COMPACT SPACES A topological space X is locally compact if every point in X has a compact neighborhood. Example 7.1:

Consider the real line R with the usual topology. Observe that each point p G R is interior to a closed interval, e.g. [p 8 , p + SI, and that the closed interval is compact by the Heine-Borel Theorem. Hence R is a locally compact space. On the other hand, R is not a compact space; for example, the class —

c,4

=

{ ..., ( - 3, — 1), ( - 2, 0), ( - 1, 1), (0, 2), (1, 3), ...1

is an open cover of R but contains no finite subcover.

Thus we see, by the above example, that a locally compact space need not be compact. On the other hand, since a topological space is always a neighborhood of each of its points, the converse is true. That is, Proposition 11.10:

Every compact space is locally compact.

156

COMPACTNESS

[CHAP. 11

COMPACTIFICATION A topological space X is said to be embedded in a topological space Y if X is homeomorphic to a subspace of Y. Furthermore, if Y is a compact space, then Y is called a compactification of X. Frequently, the compactification of a space X is accomplished by adjoining one or more points to X and then defining an appropriate topology on the enlarged set so that the enlarged space is compact and contains X as a subspace. Example 8.1:

Consider the real line R with the usual topology 'U. We adjoin two new points, denoted by 00 and —co, to R and call the enlarged set R* = R u {-00, 00} the extended real line. The order relation in R can be extended to R* by defining —00 < a < 00 for any a e R. The class of subsets of R* of the form (a, b) = {x: a < x < b}, (a, co] = {x: a < x } and [—oe,a) = {x: x G a} is a base for a topology 74* on R*. Furthermore, (E*, W') is a compact space and contains (R,11) as a subspace, and so it is a compactification of (R,11).

Recall that the real line R with the usual topology is homeomorphic to any open interval (a, b) of real numbers. The above space (R*, V') can, in fact, be show n to be homeomorphic to any closed interval [a, b] which is compact by the classical Heine-Borel Theorem. Example 8.2:

Let C denote the (x, y)-plane in Euclidian 3-space R 3 , and let S denote the sphere with center (0,0,1) on the z-axis and radius 1. The line passing through the "north pole" 00 = (0,0, 2)E S and any point p E C intersects the sphere S in exactly one point p' distinct from 00, as shown in the figure. Let f: C S be defined by Then f is, in fact, a f(p) = p'. homeomorphism from the plane C, which is not compact, onto the subset S\ {co} of the sphere S, and S is compact. Hence S is a compactification of C.

Now let (X, T) be any topological space. We shall define the Alexandrov or one-point compactification of (X, T) which we denote by (X„ T,o). Here: (1) X„, = X U {00), where co, called the point at infinity, is distinct from every other point in X.

(2) T,o consists of the following sets: (i) each member of the topology T on X, (ii) the complement in X, 0 of any closed and compact subset of X. We formally state: Proposition 1141: The above class Teo is a topology on Xop, and (X oe ,T 0c) is a compactification of (X , T).

In general, the space

may not possess properties similar to those of the original space. There does exist one important relationship between the two spaces; namely, Theorem 11.12: If (X, T) is a locally compact Hausdorff space, then (X,,„ is a compact Hausdorff space. Using Urysohn's lemma we obtain an important result used in measure and integration theory: Corollary 11.13:

Let E be a compact subset of a locally compact Hausdorff space X, and let E be a subset of an open set G X. Then there exists a continuous function f: X-* [0,1] such that f[E] = (0) and f[Gc] = t1).

CHAP. 11]

157

COMPACTNESS

COMPACTNESS IN METRIC SPACES Compactness in metric spaces can be summarized by the following Theorem 11.14: Let A be a subset of a metric space X. Then the following statements are equivalent: (i) A is compact, (ii) A is countably compact, and (iii) A is sequentially compact. Historically, metric spaces were investigated before topological spaces; hence the above theorem gives the main reason that the terms compact and sequentially compact are sometimes used synonymously. The proof of the above theorem requires the introduction of two auxiliary metric concepts which are interesting in their own right: that of a totally bounded set and that of a Lebesgue number for a cover. TOTALLY BOUNDED SETS Let A be a subset of a metric space X and let e> O. A finite set of points N e2, , ern} is called an E - net for A if for every point p G A there exists an eio E N with d(p, e20) < E. Example 9.1:

Let A = {(x, y) : x 2 + y2 < 4 } , i.e. A is the open disc centered at the origin and of radius 2. If e = 3/2, then the set

N = {(1, —1), (1,0), (1,1), (0,-1), (0,0), (0,1), (-1, —1), (-1,0), (-1, 1)1 e-net for A. On the other hand, if E = 4, then N is not an e-net for A.

is an For example, p = (44) belongs to A but the distance between p and any point in N is greater than

A is shaded

N

Recall that the diameter of A, d(A), is defined by that A is bounded if d(A) < co. Definition:

is displayed

d(A) = sup {d(a, a') : a, a' G A}

and

A subset A of a metric space X is totally bounded if A possesses an e-net for every e> O.

A totally bounded set can also be described as follows: Proposition 11.15: A set A is totally bounded if and only if for every e> there exists a decomposition of A into a finite number of sets, each with diameter less than E. We first show that a bounded set need not be totally bounded. Example 9.2:

Let A be the subset of Hilbert Space, i.e. of 12-space, consisting of the following points:

(1, 0, 0, ...)

Ci

02 =-- (0, 1, 0, CO

=

(0, 0, 1,

)

.

158

COMPACTNESS

Observe that d(e,, e)

Nf -- if i

j.

[CHAP. 11

Hence A is bounded; in fact,

d(A) = sup {d(e i, e 1): e,e 1 G A } On the other hand, A is not totally bounded. For if = 12-, the only non-empty subsets of A with diameter less than c are the singleton sets, i.e. sets with one point. Accordingly, the infinite set A cannot be decomposed into a finite number of disjoint subsets each with diameter less than

The converse of the previous statement is true.

Proposition 11.16:

Namely,

Totally bounded sets are bounded.

One relationship between compactness and total boundedness is as follows:

Lemma 11.17:

Sequentially compact sets are totally bounded.

LEBESGUE NUMBERS FOR COVERS Let cA --= (Gd ) be a cover or a subset A of a metric space X. A real number 8 > 0 is called a Lebesgue number for the cover if for each subset of A with diameter less than 8 there is a member of the cover which contains A. One relationship between compactness and Lebesgue number for a cover is as follows:

Lemma (Lebesgue) 11.18:

Every open cover of a sequentially compact subset of a metric space has a (positive) Lebesgue number.

Solved Problems COMPACT SPACES 1. Let T be the cofinite topology on any set X.

Show that (X, T) is a compact space.

Solution: Let G 7-- {G,} be an open cover of X. Choose Go E G. Since T is the cofinite topology, G ot is a (a 1 . ....a m }. Since G is a cover of X, finite set, say To for each Hence G;)

2.

G, 1

•••U

G-

a k G Go'

3

Gik E G such that

X = Go U Gr„

Go u G

at, G G ik

• • • U G, m .

Thus X is compact.

Show that any infinite subset A of a discrete topological space X is not compact. Solution: Recall that A is not compact if we can exhibit an open cover of A with no finite subcover. Consider the class 4 = {{a} : a E A} of singleton subsets of A. Observe that: (i) c.,4 is a cover of A; in fact A = U {{a} : a G A). (ii) cs1 is an open cover of A since all subsets of a discrete space are open. (iii) No proper subclass of c..4 is a cover of A. (iv) c..1 is infinite since A is infinite. Accordingly, the open cover ci/ of A contains no finite subcover, so A is not compact. Since finite sets are always compact, we have also proven that a subset of a discrete space is compact if and only if it is finite.

CHAP. 111

3.

159

COMPACTNESS

Prove Theorem 11.2: Let A be a subset of a topological space (X, T). following are equivalent: (i) A is compact with respect to T. (ii) A is compact with respect to the relative topology TA on A. Solution: (i) (ii):

Let {Gi } be a TA -open cover of A.

3 Hi G T

Then the

By definition of the relative topology, Gi = A n Hi c Hi

such that

A c u i Gi c u Hi

Hence

and therefore {H i } is a T-open cover of A. By (i), A is T-compact, so {H i } contains a finite subcover, say A c H1 u • • • u Him , Flik G {H i} But then A c An (Hii u • • u Him) (A n1111 ) u • • u (A n Him) = G il u • • • u Gim Thus {G i } contains a finite subcover (ii)

(i):

Gim } and

Let {H i } be a T-open cover of A.

is compact.

Set Gi = A

n

Hi ; then

A c An (u H i) = u i (A n Hi) 7-- u 1 G 1

Acu 1 H 1

But G i G TA, so {G i } is a TA -open cover of A. a finite subcover {G il , ..., Gim }. Accordingly, A c Gil u • • • u G im =- (A

(A,T A )

n

By hypothesis, A is TA-compact; thus {G i} contains

n Him)

u • • • u (A

Thus {H i } is reducible to a finite cover {H i ,

= A n

u • • • u H im) c H11 u • • • u H

im

H1 } and therefore A is compact with respect to T. m

4. Let (Y, T*) be a subspace of (X, T) and let A c Yc if and only if A is T*-compact.

X.

Show that A is T-compact

Solution: Let TA and Tit* be the relative topologies on A. Then, by the preceding problem, A is T- or T*-compact if and only if A is TA- or TA* -compact; but T A = T.

5.

Prove that the following statements are equivalent: (i) X is compact. (ii) For every class TO of closed subsets of X, n 1 F1 =0 finite subclass {F11 , ...,Fi.} with Fi, n ••• fl F

implies {F1 ) contains a

Solution: (1)

(ii):

Suppose n 1 F1

O.

Then, by DeMorgan's Law,

X = Øc = so {.F.} is an open cover of X, since each

3 F ,.

(n i Fi.)c

F i is closed. But by hypothesis, X is compact; hence

G {F}

such that

X= F

U•••U

Thus by DeMorgan's Law, •••

X = (F and we have shown that (i)

= F c n • • • n FC = Fit n • • • n Fim

u

(ii).

Let {Gi} be an open cover of X, i.e. =

Xc =

X =_- u 1 G1. By DeMorgan's

(u i Gi)c

Law,

160

COMPACTNESS

(CHAP. 11

Since each Gi is open, {Gf } is a class of closed sets and, by above, has an empty intersection. by hypothesis, n••• n = 3 G cti , G {G ci,} such that

'1

Hence

In

Thus by DeMorgan's Law,

X = 0c = (Gic.i n • • •

n G -m )c = Ç u • • • u GLe -= Gil u • u

Accordingly, X is compact and so (ii)

6.

Gt.

(i).

Prove Theorem 11.4: A topological space X is compact if and only if every class {Fi) of closed subsets of X which satisfies the finite intersection property has, itself, a non-empty intersection. Solution: Utilizing the preceding problem, it suffices to show that the following statements are equivalent, where {Fi} is any class of closed subsets of X: Fi n • • • n Fi

(i)

n

n

0 V

j1 , • . • , Ç, s.t. F1 n • • •

F„: = Q

n F1

=

0

But these statements are contrapositives.

COMPACTNESS AND HAUSDORFF SPACES 7.

Prove: Let A be a compact subset of a Hausdorff space X and suppose p E X\A. Then p E G, AcH, GnH= 3 open sets G,H such that Solution: Let a E A.

Since p V A, p

a.

3 open sets G a , Ha

By hypothesis, X is Hausdorff; hence such that

p G Ga , ΠE Ha , Ga

Hence A c U {H a : a G A}, i.e. {Ha : a G A} is an open cover of A.

Ham G {Ha }

3 Hai ,

H = Hai u • • •

Now let

U

Ha

and G

such that Gai

n•n

But A is compact, so

A c Hai U • • • U Ham H and G are open since they are Furthermore, A c H and p G G since

Gam .

respectively the union and finite intersection of open sets. p belongs to each Ga .

Lastly we claim that G n H = Q. Note first that G ai n Ha . = 9) Thus, by the distributive law,

GnH Thus

8.

Gn (Hai u • • uHam) = (GnHai) U

U (GnHam)

implies that

G n H = 0.

Ou•--u0 =

the proof is complete.

Let A be a compact subset of a Hausdorff space X. is an open set G such that pEGC Ac.

Hence

A, then there

Show that if p

Solution: By Problem 7 there exist open sets G and H such that p G G, A c H

9.

n Ha =

and G

n

H = Q.

GnA = 0, and pEGCAc.

Prove Theorem 11.5: closed.

Let A be a compact subset of a Hausdorff space X.

Then A is

Solution: We prove, equivalently, that Ac is open. Let p G Ac, Le. p V A. Then by Problem 8 there exists an open set G p such that p G G p c A. Hence Ac = U{G p : p G Ac } . Thus Ac is open as it is the union of open sets, or, A is closed.

CHAP. 11]

161

COMPACTNESS

10. Prove Theorem 11.6: Let A and B be disjoint compact subsets of a Hausdorff space X. Then there exist disjoint open sets G and H such that A C G and B C H. Solution: Let a G A. Then a B, for A and B are disjoint. By hypothesis, B is compact; hence by Problem 1 there exist open sets Ga and Ha such that

a G Ga, B c Ha and Since a G Ga , {Ga : a G AI is an open cover of A. of the open sets, say G a , Ga , so that A c Ga m since B is a subset of each individually.

Ga n Ha =

Since A is compact, we can select a finite number • • u G am . Furthermore, B c H a n ---n Ham

Now let G = Gai u • • u and H = Hai n • n Ha . Observe, by the above, that A c G Getn. nt and B c H. In addition, G and H are open as they are the union and finite intersection respectively of open sets. The theorem is proven if we show that G and H are disjoint. First observe that, for each i, Gai n Hai = 9) implies Ga n H = Ø. Hence, by the distributive law,

G n H =

(G

• • • U Gain)

n H = (Gai n H)

u•••u

(Gam n H) = Ou•••UO =

Thus the theorem is proven.

11. Prove Theorem 11.8: Let f be a one-one continuous function from a compact space X into a Hausdorff space Y. Then X and f[X] are homeomorphic. Solution: Now f: X —> f [X] is onto and, by hypothesis, one-one and continuous, so f 1 : f[X] —> X exists. We must show that f -1 is continuous. Recall that f -1 is continuous if, for every closed subset F of X, (f - 1 ) -1- [F] = f[F] is a closed subset of f[X]. By Theorem 11.3, the closed subset F of the compact space X is also compact. Since f is continuous, .f[F] is a compact subset of f[X]. But the subspace f[X] of the Hausdorff space Y is also Hausdorff; hence by Theorem 11.5, .f[F1 is closed. Accordingly, f -1 is continuous, so f: X —> f[X1 is a homeomorphism, and X and f[X] are homeomorphic.

12. Let (X, T) be compact and let (X, T*) be Hausdorff. Show that if T* C T, then T* = T. Solution: Consider the function f: (X, T) —> (X, T) defined by f(x) =- x, i.e. the identity function on X. Now f is one-one and onto. Furthermore, f is continuous since T* c T. Thus by the preceding = T. problem, f is a homeomorphism and therefore

SEQUENTIALLY AND COUNTABLY COMPACT SETS 13. Show that a continuous image of a sequentially compact set is sequentially compact. Solution: Let f: X Y be a continuous function and let A be a sequentially compact subset of X. We want to show that f[A] is a sequentially compact subset of Y. Let (b 1 , b2, ...) be a sequence in f[A]. Then f(a) =b, , VnG N such that 3 a„ a 2 , ... G A But A is sequentially compact, so the sequence (a 1 , a 2 , ...) contains a subsequence (ail , a12, ...) which converges to a point c/o G A. Now f is continuous and hence sequentially continuous, so

(f(ail ), f(a12), Thus f[A] is sequentially compact.

)=

b 12 , . .

• )

converges to f(a0) G f[A]

162

COMPACTNESS

[CHAP. 11

14. Let T be the topology on X which consists of 0 and the complements of countable subsets of X. Show that every infinite subset of X is not sequentially compact. Solution; Recall (Example 7.3, Page 71) that a sequence in (X, T) converges if it is of the form (a 1 , a2, • • • , an p ,

PI

PI

• • •

that is, is constant from some terni on. Hence if A is an infinite subset of X, there exists a sequence (b„) in A with distinct terms. Thus (b n ) does not contain any convergent subsequence, and A is not sequentially compact.

15. Show that: (i) a continuous image of a countably compact set need not be countably compact; (ii) a closed subset of a countably compact space is countably

compact. Solution:

(1)

Let X = (N, T) where T is the topology on the positive integers N generated by the sets {1,2 , {3, 4 } , {5, 6 } , . . By Example 6.3, X is countably compact. Let Y = (N, D) where LO is the discrete topology on N. Now Y is not countably compact. On the other hand, the function X --> Y which maps 2n and 2m — 1 onto n for n G N is continuous and maps the countably compact set X onto the non-countably compact set Y. }

(ii) Suppose X is countably compact and suppose F is a closed subset of X. Let A be an infinite subset of F. Since F c X, A is also an infinite subset of X. By hypothesis, X is countably compact; then A has an accumulation point p G X. Since A c F, p is also an accumulation point of F. But F is closed and so contains its accumulation points; hence p G F. We have shown that any infinite subset A of F has an accumulation point p G F, that is, that F is countably compact.

16. Prove: Let X be compact. Then X is also countably compact. Solution: Let A be a subset of X with no accumulation points in X. Then each point p G X belongs to an open set G1, which contains at most one point of A. Observe that the class {G, : p E X} is an open {G„ i , Gpm } . cover of the compact set X and, hence, contains a finite subcover, say Hence

A cXcG„ 1 u-•-uGpm

But each G. contains at most one point of A; hence A, a subset of G

•••

U

Gpm , can contain at

most m points, i.e. A is finite. Accordingly, every infinite subset of X contains an accumulation point in X, i.e. X is countably compact.

17. Prove: Let X be sequentially compact. Then X is also countably compact. Solution: Let A be any infinite subset of X. Then there exists a sequence (a 1 , a2, .) in A with distinct terms. Since X is sequentially compact, the sequence (an ) contains a subsequence (a1 , a12, .) (also with distinct terms) which converges to a point p E X. Hence every open neighborhood of p contains an infinite number of the terms of the convergent subsequence (ai.n). But the terms are distinct; hence every open neighborhood of p contains an infinite number of points in A. Accordingly, p E X is an accumulation point of A. In other words, X is countably compact.

Remark: Note that Problems 16 and 17 imply Theorem 11.9.

163

COMPACTNESS

CHAP. 11]

18. Prove: Let A C X be sequentially compact. Then every countable open cover of A is reducible to a finite cover. Solution: We may assume A is infinite, for otherwise the proof is trivial. We prove the contrapositive, i.e. assume 3 a countable open cover {G: with no finite subcover. We define the sequence as follows.

A n Gn1 be the least positive integer larger than n 1 such that A n

Q. Choose a l G A n G. Let n2

Let n 1 be the smallest positive integer such that

a2 E (A n Gn2)\(A

n

Q. Choose

Gni)

Such a point always exists, for otherwise Gni covers A. Continuing in this manner, we obtain the with the property that, for every i G N, sequence (a 1 , a2, ai GAnG ni ,ai Gu71-11 (A

. n Ga) .7

and

We claim that (ai) has no convergent subsequence in A.

3 Gio G {Gi} Now A n G io

ni > n i _

Let p G A.

Then

p E Gio

such that

0, since pE A n Gio ; hence 3 Jo E N

such that

Gill() =

But by the choice of the sequence (a l , a2, ...) i > J0

ai Gio

Accordingly, since Gio is an open set containing p, no subsequence of (ai) converges to p.

But p was

arbitrary, so A is not sequentially compact.

COMPACTNESS IN METRIC SPACES 19. Prove Lemma 11.17: Let A be a sequentially compact subset of a metric space X. Then A is totally bounded. Solution: We prove the contrapositive of the above statement, i.e. if A is not totally bounded, then A is not sequentially compact. If A is not totally bounded then there exists an c > 0 such that A possesses no (finite) e-net. Let al G A. Then there exists a point a2 E A with d(a„ a2 ) E, for otherwise {a / } would be an E-net for A. Similarly, there exists a point 0 3 G A with d(œ 1 , 03 ) E and d(a2 , a 3) = E, for otherwise {ai , a2} would be an e-net for A. Continuing in this manner, we arrive at E for i a sequence (a / , a2, ...) with the property that d(ai, J. Thus the sequence (an) cannot contain any subsequence which converges. In other words, A is not sequentially compact.

20. Prove Lemma (Lebesgue) 11.18: Let 0,4 = (Gil be an open cover of a sequentially compact set A. Then cA' has a (positive) Lebesgue number. Solution: Suppose c.4 does not have a Lebesgue number. Then far each positive integer n G N a subset B n of A with the property that

0 < d(Rp) G 1/n

and B

there exists

for every Gi in c."1

For each n E N, choose a point bn G B. Since A is sequentially compact, the sequence (b 1 , b 2 , ...) contains a subsequence (b ii , b52 , ...) which converges to a point p E A. Since p G A, p belongs to an open set Gp in the cover cA. Hence there exists an open sphere S(p, e), with center p and radius e, such that p G S(p,e) c G. Since (b 5 ) converges to p, there exists a positive integer 4,0 such that d(p, bind <

bino G Bino and

d(B;n0) <

Using the Triangle Inequality we get Bino c S(p, e) c G. But this contradicts the fact that Blno Gi for every Gi in the cover c4. Accordingly c.,4 does possess a Lebesgue number.

0

is shaded

[CHAP. 11

COMPACTNESS

164

21. Prove: Let A be a countably compact subset of a metric space X. sequentially compact. Solution: Let (a l , a 2 , ...) be a sequence in A. If the set B say a i , satisfies a io ai for infinitely many j G N.

Then A is also

a 2, ...} is finite, then one of the points, (a40, aio , ...) is a subsequence of (a n)

Hence

which converges to the point aio in A. On the other hand, suppose B = {a 1 , a2 , ...} is infinite. By hypothesis, A is countably compact. Hence the infinite subset B of A contains an accumulation point p in A. But X is a metric space; hence we can choose a subsequence (a il , a i2, ...) of the sequence (a n) which converges to the point p in A.

In other words, A is sequentially compact.

22. Prove Theorem 11.14: Let A be a subset of a metric space X. Then the following are equivalent: (1) A is compact, (ii) A is countably compact, and (iii) A is sequentially

compact. Solution: Recall (see Theorem 11.8) that (i) implies (ii) in every topological space; hence it is true for a metric space. In the preceding problem we proved that (ii) implies (iii). Accordingly, the theorem is proven if we show that (iii) implies (i). Let A be sequentially compact, and let es/ = { G 1 } be an open cover of A. We want to show that A is compact, i.e. that- c,,4 possesses a finite subcover. By hypothesis, A is sequentially compact; hence, by Lemma 11.18, the cover c4 possesses a Lebesgue number 8 > O. In addition, by Lemma 11.17, A is totally bounded. Hence there is a decomposition of A into a finite number of subsets, say 13 1 , , B,„ with d(B 1) < 8. But 8 is a Lebesgue number for crf; hence there are open sets G f 1 1 • • . Gtm• E oil such that B1 C

A

Accordingly,

c Gi m

.

c B 1 U B2 U • • • U B m

C G1

U Gi2 U • • •

U

Thus c..4 possesses a finite subcover {Gil , ..., Gi. }, i.e. A is compact.

Show that for any B C X there

23. Let A be a compact subset of a metric space (X, d). is a point p e A such that d(p, B) = d(A,B). Solution: Let d(A, B)

E. Since d(A, B) = inf {d(a, b) :aGA, be 13 } , for every positive integer n E N,

3 an G A, I), B

such that

E

d(a n, bn) < e+ 1/n

Now A is compact and hence sequentially compact; so the sequence (a 1 , a2, ...) d(A, B) = e. which converges to a point p G A. We claim that d(p, B) Suppose d(p, B) > e, say d(p, B) to p,

3 no G N Then

such that

E+

8

where

d(p, ano) < ;8

d(p, a„ 0) + d(ano , b.,0) <

has a subsequence

8 > O. Since a subsequence of (an) converges and

d(ano, bnd <

+ e + ;8 = e + 8

But this contradicts the Triangle Inequality; hence d(p, B)

E

d(p, B)

1/n0 < e + ;8

d(p,

d(A, B).

24. Let A be a compact subset of a metric space (X, d) and let B be a closed subset of X such that A n B = O. Show that d(A,B) > O. Solution: Suppose d(A, B) =

O. Then, by the preceding problem, 3 pG A

such that

d(p, B) = d(A, B) = 0

But B is closed and therefore contains all points whose distance from B is zero. so p GA n B. But this contradicts the hypothesis; hence d(A,B)> O.

Thus

pG B

and

CHAP. 11]

165

COMPACTNESS

25. Prove: Let f be a continuous function from a compact metric space (X, cl) into a metric space (Y, cl*). Then f is uniformly continuous, i.e. for every E > 0 there exists a 8 > 0 such that

d*(f(x), f(y)) <

(Remark: Uniform continuity is a stronger condition than continuity, in that the 8 above depends only upon the E and not also on any particular point.) Solution: Let E > O. Since f is continuous, for each point p G X there exists an open sphere S (p , 8p) such that f (x) G S(f(P), le) x G S(P , 8p) Observe that the class 0.4 = {S(p, s p ) : p E X} is an open cover of X. By hypothesis, X is compact and hence also sequentially compact. Therefore the cover c..4 possesses a Lebesgue number 8 > O. Now let x, y G X with d(x, y) < 3. member S(po, 60) of the cover A. Now

But d(x, y)

d{x, y

}

< 8 implies {x, y } is contained in a

f(x), f(y) e S(f(P0), 1E)

x,Y E S(Po, sp o)

But the sphere S(f(p 0), i-E) has diameter E. Accordingly, d*(f(x), f(y)) <

d(x, y) < 8

e

In other words, f is uniformly continuous.

Supplementary Problems COMPACT SPACES 26.

Prove: If E is compact and F is closed, then EnF is compact.

27. Let A 1 , ..., A m be compact subsets of a topological space X. compact.

Show that A l u•—u A m is also

28. Prove that compactness is a topological property.

•

29. Prove Proposition 11.11: The class T. is a topology on Xœ and (X., T.) is a compactification of (X, T). (Here (X T.) is the Alexandrov one-point compactification of (X, T).) 30.

Prove Theorem 11.12: Hausdorff space.

If (X, T) is a locally compact Hausdorff space, then (X., T.) is a compact

SEQUENTIALLY AND COUNTABLY COMPACT SPACES 31.

Show that sequential compactness is a topological property.

32.

Prove: A closed subset of a sequentially compact space is sequentially compact.

33.

Show that countable compactness is a topological property.

34. Suppose (X, T) is countably compact and T* 35.

T.

Show that (X, T*) is also countably compact.

Prove: Let X be a topological space such that every countable open cover of X is reducible to a finite cover. Then X is countably compact.

36.

(CHAP. 11

COMPACTNESS

166

Prove: Let X be a T1 - space. Then X is countably compact if and only if every countable open cover of X is reducible to a finite cover.

37. Prove: Let X be a second countable T 1 -space. Then X is compact if and only if X is countably compact.

TOTALLY BOUNDED SETS 38.

Prove Proposition 11,15: A set A is totally bounded if and only if for every E > 0 there exists a decomposition of A into a finite number of sets each with diameter less than E.

39.

Prove Proposition 11.16:

40.

Show that every subset of a totally bounded set is totally bounded.

41.

Show that if A is totally bounded then  is also totally bounded.

42.

Prove:

Totally bounded sets are bounded.

Every totally bounded metric space is separable.

COMPACTNESS AND METRIC SPACES 43.

Prove: A compact subset of a metric space X is closed and bounded.

44.

Prove: Let f: X —> Y be a continuous function from a compact space X into a metric space Y. Then f[XI is a bounded subset of Y.

45.

Prove: A subset A of the real line I? is compact if and only if A is closed and bounded.

46.

Prove:

47.

Prove: The Hilbert cube I

48.

Prove: Let A and B be compact subsets of a metric space X. such that d(a, b) = d(A,B).

Let A be a compact subset of a metric space X.

Then the derived set A' of A is compact.

{(a n) : O a = 1/n} is a compact subset of R. Then there exist a G A and b G B

LOCALLY COMPACT SPACES 49.

Show that local compactness is a topological property.

50.

Show that every discrete space is locally compact.

51.

Show that every indiscrete space is locally compact.

52.

Show that the plane R2 with the usual topology is locally compact.

53.

Prove: Let A be a closed subset of a locally compact space (X, T). Then A with the relative topology is locally compact.

Chapter 12 Product Spaces PRODUCT TOPOLOGY Let {Xi: i e /} be any class of sets and let X denote the Cartesian product of these sets, i.e. X = fJ1 X 1. Note that X consists of all points p = (a1: i e /) where ai E X. Recall that, for each jo e /, we defined the projection rjo from the product set X to the coordinate space Xi o, i.e. 7r3-0 : X X io, by 7110((ai : j E /)) =

aio

These projections are used to define the product topology.

Definition:1

Let {(X1, TO} be a collection of topological spaces and let X be the product of the sets Xi, i.e. X = fl, x-1 The coarsest topology T on X with respect to which all the projections 71 : X -3 X i are continuous is called the (Tychonoff) product topology. The product set X with the product topology T, i.e. (X, T), is called the product topological space or, simply, product space. .

In other words, the product topology T on the product set X = generated by the projections (see Chapter 7). Example

1.1:

ri, x-,

is the topology

Consider the Cartesian plane R2 = R X R. Recall that the inverses 77- 1-1 (a, b) and 7 2-1 (a, b) are infinite open strips which form a subbase for the usual topology on R2.

b

al

(a, b)

—1 2

Thus the usual topology on R2 into R.

(a b)

R2 is the topology generated by the projections from

In view of the above definition, we can state the result in Example 1.1 as follows: Theorem 12.1: The usual topology on R2 = R X R is the product topology. Example 1.2:

{X i : j E /} be a collection of Hausdorff spaces and let X be the product space, X = fIX. We show that X is also a Hausdorff space. Let p = (ai : j G q : j G I) be distinct points in X. Then p and q must differ in at least coordinate space, say X 10 , i.e. a10 * b lo. By hypothesis, X 10 Is Hausdorff; hence there exist disjoint open subsets G and H of X 30 such that ajo G G and b 0 H. By definition of the product space, the projection 774 : X X is con1 G 10

Let i.e. and one

167

PRODUCT SPACES

168 tinuous.

Accordingly 77- .70 1 [G] and

7201

[CHAP. 12

[H] are open disjoint subsets of X contain-

ing p and q respectively. Hence X is also a Hausdorff space.

G

rjo

BASE FOR A FINITE PRODUCT TOPOLOGY The Cartesian product A x B of two open finite intervals A and B is an open rectangle in R2 as illustrated below.

As noted previously, the open rectangles form a base for the usual topology on R2 which is also the product topology on R2. A similar statement is true for every finite Product topology. Namely, Proposition 12.2: Let X1 , • •, be a finite number of topological spaces, and let X be the product space, i.e. X = Xi x•• x X. Then the following subsets of the product space X, G I X G2 X • • X G. where Gi is an open subset of XI, form a base for the product topology on X. As we shall see in the next section, the above proposition is not true in the case of an infinite product space. DEFINING SUBBASE AND DEFINING BASE FOR THE PRODUCT TOPOLOGY Let {X,: i e I) be a collection of topological spaces and let X denote the product space, i.e. X X. . If Gio is an open subset of the coordinate space X50, then 7ri-o I [Giol consists of all points ' p = (a- i e I) in X such that 7r,o (p) e Gio . In other words,

n

7r.,70 1 [Gio]

H {X1:

Jo )X Gio

In particular, if we have a denumerable collection of topological spaces, say )X1, X2, ...), then the product space X = 11 X„ = X1xX2 xX3X • • • n =1 consists of all sequences .) p = (a 1, a2, a3, where an E Xn and, furthermore, Xi0+1 x 7 1 [Gi ] x • • • x Xio _ i x Gio 20 -0

169

PRODUCT SPACES

CHAP. 121

By definition, the product topology on X is the "smallest", i.e. coarsest, topology on X with respect to which all the projections are continuous, i.e. the topology generated by the projections. Accordingly, the inverse projections of open sets in the coordinate spaces form a subbase for the product topology (Theorem 7.8). Namely, Theorem 12.3: The class of subsets of a product space X = of the form 7[0 1 [G jo] =

where Gio is is called the Furthermore, since finite topology, we also have Theorem 12.4: The class of 17

1 [G1 1

]

t

nx,

xt:

Jo}

Gio

an open subset of the coordinate space X30, is a subbase and defining subbase for the product topology. intersections of the subbase elements form a base for the subsets of a product space X =

H {Xi : i

n • • • n

nx-,

/1,

of the form

..., 1.) x Gj1 x • • • x

where G ik is an open subset of the coordinate space Xik, is a base and is called the defining base for the product topology. Using the above properties, we can prove the following central facts about product spaces. Theorem 12.5: A function f from a topological space Y into X a product space X = is continuous if rt and only if, for every projection 7r t , the camposition mapping 7r1 0 f : Y ---> X, is also con-

x-,

xi

tinuous.

ni x-,

Theorem 12.6:

Every projection 7r, X X . a product space X = and continuous, i.e. bicontinuous.

Theorem 12.7:

A sequence p i , pi, ... of points in a product space X = to X. the point q in X if and only if, for every projection 7ri X -3 Xi, the sequence 77-0 1),71- 1(p 2), . converges to -71(q) in the coordinate space X.

is both open

In other words, if 2)1 = (ai i), p2 = (a2 1), ... and q = (1),) are points in a product space X= then if an i ---> bi in every coordinate space X i pn —> q in X

Hi xi

,

EXAMPLE OF A PRODUCT SPACE

Let Ri denote a copy of R, the set of real numbers with the usual topology, indexed by the closed unit interval I = [0,1]. Consider the product space X = fi (R1: i E I}. We can represent X graphically as in the adjoining figure. Here the horizontal axis denotes the index set I = 10, 1], and each vertical line through a point, say jo , in I denotes the coordinate space R10 . Consider an element p i e /) in the product space X. Observe that p assigns to each number i G I the real number ai, i.e. p is a real-valued function defined on the index set I = 10,1). In other words, the product space X is the class of all realvalued functions defined on I, i.e., X -= tp : p :1 ---> Some of the elements of X are also displayed in the figure.

Members of X

170

PRODUCT SPACES

[CHAP.

12

We now describe one of the members of the defining subbase (:-5 for the product topology on X. Recall that c-f consists of all the subsets of X of the form {R1 : i Jo ) x Gio Jo -0 where Gio is an open subset of the coordinate space R io. Suppose Gjo is the open interval (1, 2). Then 7ri-0 1 [G30 1 consists of all points p =-- (at : i e I) in X such that (tic, E Gio = (1, 2), i.e. all functions such that 1 < p(jo) < 2. Graphically, 7r 3-0 1 1G10 1 consists of all those functions passing through the open interval G, o = (1, 2) on the vertical line representing the coordinate space Rio, as illustrated in the adjacent diagram. 7r -1 [G I

Members of

v- 7 1 20

[GI

Lastly we describe one of the open sets, say B, of the defining base 6/3 for the product topology on X. Recall that B is the intersection of a finite number of the members of the defining subbase el for the product topology, say,

B

71- 3-1 1 [Gfi l ]

-_-_-

n

{R, : j

7T2 1 [G, 2

1 n

71

- 33 1

[G53 1

j1 , j2, j3 ) x G 1 x Gj2 X Gj 3

Graphically, then, B consists of all functions passing through the open sets Gi p Gi2 and G 3 which lie on the vertical lines denoting the coordinate spaces R il, and Ri3 . Some of the members of B are displayed in the diagram on the right.

Ril

R.

Ri3

Members of B

Consider the following proposition. Proposition 12.8: Let ((Xi, TO) be a collection of topological spaces and let X denote the product of the sets Xi, i.e. X = n, X i . Then the subsets of X of the form

H {G, : E r) where Gi is an open subset of the coordinate space X,, form a base for a topology on the product set X. Remark:

appearing in Proposition 12.8 is The topology on the product set X = J not always identical to the product topology on X as defined in this chapter. On the other hand, Proposition 12.2 shows that the two topologies coincide in the case of a finite product space. Historically, the topology in Proposition 12.8 appeared and was investigated first. Tychonoff is credited with defining the (Tychonoff) product topology and proving for it one of the most important and useful theorems in topology, the Tychonoff Product Theorem. It is because of this theorem that the product topology is considered the "right" topology on the product set. [

.

TYCHONOFF'S PRODUCT THEOREM A property P of a topological space is said to be product invariant if a product space X =-X1 possesses P whenever each coordinate space Xi possesses P. For example, the property P of being a Hausdorff space is product invariant since, in. view of Example 1.2, the product of Hausdorff spaces is also Hausdorff. The celebrated Tychonoff Product Theorem states that compactness is also a product invariant property:

CHAP. 121

PRODUCT SPACES

Theorem (Tychonoff) 12.9:

171

The product of compact spaces is compact.

The proof of Theorem 12.9 which is given in the solved problem section relies on the set-theoretic lemma which follows; the proof of the lemma requires Zorn's Lemma. This is not too surprising since it has been shown that the Tychonoff Product Theorem is, in fact, equivalent to Zorn's Lemma. Lemma 12.10:

Let 34 be a class of subsets of a set X with the finite intersection property. Consider the collection P of all superclasses of 34 which have the finite intersection property. Then P, ordered by class inclusion, contains a maximal element W.

Recall (see Chapter 11) that a class 0.4 = (A i : i E I} property if every finite subclass of 04, say (A,,, i.e. A i, n • • • n A i. Q.

possesses the finite intersection has a non-empty intersection,

METRIC PRODUCT SPACES Let ( (X1, di)) be a collection of metric spaces and let X denote the product of the sets Xi, i.e. X •=-- fi. X. Since the metric spaces (Xi, di) are also topological spaces, we can speak of the product topology on X. On the other hand, it is natural to ask whether or not it is possible to define a metric d on the product set X so that the topology on X induced by the metric d is identical to this product topology. The next two propositions give a positive answer to this question in the case of a finite or a denumerable collection of metric spaces. The metrics given are in no way unique. .

Proposition 12.11:

Let (Xi, d1), ..

(X., d.) be metric spaces and let p = (ai, ..., a.) and

n trl

q = (b1 ,

, bin) be arbitrary points in the product set

X --=

X.

Then each of the following functions is a metric on the product set X: di(ai, b 1)2 + • • • + dm(a., b.)2

d(p ,

max tdi(a i, b 1

=

d(p, q)

),

.

.

.

,

dm (am , b ni)}

b1) + • • • + dr.(a.„ bm)

d(p , q) -=

Moreover, the topology on X induced by each of the above metrics is the product topology. Proposition 12.12:

Let ((Xi, d1), (X2, d2), and let p = (a1, a2, product set X =

) be a denumerable collection of metric spaces and q = (b1, b2, .) be arbitrary points in the

TT X.

Then the function d defined by

d(p, q) =

dn(an, b„) 1 n-1 2n 1 + dn(an, b.)

is a metric on the product set X and the topology induced by d is the product topology. CANTOR SET We now construct a set T of real numbers, called the Cantor or ternary set, which has some remarkable properties. Trisect the closed unit interval I = [0, 1] at the points and * and then delete the open interval (i, I), called the "middle third". Let T1 denote the remainder of the points in I, i.e., T1 = [0,

u

[f, 1]

172

PRODUCT SPACES

[CHAP. 12

We now trisect each of the two segments in T 1 at and 4 and and and then delete the "middle third" from each segment, i.e. ( , .4) and G, Let T, denote the remainder of the points in T1, i.e., T2 -= [0,

kl

'I

U

U

41 ' U

1]

If we continue in this manner we obtain a descending sequence of sets

T i D T2 D T3 D • • • where Tm consists of the points in Tm-1 excluding the "middle thirds", as shown. I • 0

1

• 0 T2

• 0

T3 •

o

1

2

1 3

2

3

s

111,

• • •—•

2

3

1

•

1

110-• 4

• • •

111

Observe that I'm consists of 2m disjoint closed intervals and, if we number them consecutively from left to right, we can speak of the odd or even intervals in T m . The Cantor set T is the intersection of these sets, i.e.

T = n {Ti : i E NI.

PROPERTIES OF THE CANTOR SET We define a function f on the Cantor set T as follows:

f(x)

=

(a1, a2, ...)

where am

if x belongs to an odd interval in T.

12

if x belongs to an even interval in T m

The above sequence corresponds precisely to the "decimal expansion" of x written to the base 3, i.e. where

1 ai ( -3 ) + a2(-91)

a3 ( 0 1

Consider now a discrete space of two elements, say A copy of A indexed by i e N, the positive integers.

Proposition 12.13:

a.

{0, 2), and let A n denote a

The Cantor set T is homeomorphic to the product space

X --=

i E NI

In particular, the function f:T-->X defined above is a homeomorphism.

Remark:

The Cantor set T possesses the following interesting properties:

(1)

T is non-denumerable. For T is equivalent to the set of sequences (a 1 , a2, ), where a = 0 or 2, which has cardinality 2 14 = c.

(2)

T has "measure" zero. For the measure of the complement of T relative to I = f0, 11, i.e. the union of the middle thirds, equals 8 2 4 1 1 3 + 9 ± 27 + 81 ± • • • But the measure of I = [0,1] is also 1. Hence the measure of T is zero.

CHAP. 12 .]

173

PRODUCT SPACES

Solved Problems PRODUCT SPACES 1.

Consider the topology T = {X, 0, {a}, {b, c}} on T* = {Y, y), {u}} on Y = fu,v1.

X = {a, b, c}

and the topology

(1) Determine the defining subbase ci of the product topology on X x Y. (ii) Determine the defining base '13 for the product topology on X x Y. Solution: X x Y

Note first that

= {(a, u), (a, v), (b, u), (b, y), (c, u), (e, y)}

is the product set on which the product topology is defined.

(i)

The defining subbase GI is the class of inverse sets 11-» [G] and 7,-,T 1 [II] where G is an open subset of X and H is an open subset of Y. Computing, we have = 7 -1 [Y] ---=--

v X-1 [X1 [ (Al

7r — 1

X

7T

=

[{a,} ]

((b, u), (b, y), (c, u), (e, y)}

[{b, e}]

X

— 1 [fun

Y

w' Y— [Q] = 0 {(a, u), (a, v)} 7T

L '-.j

X

XxY

Y

=

{(a, u), (b, u), (e,

2

)}

Hence the defining subbase j consists of the subsets of X x Y above.

(ii)

The defining base '13 consists of finite intersections of members of the defining subbase J. TLat is, '3 = (X X Y, 0, {(a, u)}, {(b, u), (e, u)}, {(a, u), v)}, {(b, u), (b, y), (e, u), (c, v)}, {(a, u), (b, u), (e, u)}}

2.

Prove Theorem 12.5: A function f: Y --) X from a topological space Y into a product space X = fl X, is continuous if and only if, for every projection 7ri : X --> X i, the composition 7r,o f : Y —> X. is continuous. Solution: By definition of the product space, all projections are continuous. So if f is continuous, then 7 1 f, the composition of two continuous functions, is also continuous. On the other hand, suppose every composition function an open subset of X. Then, by continuity of 7 7 o f,

( r o f) is an open set in Y.

f

1 [G]

Let G be

[G]l

But the class of sets of the form 7T

where G is an open subset of X,

1 [G]

is the defining subbase for the product topology on X. of Y, f is a continuous function by Theorem 7.2.

3.

1 FT

vi of: Y —> X 1 is continuous.

Since their inverses under f are. open subsets

Let B be a member of the defining base for a product space X =fJ. X. the projection of B into any coordinate space is open. Solution: Since B belongs to the defining base for X,

B

=

fI Xi {

where G3k is an open subset of Xi k .

, . . . ,

jn,} x Gil x • • • x

So, for any projection

7T a ( B )

In either case, v a (B) is an open set.

:

--=

xa 1G a

7T a :

X —> X 0 ,

if a if a G

fin}

Show that

174

4.

PRODUCT SPACES

(CHAP. 12

Prove Theorem 12.6: Every projection 7r : X —> X i on a product space X = both open and continuous, i.e. bicontinuous.

ri, xi is

Solution: By definition of the product space, all projections are continuous. So we need only show that they are open. Let G be an open subset of the product space X = fi. X. For every point p G G there is a member B of the defining base of the product topology such that pEBcG. Thus, for any projection 7 1 : X —> Xi, w1 (p) G r 1 [B1 C w i [G1 By the preceding problem, 7r 1 [B] is an open set. In other words, every point 7 7- 1 (p) in v- E[G] belongs to an open set w l [B1 which is contained in v i [G]. Thus v- i [G] is an open set.

5.

Prove Theorem 12.7: A sequence pi,p2, ... of points in a product space X converges to the point q E X if and only if, for every projection 771 : X ---> X o the sequence 77-0 1 ), 7,(P 2), • • . converges to r1(q) in the coordinate space X. Solution: Suppose

pn q.

Then, since all projections are continuous,

77-1 (p n) —>

Conversely, suppose 7 1 (p) —> r 1 (q) for every projection 77i. In order to prove that p n —> q, sufficient to show that if B is a member of the defining base of the product space X = FL x, contains the point q G X, then 3 no G N

such that n> n o

that

pn G B

By definition of the defining base for the product space B =

it is

X = Ij i

n ••• n

where Gik is an open subset of the coordinate space Xj.k . Recall that q E B; hence7T 1 (q) E v-ji [B] = rim(q) G rim[B] =Gjm 3 n E N

Let v o

By hypothesis, r il(pn ) —> z- ji (q). Hence, for each i = 1, ..., m, EG .

such that n > ni

p G 77- 3:: 1 [Gii]

max (n 1 , ..., n,„). Then n> no

p n G 7 1 [Gi 1

1

n

n

77-:-1 [G• 7m

3 M.

= B

Consequently, p —> q.

TYCHONOFF THEOREM 6.

Prove Lemma 12.10: Let 34 be a class of subsets of a set X with the finite intersection property. Consider the collection P of all superclasses of 04 each with the finite intersection property. Then P, ordered by class inclusion, contains a maximal element t. Solution: Let T = {13 1} he a totally ordered subcollection of p, and let '13 = U . We show that 'B belongs to p, i.e. that '13 is a superclass of a with the finite intersection property. It will then follow that 93 is an upper bound for T and so, by Zorn's Lemma, p contains a maximal element M. Clearly '13 = u 93 ; is a superclass of a since each B is a superclass of has the finite intersection property, let {A 1 , ..., A m } be any finite subclass of hence such that A 1 ..., A m G m 3 'B y G T

a.

To show that '13 93 = U i 'B i ;

'B. But

Recall that T is totally ordered; hence one of the classes, say Bijo, contains all the sets A 1 and, furthermore, since it has the finite intersection property, AI

n

A2

n ••• n

Am

0

A n,} of B has a non-empty intersection, i.e. We have just shown that each finite subclass {A 1 , B has the finite intersection property. Consequently, B belongs to p.

CHAP. 12]

7.

PRODUCT SPACES

175

Prove: The maximal element ,91 in Lemma 12.10 possesses the following properties: (i) Every superset of a member of ,r,v also belongs to (ii) The intersection of a finite number of members of ni also belongs to M. (iii) If AnMO, for every M E TW , then A belongs to J. Solution: We only prove (ii) here.

The proofs of (i) and (iii) will be left as supplementary problems.

We prove that the intersection of any two sets A, B G w also belongs to w. The theorem will then follow by induction. Let C = A n B. If we show that ,A1 U {C} has the finite intersection property, then ,w u {C} will belong to p and, since ,14- is maximal in p, 21 = M u {C}. Thus C will belong to -11, as was to be proven.

(ii)

Let

{A i , A2, ..., A ,„} be a finite subclass of

g , 7vi u {C}.

There are two cases:

Case I.

C does not belong to {A 1 , A 77,} c ,9itu {C}. Then {A 1 , of ..1/ alone. But ,M has the finite intersection property; hence A 1 n

Case II.

C does belong to {A l, ...,A,}, say C = A l .

A, B,

O.

Then

A l n•—nA m = CnA 2 n•• • nA m = since

A.} is a finite subclass n

4 nBnA 2 n•••nA m

E M.

A2, ..

In either case, {A1, ..., A ni l has a non-empty intersection. So ,711 u {C} E stated above, C G

8.

p and, for reasons

Prove Theorem (Tychonoff) 12.9: Let (A i : i E I} be a collection of compact topological tAi: i E II is also compact. spaces. Then the product space X = Solution: Let c.,4 = {F i } be a class of closed subsets of X with the finite intersection property. theorem is proven if we show that cA has a non-empty intersection, i.e., 3 pGX

such that

The

p E F, for every F E ve1

Let ui= {M k : k G K} be a maximal superclass of c,4 with the finite intersection property (see Lemma 12.10). Define ,i1= {M k : k G K}. Observe that

Fi E c/I

F 5 = Pi

F5 G cil > F5 G

and

So if we prove that 297 has a non-empty intersection, then c. 4 will also have a non-empty intersection. In other words, the proof is complete if 3 pGX

such that p E/il k for every k E K

or, equivalently, pGX

BnM

such that p E B

Q for every k E K

(1)

where B is a member of the defining base for the product topology on X = fl 1 A 1, since p is then an accumulation point of each of the sets Mk and so is contained in Mk. Recall that 271/ = {M k : k G K } has the finite intersection property; 71 : X—> A i, the class i [Mk ] k G KI

so, for each projection

of subsets of the coordinate space A i also has the finite intersection property. Hence the class of closures {Tri(Mk k E K1 is a class of closed subsets of A i with the finite intersection property. By hypothesis, A i is compact; so {77- i [Mk ] k G KI has a non-empty intersection, i.e., 3 a l G A i such that

ai G iri [lik] for every k G K

or, equivalently, 3 ai E Ai

such that

a i G Gi

Gi n 5Ti [Mk]

where G i is any open subset of the coordinate space Ai.

for every k G K

(2)

176

(CHAP. 12

PRODUCT SPACES

Define

p

We want to show that p satisfies the condition (1). Let p G B, where X = fL 21 1 , i.e.,

(ai : i G I).

B is a member of the defining base for the product topology on B

=

7r

n '1 [G 11 '

where Gia is an open subset of Aia . Observe that p G B

(p)

a il belongs to irii [B] = Gil .

So, by (2) above,

for every k G K

Gi l n vi i (Mk1 which implies that Tr:- [G1 1 n

= (11 (Ai : i

1-

x Gi 1) n

for every Mk G

Mk

By the property (iii) of ,9t, stated in the preceding problem, Tri—i (G il l belongs to

[G 19

]

2

.

.

ri

2

also belongs to nz. But gu satisfies the finite intersection property; so

tit

i2

B n

z. Similarly,

Mk =

0

n • •• n 7T im 1 [G1] m n Mk

I

for every k G K

Thus (1) is satisfied and the theorem is proven.

CANTOR SET

9. Show that the Cantor set T is a closed subset of R. Solution: Recall that T„, is the union of 20' closed intervals. So Tm , the union of a finite number of closed sets, is also closed. But T = n {Ti N}; hence T is closed, as it is the intersection of closed sets.

10. Show that T is compact. Solution: Since T is a closed and bounded set of real numbers, it is compact.

11. Let X = ri X is compact.

{A1: j G N} where Ai = {0,2) with

Solution: Observe that A i is compact since it is finite. compact.

the discrete topology.

So, by the Tychonoff Product Theorem,

Show that X = fi 1 A i isalo

12. Let X = 11{A 1 : i G NI where A i = {0, 2) with the discrete topology. (i) Prove that the function f: X ---> T defined by a2 , ...))

=

=ai(i) 1

ai() + a2 ( ) + a3(47) + • •

is continuous.

(ii) Prove that X is homeomorphic to T. Solution: (i) Let p = (a1, 11.2, • • . G X containing p such that

and let e

> O. We need to

x G B Note that

1, (i)i converges.

implies

show that there is an open subset B of X

(x) — f (P)1 <

e

Hence 00

3 no G N

such that

<e

(

no + 1

Consider the subset

B =

x

{a 2 } x • • • x {ano}

x

A n0+1

x

A no+2

x • • •

of X. Observe that p E B and B is a member of the defining base of the product topology on X = 11A 1 and so is open. Furthermore,

CHAP. 121

177

PRODUCT SPACES

x implies

11(x)

f(P)I

...,

(b,.

lc*

=

0,

i =n0 +1

0

+1, bn0+2, ...) G B

a i)("D i

<

i=n 0 +1

e

Thus f is continuous. (ii)

The function f:X-+T is a one-one continuous function from the compact space X onto the metric space T. By Theorem 11.8, f is a homeomorphism.

Supplementary Problems PRODUCT SPACES 13. Show that the property of being a T1 -space is product invariant, i.e. the product of T 1 -spaces is a T r-space. 14.

Show that the property of being a regular space is product invariant.

15. Show that—the property of being a completely regular space is product invariant. 16.

Prove: Let p = (ai : i G I) be any point in a product space

X 513 x

{ai :

j0)

{X:

X=

is homeomorphic

I).

Then, for any

to

(In the special case of Euclidean 3-space R 3 , this theorem states that the line, say Y =

{a i } X {a 2} X R3

through p = (a 1 , a2, a 3), is homeomorphic to R.)

{(a 1 , a 2, x): x G

See Fig. (a).

R2

Fig. (b)

Fig. (a) 17.

Prove: Let p (ai : i G I) be any point in a product space X = {ai } x

:i

Jo}

is homeomorphic to

1-1

: j G I). 11

i

Then, for any

Jo E I,

jo l

(In the special case of Euclidean'3-space R3, thiS theorem states that the plane, say Y = R1 X R2 X {a3 } through p =

{(x, y, a 3) : x,y C R)

a2, a3), is- homeomorphic to R2 =-, R x R.)

See Fig. (b).

Ft xi

18.

Prove the converse of the Tychonoff Product Theorem, i.e. if a Product space X = then each coordinate space Xi is also compact.

19.

Let A be a subset of a product space X -= TI { X,: i G II and let ir • A —> X i denote the restriction • of the projection 7 1 : X —> X i to A. Show that the relative topology on A is the coarsest topology with respect to which the functions 7T iA are continuous.

is compact,

178

PRODUCT SPACES

20. (i) (ii) 21.

(CHAP. 12

Prove that a countable product of first countable spaces is first countable. Show that an arbitrary product of first countable spaces need not be first countable.

Show that an uncountable product space X = 11 1 2( i is not metrizable (unless all except a countable number of coordinate spaces are singleton sets).

22. (i) (ii)

Prove that a countable product of second countable spaces is second countable. Show that an arbitrary product of second countable spaces need not be second countable.

23.

Let A, be an arbitrary subset of a topological space X,; thus fJ A, is a subset of the product space X -=fJ 1 X 1 . Prove that (i) (ii) EL A ,c.) D (1-1 1 A L)°. Give an example to show that equality does not hold for (ii) in general.

24.

Let A i be an arbitrary subspace of X,. Show that the product topology on 11 1 A t is equal to the relative topology on A, as a subset of the product space fJ1X1.

a

a

ARBITRARY TOPOLOGIES ON PRODUCT SETS Prove Proposition 12.8:

25.

{(X,, T,): G II be a collection of topological spaces and let X be the product of the sets X i , i.e. X ---- fi. X i . Then the subsets of X of the form 1-1 {G, : i E / } where Gi is an open subset of the coordinate space Xi, form a base for a topology T on the product set X.

26.

Show that the product topology on a product set X = JJ X i is coarser than the topology T on X defined in the preceding problem (Proposition 12.8).

27.

Give an example of a topology T on a product set topology on X.

28.

Let T be the topology on a product set X = fi x, defined in Problem 25 (Proposition 12.8). that (X, T) is discrete if each coordinate space X, is discrete.

Let

X -=fi X,

which is coarser than the product

Show

FINITE PRODUCTS 29.

Prove Proposition 12.2: The subsets of a product space X = X1 X • • • X X,„ of the form G1 X • • • X G„,„ where G, is an open subset of X i, form a base for the product topology on X.

30.

Prove: If B is a base for X and 93* is a base for Y, then the product space X X Y.

31.

Prove: If B a is a local base at a, G X and Bb is a local base at bE Y, then is a local base at p = (a, b) e X X Y.

32.

Prove that the product of two first countable spaces is first countable.

33.

Prove that the product of two second countable spaces is second countable.

{G XH :GEB, HEB *}

{GXH: G

is a base for

H

'BO

34. Prove that the product of two separable spaces is separable. 35.

Prove that the product of two compact spaces is compact (without using Zorn's Lemma or its equivalents).

36.

Let B . ' be the topology on the plane R2 generated by the half-open rectangles

[ct, b) X [e, d)

{(x,y) : a.

x < b, c

y<

Furthermore, let T be the topology on the real line R generated by the closed-open intervals [a , b). Show that (R2 , T'') is the product of (R, T) with itself. .

CHAP. 121

179

PRODUCT SPACES

37.

Show, by a counter-example, that the product of two normal spaces need not be normal.

38.

Let AcX

and

BcY

and hence

(i)

AX/3 =

AXB cXXY. A XB

(ii)

Prove that

A° >< B° = (AXB)°

(Recall [see Problem 23 ] that equality does not hold in general.)

39.

40.

Let f: X -> Y and let F : X -> X X Y be defined by F(x) = (x, f(x)). Prove that f is continuous if and only if F is a homeomorphism of X with F[X] . (Recall that F[X] is called the graph of f.)

Let X be a normed vector space over = p + q is continuous.

R.

Show that the function

f:X 1< X->X

defined by

f:R X X- X

defined by

f((P, 41.

Let X be a normed vector space over R. Show that f ((k, I))) = kp is continuous.

the function

METRIC PRODUCT SPACES Every closed and bounded subset of Euclidean m-space R M is compact.

42.

Prove:

43.

Prove Proposition 12.11:

Let

(X 1 , d 1 ), .. . ,

dm )

be metric spaces and let

p = (a i , . . . , a„.,) and

HI

, bm ) be arbitrary points in the product set q = (b1 , functions is a metric on X: (i)

d(p, q)

=

Vd i (ai , b 1 ) 2 + - - • + dm (am , bm ) 2

(ii)

d(p, q)

=

max {d i (ai , b 1 ), . . . , d.„(a.„, 1).)1

(iii)

d(p, q)

=

d i (ai , b 1 ) + • • • + dm (am , bai)

X

X.

Then each of the following

Moreover, the topology on X induced by each of the above metrics is the product topology.

44.

Prove Proposition 12.12:

Let

{(X 1 , d 1 ), (X2, d2 ), . . . I

be a denumerable collection of metric spaces

and q = (1)1 , b2, .) be arbitrary points in the product set and let p = (a i , a2, function d defined by Then the

d(P, q) =

n =1

dn (a., bn ) 2n 1 + dn(an , bn )

is a metric on X and the topology induced by d is the product topology.

X=

11 X.

n1

Chapter

13

Connectedness SEPARATED SETS Two subsets A and B of a topological space X are said to be separated if (i) A and B are disjoint, and (ii) neither contains an accumulation point of the other. In other words, A and B are separated if Ani3 = 9) and An B Ø Example 1.1:

Consider the following intervals on the real line R:

A = (0,1), B = (1,2) and

C = [2,3)

/3 = [1,2 ] , and so A n

and

are empty. On the other hand, B and C are not separated since is a limit point of B; thus:

2G C

Now A and B are separated since

A = [0,1]

and

A nB

n C Example 1.2:

[1,

n [2, 3) = {2}

9)

Consider the following subsets of the plane R2:

A

B

= {(x,y) : y = sin (1/x), 0 < x

1}

1 1

A

-1Now each point in A is an accumulation point of B; hence A and B are not separated sets.

CONNECTED SETS Definition:i

A subset A of a topological space X is disconnected if there exist open subsets G and H of X such that A n G and AnH are disjoint non-empty sets whose union is A. In this case, G U H is called a disconnection of A. A set is connected if it is not disconnected.

Observe that

A = (AnG)u(AnH)

iff

AcGuH

= (AnG)n(AnH) if GnHcAc

and

Therefore G U H is a disconnection of A if and only if

AnG 0, AnH 0, AcGuH, Note that the empty set

V)

and

GnHc Ac

and singleton sets fp) are always connected.

180

CHAP. 131

181

CONNECTEDNESS

Example 21:

The following subset of the plane R2 is disconnected:

A

= {(x, y) : x2 — y2 4}

For the two open half-planes

G = {(x, y) : x < —1}

and

H = {(x, y) : x > 1}

form a disconnection of A as indicated in the diagram above. Example 2.2:

Consider the following topology on X = {a, h, c, d, e}:

T =

{X, 0, {a, b,

{c, d, e}, {c}} Now A = {a, d, el is disconnected. For let G = {a, b, c} and H = {c, d, e}; then A n G = {a} and A n H = {d, el are non-empty disjoint sets whose union is A. (Observe that G and H are not disjoint.)

The basic relationship between connectedness and separation follows: Theorem 13.1: A set is connected if and only if it is not the union of two non-empty separated sets. The following proposition is very useful. Proposition 13.2: If A and B are connected sets which are not separated, then A U B is connected. Example 2.3: Let A and B be the subsets of the plane R2 defined and illustrated in Example 1.2. We show later that A and B are each connected. But A and B are not separated; hence, by the previous proposition, A u B is a connected set.

CONNECTED SPACES Connectedness, like compactness, is an absolute property of a set; namely, Theorem 13.3: Let A be a subset of a topological space (X, T). Then A is connected with respect to T if and only if A is connected with respect to the relative topology TA on A. Accordingly, we can frequently limit our investigation of connectedness to those topological spaces which are themselves connected, i.e. to connected spaces. Example 3.1:

Let X be a topological space which is disconnected, and let G u H be a disconnection of X; then

X = (X n G) u (X n H)

and

(X n G) n (X n H) = 0

But XnG =G and X n H H; thus X is disconnected if and only if there exist non-empty open sets G and H such that

X = GuH

and

GnH=

In view of the discussion in the above example, we can give a simple characterization of connected spaces.

Theorem 13.4:

A topological space X is connected if and only if (i) X is not the union of two non-empty disjoint open sets; or, equivalently, (ii) X and Vi are the only subsets of X which are both open and closed.

182

[CHAP. 13

CONNECTEDNESS

Example 3.2:

Consider the following topology on X = {a, b, c, cl, e}:

T

=

{X, 0, {a} , {c, d} , {a, c, cl } , {b, c, cl, e}}

Now X is disconnected; for {a} and {b, c, cl, el are complements and hence both open and closed. In other words,

X = {a} u {b, c, d, el is a disconnection of X. Observe that the relative topology on the subset A = {b, d, el is {A, 0, {d}}. Accordingly, A is connected since A and 0 are the only subsets of A both open and closed in the relative topology. Example 3.3:

The real line R with the usual topology is a connected space since R and 0 are the only subsets of R which are both open and closed.

Example 3.4:

Let f be a continuous function from a connected space X into a topological space Y. Thus f: X —> f[X] is continuous (where f[X] has the relative topology). We show that f[X1 is connected. Suppose f[X] is disconnected; say G and H form a disconnection of f[Xl. Then

_V] = G u H and so

X = f — G] u f -1 [H]

and and

GnH= f —1 [G]

n f'

Since f is continuous, f -1 [G] and f — [H] are open subsets of X and hence form a disconnection of X, which is impossible. Thus if X is connected, so is f[X].

We state the result of the preceding example as a theorem.

Theorem 13.5:

Continuous images of connected sets are connected.

Example 3.5:

G u H is a disconnection of X.

Let X be a disconnected space; say, function

f(x)

0 1

if x G G if x

H

Then the

is a continuous function from X onto the dis-

crete space Y =

On the other hand, by Theorem 13.5, a continuous image of a connected space X cannot be the disconnected discrete space Y = {O, 1} . In other words,

Lemma 13.6:

A topological space X is connected if and only if the only continuous functions from X into Y = {0,1) are the constant functions, f(x) = 0 or f(x) = 1.

CONNECTEDNESS ON THE REAL LINE The connected sets of real numbers can be simply described as follows: Theorem 13.7:

A subset E of the real line R containing at least two points is connected if and only if E is an interval.

Recall that the intervals on the real line R are of the following form: (a, b), (a, b1, [a, b), [a, b],

finite intervals

(—.0, a), (-00, a], (a, .0), [a, .0), (-00, .0),

infinite intervals

An interval E can be characterized by the following property: a,beE,a<x
Theorem 13.8:

Let f: X ---> R be a real continuous function defined on a connected set X. Then f assumes as a value each number between any two of its values.

CHAP. 131

183

CONNECTEDNESS

Example 4.1:

An interesting application of the theory of connectedness is the following "fixedpoint theorem": Let I = (0, 11 and let f:I ->I be continuous; then 3p G I such that f (p) = p. This theorem can be interpreted geometrically.

Note first that the graph of

f:I ->I lies in the unit square 12 r= {(X, y):0x= 1, 0 -=y -= 1} The theorem then states that the graph of f, which connects a point on the left edge of the square to a point on the right edge of the square, must intersect the diagonal line A at, say, (p, p) as indicated in the diagram.

COMPONENTS A component E of a topological space X is a maximal connected subset of X; that is E is connected and E is not a proper subset of any connected subset of X. Clearly E is non-empty. The central facts about the components of a space are contained in the following theorem. Theorem 13.9: The components of a topological space X form a partition of X, i.e. they are disjoint and their union is X. Every connected subset of X is contained in some component. Thus each point p

e X belongs to a unique component of X, called the

Example 5.1:

If X is connected, then X has only one component X-itself.

Example 5.2:

Consider the following topology on X = { a, b, c, d, e}:

T =

component of p.

(X, 0, {a}, {c, d}, {a, e, d } , {b, e, d, e}}

The components of X are {a } and { b, c, d, e}. Any other connected subset of X, such as { b, d, el (see Example 3.2), is a subset of one of the components.

The statement in Example 5.1 is used to prove that connectedness is product invariant; that is,

Theorem 13.10:

The product of connected spaces is connected.

Corollary 13.11:

Euclidean m-space R' is connected.

LOCALLY CONNECTED SPACES A topological space X is locally connected at p e X if every open set containing p contains a connected open set containing p, i.e. if the open connected sets containing p form a local base at p. X is said to be locally connected if it is locally connected at each of its points or, equivalently, if the open connected subsets of X form a base for X. Example 6.1:

Every discrete space X is locally connected. For if p G X, then {p } is an open connected set containing p which is contained in every open set containing p. Note that X is not connected if X contains more than one point.

184

[CHAP. 13

CONNECTEDNESS

Example 6.2:

Let A and B be the subsets of the plane R2 of Example 1.2. Now A u B is a connected set. But A u B is not locally connected at p = (0,1). For example, the open disc with center p and radius I does not contain any connected neighborhood of p.

1

—1-

PATHS Let I = [0,1], the closed unit interval. A path from a point a to a point b in a topological space X is a continuous function f:I--->X with f(0) = a and f(1) = b. Here a is called the initial point and b is called the terminal point of the path. Example 7.1:

For any p G X, the constant function ep :I—>X defined by el,(s) = p is continuous and hence a path, It is called the constant path at p.

Example 7.2:

Let f: I —> X be a path from a to b.

Then the function

Î:

X defined by

(s) = f(1— s) is a path from b to a. Example 7.3:

Let f: I —> X be a path from a to b and let g: I —> X be a path from b to c. Then the juxtaposition of the two paths f and g, denoted by f * g, is the function f *g:I —>X defined by f (2s) if 0 s

(f * g)(s)

f

g(2s — 1)

- if

s

1

which is a path from a to c obtained by following the path f from a to b and then following g from b to c.

ARCWISE CONNECTED SETS A subset E of a topological space X is said to be arcwise connected if for any two points a, b G E there is a path f: I ---> X from a to b which is cOntained in E, i.e. f[i] C E. The maximal arcwise connected subsets of X, called arcwise connected components, form a partition of X. The relationship between connectedness and arcwise connectedness follows: Theorem 13.12: Arcwise connected sets are connected. The converse of this theorem is not true, as seen in the next example. Example 8.1:

Consider the following subsets of the plane R2:

A

= {(x, y): 0 x

B = {(x,0) :

x

1, y = xln, n E N} 1}

Here A consists of the points on the line segments joining the origin (0,0) to the points (1,1/n), n E N; and B consists of the points on the x-axis between -1 and 1. Now A and B are both arcwise connected, hence also connected. Furthermore, A and B are not separated since each p G B is a limit point of A; and so A u B is connected. But A u B is not arcwise connected; in fact, there exists no path from any point in A to any point in B. Example 8.2:

Let A and B be the subsets of the plane R2 defined in Example 1.2. Now A and B are continuous images of intervals and are therefore connected. Moreover, A and B are not separated sets and so A u B is connected. But A u B is not arcwise connected; in fact, there exists no path from a point in A to a point in B.

CHAP. 131

185

CONNECTEDNESS

The topology of the plane R2 is an essential part of the theory of functions of a complex variable. In this case, a region is defined as an open connected subset of the plane. The following theorem plays an important role in this theory.

Theorem 13.13:

An open connected subset of the plane R2 is arcwise connected.

HOMOTOPIC PATHS Let f:I---> X and g:I-> X be two paths with the same initial point p e X and the same terminal point q e X. Then f is said to be homotopic to g, written f g, if there exists a continuous function H:/2 -->X such that H(0,t) =-- p H(s, f(s) H(1,t) = H(s,1) = g(s) as indicated in the adjacent diagram. We then say that f can be continuously deformed into g. The function H is called a homotopy from f to g. Example 9.1:

Let X be the set of points between two concentric circles (called an annulus). Then the paths f and g in the diagram on the left below are homotopic, whereas the paths f and g' in the diagram on the right below are not homotopic.

Example 9.2:

Let f: I —> X be any path. Then f f, function H: 12 ---> X defined by

i.e. f is homotopic to itself.

For the

H(s, t) = f(s) is a homotopy from f to f. Example 9.3:

Let f = g and, say, H: 12 —> X is a homotopy from f to g. defined by r/(s, t) H(s, 1 — t)

Then the function

PI: 12 —> X

is a homotopy from g to f, and so g Example 9.4:

f.

Let f=g and g=h; say, F:P—>X is a homotopy from f to g and is a homotopy from g to h. The function H: 12 —> X defined by

H(8, t)

{F(s, 2t) G(s, 2t-1)

G:P—>X

if Ot if 1 2

t

1

is a homotopy from f to h, and so f = h. The homotopy H can be interpreted geometrically as compressing the domains of F and G into one square. h domain of G --0-

h 3

domain of F

rum

domain of H

[CHAP. 13

CONNECTEDNESS

186

The previous three relations imply the following proposition: Proposition 13.14: The homotopy relation is an equivalence relation in the collection of all paths from a to b. SIMPLY CONNECTED SPACES A path f:I-->X with the same initial and terminal point, say f(0) = f(1) = p, is called a closed path at p X. In particular, the constant path ep: I --> X defined by e n(s) = p is a closed path at p. A closed path f: I X is said to be contractable to a point if it is homotopic to the constant path.

A topological space is simply connected if every closed path in X is contractable to a point. Example 10.1:

An open disc in the plane R2 is simply connected, whereas an annulus is not simply connected since there are closed curves, as indicated in the diagram, that are not contractable to a point.

simply connected

not simply connected

Solved Problems SEPARATED SETS 1. Show that if A and B are non-empty separated sets, then A u B is disconnected. Solution: Since A and B are separated,

G and H are open and

An- 0

and

(A U B) n G = A

are non-empty disjoint sets whose union is A U and so A Li B is disconnected.

and

n B = 0.

Let

G = PC and H

AC.

Then

(A U B) n H = B

B. Thus G and H form a disconnection of A U B,

CHAP. 13]

2.

187

CONNECTEDNESS

Let GUH be a disconnection of A.

Show that A

n

G and A

n

H are separated sets.

Solution: Now A n G and A n H are disjoint; hence we need only show that each set contains no accumulation point of the other, Let p be an accumulation point of A n G, and suppose pGAn H. Then H is an open set containing p and so H contains a point of A n G distinct from p, i.e. (A n G) n H 0. But

(A

n G) n (A n H) _=

= (A n G) n H

Accordingly, p A n H. Similarly, if p is an accumulation point of A are separated sets.

3.

n H,

then p

A n G.

Thus A n G and A

nH

Prove Theorem 13.1: A set A is connected if and only if A is not the union of two non-empty separated sets. Solution: We show, equivalently, that A is disconnected if and only if A is the union of two non-empty separated sets. Suppose A is disconnected, and let G u H be a disconnection of A. Then A is the union of non-empty sets A n G and A n H which are, by the preceding problem, separated. On the other hand, if A is the union of two non-empty separated sets, then A is disconnected by Problem 1.

CONNECTED SETS 4.

Let G U H be a disconnection of A and let B be a connected subset of A. Show that either BnH = 9) or BnG = 9), and so either B c G or B c H. Solution: Now B c A, and so

AcGuH

BcGuH

and

GnHcBc

GnHcAc

Thus if both B n G and B n H are non-empty, then G u H forms a disconnection of B. connected; hence the conclusion follows.

5.

But B is

Prove Proposition 13.2: If A and B are connected sets which are not separated, then A U B is connected. Solution: Suppose A u B is disconnected and suppose G U H is a disconnection of A u B. Since A is a connected subset of A u B, either A c G or A c H by the preceding problem. Similarly, either B c G or B c H. Now if A c G and B c H (or B c G and A c H), then, by Problem 2,

(A u B) n G = A

and

(A u B) n H

B

are separated sets. But this contradicts the hypothesis; hence either AuBcG or and so G u H is not a disconnection of A u B. In other words, A u B is connected.

6.

AUB'C H,

Prove: Let 0.4 = {Ai } be a class of connected subsets of X such that no two members of c.,/ are separated. Then B = U i A i is connected. Solution: Suppose B is not connected and G u H is a disconnection of B. Now each A i G c4 is connected and so (Problem 4) is contained in either G or H and disjoint from the other. Furthermore, any two members A il , A 2 G cA are not separated and so, by Proposition 13.2, A il u A i2 is connected; then A A contained in G or H and disjoint from the other. Accordingly, all the members of 0. 4 , and hence B = u j A i, must be contained in either G or H and disjoint from the other. But this contradicts the fact that G u H is a disconnection of B; hence B is connected.

188

7.

[CHAP. 13

CONNECTEDNESS

Prove: Let a = {Ai} be a class of connected subsets of X with a non-empty intersection. Then B = UA i is connected. Solution: Since n i Ai 0, any two members of e4 are not disjoint and so are not separated; hence, by the preceding problem, B = u i A t is connected.

8.

Let A be a connected subset of X and let ACBC A. hence, in particular, A is connected.

Show that B is connected and

Solution: Suppose B is disconnected and suppose G uH is a disconnection of B. Now A is a connected subset of B and so, by Problem 4, either A n H = or AnG = 0; say, AnH = 0. Then He is a closed superset of A and therefore AcBcAcHc. Consequently, BnH= Ø. But this contradicts the fact that G u H is a disconnection of B; hence B is connected.

CONNECTED SPACES 9. Let X be a topological space. (i) X is disconnected.

Show that the following conditions are equivalent:

(ii) There exists a non-empty proper subset of X which is both open and closed. Solution:

(i)

(ii): Suppose X=-GuH where G and H are non-empty and open. proper subset of X and, since G = He, G is both open and closed.

(ii)

(i): Suppose A is a non-empty proper subset of X which is both open and closed. Then Ac is also non-empty and open, and X = AuAc. Accordingly, X is disconnected.

Then G is a non-empty

10. Prove Theorem 13.3: Let A be a subset of a topological space (X, T) and let TA be the relative topology on A. Then A is T-connected if and only if A is TA -connected. Solution: Suppose A is disconnected with G U H forming a T-disconnection of A. Now G,H G T and so AnG,AnH e TA. Accordingly, AnG and AnH form a TA-disconnection of A; hence A is TA -disconnected. On the other hand, suppose A is TA-disconnected, say G* and H* form a TA-disconnection of A. Then G*, H' E TA and so

3 G, H But

T

AnG* AnAnG

such that

--= AnG

H* -- AnH

G*

AnG

and

AnH* = AnAnH=r-AnH

and

Hence G u H is a T-disconnection of A and so A is T-disconnected.

of X are said to form a simple (finite) chain 11. Let p, q e X. The subsets A1, p, Am (and only Am) contains q, and A 1 (and only contains A1) joining p to q if > 1. A nA = Q if Prove: Let X be connected and let a be an open cover of X. Then any pair of points in X can be joined by a simple chain consisting of members of c.f. Solution: Let p be any arbitrary point in X and let H consist of those points in X which can be joined to p by some simple chain consisting of members of efl. Now since p e H. We claim that H is both open and closed and so H = X since X is connected.

CHAP. 13]

CONNECTEDNESS

189

Let h G H. Then 3 G I , ..., G E c.,/ which form a simple chain from h to p. But if x E G I \ G2, then G I , ..., Gm form a simple chain from x to p; and if y E G 1 n G2, then G2, Gm form a simple chain from y to p, as indicated in the diagram below. ...,

G1 Thus G 1 is a subset of H, i.e. so H is open.

h c G 1 c H.

Hence H is a neighborhood of each of its points, and

Hr. Since c/6/ is a cover of X, 3 G E cif such that g E G, and G is open. If 0, 3hEGnHcH and so 3 G I , ..., G E e4 forming a simple chain from h to p. But then either G, Gk, , Gm , where we consider the maximum k for which G intersects Gk, or GI, Gm form a simple chain from g to p, and so g G H, a contradiction. Hence G n H = 0, and so gEGc He. Thus He is an open set, and so Hee = H is closed.

Now let g

GnH

12. Prove Theorem 13.7: Let E be a subset of the real line R containing at least two points. Then E is connected if and only if E is an interval. Solution: Suppose E is not an interval; then

3 a, b E E, p

E

such that

a < p
Then' a E G and b E H, and hence E n G and E n H are Set G = (— of), p) and H = (p, co). non-empty disjoint sets whose union is E. Thus E is disconnected. Now suppose E is an interval and, furthermore, assume E is disconnected; say, G and H form a disconnection of E. Set A=EnG and B=EnH; then E=AuB. Now A and B are nonempty; say, a E A, b G B, a < b and p = sup {A n [a, N}. Since [a, b] is a closed set, p E [a, b] and hence p E E. Suppose pEA-= E n G.

3

Then p
>0

p

such that

Hence p G E and so p + 8 E A. Therefore p E A.

E G and

p+
But this contradicts the definition of p, i.e. p = sup {A n [a, bp.

On the other hand, suppose pGB:=EnH. Then, in particular, p E H.

3 S* > 0

such that

[p — 8*, /3] c H

Since H is an open set,

and a < p —

Accordingly, [p — 8*, p] n A = Ø. But then Hence [p — 6* , p] c E and so [p — 8*, p] C B. p — 8* is an upper bound for A n [a, b], which is impossible since p = sup {A n [a, N}. Hence p B. But this contradicts the fact that p E E, and so E is connected.

13. Prove (see Example 4.1): such that f (p) = p.

Let I -= [0,1] and let

Solution: If f(0) = 0 or f(1) = 1, the theorem follows; hence we can assume that f(0) > 0 and f(1) < 1. Since f is continuous, the graph of the function

F: I -4 R2 defined by is also continuous. Set G = {(x, Y)

F(x) =

I be continuous. Then 3 p E

G

(x, f (x))

then x < Y}, H = ((x,Y): < x); Hence if F[/1 does not contain a

(0, f(0)) E G, (1, f(1)) E H.

H

point of the diagonal A -= {(x, y) : x

y} = R2 \ (G u H)

then G U H is a disconnection of F[/1. But this contradicts the fact that F[/], the continuous image of a connected set, is connected; hence F[/1 contains a point (p, p) E A, and so f(p) = P.

FE/1

190

[CHAP. 13

CONNECTEDNESS

COMPONENTS 14. Show that every component E is closed. Solution; Now E is connected and so, by Problem 6, is È- connected, E c maximal connected set; hence E = k, and so E is closed.

R.

But E, a component, is a

15. Prove: Let p e X and let CA p {A,} be the class of connected subsets of X containing p. Furthermore, let C, = Ui Ai. Then: (i) C is connected. (ii) If B is a connected subset of X containing p, then B C Cp . (iii) Cp is a maximal connected subset of X, i.e. a component. Solution: (i) Since each

p G n i A l and so, by Problem 7, Cp u i A i is connected. a connected subset of X containing p, then B G c/Tp and so B c Cp u {A i : A i G c D, where D is connected. Then p G D and hence, by (ii), D c Cp; that is, Cp = D.

(ii)

If B is

(iii)

Let Cp Therefore C

AiG

0Ap

contains p,

is a component.

16. Prove Theorem 13.9: The components of X form a partition of X. subset of X is contained in some component.

Every connected

Solution; Consider the class C -= {Cp : p G X} where C p is defined as in the preceding problem. We claim that C consists of the components of X. By the preceding problem, each C p G c is a component. On the other hand, if D is a componeqt, then D contains some point p o G X and so D c Cpo . But D is a component; hence D = Cpo . We now show that C is a partition of X. Clearly, X = u {C p : p G X ; hence we need only show that distinct components are disjoint or, equivalently, if Cp n C0 0, then C P C 9 ' Let a G C p n C q• Then Cp c Ca and Co c Ca , since Cp and Co are connected sets containing a. But Cp and Co are components; hence Cp = Ca -= }

Lastly, if E is a non-empty connected subset of X, then E contains a point p o G X and so E c Cp0 by the preceding problem. If E = V), then E is contained in every component.

17. Show that if X and Y are connected spaces, then X product of connected spaces is connected.

X

Y is connected. Hence a finite

Solution; Let (xi, yi) and q = (x2 , Y2) be any pair of points in X>< Y. Now {x 1 } X Y morphic to Y and is therefore connected. Similarly, X X {y2 } is connected.

is homeo-

But {x 1 } xYnXX {Y2 } = foc o y2)); hence {x i } xYuXx {y2 } is connected. Accordingly, and q belong to the same component. But p and q were arbitrary; hence X X Y has one component and is therefore connected.

p

18. Prove Theorem 13.10: The product of connected spaces is connected, i.e. connectedness is a product invariant property. Solution; Let {X: E I} be a collection of connected spaces and let X = LT be the product space. i G E c X I) G X and let be component of p. Furthermore, let p = (a,: the We claim that every point x = (xi : i G I) G X belongs to the closure of E and hence belongs to E since E is closed. Now let

xi

G

=

be any basic open set containing x G X.

H

x

{Xi :

Gil

x • - - x Gim

Now

= jj {{ai } : i

X X ii X •• - X X i.

is homeomorphic to X il X • • • X Xi. and hence connected. Furthermore, p G H and so H is a subset of E, the component of p. But G n H is non-empty; hence G contains a point of E. Accordingly,

x C Ê = E.

Thus X has one component and is therefore connected.

CHAP. 13]

191

CONNECTEDNESS

ARCWISE CONNECTED SETS 19. Let

X be any path in X.

Show that

/14

the range of f, is connected.

Solution: I = [0,1] is connected and f is continuous; hence, by Theorem 13,5, f[i] is connected.

20. Prove:

Continuous images of arcwise connected sets are arcwise connected.

Solution: Let E C X be arcwise connected and let f:X--0Y be continuous. We claim that f[E] is arcwise connected. For let p,q C f[E], Then ] 23 4 ,q 1`C E such that f(p 4 ) = 13 and f(e) q. But E is arcwise connected and so

3 a path g: I --0 X

such that

g(0) = p 4 , g(l) = q4 and g[I] c E

Now the composition of continuous functions is continuous and so fog:I-4Y is continuous. Furthermore, fo g(0) f(p4) = p , fog(1) = f(e) = and f g]I] = c f[E] Thus f[E] is arcwise connected.

21. Prove Theorem 13.12:

Every arcwise connected set A is connected.

Solution: If A is empty, then A is connected. Suppose A is not empty; say, p G A. Now A is arcwise connected and so, for each ctEA, there is a path fa :I->A from p to a. Furthermore, a E fa ['I c A

and so

A -= Uffa [1 ]: a G AI

But p E fa[i], for every a G A; hence n{f a[1]: a connected and so, by Problem 7, A is connected.

e A}

is non-empty. Moreover, each f a r11 is

be a class of arcwise connected subsets of X with a non-empty intersection, Then B = U{A: A E c/ } is arcwise connected.

22. Prove: Let

C/4

Solution: Let a, b C B.

Then

3 A d, Ab E c4

such that

a

e

Aa , b G A b

Now e// has a non-empty intersection; say, p G NA: A G 0,0. Then p C A d and, since A, is arcwise connected, there is a path f: I A d CB from a to p. Similarly, there is a path g : I -0 Ab C B from p to b. The juxtaposition of the two paths (see Example 7.3) is a path from a to b contained in B. Hence B is arcwise connected.

23. Show that an open disc D in the plane R2 is arcwise connected. Solution: Let p

b i ), q = (a2 , b 2) G D. f(t) =

The function f: I -0 R2 defined by

(a1 + t(a2

is a path from p to q which is contained in D. p and q.) Hence D is arcwise connected.

-a 1 ), b 1 + t(b2 -b 1 )) (Geometrically, f[/] is the line segment connecting

24. Prove Theorem 13,13: Let E be a non-empty open connected subset of the plane R2. Then E is arcwise connected. Solution:

Method 1. Let p G E and let G consist of those points in E which can be joined to p by a path in E. We claim that G is open. For let qEGcE. Now E is open and so 3 an open disc D with center q such that qEDCE. But D is arcwise connected; hence each point x E D can be joined to q which can be joined to p. Hence each point x E D can be joined to p, and so qED c G. Accordingly, G is open.

192

CONNECTEDNESS

[CHAP. 13

Now set H = E\G, i.e. H consists of those points in E which cannot be joined to E by a path in E. We claim that H is open. For let q* G H C E. Since E is open, 3 an open disc D* with center q* such that q* G D* C E. Since D* is arcwise connected, each x G D* cannot be joined to p with a path in E, and so q* G D c H. Hence H is open. But E is connected and therefore E cannot be the union of two non-empty disjoint open sets. Then H = , and so E = G is arcwise connected.

Method 2. Since E is open, E is the union of open discs. But E is connected; hence, by Problem 11, ] open discs S I , ...,S, c E which form a simple chain joining any p G E to any q G E. Let ai be the center of Si and let b i e Si n si+ Then the polygonal arc joining p to al to b 1 to a2 , etc., is contained in the union of the discs and hence is contained in E. Thus E is arcwise connected.

TOTALLY DISCONNECTED SPACES 25. A topological space X is said to be totally disconnected if for each pair of points p,q E X there exists a disconnection G U H of X with p E G and q E H. Show that the real line R with the topology T generated by the open closed intervals (a, bl is totally disconnected. Solution: Let p, q G R; say, p < q. Then G = °,p] and H = (p, .0) are open disjoint sets whose union is R, i.e. G u H is a disconnection of R. But p e G and q G H; hence (R, T) is totally disconnected.

26. Show that the set Q of rational numbers with the relative usual topology is totally disconnected. Solution: Let p, q E Q; say, p < q.

Now there exists an irrational number a such that p
Set G = {x G Q : x < a} and H = {x E Q : x > a}. Then G u H is a disconnection of Q, and p G G and q G H. Thus Q is totally disconnected.

27. Prove: The components of a totally disconnected space X are the singleton subsets of X. Solution: Let E be a component of X and suppose p,q G E there exists a disconnection G u H of X such that E n H are non-empty and so G u H is a disconnection a component and so is connected. Hence E consists of

with p q. Since X is totally disconnected, p E G and q G H. Consequently, E n G and of E. But this contradicts the fact that E is exactly one point.

LOCALLY CONNECTED SPACES 28. Prove: Let E be a component in a locally connected space X.

Then E is open.

Solution: Let p G E. Since X is locally connected, p belongs to at least one open connected set G. But E is the component of p; hence p G Gp c E

and so

Therefore E is open, as it is the union of open sets.

E

U{Gp : p G E}

CHAP. 131

CONNECTEDNESS

193

29. Prove: Let X and Y be locally connected. Then X

X Y is locally connected. Solution: Now X is locally connected iff X possesses a base '73 consisting of connected sets. Similarly, Y possesses a base '73* consisting of connected sets. But X X Y is a finite product; hence {GXH: GEB,HGB* } is a base for the product space X X Y. Now each G X H is connected since G and H are connected. In other words, X x Y possesses a base consisting of connected sets and so X X Y is locally connected.

Xi} be a collection of connected locally connected spaces. Then the product space X = is locally connected.

30. Prove: Let

n x-,

Solution: Let G be an open subset of X containing p the defining base

= such that

G1 1

(a i : i G G X.

X • • • X Gi X H

Then there exists a member of

{Xi :

pEBcG, and so a,,k E Gfk. Now each coordinate space is locally connected, and so there

exists connected open subsets Hik C Xik such that ai. E H-,. c G,

a, i H, 1 c G il , Set

H = 1111

X • • • X

Hi.

x

nt

{X1 :

Since each xi is connected and each H-. is connected, H is also connected. Furthermore, H is open and pEllcBcG. Accordingly, X is locally connected.

Supplementary Problems CONNECTED SPACES 31.

Show that if (X, T) is connected and T* < T, then (X, T*) is connected.

32.

Show that if (X, T) is disconnected and T < T*, then (X, T*) is disconnected.

33.

Show that every indiscrete space is connected.

34.

Show, by a counterexample, that connectedness is not a hereditary property.

35.

Prove: If A 1, A2, .. is a sequence of connected sets such that A 1 and A2 are not separated, A2 and A 3 are not separated, etc., then A 1 U A2 U • • • is connected.

36.

Prove: Let E be a connected subset of a T 1 -space containing more than one element. Then E is infinite.

37.

Prove: A topological space X is connected if and only if every non-empty proper subset of X has a non-empty boundary.

COMPONENTS 38.

Determine the components of a discrete space.

39.

Determine the components of a cofinite space.

40.

Show that any pair of components are separated.

194

[CHAP, 13

CONNECTEDNESS

41. Prove: If X has a finite number of components, then each component is both open and closed. 42. Prove: If E is a non-empty connected subset of X which is both open and closed, then E is a component. 43. Prove: Let E be a component of Y and let f: X -4 Y be continuous. Then f l [E] is a union of components of X.

44. Prove: Let X be a compact space. If the components of X are open, then there are only a finite number of them.

ARCWISE CONNECTED SETS 45. Show that an indiscrete space is arcwise connected. 46. Prove: The arcwise connected components of X form a partition of X. 47. Prove:

Every component of X is partitioned by arcwise connected components.

MISCELLANEOUS PROBLEMS

48. Show that an indiscrete space is simply connected. 49. Show that a totally disconnected space is Hausdorff. 50. Prove: Let G be an open subset of a locally connected space X.

Then G is locally connected.

51. Let A = b} be discrete and let I = [0, 11. Show that the product space X = is not locally connected. Hence locally connectedness is not product invariant. -

-

1-1 fA i : A i =---• A, i E I)

52. Show that "simply connected" is a topological property. 53. Prove: Let X be locally connected. Then 'X is connected if and only if there exists a simple chain of connected sets joining any pair of points in X.

Chapter

14

Complete Metric Spaces CAUCHY SEQUENCES Let X be a metric space. every E > 0,

3 no E N

A sequence (a 1 , a2, such that

) in X is a Cauchy sequence if for

n, m > no

d(an, am) <

Hence, in the case that X is a normed space, (an) is a Cauchy sequence if for every €> 0,

5 no G N Example 1.1:

such that

<

n, m > no

Let (an ) be a convergent sequence; say a n -4 p. sequence since, for every E > 0,

3 n0 EN

such that

Then (an) is necessarily a Cauchy

n > no

p) <

Hence, by the Triangle Inequality,

n, m> n o

d(a.,„, am ) -=

p) d(am , p) < ie

E

In other words, (an) is a Cauchy sequence.

We state the result of Example 1.1 as a proposition. Proposition 14.1: Every convergent sequence in a metric space is a Cauchy sequence.

The converse of Proposition 14.1 is not true, as seen in the next example. Example 1.2:

Let X = (0,1) with the usual metric. Then (I, A, t, ...) is a sequence in X which is Cauchy but which does not converge in X.

Example 1.3:

Let d be the trivial metric on any set X and let (an) be a Cauchy sequence in (X, d). Recall that d is defined by 0 if a -= b

d(a, b)

Let E

1.

1

Then, since (an) is Cauchy, 3 no G N such that

n, m > no

d(ar,, am ) <

In other words, (a n ) is of the form (a 1 , a2, some term on. Example 1.4:

b

if a

an -= am ano,

p, p, p, . .), Le. constant from

Let (p i, p2, ...) be a Cauchy sequence in Euclidean m-space Rm; say, Pi = (a (1 1) ,

a i(m) ), P2 -=

(C(

...

1) ,

The projections of (AO into each of the m coordinate spaces, i.e., (1) (a" ) , a(1) 2 ' a3

(a (m), a 2( m )3' (m) a3

are Cauchy sequences in R, for, let e> 0.

r,s

> no

d(Pr, Ps) 2 =-

)

Since (p„) is Cauchy, 3 no E N such that

ar(1) _ a

)12

.. .

a (m )

as( m )1 2 <

Hence, in particular,

r, s > no

ar(1) — asu)12 < f2,

! ct, r(m) _ a (m) i2 <

In other words, each of the m sequences in (1) is a Cauchy sequence.

195

(1)

COMPLETE METRIC SPACES

196

[CHAP. 14

COMPLETE METRIC SPACES Definition:] A metric space (X, d) is complete if every Cauchy sequence (an) in X converges to a point p E X. Example 2.1:

By the fundamental Cauchy Convergence Theorem (see Page 52), the real line R with the usual metric is complete.

Example 2.2:

Let d be the trivial metric on any set X. Now (see Example 1.3) a sequence (an ) in X is Cauchy ill' it is of the form (a 1 , a2, ano , p, p, p, .), which clearly converges to p G X. Thus every trivial metric space is complete.

Example 2.3:

The open unit interval X (0,1) with the usual metric is not complete since (see Example 1.2) the sequence (i-,1-, t, ...) in X is Cauchy but does not converge to a point in X.

Remark:

Examples 2.1 and 2.3 show that completeness is not a topological property; for R is homeomorphic to (0, 1) even though R is complete and (0, 1) is not.

Example 2.4:

Euclidean m-space Itm is complete. IV where

For, let

(p i, P2,

be a Cauchy sequence in

= (aV ), • a4 m) ), P2 =(cq l) , • • •

Then (see Example 1,4) the projections of (pn) into the m coordinate spaces are Cauchy; and since R is complete, they converge: ...)

(arn ) ,a 7' ) , ...)

b 1,

br„,

, bm ) G Rm, since each of the m proThus (p a) converges to the point q = (b1 , jections converges to the projection of q (see Page 169, Theorem 12.7).

PRINCIPLE OF NESTED CLOSED SETS Recall that the diameter of a subset A of a metric space X, denoted by d(A), is defined by d(A) = sup {d(a, a') : a,a' E A} and that a sequence of sets, Ai, A2, . ., is said to be nested if A1 D A2 D • • • . The next theorem gives a characterization of complete metric spaces analogous to the Nested Interval Theorem for the real numbers. Theorem 14.2: A metric space X is complete if and only if every nested sequence of non-empty closed sets whose diameters tend to zero has a non-empty intersection. In other words, if A1 D A2 D • • • are non-empty closed subsets of a complete metric space X such that lim d(An) = 0, then nz=c, A, 0; and vice versa. The next examples show that the conditions closed, are both necessary in Theorem 14.2.

lim d(A n) = 0

n-0.0

and that the Ai are

Example 3.1:

Let X be the real line R and let A n [n, *0). Now X is complete, the A n are closed, But fl 1 A n is empty. Observe that lim d(A n) and A I D A2 D - • • . O.

Example 3.2:

Let X be the real line R and let A n = (0,1/4 Now X is complete, A1 D A 2 D " • and lim d(A n) = 0. But n r,°*=1 A, is empty. Observe that the A not closed. n

COMPLETENESS AND CONTRACTING MAPPINGS Let X be a metric space. A function f: X --> X is called a contracting mapping if there exists a real number a, 0 a < 1, such that, for every p,q E X, d(f(p), f(q))

d(p, q) < d(p, q)

Thus, in a contracting mapping, the distance between the images of any two points is less than the distance between the points.

CHAP. 14]

197

COMPLETE METRIC SPACES

Example 4.1:

Let f be the function on Euclidean 2-space R2 , i.e. f: R2 -9. R2 , defined by f(p) = ip. Then f is contracting, for d(f(p), f(q)) = 111(p) —

=

f(q)1

— q11 =

=ILi - iql d(p, q)

If X is a complete metric space, then we have the following "fixed point" theorem which has many applications in analysis. If f is a contracting mapping on a complete metric space X, then there exists a unique point p e X such that f(p) = p.

Theorem 14.3:

COMPLETIONS

A metric space X* is called a completion of a metric space X if X* is complete and X is isometric to a dense subset of X*. Example 5.1:

The set R of real numbers is a completion of the set Q of rational numbers, since R is complete and Q is a dense subset of R.

We now outline one particular construction of a completion of an arbitrary metric space X. Let C[X] denote the collection of all Cauchy sequences in X and let be the relation in C[X] defined by (a.) — (b.) if lim d(a., b.) = 0 Thus, under "—j" we identify those Cauchy sequences which "should" have the same

"limit". Lemma 14.4:

The relation — is an equivalence relation in C[X].

Now let X* denote the quotient set C[X]/—, i.e. X* consists of equivalence classes (a, ] of Cauchy sequences (a.) E C[X]. Let e be the function defined by [

)

e([(a.)],[(b.))) = lim d(a., b.) where t(a)1, [(b.)1 G X.

Lemma 14.5:

The function e is well-defined, i.e. (a.) — (a:) and (b.) — (1):) implies lim d(a.,b.) = lim d(a:, b:).

In other words, e does not depend upon the particular Cauchy sequence chosen to represent any equivalence class. Furthermore, Lemma 14.6:

The function e is a metric on X.

Now for each p e X, the sequence (p,p,p,

[(p, p,

..)1

and

.) e C[X], i.e. is Cauchy. Set 1= : p E X)

Then If is a subset of X*.

1,

and X1/4 is dense in X*.

Lemma 14.7:

X is isometric to

Lemma 14.8:

Every Cauchy sequence in X* converges, and so X* is a completion of X.

198

[CHAP. 14

COMPLETE METRIC SPACES

Lastly, we show that Lemma 14.9: If Y* is any completion of X, then Y* is isometric to X. The previous lemmas imply the following fundamental result. Theorem 14.10: Every metric space X has a completion and all completions of X are isometric. In other words, up to isometry, there exists a unique completion of any metric space. BAIRE'S CATEGORY THEOREM Recall that a subset A of a topological space X is nowhere dense in X if the interior

of the closure of A is empty: int (A) --, 9) Example 6.1:

The set Z of integers is a nowhere dense subset of the real line R. i.e. Z = 2, and its interior is empty; hence

For Z is closed,

int (i) = int (Z) = 0 Similarly every finite subset of R is nowhere dense in R. On the other hand, the set Q of rational numbers is not nowhere dense in R since the closure of Q is R and so

int (4) = int (R) = R #

(z)

A topological space X is said to be of first category (or meager or thin) if X is the countable union of nowhere dense subsets of X. Otherwise X is said to be of second category (or non-meager or thick). Example 6.2:

The set Q of rational numbers is of first category since the singleton subsets {p} of Q are nowhere dense in Q, and Q is the countable union of singleton sets.

In view of Baire's Category Theorem, which follows, the real line R is of second category. Theorem (Haire) 14.11: Every complete metric space X is of second category. COMPLETENESS AND COMPACTNESS

Let A be a subset of a metric space X. Now A is compact if A is sequentially compact if every sequence (a n) in A has a convergent subsequence ci,in ). But, by Example 1.1, (ain) is a Cauchy sequence. Hence it is reasonable to expect that the notion of completeness is related to the notion of compactness and its related concept: total boundedness. We state two such relationships: Theorem 14.12: A metric space X is compact if and only if it is complete and totally bounded. Theorem 14.13:

Let X be a complete metric space. Then A C X is compact if and only if A is closed and totally bounded.

CONSTRUCTION OF THE REAL NUMBERS

The real numbers can be constructed from the rational numbers by the method described in this chapter. Specifically, let Q be the set of rational numbers and let R be the collection of equivalence classes of Cauchy sequences in Q:

CHAP. 14]

199

COMPLETE METRIC SPACES

R = Wan)] : (an) is a Cauchy sequence in tl) Now R with the appropriate metric is a complete metric space.

Let X be a normed vector space. The construction in this chapter gives us a complete metric space X*. We can then define the following operations of vector addition, scalar multiplication and norm in X* so that X* is, in fact, a complete normed vector space, canal a Banach space: (i) [(an)] + [(b.)] =--- 1(an + b n)] (ii) k I(an)]E--- [(lean)] (iii) il [(ad --- l im Ila nl I

Remark:

n -. 09

Solved Problems CAUCHY SEQUENCES 1. Show that every Cauchy sequence (a„) in a metric space X is totally bounded (hence also bounded). Solution: Let f > O. We want to show that there is a decomposition of {a n } into a finite number of sets, each with diameter less than E. Since (an) is Cauchy, 3 no G N such that

d(an, am ) < c

n, m > no

Accordingly, B = {ano+ 1 , ano+ 2, . . . } has diameter at most E. Thus {a1 }, ..., {a no }, B decomposition of {an } into sets with diameter less than E, and so (an) is totally bounded.

is a finite

2. Let (ai, a2, ... ) be a sequence in a metric space X, and let Al =- {al, a2, . . . ),

A2 --= {a2, a3, . . . ),

143 --= {a3, a4, . . . },

Show that (an) is a Cauchy sequence if and only if the diameters of the i.e.

lim d(21) =

An

tend to zero,

O.

n -...

Solution: Suppose (an ) is a Cauchy sequence.

3 no E N Accordingly,

Let E > O.

such that

d(A n) <

n> no

n, m > no and so

E

On the other hand, suppose lim d(A n) = O. n -4

cc

3 no G N n, m > no

Hence

Then

Let

e

lim (A n) -= 0 E>

such that

d(a,„ am ) <

O. Then

d(A no+ i ) <

an, a. G A n +1

€

d(a„, am ) <

f

and so (an) is a Cauchy sequence.

3. Let (al , a2, . . . ) be a Cauchy sequence in X and let (ai1 , ai,, . . . ) be a subsequence of (an ). Show that lim d(an, a n) = 0. n -. 0,

Solution: Let e> O.

Since (an) is a Cauchy sequence,

3 no G N Now i

no

such that

n,m > no — 1

•=-- no > no — 1 and therefore d(a,„0, cu no) <

E.

d(a„, am ) <

E

In other words, link d(a„, a in) = O. n -+ .

200 4.

[CHAP. 14

COMPLETE METRIC SPACES

Let (a1, a2) • ) be a Cauchy sequence in X and let (a, i , (an) converging to p E X. Show that (an) also converges to p. Solution: By the Triangle Inequality, d(a,„ p)

lim d(a„, p)

lim d(an ,

n—roo

Since

+

d(a„, n -+

d(ain , p)

and therefore

d(ai, + Tilim •-+ n

lirn d(a, p) = 0 and, by the preceding problem,

p,

be a subsequence of

n—roo

lim d(a„, atn )

O. Then

n- , co

lim d(a„, p) = 0

and so a0 -*p

n

5.

Let (1)1, b2, be a Cauchy sequence in a metric space X, and let (ai, a2, sequence in X such that d(an,bn) < 1/n for every n E N. (i) Show that (an) is also a Cauchy sequence in X. (ii) Show that (an) converges to, say, p E X if and only if (bn) converges to p.

) be a

Solution: (i) By the Triangle Inequality, d(br,„ b 0 ) + d(bn , an)

d(a„,, an) = d(a,n , bm ) Let

E>

O. Then 3 n 1 C N such that 1/n1 < E/3.

Hence

d(am , an) < €13 + d(b,n, b0 )

n, m > n 1

f1 3

By hypothesis, (b 1 , b 2, ...) is a Cauchy sequence; hence

3 n2 G N Set no = max {n 1 , n2 .

such that

d(bm , b n) < €13

n, m > n2

Then

}

€13 + €13

d(am , a,,) < f/3

n,m > no Thus (an) is a Cauchy sequence. (ii)

By the Triangle Inequality, d(bn , p)

d(b„, an) -I- d(an, p); hence lim d(b n , an) in

lirn d(b.,„ p)

n

lim d(b n , an )

But

oo

lirn (an , p) fl -+

oO

lirn (1/n) = O. Hence, if a0 p, lim d(b p) 0.-“o

ro.co

n

lim

p)

0 and so

(1) 0 ) also converges to p. Similarly, if b„ p then a. -* p.

COMPLETE SPACES 6.

Prove Theorem 14.2: The following are equivalent: (i) X is a complete metric space. (ii) Every nested sequence of non-empty closed sets whose diameters tend to zero has a non-empty intersection. Solution:

(1 )

(ii): Let

A1 D

A2 D • • •

be non-empty closed subsets of . X such that

lim d(A) 7E

prove that

fl y,

An

0.

Since each A i is non-empty, we can choose a sequence

(a 1 , a2, ..)

such that

We claim that oft00 is a Cauchy sequence. Let

3 no G N

E>

a l G A 1, a2 G O. Since

such that

A2, ...

lirn d(A)== O,

d(A 00 ) <

E

But the A i are nested; hence

n, m > no Thus (a.) is Cauchy.

A n, Am c A no

a., am G kno

d(an , a.) <

O. We want to

CHAP. 14]

201

COMPLETE METRIC SPACES

Now X is complete and so (a n ) converges to, say, p G X. We claim that not, Le. suppose 3 k EN such that p e Ak Since Ak is a closed set, the distance between p and Ak is non-zero; say, Ak and the open sphere S = S(p, S) are disjoint . Hence n> k

d(p, A k ) =

Suppose

> O. Then

n n A n is non-empty.

and so

(i): be a Cauchy sequence in X.

Let (a l , a2, ...)

A1 — i.e.

n y, A n.

an e S(p, cl)

an --* p. In other words, p G n„ A.

This is impossible since

(ii)

an E Ak

p E

:n

Ak

since d(ii) = d(A),

kl.

A

Set

A-2 — {a 2 , a3 , • -

.1,

A 1 D A2 D • • •

Then

where

a2,

We want to show that (an ) converges.

and, by Problem 2, lim d(A n ) = O. Furthermore,

A, A,

is the closure of

D A2D •••

n„ A n

sets whose diameters tend to zero . Therefore, by hypothesis, that the Cauchy sequence (a n ) converges to /3.

is L.'-'séquence of non-empty closed 0; say, p G n n A n. We claim

lim d(A) = 0,

Let E > O. Since

3 no E N

such that

an, p G A ,,

n> no

and so

d(ii. no ) <

E

d(an ,p) <

E

In other words, (an) converges to p,

7. Let X be a metric space and let f:X--*X be a contracting mapping on X, i.e. there a d(p, q). Show exists a E R, 0 a < 1, such that, for every p, q E X, d(f(p), f (q)) that f is continuous. Solution: We show that f is continuous at each point

x o C X.

d(f (x), f(x 0 ))

d(x, x 0) <

Let

E>

a d(x, x 0 )

O. Then

<

(YE

and so f is continuous.

8. Prove Theorem 14.3: Let f be a contracting mapping on a complete metric space X, say d(f (a), f (b))

a

0

d(a, b),

a<

1

Then there exists one and only one point p e X such that f (p) = p. Solution: Let ao be any point in X.

Set

= f(ao), a2 = Pal) = f2 (a0),

= f(an-i) = Nap),

We claim that (ai , a2, ...) is a Cauchy sequence. First notice that d(f s 4 t (ao), ft(ao))

= But

at

d(f s+

[d(ao, Pao)) + d(f(a0), f2 (a0 )) + • • • + cl(fs-1 (an), fs (ao))]

d(fi+ (a0), Mao)) =- at d(f(a n), ao )

and so

a t d(f(a0), ao) (1 ±

d(fs+ t (ao), ft(a0 )) = since

(1 -I-

a -I-

at d(fs(a0), ao)

(ao), f t-1 (ao))

a2 -I- • • • -I- as -1 )

a

± a2 ± • • • -I- as -1 )

at d(f(ao), ao ) [1 7 (1 — a)]

1/(1 — a).

202

[CHAP. 14

COMPLETE METRIC SPACES

Now let e > 0 and set

f(1 — a)

6

< 1,

Hence if r

3 no C N

ao)

if d(f(ao), ao)

— aild(f (ao), ao)

{ Since a

if d(f

0

ano <

such that

8 > no , a s [1 1 (1 —

d(a„ ar)

a)]

[1/(1 — a)] d (Pao), ao)

d(f(ao), ao) <

and so (an) is a Cauchy sequence. Now X is complete and so (a.) converges to, say, tinuous and hence sequentially continuous, and so

Do

lim an + j = p

n ■- ■

Lastly, we show that p is unique. Suppose 1(p) = p and

d(p, q) = d(f (p), fig))

f(p) = p; for f is con-

We claim that

Ihn an ) = lim f(a) =

f

f(P)

p G X.

1(g) = g; then

a d(p, q)

But a < 1; hence d(p, q) = 0, i.e. p = q.

COMPLETIONS 9. Show that (a.) — (b„) if and only if they are both subsequences of some Cauchy sequence Solution: Suppose (an) — (b.), i.e.

lim d(an, bn ) = O. Define (ca) by

n

oc

en =

Thus

(e n ) = (b 1 , a o b2 , a2 , ...).

awn if n is even lb,/ . +1) if n is odd

3 n1 G N

such that

m, n> n 1

d(a„, a„) <

3 n2 G N

such that

m, n > n2

d(bm , b n) <

3 n3 G N

such that

n > n3

Set n o = max (n o n2 , n3).

now

d(an , bn) < le

We claim that

d(c„„ cn) <

m, n > 2no

d(cm , en) < if <

cm = a1/2m , c. = a 1/2 .

m, n even C

m, n odd

Cm

E

d(c m , en) < e <e

b1/2 (n + ), en

m even, n odd

E

> n o n3 ; (m -I- 1) > n 2 , n3

m> 2n0

Note that Thus

E > 0;

We claim (c„) is a Cauchy sequence. For, let

a1/2m, en =" b 1/2(n +1)

d(c m , en ) -= d(a1/2m , b1/2 m )

d(bv2m , b1/2 (n+ 1)) <

e -I-

E

and so (c.) is a Cauchy sequence, Conversely, if there exists a Cauchy sequence (en) for which (an) = (c).)

and

(b„) = (ck.), then

lim d(an , b n) = lirn d(c ck n)

n —•

since (c.) is Cauchy and

n

7t..

co implies j,, k n

00 .

10. Prove Lemma 14.5: The function e is well-defined, i.e. (an) — (a:) and (b.) implies lim d(an,bn) = lim d(a:,b:). Solution: Set r

lim d(an , bn)

and r*

lim d(a:, b:), and let

e > O. Note that

(1):)

CHAP. 14]

d(a„, b„) Now

203

COMPLETE METRIC SPACES

d(an , a) +

d(b:, bn)

3 n1 G N

such that

n > n1

d(ct„,,,a:) < E/3

3 n2 E N

such that

n> n2

d(b„, b ) < E/3

3 n3 EN

such that

n> n3

Id(a7t , b:)

r * < E/ 3

Accordingly, if n > max (n 1 , n2 , n3), then d(a„, b„) < r* +E But this inequality holds for every r; thus r r*.

lim d(a„, b„) = r

and so

E > 0;

r* + E

n.-+X

hence

r r*.

11. Let (a.) be a Cauchy sequence in X. Show that sequence (ai, a2, ) in 1. (Here / = { P= [(A

In the same manner we may show that

e X* is the limit of the .)] pe XI.)

= [(a.)]

a

P, • •

Solution: Since (a„) is a Cauchy sequence in X, lim e(est,„ , a)

lim

Ern

n

d(am , a„))=

—4 o0

lim d(am , an )

0

m -4 .9 It

Accordingly, (C.Î.n ) 4 a.

12. Prove Lemma 14.7: X is isometric to Solution:

2,

and / is dense in X*.

For every p,q G X, e(; and so X is isometric to

5? .

= lim d(p, q)

d(p, q)

We show that 1" is dense in X* by showing that every point in X* is the

limit of a sequence in 3?. Let a

=

[(a l , a2, ...)1 be an arbitrary point in X*.

Then (a„) is a Cauchy AA ( a l , a2 , ...) in X.

sequence in X and so, by the preceding problem, a is the limit of the sequence Thus X is dense in X*.

13. Prove Lemma 14.8: Every Cauchy sequence in (X*, e) converges, and so (X*, e) is a completion of X. Solution: Let ( a l , a2, ...) be a Cauchy sequence in X*.

3

ein E 5.e

Since 5? is dense in X*, for every

such that

n E N,

e(ein , an) < 1/n

Then (Problem 5) ( 'C‘ti g2 , ) is also a Cauchy sequence and, by Problem 12, (i g2, ) converges to p = [(ct i , a2, ...)] G X*. Hence (Problem 5) (a„) also converges to p and therefore (X*, e) is complete.

14. Prove Lemma 14.9: Solution:

If Y* is a completion of X, then Y* is isometric to X*.

We can assume X is a subspace of Y".. Hence, for every p E Y*, there exists a sequence ...) in X converging to p; and in particular, (an) is a Cauchy sequence. Let f; Y* X* be defined by )] a2, f(P) = Now if (45t, a:,

.) G X also converges to p, then Urn d(a„, a*) = rt

n

and so

[(an)] = [(a:)]

In other words, f is well-defined. Furthermore, f is onto . For if [(b i , b2, ...)] G X* , then (b 1 , b2, ...) is a Cauchy sequence in X c Y* and, since 17 * is complete, (bn) converges to, say, q G Y*. Accordingly, f(q) = [(b„)].

[CHAP. 14

COMPLETE METRIC SPACES

204

Now let p,q E Y* with, say, sequences (an) and (b„) in X converging, respectively, to p and q. Then lim d(an , b n) = d ( urn a, lim b n ) =- d(p,

e(f(p), f (q)) = ea(an)i, [(bn )l)

Consequently, f is an isometry between Y* and X*.

HAIRE'S CATEGORY THEOREM 15. Let N be a nowhere dense subset of X.

Show that

Solution: Suppose 1Vc is not dense in X, i.e. 3 p E X pEG

/Ve

is dense in X.

and an open set G such that

and

G n A7c

Then peGc A— - ,7 and so p E int (Kr). But this is impossible since int (N7) 0. Therefore N is dense in X.

N is nowhere dense in X, i.e.

16. Let G be an open subset of the metric space X and let N be nowhere dense in X. Show that there exist p E X and 8 > 0 such that S(p, 8) c G and S(p, 8) n N = Ç. Solution: Set H = G n /Ve. Then H C G and H n N = 0. Furthermore, H is non-empty since G is open and /Ç7C - is dense in X; say, p G H. But H is open since G and Kre are open; hence 38 > 0 such that S(p, 8) C H. Consequently, S(p, 8) C G and S(p, 8) n N = Ø.

17. Prove Theorem

Every complete metric space X is of second category.

14.11:

Solution: X, i.e. 3 p G X such Let M c X and let M be of first category. We want to show that M that p (% M. Since M is of first category, M = N i U N2 U • • • where each N, is nowhere dense in X. Since N 1 is nowhere dense in X, 81/2. Then Now S(a i,

i)

8 2 /2

and

S(a i ,f i) n

NI

S i > 0 such that

S(a i, 8 1 ) n

N 1 =-- 9).

Set

= 0

is open and N2 is nowhere dense in X, and so, by Problem 16,

3 a2 e X and 82 > 0 Set E2

3 al G X

such that

S(a2, 8 9) c S(a i , f i) c S(a i, f i )

and S(a 2 , 8 2) n N2 —

E I /2 = 8 i /4. Then S(a2, E2) c S(a i, E i )

S(a2 , E2) n N2 = 0

and

Continuing in this manner, we obtain a nested sequence of closed sets S(12 1 , E 1) such that, for every n G Thus lim E7, 'It*

N,

D

S(a2,f 2)

S(a„, E.) n N n = 0

lim 8 i/2 7' = 0

D

S(a3, f3)

and

D

•

8 1/2 n

En

and so, by Theorem 14.2,

DO

3 p GX Furthermore, for every n E

N, p

such that

p G n

1 S(an, 6,0

Nn and so p M.

COMPLETENESS AND COMPACTNESS 18. Show that every compact metric space X is complete. Solution: Let (a1 , a2 , .) be a Cauchy sequence in X. Now X is compact and so sequentially compact; hence (a n) contains a subsequence (a 1 , a 2 , .) which converges to, say, p G X. But (Problem 4) (a,,) also converges to p. Hence X is complete.

CHAP. 14]

205

COMPLETE METRIC SPACES

19. Let E be a totally bounded subset of a metric space X. (an) in E contains a Cauchy subsequence.

Show that every sequence

Solution: Since E is totally bounded, we can decompose E into a finite number of subsets of diameter less than e i = 1. One of these sets, call it A1, must contain an infinite number of the terms of the sequence; hence a G A1 3 j1 G N such that Now A 1 is totally bounded and can be decomposed into a finite number of subsets of diameter less than E2 Similarly, one of these sets, call it A 2 , must contain an infinite number of the terms of the sequence; hence

3 i2 G N Furthermore,

A2 C

such that

1 2 > i 1 and a12 G A2

A 1.

We continue in this manner and obtain a nested sequence of sets

E

E > 0;

A1

D A2 D • • •

with d(A n) < 1/n

) of (an) with a,. G A. We claim that (a i ) is a Cauchy sequence.

and a subsequence (ay For, let

D

then

3 no G N Therefore

20. Prove Theorem 14.12: totally bounded.

in ,

>ino

such that

1/n0 < GAn

E

and so d(A n ) <

d(a,m ,

E

<

A metric space X is compact if and only if X is complete and

Solution: Suppose X is compact. Then, by Problem 15, X is complete and, by Lemma 11.17, Page 158, X is totally bounded. On the other hand, suppose X is complete and totally bounded. Let (a1 , a2 , ...) be a sequence in X. Then, by the preceding problem, (a„,) contains a Cauchy subsequence (a,n) which converges since X is complete. Thus X is sequentially compact and therefore compact.

21. Prove Theorem 14.13: Let A be a subset of a complete metric space X. Then the following are equivalent: (i) A is compact. (ii) A is closed and totally bounded. Solution: If A is compact, then by Theorem 11.5 and Lemma 11.17 it is closed and totally bounded. Conversely, suppose A is closed and totally bounded. Now a closed subset of a complete space is complete, and so A is complete and totally bounded. Hence, by the preceding problem, A is compact.

206

COMPLETE METRIC SPACES

[CHAP. 14

Supplementary Problems COMPLETE METRIC SPACES

22. Let (X, (I) be a metric space and let e be the metric on X defined by e(a, b) = min {1, d(a, b)}. that (an) is a Cauchy sequence in (X, d) if and only if (an) is a Cauchy sequence in (X, e).

Show

23. Show that every finite metric space is complete.

24. Prove: Every closed subspace of a complete metric space is complete.

25.

Prove that Hilbert Space (1 2-space) is complete.

26. Prove: Let c13(X, R) be the collection of bounded real-valued functions defined on X with norm Then c13 (X , R) is complete.

Ilill

—

sup{

f(4

:

x

e X}

27. Prove: A metric space X is complete if and only if every infinite totally bounded subset of X has an accumulation point.

28. Show that a countable union of first category sets is of first category.

29. Show that a metric space X is totally bounded if and only if every sequence in X contains a Cauchy subsequence.

30. Show that if X is isometric to Y and X is complete, then Y is complete.

MISCELLANEOUS PROBLEM Every normed vector space X can be densely embedded in a Banach space, i.e. a complete normed vector space. (Hint: See Remark on Page 199).

31. Prove:

Chapter

15

Function Spaces FUNCTION SPACES Let X and Y be arbitrary sets, and let T(X, Y) denote the collection of all functions from X into Y. Any subcollection of T(X, Y) with some topology T is called a function space. We can identify T(X, Y) with a product set as follows: Let Y. denote a copy of Y indexed by x e X, and let F denote the product of the sets Yx, i.e., : x E 2f)

F =

Recall that F consists of all points p--= (cti: x E X) which assign to each x E X the element ax E Yx = Y, i.e. F consists of all functions from X into Y, and so F = T(X, Y). Now for each element x E X, the mapping ex from the function set T(X, Y) into Y defined by ex(f) = f(x) is called the evaluation mapping at x. (Here f is any function in px, Y), i.e. f: X -> Y.) Under our identification of T(X, Y) with F, the evaluation mapping ex is precisely the projection mapping 7r1 from F into the coordinate space 171 = Y. Example 1.1:

Let T(/, R) be the collection of all real-valued functions defined on / =- [0,1], and let f,g,h G R) be the functions

f(x) = x 2, g(x) = 2x

-I-

1, h( x)

Consider the evaluation function Then at, say, j =

sin /Tx

T(I,R)- 3R

e(f) = 1(1) = f ( ) = ei (g) = g(j) = g(i) = 2 e(h) = h(j) = h(4) = 1 Graphically, eg), e(g) and e(h) are the points where the graphs of f, g and h intersect the vertical line Ri through x = j.

POINT OPEN TOPOLOGY Let X be an arbitrary set and let Y be a topological space. We first investigate Y) where we identify pX, Y) with the product set the product topology T on of the product topology F= { Yx: x E X) as above. Recall that the defining subbase on F consists of all subsets of F of the form

fi

7;0 1

[G] = {f: 710 (f) E G)

where X0 G X and G is an open subset of the coordinate space Y1 0 = Y. But 71 0 (f) e10(f) = f (x0), where exo is the evaluation mapping at x o e X. Hence the defining subbase ef of the product topology T on T(X, Y) consists of all subsets of T(X, Y) of the form 207

208

FUNCTION SPACES

[CHAP. 15

[f: f(x 0) e G), i.e. all functions which map an arbitrary point xo E X into an arbitrary open set G of Y. We call this product topology on f(X, Y), appropriately, the point open topology.

Alternatively, we can define the point open topology on ]'(X, Y) to be the coarsest topology on f(X, Y) with respect to which the evaluation functions ex : f(X, Y) Y are continuous. This definition corresponds directly to the definition of the product topology. Example 2.1:

Let T be the point open topology on T(/, R) where I = [0, 1]. As above, members of the defining subbase of T are of the form

{f

Rio) E G}

where 50 e I and G is an open subset of R. Graphically, the above subbase element consists of all functions passing through the open set G on the vertical real line R through the point :7 0 on the horizontal axis. Recall that this is identical to the subbase element of the product space

E I}

X=

illustrated in Chapter 12, Page 170. Example 2.2:

If A

is

a

subset of a product space then A is a subset of the product of its projections, i.e.

fi {Xi : i E I}, A

cn

:

i

:iE

(as indicated in the diagram). Thus A c Fl {ir[A] : i E I} where is the closure of 7,[A]. Accordingly, if c.4 = e4(X,Y) is a subcollection of T(X, Y), then

7 1 [A]

i [A]

c4 c 11

{77.,[c4] x

e

= FI fe x [eAr] :

x

E

and e[c-4] = {f (x) : f E c// } . Hy the Tychonoff Product Theorem, if {f (x): x E is compact for every x E X, then 11 {irx [e.4] : x E X} is a compact subset of the product space

II {Yx :

x E X}.

Recall that a closed subset of a compact set is compact. Hence the result of Example 2.2 implies Theorem 15.1: Let a be a subcollection of T(X, Y). Then a is compact with respect to the point open topology on T(X, Y) if (i) 0,4 is a closed subset of T(X, Y) and (ii) for every x E X, {f(x) f e c4 } is compact in Y. In the case that Y is Hausdorff we have the following stronger result: Theorem 15.2: Let Y be a Hausdorff space and let a c T(X, Y). Then cis compact with respect to the point open topology if and only if 0,4 is closed and, for every x E X, tf(x): f E 0. 4) is compact.

POINTWISE CONVERGENCE

Let (f i , f2, ) be a sequence of functions from an arbitrary set X into a topological space Y. The sequence (f.) is said to converge pointwise to a function g: X Y if, for every x0 G X, i.e. lim f.(xo) = g(x0) (fi (X0), f2 (X0), . . . ) converges to g(x0), n

0.0

CHAP. 15]

FUNCTION SPACES

209

In particular, if Y is a metric space then (fn) converges pointwise to g if for every and every xo e X, 3 no = no(xo, €) E N such that n > no d(f(x o), g(x0)) < € Note that the no depends upon the E and also upon the point xo. Example 3.1:

E>

0

Let (i1•f2, • ..) be the sequence of functions from 1 = [0,11

into R defined by

f i(x) = x, f 2 (x) = x 2 , f 3 (x) Then (fa) converges pointwise to the function g: 1 --> R defined by

0 1

if 0 = x < 1 if x = 1

Observe that the limit function g is not continuous even though each of the functions fi is continuous.

The notion of pointwise convergence is related to the point open topology as follows: Theorem 15.3: A sequence of functions (f1,12, . . . ) in T(X, Y) converges to g E T(X, Y) with respect to the point open topology on T(X, Y) if and only if (fa) converges pointwise to g. In view of the above theorem, the point open topology on T(X, Y) is also called the topology of pointwise convergence.

Remark:

Recall that metrizability is not invariant under uncountable products; therefore, the topology of pointwise convergence of real-valued functions defined on [0, 1] is not a metric topology. The theory of topological spaces, as a generalization of metric spaces, was first motivated by the study of pointwise convergence of functions.

UNIFORM CONVERGENCE Let (fi, f2, . . .) be a sequence of functions from an arbitrary set X into a metric space (Y, d). Then (fn) is said to converge uniformly to a function g:X->Y if, for every E>0, 3 no = no(E) E N

such that

n> n o

d(f.(x), g(x)) <

E,

VXEX

In particular, (fn) converges pointwise to g; that is, uniform convergence implies pointwise convergence. Observe that the no depends only on the E, whereas, in pointwise convergence, the no depends on both the E and the point x. In the case where X is a topological space, we have the following classical result: Proposition 15.4: Let (f 1, f2, . . .) be a sequence of continuous functions from a topological space X into a metric space Y. If (f.) converges uniformly to g: X --> Y, then g is continuous. Example 4.1:

Let 11,/2, . . . be the following continuous functions from 1 = [0, 1] into R:

fi(x) = x, f2(x) = x 2, f 3 (x) — x3, . . . Now, by Example 3.1, (f,) converges pointwise to g: I --> It defined by g(X)

0

if 0 -= x < 1 if x =-- 1

Since g is not continuous, (fn) does not converge uniformly to g.

210

[CHAP. 15

FUNCTION SPACES

Example 4.2:

Let (fi, f 2, • . .) be the following sequence of functions in j. (R, R):

I i — —1jx1 n

if

x
if

xl

n 10

Now (f„) converges pointwise to the con1, But (f,) does not stant function g(x) converge uniformly to g. For, let = Note that, for every n G N, there exist points x o G R with f„(x 0) = 0 and so f„(x o) — g (X 0 ) 1 1 > f,

Let 673(X , Y) denote the collection of all bounded functions from an arbitrary set X into a metric space (Y, d), and let e be the metric on q3(X, Y) defined by e(f,, g)

= sup fcl(f (x), g(x)) : x

e X}

This metric has the following property: Theorem 15.5: Let (fl, f2, . . .) be a sequence of functions in 93(X, Y). Then (f) converges to g E 673(X , Y) with respect to the metric e if and only if (f9) converges uniformly to g.

In view of the above theorem, the topology on cB(X, Y) induced by the above metric is called the topology of uniform convergence. Remark:

The concept of uniform convergence defined in the case of a metric space Y cannot be defined for a general topological space. However, the notion of uniform convergence can be generalized to a collection of spaces, called uniform spaces, which lie between topological spaces and metric spaces.

THE FUNCTION SPACE C[0, 1]

The vector space C[0, 1] of all continuous functions from / = [0, 1] into R with norm defined by sup { f (x) : x G

I}

is one of the most important function spaces in analysis. induces the topology of uniform convergence.

Note that the above norm

Since / = [0, 1] is compact, each f e C[0,1 ] is uniformly continuous; that is, Proposition 15.6:

Let f: [0, 1] -> R be continuous. Then for every €> 0, 3 8 = 8(E) > 0

such that

;xo — x11 < 8

If(x0)—f(x1)1

<E

Uniform continuity (like uniform convergence) is stronger than continuity in that the S depends only on E and not on any particular point.

One consequence of Proposition 15.4 follows: Theorem 15.7: C[0, 1] is a complete normed vector space. We shall use the Baire Category Theorem for complete metric spaces to prove the following interesting result:

Proposition 15.8: Remark:

There exists a continuous function differentiable.

f: [0, 1] --> R which is nowhere

All the results proven here for C[0, 1] are also true for the space C[a,b] of all continuous functions on the closed interval [a, b].

CHAP. 15]

FUNCTION SPACES

211

UNIFORM BOUNDEDNESS

In establishing necessary and sufficient conditions for subsets of function spaces to be compact, we are led to the concepts of uniform boundedness and equicontinuity which are interesting in their own right. A collection of real valued functions e.4 = { f: X -> R} defined on an arbitrary set X is said to be uniformly bounded if lf(x) 1- M, VfecA, VxeX 3 MER such that That is, each function f E cA is bounded and there is one bound which holds for all of the functions. In particular if a C C[0,1], then uniform boundedness is equivalent to ] M E R such that If 1 M, V f 0_// or, cA is a bounded subset of CEO, 11. Example 5.1:

Let c.4 be the following subset of T(R,R):

= {fi (x) = sin x, f 2 (x) = sin 2x, • • • } Then cAl is uniformly bounded. For, let M = 1; then, for every f E c..4 and every x G R, f (x)1 M. See Fig. (a) below.

Fig. (a) Example 5.2:

Let c/I C CEO, 1] be defined as follows (see Fig. (b) above): c,T

=

{f i (x) = x, f2(x) = 2x, 13(x) = 3x,

is bounded, c,4 is not Although each function in C[0,1], and in particular in uniformly bounded. For if M is any real number, however large, 3 no G N with n o > M and hence fflo(1) = no > M.

EQUICONTINUITY. ASCOLI'S THEOREM A collection of real-valued functions 0_A = { f, : X -> R} defined on an arbitrary metric space X is said to be equicontinuous if for every €> 0, 3 8 = 8(0 > 0 d(x o, xi) < 8 such that 11(4) — f(xi) < E, V f E cA

Note that 8 depends only on E and not on any particular point or function. It is clear that each f E cA is uniformly continuous. Theorem (Ascoli) 15.9: Let 0.A be a closed subset of the function space 00, 11. Then a is compact if and only if a is uniformly bounded and equicontinuous. COMPACT OPEN TOPOLOGY Let X and Y be arbitrary sets and let A C X and B C Y. We shall write F(A,B) for the class of functions from X into Y which carry A into B: F(A,B) = {f E T(X,Y)

: f[A] C B}

212

[CHAP. 15

FUNCTION SPACES

Example 6.1:

Let of be the defining subbase for the point open topology on T(X, Y). the members of ef are of the form (f

e px,

Recall that

(x) E GI, where x E X, G an open subset of Y

Y) :

Following the above notation, we denote this set by F(x,G) and we can then define

cj: by

=

{F(x, G) : x G X, G c Y open}

Now let X and Y be topological spaces and let cA be the class of compact subsets of be the class of open subsets of Y. The topology T on T(X, Y) generated by

X and

(-;

= {F(A, G) : A E

GE

is called the compact open topology on T(X, Y), and g is a defining subbase for T. Since singleton subsets of X are compact, c4 contains the members of the defining subbase for the point open topology on T(X, Y). Thus:

Theorem 15.10:

The point open topology on T(X, Y) is coarser than the compact open topology on T(X, Y).

Recall that the point open topology is the coarsest topology with respect to which the evaluation mappings are continuous. Hence,

Corollary 15.11:

The evaluation functions er:T(X,Y) > Y are continuous relative to the compact open topology on .T(X, Y). -

TOPOLOGY OF COMPACT CONVERGENCE

Let (fi, f2, ) be a sequence of functions from a topological space X into a metric space (Y, d). The sequence (fa) is said to converge uniformly on compacta to g: X --17 if for every compact subset E C X and every E > 0, 3 no = no(E , E) E N

such that

n > no

d(f n(x), g(x)) < E, VX E E

In other words, (f.) converges uniformly on compacta to g if, for every compact subset E c X, the restriction of (to to E converges uniformly to the restriction of g to E, i.e., (1'11E, f2 E, . . .)

converges uniformly to g 1E

Now uniform convergence implies uniform convergence on compacta and, since singleton sets are compact, uniform convergence on compacta implies pointwise convergence. Example 7.1:

Let (fi, /2,

• ..)

be the sequence in T(R,R) defined by

1 f(x)

— n

-

lx1

< n

if

if

lx1

n

Now (f.,,) converges pointwise to the constant function g(x) -= 1 but (f„) does not converge uniformly to g (see Example 4.2). However, since every compact subset E of R is bounded, does converge uniformly on compacta to g.

-3

-2

CHAP.

15]

213

FUNCTION SPACES

Theorem 15.12:

Let C(X, Y) be the collection of continuous functions from a topological space X into a metric space (Y, d). Then a sequence of functions (f.) in C(X, Y) converges to g G C(X , Y) with respect to the compact open topology if and only if (J..) converges uniformly on compacta to g.

In view of the preceding theorem, the compact open topology is also called the topology of compact convergence.

FUNCTIONALS ON NORMED SPACES Let X be a normed vector space (over R). f: X --> R, is called a functional. Definition:1

A real-valued function f with domain X, i.e.

A functional f on X is linear if

(i) f(x + y) = f(x)+ f(y), V x,y

G

(ii) f(kx)= k[f(x)1, V x e X, k e R

X, and

A linear functional f on X is bounded if

3M>0

m Hxd, vx E x

lf(x) I

such that

Here M is called a bound for f. Example

8.1:

Let X be the space of all continuous real-valued functions on [a, b] with norm = sup {If(x)1 x E [a, b]}, i.e. X = C[a, N. Let I: X -4 R be defined by

II/II

I(f) = f

f dt

Then I is a linear functional; for

I(f + g)

=

f b (PO+ g(t)) dt =

S I' f(t) dt a

a

1(kf) =

J'b a

Furthermore, M

+ i

g(c) dt = 1(f) + 1(g)

a

(kf)(t) dt = j k[f (t)] dt

b

=

k f f(t) dt

k1(f)

a is a bound for I since

1(f) = f f(t) dt

M sup {PO

=

a

Proposition 15.13:

=

a

a

b—

.b

M

HfI

Let f and g be bounded linear finctionals on X and let k E R. Then f +g and k • f are also bounded linear functionals on X.

Thus (by Proposition 8.14, Page 119) the collection X* of all bounded linear functionals on X is a linear vector space. Proposition 15.14:

The following function on X* is a norm:

I If =

sup

tlf(x)1/114 :

x

o)

Observe that if M is a bound for f, i.e. !f(x)1 ---- MI!x11, V x e X, then in particular, for x 0, If(x)1/14 - - M and so Hf I — M. In fact, In could have been defined equiva-

lently by

Remark:

Hf H = inf tM : M is a bound for f) The normed space of all bounded linear functionals on X is called the dual space of X.

214

FUNCTION SPACES

[CHAP. 15

Solved Problems POINTWISE CONVERGENCE, POINT OPEN TOPOLOGY

1. Let (fi, f2, • . . ) be the sequence of functions in where I = [0, 1], defined by 4n2x —4n2x + 4n 0

(I ,

2n

R),

if 0 x 1/2n if 1/2n < x < 1/n if 1/n x 1

Show that (fa) converges pointwise to the constant function g(x) = 0. Solution: then

Now f(0) = 0 for every n G N, and so urn f(0) = g(0) = O. 3 n o G N such that 1/n 0 < x o ; hence

n >

0

f (x 0) = 0

lirri

f n (X0)

On the other hand, if x o

O

= g(X0)

Thus (f 7,) converges pointwise to the zero function.

./ Observe that

11 f n (x) dx = 1, for every n G N,

and

g(x) dx =

Thus, in this case, the limit of the integrals does not equal the integral of the limit, i.e.,

c urn f f (x) dx n-+

0

Urn

n (x) dx

.1 1:1 n

n

2. Let C(I, R) denote the class of continuous real valued functions on

I =

[0,1] with norm

r

ilf11

If(x)1 dx

Give an example of a sequence (f 1, f2, .) in C(I, R) such that f, -0 g in the above norm but (f a) does not converge to g pointwise. Solution: Let (fn) be defined by f n (x) = xn.

lirn IfH

=

Then lim f xn dx

Um 11 (n + 1) =

n-k.x

n-4, 00

Hence (fn) converges to the zero function g(x) = 0 in the above norm. On the other hand, (f x < 1 and converges pointwise (see Example 3.1) to the function f defined by 1(x) = 0 if 0 f(x) = 1 if X = 1. Note f g.

3. Show that if Y is Ti , T2, regular, or connected, then

/(X,

Y) with the point open

topology also has that property. Solution: Since the point open topology on T(X,Y) is the product topology, T(X, Y) inherits any product invariant property of Y. By previous results, the above properties are product invariant.

CHAP. 15]

4.

215

FUNCTION SPACES

Prove Theorem 15.2: Let Y be Hausdorff and let a be a subset of , Y) with the point open topology. Then the following are equivalent: (i) a is compact. (ii) a is closed and ff(x): f E cA} is compact in Y, for every x e X. Solution: By Theorem 15.1, (ii) (i) and so we need only show that (i) (ii). Since Y is Hausdorff and T2 is product invariant, T(X, Y) is also Hausdorff. Now by Theorem 11.5 a compact subset of a Hausdorff space is closed; hence c4 is closed. Furthermore, each evaluation map e x : T(X, Y) -4 Y is continuous with respect to the point open topology; hence, for each x E X,

ex [c./1]

{f(x) : f C c4}

=

is compact in Y and, since Y is Hausdorff, closed. is compact.

In other words,

{f (x): f G

cif}

= {f(x): f G

5. Prove Theorem 15.3: Let T be the point open topology on F(X, Y) and let (f1,f2, ..) be a sequence in T(X, Y). Then the following are equivalent: (i) (fo converges to g e T(X, Y) with respect to T. (ii) (fn) converges pointwise to g. Solution: Method 1. We identify T(X, Y) with the product set F = fl {Yx : x G X} and T with the product topology. Then by Theorem 12.7 the sequence (f„) in F converges to g C F if and only if, for every projection 1TX ,

= (ex (f n )) = (fo (x)) In other words,

converges to

ir(g) = ex (g) = g(x)

f„--> g with respect to T Hf lim f n (x) = g(x), VxGX

i.e. HT (f n) converges pointwise to g. Method 2.

(ii):

(i) i.e. g(x 0)

e G.

and so F(x 0 , G) to T; hence

Let x o be an arbitrary point in X and let G be an open subset of Y containing g(x 0 ), Then

g e F(xo , G) =

tf e T(X, Y) : f(x 0 ) E G}

is a T-open subset of T(X, Y) containing g. By (i), (f a) converges to g with respect 3 no C N n > no

Accordingly,

n> no

such that

fn (oco) E G

f G F(xo , G)

lim n(x0) = g(x 0 )

But x o was arbitrary; hence (fn) converges pointwise to g. (ii) contains g.

(i):

Let F(x o , G) = {f : f(x0) e G} be any member of the defining subbase for T which G. By (ii), (f ) converges pointwise to g; hence

Then g(x 0) G

3 no C N and so

n > no

n > no

such that In G F(X 0 , G)

(fn) T-converges to g

UNIFORM CONVERGENCE 6.

be a sequence of continuous functions from Prove Proposition 15.4: Let (1'1, f2, a topological space X into a metric space Y, and let (fn) converge uniformly to g: X Y. Then g is continuous. Solution:

Let x o G X

and let e > O. Then g is continuous at x o if 3 an open set G c X containing xo

such that

x E G

d(g(x),g(x 0)) <

f

216

[CHAP. 15

FUNCTION SPACES

Now JO converges uniformly to g, and so

3 mG N

such that

d(f in (x), g(x)) < se, Yoe C X

Hence, by the Triangle Inequality,

m (x), f nt (x () )) + d(f m(xo), g(xo)) < d(f m(x), f m(x0)) +

d(g(x), f nt (x))

d(g(x), g(x 0))

Since fm is continuous, 3 an open set G

C X containing xo such that

d(f ,n (x), f m (x 0 )) < We

G

x G G

and so

d(g(x), g(x 0 )) <

e

Thus g is continuous.

7.

Let (fi, f2, be a sequence of real, continuous functions defined on [a, b] and converging uniformly to g: [a, b] > R. Show that -

liM n-÷eo

f(x) dx =

a

g (x) dx

Observe (Problem 1) that this statement is not true in the case of pointwise convergence. Solution: Let e>

O. We need to show that 3 no E N

such that

n > no

fab f n (x) dx —

g(x) dx1 < e a

Now (fa) converges uniformly to g, and so 3 no G N such that n (x) — g(x)I < el(b — a),

n> no

jab Hence, if n > no ,

V x G [a, b]

çb

(f n (x) — g(x)) dx

f n (x) dx — f g(x) dx a

fb a

Prove Theorem 15.5:

g(x)1 dx

el(b — a) dx

f 8.

Ifn (X)

) be a sequence in 473(X, Y) with metric

Let (f i, f2,

e(f,, g) = sup td(f (x), g (x)) : x e X) Then the following are equivalent: (ii) (fn) converges uniformly to g.

(i) (f 7,) converges to g E T(X, Y) with respect to e.

Solution:

(i)

(ii):

Let e > O. Since (f a) converges to g with respect to e,

3 no C N

such that

e(f, n, g) <e

n> no

Therefore,

n > no

d(f n (x), g(x))

sup {d(f n (x), g(x)) x

= e(f n, g) <e, VxCX

that is, (f,) converges uniformly to g.

(ii)

(i):

Let e > O. Since (fn) converges uniformly to g,

3 no C N Therefore,

such that

n> no

n > no

d(in(x),g(x)) < e/2, VxE X

sup {d(f n (x), g(x)) : x

e

e/2 < e

n , g) < 6, and so JO converges to g with respect to e. that is, n > no implies e(f,

CHAP. 151

217

FUNCTION SPACES

THE FUNCTION SPACE C[0, 1] 9. Prove Proposition 15.6: Let f:I->ft be continuous on I= [0, 11. Then for every > 0,

Ix-

such that

3 8 = 8 (E) > 0

<

i.e. f is uniformly continuous. Solution: Let e > O. Since f is continuous, for every p G I,

3 S p >0

Ix — pi < sp

such that

If(x) —

f(P)I <

(1)

For each p G I, set S p = I n (p — p +18 p). Then {S p : p G I} is an open cover of I and, since I is compact, a finite number of the Sp also cover I; say, I = Sp1 U • • • U S. Set 5 = I min

Suppose 1QC

YI

< 8. Then

x

<

Y Pki : y — xl + XpkI

Hence by (1),

8 + 4-8,k

and

f(x) — f(POI < 4-E

< 8 1,k and

<

S p?, for some k, and so ix

8 Pk 8 i

Pk = 8 Pk

f(y) f(POI <

Thus by the Triangle Inequality, 11(x)

11(x) —

f(Pk) + If(Pk) — RP)

) be a Cauchy sequence in C[0, 1 1. is a Cauchy sequence in R. (f i(xo), f2(x0),

10. Let (fi, f2, Solution: Let xo

e I and let

e > 0.

Show that, for each xo e I = [0, 1],

Since (1,) is Cauchy, 3 no C N such that

m, n > no

— fil

=

sup (1/.(x) — f m (x)1: x e

<

Ifn( X0) fm ( X0)1 < e Hence 1 n(x0)> is a Cauchy sequence.

11. Prove Theorem 15.7: C[0, 1] is a complete normed vector space. Solution: Let 41,12, . ) be a Cauchy sequence in C[0,1]. Then, for every x o e I, (f i,(x 0 )) is a Cauchy Define g: I -4 R by g(x) lim f n (x). Then sequence in R and, since R is complete, converges. n—* .o

(see Problem 32) (f.) converges uniformly to g. g C CEO, 1 1 ;

But, by Proposition 15.4, g is continuous, i.e.

hence C[0,1] is complete.

12. Let f E C[0, 1] and let E > 0. Show that 3 no E N

and points Po = (0, €163/5), • • • pi

= (i/n o, dei/5),

pno = (1, €1c7,0/5) where ko, , kn o are integers such that, if g is the polygonal arc connecting the pi , then < € (see adjacent diagram). In other words, the piecewise linear (or polygonal) functions are dense in C[0, 1 1.

—e/5 —2e/5

Solution: Now f is uniformly continuous on [0,1] and so

3 n0 EN

such that

la —

1/n0

( — f ( b)I <€15

(1)

218

FUNCTION SPACES

Consider the following subset of I

yi) e

pi =

Then

R:

y = kel5 where j: 0, ..., no; k E Z1

-= {(x, y ) : x =

A

Choose

x

such that

A

[CHAP. 15

Yi

f(xi) < yi + €15 and by (1),

Iftr0 — g(01 = If(x) — y < f 1 5

f(x)

f(xi+1)1 < e15

as indicated in the diagram above. Observe that

If(x i)—f(x i+1)1+1f(xi+i)— g(x i+

g(xt)— f(x i)1

< e15

e15

e15 = 3e/5

Since g is linear between x i and xi+1 , xi

z

Now for any point zJ

11(z) — g(z) i

,

g(x i+i)I < 3e/5

Ig(xi) — g(z)i

xi+

x k satisfying x k z xk +1 .

Hence

f (x0I + If(xk) g(xk)I + Ig(x) - g(z) I < E/5 point in I; hence < E. — 1 1(z)

But z was an arbitrary

€15 + 3e/5 =

13. Let m be an arbitrary positive integer and let Am c C[0, 1] consist of those functions f with the property that If(xo + h) f(x0) m, y h 1 3 x o e 0, 1 – — such that G 0, 1) mj h Show that Am is a closed subset of C[0,1]. (Notice that every function f in C[0,1] which is differentiable at a point belongs to some A m for m sufficiently large.) –

Solution: We want to show that g E A m , i.e. A m = A m . Since g G A m , there exists a sequence !fi, 12, •..) in A m converging to g. Now for each fi there exists a point x i such that Let

g E A m.

xi e But

Lo,

– 9J1i

1

and

(x,i) is a sequence in a compact set

fi (x i h)

fi (x)

h

1 0, 1 — —

m, V h E (0, -1)

(1)

and so has a subsequence (x in) which converges

1—

to, say, x o [0, 1 — —

Now

g implies fin —> g, and so (Problem 30), passing to the limit in (1), gives

g(x 0 + h) — g(x 0)

Vh

(0, 272 1)

Hence g E A m , and A m is closed.

14. Let Am C C[0,1] be defined as in Problem 13. C[0, 1]. Solution: A m is nowhere dense in C[0,1] HT int (A)

=

int (A)

Show that Am is nowhere dense in

=.

Let S -= S(f, 8) be any open sphere in C[0,11. We claim that S contains a point not belonging to A m , and so int (A m )

By Problem 12, there exists a polygonal arc p G C[0,1] such that IV — PH < l. Let g be a saw-tooth function with magnitude less than l and slope sufficiently large (Problem 33). Then the function h -= g belongs to C[0, 1] but does not belong to Am . Furthermore,

lf– hH

l if — pH +1 gll < 8 +FS =

so h G S and the proof is complete.

Iiiiiiiii/iiAliAtAiiiiiiiiiiiAi

CHAP.

15]

219

FUNCTION SPACES

15. Let A. C C[0, 11 be defined as in Problem 13.

u:=, A..

Show that C[0, 1]

Solution: Since A m is nowhere dense in C[0,1], B =U:=1 A m is of the first category. But, by Category Theorem, C[0,1], a complete space, is of the second category. Hence C[0, 1 1 B.

There exists a continuous function

16. Prove Proposition 15.8: nowhere differentiable. Solution: Let f

G C[0, 1 1 have a derivative at, say, xo and suppose Inx0)1 3 €> 0

f(x o + h) — f (x 0) h

such that

t.

R which is

Then

1, V h

Now choose mo G N so that t 1 mo and 1/m0 < E. Then f all functions which are differentiable at some point of I.

EAm .

u:_ 1 A m contains

Thus

u: =1 A m and so there exists a function in C[0,1] which

C[0,1]

But by the preceding problem, is nowhere differentiable.

f: [0, 11

Baire's

17. Prove Theorem (Ascoli) 15.9: Let a be a closed subset of C[0, 1]. Then the following are equivalent: (i) eA is compact. (ii) a is uniformly bounded and equicontinuous. Solution:

(i) (ii): Since e4 is compact it is a bounded subset of C[0,1] and is thus uniformly bounded as a set of functions. Now we need only show that cA is equicontinuous. Let e > O. Since c.4 is compact, it has a finite

3 fio C 4/3 Therefore, for any x,y E

If(x)

/ =- [0,1], f 0 (y) + f 0 (y)

I11(x) — f(x) + fio (x)

f(y)

f(x)

fio (x)I + fio (x) (x)

—

fio

fio (Y)I +

e/ 3

f(Y) I

+ Ifio (y) — f(Y)1 = ifio (x)

f0(y) + 2e1 3

fi G 93 is uniformly continuous and so 3 (S i > 0

8= min {8 1 , ...,8 t}• Ix

Thus

=

—

e/3

Set

ft}. Hence, for any f E cA,

{f 1 ,

I If fio ll--- sup tlf(x) — fio (x)1: x

such that

=

Now each

e/3-net, say, 93

such that

Then, for any f f(Y)1

<8

Ix

— yI< Si

Ifi(x) — fi(Y)I < e/3

E c.4 ,

If io (x)

fi o (Y)1 + 2e 1 3 < e/3

2E/3

c4 is equicontinuous.

(ii) (i): Since cA is a closed subset of the complete space is totally bounded. Let e > O. Since cA is equicontinuous, 3 n0 E N

such that

abl < 1/no

C[0,1], we need only show that cA

I f (a) — f (b)1 < e15, V f E cA

Now for each f G c4, we can construct, by Problem 12, a polygonal arc pf such that Ilf — P f t p f connects points belonging to

A

<

and

= {(x, y) : x= 0, 1/no, 2/N, ..., 1; y = 71 E15, n Z}

G cAl is finite and hence a finite e-net for c-4. Now cA is uniformly bounded, and so 93 is uniformly bounded. Therefore only a finite number of

We claim that 93 = fp f f

the points in A will appear in the polygonal arcs in 'B. Hence there can only be a finite number of arcs in 93. Thus ‘13 is a finite E-net for c-1, and so cA is totally bounded.

[CHAP. 15

FUNCTION SPACES

220

COMPACT CONVERGENCE 18. Let (fi, f, ...) in T(R, R) be defined by fn(x) =

1 --i 2 lx1

0

if Ix' < n if xl ---- n

Show that (fa) converges uniformly on compacta to the constant function g (x) = 1. Solution: Let E be a compact subset of R and let E c (—M, M) for M > O. Now

3 no G N Therefore,

such that

i..(x)__

n > no

0 < e < 1.

Since E is compact, it is bounded; say,

no > M/e, or, Mini) < E

g(x)1 = n1 X <Mlno < c,

lixEE

Hence (f a ) converges uniformly to g on E.

19. Show: If 1 is Hausdorff, then the compact open topology on T(X, Y) is also Hausdorff. Solution: Method 1. Let f,g E T(X, Y) with f g. Then 3 p e X such that f(p) g(p). Now Y is Hausdorff, hence 3 open subsets G and H of Y such that f(p) G G, g(p) G H and G n H = Ø. Hence

f e F(p, G), g E F(p, H)

and

F(p, G) n F(p, H) = 0

But the singleton set {p} is compact, and so F(p, G) and F(p, H) belong to the compact open topology on T(X, Y). Accordingly, y,•(x, Y) is Hausdorff.

Method 2. The compact open topology is finer than the point open topology, which is Hausdorff since T2 is a product invariant property. Hence the compact open topology is also Hausdorff.

20. Prove Theorem 15.12: Let (fi, f2, ...) be a sequence in C(X, Y), the collection of all continuous functions from a topological space X into a metric space (Y, d). Then the following are equivalent: (i) yn) converges uniformly on compacta to g C C(X, Y). (ii) (f 71) converges to g with respect to the compact open topology T on C(X, Y). Solution:

(1)

(if):

Let F(E , G) be an open subbase element of T containing g; hence g[E] c G where E is compact and G is open. Since g is continuous, g[E] is compact. Furthermore, g[E] n Ge = 0 and so (see Page 164) the distance between the compact set g[E] and the closed set GC is greater than zero; say, d(g[E],Gc)-= c > O. Since (fn) converges uniformly on compacta to g,

3 no G N

such that

n > no

d(f.„(x),g(x)) < e,

d(f.„(x),g[E]) -- d(fn (x),g(x)) < e,

Therefore,

VxGE

VxGE

and so, for every x G E, f n (x) (Z G. In other words,

n> no

f.n [E] cG

f n e F(E, G)

Accordingly, (fn) converges to g with respect to the compact open topology T.

(ii)

(i): Let E be a compact subset of X and let E > O. We want to show that (fo) converges uniformly

on E to g, i.e.,

3 no G N

such that

n > no

d(f n (x),g(x)) < c,

VxGE

CHAP. 15]

221

FUNCTION SPACES

Since E is compact and g is continuous, g[E] is compact. Let for g[E]. Consider the open spheres

S1 =

S(Pi, ei3),

S(Pt, E/3)

•, St

Hence 1cG1.....ScG 1 .

Gt

S(p t , 2e13)

Furthermore, since 93 is an E/3-net for g[E],

g[E] c gi u ••• ugt E i = E n g1 [1

Now set

Gi = S(P1, 261 3),

and

be a finite c/3-net

93 = {pi, ...,pt}

E c g - lrg]u • • u g- lrSd

and so

E = E l u • • • u Et and g[E] C C

and so

compact. For g is continuous and so g -1 [gi], the inverse of a closed set, is We claim that the E. closed; hence Ei = E n g -1 [gib the intersection of a compact and a closed set, is compact.

g[E11 c G, and so the F(E„ GO are T-open subsets of T(X, Y) containing g; hence F(E„ G1 ) is also a T-open set containing g. But (fn ) converges to g with respect to T; hence

Now

3 no E N Now let x E E.

G n =1 F(Ei,G,)

n > no

such that

Then x G E10 and so, for n>

f n [E 1 ] cG1 , .

no, d(f(x), p10) < 2c/3

f n (x) E f n [Eto ] C Gio g(x) G g[E10 ] c S

and

fn [Et ]cGt

E/3

Therefore, by the Triangle Inequality,

n > no

d(p io , g(x)) < 2613

d(fn (x),p, o )

d(f 7, (x), g(x))

FUNCTIONALS ON NORMED SPACES 21. Show that if f is a linear functional on X, then f(0)

E, Vx EE

E/3

O.

Solution t Since f is linear and 0 = O ± 0, f( 0 ) = f (0 + 0) = f (0)

f(0)

Adding — f(0) to both sides gives f(0) = O.

22. Show that a bounded linear functional f on X is uniformly continuous. Solution: Let M be a bound for f and let e > O. X y,

Set = 61M,

< s

Then y)\

f(Y)

M

<

23. Prove Proposition 15.13: Let f and g be bounded linear functionals on X and let c E R. Then f + g and c • f are also bounded linear functionals on X. Solution: Let M and M* be bounds for f and g respectively. Then

(f g)(x + y) = f(x ± y) ± g(x + y) = f(x) f(y) g(x) (f g)(kx) = f(kx)

k 1(x)

g(kx)

g)(x)i = f(x)+g(x)

LE

f(x)

Thus f g is a bounded linear functional.

g(y) = (f g)(x)

(f g)(y)

k g(x) = k[f(x) g(x)] = k (f g)(x)

Ig(x) 1

M !x11

M* x = (M M*)

222

[CHAP. 15

FUNCTION SPACES

Furthermore,

(c • f)(x +

f(x + y) = c [f(x) + (y)] = c f (x) + c f (y) = (c • f)(x) + (c • f)(y) (c • f)(kx) = e f(kx) = ck f(x) = kc f(x) = k (c • f)(x) (e • f)(x) = te f(x)f = !cl !Ave)!

(*if) = (H 111)fxH

and so c • f is a bounded linear functional.

The following function on X* is a norm:

24. Prove Proposition 15.14:

= sup {f(x)/x: x Solution: If f -= 0, then f(x) 0, V x E X, and so if 0, and so that f(x 0 ) sup {if(x)InxIl} 11f11

= sup {0} -= 0.

If f

0, then

3 xo 0 such

0

Thus the axiom [Ni ] (see Page 118) is satisfied. Now

I

k

• ill

=

sup sup

I /1Ix I 1 {Vel If(x)7114}

sup {1 1c[f(x)]i/lixlil

{I(k•f)(x)

=

1.1c

sup {if(x)A40

Hence axiom [N2 ] is satisfied. Furthermore, sup { 1(x) + g(41/11x

Ilf +g11

11

sup {01*(4 + Ig(x)i)/11x11}

sup {1.((x)1/11xil} + sup flg(x)1/11x0

11111 + 110

and so axiom [N3 ] is satisfied.

Supplementary Problems CONVERGENCE OF SEQUENCES OF FUNCTIONS

(f 1 ,12 , ...) f n (x) = x' /n.

25. Let

be the sequence

of real-valued functions with domain

Show that (fn) converges pointwise to the constant function lim f n (x) O.

I = [0,1]

g(x) = 0,

defined by

Le. for every x G I,

-7 -

n

(ii)

Show that

d fn(x) ax

n

d lim, ax

lim f n (x)

n .

26. Let (fi, f 2 , .) be a sequence of real-valued differentiable functions with domain [a, bl which converge uniformly to g. Prove:

d

.

d fn(x)

fn(x)

(Observe, by the preceding problem, that this result does not hold in the case of pointwise convergence.)

27. Let f n :

R 4R be defined by -

n 1

f„(x) 0

vn2 _ x2

if xl < n if lx1

(i)

Show that (In) does not converge uniformly to the constant function g(x) = 1.

(ii)

Prove that (fa ) converges uniformly on compacta to the constant function g(x) = 1.

CHAP. 15 1

223

FUNCTION SPACES

28. Let (f 1 , f2, .) be the sequence of functions with domain I = [0,1] defined by fn(x) = nx(1 — x)n. (i)

Show that (f n ) converges pointwise to the constant function g(x) = 0.

(ii)

Show that (f.„) does not converge uniformly to g(x) = O.

(iii)

Show that, in this case, ci

lim 501 f n (x) dx -=

Urn f n (x)idx [

n+1

29. Let (f1, f2, • ) be the sequence in T(R, R) defined by f n (x)

30.

X.

(i)

Show that (f a) converges uniformly on compacta to the function

(ii)

Show that (f„) does not converge uniformly to g(x) x.

Let (f 1 , 2 , ...) be a sequence of (Riemann) integrable functions on said to converge in the mean to the function g if

lim

0

If, (x)

g(x)1 2 dx

g(x) = x.

[0, 1].

I

The sequence (f n ) is

0

Show that if (f n) converges uniformly to g, then (f„) converges in the mean to g.

(ii)

Show, by a counterexample, that convergence in the mean does not necessarily imply pointwise convergence.

THE FUNCTION SPACE C[0,1] 31.

Show that C[a, b] is isometric and hence homeomorphic to C[0, 1].

32.

Prove: Let (f„) converge to g in C[0,11 and let xn x o .

33.

Let p be a polygonal arc in C[0, 1] and let 8 > O. Show that there exists a sawtooth function g with magnitude less than FS, i.e. I < IS, such that p+ g does not belong to A 1 (see Problem 14).

34.

Let (f n ) be a Cauchy sequence in C[0,1] and let (f a) converge pointwise to g. Then (f a) converges uniformly to g.

Then

lirn f n (x n)

g(x 0 ).

OA

IiiiiiilAkiitilAAALAILIL1Al

UNIFORM CONTINUITY is not uniformly continuous on the open interval (0, 1).

35.

Show that f(x)

36.

Define uniform continuity for a function

37.

Prove: Let f be a continuous function from a compact metric space X into a metric space Y. Then f is uniformly continuous.

f:X—>Y

where X and Y are arbitrary metric spaces.

FUNCTIONALS ON NORMED SPACES 38.

Let f be a bounded linear functional on a normed space X. sup {If(x)1/114

x

0 } = inf {M

Show that M is a bound for f}

39.

Show that if f is a continuous linear functional on X then f is bounded.

40.

Prove: The dual space X* of any normed space X is complete.

APPENDIX Properties of the Real Numbers FIELD AXIOMS The set of real numbers, denoted by R, plays a dominant role in mathematics and, in particular, in analysis. In fact, many concepts in topology are abstractions of properties of sets of real numbers. The set R can be characterized by the statement that R is a complete, Arch,imedian ordered field. In this appendix we investigate the order relation in R which is used in defining the usual topology on R (see Chapter 4). We now state the field axioms of R which, with their consequences, are assumed throughout the text. Definition:1 A set F of two or more elements, together with two operations called addition (+) and multiplication ( • ), is a field if it satisfies the following axioms: [ A l]

Closure: a, b EF

a+bEF

[A2]

Associative Law: a, b, c E F

[A3]

3 OEF such that 0 + a = a +0 = a, Va E F 3 —a E F such that a + (—a) = (—a) + a = 0 (Additive) Inverse: a E F a+b=b+a Commutative Law: a,bEF a•bEF Closure: a,bEF Associative Law: a, b, c E F (a • b) • c = a • (b • c) (Multiplicative) Identity: 3 1 E F, 1 0 such that 1 • a = a • 1 = a, Va EF 3 a - E F such that a • a -1 = a - ' • a = 1 (Multiplicative) Inverse: a E F, a 0 a•b --= b • a Commutative Law: a, b EF a•(b+c) Left Distributive Law: a, b,c E F a•b + a•c (b+c)•a = b•a + c•a Right Distributive Law: a,b,c E F

(a + b) + c = a + (b + c)

(Additive) Identity:

[A4] [A5] [MI]

[Md [M3] [M4]

IM 5 1 [D1] [D2]

Here 3 reads "there exists", V reads "for every", and

reads "implies".

The following algebraic properties of the real numbers follow directly from the field axioms. Proposition Ai: Let F be a field.

Then:

(i)

The identity elements 0 and 1 are unique.

(ii)

The following cancellation laws hold:

b=c,

(1) a+b=a+c

(2) a•b=a•c,a0

(iii)

The inverse elements —a and a -1 are unique.

(iv)

For every a, b E F,

b=e

(1) a • 0 = 0, (2) a • (—b) = (—a) • b = —(a • b), (3) (—a)• (—b) = a • b Subtraction and division (by a non-zero element) are defined in a field as follows:

b — a - -E b + (—a)

and

225

— a

b • a -- '

PROPERTIES OF THE REAL NUMBERS

226

Remark:

[APPENDIX

A non-empty set together with two operations which satisfy all the axioms of a field except possibly Ind, IM41 and 1M..5 1 is called a ring. The set Z of integers under addition and multiplication, for example, is a ring but not a field.

REAL LINE We assume the reader is familiar with the geometric representation of R by means of points on a straight line as in the figure below. Notice that a point, called the origin, is chosen to represent 0 and another point, usually to the right of 0, is chosen to represent 1. Then there is a natural way to pair off the points on the line and the real numbers, i.e. each point will represent a unique real number and each real number will be represented by a unique point. For this reason we refer to R as the real line and use the words point and number interchangeably.

0

2

The real line R

SUBSETS OF R The symbols Z and N are used to denote the following subsets of R: Z = { . . ., —3, —2, —1, 0, 1, 2, 3, . . . } ,

N = 11, 2, 3, 4, . . .1

The elements in Z are called rational integers or, simply, integers; and the elements in N are called positive integers or natural numbers. The symbol Q is used to denote the set of rational numbers. The rational numbers are those real numbers which can be expressed as the ratio of two integers provided the second is non-zero: Q= Ix ER: x p/q; p,q EZ, 0} Now each integer is also a rational number since, e.g., —5 = 5/-1; hence Z is a subset of Q. In fact we have the following hierarchy of sets: N C Z CQ CR The irrational numbers are those real numbers which are not rational; thus Qc, the complement ( relative to It) of the set Q of rational numbers, denotes the set of irrational numbers.

POSITIVE NUMBERS Those numbers to the right of 0 on the real line R, i.e. on the same side as 1, are the positive numbers; those numbers to the left of 0 are the negative numbers. The following axioms completely characterize the set of positive numbers: Wu] If a G R, then exactly one of the following is true: a is positive; positive. [P2]

If a, b e R are positive, then their sum positive.

a+b

a = 0; —a is

and their product a • b

It follows that a is positive if and only if —a is negative.

are also

227

PROPERTIES OF THE REAL NUMBERS

APPENDIX]

Example 1.1:

We show, using only [P1 ] and [P2 ], that the real number 1 is positive. By [P1 ], either 1 or —1 is positive. Assume that —1 is positive and so, by [P2 ], the product (-1)(-1) = 1 is also positive. But this contradicts [P1 ] which states that 1 and —1 cannot both be positive. Hence the assumption that —1 is positive is false, and 1 is positive.

Example 1.2:

The real number —2 is negative. For, by Example 1.1, 1 is positive and so, by [P2 ], the sum 1 +1 = 2 is positive. Therefore, by [P 1 ], —2 is not positive, i.e. —2 is negative.

Example 1.3:

We show that the product a • b of a positive number a and a negative number b is negative. For if b is negative then, by [P1 ], —b is positive and so, by [P2 ], the product a • (—b) is also positive. But a • (—b) = —(a • b). Thus —(a • b) is positive and so, by [P 1 ], a • b is negative.

ORDER We define an order relation in R, using the concept of positiveness.

Definition: I The real number a is less than the real number b, written a < b, if the difference b — a is positive. Geometrically speaking, if a < b then the point a on the real line lies to the left of the point b.

The following notation is also used: b > a, read b is greater than a, a

b,

means a < b means a < b or a=b means a b b

read a is less than or equal to b, read b is greater than or equal to a,

a, Example 2.1:

2 < 5; —6 :FL —3; 4 4; 5 > —8

Example 2.2:

A real number x is positive if x > 0, and x is negative if x < O.

Example 2.3:

The notation 2 G x < means 2 < ancl also x < I; hence x will lie between 2 and 7 on the real line.

The axioms [Pi] and [P2 ] which define the positive real numbers are used to prove the following theorem. Theorem A.2: Let a, b and c be real numbers. Then: (i) either a < b, a = b or b < a; (ii) if a < b and b < c, then a < c; (iii) if a < b, then a + c < b + c; (iv) if a < b and c is positive, then ac < be; and (v) if a < b and c is negative, then ac > be. Corollary A.3:

The set R of real numbers is totally ordered by the relation a

b.

ABSOLUTE VALUE The absolute value of a real number x, denoted by xj is defined by ,

Ix!

if x 0 1—x if x < 0 x

Observe that the absolute value of any number is always non-negative, i.e. every x E R.

Ix!

0 for

228

PROPERTIES OF THE REAL NUMBERS

[APPENDIX

Geometrically speaking, the absolute value of x is the distance between the point x on the real line and the origin, i.e. the point 0. Furthermore, the distance between any two points a, b E R is 'a — bl = lb — al.

!—Ifi I

2, l I = 7, 1, —H =

Example 3.1:

I

Example 3.2:

13-8

Example 3.3:

The statement Ix < 5 can be interpreted to mean that the distance between x and the origin is less than 5; hence x must lie between —5 and 5 on the real line. In other words, <5 and —5 < x < 5

1 -5 = 5 and

8 — 3 = 151 = 5

have identical meaning and, similarly,

5

and

—5 = x -= 5

have identical meaning.

The graph of the function f(x) = x i.e. the absolute value function, lies entirely in the upper half plane since f(x) 1.- 0 for every x E R (see diagram below). ,

-2

Graph of f(x)

Ix

The central facts about the absolute value function are the following: Proposition A.4: Let a, b and c be real numbers. Then: al 0, and a =0 if a = 0; ab =

lb a + 1b1;

a+b

a

a—b

)a — cl

b ; and

la —1) 1 + )b — cl.

LEAST UPPER BOUND AXIOM Chapter 14 discusses the concept of completeness for general metric spaces. For the real line R, we may use the definition: R is complete means that R satisfies the following

axiom: [LUB1

(Least Upper Bound Axiom): If A is a set of real numbers bounded from above, then A has a least upper bound, i.e. sup (A) exists. Example 4.1:

The set Q of rational numbers does not satisfy the Least Upper Bound Axiom. For let A = {qEQ: q> 0, q2 < 2} i.e., A consists of those rational numbers which are greater than 0 and less than Now A is bounded from above, e.g. 5 is an upper bound for A. But A does not have a least upper bound, i.e. there exists no rational number m such that ni sup (A). Observe that m cannot be A/i since Vi does not belong to Q.

a

PROPERTIES OF THE REAL NUMBERS

APPENDIX]

229

We use the Least Upper Bound Axiom to prove that R is Archimedean ordered: Theorem (Archimedean Order Axiom) A.5:

The set N = (1, 2,3, ... } of positive integers

is not bounded from above. In other words, there exists no real number which is greater than every positive integer. One consequence of this theorem is:

Corollary A.6:

There is a rational number between any two distinct real numbers.

NESTED INTERVAL PROPERTY The nested interval property of R, contained in the next theorem, is an important consequence of the Least Upper Bound Axiom, i.e. the completeness of R.

Theorem (Nested Interval Property) A.7:

Let I = [al, b1], 12 = raz, b21, . .. be a sequence of nested closed (bounded) intervals, i.e. II D D . . . . Then there exists at least one point common to every interval, i.e. n /,

It is necessary that the intervals in the theorem be closed and bounded, or else the theorem is not true as seen by the following two examples. Example 5.1:

Let A 1 , A2, . , be the following sequence of open-closed intervals: A 1 = (0, 1], A2 = (0,1/2],

•

Ak= (0,1/k],

Now the sequence of intervals is nested, i.e. each interval contains the succeeding interval: A 1 D A2 D • • • . But the intersection of the intervals is empty, i.e., A,

n A2 n • • • n Ak n • • • =

Thus there exists no point common to every interval. Example 5.2:

Let A i , A2,

..

be the following sequence of closed infinite intervals: A 1 =-- [1, 00), A2 = [2, .0),

A,, = [k, ) ,

Now A 1 D A 2 D • • •, i.e. the sequence of intervals is nested. point common to every interval, i.e., A, n A 2

.

But there exists no

n • • • n Ak n • • • = (2)

Solved Problems FIELD AXIOMS

I. Prove Proposition A.1(iv):

For every a, b e F,

(1) a0 = 0, (2) a(—b) = (—a)b = — ab, (3) (—a)(—b)=-- ab Solution: (1)

a0 = a(0 0) = a0 a0.

(2)

0 = a0

Adding —a0 to both sides gives

0 = a0.

a(b (—b)) = ab ± a(—b). Hence a(—b) is the negative of ab, that is, a(—b) = — ab.

Similarly, (—a)b =- —ab. (3)

0 = (—a)0 --= (—a)(b (—b)) = (—a)b (—a)(—b) = — ab (—a)(—b).

gives ab

(—a)(—b).

Adding ab to both sides

230

PROPERTIES OF THE REAL NUMBERS

[APPENDIX

2. Show that multiplication distributes over subtraction in a field F, i.e. a(b — c) = ab — ac. Solution:

a(b — c) = a(b + (—c)) = ab + a(—c) = ab + (—ac) = ab — ac

3. Show that a field F has no zero divisors, i.e. ab = 0

a =.0 or b = 0.

Solution: Suppose ab = 0 and a

O. Then a -1 exists and so

b = lb = (a -1 a)b = a -1 (ab) = a -1 0 = O.

INEQUALITIES AND POSITIVE NUMBERS 4. Rewrite so that x is alone between the inequality signs: (i) 3 < 2x-5 < 7, (ii) —7 < —2x+3 < 5. Solution: We use Theorem A.2:

5.

(i)

By (iii), we can add 5 to each side of 3< 2x — 5 <7 to get 8< 2x <12. By (iv), we can multiply each side by 4 to obtain 4 < x < 6.

(ii)

Add —3 to each side to get —10 < —2x < 2. reverse the inequalities to obtain —1 < x < 5.

By (N), we can multiply each side by — 4 and

Prove that 4 is a positive number. Solution: By [P1 ], either — 4 is positive or 4 is positive. Suppose — 4 is positive and so, by [P2 ], (— + (— —1 is also positive. But by Example 1.1, 1 is positive and not —1. Thus we have a contradiction, and so is positive.

6. Prove Theorem A.2(ii):

If a
Solution: By definition, a < b means b — a is positive; and b < c means c — b is positive. Now, by ] c — a is positive and so, by definition, a < c. , [P2 the sum (b — a) + (c — b)

7. Prove Theorem A.2(v):

If a < b and c is negative, then ac > be.

Solution: By definition, a < b means b — a is positive. By [P 1 ], if c is negative then —c is positive, and so, by [P2], the product (b — a)(—c) = ac — be is also positive. Hence, by definition, be < ac or, equivalently, ac > bc.

8. Determine all real numbers x such that (x —1)(x +2) < 0. Solution: (x — 1)(x + 2) We must find all values of x such that y is negative. Since the product of two numbers is negative if one is positive and the other is negative, y is negative if

(i) x — 1 < 0 and x + 2 > 0, or (ii) x — 1 > 0 and x + 2 < 0. If x 1 > 0 and x 2 < 0, then x > 1 and x < —2, which is impossible. Thus y is negative if x — 1 < 0 and x + 2 > 0, or x < 1 and z > —2, that is, if —2 < x < 1. Observe that the graph of y ----- (x — 1)(x + 2) crosses 1 and x = —2 (as shown on the right). the x-axis at x Furthermore, the graph lies below the x-axis if y is negative, that is, if —2 < x < 1.

APPENDIX]

231

PROPERTIES OF THE REAL NUMBERS

ABSOLUTE VALUES

9.

(i) 1 3 + —7, (ii) 1-1 — 4 — 3 — 3 — 5

Evaluate:

,

(iii)

1-21 — —6

Solution:

(i) (ii)

1 1 3 + 7 = —2 + 1-71 = 2 + 7 = 9 1 -1-4 —3-3-5 1 1-51 — 3 — 1-21 = 5 — 3 — 2

(iii)

11 - 21 — 1 - 611 = 12 — 61 = 1 - 41 = 4

10. Rewrite without the absolute value sign:

0

(i) lx — 2 < 5, (ii) 2x + 3 <7.

Solution:

(i) (ii)

—5 < x — 2 < 5 or —3 < x < 7 —7 < 2x + 3 < 7 or —1 0 < 2x < 4 or —5 < x < 2

11. Rewrite using the absolute value sign:

(i) —2 < x < 6, (ii) 4 < x < 10.

Solution: First rewrite each inequality so that a number and its negative appear at the ends of the inequality: (i)

Add —2 to each side of x -2 <4.

—2 < x < 6 to obtain

—4 < x — 2 < 4 which is equivalent to

(ii)

Add —7 to each side of x - 71 < 3.

4 < x < 10 to obtain

—3 < x 7 < 3 which is equivalent to

12. Prove Proposition A.4(iii): a + bl

a

b 1.

Solution:

Method 1. Since

= ± a,

al

=

also —lb

a -=

+ lb )

la +b

Therefore, since I al +

bI

b

Ibl.

a + b

I lal + Ibl I

Then, adding,

lal + lb al + Ibl

0.

Method 2. Now ab lab (a But

= la l Ib l

implies 2ab

2 la lbl, and so

a2 + 2 la Ibl + b 2 =- 1 a 1 2 + 2 al Ibl + I bI 2 =

+ b) 2 =- a2 + 2ab + b2

la + i b l.

),/(a + b) 2 = la + bl and so, by the square root of the above, la + b

13. Prove Proposition A.4(v): Solution:

la —cl

a—c

a—b +b— c

=

(a — b) + (b — c)I = la — b

+

b

(lai + I b 1) 2

—

LEAST UPPER BOUND AXIOM 14. Prove Theorem (Archimedean Order Axiom) A.5: is not bounded from above.

The subset N = {1, 2, 3, ... } of R

Solution: Suppose

b = sup (N).

N is bounded from above. By the Least Upper Bound Axiom, sup (N) exists, say Then b — 1 is not an upper bound for N and so 3 no E N such that b — 1 < n o or b < no + 1

But n o G N implies n o + 1 E N, and so b is not an upper bound for N, a contradiction. not bounded from above.

Hence N is

232

PROPERTIES OF THE REAL NUMBERS

[APPENDIX

15. Prove: Let a and b be positive real numbers. Then there exists a positive integer no e N such that b < n oa. In other words, some multiple of a is greater than b. Solution: Suppose no does not exist, that is, na < b for every n G N. Then, since a is positive, n < bla for every 7/ E N, and so bla is an upper bound for N. This contradicts Theorem A.5 (Problem 14), and so n o does exist.

16. Prove: If a is a positive real number, i.e. 0 < a, then there exists a positive integer no E N such that 0 <1/no < a. Solution: Suppose n o does not exist, i.e. a 1/n for every n E N. Then, multiplying both sides by the positive number nla, we have n 1/a for every n G N. Hence N is bounded by 1/a, an impossibility. Consequently, n o does exist.

17. Prove Corollary A.6: There is a rational number q between any two distinct real numbers a and b. Solution: One of the real numbers, say a, is less than the other, i.e. a < b. If a is negative and b is positive, then the rational number 0 lies between them, i.e. a < 0 < b. We now prove the corollary for the case where a and b are both positive; the case where a and b are negative is proven similarly, and the case where a or b is zero follows from Problem 16. Now a < b means b

a is positive and so, by the preceding problem,

3 n 0 EN

0< 1/n0 < b — a or

such that

a + (1/n o) < b

We claim that there is an integral multiple of no which lies between a and b. Notice that 1/n 0 < b since 1/n 0 < a + (1/n 0) < b. By Problem 15, some multiple of 1/n 0 is greater than b. Let mo be the least positive integer such that m0/n0 b; hence (mo — 1)/n 0 < b. We claim that a<

mo — 1

Otherwise

a

no

< b

mo mo — 1 1 + — = no no no

and so

no

mo — 1


Thus (mo — 1)/n 0 is a rational number between a and b.

which contradicts the definition of mo.

NESTED INTERVAL PROPERTY 18. Prove Theorem A.7 (Nested Interval Property): Let I = [al, b1], a sequence of nested closed (bounded) intervals, i.e. I D I, D • • • . at least one point common to every interval. Solution: Now /1

D

1 2 D • • • implies that a l

a2

• • • and

am < b n for every

•

p

b7,

b1.

We claim that

m,n E N

for, m > n implies a m < b m = b, and m n implies am bound for the set A 7-- {a 1 , a2, , ,} of left end points. By b., for every n sup (A) exists; say, p = sup (A). Now p for A and p is the least upper bound. Furthermore, a n p bound for A = a2, . .}. But an

b2

[a2, b2], „ be Then there exists

pE

a , < b.. Thus each b. is an the Least Upper Bound Axiom C N, since each b n is an upper for every n G N, since p is an

upper of R, bound upper

[a,,, b,,]

Hence p is common to every interval.

19. Suppose, in the preceding problem, that the lengths of the intervals tend to zero, i.e. lim (b n — a„) = O. Show that there would then exist exactly one point common n

to every interval.

Recall that lim (b n — an) = 0 means that, for every n

3 no E

N

such that n > n o

(bn — an) <

E

E>

0,

APPENDIX]

PROPERTIES OF THE REAL NUMBERS

233

Solution: Suppose p i and p2 belong to every interval. If

lim (b„— a n ) = 0, the distance

p i p2,

there exists an interval / no = [ano , b„0]

1IP1 —

then

8 > 0.

P2

such that the length of I

less than

1p 1 —p 2 ----- 8 between p i and p2. Accordingly, p i and p2 cannot both belong to

contradiction.

Thus

Since

a

p1 = p 2 , i.e. only one point can belong to every interval.

Supplementary Problems FIELD AXIOMS 20.

Show that the Right Distributive Law [D2 ] is a consequence of the Left Distributive Law [D 1 ] and the Commutative Law [M5 ].

21.

Show that the set Q of rational numbers under addition and multiplication is a field.

22.

Show that the following set A of real numbers under addition and multiplication is a field:

A = {a + 23.

a, b rational}

Show that the set A = {,, , —4, —2, 0, 2, 4, , } of even integers under addition and multiplication satisfies all the axioms of a field except [M3], [M4] and [M5 ], that is, is a ring.

INEQUALITIES AND POSITIVE NUMBERS 24.

Rewrite so that x is alone between the inequality signs:

(i) 4 < —2x < 10,

(iii) — 3 < 5— 2x < 7.

(ii) —1 < 2x — 3 < 5,

The product of any two negative numbers is positive.

25.

Prove:

26.

Prove Theorem A.2(iii):

If a < b, then a + e

27.

Prove Theorem A.2(iv):

If a < b

28, Prove Corollary A.3:

ac

< be.

The set R of real numbers is totally ordered by the relation

If a < b and c is positive, then:

30.

fca) =

31. Prove:

b + c.

and c is positive, then

29. Prove: Prove:

<

(a -I- b)/2.

a

b

(i) — < — , C

More generally, prove

Let a and b be real numbers such that

C

b.

c

(ii) — < — . b a

Va1 a2 •• •an

(a1

a < b+ f for every

(i) x3 + x2 — tlx > 0,

32. Determine all real values of x such that:

a

(ii)

a2 -I- • • • ± an)/n.

> 0.

E

Then a

(x —1)(x -I- 3) 2

b.

0.

ABSOLUTE VALUES

(i) 1 - 21 ± 11 — 41, (ii) 13 — 81 — 11 — 91, (iii) 11 - 4 — 12 — 7 11-

33.

Evaluate:

34.

Rewrite, using the absolute value sign:

35.

Prove:

(i) 1—al =

, (ii) a2

(i) —3 < x < 9, (ii) 2

al 2, (iii)

= ArcT,2,

(iv)

x

8, (iii) —7 < x < —1.


PROPERTIES OF THE REAL NUMBERS

234

[APPENDIX

36. Prove Proposition A.4(ii): l ab 1

—

—

37. Prove Proposition A.4(iv):

LEAST UPPER BOUND AXIOM

38. Prove: Let A be a set of real numbers bounded from below. Then A has a greatest lower bound, i.e. inf (A) exists. 39. Prove:

(i) Let x C R such that x2 < 2; then anGN such that (x

1/1'

)2

(ii) Let xC R such that x2 >2; then 3 3/ E N such that (x — 1/n) 2

<

2.

> 2.

40. Prove:

There exists a real number a G R such that a2

2.

41. Prove:

Between any two positive real numbers lies a number of the form r2, where r is rational.

42. Prove:

Between any two real numbers there is an irrational number.

Answers to Supplementary Problems 24. (i) —5 < x < —2

(iii) —1 < x < 4

(ii) 1 < x < 4

32. (i) —3 < x < 0 or x > 2, i.e. x G (-3, 0) u (2, co) 33. (i) 5 34. (i) x

(ii) —3

-

3<6

(iii) 1

(ii) x

-

5

3

(iii)

+ 41 <3

(ii)

x

1

INDEX Absolute property, 152 Absolute value, 227 Accumulation point, 48, 53, 68 Adherent point, 69 Aleph-null, 32 Alexandrov compactification, 156 Algebra, of real-valued functions, 21 of sets, 4 Algebraic numbers, 38 Almost all, 50 Anti-symmetric relation, 35 Arbitrary closeness, 98 Archimedean order axiom, 229, 231 Arcwise connected, 184 Ascoli's theorem, 211, 219 Axiom of choice, 37

Collection, 2 Compact, countably, 155 locally, 155 sequentially, 154 sets, 151 spaces, 152 Compact open topology, 211 Compactification, 156 Compactness, 151 Complement, 3 Completely regular space, 142 Completeness, 51, 196, 228 Completion, 197 Components, 183 Composition, of functions, 18 of relations, 7 Connected, arcwise, 184 locally, 183 sets, 101, 180 simply, 186 spaces, 181 Constant function, 17 Continuous, at a point, 52, 99, 118 function, 52, 54, 97 uniformly, 165 Continuum, 33 Contracting mapping, 196 Convergence, compact, 214 pointwise, 208 uniform, 209 Convergent sequence, 50, 54, 71, 118 Coordinate set, 19 Countability, 131 Countable set, 32 Countably compact, 155 Cover, 49, 151

Baire's category theorem, 198, 204 Ball, 113 Banach space, 199 Base, for a topology, 87 local, 89 Bicompact, 154 Bicontinuous function, 100 Binary relation, 5 Bolzano-Weierstrass theorem, 48, 56, 155 Boundary, 70 Bounded, function, 22 set, 36, 112 totally, 157 uniformly, 211 Cantor, set, 171 theorem, 34, 42 Cardinal number, 34, 45 Cardinality, 33 Cartesian, plane, 4 product, 19 Category, 198 Cauchy sequence, 51, 195 Cauchy-Schwarz inequality, 125 Characteristic function, 30 Class, 2 Closed, function, 99 interval, 1 path, 186 sets, 48, 53, 67 Closure, 68 Cluster point, 67 Coarser topology, 71 Co-domain, 17 Cofinite topology, 66

Defining, base, 168 subbase, 101, 168 De Morgan's laws, 4, 20 Dense, 69 Denumerable, 32 Derived, point, 67 set, 67 Diameter, 112 Difference of sets, 3 Disconnected, 101, 180, 192 Disconnection, 180 Discrete (topological) space, 66 Disjoint sets, 3

235

236 Distance, 111, 112 Domain, of a function, 17 of a relation, 5 Dominates, 35 Dual space, 213 Element, 1 Embedded, 156 Empty set, 2 Equality, of functions, 17 of sets, 1

Equicontinuity, 211 Equivalence, class, 6 relation, 6 Equivalent, metrics, 115 sets, 32 topologically, 100 Euclidean, metric, 117 norm, 118 space, 22, 117 Evaluation mapping (function), 207 Extended real line, 156 Extension of a function, 18 Exterior, 70 Family, 2 Field axioms, 225 Filter, 140 Finer topology, 71 Finite, intersection property, 153 sets, 1, 32 First, axiom of countability, 131 category, 198 countable space, 131 element, 36 Follows, 35 Function, 17 projection, 19 set, 20 space, 207

Functionals, 213 Graph, 17 Greatest lower bound, 36 Hausdorff space, 131 Heine-Borel theorem, 49, 58, 151 Hereditary property, 133 Hilbert, cube, 129 space, 117 Homeomorphic spaces, 100

Homeomorphism, 100 Homotopic, 185 Homotopy, 185

I N DEX Identity, function, 18 relation, 6 Image, 17, 20 Inclusion function, 29 Indexed sets, 19 Indiscrete (topological) space, 66 Induced, metric, 118 topology, 114

Infimum (inf), 36 Infinite sets, 1, 32 Initial point, 184 Integers, 226 Interior, function, 99 of a set, 70 point, 47, 53, 69 Intersection, 3, 19 Intervals, 1, 182 Inverse, function, 18 image, 20 relation, 5 Irrational numbers, 226 Isometric, 116

Kuratowski's closure axioms, 72 4-metric, 117 12-norm, 119 12-space, 117 Larger, element, 35 topology, 71 Lattice, 83 Least upper bound, 36 axiom, 228 Lebesgue number, 158 Lexicographically ordered, 35 Limit, of a sequence, 50, 71 point, 48, 53, 67

Lindeled, space, 132, 138 theorems, 132, 135 Linear (vector) space, 22 Linearly ordered, 35 Local base, 89 Locally, compact, 155 connected, 183 Lower, bound, 36 limit topology, 88 Mapping, 17 Maximal element, 36 Meager, 198 Member, 1 Metric, 111 product space, 171 space, 114

237

INDEX

Metric (cont.) subspace, 114 topology, 114

Metrizable, 115 Metrization problem, 116 Minimal element, 36 Minkowski's inequality, 126 N, positive integers, 2, 226 Natural, numbers, 226 order, 35 Negative numbers, 226 Neighborhood, 70 Nested, interval property, 229, 232 local base, 131 Net, 140, 157 Non-denumerable, 33 Norm, 118 Normal space, 141 Normed space, 118 Nowhere dense, 70 Null set, 2 One-one function, 18 One-point compactification, 156 One-to-one, correspondence, 32 function, 18 Onto function, 18 Open, cover, 151 disc, 53 function, 99 interval, 1 neighborhood, 67 set, 47, 53, 66 sphere, 113 Order, inverse, 35 natural, 35 on the real line, 227 partial, 34 topology, 89 Ordered, linearly, 35 pair, 5 subsets, 35 totally, 35 Partial order, 34 Partition, 7 Path, 184 Plane, 4 Point open topology, 207 Pointwise convergence, 208 Positive, integers, 226 numbers, 226 Power, of the continuum, 33 set, 3

Product, Cartesian, 19 invariant, 170 metric space, 171 of functions, 18 set, 4 space, 167 topology, 167 Projection function, 19 Proper subset, 2

Pseudometric, 112 Q, rational numbers, 2, 226 Quotient set, 6 R, real numbers, 2, 225 Rm , Euclidean 7n-space, 117 12-space, 117 Range, of a function, 17 of a relation, 5 Rational numbers, 226 Real, line, 47, 226 numbers, 198 Real-valued functions, 17, 21 Reflexive relation, 6 Region, 185 Regular space, 140 Relation, 5 Relative topology, 72 Restriction of a function, 18 Ring, 226 Schroeder-Bernstein theorem, 33 Second, axiom of countability, 131 category, 198 countable space, 131 Separable, 132 Separated sets, 180 Separation axioms, 115, 139 Sequence, 32, 49 Cauchy, 51, 195 convergent, 50, 54, 71 of sets, 19 Sequentially, compact, 154 continuous, 99 Set, 1, 4 Set functions, 20 Simply connected, 186 Smaller, element, 35 topology, 71 Space, completely regular, 142 Hausdorff, 131

Lindel6f, 132, 138 linear, 22 metric, 114 normal, 141 normed, 118

238 Space (cont.) regular, 140 topological, 66 Tychonoff, 142 vector, 22 Sphere, 113 Subbase, 88 Subsequence, 51 Subset, 2 Subspace (topological), 72 Supremum (sup), 36 Symmetric relation, 6 To-space, 149 T1 -space, 139 T2-space, 139 T3-space, 141 T31/2 -space, 142 Ti-space, 141

Terminal point, 184 Ternary set, 171 Thick, 198 Thin, 198 Topological, function, 100 property, 100 space, 66 subspace, 72 Topologically equivalent, 100 Topology, 66 of compact convergence, 213 of pointwise convergence, 209 of uniform convergence, 210 Totally, bounded, 157 disconnected, 192 ordered sets, 35

INDEX

Transcendental numbers, 45

Transitive relation, 6 Triangle inequality, 111

Trivial metric, 111 Tychonoff, product theorem, 170, 175 product topology, 167 space, 142 Unbounded set, 112

Uniform, boundedness, 211 continuity, 165 convergence, 209

convergence on compacta, 210 Union, 3, 19 Universal set, 2 Upper, bound, 36 limit topology, 88 Urysohn, lemma, 142, 146 metrization theorem, 142, 147 Usual, metric for real numbers, 111 topology for real numbers, 66 Vector space, 22 Venn diagram, 3 Weaker topology, 71 Weierstrass intermediate value theorem, 53, 64 Z, the integers, 2, 226 Zorn's lemma, 37

Index of Symbols

(X, T)

topological space, 66

ext (A)

exterior of A, 70

(X, d)

metric space, 114

ex

evaluation mapping, 207

11'*•11

norm, 118

T(X, Y)

class of functions from X into Y, 207

-<

coarser, 71

F(A,B)

class of functions from A into B, 211

if

if and only if

inf (A)

infimum of A, 36

int(A)

interior of A, 70

N

the set of positive integers, 2,

implies, 7

3

there exists, 7

V

for all, 7

\

(e.g. A\B), difference, 3

A'

derived set of A, 67

A

226

or A —

closure of A, 68

Np

neighborhood system of p, 70

A or A°

interior of A, 70

T(A)

power set of A, 3

Ac

complement of A, 3

Q

the set of rational numbers,

n (A i : i G I}

Cartesian product, 19

2, 226 R

or FL E/ Ai or

the set of real numbers, 2,

225

rbili Rm

Euclidean m-space,'117

R c°

4-space, 117

S(P, 8)

open sphere, 113

s.t.

such that, 7

sup (A)

supremum of A, 36

[0,1], 111, 210

T

topology, 66

d(A)

diameter of A, 112

TA

relative topology on A, 72

d(a,b)

distance from a to b, 111

71

usual topology, 66

discrete topology, 66

Z

the set of integers, 2, 226

lri

ith projection function, 19

(• .., . • .)

ordered pair, 5

0

empty set, 2

b(A)

boundary of A, 70

C[0,1]

continuous functions on

o

.7)

239