Introduction to Topology

Third Edition

by Bert Mendelson

DOVER PUBLICATIONS, INC.

NEW YORK

Copyright

Copyright © 1962, 1968, 1975 by Marion Mendelson.

Copyright © renewed 1990 by Marion Mendelson.

All rights reserved.

Bibliographical Note

This Dover edition, first published in 1990, is an unabridged, unaltered republication of the work originally published by Allyn and Bacon, Inc., Boston, 1975 (First Edition, 1962; Second Edition, 1968).

Library of Congress Cataloging-in-Publication Data

Mendelson, Bert, 1926–1988

Introduction to topology / by Bert Mendelson. — 3rd ed.

p.   cm.

“Unabridged, unaltered republication of the work originally published by Allyn and Bacon, Inc., Boston, 1975”—T.p. verso.

Includes bibliographical references (p. ).

ISBN-13: 978-0-486-66352-4

ISBN-10: 0-486-66352-3

1. Topology. I. Title.

[QA611.M39 1990]

514—dc20

90-32980

CIP  

Manufactured in the United States by Courier Corporation

66352313

www.doverpublications.com

Contents

Preface

1Theory of Sets

1Introduction

2Sets and subsets

3Set operations: union, intersection, and complement

4Indexed families of sets

5Products of sets

6Functions

7Relations

8Composition of functions and diagrams

9Inverse functions, extensions, and restrictions

10Arbitrary products

2Metric Spaces

1Introduction

2Metric spaces

3Continuity

4Open balls and neighborhoods

5Limits

6Open sets and closed sets

7Subspaces and equivalence of metric spaces

8An infinite dimensional Euclidean space

3Topological Spaces

1Introduction

2Topological spaces

3Neighborhoods and neighborhood spaces

4Closure, interior, boundary

5Functions, continuity, homeomorphism

6Subspaces

7Products

8Identification topologies

9Categories and functors

4Connectedness

1Introduction

2Connectedness

3Connectedness on the real line

4Some applications of connectedness

5Components and local connectedness

6Path-connected topological spaces

7Homotopic paths and the fundamental group

8Simple connectedness

5Compactness

1Introduction

2Compact topological spaces

3Compact subsets of the real line

4Products of compact spaces

5Compact metric spaces

6Compactness and the Bolzano-Weierstrass property

7Surfaces by identification

Bibliography

Index

Preface to the Third Edition

The first edition of this text was based on lecture notes prepared for a one-semester undergraduate course given at Smith College. The aim was to present a simple, thorough survey of elementary topics to students whose preparation included a calculus sequence in which some attention had been paid to definitions and proofs of theorems. With this in mind, I have attempted to resist the temptation to include more topics. There are many excellent introductory topology texts which are first-year graduate school level texts and it was not my original intention, nor is it now, to write at that level.

The main outlines of the text have not been changed. The first chapter is an informal discussion of set theory. The concept of countability has been postponed until Chapter 5, where it appears in the context of compactness.

The second chapter is a discussion of metric spaces. The topological terms “open set,” “neighborhood,” etc., are introduced. Particular attention is paid to various distance functions which may be defined on Euclidean n-space and which lead to the ordinary topology.

In taking up topological spaces in Chapter 3, the transition from the particular to the general has been maintained, so that the concept of topological space is viewed as a generalization of the concept of metric space. Thus there is a similarity or, perhaps, a redundancy in the presentation of these two topics. A great deal of attention has been paid to alternate procedures for the creation of a topological space, using neighborhoods, etc., in the hope that this seemingly trivial, but subtle, point would be clarified.

Chapters 4 and 5 are devoted to a discussion of the two most important topological properties: connectedness and compactness. Some of this material could lead to further discussion of topics related to analysis, function spaces, separation axioms, metrization theorems, to name a few. On the other hand, material such as homotopy and two-dimensional closed surfaces could lead to further discussion of topics related to algebraic topology.

In conclusion, it is a pleasure to express in print my gratitude to those mathematicians under whom I studied and who helped make this book possible. In particular I should like to mention Professors C. Chevalley, S. Eilenberg, I. James, H. Riberio, P. Smith, and E. Thomas.

B. M.

Introduction to Topology

CHAPTER 1

Theory of Sets

1  INTRODUCTION

As in any other branch of mathematics today, topology consists of the study of collections of objects that possess a mathematical structure. This remark should not be construed as an attempt to define mathematics, especially since the phrase “mathematical structure” is itself a vague term. We may, however, illustrate this point by an example.

The set of positive integers or natural numbers is a collection of objects N on which there is defined a function s, called the successor function, satisfying the conditions:

1.  For each object x in N, there is one and only one object y in N such that y = s(x);

2.  Given objects x and y in S such that s(x) = s(y), then x = y;

3.  There is one and only one object in N, denoted by 1, which is not the successor of an object in N, i.e., 1 ≠ s(x) for each x in N;

4.  Given a collection T of objects in N such that 1 is in T and for each x in T, s(x) is also in T, then T = N.

The four conditions enumerated above are referred to as Peano’s axioms for the natural numbers. The fourth condition is called the principle of mathematical induction. One defines addition of natural numbers in such a manner that s(x) = x + 1, for each x in N, which explains the use of the word “successor” for the function s. What is significant at the moment is the conception of the natural numbers as constituting a certain collection of objects N with an additional mathematical structure, namely the function s.

We shall describe a topological space in the same terms, that is, a collection of objects together with a specified structure. A topological space is a collection of objects (these objects usually being referred to as points), and a structure that endows this collection of points with some coherence, in the sense that we may speak of nearby points or points that in some sense are close together. This structure can be prescribed by means of a collection of subcollections of points called open sets. As we shall see, the major use of the concept of a topological space is that it provides us with an exact, yet exceedingly general setting for discussions that involve the concept of continuity.

By now the point should have been made that topology, as well as other branches of mathematics, is concerned with the study of collections of objects with certain prescribed structures. We therefore begin the study of topology by first studying collections of objects, or, as we shall call them, sets.

2  SETS AND SUBSETS

We shall assume that the terms “object,” “set,” and the relation “is a member of” are familiar concepts. We shall be concerned with using these concepts in a manner that is in agreement with the ordinary usage of these terms.

If an object A belongs to a set S we shall write (read, “A in S”). If an object A does not belong to a set S we shall write (read, “A not in S”). If A₁, . . . , An are objects, the set consisting of precisely these objects will be written {A₁, . . . , An}. For purposes of logical precision it is often necessary to distinguish the set {A}, consisting of precisely one object A, from the object A itself. Thus is a true statement, whereas A = {A} is a false statement. It is also necessary that there be a set that has no members, the so-called null or empty set. The symbol for this set is Ø.

Let A and B be sets. If for each object , it is true that , we say that A is a subset of B. In this event, we shall also say that A is contained in B, which we write , or that B contains A, which we write .

In accordance with the definition of subset, a set A is always a subset of itself. It is also true that the empty set is a subset of A. These two subsets, A and Ø, of A are called improper subsets, whereas any other subset is called a proper subset.

There are certain subsets of the real numbers that are frequently considered in calculus. For each pair of real numbers a, b with a < b, the set of all real numbers x such that is called the closed interval from a to b and is denoted by [a, b]. Similarly, the set of all real numbers x such that a < x < b is called the open interval from a to b and is denoted by (a, b). We thus have , where R is the set of real numbers.

Two sets are identical if they have precisely the same members. Thus, if A and B are sets, A = B if and only if both and . Frequent use is made of this fact in proving the equality of two sets.

Sets may themselves be objects belonging to other sets. For example, {{1, 3, 5, 7}, {2, 4, 6}} is a set to which there belong two objects, these two objects being the set of odd positive integers less than 8 and the set of even positive integers less than 8. If A is any set, there is available as objects with which to constitute a new set, the collection of subsets of A. In particular, for each set A, there is a set we denote by 2A whose members are the subsets of A. Thus, for each set A, we have if and only if .

EXERCISES

1.  Determine whether each of the following statements is true or false:

(a)  For each set .

(b)  For each set .

(c)  For each set .

(d)  For each set .

(e)  For each set .

(f)  There are no members of the set {Ø}.

(g)  Let A and B be sets. If , then .

(h)  There are two distinct objects that belong to the set {Ø, {Ø}}.

2.  Let A, B, C be sets. Prove that if and , then .

3.  Let A₁, . . . , An be sets. Prove that if and , then A₁ = A₂ = . . . = An.

3  SET OPERATIONS: UNION, INTERSECTION, AND COMPLEMENT

If x is an object, A a set, and , we shall say that x is an element, member, or point of A. Let A and B be sets. The intersection of the sets A and B is the set whose members are those objects x such that and . The intersection of A and B is denoted by (read, “A intersect B”). The union of the sets A and B is the set whose members are those objects x such that x belongs to at least one of the two sets A, B; that is, either or .* The union of A and B is denoted by (read, “A union B”).

The operations of set union and set intersection may be represented pictorially (by Venn diagrams). In Figure 1, let the elements of the set A be the points in the region shaded by lines running from the lower left-hand part of the page to the upper right-hand part of the page, and let the elements of the set B be the points in the region shaded by lines sloping in the opposite direction. Then the elements of are the points in either shaded region and the elements of are the points in the cross-hatched region.

Figure 1

Let . The complement of A in S is the set of elements that belong to S but not to A. The complement of A in S is denoted by Cs(A), S/A, or by S − A. The set S may be fixed throughout a given discussion, in which case the complement of A in S may simply be called the complement of A and be denoted by C(A). C(A) is again a subset of S and one may take its complement. The complement of the complement of A is A; that is, C(C(A)) = A.

There are many formulas relating the set operations of intersection, union, and complementation. Frequent use is made of the following two formulas.

THEOREM  (DeMorgan’s Laws). Let , . Then

(3.1)   ,

(3.2)   .

Proof. Suppose . Then and . Thus, and , or and . Therefore and, consequently,

Conversely, suppose . Then and and . Thus, and , and therefore . It follows that and, consequently,

We have thus shown that

One may prove Formula 3.2 in much the same manner as 3.1 was proved. A shorter proof is obtained if we apply 3.1 to the two subsets C(A) and C(B) of S, thus

Taking complements again, we have

EXERCISES

1.  Let , . Prove the following:

(a)   if and only if .

(b)   if and only if .

(c)   if and only if .

(d)   if and only if .

(e)   if and only if .

(f)   if and only if .

2.  Let . Prove the following:

(a)  .

(b)  .

4  INDEXED FAMILIES OF SETS

Let I be a set. For each , let Aα be a subset of a given set S. We call I an indexing set and the collection of subsets of S indexed by the elements of I is called an indexed family of subsets of S. We denote this indexed family of subsets of S by (read, “A sub-alpha, alpha in I”).

Let be an indexed family of subsets of a set S. The union of this indexed family, written, , (read “union over α in I of Aα”) is the set of all elements such that for at least one index . The intersection of this indexed family, written (read “intersection over α in I of Aα”) is the set of all elements such that for each . [Note that , for which reason the two occurrences of “α” in the expression are referred to as dummy indices.]

As an example, let A₁, A₂, A₃, A₄ be respectively the set of freshmen, sophomores, juniors, and seniors in some specified college. Here we have I = {1, 2, 3, 4} as an indexing set, and is the set of undergraduates while .

If the indexing set I contains precisely two distinct indices, then the union over α in I of Aα is the same as the union of two sets as defined in the previous section; that is,

Similarly,

We have allowed for the possibility that the indexing set I is the empty set in which case a careful reading of the definition shows that

DeMorgan’s laws are applicable to unions and intersections of indexed families of subsets of a set S.

THEOREM  Let be an indexed family of subsets of a set S. Then

(4.1)  ,

(4.2)  .

Proof. Suppose . Then ; that is, for each index . Thus for each index and . Therefore,

Conversely, suppose that . Then for each index . Thus for each index ; that is, . Therefore, and

This proves 4.1. The proof of 4.2 is left as an exercise.

Given any collection of subsets of a set S, the concept of indexed family of subsets allows us to define the union or intersection of the aforementioned subsets. We need only construct some convenient indexing set. In the event that the collection of subsets is finite, the finite set {1, 2, . . . , n} of integers is a convenient indexing set. Given subsets A₁, A₂, . . . , An of S, we shall often write for and, similarly, .

EXERCISES

1.  Let be two indexed families of subsets of a set S. Prove the following:

(a)  For each .

(b)  For each .

(c)  .

(d)  .

(e)  If for each then

(f)  Let . Then

2.  Let . Then

3.  Let be an indexed family of subsets of a set S. Let . Prove that

(a)  .

(b)  .

4.  Let be an indexed family of subsets of a set S. Let . Prove that

(a)   if and only if for each .

(b)   if and only if for each .

5.  Let I be the set of real numbers that are greater than 0. For each , let Ax be the open interval (0, x). Prove that , . For each , let Bx be the closed interval [0, x]. Prove that .

5  PRODUCTS OF SETS

Let x and y be objects. The ordered pair (x, y)* is a sequence of two objects, the first object of the sequence being x and the second object of the sequence being y. Let A and B be sets. The Cartesian product of A and B, written A × B, (read “A cross B”) is the set whose elements are all the ordered pairs (x, y) such that and .

The Cartesian product of two sets is a familiar notion. The coordinate plane of analytic geometry is the Cartesian product of two lines. The possible outcomes of the throw of a pair of dice is the Cartesian product of two sets, A and B, where A = B = {1, 2, 3, 4, 5, 6}. Unless A = B, the two Cartesian products A × B and B × A are distinct.

A generalization of the Cartesian product of two sets is the direct product of a sequence of sets. Let A₁, A₂, . . . , An be a finite sequence of sets, indexed by {1, 2, . . . , n}. The direct product of A₁, A₂, . . . , An, written

(read “product i equals one to n of Ai”) is the set consisting of all sequences (a₁, a₂, . . . , an) such that . In particular,

For this reason we shall often write

for .

The concept of direct product may be extended to an infinite sequence A₁, A₂, . . . , An, . . . of sets, indexed by the positive integers. The direct product of A₁, A₂, . . . , An, . . . , written

or

is the set whose elements are all infinite sequences (a₁, a₂, . . . , an, . . .) such that for each positive integer i.

The set of points of Euclidean n-space yields an example of a direct product of sets. If for i = 1, 2, . . . , n we have Ai = R, where R is the set of real numbers, then

is the set of points of a Euclidean n-space. An element is a sequence x = (x₁, x₂, . . . , xn) of real numbers. In general, if the sets A₁, A₂, . . . , An are all equal to the same set A, we write

and call an element an n-tuple.

EXERCISES

1.  Let . Prove that

2.  Prove that if A has precisely n distinct elements and B has precisely m distinct elements, where m and n are positive integers, then A × B has precisely mn distinct elements.

3.  Let A and B be sets, both of which have at least two distinct members. Prove that there is a subset that is not the Cartesian product of a subset of A with a subset of B. [Thus, not every subset of a Cartesian product is the Cartesian product of a pair of subsets.]

6  FUNCTIONS

The most familiar example of a function in mathematics is a correspondence that associates with each real number x a real number f(x). The purpose of marking an examination may be described as the construction of a marking function that makes correspond to each student taking the examination some integer between zero and one hundred. Integration of a continuous function defined on some closed interval [a, b] is another example of a function, namely the correspondence that associates with each object f in this given set of objects the real number

The concept of function or correspondence need not be restricted to the realm of numerical quantities. The correspondence that associates with each undergraduate in college one of the four adjectives freshman, sophomore, junior, or senior is also an example of a function using correspondence as an undefined concept.

DEFINITION  Let A and B be sets. A correspondence that associates with each element a unique element is called a function from A to B, which we shall write

or

DEFINITION  Let f:A → B. The subset , which consists of all ordered pairs of the form (a, f(a)) is called the graph of f:A → B.

The graph Γf of a function f:X → Y is the subset of X × Y consisting of precisely those points (x, y) for which the statement f(x) = y is true. This set is sometimes written

This notation, called the set builder notation, is of the general form {z | P(z) }, where P(z) is some statement which may or may not be true of z. The resulting set is the set of all z, in an appropriate universe, for which P(z) is true.

Let A and B be sets. Given a subset Γ of A × B there is a function f:A → B such that Γ is the graph of f:A → B if, for each , there is one and only one element of the form .

(Thus the equivalent definition of a function as a subset with the aforesaid property is frequently employed, in which case for each , the function Γ makes correspond to x the element such that .)

DEFINITION  Let f:A → B be given. For each subset X of A, the subset of B whose elements are the points f(x) such that is denoted by f(X). f(X) is called the image of X. For each subset Y of B, the subset of A whose elements are the points such that is denoted by f⁻¹(Y). f⁻¹(Y) is called the inverse image of Y, counter image of Y, or f inverse of Y.

DEFINITION  Let f:A → B be given. A is called the domain of f. B is called the range of f.

EXAMPLE  Let f:R → R, R the set of real numbers, be the function such that for each . If X is the closed interval [1, 2], then f(X) = [−2, 0]. If Z is the open interval (−1, 1), then is the set of roots of the polynomial x² − x − 2. f⁻¹([−5, −4]) = Ø.

A function f:A → B is also called a mapping or transformation of A into B. We may think of such a function as carrying each point into its corresponding point .

DEFINITION  A function f:A → B is called one-one if whenever f(a) = f(a′) for , then a = a′.

Thus, f:A → B is one-one if for each there is only one such that f(a) = b.

DEFINITION  A function f:A → B is called onto if B = f(A).

A one-one function is sometimes called injective and an onto function is sometimes called surjective. A function which is both one-one and onto is sometimes called bijective.

DEFINITION  A function f:A → B is called a constant function if there is a point such that f(x) = b for each .

DEFINITION  A function f:A → A is called the identity function (on A) if f(x) = x for each .

EXERCISES

1.  Let f:A → B be given. Prove the following:

(a)  For each subset .

(b)  For each subset .

(c)  If f:A → B is one-one, then for each subset ,

(d)  If f:A → B is onto, then for each subset ,

2.  Let A = {a₁, a₂} and B = {b₁, b₂} be two sets, each having precisely two distinct elements. Let f:A → B be the constant function such that f(a) = b₁ for each .

(a)  Prove that f⁻¹(f({a₁})) ≠ {a₁}. [Thus it is usually the case that f⁻¹(f(X)) and X are not equal.]

(b)  Prove that f(f⁻¹(B)) ≠ B. [Thus it is usually the case that f(f⁻¹(B)) and B are not equal.]

(c)  Prove that . [Thus it is usually the case that are are not equal.]

3.  Let f:A → B be given and let be an indexed family of subsets of A. Prove:

(a)  .

(b)  .

(c)  If f:A → B is one-one, then .

4.  Let f:A → B be given and let be an indexed family of subsets of B. Prove:

(a)  .

(b)  .

(c)  If X is a subset of B then f⁻¹(C(X)) = C(f⁻¹(X)).

(d)  If X is a subset of A, and Y is a subset of B, then .

5.  Let A and B be sets. The correspondence that associates with each element the element p₁(a, b) = a is a function called the first projection. The correspondence that associates with each element the element p₂(a, b) = b is a function called the second projection. Prove that if B ≠ Ø, then p₁:A × B → A is onto and if A ≠ Ø, then p₂:A × B → A is onto. Under what circumstances is p₁ or p₂ one-one? What is for an element ?

6.  Let A and B be sets, with B ≠ Ø. For each the correspondence that associates with each element the element is a function. Prove that for each is one-one. What is for a subset ?

7.  Let A be a set and . The function χE:A → {0, 1} defined by if and if is called the characteristic function of E. Let E and F be subsets of A, show:

(a)  ;

(b)   and find a similar expression for .

8.  Let A be a set to which there belong precisely n distinct objects. Prove that there are precisely 2n distinct objects that belong to 2A.

7  RELATIONS

A function may be viewed as a special case of what is called a relation. We are accustomed to thinking of one object being in a given relation to another; for example, Jeanne is the sister of Sam or silk purses are more expensive than sows’ ears. To say that the number 2 is less than the number 3, or 2 < 3, is thus to say that (2, 3) is one of the number pairs (x, y) for which the relation “less than” is true.

DEFINITION  A relation R from the elements of a set A to the elements of a set B is a subset of A × B. A relation R on a set E is a subset of E × E.

If , one frequently writes aRb. We wish to distinguish certain properties that a relation on a set E may or may not have.

DEFINITION  A relation R on a set E is called reflexive if aRa is true for all . It is called symmetric if, whenever aRb, also bRa. It is called transitive if, whenever aRb and bRc, then aRc.

Let < be the pairs of real numbers (x, y) such that x < y. Then < is a transitive relation on the set E of real numbers, but < is not reflexive and not symmetric. Let R be the pairs of real numbers (x, y) such that |x − y| < 1. Then R is reflexive and symmetric, but not transitive. Let Λ be the pairs of real numbers (x, y) such that x − y is an integer. Then Λ is reflexive, symmetric, and transitive.

DEFINITION  A relation R on a set E which is reflexive, symmetric, and transitive is called an equivalence relation.

DEFINITION  Let R be an equivalence relation on a set E. For each , the equivalence class of a, denoted by π(a), is the subset of E consisting of all x such that aRx.

Two equivalence classes are either disjoint or identical.

LEMMA  Let R be an equivalence relation on a set E and let for . Then π(a) = π(b).

Proof. Let . Then aRc and bRc. Suppose so that aRx. cRa by symmetry, so cRx by transitivity. Another application of transitivity yields bRx, so . Thus . Similarly .

By the reflexive property is always true, so the equivalence classes are non-empty and disjoint. Let E/R be the set of equivalence classes, then π:E → E/R is an onto function. E/R is sometimes called the quotient of E by the relation R, and π is called the projection.

EXERCISES

1.  Let P be a subset of the real numbers R such that (i) (ii) if then , and (iii) for each , one and only one of the three statements, , or is true. Define and . Prove that Q is a transitive relation.

2.  Let f:X → Y be a function from a set X onto a set Y. Let R be the subset of X × X consisting of those pairs (x, x′) such that f(x) = f(x′). Prove that R is an equivalence relation. Let π:X → X/R be the projection. Verify that, if is an equivalence class, to define F(α) = f(a), whenever α = π(a), establishes a well-defined function F:X/R → Y which is one-one and onto.

3.  Let f:X → X be a one-one function of a set X into itself. Define a sequence of functions f⁰, f¹, f², . . ., fn, . . .:X → X by letting f⁰ be the identity, f¹ = f, and inductively fn(x) = f(fn−1(x)). Prove that each of these functions is one-one. Let R be the subset of X × X consisting of those pairs (a, b) such that b = fk(a) for some integer k or a = fi(b) for some integer j. Prove that R is an equivalence relation.

4.  Let X be the set of functions from the real numbers into the real numbers possessing continuous derivatives. Let R be the subset of X × X consisting of those pairs (f, g) such that Df = Dg where D maps a function into its derivative. Prove that R is an equivalence relation and describe an equivalence set π(f).

5.  Let E be the set of all functions from a set X into a set Y. Let and let R be the subset of E × E consisting of those pairs (f, g) such that f(b) = g(b). Prove that R is an equivalence relation. Define a one-one onto function eb:E/R → Y.

8  COMPOSITION OF FUNCTIONS AND DIAGRAMS

DEFINITION  Let f:A → B and g:B → C be given. The composition of f:A → B and g:B → C is the correspondence that associates with each element , the element . This function is written .

A function h:A → C is, therefore, the composition of f:A → B and g:B → C (often abbreviated by writing h = gf) if for each . In a pictorial representation of these functions, we have h = gf when these functions behave in the manner indicated in Figure 2.

Figure 2

The concept of the composition of functions can be extended to the composition of a finite number of functions.

DEFINITION  Let f₁:A₁ → A₂, f₂:A₂ → A₃, . . . , fn:An → An+1 be given. The composition of f₁:A₁ → A₂, f₂:A₂ → A₃, . . . , and fn:An → An+1 is the correspondence that associates with each element the element . This function is written

or

Let three functions f:A → B, g:B → C, and h:C → D be given. We may form hgf:A → D. We may also form gf:A → C and compose this function with h:C → D to obtain h(gf):A → D. Similarly, we may form (hg)f:A → D. We thus have three functions hgf, h(gf), (hg)f:A → D. But

Thus, these three functions are the same. This observation provides a basis for the justification of the removal or replacement of parentheses in expressions such as (f₄f₃)(f₂f₁), etc.

Suppose we are given three functions f:A → B, g:B → C, and k:A → C. The existence of these three functions may be indicated, as in Figure 3, by what we shall call a diagram. The letters A, B, C stand for the various sets, and an arrow leading

Figure 3

from one set to another indicates a function from the first set to the second, namely, the function that carries each element x of the first set into the element t(x) of the second set, where t stands for the symbol closest to the middle of the arrow. The fact that we may form the composition of two functions (such as gf:A → C in the above diagram) is represented by a path in the direction of the arrows that goes from one set to a second and from the second set to a third. (In the above diagram we say, “We may go from A to B via f and from B to C via g.”)

We shall desire to diagram more complex situations than the one indicated in Figure 3. Let us say that by a diagram we shall mean a figure consisting of several symbols denoting sets and arrows leading from one symbol to another, each arrow leading from a set X to a set Y having an associated symbol t, the arrow and its symbol representing a given function t:X → Y. For example, diagram (8.1) indicates the existence of given functions f:A → B, g:A → C, k:B → D, h:C → D. This diagram shows us

that by composing functions we may obtain two functions from A to D, namely, kf, hg:A → D. In any diagram, a path from X to Y consisting of a sequence of arrows leading from X to Y indicates the existence of a function from X to Y obtained by composing the functions represented by these arrows in the order of their occurrence, starting at X and terminating at Y.

In diagram (8.1) it may or may not be true that kf = hg. In the event that kf = hg we will say that diagram (8.1) is commutative. In general, a diagram is said to be commutative if for each X and Y in the diagram that represent sets, and for any two paths in the diagram beginning at X and ending at Y, the two functions from X to Y so represented are equal. For example, the statement that diagram (8.2) is commutative means that f = jh, k = gj, and kh = gjh = gf (note that the first two equalities imply the third).

A given set A may occur more than once in a diagram. For example, let A be the set of positive real numbers and R the set of real numbers. Let f:A → R be defined by the correspondence , and let g:R → A be defined by the correspondence . Let i:A → A be the identity function. Then the diagram (8.3) is commutative, for for .

EXERCISES

1.  Using the functions defined by the correspondences g(x) = x² and , construct an example of a commutative diagram similar to diagram (8.3).

2.  Let f:R × R → R be the function defined by the correspondence f(x, y) = x² + y² and let g:R × R → R be the function defined by the correspondence g(x, y) = x + y. Let h:R → R be the function defined by the correspondence h(x) = x². Is the diagram

commutative?

3.  Let i:A → A be the identity function. Let the diagram

be commutative. Prove that g:B → A is onto and that f:A → B is one-one.

4.  Let f:A → B, g:B → C. Prove that for .

9  INVERSE FUNCTIONS, EXTENSIONS, AND RESTRICTIONS

DEFINITION  Let f:A → B and g:B → A be given. The function f:A → B is called the inverse of g:B → A and the function g:B → A is called the inverse of f:A → B if g(f(a)) = a for each and f(g(b)) = b for each .

In this event we shall also say that f:A → B and g:B → A are inverse functions and that each of them is invertible.

Let iA:A → A and iB:B → B be identity functions. The statement that f:A → B and g:B → A are inverse functions is equivalent to the statement that the two diagrams

are commutative.

THEOREM  Let f:A → B and g:B → A be inverse functions, then both functions are one-one and onto.

Proof. Suppose . Then x = g(f(x)) = g(f(y)) = y and therefore f is one-one. To show that f is onto, let . We have and f(g(b)) = b, therefore if we set a = g(b), b = f(a) and f is onto. The roles of the two functions may be interchanged, since the definition of inverse functions imposes conditions symmetrical with regard to the two functions. Therefore, g:B → A is also one-one and onto.

We have shown that, given a function h:X → Y, a necessary condition that this function be invertible is that the function be one-one and onto. This condition is also sufficient.

THEOREM  Let f:A → B be one-one and onto. Then there exists a function g:B → A such that these two functions are inverse functions.

Proof. We shall first define g:B → A. Given , we may write b = f(a) for some since f is onto. Furthermore, f is one-one; hence there is only one element such that f(a) = b. We define g(b) = a. The correspondence that associates with each the element , as defined above, is a function g:B → A.f(g(b)) = b for each by the definition of g:B → A. Given , let a′ = g(f(a)). Then f(a′) = f(g(f(a))) = f(a) by the remark just made. Since f:A → B is one-one, a = a′ = g(f(a)). Thus, f:A → B and g:B → A are inverse functions.

The last two theorems may be combined in the statement: given f:A → B, a necessary and sufficient condition that there be a function g:B → A such that these two functions are inverse functions is that f:A → B be one-one and onto. Furthermore, in this event, the function g:B → A is uniquely determined.

THEOREM  Let f:A → B, g:B → A be inverse functions and let f:A → B and g′:B → A be inverse functions. Then g:B → A and g′:B → A are equal.

Proof. We must prove that g(b) = g′(b) for each . But b = f(g(b)) and therefore g′(b) = g′(f(g(b))) = g(b), since g′(f(a)) = a for each .

The proof of this last theorem may also be viewed as a direct consequence of the commutativity of the diagram

which yields g′(b) = g′(iB(b)) .= g′(f(g(b))) = iA(g(b)) = g(b).

DEFINITION  Let . Let f:A → Y and F:X → Y. If for each , we say that F is an extension of f to X or that f is a restriction of F to A. In this event we shall write f = F | A.

EXAMPLE  Let A be the open interval (0, π/2). For each , let Δθ be a right triangle one of whose acute angles is θ radians, and let f(θ) be the ratio of the length of the side of this triangle opposite the angle of magnitude θ to the length of the hypotenuse of Δθ, (more familiarly,

Thus f:A → R. For each , let (a, b)θ be the point of the plane R² whose distance from the origin is 1 and such that the rotation about the origin of the line segment whose end points are the origin and (1, 0) to the position of the line segment whose end points are the origin and (a, b)θ represents an angle of magnitude θ radians. Define F(θ) = b. Then F:R → R. F is an extension of f to R as is easily seen if one recognizes f:A → R as the sine function defined for acute angles by means of right triangles and F:R → R as the sine function defined for angles of arbitrary magnitude by means of the unit circle.

DEFINITION  Let . The function i:A → X, which is defined by the correspondence i(x) = x for each is called an inclusion mapping or function.

Let and F:X → Y. Then F is an extension of f if and only if the diagram

is commutative, where i:A → X is an inclusion mapping.

Given F:X → Y, there are as many restrictions of F:X → Y as there are subsets of X. Given a subset , we may obtain the restriction of F to A by forming the composition of the inclusion mapping i:A → X and F:X → Y. Thus, we may write F | A = Fi.

EXERCISES

1.  Let A be the set of all functions f:[a, b] → R that are continuous on [a, b]. Let B be the subset of A consisting of all functions possessing a continuous derivative on [a, b]. Let C be the subset of B consisting of all functions whose value at a is 0. Let d:B → A be the correspondence that associates with each function in B its derivative. Is the function d:B → A invertible?

To each , let h(f) be the function defined by

for . Verify that h:A → C. Find the function g:C → A such that these two functions are inverse functions.

2.  Let R be the real numbers and ∞ an object not in R. Define a set . Let a, b, c, d be real numbers. Let f:R* → R* be a function defined by f(x) = (ax + b)/(cx + d) when x ≠ −d/c, ∞, while f(−d/c) = ∞ and f(∞) = a/c. [In the event that c = 0, f is linear and f(x) = (ax + b)/d when x ≠ ∞ and f(∞) = ∞.] Prove that f has an inverse provided ad − bc ≠ 0.

3.  Let . Let f:A → Y, g:B → Y, and F:X → Y. Prove that if g is an extension of f to B and F is an extension of g to X, then F is an extension of f to X.

4.  Let m, n be positive integers. Let X be a set with m distinct elements and Y a set with n distinct elements. How many distinct functions are there from X to Y? Let A be a subset of X with r distinct elements, 0 ≦ r < m and f:A → Y. How many distinct extensions of f to X are there?

10  ARBITRARY PRODUCTS

Let X₁, . . . , Xn be sets. We have defined a point

as an ordered sequence such that . Given such a point, by setting x(i) = xi we obtain a function x which associates to each integer i, , the element . Conversely, given a function x which associates to each integer , an element , we obtain the point

It is easily seen that this correspondence between points of and functions of the above type is one-one and onto, so that a point of may also be defined as a function x which associates to each integer , a point . The advantage of this second approach is that it allows us to define the product of an arbitrary family of sets.

DEFINITION  Let {Xα}α∈I be an indexed family of sets. The product of the sets {Xα}α∈I, written , consists of all functions x with domain the indexing set I having the property that for each .

Given a point , one may refer to x(α) as the αth coordinate of x. However, unless the indexing set has been ordered in some fashion (as is the case with finite products in our earlier discussion), there is no first coordinate, second coordinate, and so on.

DEFINITION  Let . The function defined by pα(x) = x(α) is called the αth projection.

Clearly two points are identical if and only if, for each , that is, x(α) = x′(α).

In dealing with product spaces use is frequently made of a principle, called the axiom of choice, whereby we assume that if for each we can choose a point , then we may construct a point or function by setting x(α) = xα. This is equivalent to the statement that the product of non-empty sets is non-empty. Using the axiom of choice we may prove

PROPOSITION  If for each is non-empty, then each of the projection maps is onto.

Proof. Let be given. Set x(α) = xα. Suppose . Since Xβ is non-empty we may choose a point . Then and pα(x) = x(α) = xα, hence pα is onto.

In the above proof we have obtained a point , that is, a point whose αth coordinate is xα and whose other coordinates are unrestricted. If then to say that is to restrict the αth coordinate of x to lie in B and leave all other coordinates unrestricted.

EXERCISES

1.  Let A be a set. For each , let Xα. Verify that is the set of all functions from the set I to the set A. This set of functions is denoted by AI. Suppose A = {0, 1}. If I is finite how many elements are there in AI? Verify that AI in this case is the set of all characteristic functions defined on I.

2.  Let be two indexed families of sets with the same indexing set I. For each . Prove that there is a unique function such that pαf = fαpα for each , where pα is the appropriate projection map. This function f is denoted by . Given a third indexed family of sets and functions gα: Yα → Zα for each , show that . Suppose that each fα has an inverse kα. Prove that has the inverse .

3.  Let be an indexed family of sets and let I = I₁ ∪ I₂, where I₁ ∩ I₂ = Ø. Show that there is a one-one mapping of onto . More generally, let be a partition of I, that is for γ₁ ≠ γ₂, each Iγ ≠ Ø. Show that there is a one-one mapping of onto .

4.  Let N be the set of positive integers. In the notation of Problem 1, an infinite sequence x₁, x₂, . . . of points of a set X may be viewed as an element . If j:N → N is a function such that j(i) < j(i + 1) for , then the infinite sequence xj is a subsequence of the sequence x. Prove that a subsequence of xj is a subsequence of x.

For further reading, the books by Halmos, Naive Set Theory, and Kaplansky, Set Theory and Metric Spaces are both excellent sources.

* The logical connective “or” is used in mathematics (and also in logic) in the inclusive sense. Thus, a compound statement “P or Q” is true in each of the three cases: (1) P true, Q false; (2) P false, Q true; (3) P true, Q true, whereas “P or Q” is false only if both P and Q are false.

* If x and y are real numbers, the symbol (x, y) is ambiguous, for it may stand for either the ordered pair whose first element is x and the second y, or for the open interval (x, y). It is hoped that this ambiguity will be resolved by the context in which the symbol occurs.

CHAPTER 2

Metric Spaces

1  INTRODUCTION

A metric space is a set of points and a prescribed quantitative measure of the degree of closeness of pairs of points in this space. The real number system and the coordinate plane of analytic geometry are familiar examples of metric spaces. Starting from the vague characterization of a continuous function as one that transforms nearby points into points that are themselves nearby, we can, in a metric space, formulate a precise definition of continuity. Although this definition may be stated in the so-called “ε, δ” terminology, there are other, equivalent formulations available in a metric space. These include characterizations of continuity in terms of the behavior of a function with respect to certain subsets called neighborhoods of a point, or with respect to certain subsets called open sets.

2  METRIC SPACES

Given two real numbers a and b, there is determined a non-negative real number, |a − b|, called the distance between a and b. Since to each ordered pair (a, b) of real numbers there is associated the real number |a − b|, we may write this correspondence in functional notation by setting

Thus we have a function d:R × R → R, where R is the set of real numbers. This function has four important properties, which the reader should verify:

1.  

2.  d(x, y) = 0 if and only if x = y;

3.  d(x, y) = d(y, x);

4.  

for . For the purposes of discussing “continuity” of functions, these four properties of “distance” are sufficient. This fact suggests the possibility of examining “continuity” in a more general setting; namely, in terms of any set of points for which there is defined a “distance function” such as the function d:R × R → R above.

DEFINITION 2.1  A pair of objects (X, d) consisting of a non-empty set X and a function d:X × X → R, where R is the set of real numbers, is called a metric space provided that:

1.  

2.  d(x, y) = 0 if and only if ;

3.  d(x, y) = d(y, x),

4.  

The function d is called a distance function or metric on X and the set X is called the underlying set.

[A more precise notation for a metric space would be (X, d:X × X → R) and for a distance function d:X × X → R. We shall, however, frequently delete the sets and arrow in the symbol for a function, when, in a given context, it is clear which sets are involved.]

We may think of the distance function d as providing a quantitative measure of the degree of closeness of two points. In particular, the inequality may be thought of as asserting the transitivity of closeness; that is, if x is close to y and y is close to z, then x is close to z.

Let where R is the set of real numbers. The verification that the function d(a, b) = |a − b| satisfies the four properties enumerated in Definition 2.1 establishes:

THEOREM 2.2  (R, d) is a metric space, where d is the function defined by the correspondence d(a, b) = |a − b|, for

Given a finite collection (X₁, d₁), (X₂, d₂), … , (Xn, dn) of metric spaces, there is a standard procedure for converting the set

into a metric space; that is, for defining a distance function on X.

THEOREM 2.3  Let metric spaces (X₁, d₁), (X₂, d₂), … , (Xn, dn) be given and set

For each pair of points , let d:X × X → R be the function defined by the correspondence

Then (X, d) is a metric space.

Proof. With x and y as above, for , and therefore . If d(x, y) = 0, then di(xi, yi) = 0 for and therefore xi = yi for each i. Consequently, x = y. Conversely, if x = y, then di(xi, yi) = 0 for each i, and d(x, y) = 0. Since di(xi, yi) = di(yi, xi) for . Finally, let . Let j and k be integers such that d(x, y) = dj(xj, yj) and d(y, z) = dk(yk, zk). Thus, for , and

As an immediate application of this theorem, we have:

COROLLARY 2.4  (Rn, d) is a metric space, where d:Rn × Rn → R is the function defined by the correspondence

It is interesting to compare the metric space (R², d) that we obtain in the above manner with what might be considered a more natural model of the coordinate plane. In (R², d) as defined above, the distance from the point (1, 2) to the point (3, 1) is 2, since maximum {|1 − 3|, |2 − 1|} = 2. The distance function d′ used in analytical geometry would yield

If, for each pair of points we define

then we are constructing a new metric space (R², d′), (provided, of course, that d′ is a distance function), which must be distinguished from the metric space (R², d) where

For example, in (R², d) the set M of points x such that for a fixed point is a square of width 2 whose center is at a and whose sides are parallel to the coordinate axes, whereas in (R², d′) the set of points x such that for a fixed point is a circular disc whose center is a and whose radius is 1 (see Figure 4).

Figure 4

The formula used to define the function d′ may be generalized to yield a distance function for Rn, often referred to as the Euclidean distance function.

THEOREM 2.5  (Rn, d′) is a metric space, where d′ is the function defined by the correspondence

The proof of this theorem will be found in Section 8.

The fact that we have metric spaces (Rn, d) and (Rn, d′), with d and d′ defined as above, serves to emphasize the fact that a metric space consists of two objects, a set and a distance function. Two metric spaces may be distinct even though the underlying sets of points of the two spaces are the same.

EXERCISES

1.  Let (X, d) be a metric space. Let k be a positive real number and set dk(x, y) = k·d(x, y). Prove that (X, dk) is a metric space.

2.  Prove that (Rn, d″) is a metric space, where the function d″ is defined by the correspondence

for . In (R², d″) determine the shape and position of the set of points x such that for a point .

3.  Let d be the distance function defined on Rn by using Theorem 2.3, let d′ be the Euclidean distance function, and let d″ be the distance function defined in Problem 2 above. Prove that for each pair of points ,

4.  Let X be the set of all continuous functions f:[a, b] → R. For f, , define

Using appropriate theorems from Calculus, prove that (X, d) is a metric space.

5.  Let . A function f:S → R is called bounded if there is a real number K such that (or equivalently, Let X′ be the set of all bounded functions f:[a, b] → R. For f, define

(l.u.b. is an abbreviation of least upper bound, see Definition 5.5 of this chapter). Prove that (X′, d′) is a metric space.

6.  Let f, g:[a, b] → R be two functions that are both continuous and bounded. Compare d(f, g) and d′(f, g), where d and d′ are defined as in Problems 4 and 5 respectively.

7.  Let X be a set. For x, define the function d by

and

if x ≠ y. Prove that (X, d) is a metric space.

8.  Let Z be the set of integers. Let p be a positive prime integer. Given distinct integers m, n there is a unique integer t = t(m, n) such that m − n = pt·k, where k is an integer not divisible by p. Define a function d:Z × Z → R by the correspondence d(m, m) = 0 and

for m ≠ n. Prove that (Z, d) is a metric space. [Hint: for a, b, Let p = 3. What is the set of elements such that d(0, x) < 1? What is the set of elements such that d(0, x) < ?

3  CONTINUITY

In calculus, the first occurrence of the word “continuity” is with reference to a function f:R → R, R the set of real numbers. To decide which condition or conditions this function must satisfy for us to say, “the function f is continuous at a point ,” we try to decide upon a precise formulation of the statement “a number f(x) will be close to the number f(a) whenever the number x is close to a.” Having defined a distance function for the real numbers R, we have a quantitative measure of the degree of closeness of two numbers. But how close must f(x) be to f(a)? Instead of specifying some particular degree of closeness of f(x) to f(a), let us think, rather, of requiring that no matter what choice is made for the degree of closeness of f(x) to f(a), it can be so arranged that this degree of closeness is achieved. By the phrase “arrange matters” we mean that we can find a corresponding degree of closeness so that whenever x is within this corresponding degree of closeness to a, then f(x) is within the prescribed degree of closeness to f(a). We have now arrived at the following formulation, “the function f:R → R is continuous at the number , if given a prescribed degree of closeness, f(x) will be within this prescribed degree of closeness to f(a), whenever x is within some corresponding degree of closeness to a.” To put this statement in its final form, we shall substitute for “a prescribed degree of closeness” the symbol “ε,“ and for the phrase “some corresponding degree of closeness” the symbol “δ,“ and use the distance function to measure the degree of closeness.

DEFINITION 3.1  Let f:R → R. The function f is said to be continuous at the point , if given ε > 0, there is a δ > 0, such that

whenever

The function f is said to be continuous if it is continuous at each point of R.

Because we initially formulated the definition of continuity in terms of the phrase “degree of closeness,” we may easily devise a definition of “continuity” applicable to metric spaces in general, since we need only use the distance functions of these metric spaces to measure “degree of closeness.”

DEFINITION 3.2  Let (X, d) and (Y, d′) be metric spaces, and let A function f:X → Y is said to be continuous at the point if given ε > 0, there is a δ > 0, such that

whenever and

The function f:X → Y is said to be continuous if it is continuous at each point of X.

Definitions, such as those given above, are created to serve two purposes. First of all, they are abbreviations. Thus, the statement that begins, “given ε > 0, there is … ,” is replaced by the shorter statement, “f:X → Y is continuous at the point .” Second, these definitions are attempts to formulate precise characterizations of what we feel are significant properties; in this case, the property of being continuous at a point. We have tried to indicate in the discussion preceding these definitions that they do provide a precise characterization of our intuitive, and perhaps vague, concept of continuity. There are, in a certain sense, tests that we may apply to see whether or not they do so. As an illustration, there are certain functions that we “know” are “continuous,” that is, we are sure that they possess this property we are trying to characterize. If it should turn out that a function we “know” to be “continuous” is not continuous in accordance with these definitions, then, although these definitions may be precise, they would not furnish a precise characterization of the property we have in mind when we say a function is “continuous.” This type of testing of a definition thus takes the form of proving theorems to the effect that certain functions are continuous. For example:

THEOREM 3.3  Let (X, d) and (Y, d′) be metric spaces. Let f:X → Y be a constant function, then f is continuous.

Proof. Let a point and ε > 0 be given. Choose any δ > 0, say δ = 1. Then whenever d(x, a) < δ, we have d′(f(x), f(a)) = 0 < ε.

THEOREM 3.4  Let (X, d) be a metric space. Then the identity function i:X → X is continuous.

Proof. Suppose Let ε > 0 be given. Choose δ = ε, then whenever d(x, a) < δ we have d(i(x), i(a)) = d(x, a) < ε.

Note that in the above proof we could have equally well chosen δ to be any positive number, provided only that and the proof would still be valid. The choice of δ need not be a very efficient choice; all that is required is that it “do the job.”

There is one situation we shall have to consider for which the notation f:X → Y that we have adopted for a function from a metric space (X, d) into a metric space (Y, d′) is ambiguous. Consider metric spaces (X, d) and (X, d′) with the same underlying set. If we simply write f:X → X for a function, it is impossible to tell which metric space is denoted by the first occurrence of X and which by the second. For this reason, when considering one set X with two different distance functions, we shall write f:(X, d) → (X, d′) if we intend to think of f:X → X as a function from the metric space (X, d) into the metric space (X, d′). As an illustration, we shall prove:

THEOREM  3.5  Let i:Rn → Rn be the identity function. Then

and

are continuous, where the distance function d is the maximum distance between corresponding coordinates (as defined in Section 2) and d′ is the Euclidean distance.

Proof. Let We shall first prove that i:(Rn, d) → (Rn, d′) is continuous. Let ε > 0 be given. Choose . Suppose x = (x₁, x₂, … , xn) is such that d(x, a) < δ; that is, . Then

Therefore, given ε > 0, there is a a δ > 0 such that d′(i(x), i(a)) < ε whenever d(x, a) < δ.

We now prove that i:(Rn, d′) → (Rn, d) is continuous. Let ε > 0 be given. Choose δ = ε. Suppose that x = (x₁, x₂, … , xn) is such that d′(x, a) < δ. Then

and therefore for each i, (ai − xi)² < δ², or |ai − xi| < δ = ε. Consequently, d(x, a) < ε. Thus, given ε > 0, there is a δ > 0, such that d(i(x), i(a)) < ε whenever d′(x, a) < δ.

One of the most important elementary theorems about continuous functions is the statement that the composition of two continuous functions is again a continuous function.

THEOREM 3.6  Let (X, d), (Y, d′), (Z, d″) be metric spaces. Let f:X → Y be continuous at the point and let g:Y → Z be continuous at the point Then gf:X → Z is continuous at the point .

Proof. Let ε > 0 be given. We must find a δ > 0 such that whenever and d(x, a) < δ, then d″(g(f(x)), g(f(a))) < ε. Since g is continuous at f(a), there is an η > 0, such that whenever and d′(y, f(a)) < η, then d″(g(y), g(f(a))) < ε. Using the fact that f is continuous at a, we know that given η > 0, there is a δ > 0, such that and d(x, a) < δ imply that d′(f(x), f(a)) < η and hence d″(g(f(x)), g(f(a))) < ε.

COROLLARY 3.7  Let (X, d), (Y, d′), (Z, d″) be metric spaces. Let f:X → Y and g:Y → Z be continuous. Then gf:X → Z is continuous.

EXERCISES

1.  Let X be the set of continuous functions f:[a, b] → R. Let d* be the distance function on X defined by

for . For each , set

Prove that the function I:(X, d*) → (R, d) is continuous.

2.  Let (Xi, di), (Yi, d′i), i = 1, … , n be metric spaces. Let fi:Xi → Yi, i = 1, … , n be continuous functions. Let

and convert X and Y into metric spaces in the standard manner. Define the function F:X → Y by

Prove that F is continuous.

3.  Define the function f:R² → R by f(x₁, x₂) = x₁ + x₂. Prove that f is continuous, where the distance function on R² is either d or d′.

4.  Define functions g, h, k, m as follows: g:R² → R² × R² by g(x, y) = ((x, y), (x, y)); h:R² × R² → R × R by h((a, b), (c, d)) = (a + b, c − d); k:R × R → R × R by k(u, v) = (u², v²); m: R × R → R by . Prove that all these functions are continuous and that xy = mkhg(x, y).

4  OPEN BALLS AND NEIGHBORHOODS

In the definition of continuity of a function f at a point a in a metric space (X, d), we are concerned with how f transforms those points such that d(x, a) < δ. If we give a name to this particular collection of points in X we shall be able to cast the definition of continuity in a more compact form.

DEFINITION 4.1  Let (X, d) be a metric space. Let and δ > 0 be given. The subset of X consisting of those points such that d(a, x) < δ is called the open ball about a of radius δ and is denoted by

Thus, if and only if and d(x, a) < δ. Similarly, if (Y, d′) is another metric space and f:X → Y, we have if and only if and d′(y, f(a)) < ε. Thus:

THEOREM 4.2  A function f:(X, d) → (Y, d′) is continuous at a point if and only if given ε > 0 there is a δ > 0 such that

For a function f:X → Y we have if and only if , where U and V are subsets of X and Y respectively. Therefore:

THEOREM 4.3  A function f:(X, d) → (Y, d′) is continuous at a point if and only if given ε > 0 there is a δ > 0 such that

Given a point a in a metric space (X, d), the subset B(a; δ) of X, for each δ > 0, is an example of the type of subset of X that is called a neighborhood of a.

DEFINITION 4.4  Let (X, d) be a metric space and . A subset N of X is called a neighborhood of a if there is a δ > 0 such that

The collection of all neighborhoods of a point is called a complete system of neighborhoods of the point a.

A neighborhood of a point may be thought of as containing all the points of X that are sufficiently close to a or as “enclosing” a by virtue of the fact that it contains some open ball about a. In particular, for each δ > 0, B(a; δ) is a neighborhood of a. These open balls have the property that they are neighborhoods of each of their points.

LEMMA 4.5  Let (X, d) be a metric space and . For each δ > 0, the open ball B(a; δ) is a neighborhood of each of its points.

Proof. Let . In order to show that B(a; δ) is a neighborhood of b we must show that there is an η > 0 such that . Since , d(a, b) < δ. Choose η < δ − d(a, b). If then

and therefore . Thus and B(a; δ) is a neighborhood of b.

We may describe this proof pictorially. We have started with an open ball B(a; δ) about a. We choose a point . Then the minimum distance from b to points not in B(a; δ) is at least δ − d(a, b), as indicated in Figure 5, so that a ball about b of radius η < δ − d(a, b) is contained in B(a; δ).

Figure 5

The complete system of neighborhoods of a point may be used to characterize continuity of a function at a point.

THEOREM 4.6  Let f:(X, d) → (Y, d′). f is continuous at a point if and only if for each neighborhood M of f(a) there is a corresponding neighborhood N of a, such that

or equivalently,

Proof. First suppose that f is continuous at the point . We must show that, given a neighborhood M of f(a), we can find a neighborhood N of a such that . Since M is a neighborhood of f(a), there is an ε > 0 such that . Since f is continuous at a, there is a δ > 0 such that . But N = B(a; δ) is a neighborhood of a, therefore

Conversely, suppose that f satisfies the property that for each neighborhood M of f(a), there is a corresponding neighborhood N of a, such that . Let ε > 0 be given. To prove that f is continuous at a we must show that there is a δ > 0 such that

But B(f(a); ε) = M is a neighborhood of f(a) and therefore there is a neighborhood N of a such that . Since N is a neighborhood of a, there is a δ > 0 such that . Therefore

The proof of the first part of the above theorem may be represented pictorially by considering an arbitrary neighborhood M of f(a) (as indicated in Figure 6). Since M is a neigh-borhood of f(a), it contains an open ball B(f(a); ε) for some ε > 0. Since f is continuous at a, for some δ > 0 the neighborhood N = B(a; δ) is carried into M by f. Similarly, the proof of the second part of the theorem may be depicted by Figure 7. We start with a neighborhood M = B(f(a); ε) of f(a). The assumed property of f allows us to assert that there is a neighborhood N of a that is carried into M by f. Since N is a neighborhood of a we have an open ball B(a; δ) contained in N, which must also be carried into M.

Figure 6

Figure 7

If N is a neighborhood of a point a in a metric space (X, d) and N′ is a subset of X that contains N, then N′ contains the same open ball about a that N does and therefore N′ is also a neighborhood of a. Thus, the previous theorem becomes:

THEOREM 4.7  Let f:(X, d) → (Y, d′). f is continuous at a point if and only if for each neighborhood M of f(a), f⁻¹(M) is a neighborhood of a.

The collections of neighborhoods of points in a metric space possess five properties that will be of significance in the next chapter.

THEOREM 4.8  Let (X, d) be a metric space.

N1. For each point , there exists at least one neighborhood of a.

N2. For each point and each neighborhood N of a, .

N3. For each point , if N is a neighborhood of a and , then N′ is a neighborhood of a.

N4. For each point and each pair N, M of neighborhoods of a, N ∩ M is also a neighborhood of a.

N5. For each point and each neighborhood N of a, there exists a neighborhood O of a such that and O is a neighborhood of each of its points.

Proof. For , X is a neighborhood of a, thus N1 is true. N2 is trivial and N3 has already been discussed. To prove N4, let N and M be neighborhoods of a. Then N and M contain open balls B(a; δ₁) and B(a; δ₂) respectively and therefore N ∩ M contains the open ball B(a; δ), where δ = minimum {δ₁, δ₂}. To prove N5, let N be a neighborhood of a. Then N contains an open ball B(a; δ) and by Lemma 4.5, O = B(a; δ) is a neighborhood of each of its points.

For a given point a in a metric space X, the collection of open balls with center a has been used to generate the complete system of neighborhoods at a, in the sense that the neighborhoods of a are precisely those subsets of X which contain one of these open balls.

DEFINITION 4.9  Let a be a point in a metric space X. A collection of neighborhoods of a is called a basis for the neighborhood system at a if every neighborhood N of a contains some element B of .

As an example, if a is a point on the real line R, a basis for the neighborhood system at a is the collection of open intervals containing a.

EXERCISES

1.  Let (X, d) be a metric space such that d(x, y) = 1 whenever x ≠ y. Let . Prove that {a} is a neighborhood of a and constitutes a basis for the system of neighborhoods at a. Prove that every subset of X is a neighborhood of each of its points.

2.  Let and f:R → R be defined by for x > a. Prove that f is not continuous at a, but is continuous at all other points.

3.  Let f:X → Y be a function from a metric space X into a metric space Y. Let and let be a basis for the neighborhood system at f(a). Prove that f is continuous at a if and only if for each is a neighborhood of a.

4.  Let a be a point on the real line R. Prove that each of the following collections of subsets of R constitute a basis for the system of neighborhoods at a:

i)

All closed intervals of the form [a − ε, a + ε], ε > 0;

ii)

All open balls B(a; ε), ε a positive rational number;

iii)

All open balls , n a positive integer;

iv)

All open balls , n a positive integer larger than some fixed integer k.

Show that no finite collection of subsets of R can be a basis for the system of neighborhoods at a.

5.  Let a be a point in a metric space X. Let N be the set of positive integers. Prove that there is a collection of neighborhoods of a which constitutes a basis for the system of neighborhoods at a.

6.  Let a and b be distinct points of a metric space X. Prove that there are neighborhoods Na and Nb of a and b respectively such that Na ∩ Nb = Ø.

7.  Let (X₁, d₁), (X₂, d₂), … , (Xn, dn) be metric spaces and convert

into a metric space (X, d) in the standard manner. Prove that an open ball in (X, d) is the product of open balls from X₁, X₂, … , Xn respectively. Let , and let be a basis for the neighborhood system at ai. Let be the collection of all sets of the form . Prove that is a basis for the neighborhood system at . Let pi: X → Xi, i = 1, 2, … , n, be the projection that maps pi(a) = ai. Prove that each pi is continuous. Let Y be a metric space and f: Y → X a function. Prove that f is continuous if and only if each of the n functions pif is continuous.

8.  Let R be the real numbers and f:R → R a continuous function. Suppose that for some number . Prove that there is a positive number k and a closed interval F = [a − δ, a + δ] for some δ > 0 such that for .

5  LIMITS

The concept of limit of a sequence of real numbers may be generalized to an arbitrary metric space. First, let us recall the appropriate definition in the real line.

DEFINITION 5.1  Let a₁, a₂, … be a sequence of real numbers. A real number a is said to be the limit of the sequence a₁, a₂, … if, given ε > 0, there is a positive integer N such that, whenever n > N, |a − an| < ε. In this event we shall also say that the sequence a₁, a₂, … converges to a and write limn an = a.

Interpreting ε as an “arbitrary degree of closeness” and N as “sufficiently far out in the sequence,” we see that we have defined limn an = a in the event that an may be made arbitrarily close to a by requiring that an be sufficiently far out in the sequence.

Now, suppose that we have a metric space (X, d) and a sequence a₁, a₂, … of points of X. Given a point we measure the distance from a to the successive points of the sequence, by the sequence of real numbers d(a, a₁), d(a, a₂), … . It is natural to say that the limit of the sequence a₁, a₂, … of points of X is the point a if the limit of the sequence of real numbers d(a, a₁), d(a, a₂), … is the real number 0.

DEFINITION 5.2  Let (X, d) be a metric space. Let a₁, a₂, … be a sequence of points of X. A point is said to be the limit of the sequence a₁, a₂, … if limn d(a, an) = 0. Again, in this event, we shall say that the sequence a₁, a₂, … converges to a and write limn an = a.

COROLLARY 5.3  Let (X, d) be a metric space and a₁, a₂, … be a sequence of points of X. Then limn an = a for a point if and only if for each neighborhood V of a there is an integer N such that whenever n > N.

Proof. Let V be a neighborhood of a. For some . Thus if limn an = a there is an integer N such that whenever n > N, d(a, an) < ε and hence . Conversely, given ε > 0, B(a; ε) is a neighborhood of a. If there is an integer N such that for n > N, , then d(a, an) < ε and limn an = a.

If S is a set of infinite points, and there is at most a finite number of elements of S for which a certain statement is false, then the statement is said to be true for almost all the elements of S. Thus limn an = a if for each neighborhood V of a almost all the points an are in V.

Continuity may be characterized in terms of limits of sequences in accordance with the following theorem.

THEOREM 5.4  Let (X, d), (Y, d′) be metric spaces. A function f:X → Y is continuous at a point if and only if, whenever limn an = a for a sequence a₁, a₂, … of points of X, limn f(an) = f(a).

Proof. Suppose f is continuous at a and limn an = a. Let V be a neighborhood of f(a). Then f⁻¹(V) is a neighborhood of a, so by Corollary 5.3 there is an integer N such that whenever n > N. Consequently, whenever n > N. Thus, for each neighborhood V of f(a) there is an integer N such that whenever n > N and again, applying Corollary 5.3, limn f(an) = f(a).

To prove the “if” part of this theorem, we shall prove that if f is not continuous at a, then there is at least one sequence a₁, a₂, … of points of X, such that limn an = a, but limn f(an) = f(a) is false. Since f is not continuous at a, there is a neighborhood V of f(a) such that for each neighborhood U of . In particular, for each neighborhood , n = 1, 2, … . Thus, for each positive integer n, there is a point an with and . Now and therefore limn an = a, whereas, limn f(an) = f(a) is impossible, since for all n.

If limn an = a, we can write limn f(an) = f(a) as limn f(an) = f(limn an). We may therefore describe a continuous function as one that commutes with the operation of taking limits. It is worth noting that in proving f is continuous whenever f commutes with the operation of taking limits we have used the fact that the sequence of neighborhoods , n a positive integer, constitutes a basis for the neighborhood system at a.

In order to introduce the concept of distance from a point to a subset we shall recall some facts about the real number system.

DEFINITION 5.5  Let A be a set of real numbers. A number b is called an upper bound of A if for each . A number c is called a lower bound of A if for each . If A has both an upper and lower bound A is said to be bounded.

An upper bound b* of A is called a least upper bound (abbreviated l.u.b.) of A if for each upper bound b of A, . A lower bound c* of A is called a greatest lower bound (abbreviated g.l.b.) of A if for each lower bound c of A, .

Not every set of real numbers has an upper bound. One of the properties of the real number system, usually referred to as the completeness postulate, is that a non-empty set A of real numbers which has an upper bound necessarily has a l.u.b. Given a non-empty set B of real numbers which has a lower bound, the set of negatives of elements of B has an upper bound, hence a l.u.b. whose negative is a g.l.b. of B. Thus it follows that every non-empty set B of real numbers which has a lower bound has a g.l.b.

The greatest lower bound of a set A of real numbers may or may not be an element of A. For example, 0 is a g.l.b. of [0, 1] and , whereas 0 is also a g.l.b. of (0, 1) but . In any event, the g.l.b. of a set of real numbers must be arbitrarily close to that set.

LEMMA 5.6  Let b be a greatest lower bound of the non-empty subset A. Then, for each ε > 0, there is an element such that

Proof. Suppose there were an ε > 0 such that for each . Then for each and b + ε would be a lower bound of A. Since b is a g.l.b. of A, we obtain the contradiction .

COROLLARY 5.7  Let b be a greatest lower bound of the non-empty subset A of real numbers. Then there is a sequence a₁, a₂, … of real numbers such that for each n and limn an = b.

Proof. For we obtain an element such that . Since b is a lower bound of A, . Therefore limn an = b.

DEFINITION 5.8  Let (X, d) be a metric space. Let and let A be a non-empty subset of X. The greatest lower bound of the set of numbers of the form d(a, x) for is called the distance between a and A and is denoted by d(a, A).

From Corollary 5.7 we obtain:

COROLLARY 5.9  Let (X, d) be a metric space, , and A a non-empty subset of X. Then there is a sequence a₁, a₂, … of points of A such that limn d(a, an) = d(a, A).

EXERCISES

1.  Let X₁, X₂, … , Xk be metric spaces and convert into a metric space in the standard manner. Each of the points a₁, a₂, … of a sequence of points of X has k coordinates; that is an = Let . Prove that limn an = c if and only if limn .

2.  In each of the three metric spaces (Rk, d), (Rk, d′), (Rk, d″) considered in Section 2, prove that limits of sequences are the same.

3.  Prove that a subsequence of a convergent sequence is convergent and converges to the same limit as the original sequence.

4.  A sequence of real numbers a₁, a₂, … is called monotone non-decreasing if , for each i and called monotone non-increasing if , for each i. A sequence which is either monotone non-decreasing or monotone non-increasing is said to be monotone. The sequence is said to be bounded above if there is a number K such that for each i and bounded below if there is a number M such that for each i. A sequence which is both bounded above and bounded below is called bounded. Prove that a convergent sequence of real numbers is bounded. Prove that a monotone non-decreasing sequence of real numbers which is bounded above converges to a limit a and that a is the l.u.b. of the set {a₁, a₂, …}. Similarly prove that a monotone non-increasing sequence which is bounded below converges to a limit b and that b is the g.l.b. of the set {a₁, a₂, …}.

5.  Let a₁, a₂, … be a bounded sequence of real numbers. Since each of the sets An = {an, an+1, …} is bounded we may set vn = g.l.b. An, un = l.u.b. An. Observe that , … is monotone non-decreasing and bounded above; and u₁, u₂, … is monotone non-increasing and bounded below. Let V = limn un and U = limn un. Prove that there are subsequences of a₁, a₂, … which converge to U and V respectively (thus a bounded sequence of real numbers has a convergent subsequence). Prove that a₁, a₂, … converges if and only if U = V.

6.  Let (X, d) be a metric space and A a non-empty subset of X. For , prove that .

7.  Let A be a non-empty subset of a metric space (X, d). Define the function f:X → R by f(x) = d(x, A). Prove that f is continuous.

8.  Let A be a non-empty subset of a metric space (X, d) and let . Prove that d(x, A) = 0 if and only if every neighborhood of x contains a point of A.

9.  Let (X, d) be a metric space. Define a distance function d* on X × X by the method of Theorem 2.3. Prove that the function d:(X × X, d*) → (R, d) is continuous.

6  OPEN SETS AND CLOSED SETS

In a metric space, the open ball B(a; δ) is a neighborhood of each of its points (Lemma 4.5). The collection of subsets possessing this property plays a fundamental role in topology.

DEFINITION 6.1  A subset O of a metric space is said to be open if O is a neighborhood of each of its points.

Open sets may be characterized directly in terms of open balls.

THEOREM 6.2  A subset O of a metric space (X, d) is an open set if and only if it is a union of open balls.

Proof. Suppose O is open. Then for each , there is an open ball . Therefore 0 = is a union of open balls. Conversely, if O is a union of open balls, then using the centers of these balls as the elements of an indexing set we can write . If , then for some . B(a; δa) is a neighborhood of x and since , by N3, O is a neighborhood of x. Thus O is a neighborhood of each of its points, and by Definition 6.1, O is open.

Most of the functions considered in topology are continuous. Open sets provide a simple characterization of continuity.

THEOREM 6.3  Let f:(X, d) → (Y, d′). Then f is continuous if and only if for each open set O of Y, the subset f⁻¹(O) is an open subset of X.

Proof. First, suppose f is continuous. Let be open. We must show that f⁻¹(O) is open; that is, f⁻¹(O) is a neighborhood of each of its points. To this end, let , then and O is a neighborhood of f(a). Since f is continuous at a, Theorem 4.7 may be applied, yielding f⁻¹(O) is a neighborhood of a.

Conversely, suppose for each open set , f⁻¹(O) is open. Let and let M be a neighborhood of f(a). Then there is an ε > 0 such that . But B(f(a); ε) is open and therefore f⁻¹(B(f(a); ε)) is open. Since , this subset is a neighborhood of a. Therefore f⁻¹(M) contains a neighborhood of a and f is continuous at a. Since a was arbitrary, f is continuous.

Just as the collections of neighborhoods of points in a metric space possess certain significant properties so do the collection of open sets in a metric space.

THEOREM 6.4  Let (X, d) be a metric space.

O1.  The empty set is open.

O2.  X is open.

O3.  If O₁, O₂, … , On are open, then O₁ ∩ O₂ ∩ … ∩ On is open.

O4.  If for each , Oα is an open set, then is open.

Proof. The empty set is open, for in order for it not to be open there would have to be a point . Given a point , for any δ > 0, and therefore X is a neighborhood of each of its points; that is, X is open. To prove O3, let , where for i = 1, 2, … , n, Oi is open. Then each Oi is a neighborhood of a. By N4, the intersection of two neighborhoods of a is again a neighborhood of a, and hence by induction, the intersection of a finite number of neighborhoods of a is again a neighborhood of a. Therefore O₁ ∩ O₂ ∩ … ∩ On is a neighborhood of each of its points. Finally, to prove O4, let , where for each , Oα is open. Then for some and Oβ is a neighborhood of a. Since , by N3, O is a neighborhood of a. Therefore O is a neighborhood of each of its points.

DEFINITION 6.5  A subset F of a metric space is said to be closed if its complement, C(F), is open.

In the real number system, a closed interval [a, b] is a closed set, for its complement is the union of the two open sets O₁ and O₂, where O₁ is the set of real numbers x such that x < a and O₂ is the set of real numbers x such that x > b. A common mistake is the assumption that a set cannot be both open and closed. In any metric space (X, d), the two sets Ø and X are open, and therefore their complements X and Ø are closed. Thus, X and also Ø are both open and both closed. Whether or not, in a given metric space, there are other subsets that are simultaneously open and closed, is a significant topological property, which we shall subsequently describe by the adjective “connected.” In any event, the adjectives open and closed are not mutually exclusive. Nor, for that matter, are they all-inclusive, for we shall shortly give an example of a subset of the real number system that is neither open nor closed.

DEFINITION 6.6  Let A be a subset of a metric space X. A point is called a limit point of A if every neighborhood of b contains a point of A different from b.

If b is a limit point of A then each of the open balls contains a point and limn an = b. Thus a limit point of a set is the limit of a convergent sequence of points of A. The converse is false, for the point b may be a point of A while for some δ, B(b; δ) contains no point of A other than b. Thus b is not a limit point of A although the sequence b, b, … converges to b. In this latter case b is called an isolated point of A.

THEOREM 6.7  In a metric space X, a set is closed if and only if F contains all its limit points.

Proof. Let F′ denote the set of limit points of F. First suppose F is closed and consequently C(F) is open. Choose a point . Since C(F) is open there is a δ > 0 such that or . Hence and .

Conversely, suppose , or equivalently, . If , then . It follows that for some , or . Hence C(F) is open and F is closed.

THEOREM 6.8  In a metric space (X, d), a set is closed if and only if for each sequence a₁, a₂, … of points of F that converges to a point we have .

Proof. First, let F be closed. Suppose limn an = a and for n = 1, 2, … . If the set of points {a₁, a₂, …} is infinite then every neighborhood of a contains infinitely many points of F, a is a limit point of F, and so by Theorem 6.7, . If this set of points is finite, then for some integer N, an = am whenever n, m > N. Since limn an = a, d(an, a) = 0 for n > N or an = a, whence . Conversely, suppose that F is a set such that for each sequence with limn an = a and for all n, we have . If b is a limit point of F then b is the limit of a convergent sequence of points of F and . Thus by Theorem 6.7 F is closed.

Finally, we may characterize closed sets in terms of distance from a point to a set.

THEOREM 6.9  A subset F of a metric space (X, d) is closed if and only if for each point , d(x, F) = 0 implies .

Proof. First, suppose F is closed. Let be such that d(x, F) = 0. By Corollary 5.9 there is a sequence of points of F such that limn d(x, an) = 0. Thus, every neighborhood of x contains points of F. If some an = x, x is in F. Otherwise each an is different from x, so that x is a limit point of the sequence and hence of F. Thus, by Theorem 6.7, . Conversely, suppose that F is such that d(x, F) = 0 implies . If x is a limit point of F then d(x, F) = 0. Thus in this case F contains all its limit points and is closed.

Continuity may be characterized by means of closed sets.

THEOREM 6.10  Let (X, d), (Y, d′) be metric spaces. A function f:X → Y is continuous if and only if for each closed subset A of Y, the set f⁻¹(A) is a closed subset of X.

Proof. For , we have C(f⁻¹(A)) = f⁻¹(C(A)). But f is continuous if and only if the inverse image of each open set is an open set, and this is true if and only if the inverse image of each closed set is a closed set.

As a final result in this section we record the following facts about closed sets.

THEOREM 6.11  Let (X, d) be a metric space.

C1. X is closed.

C2. Ø is closed.

C3. The union of a finite collection of closed sets is closed.

C4. The intersection of a family of closed sets is closed.

Proof. C1 and C2 have already been discussed. C3 and C4 follow from the application of DeMorgan’s formulas to the corresponding properties O3 and O4 of open sets.

The union of closed sets need not, in general, be a closed set, as may be seen by the following example. For each positive integer n let Fn be the closed interval . Then , where (0, 1] is the set of real numbers x such that . The set (0, 1] is not closed, for 0 is a limit point of the set but is not in the set.

EXERCISES

1.  Let (Xi, di), i = 1, 2, … , n be metric spaces. Let , and let (X, d) be the metric space defined in the standard manner by Theorem 2.3. For i = 1, 2, … , n, let Oi be an open subset of Xi. Prove that the subset O₁ × O₂ × … × On of X is open and that each open subset of X is a union of sets of this form. [A collection of open sets of a metric space is called a basis for the open sets if each open set is a union of sets in this collection. For example, the open balls in a metric space form a basis for the open sets.]

2.  Let X be a set and d the distance function on X defined by d(x, x) = 0, d(x, y) = 1 for x ≠ y. Prove that each subset of (X, d) is open.

3.  Let (X, d₁), (Y, d₂) be metric spaces. Let f:X → Y be continuous. Define a distance function d on X × Y in the standard manner. Prove that the graph Γf of f is a closed subset of (X × Y, d).

4.  Let f:R → R be defined by

Prove that the graph Γf is a closed subset of (R², d), but that f is not continuous.

5.  Let A be a closed, non-empty subset of the real numbers that has a lower bound. Prove that A contains its greatest lower bound.

6.  Let A be a subset of a metric space. Let A′ be the set of limit points of A and Ai the set of isolated points of A. Prove that A′ ∩ Ai = Ø and . The set is called the closure of A. Prove that if and only if there is a sequence of points of A which converges to x. Prove that if F is a closed set such that then . Prove that Ā is the intersection of all such closed sets F and hence is closed.

7  SUBSPACES AND EQUIVALENCE OF METRIC SPACES

Let (X, d) be a metric space. Given a non-empty subset Y of X we may convert Y into a metric space by restricting the distance function d to Y × Y. In this manner each non-empty subset Y of X gives rise to a new metric space (Y, d | Y × Y). On the other hand, we may be given two metric spaces (X, d) and (Y, d′). If , it makes sense to ask whether or not d′ is the restriction of d.

DEFINITION 7.1   Let (X, d) and (Y, d′) be metric spaces. We say that (Y, d′) is a subspace of (X, d) if:

1.  

2.  d′ = d | Y × Y.

Let and i:Y → X be an inclusion mapping. Denote by i × i:Y × Y → X × X the inclusion mapping defined by (i × i)(y₁, y₂) = (y₁, y₂). Then (Y, d′) is a subspace of (X, d) if the diagram

is commutative. There are as many subspaces of a metric space (X, d) as there are non-empty subsets of X.

EXAMPLE 1  Let Q be the set of rational numbers. Define dQ:Q × Q → R by dQ(a, b) = |a − b|. Then (Q, dQ) is a subspace of (R, d).

EXAMPLE 2  Let In (the unit n-cube) be the set of all n-tuples (x₁, x₂, … , xn) of real numbers such that , for i = 1, 2, … , n. Define dc:In × In → R by dc((x₁, x₂, … , xn), . Then (In, dc) is a subspace of (Rn, d).

EXAMPLE 3  Let Sn (the n-sphere) be the set of all (n + 1)-tuples (x₁, x₂, … , xn+1) of real numbers such that . Define ds:Sn × Sn → R by

Then (Sn, ds) is a subspace of the Euclidean space (Rn+1, d′).

EXAMPLE 4  Let A be the set of all (n + 1)-tuples (x₁, x₂, … , xn+1) of real numbers such that xn+1 = 0. Define dA:A × A → R by

Then (A, dA) is a subspace of (Rn+1, d).

THEOREM 7.2  Let (Y, d′) be a subspace of (X, d). Then the inclusion mapping i:Y → X is continuous.

Proof. Given and ε > 0, choose δ = ε. If d′(a, y) < δ, then d(i(a), i(y)) = d(a, y) = d′(a, y) < δ = ε.

The metric space (A, dA) of Example 4 is in most respects a copy of the metric space (Rn, d). The only distinction between (Rn, d) and (A, dA) is that a point of Rn is an n-tuple of real numbers, whereas a point of A is an (n + 1)-tuple of real numbers of which the last one is zero. The relationship between the metric spaces (Rn, d) and (A, dA) is an example of the relationship called “metric equivalence” or “isometry.”

DEFINITION 7.3  Two metric spaces (A, dA) and (B, dB) are said to be metrically equivalent or isometric if there are inverse functions f:A → B and g:B → A such that, for each , and for each u, . In this event we shall say that the metric equivalence or isometry is defined by f and g.

THEOREM 7.4  A necessary and sufficient condition that two metric spaces (A, dA) and (B, dB) be metrically equivalent is that there exist a function f:A → B such that:

1.  f is one-one;

2.  f is onto;

3.  for each

Proof. The stated conditions are necessary, for if (A, dA) and (B, dB) are metrically equivalent, there are inverse functions f:A → B and g:B → A, and therefore f is one-one and onto. Conversely, suppose a function f:A → B with the stated properties exists. Then f is invertible and the function g:B → A such that f and g are inverse functions is determined by setting g(b) = a if f(a) = b. For u, , let x = g(u), y = g(v). Then

Given metric spaces (A, dA) and (B, dB) and functions f:A → B and g:B → A, let us denote by f × f:A × A → B × B the function defined by setting (f × f)(x, y) = (f(x), f(y)) for and, similarly, let g × g:B × B → A × A be defined by setting (g × g)(u, v) = (g(u), g(v)) for . The statement that dB(f(x), f(y)) = dA(x, y) for is equivalent to the statement that the diagram

is commutative (one may also describe this relation by saying that the function f:A → B is “distance preserving”). In terms of diagrams, the statement that (A, dA) and (B, dB) are metrically equivalent is the statement that there exist functions f:A → B, g:B → A such that the four diagrams

are commutative (where iA:A → A and iB:B → B are identity mappings). The first two diagrams express the fact that f and g are inverse functions and the last two diagrams express the fact that f and g “preserve distances.” Since the distance between x and y in A is the same as the distance between f(x) and f(y) in B, f is continuous. Similarly, g is continuous. Thus:

LEMMA 7.5  Let a metric equivalence between (A, dA) and (B, dB) be defined by inverse functions f:A → B and g:B → A. Then both f and g are continuous.

From the point of view of considerations that relate only to the concept of continuity, the relationship of metric equivalence is too narrow. We are led to define a broader concept of equivalence in which we drop the requirement of “preservation of distance”; that is, the commutativity of the last pair of diagrams, and merely require that the first two diagrams be commutative and the functions in these diagrams be continuous.

DEFINITION 7.6  Two metric spaces (A, dA) and (B, dB) are said to be topologically equivalent if there are inverse functions f:A → B and g:B → A such that f and g are continuous. In this event we say that the topological equivalence is defined by f and g.

As a corollary to Lemma 7.5 we obtain:

COROLLARY 7.7  Two metric spaces that are metrically equivalent are topologically equivalent.

The converse of this corollary is false; that is, there are metric spaces that are topologically equivalent, but are not metrically equivalent. For example, a circle of radius 1 is topologically equivalent to a circle of radius 2 (considered as subspaces of (R², d)), but the two are not metrically equivalent.

The following two results furnish a sufficient condition for the topological equivalence of two metric spaces with the same underlying sets.

LEMMA 7.8  Let (X, d₁) and (X, d₂) be two metric spaces. If there exists a number K > 0 such that for each , then the identity mapping

is continuous.

Proof. Given ε > 0 and , set δ = ε/K. If d₁(x, a) < δ then .

COROLLARY 7.9  Let (X, d) and (X, d′) be two metric spaces with the same underlying set. If there exist positive numbers K and K′ such that for each ,

then the identity mappings define a topological equivalence between (X, d) and (X, d′).

We have discussed the two metric spaces (Rn, d) and (Rn, d′), where the distance function d is determined by the maximum distance between coordinates, and the distance function d′ is what is called the Euclidean distance function. For each pair of points , the inequality holds. It therefore follows from Corollary 7.9 that the metric spaces (Rn, d) and (Rn, d′) are topologically equivalent.

THEOREM 7.10  Let (X, d) and (Y, d′) be two metric spaces. Let f:X → Y and g:Y → X be inverse functions. Then the following four statements are equivalent:

1.  f and g are continuous;

2.  A subset O of X is open if and only if f(O) is an open subset of Y;

3.  A subset F of X is closed if and only if f(F) is a closed subset of Y;

4.  For each and subset N of X, N is a neighborhood of a if and only if f(N) is a neighborhood of f(a).

Proof. 1 ⇒ 2. Let O be an open subset of X. Then f(O) = g⁻¹(O) is open since g is continuous. Conversely, if f(O) is an open subset of Y, then f⁻¹(f(O)) = O is open since f is continuous.

2 ⇒ 4. For each and , N is a neighborhood of a if and only if N contains an open set O containing a if and only if f(N) contains an open set O′ = f(O) containing f(a) if and only if f(N) is a neighborhood of f(a).

4 ⇒ 1. Let and let U be a neighborhood of f(a). Then f⁻¹(U) is a neighborhood of a, for U = f(f⁻¹(U)) is a neighborhood of f(a). Thus f is continuous. Similarly, let and let V be a neighborhood of g(b). Then g⁻¹(V) = f(V) is a neighborhood of f(g(b)) = b, and g is continuous.

Thus, statements 1, 2, and 4 are equivalent. We leave it to the reader to verify that statements 2 and 3 are equivalent.

Statement 1 in Theorem 7.10 is, of course, the statement that the metric spaces (X, d) and (Y, d′) are topologically equivalent. Consequently, Theorem 7.10 asserts that two metric spaces are topologically equivalent if and only if there exist inverse functions that establish either a one-one correspondence between the open sets of the two spaces, a one-one correspondence between the closed sets of the two spaces, or a one-one correspondence between the complete systems of neighborhoods of the two spaces.

Both metrically equivalent and topologically equivalent are equivalence relations defined on a collection of metric spaces. By Corollary 7.7, each equivalence class of metrically equivalent metric spaces is contained in an equivalence class of topologically equivalent metric spaces. Distinguishing which topologically equivalent equivalence class a metric space belongs to is a coarser, but consequently more fundamental, distinction. By Theorem 7.10, this is determined by the collection of open sets, or the “topology” of the space.

EXERCISES

1.  For each pair of points , prove that there is a topological equivalence between (Rn, d) and itself defined by inverse functions f:Rn → Rn and g:Rn → Rn such that f(a) = b. [Hint: If a = (a₁, a₂, … , an), b = (b₁, b₂, … , bn), define f by setting f(x₁, x₂, … , xn) = (x₁ + b₁ − a₁, x₂ + b₂ − a₂, … , xn + bn − an).]

2.  Prove that the open interval (−π/2, π/2), considered as a subspace of the real number system, is topologically equivalent to the real number system. Prove that any two open intervals, considered as subspaces of the real number system, are topologically equivalent. Prove that any open interval, considered as a subspace of the real number system, is topologically equivalent to the real number system.

3.  For i = 1, 2, … , n, let the metric space (Xi, di) be topologically equivalent to the metric space (Yi, d′i). Prove that if

are converted into metric spaces in the standard manner, then these two metric spaces are topologically equivalent.

4.  The open n-cube is the set of all points such that 0 < xi < 1 for i = 1, 2, … , n. Prove that the open n-cube, considered as a subspace of (Rn, d), is topologically equivalent to (Rn, d). [Hint: Use the results of Problems 2 and 3.]

5.  Let XRY mean that the metric space X is isometric to the metric space Y. Prove that: (i) XRX; (ii) if XRY then YRX; and (iii) if XRY and YRZ then XRZ. Do the same if XRY means that the metric space X is topologically equivalent to the metric space Y.

6.  Let (Y, d′) be a subspace of the metric space (X, d). Prove that a subset is an open subset of (Y, d′) if and only if there is an open subset O of (X, d) such that O′ = Y ∩ O. Prove that a subset is a closed subset of (Y, d′) if and only if there is a closed subset F of (X, d) such that F′ = Y ∩ F. For a point , prove that a subset is a neighborhood of a if and only if there is a neighborhood N of a in (X, d) such that N′ = Y ∩ N.

7.  Let (Y, d′) be a subspace of (X, d). Let a₁, a₂, … be a sequence of points of Y and let . Prove that if limn an = a in (Y, d′), then limn an = a in (X, d). [The converse is false unless one assumes that all the points mentioned lie in Y; see the next problem.]

8.  Consider the subspace (Q, dQ) (the rational numbers) of (R, d). Let a₁, a₂, … be a sequence of rational numbers such that . Prove that, given ε > 0, there is a positive integer N such that for n, m > N, |an − am| < ε. Does the sequence a₁, a₂, … converge when considered to be a sequence of points of (Q, dQ)?

8  AN INFINITE DIMENSIONAL EUCLIDEAN SPACE

In this section we shall define a metric space H, sometimes called Hilbert space, which contains as subspaces isometric copies of the various Euclidean spaces (Rn, d′). A point u of H is a sequence u₁, u₂, … of real numbers such that the series is convergent.

Let u = (u₁, u₂, …) and v = (v₁, v₂, …) be in H. Our intention is to define a metric on H by setting

In order to do this we must first know that the series in brackets converges. To accomplish this we shall make use of the following result, which is frequently referred to as Schwarz’s lemma or Cauchy’s inequality.

LEMMA 8.1  Let (u₁, u₂, … , un), (v₁, v₂, … , vn) be n-tuples of real numbers, then

Proof. It suffices to prove that

To this end, we consider, for an arbitrary real number λ, the expression . We have,

Therefore, the quadratic equation in λ,

can have at most one real solution. Consequently,

or

COROLLARY 8.2  Let u = (u₁, u₂, …), v = (v₁, v₂, …) be in H with Than the series absolutely convergent and .

Proof. For each positive integer n

Thus the partial sums of this series of positive terms are bounded and the series converges to a limit not greater than U1/2V1/2.

Furthermore, if α and β are real numbers and we set αu + βv = (αu₁ + βv₁, αu₂ + βv₂, …) then αu + βv is also in H for is the sum of three absolutely convergent series. In particular and

Taking square roots we obtain

COROLLARY 8.3  .

THEOREM 8.4  (H, d) is a metric space, where d is defined by d(u,v) = .

Proof. It is readily apparent that d satisfies all the properties of a distance function with the exception of the property that . Let a = (a₁, a₂, …), b = (b₁, b₂, …), c = (c₁, c₂, …). Set u = a − c, v = c − b so that ui = ai − ci, vi = ci − bi. Then ui + vi = ai − bi and Corollary 8.3 yields the desired inequality.

Let En be the collection of points such that uj = 0 for j > n. To each point we can associate the point . Clearly h is a one-one mapping of Rn onto the subspace En of H. Using . Since En is a metric space, (Rn, d′) is a metric space and h is an isometry of (Rn, d′) with (En, d|En).

EXERCISES

1.  Let V be a vector space with the real numbers R as scalars. A function A:V × V → R is called a bilinear form if A (αa + βb, c) = αA(a, c) + βA(b, c) and A(a, βb + γc) = βA(a, b) + γA(a, c) for scalars α, β, and and vectors a, b, and . A bilinear form is called positive definite if A (x, x) > 0, unless x is the zero vector. Define a vector space structure on Hilbert space H and show that for u = (u₁, u₂, …) and yields a positive definite bilinear form.

2.  Let V be a vector space with the real numbers R as scalars. A norm on V is a function N:V → R such that (i) for all ; (ii) N(v) = 0 if and only if v = 0; (iii) for all u, ; (iv) for all . Prove that if A is a positive definite bilinear form on V, then N(v) = (A(v, v))1/2 defines a norm on V.

3.  Let N be a norm on a vector space V as defined in the previous problem. Set d(u, v) = N(u − v) for u, . Prove that (V, d) is a metric space. Prove that the following functions are continuous: (i) a:V × V → V defined by a(u, v) = u + v; (ii) b:V → V defined by b(v) = −v; (iii) c:R × V → V defined by c(α, v) = αv.

For further reading, Kaplansky, Set Theory and Metric Spaces, Kolmogorov and Fomin, Elements of the Theory of Functions and Functional Analysis, and Simmons, Introduction to Topology and Modern Analysis all have excellent chapters on metric spaces.

CHAPTER 3

Topological Spaces

1  INTRODUCTION

In the context of metric spaces, the various topological concepts such as continuity, neighborhood, and so on, may be characterized by means of open sets. Discarding the distance function and retaining the open sets of a metric space gives rise to a new mathematical object, called a topological space. The topological concepts that have been studied in Chapter 2 must be reintroduced in the context of topological spaces. The procedure for formulating the appropriate definitions of these terms in a topological space is to find, in a metric space, the characterization of the term by means of open sets, using in most cases what is a theorem in a metric space as a definition in a topological space. There are other ways of introducing topological spaces. For example, if, upon discarding the distance function of a metric space, we were to retain the systems of neighborhoods of the points of the metric space, we obtain what we shall call a neighborhood space. We shall indicate the equivalence between the concept of a neighborhood space and the concept of a topological space. Certain new topological concepts are also introduced; namely, the closure, interior, and boundary of a set (these concepts could have been introduced in metric spaces). In many respects the elementary material in this chapter is a repetition of material from Chapter 2, but in a different context. The concept of a topological space is one of the most fruitful concepts of modern mathematics. It is the proper setting for discussions based on considerations of continuity.

2  TOPOLOGICAL SPACES

DEFINITION 2.1  Let X be a non-empty set and a collection of subsets of X such that:

O1. .

O2. .

O3. If , then

O4. If for each , then .

The pair of objects is called a topological space. The set X is called the underlying set, the collection is called the topology on the set X, and the members of are called open sets.

By virtue of Theorem 6.4, Chapter 2, if is the collection of open sets of a metric space (X, d), then is a topological space, called the topological space associated with the metric space (X, d), and the metric space (X, d) is said to give rise to the topological space . We are therefore in a position to give many examples of topological spaces; namely, for each metric space its associated topological space. On the other hand, any set X and collection of subsets satisfying O1, O2, O3, O4 is an example of a topological space, and we shall see that not every such example arises from a metric space.

EXAMPLES

1.  The real line, that is, the topological space that arises from the metric space consisting of the real number system and the distance function d(a, b) = |a − b|.

2.  The topological space that arises from the metric space (Rn, d). We shall call this topological space Euclidean n-space with the usual topology.

3.  Let X be an arbitrary set. Let . Then is a topological space.

4.  Let X be a set containing precisely two distinct elements a and b. Let . Then are four distinct topological spaces with the same underlying set.

5.  Let X be an arbitrary set. Let be the collection of all subsets of X, i.e., . Then is a topological space. Of all the various topologies that one may place on a set X, this one contains the largest number of elements and is called the discrete topology.

6.  Let X be an arbitrary set. Let be the collection of all subsets of X whose complements are either finite or all of X. Then is a topological space.

7.  Let Z be the set of positive integers. For each positive integer n, let On = {n, n + 1, n + 2, . . .}. Let . Then is a topological space.

To verify that is a topological space, one verifies that the specified collection of subsets, , is a topology; that is, that satisfies conditions O1, O2, O3, O4. For example, let X and be as in Example 6. Then , for its complement Ø = C(X) is certainly finite. Also , since C(Ø) = X. Thus, satisfies conditions O1 and O2. Next, let O₁, O₂, . . . , On be subsets of X, each of whose complements is finite or all of X. To show that we must show that is either finite or all of X. But . Either this set is a union of finite sets and hence finite, or for some i, C(Oi) = X and the union is all of X. Finally, for each , let , so that C(Oα) is either finite or X. Then . Either each of the sets, C(Oα) = X, in which case the intersection is all of X, or at least one of them is finite, in which case the intersection is a subset of a finite set and hence finite. Thus is a topological space. The reader should verify that the remaining examples do, in fact, constitute examples of topological spaces.

The relationship between the totality of metric spaces and the totality of topological spaces is indicated in Figure 8. We shall see that two distinct metric spaces (X, d) and (X, d′) may give rise to the same topological space . Also there are topological spaces , such as Example 7 above, which could not have arisen from a metric space. The subcollection of topological spaces that arise from metric spaces is called the collection of metrizable topological spaces. In passing from a metric space to its associated topological space, we may say that the “open” sets have been “preserved.”

Figure 8

DEFINITION 2.2  Given a topological space , a subset N of X is called a neighborhood of a point if N contains an open set that contains a.

This definition has been formulated so that a subset N of a metric space (X, d) is a neighborhood of a point if and only if N is a neighborhood of a in the associated topological space. Thus, in passing from a metric space to a topological space, neighborhoods have also been “preserved.”

COROLLARY 2.3  Let be a topological space. A subset O of X is open if and only if O is a neighborhood of each of its points.

Proof. First, suppose that O is open. Then, for each O contains an open set containing x; namely, O itself. Conversely, suppose O is a neighborhood of each of its points. Then for each , there is an open set Ox such that . Consequently, is a union of open sets and hence open.

DEFINITION 2.4  Given a topological space , a subset F of X is called a closed set if the complement, C(F), is an open set.

EXERCISES

1.  Let be a topological space that is metrizable. Prove that for each pair a, b of distinct points of X, there are open sets Oa and Ob containing a and b respectively, such that Oa ∩ Ob = Ø. Prove that the topological space of Example 7 is not metrizable.

2.  Prove that for each set X, the topological space (X, 2X) is metrizable. [Hint: See Exercise 2, Chapter 2, Section 6.]

3.  Let (Rn, d) and (Rn, d′) be defined as in Chapter 2 so that for x = (x₁, x₂, . . . , xn) and ,

Prove that the two metric spaces (Rn, d) and (Rn, d′) give rise to the same topological space.

4.  Let be a topological space. Prove that Ø, X are closed sets, that a finite union of closed sets is a closed set, and that an arbitrary intersection of closed sets is a closed set.

5.  Let be a topological space that is metrizable. Prove that each neighborhood N of a point contains a neighborhood V of a such that V is a closed set.

6.  Prove that in a discrete topological space, each subset is simultaneously open and closed.

3  NEIGHBORHOODS AND NEIGHBORHOOD SPACES

Theorem 4.8, Chapter 2, in which are stated certain properties of neighborhoods in a metric space, corresponds to a theorem in topological spaces.

THEOREM 3.1  Let be a topological space.

N1. For each point , there is at least one neighborhood N of x.

N2. For each point and each neighborhood N of .

N3. For each point , if N is a neighborhood of x and , then N′ is a neighborhood of x.

N4. For each point and each pair N, M of neighborhoods of is also a neighborhood of x.

N5. For each point and each neighborhood N of x, there exists a neighborhood O of x such that and O is a neighborhood of each of its points.

Proof. For each point , X is a neighborhood of x, thus N1 is true. N2 and N3 follow easily from the definition of neighborhood in a topological space. To verify N4, let N, M be neighborhoods of x. Then there are open sets O and O′ such that and . Thus, N ∩ M contains the open set O ∩ O′, which contains x, and, consequently, N ∩ M is a neighborhood of x. Finally, for a point , let N be a neighborhood of x. Then N contains an open set O containing x. In particular, O is a neighborhood of x and by Corollary 2.3, O is a neighborhood of each of its points.

In a topological space, as in a metric space, we lay down the definition:

DEFINITION 3.2  For each point x in a topological space , the collection of all neighborhoods of x is called a complete system of neighborhoods at the point x.

One may paraphrase the properties N1–N5 of neighborhoods in terms of the complete system of neighborhoods at the points :

N1. For each ;

N2. For each and ;

N3. For each and , if then ;

N4. For each and

N5. For each and , there exists an such that and for each .

The proof of Theorem 3.1 was, in most respects, similar to the proof of the corresponding theorem in metric spaces, Theorem 4.8, Chapter 2. However, it was necessary to supply a proof of Theorem 3.1 above, for in the proof of 4.8, Chapter 2, use was made of the concept of open balls, a concept which does not occur in a topological space. Though a comparison of these two theorems might lead one to believe that statements about neighborhoods that are true in a metric space are also true in a topological space, this is not always the case. Given two distinct points x and y in a metric space (X, d) there are neighborhoods N and M of x and y respectively, such that N ∩ M = Ø. This statement is false in many topological spaces. For example, let Y = {a, b}, a ≠ b, and let , so that is a topological space. Then the only neighborhood of b is Y. Thus, for each neighborhood N of a and each neighborhood M of b, N ∩ M = N ∩ Y = N ≠ Ø.

DEFINITION 3.3  A topological space is called a Hausdorff space or is said to satisfy the Hausdorff axiom, if for each pair a, b of distinct points of X, there are neighborhoods N and M of a and b respectively, such that N ∩ M = Ø.

Some authors use the term “separated space” instead of Hausdorff space. Many of the significant topological spaces are Hausdorff spaces. For this reason certain authors require a topological space to be a Hausdorff space and use the two terms synonymously; that is, they add to the list O1–O4 of properties of open sets in the definition of a topological space, the property, for each pair x, y of distinct points there are open sets Ox and Oy containing x and y respectively, such that Ox ∩ Oy = Ø.

Suppose we have a metric space (X, d) and we discard the distance function, retaining only the neighborhoods of the points in X. Then for each point , we have a collection of subsets of X; namely the complete system of neighborhoods at x. These neighborhoods satisfy certain properties. We may select some of these properties and use them as a set of axioms for what we might naturally call a “neighborhood space.”

DEFINITION 3.4  Let X be a set. For each , let there be given a collection of subsets of X (called the neighborhoods of x), satisfying the conditions N1–N5 of Theorem 3.1. This object is called a neighborhood space.

In a neighborhood space, the appropriate definition of open set is obtained from Corollary 2.3.

DEFINITION 3.5  In a neighborhood space, a subset O is said to be open if it is a neighborhood of each of its points.

It is important to realize that the mathematical object neighborhood space, although closely connected with the concept of a topological space, is a new object, and until we have defined the term open set in a neighborhood space, that term in a neighborhood space is meaningless.

LEMMA 3.6  In a neighborhood space, the empty set and the whole space are open, a finite intersection of open sets is open, and an arbitrary union of open sets is open.

Proof. [Since we are concerned with neighborhood spaces, we may use only the properties N1–N5 of neighborhoods and, of course, Definition 3.5 of open sets.] The empty set is open, for in order for it not to be open it would have to contain a point x of which it was not a neighborhood. Given a point x, there is some neighborhood N of x, so by N3, the whole space is a neighborhood of x. Thus, the whole space is a neighborhood of each of its points and hence open. If O and O′ are open, then O ∩ O′ is also open, for by N4, given , O and O′ are neighborhoods of x, hence so is O ∩ O′. Thus the intersection of two open sets is a neighborhood of each of its points, and, consequently, by induction, any finite intersection of open sets is open. Finally, suppose for each , Oα is open. If , then for some . But Oβ is a neighborhood of x and , thus by N3, is a neighborhood of x and is therefore open.

If we start with a topological space and define neighborhoods by Definition 2.2, Theorem 3.1 tells us that the underlying set and the complete systems of neighborhoods of the points of the set yield a neighborhood space. On the other hand, if we start with a neighborhood space and define open sets by Definition 3.5, Lemma 3.6 tells us that we obtain a topological space. Suppose then, we have a topological space , use the neighborhoods of to form a neighborhood space, and finally use the open sets in this neighborhood space to create a topological space . Do we end up with our original topological space ? The answer is yes. To prove this result we must show that . Now, if O is an open set in our original topological space, that is, , by Corollary 2.3, O is a neighborhood of each of its points, from which it follows that O is an open subset of the neighborhood space and hence . Conversely, if , then in the neighborhood space, O is a neighborhood of each of its points. But the neighborhoods of the neighborhood space we have created are the neighborhoods of , so that again by Corollary 2.3, O is open in or . Thus .

Logically, it would still be possible for there to be neighborhood spaces that did not arise in this manner from topological spaces. We shall now show that there are none. To do so, we need a characterization, in a neighborhood space, of neighborhoods in terms of open sets.

LEMMA 3.7  In a neighborhood space, a subset N is a neighborhood of a point x if and only if N contains an open set containing x.

Proof. First, let N contain an open set O containing x. By Definition 3.5, O is a neighborhood of x, whence, by N3, N is a neighborhood of x. Conversely, if N is a neighborhood of x, then by N5, N contains a neighborhood O of x (and by N2, O contains x), such that O is a neighborhood of each of its points.

To denote a neighborhood space, let us use the symbol , where for each , is the collection of neighborhoods of x. Now suppose that we start with a neighborhood space . We define open set in by Definition 3.5, thus obtaining a topological space . In the topological space we define neighborhood by Definition 2.2 to obtain a neighborhood space . Under these circumstances, if , by Lemma 3.7, N contains an open set O containing x, so that by Definition 2.2, N is a neighborhood of x in , or . Conversely, if , then by Definition 2.2, N contains a set , and . Since , O is open in the neighborhood space and so by Lemma 3.7, N is a neighborhood of x. Thus, for each , and the two neighborhood spaces are the same.

Collecting together the results on the correspondence between topological spaces and neighborhood spaces, we have:

THEOREM 3.8  Let neighborhood in a topological space be defined by Definition 2.2 and open set in a neighborhood space be defined by Definition 3.5. Then the neighborhoods of a topological space give rise to a neighborhood space and the open sets of a neighborhood space give rise to a topological space . Furthermore, for each topological space ,

and for each neighborhood space ,

thus establishing a one-one correspondence between the collection of all topological spaces and the collection of all neighborhood spaces.

Theorem 3.8 justifies the specification of a topological space by defining for a given set X what subsets of X are to be the neighborhoods of a point ; that is, by specifying the corresponding neighborhood space. For example, let X be the set of positive integers. Given a point and a subset U of X, let us call U a neighborhood of n if for each integer . We must then verify that these neighborhoods satisfy conditions N1–N5 so that we have a neighborhood space and consequently a topological space. The reader should verify that this corresponding topological space is the one described in Example 7 of Section 2.

EXERCISES

1.  Given a real number x, call a subset N of R a neighborhood of x if implies . Prove that this definition of neighborhood yields a neighborhood space. Describe the corresponding topological space.

2.  Given a real number x, call a subset N of R a neighborhood of x if N contains the closed interval [x, x + 1]. Prove that the neighborhoods so defined satisfy N1–N4, but not N5. Use the Definition 3.5 of open set anyway, and determine which subsets of R will be open.

3.  In a neighborhood space, a collection of neighborhoods of a point is called a basis for the complete system of neighborhoods at x, or simply a basis for the neighborhoods at x, if, for each neighborhood N of x, there is a neighborhood such that .

Prove that if for each point , is a basis for the neighborhoods at x, then:

BN1. For each , ;

BN2. For each and ;

BN3. For each and , U ∩ V contains an element ;

BN4. For each and , there is an such that and for each , O contains an element .

4.  Define a basic neighborhood space to be a set X, such that for each a collection of subsets of X satisfies the conditions BN1–BN4 of Problem 3. In a basic neighborhood space define a subset N of X to be a neighborhood of a point if for some . Prove that the neighborhoods of a basic neighborhood space yield a neighborhood space. (Thus a topological space may be constructed by specifying for each point x a basis of the neighborhoods at x satisfying BN1–BN4.) The correspondence between basic neighborhood spaces and neighborhood spaces is many-one, since there are many different bases for the neighborhoods at a point in a neighborhood space. However, prove that if and are two basic neighborhood spaces, then they give rise to the same neighborhood space if and only if for each point we have

(i)  given , there is a with , and

(ii)  given , there is a with .

Also prove that starting from a given neighborhood space , if for each , is a basis for the neighborhoods at x, then the neighborhood space that arises from the basic neighborhood space is .

4  CLOSURE, INTERIOR, BOUNDARY

In a metric space, given a point x and a subset A, we can say that there are points of A arbitrarily close to x if d(x, A) = 0. In a topological space, we can also find a characterization of “arbitrary closeness.” To indicate the proper translation from metric spaces to topological spaces of this concept, let us first prove:

LEMMA 4.1  In a metric space (X, d), for a given point x and a given subset A, d(x, A) = 0 if and only if each neighborhood N of x contains a point of A.

Proof. First, suppose that each neighborhood N of x contains a point of A. In particular, for each ε > 0, there is a point of A in B(x; ε). Thus for each ε > 0 and consequently .

Conversely, suppose that there is a neighborhood N of x that does not contain a point of A. Since N is a neighborhood of x in a metric space, there is an ε > 0 such that . It follows that a ∈ A implies that d(x, a) ≧ ε. Thus d(x, A) ≧ ε.

We shall, therefore, in a topological space, say that the points of a subset A are arbitrarily close to a given point x, if each neighborhood of x contains a point of A. Given a subset A, the collection of points that are arbitrarily close to A is called the closure of A.

DEFINITION 4.2  Let A be a subset of a topological space. A point x is said to be in the closure of A if, for each neighborhood N of x, N ∩ A ≠ Ø. The closure of A is denoted by Ā.

The purpose of the next two lemmas is to provide a description of the closure of a subset in terms of closed sets.

LEMMA 4.3  Given a subset A of a topological space and a closed set F containing A, .

Proof. Suppose , then x is in the open set C(F). Also, implies . Thus, C(F) ∩ A = Ø. Since C(F) is a neighborhood of x, . We have thus shown that or .

LEMMA 4.4  Given a subset A of a topological space and a point , then for some closed set F containing A.

Proof. If , then there is a neighborhood and hence an open set O containing x such that O ∩ A = Ø. Let F = C(O). Then F is closed and . But x ∈ 0 and therefore x ∉ F.

Combining these two lemmas, we obtain:

THEOREM 4.5  Given a subset A of a topological space, where is the family of all closed sets containing A.

Proof. By Lemma 4.3, since for each . By Lemma 4.4, for each implies that , or . Thus, .

Frequently, in introducing the concept of closure of a subset, the characterization of closure given by Theorem 4.5 is used as a definition and the statement embodied in our Definition, 4.2, is then proved as a theorem. Another possible description of the closure Ā of a subset A is the characterization of Ā as the smallest closed set containing A. For Ā is contained in each closed set containing A, while Ā, being the intersection of closed sets, is itself a closed set.

Theorem 4.5 is the characterization of closure in terms of closed sets. The next theorem characterizes closed sets in terms of closure.

THEOREM 4.6  A is closed if and only if A = Ā.

Proof. We have just seen that Ā is closed, so if A = Ā, then A is closed. Conversely, suppose A is closed. In this event A itself is a closed set containing A, so, therefore, Ā ⊂ A. On the other hand, for an arbitrary subset A, we have A ⊂ Ā, for if x ∈ A, then each neighborhood N of x contains a point of A; namely x itself. Thus, if A is closed, A = Ā.

The act of taking the closure of a set associates to each subset A of a topological space a new subset Ā. This correspondence or operation on the subsets satisfies the following five properties:

THEOREM 4.7  In a topological space ,

CL1. ;

CL2. ;

CL3. For each subset A of X, A ⊂ Ā;

CL4. For each pair of subsets A, B of X, .

CL5. For each subset A of X, .

Proof. The property CL3 has been established during the proof of Theorem 4.6. Note that CL2 follows from CL3. CL1 is true, for given a point x ∈ X and a neighborhood N of x, N ∩ Ø = Ø; thus there are no points in . To prove CL5 we note that Ā is closed, so, applying Theorem 4.6 to Ā we have . It remains for us to prove CL4. Suppose x ∈ Ā, then each neighborhood N of x contains points of A and hence points of A ∪ B. Thus . Similarly, , and, consequently, . On the other hand, and , so . Thus, is a closed set containing A ∪ B, whence .

One may use the properties CL1–CL5 as a set of axioms for what we will call a closure space and then prove that there is a “natural” one-one correspondence between the collection of topological spaces and the collection of closure spaces. An outline of how this might be done is given in Problem 11 at the end of this section.

In a topological space, we have seen that the closure of a subset A is the smallest closed set containing A. Another significant subset associated with A is the “interior” of A, which, as we shall see, is the largest open set contained in A.

DEFINITION 4.8  Given a subset A of a topological space, a point x is said to be in the interior of A if A is a neighborhood of x. The interior of A is denoted by Int (A).

LEMMA 4.9  Given a subset A of a topological space and an open set O contained in A, 0 ⊂ Int (A).

Proof. If x ∈ O, then A is a neighborhood of x, since O is open and O ⊂ A. Thus x ∈ Int (A) and O ⊂ Int (A).

LEMMA 4.10  Given a subset A of a topological space, if x ∈ Int (A), then x ∈ 0 for some open set O ⊂ A.

Proof. If x ∈ Int (A), then A is a neighborhood of x, whence A contains an open set O containing x.

In much the same manner in which Lemmas 4.3 and 4.4 combine to yield Theorem 4.5, Lemmas 4.9 and 4.10 combine to yield:

THEOREM 4.11  Given a subset A of a topological space,

where is the family of all open sets contained in A.

Thus, Int (A), being the union of open sets, is itself open, and is the largest open set contained in A. Furthermore, if is the family of open sets contained in a given set A, then is the family of closed sets containing C(A). Thus:

THEOREM 4.12  .

COROLLARY 4.13   .

For a given subset A, the set of points that are arbitrarily close to both A and C(A) is called the “boundary” of A.

DEFINITION 4.4  Given a subset A of a topological space, a point x is said to be in the boundary of A if x is in both the closure of A and the closure of the complement of A. The boundary of A is denoted by Bdry (A).

Thus, Bdry . It follows that A and C(A) have the same boundary, for Bdry . In terms of the definition of the closure of a set, we have the statement that a point x is in the boundary of a set A if and only if each neighborhood N of x contains both points of A and points of the complement of A. Since the boundary of A is the intersection of two closed sets:

COROLLARY 4.15  For each subset A, Bdry (A) is closed.

EXERCISES

1.  A family of subsets is said to be mutually disjoint if for each distinct pair β, γ of indices Aβ ∩ Aγ = Ø. Prove that for each subset A of a topological space , the three sets Int (A), Bdry (A), and Int (C(A)) are mutually disjoint and that .

2.  In a metric space (X, d), prove that for each subset A:

(a)  x ∈ Ā if and only if d(x, A) = 0;

(b)  x ∈ Int (A) if and only if d(x, C(A)) > 0;

(c)  x ∈ Bdry (A) if and only if

3.  In the real line, prove that the boundary of the open interval (a, b) and the boundary of the closed interval [a, b] is {a, b}.

4.  In Rn with the usual topology, let A be the set of points x = (x₁, x₂, . . . , xn) such that . Prove that Bdry (A) is the (n − 1)-dimensional sphere Sn−1, i.e., x ∈ Bdry (A) if and only if .

5.  In Rn+1 with the usual topology, let A be the set of points x = (x₁, x₂, . . . , xn+1) such that xn+1 = 0. Prove that Int (A) = Ø, Bdry (A) = A, Ā = A.

6.  In a topological space, each of the terms open set, closed set, neighborhood, closure of a set, interior of a set, boundary of a set, may be characterized by any other one of these terms. Construct a table containing the thirty such possible definitions or theorems in which, for example, the entry in the row labelled interior and the column labelled open set is the characterization of interior in terms of open sets (Theorem 4.17), etc.

7.  Let A be a subset of a topological space. Prove that Bdry (A) = Ø if and only if A is open and closed.

8.  A subset A of a topological space (X, ) is said to be dense in X if Ā = X. Prove that if for each open set O we have A ∩ O ≠ Ø, then A is dense in X.

9.  The “rational density theorem” for the real line states that between any two real numbers there lies a rational number. Use the rational density theorem to prove that the rational numbers are dense in the real line.

10.  The “Archimedean principle” for the real line states that if c, d > 0 then there is a positive integer N such that Nc > d. Prove the Archimedean principle for the real line and use this principle to prove the rational density theorem for the real line.

11.  Let a closure space be defined as a set X together with a correspondence which associates to each subset A of X a subset Ā of X satisfying the five conditions CL1–CL5 of Theorem 4.7. Prove that in a closure space, A ⊂ Bimplies . Define A to be closed if A = Ā. Prove that the empty set and the whole space are closed. Also that a finite union of closed sets is closed and an arbitrary intersection of closed sets is closed. Prove that for each subset A of X, , where is the family of all closed sets containing A. Now prove that there is a one-one correspondence between the collection of topological spaces and the collection of closure spaces.

12.  Prove that Ā = A ⊂ Bdry A.

13.  Let A be a subset of a topological space. Prove that A is closed if and only if Bdry (A) ⊂ A, and that A is open if and only if (A) ⊂ C(A).

5  FUNCTIONS, CONTINUITY, HOMEOMORPHISM

DEFINITION 5.1  A function f from a topological space to a topological space is a function f:X → Y.

If f is a function from a topological space to a topological space we shall write . In the event that the topologies on X and Y need not be explicitly mentioned, we may abbreviate this notation by f:X → Y or simply f.

DEFINITION 5.2  A function is said to be continuous at a point a ∈ X if for each neighborhood N of f(a), f⁻¹(N) is a neighborhood of a. f is said to be continuous if f is continuous at each point of X.

Let (X, d) and (Y, d′) be metric spaces and let their associated topological spaces be and respectively. Given a function f from the first metric space to the second, we also have a function, which we still denote by f, from the first topological space to the second. Our definition of continuity has been formulated so that for each point a ∈ X, the function f:(X, d) → (Y, d′) is continuous at a if and only if is continuous at a.

THEOREM 5.3  A function is continuous if and only if for each open subset O of Y, f⁻¹(O) is an open subset of X.

Proof. First, suppose that f is continuous and that O is an open subset of Y. For each , O is a neighborhood of f(a), therefore f⁻¹(O) is a neighborhood of a. Since f⁻¹(O) is a neighborhood of each of its points, f⁻¹(O) is an open subset of X. Conversely, suppose that for each open subset O of Y, f⁻¹(O) is an open subset of X. Let a ∈ X and a neighborhood N of f(a) be given. N contains an open set O containing f(a), so by our hypothesis, f⁻¹(N) contains the open set f⁻¹(O) containing a. Thus, f⁻¹(N) is a neighborhood of a and f is continuous at a. Since a was arbitrary, f is continuous.

For any set X, given a collection E of subsets of X, let C′(E) denote the collection of subsets of X which are complements of members of E. Also given f:X → Y and a collection E of subsets of Y, let f⁻¹(E) denote the collection of subsets of X of the form f⁻¹(E) for some E ∈ E. Theorem 5.3 states that is continuous if and only if . Let and be the closed subsets of X and Y respectively. For so that . Thus is equivalent to and we obtain:

THEOREM 5.4  A function is continuous if and only if for each closed subset F of Y, f⁻¹(F) is a closed subset of X.

It is important to remember that Theorem 5.3 says that a function f is continuous if and only if the inverse image of each open set is open. This characterization of continuity should not be confused with another property that a function may or may not possess, the property that the image of each open set is an open set (such functions are called open mappings). There are many situations in which a function has the property that for each open subset A of X, the set f(A) is an open subset of Y, and yet f is not continuous. For example, let Y be a set containing two distinct elements a and b and let each subset of Y be an open set. Let R be the real line and define f:R → Y by f(x) = a for x 0 and f(x) = b for x < 0. Every subset of Y is open so in particular for each open subset U of R, f(U) is an open subset of Y. On the other hand {a} is an open subset of Y but f⁻¹({a}), the set of non-negative real numbers, is not an open subset of the reals.

THEOREM 5.5   is continuous if and only if for each subset A of X, .

Proof. First suppose that f is continuous. Given a subset A of X, , whence . The set is closed so . Thus . Conversely, suppose that for each subset A of X, . Let F be a closed subset of Y. Then . Thus . Since it is always the case that we have ; consequently f⁻¹(F) is closed and f is continuous.

THEOREM 5.6  Let be continuous at a point a ∈ X and let be continuous at f(a). Then the composite function is continuous at a.

Proof. Let N be a neighborhood of (gf)(a) = g(f(a)). Then (gf)⁻¹(N) = f⁻¹(g⁻¹(N)). But g⁻¹(N) is a neighborhood of f(a), since g is continuous at f(a), and therefore f⁻¹(g⁻¹(N)) is a neighborhood of a, since f is continuous at a.

The equivalence relation that is appropriate to topological spaces is called homeomorphism.

DEFINITION 5.7  Topological spaces and are called homeomorphic if there exist inverse functions f:X → Y and g:Y → X such that f and g are continuous. In this event the functions f and g are said to be homeomorphisms and we say that f and g define a homeomorphism between and .

The following easily verified corollary to this definition indicates that homeomorphism is the translation from metric spaces to topological spaces of the concept of topological equivalence.

COROLLARY 5.8  Let (X, d) and (Y, d′) be metric spaces. Let and be the topological spaces associated with (X, d) and (Y, d′) respectively. Then the metric spaces (X, d) and (Y, d′) are topologically equivalent if and only if the topological spaces and are homeomorphic.

THEOREM 5.9  A necessary and sufficient condition that two topological spaces and be homeomorphic is that there exist a function f:X → Y such that:

1.  f is one-one;

2.  f is onto;

3.  A subset O of X is open if and only if f(O) is open.

Proof. Suppose that and are homeomorphic. Let the homeomorphism be defined by inverse functions f:X → Y and g:Y → X. f is invertible and consequently one-one and onto. Furthermore, given an open set O in X, the set f(O) = g⁻¹(O) is open in Y, since g is continuous. On the other hand, if f(O) = O′ is an open subset of Y, then O = f⁻¹(O′) is open in X.

Now, suppose that a function f:X → Y with the prescribed properties exists. Then f is invertible and we define g:Y → X by g(b) = a if f(a) = b, so that f and g are inverse functions. If O is an open subset of X, then f(O) = g⁻¹(O) is open in Y, so that g is continuous. Also, if O′ is an open subset of Y, then f⁻¹(O′) = O is an open subset of X and f is continuous.

EXERCISES

1.  Let a function f:X → Y be given. Prove that is always continuous, as is , where is any topology on Y and is any topology on X.

2.  Prove that a function is a homeomorphism if and only if

  (i) f is one-one;

 (ii) f is onto;

(iii) For each point x ∈ X and each subset N of X, N is a neighborhood of x if and only if f(N) is a neighborhood of f(x).

3.  Let be a homeomorphism. Let a third topological space and a function be given. Prove that h is continuous if and only if hf is continuous. Let another function be given. Prove that k is continuous if and only if fk is continuous.

4.  Let R be the real line. Prove that the function f:R → R defined by f(x) = sin x is continuous. [Hint: and Find an open interval (a, b) such that f((a, b)) is not an open interval.

6  SUBSPACES

DEFINITION 6.1  Let and be topological spaces. The topological space Y is called a subspace of the topological space X if Y ⊂ X and if the open subsets of Y are precisely the subsets O′ of the form

for some open subset O of X.

In the event that Y is a subspace of X, we may say that each open subset O′ of Y is the restriction to Y of an open subset O of X. A subset O′ that is open in Y is often called relatively open in Y or simply relatively open. A subset O of X that is open in X and is contained in Y is necessarily relatively open in Y, but the relatively open subsets of Y are in general not open in X.

We shall now prove that there are as many subspaces of a topological space X as there are non-empty subsets Y of X.

PROPOSITION 6.2  Let be a topological space and let Y be a subset of X. Define the collection of subsets of Y as the collection of subsets O′ of Y of the form

where . Then is a topological space and therefore a subspace of provided Y ≠ Ø.

Proof. We must prove that is a topology. Ø = Ø ∩ Y and Y = X ∩ Y are in . Suppose , so that for i = 1, 2, . . . , n, O′i for some . Then O′₁ ∩ O′₂ ∩ . . . ∩ O′n = (O₁ ∩ O₂ ∩ . . . ∩ On) ∩ Y is in , since O₁ ∩ O₂ ∩ . . . ∩ On is open in X. Finally, suppose that for each . Thus, for each for some . But is in , since is open in X.

Given a subset Y of a topological space , the topology of Y described in the above proposition is said to be induced by the topology on X and is called the relative topology on Y. The neighborhoods in this relative topology on Y are called neighborhoods in Y or relative neighborhoods. The following result states that the neighborhoods in Y are the restrictions of the neighborhoods in X.

THEOREM 6.3  Let Y be a subspace of a topological space X and let a ∈ Y. Then a subset N′ of Y is a relative neighborhood of a if and only if

where N is a neighborhood of a in X.

Proof. First suppose N′ is a relative neighborhood of a. Then N′ contains a relatively open set O′, which contains a. Let O′ = O ∩ Y, where O is an open subset of X. Then N = N′ ∩ O is a neighborhood of a in X and . Conversely, suppose N′ = N ∩ Y, where N is a neighborhood of a in X. Then N contains an open set O containing a and hence N′ contains the relatively open set O′ = O ∩ Y containing a. Thus N′ is a relative neighborhood of a.

EXAMPLE 1  The closed interval [a, b] of the real line with induced topology is a subspace of the real line. A relative neighborhood of the point a is any subset N of [a, b] that contains a half-open interval [a, c), where a < c and [a, c) is the set of all real numbers x such that a ≦ x < c. Similarly, a relative neighborhood of the point b is any subset M of [a, b] that contains a half-open interval (c, b], where c < b and (c, b] is the set of all real numbers x such that c < x ≦ b. If d is such that a < d < b, then a relative neighborhood of d is any subset U of [a, b] that is a neighborhood of d in the real line R.

EXAMPLE 2  Let A be the subset of Rn+1 consisting of all points x = (x₁, x₂, . . . , xn+1) such that xn+1 = 0. Let Rn+1 have the usual topology and let A have the induced topology so that A is a subspace of Rn+1. The topological space A is homeomorphic to Rn. To prove this fact we shall use the result that the relationship of subspace is “preserved” in passing from metric spaces to topological spaces.

LEMMA 6.4  Let (X, d) be a metric space and let (Y, d′) be a subspace of (X, d). If is the topological space associated with (X, d) and is the topological space associated with (Y, d′), then is a subspace of .

Proof. Since d′ is the restriction of d, an open ball in (Y, d′) is the restriction of an open ball in (X, d) to Y. Consequently a subset O′ of Y is open in Y if and only if, for each y ∈ O′, there is an εy > 0 such that . Let . Then O is open in X and O′ = O ∩ Y. Thus . Conversely, if , then O′ = O ∩ Y for some . For each we have , and O is open, so there is an εy such that . It follows that , and hence O′ is open in (Y, d′).

Returning to our example, we define f:Rn → A by setting f(x₁, x₂, . . . , xn) = (x₁, x₂, . . . , xn, 0). It is easily verified that f is one-one, onto, and that the inverse of f is the function g:A → Rn defined by g(x₁, x₂, . . . , xn, 0) = (x₁, x₂, . . . , xn). If we first consider f and g as functions defined on the metric spaces (Rn, d) and (A, d′), where (A, d′) is a subspace of (Rn+1, d), then clearly f and g are continuous. It follows that f and g are continuous functions defined on the topological spaces Rn and A, where A is considered as a subspace of Rn+1, and that they therefore define a homeomorphism.

Given a subspace of a topological space , the closed subsets of the topological space are called relatively closed in Y or simply relatively closed. Again, the relatively closed subsets are the restriction to Y of the closed subsets of X.

THEOREM 6.5  Let be a subspace of the topological space . A subset F′ of Y is relatively closed in Y if and only if

for some closed subset F of X.

Proof. First, suppose F′ is relatively closed. Then CY(F′) is relatively open. Thus, CY(F′) = O ∩ Y, where O is open in X. But then F′ = CY(O ∩ Y) = CY(O) = Cx(O) ∩ Y, where Cx(O) is a closed subset of X. Conversely, suppose F′ = F ∩ Y, where F is a closed subset of X. Then, CY(F′) = CX(F) ∩ Y; hence CY(F′) is relatively open in Y and therefore F′ is relatively closed.

EXAMPLE 3  Let a < b < c < d. Let Y = [a, b] ∪ (c, d) be considered as a subspace of the real line. Then the subset [a, b] of Y is both relatively open and relatively closed. To prove this fact we note that [a, b] = [a, b] ∩ Y so that [a, b] is relatively closed, whereas for O < ε < c − b, [a, b] = (a − ε, b + ε) ∩ Y so that [a, b] is relatively open. Since (c, d) is the complement in Y of a relatively open and relatively closed subset of Y, (c, d) is also relatively open and relatively closed in Y.

THEOREM 6.6  Let the topological space Y be a subspace of the topological space X. Then the inclusion mapping i:Y → X is continuous.

Proof. For each subset A of X, i⁻¹(A) = A ∩ Y. Thus, if O is an open subset of X, i⁻¹(O) = O ∩ Y is a relatively open subset of Y.

DEFINITION 6.7  Let and be two topologies on a set Y. The topology is said to be weaker than if .

If Y is a subset of a topological space then the relative topology on Y is the weakest topology such that the inclusion map i:Y → X is continuous, for if ₁ is another topology on Y such that is continuous then given , O′ = i⁻¹(O) with . Thus and .

Let X and Y be topological spaces and f:Y → X be a function which is not necessarily continuous. The function f induces a function f′:Y → f(Y) which agrees with f and is onto. Viewing f(Y) as a subspace of X we have:

LEMMA 6.8  f:Y → X is continuous if and only if f′: Y → f(Y) is continuous.

Proof. Since the inclusion map i:f(Y) → X is continuous, the continuity of f′ yields the continuity of f = if′. Conversely, if O′ is a relatively open set in f(Y), then O′ = O ∩ f(Y), where O is open in X. If f is continuous f⁻¹(O) = f′⁻¹(O′) is open in Y and f′ is continuous.

EXERCISES

1.  If Y is a subspace of X and Z is a subspace of Y, then Z is a subspace of X.

2.  Let O be an open subset of a topological space X. Prove that a subset A of O is relatively open in O if and only if it is an open subset of X.

3.  Let F be a closed subset of a topological space X. Prove that a subset A of F is relatively closed in F if and only if it is a closed subset of X.

4.  Prove that a subspace of a Hausdorff space is a Hausdorff space.

5.  Prove that a subspace of a metrizable space is a metrizable space.

6.  Prove that an open interval (a, b) considered as a subspace of the real line is homeomorphic to the real line.

7.  Let Y be a subspace of X and let A be a subset of Y. Denote by IntX (A) the interior of A in the topological space X and by IntY (A) the interior of A in the topological space Y. Prove that . Illustrate by an example the fact that in general IntX (A) ≠ IntY (A).

8.  Let Y be a subspace of X and let A be a subset of Y. Denote by ĀX the closure of A in the topological space X and by ĀY the closure of A in the topological space Y. Prove that . Show that in general ĀY ≠ ĀX.

7  PRODUCTS

Throughout this section let be topological spaces and let . We wish to define a topology on X that may be regarded as the product of the topologies on the factors of X. Again our guide is the corresponding situation in metric spaces. If these topological spaces were metrizable, then there is a standard procedure for converting the product of the corresponding metric spaces into a metric space. In this resulting metric space, the open subsets of X are the unions of sets of the form O₁ × O₂ × . . . × On, where each Oi is an open subset of Xi. In the general situation, where may not be metrizable, one can show that the unions of the products of open sets will constitute a topology. This result is based on the following lemma.

LEMMA 7.1  Let be a collection of subsets of a set X with the property that , and a finite intersection of elements of is again in . Then the collection of all subsets of X which are unions of elements of is a topology.

Proof. Clearly Ø and X are in . Suppose O and O′ are in . Then , where for and for . Thus for and hence

is in . Finally a union of sets each of which is a union of sets of is again a union of sets of so that is a topology.

Since in the product set X the collection of subsets of X that are unions of sets of the form O₁ × O₂ × . . . × On, each Oi an open subset of Xi, satisfies the conditions of this lemma we may state:

DEFINITION 7.2  The topological space , where is the collection of subsets of X that are unions of sets of the form

each Oi an open subset of Xi, is called the product of the topological spaces , i = 1, 2, . . . , n.

In the future we shall often denote a topological space simply by X. Thus, if we say, let X₁, X₂, . . . , Xn be topological spaces and , we shall mean that X is to be considered as the product of the topological spaces.

As was the case with metric spaces, the sets of the form O₁ × O₂ × . . . × On, Oi open in Xi, have been used as a “basis” for the open sets of X.

DEFINITION 7.3  Let X be a topological space and a collection of open sets in is called a basis for the open sets of X if each open set is a union of members of .

The next proposition characterizes the neighborhoods in the product space.

PROPOSITION 7.4  In a topological space , a subset N is a neighborhood of a point a = (a₁, a₁, … an ∈ N) if and only if N contains a subset of the form N₁ × N₂ × . . . × Nn, where each Ni is a neighborhood of ai.

Proof. First suppose that N₁ × N₂ × . . . × Nn ⊂ N where each Ni is a neighborhood of ai. By the definition of neighborhood in a topological space, each Ni contains an open set Oi containing ai hence N contains the open set O₁ × O₂ × . . . × On containing a, and therefore N is a neighborhood of a. Conversely, suppose N is a neighborhood of a. Then N contains an open set O containing a. Since O is an open subset of the product space , we may write , where for each i and each is an open subset of Xi. Since , for some , hence for i = 1, 2, . . . , n. But Oβ,i is open. Thus, if we set Ni = Oβ,i, i = 1, 2, . . . , n, Ni is a neighborhood of ai and .

DEFINITION 7.5  Let X be a topological space and . A collection of neighborhoods of a is called a basis for the neighborhoods at a if each neighborhood N of a contains a member of .

Thus, if , a basis for the neighborhoods at a is the collection consisting of all subsets of the form N₁ × N₂ × . . . × Nn, where each Ni is a neighborhood of ai.

Recall that in a product space the ith projection pi: X → Xi is the function such that pi(a) = ai. If , then

Since this set is an open subset of X the projection maps are continuous.

A subset O₁ × O₂ × . . . × On of X can be written as so that we have a guide to the appropriate topology on an arbitrary product of topological spaces.

DEFINITION 7.6  Let be an indexed family of topological spaces. The topological product of this family is the set with the topology consisting of all unions of sets of the form

We have used as a basis for the topology the collection of sets of the form . That is a topology follows from the fact that , and a finite intersection of elements of is again in . Clearly this topology makes the projection maps continuous. Since any topology on X which makes the projection maps continuous must contain the sets of this form, the product topology is the weakest topology consistent with the continuity of the projection maps.

It is easily seen that, analogous to Proposition 7.4, a basis for the neighborhoods at a point x is the collection of sets of the form , where Nαi is a neighborhood of . In effect then, in the product space X we are saying that a point y is in a given neighborhood of x or is close to x if there is a finite set of indices {α₁, . . . , αk} such that y(αi) is close to x(αi).

EXERCISES

1.  Prove that a subset is closed if and only if F is an intersection of sets, each of which is a finite union of sets of the form F₁ × F₂ × . . . × Fn, where each Fi is a closed subset of Xi. Formulate the corresponding statement in an arbitrary product of topological spaces.

2.  Let be the topological product of the family of spaces . Prove that a function f: Y → X from a space Y into the product X is continuous if and only if for each the function fα = pαf: Y → Xα is continuous.

3.  Let be a family of spaces and let , where J and K are disjoint and non-empty. Let be given. Define a function by setting for each and . Prove that φx is continuous.

4.  Let and be two families of spaces indexed by the same indexing set I. For each , let fα: Xα → Yα be a continuous function. Define by (f(x))(α) = fα(x(α)). Prove that f is continuous.

5.  Let N be the set of positive integers. For each n ∈ N let Xn = {0, 2} with the discrete topology. Let . Define a function f:X → [0, 1] by setting . Prove that f is one-one and continuous. The image f(X) is called the Cantor set D and consists of all real numbers a ∈ [0, 1] which can be represented as triadic decimals such that for all n. Given a ∈ D define (g(a))(n) = an so that g(a) ∈ X. Prove that g is a homeomorphism of D with X.

6.  Prove that the family of open intervals with rational end points is a basis for the topology of the real line.

8  IDENTIFICATION TOPOLOGIES

Let R be the real line and S the unit circle defined by . The function p:R → S defined by p(t) = (cos 2πt, sin 2πt) maps R continuously onto S so that p(t) = p(t′), provided t − t′ is an integer. One may think of p as wrapping the real line around the circle so that the points which differ by an integer are identified or superimposed on each other. Furthermore, we shall see that the topology of S may be obtained from the topology of R in such a way as to make the mapping p an identification.

DEFINITION 8.1  Let p: E → B be a continuous function mapping the topological space E onto the topological space B. p is called an identification if for each subset U of B, p⁻¹(U) open in E implies that U is open in B.

If p: E → B is an identification and g: B → Y is a continuous function defined on B, then g induces a continuous function gp:E → Y. It turns out that frequently the reverse is true, that is, a continuous function G:E → Y will induce a continuous function g:B → Y.

THEOREM 8.2  Let p:E → B be an identification and let G:E → Y be a continuous function such that for each with p(x) = p(x′), we also have G(x) = G(x′). Then for each b ∈ B we may choose any , define g(b) = G(x), and the resulting function g is continuous.

Proof. First the definition of g(b) does not depend on the choice of , for if then p(x) = p(x′) and G(x) = G(x′). Note that g is defined so that gp = G. Hence G⁻¹ = p⁻¹g⁻¹. If O is an open subset of Y, then G⁻¹(O) is open in E. But G⁻¹(O) = p⁻¹(g⁻¹(O)). Since p is an identification, g⁻¹(O) is open in B and therefore g is continuous.

The hypothesis on the function G is that Gp⁻¹ be well-defined or single-valued. The conclusion is then that the function g may be inserted in the diagram of Figure 9 and that commutativity will hold.

Figure 9

One may use an onto function p:X → Y from a topological space X to a set Y (without a topology) to construct a topology for Y so that p becomes an identification.

DEFINITION 8.3  Let p:X → Y be a function from a topological space X onto a set Y. The identification topology on Y determined by p consists of those sets U such that p⁻¹(U) is open in X.

Verification of the fact that this collection of sets is a topology depends on the behavior of p⁻¹ with respect to unions and intersections. Once Y has been given the identification topology determined by p, p is an identification.

Let f:X → Y be a function from a set X to a set Y. Let ~f be the relation defined on X by x~fx′ if f(x) = f(x′). ~f is an equivalence relation. Let X/~f be the collection of equivalence sets under this relation and let be the function which maps each x ∈ X into its equivalence class. πf is an onto function. Now suppose that X is a topological space and give X/~f the identification topology determined by πf. Let Y also be a topological space. Since πf(x) = πf(x′) if and only if f(x) = f(x′), f induces a continuous function such that f = f*πf. Furthermore f* is one-one, for if f*(u) = f*(u′), with u, , then for . Thus or u = πf(x) = πf(x′) = u′.

Consider the diagram

Let be the topology on X/~f and let be the topology on Y. Since f* is continuous, , or equivalently, since f* is one-one, . If were some other topology on Y so that f were continuous we would again have . Thus the topology carried over to Y by f* is the weakest or smallest topology such that f is continuous. Introducing the topologies into the diagram we obtain Figure 10, in which the inclusion map is continuous.

Figure 10

We shall conclude this section by considering some examples.

EXAMPLE 1  (The covering of the circle by the real line.) Let p(t) = (cos 2πt, sin 2πt) so that p:R → S is a continuous mapping of the real line onto the circle. To show that p is an identification mapping, we must show that if U ⊂ S is such that p⁻¹(U) is open, then U is open. Let and s = p(x). x is the center of an open interval of length 2ε < 1, which under p is mapped into an arc of S centered at s of length 4πε and contained in U. This arc is an open ball in S with center s; hence U is open.

The function defined by g(t) = (cos 2πt, sin 2πt, t) is a homeomorphism of the real line with a helix H in R³. Let S be taken to be the set of points defined by x² + y² = 1, z = 0. Then the projection of H onto S defined by (cos 2πt, sin 2πt, t) → (cos 2πt, sin 2πt, 0) is also an identification. This projection accounts for the literal sense in which the real line may be thought of as covering the circle.

Let f be any function defined and continuous on R. f is called periodic of period 1 if f(t + 1) = f(t) for all t ∈ R. It follows that f(t) = f(t′), provided t − t′ is an integer, so that f induces a continuous function f*, defined on the circle S such that f*(p(t)) = f(t).

EXAMPLE 2  (Shrinking a subset to a point.) Let X be a topological space and A a non-empty subset of X. Define a new untopologized set X/A as the union of X − A and a new point a*. Define a function f:X → X/A by f(x) = x for , f(x) = a* for x ∈ A. Now give X/A the identification topology determined by f. This space is the space obtained by shrinking A to a point.

Let be the boundary of the unit interval I = [0, 1]. Then is homeomorphic to a circle. In fact, by Theorem 8.2, the function p(t) = (cos 2πt, sin 2πt), defined now for t ∈ I, must induce a continuous function is one-one and a basis for the open sets containing a* is the totality of images of sets of the form .

Shrinking the boundary of I to a point amounts to pasting the two end points together to make the single point a* out of the boundary. If the boundary of a square is shrunk to a point, the resulting space turns out to be homeomorphic to the surface of a globe or a 2-sphere. One can even visualize this shrinking as a process in which an elastic sheet having a string in its boundary is deformed into a 2-sphere by gathering the string to a point.

EXAMPLE 3  (Attaching a space X to a space Y.) Let X and Y be topological spaces and let A be a non-empty closed subset of X. Assume that X and Y are disjoint and that a continuous function f:A → Y is given. Form the set and define a function by for for , and for . Give the topology in which a set is open (or closed) if and only if its intersections with both X and Y are open (or closed). φ is onto. Let be the set with the identification topology determined by φ.

If Y is a single point a*, then attaching X to a* by a function f:A → a* is the same as shrinking A to a point. Let I² be the unit square in R² and let A be the union of its two vertical edges so that and either or . Let Y = [0, 1] be the unit interval. Define f:I² → Y by f(x, y) = y. Then is a cylinder formed by identifying the two vertical edges of I².

EXERCISES

1.  Let n be an integer. Let φn:R → R be the function from the real line into itself defined by φn(x) = nx. Let p(t) = (cos 2πt, sin 2πt) as before. Show that φn induces a function φn: S → S of the circle into itself so that φnp = pφn. [φn is said to wrap the circle around itself n times for positive n.]

2.  A torus is the surface of a donut or an inflatable inner tube. It can be thought of as being generated by rotating a circle about a line in the plane of the circle, provided the circle and the plane do. not intersect. Prove that if C is a circular cylinder with S₁ and S₂ as its boundary circles and S₁ and S₂ are identified by mapping them both homeomorphically onto some third circle S, giving a map , then is a torus.

Define a relation in the plane R² by provided x − x′ and y − y′ are integers. Prove that ~ is an equivalence relation. Let T be the collection of equivalence sets and φ:R² → T the mapping carrying each point into its equivalence set. Give T the identification topology determined by φ. Show that T is homeomorphic to a torus.

3.  The unit disc is the set of points in R² given by . Its boundary in R² is the unit circle S. Let A be the subset of the circular cylinder S × [0, 1] given by S × {1}. Prove that S × [0, 1]/A is homeomorphic to the disc D. (S × [0, 1]/A is the cone over S, see the next problem.)

4.  Let X be a topological space and A the subset of X × [0, 1] given by X × {1}. The space X × [0, 1]/A is called the cone over X. Denote this space by TX. Prove that if X and X′ are homeomorphic, then so are TX and TX′.

5.  Let X be a topological space and let p₀ and p₁ be two points not in X × [−1, 1]. Let f(x, −1) = p₀ and f(x, 1) = p₁ for x ∈ X define a mapping of . Let Y = {p₀, p₁} have the discrete topology so that f is continuous. is called the suspension of X, and is denoted by SX. The equator is the image of X × {0} in SX. Prove that this subject of SX is homeomorphic to X. Prove that the image of X × [0, 1] in SX is homeomorphic to the cone over X, and that therefore the suspension of X is two cones over X identified along the equator. Prove that the suspension of a circle is homeomorphic to the 2-sphere S².

9  CATEGORIES AND FUNCTORS

A great deal of the more recent work in topological spaces has involved the consideration of a collection of topological spaces and collections of continuous mappings between these spaces. It has proven to be extremely fruitful to formulate an abstract definition of the structure involved.

DEFINITION 9.1  A category C is a collection of objects A whose members are called the objects of the category and for each ordered pair (X, Y) of objects of the category a set H(X, Y) called the maps of X into Y together with a rule of composition which associates to each and a map . This composition is associative, that is, if , then h(gf) = (hg)f and identities exist, that is, for each object X ∈ A there is an element such that for all and for all 1xh = h.

In Chapter 1 we were concerned with the category Cs of sets and functions. That is, As is the class of all sets and for X, is the set of all functions from X to Y. For X ∈ As, 1X is the identity mapping of X onto itself. In an obvious fashion one may obtain what we would call subcategories C′ of Cs by taking as objects A′ some specified collection of sets and for to be some specified set of functions from X to Y provided that we always include the identity mapping 1x in H(X, X) for each X ∈ A′ and for each ordered pair (X, Y) of A′ include in H′(X, Y) all functions f which can be written in the form hg for . For example A′ might be all finite sets and H′(X, Y) all functions from X to Y. In particular A′ could contain a single set X and H′(X, X) could be all invertible functions.

In Chapter 2 the appropriate category was the category CM of all metric spaces and continuous functions. Chapter 3 furnished us with our main example, namely the category CT of all topological spaces and continuous mappings.

We shall include in some detail one more example of a category of algebraic objects.

DEFINITION 9.2  A group G is a set G together with a function which associates to each ordered pair g₁, g₂ of elements of G an element such that:

  (i)  g₁(g₂g₃) = (g₁g₂)g₃ for ;

 (ii)  there is an element e ∈ G, called the identity such that for all e ∈ G, eg = ge = g;

(iii)  for each e ∈ G there is an element called the inverse of g, such that gg⁻¹ = g⁻¹g = e.

A homomorphism f from a group G to a group K is a function f:G → K such that f(e) = e′ if e and e′ are identities in G and K respectively and for all .

Let be a collection of groups and for G, let H(G, K) be the set of all homomorphisms of G into K. If we use the ordinary composition of functions to define for and an element , it is easily verified that we have constructed a category of groups and homomorphisms.

In Chapter 4 we shall associate to certain topological spaces a group called the fundamental group of the space.

A transformation from one category to another which preserves the structure of a category is called a “functor.”

DEFINITION 9.3  Let C and C′ be categories with objects A and A′ respectively. A functor* F:C → C′ is a pair of functions F₁ and F₂ such that F₁:A → A′ and for each ordered pair X, Y of objects of A,

so that and F₂(gf) = F₂(g)F₂(f) for .

In keeping with the notation of the examples let us denote an element by . If F:C → C′ is a functor we have

F₂ preserves identities, and if

is commutative, then so is

that is, F carries commutative diagrams into commutative diagrams.

The passage from a metric space (X, d) to its associated topological space is an example of a functor from CM to CT. As another example of a functor, this time from CT to itself, let Z be a fixed topological space. To each topological space X ∈ CT associate the topological space F₁(X) = X × Z and to each continuous function associate the function F₂(f) defined by (F₂(f))(x, z) = (f(x), z) for . Then F₂(f):F₁(X) → F₁(Y) is continuous and it is easily verified that F = (F₁, F₂) is a functor.

EXERCISES

1.  Let be an indexed family of categories with objects and maps . Let . For let . For define gf by gf(α) = g(α)f(α). Prove that this yields a category with objects A and maps H(U, V).

2.  Let C be a category with objects A. Let be such that ge = g, gf = g for all g ∈ H(X, Y) and eh = h, fh = h for all h ∈ H(W, X). Prove that e = f and that therefore the identities are unique. Let f ∈ H(X, Y) be such that there are maps g, g′ ∈ H(Y, X) with gf = 1x and fg′ = 1y. Prove that g = g′ and that therefore f has a two-sided inverse f⁻¹ = g. Such an f is called an equivalence. Prove that 1x is an equivalence for all X ∈ A, if f is an equivalence so is f⁻¹, and f ∈ H(X, Y) and f′′ ∈ H(Y, Z) are equivalences so is f′′f. Verify that in the category CT of topological spaces and continuous mappings the equivalences are the homeomorphisms. Prove that a functor carries equivalences into equivalences.

3.  Let Cs be the category of sets and functions. Verify that the set of equivalences in H(X, X) with the same rule of composition as in Cs is the group of one-one mappings of X onto itself. In general, verify that in any category C for each object X, the set of equivalences in H(X, X) with the same rule of composition is a group.

4.  Let A be a collection of pairs (X, Y) such that X is a topological space and Y is a subspace of X. Given (X, Y) and let H((X, Y), (X′, Y′)) be the set of all continuous functions f:X → X′ such that . Construct a category with objects A and these maps. Verify that if for (X, Y) ∈ A we set F₁(X, Y) = Y and for we set F₂(f) = f | Y then (F₁, F₂) is a functor.

5.  Let C be a category whose objects are pairs (X, A) where X is a topological space and A is a non-empty closed subset of X, and whose maps H((X, A), (Y, B)) are continuous functions f:X → Y with . Define F₁(X, A) = X/A. Let p(X, A) : X → X/A be the identification map. Prove that if then there is a continuous function f*:X/A → Y/B such that the diagram

is commutative. Define F₂(f) = f* and prove that F = (F₁, F₂) is a functor from the category C to the category CT of topological spaces and continuous functions.

For further reading there are many excellent general texts including Kelley, General Topology, Dugundji, Topology, and Pervin, Foundations of General Topology.

* Definition 9.3 defines a covariant functor. In further work one also needs to consider contravariant functors in which F₂:H(X, Y) → H′(F₁(Y), F₁(X)) and F₂(gf) = F₂(f)F₂(g).

CHAPTER 4

Connectedness

1  INTRODUCTION

A subspace of a topological space is “connected” if it is all “of one piece”; that is, if it is impossible to decompose the subspace into two disjoint non-empty open sets. The non-empty connected subsets of the real line are single points and intervals. The continuous image of a connected set is necessarily a connected set. A consequence of these two facts is the intermediate value theorem; that is, a continuous function f:[a, b] → R must assume all values between f(a) and f(b). A second type of connectedness is called “path-connectedness,” by which it is meant that each pair of points may be “connected” by a “path” or “arc.” Path-connectedness is a stronger condition than connectedness, since each path-connected topological space is connected, whereas the converse is false. A third type of connectedness that we shall consider is “simple connectedness.” A topological space is simply connected if there are no holes in it to prevent the continuous shrinking of each closed arc to a point. The degree to which a given topological space fails to be simply connected may be measured by an algebraic topological invariant called the fundamental group of the space.

2  CONNECTEDNESS

DEFINITION 2.1  A topological space X is said to be connected if the only two subsets of X that are simultaneously open and closed are X itself and the empty set Ø. A topological space which is not connected is said to be disconnected.

Thus, a topological space X is disconnected if and only if there are two non-empty open subsets P and Q whose union is X and whose intersection is empty, for in this event P is the complement of Q and therefore both open and closed, whereas P is neither X nor Ø. Similarly, a topological space X is disconnected if and only if there are two non-empty closed subsets F and G whose union is X and whose intersection is empty.

Every subset A of a topological space X is itself a topological space in the relative topology. We say that the subset A is connected if the topological space A with the relative topology is connected, or what amounts to the same thing,

DEFINITION 2.2  A subset A of a topological space X is said to be connected if the only two subsets of A that are simultaneously relatively open and relatively closed in A are A and Ø.

Thus, the statement, A is connected, has the same meaning whether the reference is to A as a topological space or as a subspace of some larger topological space.

We shall shortly see that intervals such as [a, b] and (a, b) are connected subsets of the real line R. As an example of a subset of the real line that is disconnected, let A = [0,1] ∪ (2,3). [0, 1] is a relatively closed subset of A since [0, 1] is closed in R. At the same time [0, 1] is a relatively open subset of A, since . Finally, [0, 1] ≠ Ø and [0, 1] ≠ A, hence A is disconnected. By the same token, the “open interval” (2, 3) is also both relatively open and relatively closed in A.

It will be useful to have the following formulation of connectedness, or more precisely, disconnectedness.

LEMMA 2.3  Let A be a subspace of a topological space X. Then A is disconnected if and only if there exist two open subsets P and Q of X such that

and P ∩ A ≠ Ø, Q ∩ A ≠ Ø

Proof. First, suppose that A is disconnected. Then there is a subset P′ of A that is different from Ø and from A and is both relatively open and relatively closed. This implies that the complement of P′ in A, CA(P′), is also different from Ø and from A and relatively open. Thus P′ = P ∩ A and CA(P′) = Q ∩ A, where P and Q are open subsets of X. We therefore have that , for and , and also P ∩ Q ∩ A = (P ∩ A) ∩ (Q ∩ A) = P′ ∩ CA(P′) = Ø so that . Finally, P′ = P ∩ A and CA(P′) = Q ∩ A are non-empty.

Conversely, given open sets P and Q satisfying the stated conditions, set P′ = P ∩ A and Q′ = Q ∩ A. Then and P′ ∩ Q′ = (A ∩ P) ∩ (A ∩ Q) = Ø. Thus P′ = CA(Q′), and P′ is both relatively open and relatively closed in A. Since P′ ≠ Ø and P′ ≠ A (for Q′ is non-empty), A is disconnected.

A corresponding result also holds, using closed sets.

LEMMA 2.4  Let A be a subspace of a topological space X. Then A is disconnected if and only if there exist two closed subsets F and G of X such that

and F ∩ A ≠ Ø, G ≠ A ≠ Ø

The next theorem asserts that connectedness is preserved under continuous mappings.

THEOREM 2.5  Let X and Y be topological spaces and let f:X → Y be continuous. If A is a connected subset of X, then f(A) is a connected subset of Y.

Proof. Suppose f(A) is not connected. Then there are open subsets P′ and Q′ of Y such that , and P′ ∩ f(A) ≠ Ø, Q′ ∩ f(A) ≠ Ø. Since f is continuous, P = f⁻¹(P′) and Q = f⁻¹(Q′) are open subsets of X. But . Also . Finally, P ∩ A ≠ Ø, Q ∩ A ≠ Ø. Thus, A is not connected. It follows that if A is connected then f(A) must also be connected.

COROLLARY 2.6  Let X and Y be topological spaces, let f: X → Y be a continuous mapping of X onto Y, and let X be connected; then Y is connected.

COROLLARY 2.7  Let X and Y be homeomorphic topological spaces, then X is connected if and only if Y is connected.

A property of a topological space is said to be a topological property if each topological space homeomorphic to the given space must also possess this property. Thus, Corollary 2.7 states that connectedness is a topological property.

Lemma 2.8 supplies an interesting characterization of connectedness, which will facilitate our proving that the product of two connected spaces is itself connected.

LEMMA 2.8  Let Y = {0, 1}. A topological space X is connected if and only if the only continuous mappings f: X → Y are the constant mappings.

Proof. Let f: X → Y be a continuous non-constant mapping. Then P = f⁻¹({0}) and Q = f⁻¹({1}) are both non-empty. Thus, P ≠ Ø and P ≠ X. {0} and {1} are open subsets of Y and f is continuous, therefore P and Q are open subsets of X. But P = C(Q), so P is both open and closed and consequently X is disconnected. Thus, if X is connected, the only continuous mappings f: X → Y are constant mappings.

Conversely, suppose X is disconnected. Then there are non-empty open subsets P, Q of X such that P ∩ Q = Ø and P ∪ Q = X. Define a mapping f: X → Y as follows: If , set f(x) = 0; if , set f(x) = 1. f is continuous, for there are four open subsets, Ø, {0}, {1}, and Y of Y and f⁻¹(Ø) = Ø, f⁻¹({0}) = P, f⁻¹({1}) = Q, and f⁻¹(Y) = X, so that the inverse image of an open set is open.

Clearly, the role of the space Y = {0, 1} in the above result could be played by any other topological space Z consisting of two points in which all subsets are open.

THEOREM 2.9  Let X and Y be connected topological spaces. Then X × Y is connected.

Proof. We shall show that the only continuous mappings f: X × Y → {0, 1} are constant mappings. Suppose, on the contrary, that there is a continuous mapping f: X × Y → {0, 1} that is not constant. Then there are points (x₀, y₀), such that f(x₀, y₀) = 0, f(x₁, y₁) = 1. If we picture f(x, y) as a number attached to the point (x, y), then we have the situation depicted in Figure 11. Suppose f(x₁, y₀) = 0. We then define an “imbedding” ix1: Y → X × Y by ix1(y) = (x₁, y). ix₁ is continuous, hence the composite mapping fix₁ : Y → {0,1} is continuous (fix₁ may be thought of as essentially f restricted to the points of the form (x₁, y); that is, the points lying above x₁ in Figure 11.) But and . Thus, in this case, there is a non-constant mapping of Y into {0, 1}, contradicting the connectedness of Y. Similarly, if f(x₁, y₀) = 1, we define an imbedding by setting and obtain a non-constant mapping , contradicting the connectedness of X. It follows that there are no non-constant mappings of X × Y into {0, 1} and that therefore X × Y is connected.

Figure 11

COROLLARY 2.10  If X₁, X₂, . . . , Xn are connected topological spaces, then is a connected topological space.

The main idea in the proof of Theorem 2.9 is that f:X × Y → {0, 1} must remain constant on each of the connected subsets {x₀} × Y and X × {y₁}. The same procedure allows us to show that in an arbitrary product of connected spaces, altering a finite set of coordinates can not change the value of a continuous function f:X → {0, 1}.

LEMMA 2.11  Let be an indexed family of topological spaces each of which is connected. Let x and x′ be two points of such that x(a) = x′(a) except on a finite set of indices and let f:X → {0, 1} be continuous. Then f(x) = f(x′).

Proof. We shall define an “imbedding” of into X. Let J = I − I′ so that x(α) = x′(α) for . Given , set (j(z))(α) = z(α) for and (j(z))(α) = x(α) for . Then and j is continuous, for each of the functions pαj is continuous (in fact, pαj = pα for , pαj is a constant function for . Both x and x′ are in the image of the connected set so that f(x) = f(x′).

THEOREM 2.12   is connected if each Xα is connected.

Proof. Again let f:X → {0, 1} be continuous and let w, x ∈ X be such that f(w) = 0. We will show that f(x) = 0. {0} is a neighborhood of 0, hence there is a neighborhood N of w such that f(N) = 0. It follows that there is a finite set of indices I′ = {α₁, . . . , αk} and neighborhoods Nαi of w(αi) in Xαi, i = 1, . . . , k, such that . Define a point by setting x′(αi) = w(αi), i = 1, . . . , k, x′(α) = x(α) for all other . Then so f(x′) = 0. Since x(α) = x′(α) except for , f(x) = 0.

EXERCISES

1.  On the real line, prove that the set of non-zero numbers is not a connected set.

2.  Let A and B be subsets of a topological space X. If A is connected, B is open and closed, and A ∩ B ≠ Ø, prove that . [Hint: Assume and use the sets P = A ∩ B and Q = A ∩ C(B) to prove that A is not connected.]

3.  Let A and B be connected subsets of a topological space X. If A ∩ B ≠ Ø, prove that A ∪ B is connected. [Hint: in the topological space A ∪ B, show by using the result of Problem 2 that the only non-empty open and closed subset is A ∪ B.]

4.  Let A and B be non-empty subsets of a space X. Prove that A ∪ B is disconnected if . Prove that X is connected if and only if for every pair of non-empty subsets A and B of X such that X = A ∪ B we have .

5.  Prove that a space X is connected if and only if for every non-empty subset A of X different from X we have Bdry (A) ≠ Ø.

3  CONNECTEDNESS ON THE REAL LINE

In this section we shall define the term “interval” and prove that a non-empty subset of the real line is connected if and only if it is either a single point or an interval.

DEFINITION 3.1  A subset A of the real line is called an interval if A contains at least two distinct points, and if given points a, b ∈ A with a < b, then for each point x such that a < x < b, it follows that x ∈ A.

Thus, an interval contains all points between any two of its points. It is a simple matter to verify that a closed interval [a, b] or an open interval (a, b) is an interval in the sense of Definition 3.1. Other subsets of the real line that are intervals are defined in Definition 3.2.

DEFINITION 3.2  Let a be a real number. The subset of R consisting of all real numbers x such that a < x is denoted by (a, + ∞). The subset of R consisting of all real numbers x such that is denoted by [a, + ∞). The subset of R consisting of all real numbers x such that x < a is denoted by (−∞, a). The subset of R consisting of all real numbers x such that is denoted by (−∞, a].

Let b be a second real number with a < b. The subset of R consisting of all real numbers x such that is denoted by (a, b]. The subset of R consisting of all real numbers x such that is denoted by [a, b).

We shall also denote R itself by (−∞, + ∞).

The subsets of R that have been mentioned in this section exhaust the collection of intervals.

THEOREM 3.3  A subset A of the real numbers is an interval if and only if it is of one of the following forms: (a, b); [a, b); (a, b]; [a, b]; (−∞, a); (−∞, a]; (a, + ∞); [a, + ∞); (−∞, +∞).

Proof. We leave it to the reader to verify that each of these nine types of sets is an interval and shall prove the “only if” part of the theorem. Suppose A is an interval. We first note that if a point , then either x is a lower bound of A or an upper bound of A, for otherwise there would be points with a < x < b and we would obtain the contradiction . We shall, consequently, distinguish four cases.

Case 1. A has neither an upper bound nor a lower bound. In this case C(A) must be empty so that A = (−∞, +∞).

Case 2. A has an upper bound but no lower bound. Since an interval is non-empty, A has a least upper bound a. We claim that if x < a, then . For, suppose x < a, then there is a point with (for otherwise a would not be a least upper bound). Since x cannot be a lower bound of A there is a point with b < x. But b < x < a′ and imply that . We have thus shown that . On the other hand, for x > a, . It follows that A is either of the form (−∞, a] or (−∞, a), depending on whether or .

Case 3. A has a lower bound but no upper bound. By reasoning similar to that of Case 2, one shows that A is either of the form [a, +∞) or (a, +∞), where a is the greatest lower bound of A.

Case 4. A has a lower bound and an upper bound. Let a be the greatest lower bound of A and let b be the least upper bound of A. Since A contains at least two distinct points, a < b. A point x, if it is to lie in A, must therefore lie in [a, b], so that . We claim that a < x < b implies that . This implication follows from the fact that for any such point x, there must be points a′ and b′ with and . Hence . Consequently, A must be of one of the four forms (a, b), [a, b), (a, b], or [a, b], depending on which, if any, of the two points a, b belong to A.

We shall now prove that apart from the empty set and single points, the only connected subsets of the real line are intervals.

THEOREM 3.4  A subset A of the real line that contains at least two distinct points is connected if and only if it is an interval.

Proof. We shall first show that if A is not an interval then it is not connected. If A is not an interval, then there are points a, b, c with a < c < b and , whereas . Let P = (−∞, c), Q = (c, +∞). P and Q are open subsets of the real line that satisfy the conditions of Lemma 2.3; hence A is not connected.

Conversely, we shall show that if A is not connected then A is not an interval. If A is not connected, by Lemma 2.4, there are closed subsets F and G of the real line such that and both F and G contain a point of A. Assume that the notation is such that there is a point and a point with a < b. We shall find a point between a and b that is not in A. Let G′ = G ∩ [a, b]. Then G′ is a closed non-empty subset of the real line and, consequently, contains its greatest lower bound c. We cannot have a = c, for then A ∩ F ∩ G ≠ Ø, contradicting . Thus, a < c. Next, let F′ = F ∩ [a, c]. F′ is also a closed non-empty subset of the real line and therefore contains its least upper bound d. In the event that c = d we have , hence and A is not an interval. Otherwise d < c and (d, c) ∩ (F ∪ G) = Ø so that (d, c) ∩ A = Ø, and again A does not contain a point between a and b and is therefore not an interval.

EXERCISES

1.  Let f:R → R be continuous. Prove that the image under f of each interval is either a single point or an interval.

2.  Prove that a homeomorphism f:[a, b] → [a, b] carries end points into end points.

3.  Let A and B be subsets of R. A function f:A → B is called monotone increasing if and x < y imply f(x) < f(y).

(a)  Let f:A → B be monotone increasing. Prove that f:A → B is one-one.

(b)  Let f:[a, b] → [f(a), f(b)] be monotone increasing and continuous. Prove that f is a homeomorphism.

4  SOME APPLICATIONS OF CONNECTEDNESS

THEOREM 4.1  (Intermediate-Value Theorem). Let f:[a, b] → R be continuous and let f(a) ≠ f(b). Then for each number V between f(a) and f(b) there is a point such that f(v) = V.

Proof. [a, b] is connected, hence f([a, b]) is connected and is therefore an interval. Now, . Thus if V is between f(a) and f(b), since f([a, b]) is an interval, ; that is, there is a such that f(v) = V.

Theorem 4.1 states that for each V between f(a) and f(b), the horizontal line y = V intersects the graph of y = f(x) at some point (v, V) with a < v < b, as indicated in Figure 12.

If the domain of a continuous real-valued function contains an interval [a, b], then its restriction to [a, b] is continuous and we can assert that f must assume at least once each value between f(a) and f(b) over the interval [a, b].

Figure 12

As a special case of the intermediate-value theorem, namely V = 0, we have

COROLLARY 4.2  Let f: [a, b] → R be continuous. If f(a)f(b) < 0, then there is an such that f(x) = 0.

COROLLARY 4.3  (Fixed-Point Theorem). Let f:[0, 1] → [0, 1] be continuous. Then there is a such that f(z) = z.

Proof. In the event that f(0) = 0 or f(1) = 1, the theorem is certainly true. Thus, it suffices to consider the case in which f(0) > 0 and f(1) < 1. Let g:[0, 1] → R be defined by

g(x) = x − f(x)

(therefore, if g(z) = 0, f(z) = z). g is continuous and g(0) = − f(0) < 0, whereas g(1) = 1 − f(1) > 0. Consequently, by Corollary 4.2, there is a such that g(z) = 0, whence f(z) = z.

We may interpret this theorem geometrically. Since f: [0, 1] → [0, 1], the graph of y = f(x) is contained in the unit square defined by . The point (z, f(z)) given by the theorem lies on both the graph of y = f(x) and the line y = x. Hence the theorem asserts that the graph of y = f(x) intersects the line y = x in this square (see Figure 13), or equivalently, that in order for the curve which constitutes the graph to connect a point on the left-hand edge of the square with a point on the right-hand edge of the square, the curve must intersect the diagonal of the square pictured in Figure 13.

Figure 13

The reason for calling Theorem 4.3 a fixed-point theorem is that, if we think of f: [0, 1] → [0, 1] as a transformation that carries each point x of [0, 1] into the point f(x) of [0, 1], then to say that f(z) = z is to say that the transformation f leaves z “fixed.”

There are many so-called “fixed-point” theorems, of which Corollary 4.3 is undoubtedly the simplest. In general, a fixed-point theorem is one that states that for a specified topological space X each continuous function f:X → X possesses a fixed point; that is, there is necessarily a such that f(z) = z. One of the convenient facts about a fixed-point theorem is that if X and Y are homeomorphic topological spaces and a fixed-point theorem is true for X, then it is also true for Y.

THEOREM 4.4  Let X and Y be homeomorphic topological spaces. Then each continuous function h: X → X possesses a fixed point if and only if each continuous function k: Y → Y possesses a fixed point.

Proof. Let f:X → Y and g:Y → X be a pair of continuous inverse functions. Let k: Y → Y be a continuous function so that we have the diagram

and suppose that each continuous function h: X → X possesses a fixed point. Then the function h = gkf:X → X is continuous and there is such that h(z) = z. Let w = f(z). We have

Thus, w is a fixed point of k. Since the hypotheses are symmetric with regard to X and Y, it also follows that if each continuous function k:Y → Y has a fixed point then so does each continuous function h:X → X.

Any two closed intervals [a, b] and [c, d] are homeomorphic Since a fixed-point theorem holds for [0, 1], we obtain

COROLLARY 4.5  Let f:[a, b] → [a, b] be continuous. Then there is a such that f(z) = z.

Theorem 4.3 is a special case of the “Brouwer Fixed-Point Theorem,” which we shall now state. Recall that in Rn, the unit n-cube In is defined as the set of points (x₁, x₂, . . . , xn) whose coordinates satisfy the inequalities for i = 1, 2, . . . , n.

THEOREM 4.6  (Brouwer Fixed-Point Theorem). Let f:In → In be continuous. Then there is a point such that f(z) = z.

We shall not prove this theorem. However, one can supply a very suggestive argument for the truth of the theorem in the case n = 2. To this end we may, on the basis of Theorem 4.5, work with a topological space homeomorphic to I². If we think of I² as being a surface constructed of elastic material, we may conceive of a deformation or stretching by which we obtain a surface that is a disc; that is, the set of points (x₁, x₂) in the plane whose coordinates satisfy the inequality . Thus, the disc is homeomorphic with I², and we may argue the validity of the fixed-point theorem with regard to the disc.

Let g be a continuous transformation of this disc into itself. Suppose that it were possible that for each point x of the disc, we had g(x) ≠ x. Then for each point x in the disc, there would be a unique half-line Lx emanating from g(x) and passing through x (see Figure 14). The half-line Lx will contain a point on the boundary of the disc other than g(x). Let us call this point h(x). In particular, if y is a boundary point of the disc, then h(y) = y. This is true even if g(y) itself is a boundary point, as may be seen by examining the various cases depicted in Figure 14. Using the given transformation g we have thus constructed a new transformation h, which has the property that it carries each point of the disc into a boundary point and leaves each boundary point fixed (h is called a “retraction” since it retracts or pulls the interior of the disc onto its boundary while leaving the boundary fixed).

Figure 14

We next argue that the transformation h is continuous, for the image h(x) will vary by a small amount if we suitably restrict the variation of x. Though it is by no means simple to prove that no continuous transformation such as h can exist, our intuition should tell us that none can. For if there were a function such as h we should be able to retract the head of a drum onto the rim, although intuitively we know that we can do so only by ripping the drum head someplace; that is, by introducing a discontinuity. Since there is no function such as the retraction h, we have obtained a contradiction, and therefore our supposition that g did not have a fixed point is untenable.

Another application of the intermediate-value theorem relates to the concept of antipodal points on spheres. Let us recall that the n-sphere, Sn, is the set of points (x₁, x₂, . . . , xn+1) in Rn+1 satisfying the equation , the topology of Sn being the relative topology. If (x₁, x₂, . . . , xn+1) is in Sn, the pair of points (x₁, x₂, . . . , xn+1) and (−x₁, −x₂, . . . , −xn+1) are called a pair of antipodal points. Given , it is convenient to denote (−x₁, −x₂, . . . , −xn+1) by − x and call −x the antipodal point of x. A pair x, −x of antipodal points is the pair of end points of a diameter of the sphere. We shall be particularly interested in the 1-sphere, S¹, which is a circle.

Figure 15

Consider a continuous function f:S¹ → R. If we define F:S¹ → R by F(x) = f(x) − f(−x) for , we can show that F(z) = 0 for some ; that is, f(z) = f(−z), or f has the same value at one or more pairs of antipodal points. The proof of this fact is motivated by the consideration that a value of F is determined by a diameter of the circle and a designation of one of its extremities as x and the other as −x. If we rotate this diameter through π radians, as indicated in Figure 15, then the initial value of F corresponding to Figure 15a is opposite in sign to the final value of F corresponding to Figure 15c. But F is continuous, so its value must be zero for some position of the diameter corresponding to a value of θ with , where θ is the angle through which the diameter has been rotated. Thus F(z) = 0 and

THEOREM 4.7  Let f: S¹ → R be continuous, then there exists a pair of antipodal points such that f(z) = f(−z).

Theorem 4.7 is the case n = 1 of the Borsuk-Ulam Theorem.

THEOREM 4.8  Let f: Sn → Rn be continuous. Then there exists a pair of antipodal points such that f(z) = f(−z).

We shall not prove this theorem. The case n = 2 answers a question about map making. The 2-sphere S² may be thought of as the surface of a globe. In this case the Borsuk-Ulam Theorem gives a negative answer to the question, “Is it possible to draw a map of the surface of the earth on a flat sheet of paper so that distinct points on the surface of the earth correspond to distinct points on the map, and nearby points on the surface of the earth correspond to nearby points on the map?” The reason the answer to this question is “no” is that otherwise the existence of such a correspondence would imply the existence of a continuous function f: S² → R² that was one-one, and this possibility is ruled out by the Borsuk-Ulam Theorem.

EXERCISES

1.  Prove that a polynomial of odd degree considered as a function from the reals to the reals has at least one real root.

2.  Let f: [a, b] → R be a continuous function from a closed interval into the reals. Let U = f(u) and V = f(v) be such that for all . Prove that there is a w between u and v such that .

3.  Let F: R² → R be a real-valued function defined and continuous on the plane. For each continuous function f: [a, b] → R we may define a new continuous function Kf: [a, b] → R by setting . Thus, if S is the set of continuous real-valued functions defined on [a, b], K defines a transformation of S into itself. Prove that an element is a fixed point of K if and only if g satisfies the differential equation g′(x) = F(x, g(x)) with initial condition g(a) = 0.

4.  Prove that the mapping p: R → S¹ defined by p(t) = (cos t, sin t) for is continuous and that therefore, for each continuous function g: S¹ → R, there is a continuous function φ:R → R such that the diagram

is commutative. Let f: S¹ → R be continuous and define F: S¹ → R by F(x) = f(x) − f(−x) for . Prove that (Fp)(t) = − (Fp)(t + π), and that therefore there is a such that (Fp)(z) = 0. Then show that if x = p(z), f(x) = f(−x), thereby proving Theorem 4.7.

5  COMPONENTS AND LOCAL CONNECTEDNESS

In any topological space X, each point belongs to a maximal connected subset of X called the “component of a.”

THEOREM 5.1  Let X be a topological space. For each point there is a non-empty subset Cmp(a), called the component of a, with the property that Cmp(a) is connected and if D is any connected subset of X containing a, then .

Proof. There are connected subsets of X containing a for {a} is such a subset. Let I be an indexing set for the family of connected subsets containing a. We set . Thus, if D is any connected subset of X containing a, D = Dβ for some , whence . It remains to prove that Cmp(a) is connected. Suppose it is not. Then there are nonempty relatively open subsets A and B of Cmp(a) such that A ∩ B = Ø and . Assume the notation is such that and let b be a point of B. Since for some connected subset Dγ of X containing a. Now , hence A′ = A ∩ Dγ and B′ = B ∩ Dγ are non-empty relatively open subsets of Dγ. Furthermore, and . Consequently, the supposition that Cmp(a) is not connected yields the contradiction that Dγ is not connected. Therefore Cmp(a) is connected.

LEMMA 5.2  In a topological space X, let . Then Cmp(b) = Cmp(a).

Proof. Since and Cmp(a) is a connected set containing b, by Theorem 5.1, . But , hence , so, by the same reasoning it follows that and therefore Cmp(a) = Cmp(b).

COROLLARY 5.3  In a topological space X, define . Then ~ is an equivalence relation.

Proof. Since {a} is connected, so a ~ a If a ~ b or , then by 5.2, Cmp(a) = Cmp(b). We have already seen that , so and b ~ a. Finally, if a ~ b and b ~ c, then as before, Cmp(a) = Cmp(b) = Cmp(c), whence and a ~ c.

A subset of X that is a component of some point is called a component of X. The components are the equivalence sets under the relation . They constitute a partition of X into maximal connected subsets in the sense of the following definition.

DEFINITION 5.4  Let X be a set and an indexed family of nonempty subsets of X. is called a partition of X if:

 (i)  ;

(ii)  .

THEOREM 5.5  Let A be a connected subset of a topological space X and let . Then B is also connected.

Proof. We shall show that if B is not connected then A is not connected. For suppose there are open subsets P, Q of X such that , and Q ∩ B ≠ Ø. It would follow that and since . To prove that A is not connected we must show that P ∩ A ≠ Ø and Q ∩ A ≠ Ø If P ∩ A = Ø, then A would be contained in the closed set C(P), hence or . But this last relation would imply that P ∩ B = Ø. Thus, P ∩ A ≠ Ø. Similarly, Q ∩ A ≠ Ø.

COROLLARY 5.6  The closure of a connected set is connected.

COROLLARY 5.7  In a topological space, each component is a closed set.

Proof. Let A be a component, say A = Cmp(a). Then Ā is a connected set containing a and therefore . But , hence in this case A = Ā and A is closed.

It might be thought that a component must also be an open set, but it need not be as the following example will indicate. Let X be the subspace of the real line consisting of the points 0 and all numbers of the form a positive integer. The only connected set containing 0 is {0}, thus Cmp(0) = {0}. On the other hand {0} is not a neighborhood of 0 in X and hence {0} is not an open subset of X.

A sufficient condition for the components in a space to be open is that the space be “locally connected.”

DEFINITION 5.8  A topological space X is said to be locally connected at a point if each neighborhood N of a contains a connected neighborhood U of a. A topological space X is said to be locally connected if it is locally connected at each of its points.

LEMMA 5.9  Let X be a locally connected topological space and let Q be a component. Then Q is an open set.

Proof. Let . Since X is locally connected there is at least one connected neighborhood N of a. But Q = Cmp(a), hence by Theorem 5.1, , which, in turn, implies that Q is a neighborhood of a. Thus, Q is a neighborhood of each of its points and therefore Q is open.

If X is locally connected at a then there are “arbitrarily small” connected neighborhoods of a, for, given any neighborhood N of a, there is a connected neighborhood that is at least as “small” as N. An equivalent formulation of local connectedness is obtained by utilizing the concept of basis for the neighborhoods at a.

LEMMA 5.10  A topological space is locally connected at a point if and only if there is a basis for the neighborhoods at a composed of connected subsets of X.

Proof. First, suppose that X is locally connected at a and let Ua be the collection of connected neighborhoods of a. Since every neighborhood N of a contains an element of Ua, Ua is a basis for the neighborhoods at a. Conversely, if there is a basis Ua for the neighborhoods of a consisting of connected sets, each neighborhood N must contain an element of Ua and therefore X is locally connected at a.

EXERCISES

1.  Prove that a non-empty connected subset of a topological space that is both open and closed is a component.

2.  Let X be a topological space that has a finite number of components. Prove that each component of X is both open and closed.

3.  Verify that local connectedness is a topological property, but the continuous image of a locally connected space need not be locally connected.

4.  Let X and Y be homeomorphic topological spaces. Prove that any homeomorphism f:X → Y establishes a one-one correspondence between the components of X and the components of Y.

5.  Prove that the product of two locally connected topological spaces is locally connected.

6.  Prove that Euclidean n-space Rn and the standard n-cube In are locally connected.

6  PATH-CONNECTED TOPOLOGICAL SPACES

In the three-dimensional geometry of the calculus, one often discusses a curve in terms of a parametric representation, usually written x = f(t), y = g(t), z = h(t). If not stated explicitly, it is generally understood that the three functions f, g, h are at least continuous, if not differentiable over some common interval [a, b] as their domain, and therefore F(t) = (f(t), g(t), h(t)) defines a continuous function F:[a, b] → R³. The curve in question is, from this viewpoint, the image of [a, b] under F; that is, F([a, b]). We may think of this curve as “connecting” the two points F(a) = (f(a), g(a), h(a)) and F(b) = (f(b), g(b), h(b)). Given two points , the question of whether or not there is a curve “connecting” A and B is therefore seen to be the same as the question of whether or not there is a continuous function F:[a, b] → R³ such that F(a) = A and F(b) = B. Furthermore, the interval [a, b] may just as well be restricted to [0, 1], for using any homeomorphism φ:[0, 1] → [a, b], one may show that the required F:[a, b] → R³ exists if and only if a corresponding G = Fφ:[0, 1] → R³ exists. These observations motivate the following two definitions:

DEFINITION 6.1  Let X be a topological space. A continuous function f:[0, 1] → X is called a path in X. The path f is said to connect or join the point f(0) to the point f(1). f(0) is called the initial point and f(1) is called the terminal point of the path f.

If f is a path in X, f([0, 1]) is called a curve in X.

DEFINITION 6.2  A topological space X is said to be path-connected if, for each pair of points , there is a path f connecting u to v.

A non-empty subset A of a topological space X is said to be path-connected if the topological space A in the relative topology is path-connected.

The real line R is a path-connected space, for if a, b are two real numbers, the path f:[0, 1] → R defined by

for connects f(0) = a and f(1) = b. Rn is also path-connected. This may be seen by either joining a given pair x, y of points of Rn by a path, or by using the general result that if X and Y are path-connected spaces, then so is X × Y (see Exercise 5 of this section). Another significant class of path-connected spaces is the spheres, Sn, n > 0.

A path f in a topological space X whose initial and terminal points coincide is called a closed path or a loop in X. Though such paths play a significant role in topology, we shall not be concerned with them in this section.

If f is a path in a topological space X and g is a continuous mapping of X into a second topological space Y, then the composite function gf:[0, 1] → Y is a path in Y.

THEOREM 6.3  Let Y be a topological space. If there exists a path-connected topological space X and a continuous mapping g:X → Y, which is onto, then Y is path-connected.

Proof. Let . Since g:X → Y is onto, there are points such that g(a′) = a, g(b′) = b. Since X is path-connected, there is a path f in X joining a′ to b′ and, consequently, the path gf joins a to b.

Note the necessity of the requirement that g:X → Y be onto. It follows that given homeomorphic topological spaces X and Y, X is path-connected if and only if Y is path-connected. Thus, path-connectedness is a topological property.

Path-connectedness is a stronger property than connectedness; that is, if a topological space X is path-connected then X is connected.

THEOREM 6.4  Let X be a path-connected topological space, then X is connected.

Proof. Suppose X were not connected. Then there is a proper subset P of X which is both open and closed. Since P is proper, there is a point and a point . Let f:[0, 1] → X be a path from a to b. f⁻¹(P) is a proper subset of [0, 1] for . Since f is continuous, f⁻¹(P) is both open and closed. But this contradicts the fact that [0, 1] is connected. Therefore X is connected.

The converse of Theorem 6.4 is false. A counter-example to the converse, that is, a topological space that is connected but not path-connected, is the subspace of the plane consisting of the set of points (x, y) such that either

or

One may obtain some idea of this space by referring to Figure 15, where we have tried to show the main characteristics of this space. It is impossible to picture this space completely, for, as the values of x approach 0, the oscillation of the graph becomes more and more rapid.

Figure 16

It is not difficult to prove that this space is connected. First of all let us decompose this space into two subsets Z₁ and Z₂, where Z₁ is the set of points on the Y-axis, and Z₂ is the complementary set consisting of those points and . The function defines a continuous mapping of the connected interval (0, 1] onto Z₂, hence Z₂ is connected. To prove that the entire space is connected, we shall prove that ; that is, . This is so because there are points of Z₂ arbitrarily close to each point of Z₁. For, let and let ε > 0 be given. We may find an even integer N sufficiently large so that . Now and , hence by the intermediatevalue theorem there is a number such that . The point is in Z₂ and its distance from (0, y) is less than ε. Thus and is the entire space Z. By Corollary 5.6, Z is connected.

Now suppose there was a path F:[0, 1] → Z with initial point and terminal point . Let us write F(t) = (F₁(t), F₂(t)). Then F₁ and F₂ are continuous functions and F₁(0) = 0, F₁(1) = 1. The set U = F₁⁻¹ ({0}) is a closed bounded subset of the real numbers and hence contains its least upper bound t*. Since F₁(1) ≠ 0, t* < 1. We shall show that because of the oscillation of for values of x close to zero, the function F₂ cannot be continuous at t*. For each value of t such that we have F₁(t) > 0, hence and . We shall show that for each δ > 0 with , there is a value of t such that |t* − t| < δ whereas . First F₁(t* + δ) > 0, hence there is an even integer N sufficiently large so that . By the intermediate-value theorem, we may find such that . Since u, v > t* we have . Thus, if , whereas if . This contradicts the continuity of F₂ at t*. Thus no path such as F exists and therefore our space Z is not path-connected.

EXERCISES

1.  Prove directly by constructing appropriate paths that the topological spaces Rn, In (the unit cube), and Sn(n > 0) are pathconnected.

2.  Verify that in a topological space X

  (i)  if there is a path with initial point A and terminal point B, then there is a path with initial point B and terminal point A, and

 (ii)  if there is a path connecting points A and B and a path connecting points B and C, then there is a path connecting points A and C.

3.  The path component of a point x in a topological space X is the set of all points of X that may be connected to x by a path in X. Denote this subset by PCmp(x). Verify:

  (i)  for each ;

 (ii)  for each , if , then ;

(iii)  for each and , then ;

(iv)  for each is path-connected;

 (v)  if A is a path-connected subset of X, then for some .

(vi)  X is path-connected if and only if X, then for some .

4.  If A and B are path-connected subsets of a topological space X and A ∩ B ≠ Ø, then A ∪ B is path-connected.

5.  Let be an indexed family of topological spaces and set . For each let fα:I → Xα be a path in Xα. Set (fA(t))(α) = fα(t) so that fA:I → X. Prove that fA is a path in X. Prove that if each Xα is path-connected, so is X.

6.  Let X be a topological space, and let TX and SX be the cone over X and the suspension of X respectively. Prove that TX and SX are both path-connected.

7  HOMOTOPIC PATHS AND THE FUNDAMENTAL GROUP

The collection of points on and between the two concentric circles x² + y² = 1 and x² + y² = 2 is called an annulus. It is easy to see that this annulus is path-connected. For example, given two points p₀ = (x₀, y₀) and p₁ = (x₁, y₁), one may construct a path from p₀ to p₁ by first traversing the radius on which p₀ lies until we reach a point whose distance from the origin is the same as that of p₁ and then traversing in a clockwise direction the circular arc from this point to p₁ (see Figure 17). Let us call this path F₀. Alternately, one may construct a second path F₁ from p₀ to p₁, by first traversing in a clockwise direction a circular arc from p₀ to the radius on which p₁ lies and then traversing this radius until p₁ is reached. If, for the moment, we think of each of these two paths F₀ and F₁ as being represented by elastic strings with initial point p₀ and terminal point p₁, it is clear that in a given unit of time it would be possible to smoothly deform the path F₀ into the path F₁ (keeping p₀ and p₁ fixed throughout the deformation). This deformation might be carried out so that at time the string lies over the curve F1/4 of Figure 18, at time the string lies over F1/2, and at time the string lies over F3/4. We may thus conceive of the deformation of the path F₀ into the path F₁ as being accomplished by constructing an entire family of paths Ft for , such that if t and t′ are close then the paths Ft and Ft′ are “close.”

Figure 17

Figure 18

The concept of regarding two paths as being “close” implies the introduction of some sort of topology in this set of paths. Although this topology might be introduced directly by defining open set or neighborhood in the set of paths, an easier procedure is first to regard the unit of time as a unit interval on a line. Instead of viewing the two original paths F₀ and F₁ as being defined on the same unit interval, let us view F₀ as temporarily being defined on the homeomorphic image I₀ of the unit interval, where I₀ is the set of points (x, 0) in the plane with (see Figure 19). Similarly, let us view F₁ as being defined on I₁, where I₁ is the set of points (x, 1), . For each value of t, , we may view the path Ft as being defined on the homeomorphic image of the unit interval It, where It is the set of points (x, t), . If we have such a situation, we may define a function H:I² → X, where I² is the unit square and X is our annulus, by setting H(x, t) = Ft(x, t), as depicted in Figure 19. Equivalently, if we insist on viewing each path Ft as being defined on the same unit interval I, we may still obtain the same function H by setting H(x, t) = Ft(x). We now introduce the concept of closeness amongst paths by requiring that the function H:I² → X be continuous.

Figure 19

DEFINITION 7.1  Let F₀, F₁ be two paths in a topological space X with the same initial point p₀ = F₀(0) = F₁(0) and the same terminal point p₁ = F₀(1) = F₁(1). F₀ is said to be homotopic to F₁ if there is a continuous function H:I² → X such that

The function H is called a homotopy connecting F₀ to F₁.

In this event we say that the path F₀ is deformable into the path F₁ with fixed end points. One may illustrate the fact that a path F₀ is homotopic to F₁ by indicating that I² is the domain of the homotopy H, where the boundary of I² is mapped in agreement with the conditions of Definition 7.1 (see Figure 20).

Figure 20

The relation of homotopy between paths satisfies the following three properties.

THEOREM 7.2  Let F₀, F₁, F₂ be three paths in a topological space X with the same initial point p₀ and the same terminal point p₁.

   (i)  F₀ is homotopic to itself.

 (ii)  If F₀ is homotopic to F₁ then F₁ is homotopic to F₀.

(iii)  If F₀ is homotopic to F₁ and F₁ is homotopic to F₂ then F₀ is homotopic to F₂.

Proof. To show that F₀ is homotopic to itself we need only define H:I² → X by H(x, t) = F₀(x). Next, suppose that F₀ is homotopic to F₁ so that there is a homotopy H:I² → X from F₀ to F₁. For each , set H′(x, t) = H(x, 1 − t). Then H′ is easily seen to be a homotopy from F₁ to F₀. To prove (iii), first let G be a homotopy from F₀ to F₁ and let H be a homotopy from F₁ to F₂. We may construct a homotopy from F₀ to F₂ in stages. First, we alter H to a function H′ defined for (x, t′) with so that G and H′ together constitute a function K′ defined for (x, t) with . Finally, we compress K′ to a function K again defined on I². The diagrams of Figure 21 depict this process. To this end let . We then have two functions G and H′, G defined on the subset A = I² of the plane and H′ defined on the subset B consisting of the points (x, t) such that and . The set A ∩ B consists of the points (x, 1), and therefore we have G(x, 1) = F₁(x), H′(x, 1) = H(x, 0) = F₁(x); that is, G and H′ agree in their common domain of definition. We shall now prove a lemma that asserts that together G and H′ define a continuous function .

Figure 21

LEMMA 7.3  Let A, B be closed subsets of a topological space Z. Let g:A → X and h:B → X be continuous functions with the property that for . Then the function k:A ∪ B → X defined by k(z) = g(z), , k(z) = h(z), , is a continuous extension of g and h.

Proof. Let U be a closed subset of X. Then g⁻¹(U) is a relatively closed subset of A and, since A is closed, g⁻¹(U) is a closed subset of Z. Similarly, h⁻¹(U) is a closed subset of Z. But , hence k⁻¹(U) is closed and k is continuous.

Continuing now with the proof of Theorem 7.2, the function K′: A ∪ B → X defined by is continuous. We finally “compress” K′ to the function K:I² → X defined by . To recapitulate, for with , we have

whereas for , we have

From these two equations it follows that K(0, t) is the initial point of F₀ and F₂, K(1, t) is the terminal point of F₀ and F₂, and that K(x, 0) = G(x, 0) = F₀(x), whereas K(x, 1) = H(x, 1) = F₂(x). Therefore K is a homotopy from F₀ to F₂. This completes the proof of Theorem 7.2.

If F is a path that is homotopic to a path G we shall write F ≅ G. Theorem 7.2 then states: (i) F₀ ≅ F₀ (ii) if F₀ ≅ F₁ then F₁ ≅ F₀; (iii) if F₀ ≅ F₁ and F₁ ≅ F₂ then F₀ ≅ F₂. Thus ≅ is an equivalence relation. We shall denote the equivalence class of a path F by [F].

DEFINITION 7.4  An equivalence set of homotopic paths is called a homotopy class of paths. At a point z in a topological space Z the collection of homotopy classes of closed paths at z is denoted by Π(Z, z). Among these homotopy classes there is the homotopy class 〚ez〛, where ez is the constant path defined by .

The remainder of this section will be devoted to showing that there is a natural procedure whereby II(Z, z) may be converted into a group with [ez] as its identity.

DEFINITION 7.5  Let F, G:I → Z be closed paths at . Define F · G:I → Z by

Since F(1) = G(0) = z, by Lemma 7.3 F · G is a closed path at z. F · G is called the product or concatenation of F and G or F followed by G. We will now show that this product induces a product in II(Z, z).

LEMMA 7.6  In II(Z, z) let 〚F〛 = 〚F′〛 and 〚G〛 = 〚G′〛, then 〚F · G〛 = 〚F′ · G′〛.

Proof. We are given homotopies K, L:I² → Z connecting F to F′ and G to G′ respectively. In effect, a concatenation of the homotopies K and L yields a homotopy connecting F · G to F′ · G′. Let H(t, s) = K(2t, s), , and H(t, s) = L(2t − 1, s), . Since K(1, s) = L(0, s) = z, Lemma 7.3 shows that H is continuous, while H(t, 0) = (F · G)(t), H(t, 1) = (F′ · G′)(t), and H(0, s) = H(1, s) = z.

DEFINITION 7.7  In II(Z, z) let 〚F〛·〚G〛 = 〚F · G〛.

LEMMA 7.8  〚F〛 · 〚ez〛 = 〚ez〛·〚F〛 = 〚F〛 for all .

Proof. We shall first show that 〚F〛 · 〚ez〛 = 〚F〛. Define H:I² → Z by

If and s = 2t − 1, then . Thus, by Lemma 7.3, H is continuous. If s = 1 then and H(t, 1) = F(t). If s = 0 then for we have so that H(t, 0) = F(2t) while for we have so that H(t, 0) = ez(t) = z. Therefore H(t, 0) = (F · ez)(t) and H connects F · ez to F.

To show that [ez] · [F] = [F], we define H(t, s) = z for and for and in similar fashion show that H connects Fz to ez · F.

The apparent complexity of the expressions for the homotopies H in these two cases is explained by the use of the following two Figures. In Figure 22 we have projected the point with onto the point from the point (0, −1). In this fashion as s goes from 0 to 1 the interval is gradually enlarged until it becomes the interval . By analytic geometry . By setting we have arranged matters so that for a fixed s the interval is mapped in such a way as to trace out the same path as F. Finally the interval is mapped into z. Thus we have started out along the interval , mapping this horizontal interval by F · ez and gradually, as s increases, increased the length of the horizontal interval mapped using F and decreased to zero the length of the horizontal interval mapped by ez.

Figure 22

Figure 23

Similarly, using Figure 23, the projection of the point with onto the point from the point (1, 2) results in gradually contracting the interval , into the interval .

DEFINITION 7.9  Let F:I → Z be a path. Define F⁻¹:I → Z by F⁻¹(t) = F(1 − t).

If F is a path from z to y then F⁻¹ is a path from y to z. In particular if F is a closed path at z then F⁻¹ is also a closed path at z which may be thought of as F traversed in the opposite sense.

LEMMA 7.10  For each , 〚F〛 · 〚F⁻¹〛 = 〚F⁻¹〛 · 〚F〛 = 〚ez〛.

Proof. We must show that F · F⁻¹ ≅ ez ≅ F⁻¹ · F. To show that F · F⁻¹ ≅ ez define H:I² → Z as follows:

Since the various definitions of H agree when s = 2t and s = −2t + 2, H is continuous. By setting s = 0 and s = 1, H is easily seen to be a homotopy connecting ez to F·F⁻¹. Interchanging the roles of F and F⁻¹ yields a homotopy connecting ez to F⁻¹ · F.

Figure 24 may be used to explain the construction of the homotopy H. We have marked the upper and lower edges of I² with the symbols for paths to indicate how the mapping H behaves along these edges. Let Fs(t) = H(t, s). The path Fs starts out by tracing the path of F at twice its normal rate until s = 2t whereupon it remains stationary at F(s) until s = −2t + 2. The path Fs then returns to z backwards along this portion of the path of F, again at twice the normal rate.

Figure 24

We have thus shown that every element of Π(Z, z) has an inverse. To complete the proof that Π(Z, z) is a group we must show that the product is associative.

LEMMA 7.11  (〚F〛 · 〚G〛) · 〚K〛 = 〚F〛 · (〚G〛 · 〚K〛) for all .

Proof. We must show that (F · G) · K ≅ F · (G · K). We define H:I² → Z as follows:

The various definitions of H agree when s = 4t − 1 and s = 4t − 2 so H is continuous. Again by setting s = 0 and s = 1 it is easily seen that H is a homotopy connecting (F · G) · K with F · (G · K).

In Figure 25 we have illustrated the homotopy H. A point (t, s) with is projected from (0, − 1) onto the point which by (F · G) · K is mapped into . Points (t, s) with are parallel projected onto which by (F · G) ·K are mapped into . Finally a point (t, x) with is projected from (1, 2) onto the point which by (F · G) · K is mapped into .

Figure 25

EXERCISES

1.  Let X, Y be topological spaces and f:X → Y be a continuous function with f(x) = y. Let g and g′ be closed paths at . Prove that fg ≅ fg′ whenever g ≅ g′. Set f*〚g〛 = 〚fg〛. Prove that f* is a homomorphism from Π(X, x) to Π(Y, y).

2.  The category of topological spaces with base points has as its objects pairs of the form (X, x) where X is a topological space and and has as its mappings functions f: (X, x) → (Y, y) such that f: X → Y is a continuous function and f(x) = y. Set F₁(X, x) = Π(X, x) and F₂(f) = f* as defined in Exercise 1. Prove that (F₁, F₂) is a functor from the category of topological spaces with base points to the category of groups and homomorphisms as defined in Section 9, Chapter 3.

3.  Two groups G and G′ are called isomorphic if there are homomorphisms h:G → G′ and h′:G′ → G such that h′h is the identity mapping of G and hh′ is the identity mapping of G′. Prove that if f:X → Y is a homeomorphism of the topological space X with the space Y such that f(x) = y then Π(X, x) is isomorphic to Π(Y, y).

4.  Let A be a subspace of a topological space X. If there is a continuous function r:X → A such that r(a) = a for all , A is called a retract of X and r is called a retraction, i.e., ri = 1A where i:A → X is the inclusion map and 1A is the identity mapping of A. Let . Prove that if r:X → A is a retraction then i*:Π(A, a₀) → Π(X, a₀) is one-one and r*:Π(X, a₀) → Π(A, a₀) is onto. Prove that a circle on the boundary of an annulus is a retract of the annulus.

5.  Two continuous functions f, g: X → Y are said to be homotopic if there is a continuous function H:X × I → Y such that H(x, 0) = f(x), H(x, 1) = g(x). If furthermore for some we have , they are said to be homotopic rel x₀. Let f, g:X → Y be homotopic rel x₀ and let p:I → X be a closed path at x₀. Set K(t, s) = H(p(t), s). Prove that K is a homotopy connecting fp to gp and that therefore f* = g*.

6.  A subspace A of a topological space X is called a deformation retract rel x₀ of X if there is a retraction r:X → A such that ir:X → X is homotopic rel x₀ to the identity map of X where i is the inclusion map. Prove that in this case Π(A, a₀) and Π(X, a₀) are isomorphic. Prove that the center x₀ of a closed disc is a deformation retract rel x₀ of the disc. Let C be the circular boundary of a closed disc D. Prove that if Π(C, c) contains more than one element then C cannot be a retract of D.

8  SIMPLE CONNECTEDNESS

DEFINITION 8.1  A topological space Z is said to be simply connected if at each point there is only one homotopy class of closed paths.

Thus if Z is simply connected, at each point the fundamental group Π(Z, z) consists of precisely the identity element 〚ez〛. In this case there is for each closed path f at z a homotopy H: I² → Z which deforms f into the constant path ez, as depicted in Figure 26. The possibility of carrying out the deformation corresponds to the fact that the curve traced out by f does not enclose any holes in the space Z.

Figure 26

One can prove that an annulus is not simply connected, for, although a closed path such as C₁ (see Figure 27) is homotopic to a constant path, a closed path such as C₂ is not homotopic to a constant path.

Figure 27

THEOREM 8.2  Let Z be a path-connected topological space and let . Z is simply connected if and only if there is exactly one homotopy class of closed paths at z.

In order to prove this theorem we must develop a procedure for comparing the homotopy classes of closed paths at different points.

DEFINITION 8.3  Let f be a path in a topological space Z with z = f(0) and y = f(1). Let g be a closed path at y. Define gf:I → Z by

gf is a closed path at z which is constructed in accordance with Figure 28. In particular if g is the constant path ey the

Figure 28

same homotopy used in the proof of Theorem 7.10 shows that (ey)f ≅ ez.

LEMMA 8.4  Let , then .

Proof. Let K:I² → Z be the homotopy connecting g to g′. Define H:I² → Z as follows:

In the usual fashion one verifies that H is a homotopy connecting gf to g′f.

We may picture the homotopy H as being constructed in accordance with Figure 29. The homotopy K has been contracted by a factor of 3 on the t-axis so that it can occupy the middle strip of I², while the first and third segments of each horizontal line are appropriately contracted repetitions of f and f⁻¹. In fact, if f′ is homotopic to f it is easily seen that using a contraction of this homotopy to map the first strip and the reverse to map the third strip we can obtain a homotopy connecting gf and g′f′. We have thus shown:

LEMMA 8.5  Let f and f′ be homotopic paths with f(0) = f′(0) = z and f(1) = f′(1) = y. Then for .

Figure 29

DEFINITION 8.6  Let f:I → Z be a path with z = f(0) and y = f(1). For set af(〚g〛) = 〚gf〛.

PROPOSITION 8.7  af:Π(Z, y) → Π(Z, z) is a homomorphism and if f ≅ f′ then af = af′.

Proof. Since (ey)f ≅ ez, af carries the identity of Π(Z, y) into that of Π(Z, z). We must show that

for . Now af(〚g〛 · 〚h〛) = af(〚g · h〛) = 〚(g · h)f〛 and (af(〚g〛)) · (af(〚h〛)) = 〚gf〛 · 〚hf〛 = 〚gf · hf〛.

Figure 30

Thus we must show that (g · h)f ≅ gf · hf. In Figure 30 we have indicated how I² can be mapped along the lower edge by (g · h)f and along the upper edge by gf · hf. By now the procedure for constructing the appropriate homotopy should be clear. The last part follows from Lemma 8.5.

THEOREM 8.8  af:Π(Z, y) → Π(Z, z) and af−1:Π(Z, z) → Π(Z, y) are inverse functions.

Proof. Suppose . Figure 31 shows how (gf)f−1 is defined. Again a slight modification of the construction used in Lemma 7.10 provides a homotopy

Figure 31

connecting (gf)f−1 to g. Thus . Similarly afaf− is the identity.

If Z is path-connected and Π(Z, y) consists of a single element 〚ey〛, then for any point there is a path f from z to y and Π(Z, z) = af(Π(Z, y)) is also a single element. Thus Theorem 8.2 is a corollary to Theorem 8.8.

A homomorphism a:G → G′ of a group G into a group G′ which has an inverse is called an isomorphism and G and G′ are said to be isomorphic. In this event a is one-one and onto. The relation G is isomorphic to G′ is an equivalence relation. Theorem 8.8 therefore states that in a path-connected space the fundamental groups at any two points are isomorphic.

EXERCISES

1.  An isomorphism of a group G with itself is called an automorphism. Let f and f′ be paths in a space Z with f(0) = f′(1) = z and f(1) = f′(0) = y. Let f′ · f⁻¹ be the path defined by . Prove that af′af is an automorphism of Π(Z, y) such that af′af(〚g〛) = 〚f′ · f⁻¹〛 · 〚g〛 · 〚f′ · f⁻¹〛⁻¹.

2.  The fundamental groupoid of a space Z has as its objects the points of Z and as its maps H(z, y) the homotopy classes of paths from z to y. Define a rule of composition H(z, y) × H(y, w) → H(z, w) so that the fundamental groupoid of Z becomes a category. Prove that for each there is an element with α⁻¹α = 1z and αα⁻¹ = 1y. Let f:Z → W be continuous. Let F₁(z) = f(z) and F₂(〚g〛) = 〚fg〛 for . Prove that (F₁, F₂) is a functor from the fundamental groupoid of Z to that of W.

3.  Prove that a product of simply connected spaces is simply connected.

4.  Prove that for each positive integer n,Rn and In are simply connected.

For further reading, in addition to the more general texts, Wall, A Geometric Introduction to Topology, Chinn and Steenrod, First Concepts of Topology, Massey, Algebraic Topology: An Introduction, and Wallace, Introduction to Algebraic Topology are highly recommended.

CHAPTER 5

Compactness

1  INTRODUCTION

A closed and bounded subset A of the real line R is characterized by the fact that for each collection of open subsets of R such that , there is a finite subcollection with . This second property is stated in terms that are applicable to any topological space. If this property holds in a particular topological space, the space is said to be “compact.” The closed and bounded subsets of Rn are precisely the compact subspaces of Rn. This fact can be either proved directly or established by proving that the product of two compact spaces is itself compact. In metrizable spaces there is an alternate formulation of compactness; namely, that each infinite subset has a “point of accumulation.”

Compactness, like connectedness and arcwise connectedness, is a “global” property, in that it depends on the nature of the entire space. The advantage in compact spaces is that one may study the whole space by studying a finite number of open subsets. We shall see this when we prove that a continuous function f:X → Y from a compact metric space X to a metric space Y is “uniformly continuous.” In conclusion we shall examine some compact surfaces that may be formed by “identifying” edges of a rectangle.

2  COMPACT TOPOLOGICAL SPACES

DEFINITION 2.1  Let X be a set, B a subset of X, and an indexed family of subsets of X. The collection is called a covering of B or is said to cover B if . If, in addition, the indexing set I is finite, is called a finite covering of B.

Let X be a topological space and for each let Nx be a neighborhood of x. Then is a covering of X. For each integer n, let An = [n, n + 1]. Then , where Z is the set of integers, is a covering of the set R of real numbers. Similarly, if for each ordered pair (m, n) of integers we let Am, n be the set of points , such then , is a covering of R². As a final example of a covering, let X = R and let B = (0, 1]. If we set , and in general, for each positive integer n > 1, set , then , where N is the set of natural numbers, is a covering of B.

DEFINITION 2.2  Let X be a set and let be two coverings of a subset C of X. If for each for some , then the covering is called a subcovering of the covering .

Thus is a subcovering of if “every Aα is a Bβ.” In particular, if is a covering of a subset C, and I is a subset of J such that is also a covering of C, then is a subcovering of . Let Q be the set of rational numbers and for each set Bq = [q, q + 1]. Then is a covering of the real numbers R. If again we let Z be the set of integers and An = [n, n + 1], then is a subcovering of .

Suppose that f:X → Y is a continuous function from a topological space X to a metric space Y. Given ε > 0, the continuity of f gives rise to a covering of X in the following manner. For each , given this ε > 0, there is an open neighborhood Ux of x such that the images under f of all points of Ux are within ε of f(x), or equivalently, . The family of these subsets of X is clearly a covering of X. This covering has the additional property that it is composed of open sets. We shall, naturally, refer to such a covering as an “open” covering.

DEFINITION 2.3  Let X be a topological space and B a subset of X. A covering of B is said to be an open covering of B if for each , is an open subset of X.

DEFINITION 2.4  A topological space X is said to be compact if for each open covering of X there is a finite subcovering .

As an alternate definition of compactness we may use the criterion, X is compact if for each open covering of X there is a finite subset of indices {α₁, α₂, … , αn} such that the collection covers X.

DEFINITION 2.5  A subset C of a topological space X is said to be compact, if C is a compact topological space in the relative topology.

A topological space C may be a subspace of two distinct larger topological spaces X and Y. In this event the relative topology of C is the same whether we regard C as a subspace of X or of Y, and, consequently, the assertion C is compact depends only on C and its topology. We may relate the compactness of a subspace C of a topological space X to the topology of X by means of the following theorem.

THEOREM 2.6  A subset C of a topological space X is compact if and only if for each open covering of C, Uα open in X, there is a finite subcovering of C.

Proof. Let C be compact and let be an open covering of C. Then is a covering of C by relatively open sets. Thus there is a finite sub-covering and covers C. Conversely, suppose that for each open covering of C there is a finite subcovering. Let be a covering of C by relatively open subsets of C. For each where Uβ is open in X. Thus is an open covering of C. By our hypothesis there is a finite subcovering It follows that covers C and C is compact.

Compactness may be characterized in terms of neighborhoods.

THEOREM 2.7  A topological space X is compact if and only if, whenever for each a neighborhood Nx of x is given, there is a finite number of points x₁, x₂, … , xn of X such that .

Proof. Suppose X is compact. Let there be given for each a neighborhood Nx of x. For each x, there is an open set Ux such that and consequently the family is an open covering of X. Since X is compact there is a finite subcovering . But for each i, whence covers X.

Conversely, suppose that whenever, for each a neighborhood Nx of x is given, there is a finite number of points x₁, x₂, … , xn of X such that . Let be an open covering of X. Then, for each , there is an α = α(x) such that , and therefore Nx = Uα is a neighborhood of x. By our hypothesis, there are points x₁, x₂, … , xn of X such that , i = 1, 2, … , n, covers X, and hence X is compact.

In terms of closed sets, we have:

THEOREM 2.8  A topological space is compact if and only if whenever a family of closed sets is such that then there is a finite subset of indices {α₁, α₂, … , αn} such that .

Proof. Suppose X is compact and a family of closed sets is given such that . Then

so that is an open covering of X. Hence there is a finite subcovering . Therefore

Conversely, suppose that for each family of closed sets such that there is a finite subset of indices {α₁, α₂, … , αn} such that Let be an open covering of X. Then is a family of closed sets such that . Thus and , is a finite subcovering.

THEOREM 2.9  Let f:X → Y be continuous and let A be a compact subset of X. Then f(A) is a compact subset of Y.

Proof. Let be an open covering of f(A) Thus and consequently

so that is a covering of A. Since f is continuous, f⁻¹(Uα) is an open subset of X for each and therefore is an open covering of A. A is compact, thus there is a finite subcovering of A. But implies that was an arbitrary open covering of f(A), whence by Theorem 2.6, we have shown that f(A) is compact.

COROLLARY 2.10  Let the topological spaces X and Y be homeomorphic. Then X is compact if and only if Y is compact.

Not every subset of a compact space is itself compact. We shall see that the closed interval [0, 1] is compact, whereas the open interval (0, 1) is not compact. To show that (0, 1) is not compact, it suffices to find one open covering of (0, 1) that does not have a finite subcovering. To this end, for each integer n = 3, 4, 5, … , let . Then (Un)n=3, 4, 5, … is an open covering of (0, 1). On the other hand, for each integer k > 3 we have . Thus the union of every finite subcollection of {Un}n = 3, 4, 5, … must fail to contain some point of (0, 1), and hence there is no finite subcovering of {Un}n = 3, 4, 5, ….

We do, however, have this result.

THEOREM 2.11  Let X be compact. Then each closed subset of X is compact.

Proof. Let F be a closed subset of the compact space X. If is an open covering of F, then by adjoining the open set O = C(F) to the family we obtain an open covering of X. Since X is compact there is a finite subcovering Vβ1, Vβ2, … , Vβm of X. But each is either equal to a Uα for some or equal to O. If O occurs among Vβ1, Vβ2, … , we may delete it to obtain a finite collection of the Uα’s that covers F = C(O).

Thus, in a compact space, for each subset the property of being closed implies the property of being compact. In a Hausdorff space, the converse is also true.

THEOREM 2.12  Let X be a Hausdorff space. If a subset F of X is compact, then F is closed.

Proof. We shall show that O = C(F) is open by showing that for each point there is a neighborhood U of z contained in O, or equivalently, for which U ∩ F = Ø. To this end, with fixed, by the Hausdorff property, we may choose for each point an open neighborhood Ux of z and an open neighborhood Vx of x such that Ux ∩ Vx = Ø. The family is an open covering of F. Since F is compact, there is a finite subcovering of F. The intersection is an intersection of a finite set of neighborhoods of z and is therefore a neighborhood of z. Furthermore, U cannot intersect F since it does not intersect each element of a covering of F. Thus , from which it follows that O is a neighborhood of each of its points and F = C(O) is closed.

COROLLARY 2.13  Let X be a compact Hausdorff space. Then a subset F of X is compact if and only if it is closed.

THEOREM 2.14  Let f:X → Y be a one-one continuous mapping of the compact space X onto a Hausdorff space Y. Then f is a homeomorphism.

Proof. We define g:Y → X by setting g(y) = x if f(x) = y, so that f and g are inverse functions. It remains to prove that g is continuous. We shall prove this by proving that for each closed subset F of X, g⁻¹(F) is a closed subset of Y. Given a closed subset F of X, by Theorem 2.11, F is compact. Hence f(F) = g⁻¹(F) is a compact subset of Y. By Theorem 2.12, g⁻¹(F) is a closed subset of Y. Thus, g is continuous and f is a homeomorphism.

EXERCISES

1.  Prove that the real line R is not compact.

2.  Prove that every finite subset of a topological space is compact.

3.  Let be an open covering of [0, 1]. Define a subset P of [0, 1] as follows: a point x is in P if and only if there is a finite collection of elements of that covers [0, x]. Prove that if , then there is a neighborhood O of x such that and that therefore P is open. Prove that if , then there is a neighborhood O of x such that O ∩ P = Ø and therefore P is closed. Conclude that P = [0, 1] and that therefore [0, 1] is compact.

4.  Let X be a topological space. A family of subsets of X is said to have the finite intersection property if for each finite subset J of I, . Prove that X is compact if and only if for each family of closed subsets of X that has the finite intersection property, we have .

5.  Let X be a set and and ′ be two topologies on X. Prove that if and is compact then is compact. Prove that if is Hausdorff and is compact with , then .

6.  Let f:X → Y be a continuous mapping of a compact space X onto a Hausdorff space Y. Prove that f is an identification.

3  COMPACT SUBSETS OF THE REAL LINE

DEFINITION 3.1  A subset A of Rn is said to be bounded if there is a real number K such that for each

In particular a subset A of the real line R is bounded if A is contained in some closed interval [− K, K], K > 0. Every closed interval [a, b] is bounded for where K = maximum {|a|, |b|}.

LEMMA 3.2  If A is a compact subset of R then A is closed and bounded.

Proof. Since the real line satisfies the Hausdorff axiom, by Theorem 2.12, A is closed. For each positive integer n, let , where N is the set of natural numbers. Therefore is an open covering of A. Since A is compact, . If we set k = maximum {n₁, n₂, … , nq} then for i = 1, 2, … , q, whence . Thus and A is bounded.

LEMMA 3.3  The closed interval [0, 1] is compact.

Proof. Let be a covering of [0, 1] by open sets. Assume that there is no finite subcovering of . In this event, at least one of the two closed intervals or cannot be covered by a finite subcollection of the family . Let [a₁, b₁] denote one of these two intervals of length that cannot be covered by a finite sub-collection of the family . We may now divide the interval [a₁, b₁] into the two subintervals

and

of length and assert that at least one of these two intervals cannot be covered by a finite subcollection of the family . Let [a₂, b₂] denote one of these two intervals of length that has the property that it cannot be covered by a finite subcollection of the family . We shall now proceed to define a sequence [a₀, b₀], [a₁, b₁], [a₂, b₂], … , [an, bn], … of such intervals. Assume that for r = 0, 1, 2, … , n we have defined intervals [ar, br] such that:

1.  [a₀, b₀] = [0, 1];

2.  ;

3.  for r = 0, 1, … , n − 1, either or ;

4.  for each r = 0, 1, … , n, no finite subcollection of covers [ar, br].

We then define [an+1, bn+1] in the following manner. In view of (4) at least one of the two intervals

cannot be covered by a finite subcollection of . Denote by [an+1, bn+1] whichever of these two intervals cannot be covered by a finite subcollection of , (if neither can be, we may agree that [an+1, bn+1] is the first of the two). Then conditions (2), (3), and (4) will also hold for [an+1, bn+1]. It follows by induction that we may define a sequence [a₀, b₀], [a₁, b₁], [a₂, b₂], … of such intervals for which conditions (1) through (4) are true.

By conditions (3), . It follows that for each pair of positive integers m and . Thus each bn is an upper bound of the set {a₀, a₁, a₂, …}. Let a be the least upper bound of this set. Then for each n, and hence a is a lower bound of the set {b₀, b₁, b₂, …}. Let b be the greatest lower bound of the latter set. We therefore have . But, by the definition of a and b, we must have for each n, whence by condition (2), for each n and we conclude that a = b. We are now in a position to obtain a contradiction to condition (4), from which it will follow that our assumption that there is no finite subcovering of [0, 1] is untenable.

covers [0, 1] and . Therefore for some . Since Oβ is open there is an ε > 0 such that . Let us choose the positive integer N large enough so that . Then bN − aN < ε. Now . Therefore, and . Consequently, . Thus [aN, bN] may be covered by a finite subcollection (namely, one!) of the family . Therefore the assumption that no finite subcollection of covers [0, 1] leads to a contradiction and we must conclude that [0, 1] is compact.

It can be seen that the gist of the above argument is that if no finite subcollection of covers [0, 1], then no finite subcollection of covers a sequence of subintervals whose lengths approach zero, whereas on the other hand if the length of one of these subintervals is small enough it is contained in some Oβ.

Since each closed interval [a, b] is homeomorphic to the closed interval [0, 1] and compactness is a topological property, we obtain:

COROLLARY 3.4  Each closed interval [a, b] is compact.

The next theorem, which characterizes the compact subsets of the real line, is frequently referred to as the Heine-Borel Theorem.

THEOREM 3.5  A subset A of the real line is compact if and only if A is closed and bounded.

Proof. The “if” half of the theorem is Lemma 3.2. Conversely, if A is closed and bounded A is a closed subset of a closed interval [−K, K] for some K > 0. But [−K, K] is a compact space and therefore, by Theorem 2.11, A is compact.

EXERCISES

1.  Using the method of subdivision of Lemma 3.3, prove that the unit square I² is a compact subset of the plane and in general that the unit n-cube In is a compact subspace of Rn.

2.  Let X be a compact space and (Fn)n = 1,2,3, … a sequence of nonempty closed subsets of X such that for each n. Prove that .

3.  Let f:[a, b] → R be continuous. Prove that the set f([a, b]) has both a least upper bound M and a greatest lower bound m and that there are points such that f(u) = M, f(v) = m.

4.  Let f:[a, b] → [c, d] be a continuous increasing function such that f(a) = c, f(b) = d. Prove that f is a homeomorphism.

4  PRODUCTS OF COMPACT SPACES

The fundamental result of this section is that the product of two compact spaces is itself compact. We shall establish this fact with the aid of the next lemma, which relates compactness to coverings by members of a base for the open sets. Let us recall that a base for the open sets of a topological space Z is a collection of open subsets with the property that each open subset of Z is a union of members of the collection .

LEMMA 4.1  Let be a base for the open sets of a topological space Z. If, for each covering of Z by members of , there is a finite subcovering, then Z is compact.

Proof. We must show that, if each covering of Z by “basic” open sets has a finite subcovering, then each open covering of Z has a finite subcovering. For each , Oα is a union of members of . Let J be an indexing set for all the basic sets Bβ that occur in the expression of some Oα as a union of members of . Thus and hence is a covering of Z by members of . It follows from our hypothesis that there is a finite subcovering of Z. Since each occurs in the expression of some , as a union of members of , Consequently, must cover Z and therefore Z is compact.

Let us recall that if X and Y are topological spaces, then a base for open sets of X × Y is the collection of sets of the form U × V, where U is open in X and V is open in Y.

THEOREM 4.2  Let X and Y be compact topological spaces; then X × Y is compact.

Figure 32

Proof. By virtue of Lemma 4.1 it suffices to prove that each covering of X × Y by sets of the form U × V, U open in X, V open in Y, has a finite subcovering. Let be such a covering. As an aid to understanding the proof, let us view the product X × Y as pictured in Figure 32, where a point lies over the point and level with the point . In particular, for each , the subset of X × Y consisting of all points (x₀, y), , may be thought of as the collection of points lying over x₀. The open covering is necessarily an open covering of . But is homeomorphic to Y and hence compact. We may therefore find a finite subset of I such that covers [this finite covering of is portrayed by the small rectangles in Figure 32]. We may also assume that for each , for otherwise we may delete Uβ × Vβ and still cover . The set is a finite intersection of open sets containing x₀ and is therefore an open set containing x₀. We assert that is an open covering of . For, suppose . The point (x₀, y) is in Uβ × Vβ for some . Since for all . It follows that , proving our assertion.

Now is an open covering of the compact space X, hence there is a finite subcovering of X. Let us set and show that the finite family is a covering of X × Y. Given a point for some xi so that . By our previous assertion for some which certainly implies that for some . We have thus established that is a finite subcovering and that therefore X × Y is compact.

If X₁, X₂, … , Xn are topological spaces, one may distinguish between and , for the points of the first space are n-tuples (x₁, x₂, … , xn), whereas the points of the second space are ordered pairs ((x₁, x₂, … , xn−1), xn) whose first elements are (n − 1)-tuples. Nevertheless, these two spaces are certainly homeomorphic [the obvious homeomorphism carries a point (x₁, x₂, … , xn) into ((x₁, x₂, … , xn−1), xn)], hence by induction on n we obtain:

COROLLARY 4.3  Let X₁, X₂, … , Xn be compact topological spaces. Then is also compact.

It is true that the product of an arbitrary family of compact spaces is compact. This result, which we shall not prove, is called the Tychonoff Theorem.

Let us recall that the unit n-cube In is the subset of Rn consisting of all points x = (x₁, x₂, … , xn) such that for i = 1, 2, … , n. As a subspace of Rn, In has the same topology as the product space I × I × … × I (n-factors). Since I = [0, 1] is compact, as a special case of Corollary 4.3 we have:

COROLLARY 4.4  The unit n-cube In is compact.

THEOREM 4.5  A subset A of Rn is compact if and only if A is closed and bounded.

Proof. The proof that if A is compact then A is closed and bounded is similar to the proof of this fact for a subset of the real line as presented in the preceding section. Conversely, we shall first show that each closed “cube” is compact. The collection of points x = (x₁, x₂, … , xn) in Rn such that for i = 1, 2, … , n, which we shall denote by MK, is a “cube” of width 2K with center at the origin. MK is homeomorphic to the unit n-cube In, for the function F:In → MK defined by

is easily seen to be a homeomorphism (Theorem 2.14). Since In is compact, MK is compact. Now suppose A is closed and bounded; then A is a closed subset of the compact cube MK for some K, whence A is compact.

EXERCISES

1.  Let S be the set [0, 1] and define a subset F of S to be closed if either it is finite or is equal to S. Prove that this definition of closed set yields a topology for S. Show that S with this topology is connected, path-wise connected, and compact, but that S is not a Hausdorff space. Show that each subset of S is compact and that therefore there are compact subsets of S that are not closed.

2.  A topological space X is said to be locally compact if each point has at least one compact neighborhood. Prove that the real line and Rn are locally compact.

3.  Let X be a topological space and x* a point of X. Assume a base for the system of neighborhoods of x* consists of the complements of compact subsets of X − {x*}. Prove X is compact. Prove that if in addition X − {x*} is a locally compact Hausdorff space, then X is a compact Hausdorff space. Given a locally compact Hausdorff space Y which is not compact, show that Y is a subspace of a compact Hausdorff space that contains one more point than Y does. This space is called the one-point compactification of Y. Prove that the one-point compactification of Rn is homeomorphic to Sn.

5  COMPACT METRIC SPACES

A metric space (X, d) is said to be compact or is called a compactum if its associated topological space is compact. In this section we shall derive certain properties of compact metric spaces. A basic result is that a metric space is compact if and only if every infinite subset has at least one “point of accumulation.”

DEFINITION 5.1  Let X be a topological space and A a subset of X. A point is called an accumulation point of A if each neighborhood of a contains infinitely many distinct points of A.

In referring to the accumulation points of a set A, care must be taken to specify of which topological space A is to be considered a subset. For example, in the real line R, the subset has the accumulation point 0, whereas in the topological space (0, + ∞), the same set A has no accumulation points.

Recall that in a metric space we defined a as a limit point of a subset A if every neighborhood of a contains a point of A different from a. If we use the same definition in a topological space every accumulation point of A is also a limit point of A. In Hausdorff spaces, and hence in metric spaces, accumulation points and limit points coincide.

LEMMA 5.2  Let X be a Hausdorff space and A a subset of X. A point is an accumulation point of A if and only if a is a limit point of A.

Proof. Suppose a is not an accumulation point of A. Then there is a neighborhood N of a that contains at most a finite collection {a₁, a₂, … , ap} of points of A distinct from a. For each of these points ai, i = 1, 2, … , p, we can find neighborhoods Ui of a and neighborhoods Vi of ai such that Ui ∩ Vi = Ø. Then N ∩ U₁ ∩ U₂ ∩ … ∩ Up is a neighborhood of a that contains no points of A other than possibly a.

THEOREM 5.3  Let X be a compact space; then every infinite subset K of X has at least one limit point in X.

Proof. Suppose K is a subset of X that has no limit points. For each there is a neighborhood Nx of x such that Nx ∩ K = {x}. K is closed and hence compact. Therefore there are points x₁, x₂, … , xm such that cover K. It follows that K = {x₁, x₂, …, xm} and K is finite.

For compact Hausdorff spaces, and in particular for compact metric spaces, Theorem 5.3 becomes every infinite subset A of X has at least one point of accumulation in X. The next two lemmas are used to prove the converse for metric spaces.

LEMMA 5.4  Let (X, d) be a metric space such that every infinite subset of X has at least one accumulation point in X. Then, for each positive integer n, there is a finite set of points of X such that the collection of open balls

covers X.

Proof. Suppose there were an integer n such that no finite collection of balls of radius covered X. Choose a point certainly does not cover X, hence there is a point such that does not cover X, hence there is a point such that . Continuing in this way we may construct a sequence x₁, x₂, … , xk, … of points of X such that for k > 1,

Thus

if k ≠ k′. It follows that the set {x₁, x₂, … , xk, …} is infinite and therefore has a point of accumulation . The neighborhood contains infinitely many points of {x₁, x₂, … , xk, …} and in particular contains two points xk, xk′ with k ≠ k′. Since , we obtain the contradiction .

A similar argument yields the following result.

LEMMA 5.5  Let (X, d) be a metric space such that each infinite subset of X has at least one point of accumulation. Then for each open covering of X there is a positive number ε such that each open ball B(x; ε) is contained in an element Oβ of this covering.

Proof. We shall suppose the result is false and obtain a contradiction. If the lemma is false, for each n = 1, 2, … , there is an open ball such that for each . Let A = {x₁, x₂, …}. If A is finite, some point occurs infinitely often in the sequence x₁, x₂, … . Since covers for some Also, Oβ is open, hence there is a δ > 0 such that . We may, however, choose n so that and xn = x, in which case

a contradiction. There remains the possibility that A = {x₁, x₂, …} is infinite. Thus A has at least one point of accumulation x. Again for some so that for some δ > 0. There are infinitely many points of A in the neighborhood of x. Hence we may choose an n such that and . We then have , which is again a contradiction.

The number ε of Lemma 5.5 depends on the particular open covering considered. Given the open covering , if the number ε has the property that for each for some , then each number ε′ with 0 < ε′ < ε also has this property. The least upper bound of the set of numbers having this property is called the Lebesgue number, εL, of the open covering . We may now state:

COROLLARY 5.6  Let (X, d) be a metric space such that each infinite subset of X has an accumulation point. Then each open covering of X has a Lebesgue number εL.

A topological space X is said to have the Bolzano-Weierstrass property if each infinite subset of X has at least one point of accumulation. We may now prove that every metric space that has the Bolzano-Weierstrass property is a compact metric space.

THEOREM 5.7  Let (X, d) be a metric space that has the property that every infinite subset of X has at least one accumulation point. Then X is compact.

Proof. Let be an open covering and let εL be its Lebesque number. Let us choose n so that . By Lemma 5.4 there is a finite set {x₁, x₂, … , xp} of points of X such that the open balls cover X. Furthermore, by Lemma 5.5, for each i = 1, 2, … , p, there is a such that . It follows that the collection is a finite subcovering of .

We have now proved the main result of this section.

THEOREM 5.8  Let (X, d) be a metric space. Each infinite subset of X has at least one accumulation point if and only if X is compact.

Having proved that a subspace X of Euclidean n-space Rn is compact if and only if it is closed and bounded, we may state:

COROLLARY 5.9  Let X be a subspace of Rn. Then the following three properties are equivalent:

1.  X is compact.

2.  X is closed and bounded.

3.  Each infinite subset of X has at least one point of accumulation in X.

The existence, for each open covering of a compact metric space, of a Lebesgue number has as a consequence the fact that each continuous function defined on a compact metric space is “uniformly” continuous.

DEFINITION 5.10  Let f:(X, d) → (Y, d′) be a function from a metric space (X, d) to a metric space (Y, d′). f is said to be uniformly continuous if, for each positive number ε, there is a δ > 0, such that whenever d(x, y) < δ, then d′(f(x),f(y)) < ε.

If the function g:X → Y is continuous, then for each and each ε > 0, there is δ > 0, where δ may depend on both the choice of x and ε, such that d(x, a) < δ implies d′(g(x), g(a)) < ε. If, however, g is uniformly continuous, then given ε > 0, the number δ may be used at each point , that is, uniformly throughout X, to yield d′(g(x), g(a)) < ε if d(x, a) < δ. Thus:

COROLLARY 5.11  If f:X → Y is uniformly continuous, then f is continuous.

On the other hand a continuous function need not be uniformly continuous. As an example, consider f:(0, 1] → R defined by . Given ε = 1, we shall show that there does not exist a δ > 0 such that |x − y| < δ implies |f(x) − f(y)| < 1. For given any δ > 0 we can choose n large enough so that if we have

whereas

In view of the next result, it should be noted that in this example the interval (0, 1] is not compact.

THEOREM 5.12  Let f:(X, d) → (Y, d′) be a continuous function from a compact metric space X to a metric space Y. Then f is uniformly continuous.

Proof. Given ε > 0, for each , there is a δx > 0 such that if then . The collection is an open covering of X. Since X is compact, this open covering has a Lebesgue number. Let us choose δ to be a positive number less than this Lebesgue number. If and d(z, z′) < δ so that z and z′ are in a ball of radius less than δ, we have for some . Consequently, , whence .

EXERCISES

1.  In a metric space (X, d), a sequence a₁, a₂, … of points of X is called a Cauchy sequence if for each ε > 0 there is a positive integer N such that d(an, am) < ε whenever n, m > N. A metric space X is called complete if every Cauchy sequence in X converges to a point of X. Prove that a compact metric space is complete.

2.  In Euclidean n-space Rn, prove that every Cauchy sequence lies in a bounded closed subset of Rn. Use this fact to prove that Rn is complete.

3.  Let (X, d) be a compact metric space. Prove that X is “bounded with respect to d”; that is, there is a positive number K such that for all

4.  Let (X, d) be a compact metric space and let be a family of closed subsets of X such that . Prove that there is a positive number c such that for each for some .

5.  A subset A of a topological space X is called dense if Ā = X. Let X be a compact metric space. Prove that there is a sequence a₁, a₂, … of points of X such that the set A = {a₁, a₂, …} is dense in X.

6.  Let X be the set of continuous functions f:[a, b] → R. Let I:X → R be defined by . Define a distance function d on X by setting . Prove that I is uniformly continuous. Let f₁, f₂, … be a Cauchy sequence in (X, d). Prove that for each , … is a Cauchy sequence of real numbers. For each , denote by f(t) the limit of this sequence. Prove that the function f:[a, b] → R so defined is continuous, that limn fn = f in X, and therefore (X, d) is complete, so that in the terminology of Problems 2 and 3 in Section 8, Chapter 2, X is a complete normed vector space. [A complete normed vector space with either the real or complex numbers as scalars is called a Banach space.]

7.  Let A be any set and let RA be the set of all functions f:A → R where R is the reals. Define f + g by (f + g)(a) = f(a) + g(a) and αf by (αf)(a) = αf(a), for and . Prove that RA is a vector space with R as scalars. A function is bounded if exists. Prove that the set B of bounded functions is a normed vector space in the sense of Problem 2, Section 8, Chapter 2. Prove that B is a complete metric space. Now let A be a topological space and let C(A, R) be the set of all bounded continuous functions from A to R. Prove that C(A, R) is a closed subset of B and is complete.

6  COMPACTNESS AND THE BOLZANO-WEIERSTRASS PROPERTY

Theorem 5.8, which states that a metric space is compact if and only if each infinite subset has at least one accumulation point, raises the question as to whether or not these two properties are equivalent in an arbitrary topological space. We already know that the first implies the second for Hausdorff spaces. Since there are examples of topological spaces that are not compact, but in which each infinite subset has a point of accumulation, compactness is the stronger of the two properties. We might therefore think of the second property, which we have called the Bolzano-Weierstrass property, as a weaker type of compactness. To illustrate how much weaker the Bolzano-Weierstrass property is, we need to introduce the concept of countability.

DEFINITION 6.1  A non-empty set X is said to be countable if there is an onto function f:N → X, where N is the set of positive integers.

A finite set X = {x₁, x₂, . . . , xn} is countable, for we may construct an onto function f:N → X by setting f(i) = xi, 1 ≦ i ≦ n, and defining f(i) for i > n arbitrarily, say f(i) = x₁, i > n. A countable set that is not finite is called denumerable. In this case an onto function f:N → X gives rise to an “enumeration,” x₁ = f(1), x₂ = f(2), . . . , xn = f(n), . . . of the elements of X, so that we may write X = {x₁, x₂, . . . , xn, . . .}. Since we have not required the function f to be one-one, a given element x ∈ X may occur more than once in this enumeration. It is easy to see, however, that by deleting all but the first occurrence of any element x ∈ X and reducing the succeeding subscripts accordingly, it is possible to obtain an enumeration of X in which each element occurs one and only one time.

There are several facts about countability that are of general interest. As a simple consequence of Definition 6.1 we obtain:

COROLLARY 6.2  Let X and Y be non-empty sets. If X is countable and there is an onto function g:X → Y, then Y is countable.

Proof. Since X is countable, there is an onto function f:N → X, N the set of positive integers. The composite function gf:N → Y is onto and hence Y is countable.

COROLLARY 6.3  A non-empty subset of a countable set is countable.

Proof. Let A ⊂ X, countable, A non-empty. We may define an onto function g:X → A by setting g(a) = a for a ∈ A and defining g arbitrarily for points x ∉ A.

The set N of positive integers is countable, since the identity function i:N → N is onto. On the other hand, the collection 2N of subsets of N is not countable, since for an arbitrary set X there is no onto function f:X → 2X [see Exercise 1]. A set that is not countable is called uncountable. Another example of an uncountable set is the set R of real numbers [see Exercise 2]. Surprisingly, N × N is a countable set.

THEOREM 6.4  Let N be the set of positive integers. Then N × N is countable.

Proof. The elements of N × N may be arrayed in the form of the infinite matrix of Figure 33. We may arrange these elements in the form of a sequence,

Figure 33

x₁ = f(1), x₂ = f(2), . . . , xk = f(k), . . . , by setting x₁ = (1, 1), x₂ = (2, 1), x₃ = (1, 2), x₄ = (3, 1), . . . ; that is, having exhausted the entries on the diagonal of this matrix from (p, 1) to (1, p) we proceed to enumerate the entries on the diagonal from (p + 1, 1) to (1, p + 1). To explicitly define the onto function f:N → N × N we note that there are entries on or above the diagonal from (p, 1) to (1, p), hence if 1 ≦ j ≦ p + 1 we are setting

As a direct consequence of Theorem 6.4 and Corollary 6.2 one obtains the result that the set Q⁺ of positive rational numbers is countable, for the function h:N × N → Q⁺ defined by is onto.

COROLLARY 6.5  Let X₁, X₂, . . . , Xn, . . . , be a sequence of sets, each of which is countable. Then is a countable set.

Proof. Since each Xi is countable there is an onto function fi:N → Xi, i = 1, 2, . . . , n, . . . . We define a function by setting . F is onto, for if for some i, whence x = fi(j) = F(i, j) for some (i, j) ∈ N × N. But N × N is countable and therefore is countable.

A more direct proof of Corollary 6.5 can be given by utilizing the matrix of Figure 33 to display the elements of , entering the element in the ith row and jth column. One then enumerates the elements of in accordance with the scheme adopted in the proof of Theorem 6.4 Since any countable collection of sets may be arranged in the form of a finite or infinite sequence of sets, Corollary 6.5 states that, if X is the union of a countable collection of sets, each of which is countable, then X is countable.

In view of the fact that the set Q⁺ of positive rational numbers is countable, the set Q⁻ of negative rational numbers is also countable. Consequently, the set Q of all rational numbers is countable. Using Corollary 6.5 we may then assert that the collection B of all open intervals on the real line of the form B(p; q), q > 0, with p and q rational, is also a countable set, for it is a countable union of sets each of which is countable. This fact may be used to prove that there is a countable basis for the open sets on the real line.

Let us now return to our discussion of the relation between compactness and the Bolzano-Weierstrass property. The Bolzano-Weierstrass property implies that each countable covering has a finite subcovering.

THEOREM 6.6  Let E be a subspace of a topological space X with the property that each infinite subset of E has a point of accumulation in E. Then every countable open covering of E has a finite subcovering.

Proof. We may assume that a countable open covering of E is given in the form of a sequence O₁, O₂, . . . , On, . . . of open subsets of X such that . Suppose that no finite subcollection covers E. Then for each integer k, the open set does not cover E. Hence for each k there is a point xk ∈ E such that . The subset

of E must be infinite. Let x ∈ E be a point of accumulation of A. Since x ∈ E, x ∈ Op for some index p. Op is a neighborhood of x and therefore infinitely many of the points of A belong to Op. In particular, for some k > p we would have , contradicting the choice of xk. Therefore there must be a finite subcollection of the open sets O₁, O₂, . . . , On, . . . that covers E.

If a topological space X is such that every open covering has a countable subcovering, by virtue of Theorem 6.6, the Bolzano-Weierstrass property implies compactness. A sufficient condition for every open covering to have a countable subcovering is given by the next theorem, often called Lindelöf’s Theorem.

THEOREM 6.7  Let X be a topological space that has a countable basis for the open sets. Then each open covering has a countable subcovering.

Proof. Let be a countable basis for the open sets of X. We shall first prove that for each point x ∈ X and each open set O containing x, there is a basis element Bβ such that x ∈ Bβ ⊂ O. For, since is a basis for the open sets, O is a union of elements of , thus for some subset J′ of J. But x ∈ O, hence x ∈ Bβ for some β ∈ J′, and clearly . Now suppose that is an open covering of x. We must find a countable subset such that is a covering. For each x ∈ X and each Oα containing x, we choose a Bβ such that . The totality of sets Bβ so chosen constitute a countable subfamily of the basis and this subfamily covers X. Now, for each such Bβ with , let us choose a single index such that . The totality of sets Oα so chosen constitute a subfamily , which is also countable and must cover X, for .

COROLLARY 6.8  Let X be a topological space that has a countable basis for the open sets. Then X is compact if and only if X has the Bolzano-Weierstrass property.

Although we shall not give an example of a topological space X that has the Bolzano-Weierstrass property, but is not compact, the preceding discussion has revealed that such a space must be found among those topological spaces which are not metrizable and do not possess a countable basis for the open sets. Those spaces which possess a countable base for the open sets are called completely separable or are said to satisfy the second axiom of countability.

EXERCISES

1.  Let X be an arbitrary non-empty set and f:X → 2X an arbitrary function from X to the subsets of X. Let A be the subset of X consisting of those points x ∈ X such that x ∉ f(x). Prove that there cannot be a point a ∈ X such that A = f(a). Finally, prove that there is no onto function f:X → 2X.

2.  Let a function f:N → [0, 1] be given, N the set of positive integers. In the resulting enumeration x₁ = f(1), x₂ = f(2), . . . , of numbers in [0, 1], express each number xk in decimal notation an integer . Construct a real number y = .y₁y₂ . . . yn . . . such that , r = 1, 2, . . . , thereby obtaining the result that f cannot be onto and consequently the real numbers are not countable.

3.  Use the rational density theorem, which states that between any two real numbers there is a rational number, to prove that the collection of open intervals B(p; q), q > 0, p, q rational are a basis for the open sets of R and that therefore R satisfies the second axiom of countability.

4.  Let X and Y be topological spaces satisfying the second axiom of countability. Prove that X × Y also satisfies the second axiom of countability and hence Rn does.

5.  Let {Aα}α∈ I and {Bβ}β∈ J be families of subsets of a set X. {Aα}α∈ I is called a refinement of {Bβ}β∈ J if for each α ∈ I there is a β ∈ J such that Aα ⊂ Bβ. Suppose that {Aα}α∈ I is a refinement of {Bβ}β∈ J and that {Aα}α∈ I covers X. Prove that if I is finite there is a finite subcovering of {Bβ}β∈ J and if I is countable there is a countable subcovering of {Bβ}β∈ J.

6.  Recall that a subset A of a topological space X is called dense in X if Ā = X. A topological space X is called separable if there is a countable dense subset. Prove that X is separable if X satisfies the second axiom of countability.

7.  A topological space X is said to satisfy the first axiom of countability if at each point x ∈ X there is a countable basis for the complete system of neighborhoods at x. Prove that if X satisfies the second axiom of countability then X satisfies the first axiom of countability.

8.  Let X satisfy the first axiom of countability. Prove that for each x ∈ X there is a countable basis U₁, U₂, . . . for the neighborhoods at x such that and such that if ui ∈ Ui then limn un = x. Let f:X → Y be a function into a second topological space Y. Show that if for all sequences x₁, x₂, . . . such that x = limn xn we have f(x) = limn f(xn) then f is continuous at x.

9.  Let f:X → X be a function from a metric space X into itself. f is said to be contractive if there is a positive constant K < 1 such that d(f(x), f(x′)) < K · d(x, x′) for all x, x′ ∈ X. Prove that a contractive function is continuous. Let a ∈ X. Set a₀ = a, a₁ = f(a), a₂ = f(a₁), and in general an+1 = f(an) = fn+1(a). Prove that for such an f the following hold: d(an+1, an) < Kn−1 d(a₁, a₀); a₀, a₁, . . . , an, . . . is a Cauchy sequence. If X is a complete metric space so that limn an = a for some a ∈ X, then a is a fixed point of f, and if f(b) = b, b = a, so that every contractive mapping has a unique fixed point.

7  SURFACES BY IDENTIFICATION

In an earlier section we discussed the function p:[0, 1] → S¹ defined by p(t) = (cos 2πt, sin 2πt). p is a continuous function defined on a compact space and onto a Hausdorff space. Whenever this is the case the topology of the image space is determined by the function and the domain space.

LEMMA 7.1  Let f:X → Y be a continuous mapping of a compact space X onto a Hausdorff space Y. Then a subset B of Y is closed if and only if f⁻¹(B) is a closed subset of X.

Proof. This lemma is a weaker form of Theorem 2.14. First suppose B is closed. Then f⁻¹(B) is closed by the continuity of f. Conversely, if f⁻¹(B) is closed, then f⁻¹(B) is compact. B = f(f⁻¹(B)), hence B is compact. Being a compact subset of a Hausdorff space, B is closed.

COROLLARY 7.2  Let f:X → Y be a continuous mapping of a compact space X onto a Hausdorff space Y. Then Y has the identification topology determined by f.

As a further corollary, let πf:X → X/~f be the identification map which carries each element x ∈ X into its equivalence set determined by the relation x ~f x′ if f(x) = f(x′). ~f is continuous so X/~f is compact. By Theorem 8.2 of Chapter 3 there is a continuous map f*:X/~f → Y such that f*πf = f. As was remarked in that section, f* is one-one; hence by Theorem 2.14, f* is a homeomorphism.

COROLLARY 7.3  The mapping f*:X/~f → Y induced by a continuous function f:X → Y of a compact space onto a Hausdorff space is a homeomorphism.

One may think of a point as being represented by “pasting” together the various points in . As an example we shall consider a cylinder. We start with a rectangle with four corner vertices A, B, B′, A′ [see Figure 34a] and identify the edge AB with the edge A′B′ in such a way that A is identified with A′ and B with B′, then we obtain a surface that is homeomorphic to the cylinder in Figure 34b. We may equally well picture the cylinder as being the topological space obtained by replacing both A and A′ by a new point A*, both B and B′ by a new point B*, and similarly any pair of corresponding points C and C′ on the respective edges AB and A′B′ is replaced by a new point C* as indicated in Figure 34c.

Furthermore, a neighborhood of this new point C* would contain the interior of the small semi-circles drawn in Figure 34c. It is interesting to note that if in this figure we join C* to itself by the path represented by the horizontal line, the space consisting of the points of this line would be homeomorphic to a circle [such as the one drawn about the middle of the cylinder in Figure 34b], for it consists of an interval whose end points have been identified. This is a special case of the following general result.

Figure 34

LEMMA 7.4  Let X and Y be topological spaces, let f:X → Y be a continuous function that is onto, and let Y have the identification topology induced by f. If B ⊂ Y is such that A = f⁻¹(B) is closed, then the subspace B of Y has the identification topology induced by the restriction f | A:A → B.

Proof. We must show that a subset F of B is closed in B if and only if (f | A)⁻¹(F) is closed in A. The restriction f | A of the continuous function f to A = f⁻¹(B) is continuous, so that if F is closed in B, then (f | A)⁻¹(F) is closed in A. Conversely, suppose that (f | A)⁻¹(F) is closed in A. Then, since A is closed in X, (f | A)⁻¹(F) is closed in X. If we prove that (f | A)⁻¹(F) = f⁻¹(F), it will follow that F is closed in Y and consequently in B, for Y has the identification topology and therefore f⁻¹(F) closed in X implies F closed in Y. It remains to prove (f | A)⁻¹(F) = f⁻¹(F). Suppose that . To show that we must show that x ∈ X and . But if x ∈ f⁻⁻¹(F), then , whence . Thus x is in the domain of f | A and , hence implies that . Conversely, if , then . Now (f | A)(x) = f(x), thus fx ∈ F and x ∈ f⁻¹(F). It follows that (f | A)⁻¹(F) = f⁻¹(F), and the proof is complete.

Another surface that may be obtained by identifying some of the boundary points of a rectangle is a surface called the Möbius strip or band. Starting again with the rectangle whose vertices we shall now label in the order A, B, A′, B′ [see Figure 35a], we identify the edge AB with the edge B′A′ by first giving the rectangular strip a 180 degree twist, so that the vertices A and A′ coincide and the vertices B and B′ coincide [Figure 35b]. A topologically equivalent space is indicated in Figure 35c, where corresponding or identified pairs of points such as A, A′ have been replaced by a single new point A*. The fact that we intend to identify the two edges AB and A′B′ of Figure 35a with a twist is often indicated by labelling the edges with the same letter, such as “a,” and then placing arrowheads on these edges in such a position that the resulting identification matches up or superimposes the two arrowheads. The Möbius strip has many curious properties. The oblique line in Figure 35c joining C* to itself is homeomorphic to a circle. The upper horizontal line running from B* through D to A* is homeomorphic to an interval. However, if on the Möbius strip we trace out the curve from B* through D to A* and continue on [along the lower horizontal line of Figure 35c] through E back to B* we trace out an interval with its end-points identified, that is, a circle. Thus the Möbius strip is a surface whose bounding curve is a circle. Other interesting properties may be deduced from the representation in Figure 35c. For example, if the Möbius strip is cut down its center, the resulting surface will not be disconnected for we may still connect a point of the upper half rectangle in Figure 35c to a point of the lower half rectangle by joining both of them to the bounding curve B*DA*EB*.

Figure 35

If an arrowhead is placed on a circle we say that the circle is oriented. The sense of rotation indicated by the arrowhead is then called the positive orientation and the opposite sense of rotation is called the negative orientation. An oriented circle in the plane can be moved about in the plane in an arbitrary manner but will always be oriented in the same sense when it returns to its original position. For this reason the plane is said to be orientable. On the Möbius strip an oriented circle can be moved around the strip, say along the oblique line in Figure 35c with its center initially at C*, and when it returns to its original position the orientation will have been reversed. Surfaces with this property are called non-orientable.

So far we have considered only surfaces resulting from the identification of a pair of edges of a rectangle. If we identify the edges of a rectangle according to the scheme indicated in Figure 36a, the resulting topological space is called a torus.

A torus is topologically the surface of a donut or a rubber tire, as indicated in Figure 36b. We may view the torus as being obtained in two steps. First, we identify the two opposite edges labelled a of the rectangle to obtain a cylinder, and second, we identify the two resulting circular edges (labelled b) of the cylinder to obtain the torus. The justification for breaking the identification up into two steps is contained in the following proposition.

Figure 36

PROPOSITION 7.5  Let X, Y, Z be topological spaces, let f:X → Y and g:Y → Z be continuous and onto. If Y has the identification topology induced by f:X → Y and Z has the identification topology induced by g:Y → Z, then Z has the identification topology induced by gf:X → Z.

Proof. Clearly, if F is a closed subset of Z, then (gf)⁻¹(F) is a closed subset of X, for gf is continuous. Conversely, suppose (gf)⁻¹(F) = f⁻¹(g⁻¹(F)) is a closed subset of X. Since Y has the identification topology induced by f:X → Y, g⁻¹(F) is a closed subset of Y. Similarly, since Z has the identification topology induced by g:Y → Z, g⁻¹(F) closed in Y implies that F is closed in Z. Thus F is closed if and only if (gf)⁻¹(F) is closed; that is, Z has the identification topology induced by gf:X → Z.

Topologically, the torus is the product of two circles. An arbitrary point w of the torus may be written as w = (C, D), where C is a point of the circle b and D a point of the circle a in either Figure 36a or 36b. Furthermore, it is clear that the product of a neighborhood of C and a neighborhood of D is a neighborhood of w and conversely that a neighborhood of w contains the product of a neighborhood of C and a neighborhood of D. Thus the topology of the torus is the topology of the product of two circles. One would have anticipated this result if one viewed the torus as being generated by revolving a circle such as b in a circular path by moving it in such a way as to always have the point labelled A in contact with the circle labelled a.

There are two other surfaces resulting from the identification of opposite pairs of edges of a rectangle. One of these surfaces is called a Klein bottle. The Klein bottle may be obtained by first identifying the edges labelled a in Figure 37a in the prescribed manner to obtain a cylinder, and then identifying the two circles labelled b in either Figure 37a or 37b, not, however, in the manner of Figure 36 to obtain a torus, but with a “twist.” Unfortunately, at least from the point of view of our visualization of the Klein bottle, there is no way to identify these two circular edges of the cylinder of Figure 37b without forcing the surface of the Klein bottle to intersect or pass through itself. For this reason, it is helpful to construct the Klein bottle in several pieces.

Figure 37

In Figure 38 we have three rectangles. If the rectangles R₁ and R₂ are joined along the edge labelled c and the rectangles R₂ and R₃ are joined along the edge labelled d, we obtain the rectangle and identifications of Figure 37, so that Figure 38 also represents the Klein bottle. If, in these three rectangles, we first identify the pairs of edges labelled e, f, and g respectively, we obtain three cylinders that are homeomorphic to the three corresponding cylindrical surfaces of Figure 39, also labelled R₁, R₂, R₃. To construct the Klein bottle we need only identify these three cylinders along the pairs of circular edges labelled a, c, and d, respectively. We may join the cylinders R₁ and R₃ along the circles labelled a, so that R₃ lies inside R₁. If we then join R₁ and R₂ along the circles labelled c, we obtain the cylinder pictured in Figure 40. To complete the construction, we must identify the two circles labelled d (Figure 40) in the prescribed manner. Any attempt to literally carry out this identification will be frustrated by our inability to pass through the surface of the cylinder. We must therefore either be content, as in Figure 40, to indicate this identification, or adopt the fiction that in Figure 41 the Klein bottle does not intersect itself along the circle d, but that each point along d is to represent at the same time two points of the Klein bottle.

Figure 38

Figure 39

Figure 40

Figure 41

Figure 42

The last surface we shall consider in detail is obtained by identifying both of the pairs of opposite edges of a rectangle with a “twist.” These identifications are indicated in Figure 42. Note that in this figure all the vertices are not identified with one another, but only diagonally opposite vertices are joined together. In order to relate this surface to some of the preceding surfaces, we shall adopt the same method as the one used in the examination of the Klein bottle, [one might call this the “cut-and-paste method”]. We first separate the large rectangle into three smaller rectangles R₁, R₂, R₃, which when re-identified along the pairs of edges labelled c, d, will again give us the rectangle and the identifications of Figure 42. This operation is indicated in Figure 43. If we first join the two edges labelled f in rectangle R₂ we obtain a Möbius strip. Since we are only interested in the topological nature of this surface, we may distort [by homeomorphisms] the two rectangles R₁ and R₃ into the semicircular regions of Figure 44. If we then join the regions R₁ and R₃ along their common edge FABCDE we obtain the disc and the Möbius strip of Figure 45, with the indicated identifications. The surface we have been considering is therefore a Möbius strip whose boundary circle FGHEJKF is to be attached to the boundary circle FGHEJKF of a disc. This last surface is easily seen to be homeomorphic to one of the models of the “real projective plane,” namely, a disc with antipodal points identified.

Figure 43

Figure 44

Figure 45

An analytic model of the real projective plane is obtained in the following manner. Let A = R³ − {(0, 0, 0)} be the set of all ordered triples (x₁, x₂, x₃) of real numbers such that not all of x₁, x₂, x₃ are zero. Define an equivalence relation on A by (x₁, x₂, x₃) ~ (y₁, y₂, y₃), if there is a real number r ≠ 0 such that rx₁ = y₁, rx₂ = y₂, rx₃ = y₃. The collection of equivalence sets P is the real projective plane. A point p ∈ P is the collection of all points on a given straight line through the origin of R³ other than the origin itself. The intersection of this equivalence set p with the unit sphere S² in R³ is a pair of antipodal points. If we confine ourselves to the hemisphere of S² lying above the plane x₃ = 0, each equivalence set p meets the hemisphere in either a single point in the interior of the hemisphere or in a pair of antipodal points on the equator or boundary of the hemisphere. This upper hemisphere is a disc (view it from the point at the north pole so that it may be projected onto the equatorial plane). Identifying antipodal points on the boundary yields an identification space which is equivalent to the real projective plane.

The sphere, torus, Klein bottle, and projective plane are examples of a larger class of surfaces that may be obtained by identifying pairs of edges of a polygon with 2n sides. Such surfaces are called closed 2-manifolds. For example, in Figure 46 we have indicated a surface that can be obtained by identifying pairs of sides of an octagon. With each such figure we may associate a “surface symbol.” We do so by starting at any vertex, such as A in Figure 46, and writing down the labels of the edges in clockwise order if the arrow along that edge is also pointing in the clockwise direction or the label with an inverse sign above if the arrow points in the counterclockwise direction. Thus a surface symbol for the surface of Figure 46 would be abbc⁻¹a⁻¹cdd. Referring back to Figure 36, one can see that a surface symbol for the torus is ab⁻¹a⁻¹b.

Figure 46

By the “cut-and-paste” method one can show that each 2-manifold is homeomorphic to a 2-manifold whose surface symbol is of one of the following four forms: abb⁻¹a⁻¹; The first form indicates that the surface is homeomorphic to a sphere. The second form includes the surface symbol of a torus and in general indicates that the surface is homeomorphic to a sphere with p handles. These two classes of surface are orientable. They can all be constructed in three-dimensional Euclidean space. The third form indicates that the surface is homeomorphic to the projective plane. We have seen that the projective plane is a disc to whose circular boundary has been attached a Möbius strip. One may think of the disc as constituting the portion of the surface of a sphere obtained by removing a circular region. Attaching a Möbius strip to the circular boundary of this region is called attaching “a crosscap.” Thus the projective plane is called “a sphere with crosscap.” In the same manner, the fourth form consists of all surfaces obtained by attaching q Möbius strips or crosscaps to a sphere with q circular regions removed.

EXERCISES

1.  Prove that the triangle T with two edges identified as in Figure 47 is homeomorphic to a disc.

2.  Prove that the triangle S with two edges identified as in Figure 48 is a Möbius strip.

Figure 47

Figure 48

3.  Prove that the Klein bottle is homeomorphic to a surface with surface symbol a₁a₁a₂a₂ by cutting the rectangle of Figure 49 along the diagonal c and pasting the resulting triangles along their common edge b.

Figure 49

4.  Show that if the Klein bottle of Figure 50 is cut along the curves c and d the result is two Möbius strips and that therefore the Klein bottle is two Möbius strips joined along their circular boundaries.

Figure 50

5.  Cut the Klein bottle of Figure 51 along the curves c₁, c₂, c₃, c₄ and d₁, d₂. Show that the regions labelled S₁, S₂, S₃, S₄ are pasted together to form a surface homeomorphic to a cylinder and therefore homeomorphic to a sphere with two circular regions removed whose boundaries are the circles d₁d₂ and c₁c₂c₃c₄ respectively. Show that the region labelled M₁ is a Möbius strip whose boundary is d₁d₂ and that the regions labelled N₁ and N₂ form a second Möbius strip whose boundary is c₁c₂c₃c₄.

Figure 51

6.  Prove that the following three statements about a closed 2-manifold are equivalent: (i) M contains a Möbius strip; (ii) M is non-orientable; (iii) the surface symbol of M contains two occurrences of some symbol “a.”

For further reading in general topology we would include Kelley, Dugundji, or Simmons, while Blackett, Elementary Topology, Fréchet and Fan, Combinatorial Topology, Wallace, Introduction to Algebraic Topology, and Chinn and Steenrod, First Concepts of Topology are recommended for an introduction to topics in algebraic topology.

Bibliography

This list has been restricted to English texts and a few papers which would be appropriate to a student who has mastered the material in this book.

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Index

Accumulation point, 172

Annulus, 139

Antipodal point, 127

Archimedean principle, 87

Automorphism, 155

Banach space, 179

Basic neighborhood space, 81

Basis for neighborhood system, 45, 80, 99

Basis for open sets, 57, 98

Bolzano-Weierstrass property, 175–176, 183–184

Borsuk-Ulam theorem, 128

Bound

greatest lower, 49

least upper, 49

lower, 49

upper, 49

Boundary, 85

Brouwer fixed-point theorem, 125

Cantor set, 101

Cartesian product, 9

Category, 107

Cauchy sequence, 178

Cauchy’s inequality, 66

Characteristic function, 15

Closure, 82

Compact, 159

locally, 171

Compactum, 172

Complement, 5

Completely separable, 184

Completeness postulate, 49

Complete system of neighborhoods, 41, 76

Component, 131

of a point, 130

Cone, 106

Connected, 113, 121

locally, 132

path-, 134

Contractive, 186

Countable, 180

Covering, 158

finite, 158

open, 159

Crosscap, 198

Cylinder, 187

DeMorgan’s laws, 5, 8

Dense, 178

Denumerable, 180

Diagram, 19

commutative, 20

Direct product, 10

Disc, 126

Disconnected, 113

Distance

between a point and a set, 50

Euclidean, 33

function, 30

Equivalence class, 16

Euclidean n-space, 11, 72

Finite intersection property, 164

First axiom of countability, 185

Fixed point, 123–127

Function

bijective, 13

bounded, 34

composition of, 17, 18

constant, 14

continuous, 36, 37, 88

continuous at a point, 36, 88

domain of a, 13

extension of a, 23

identity, 14

inclusion, 24

injective, 13

inverse, 21, 22

invertible, 21

one-one, 13, 22

onto, 13, 22

range of a, 13

restriction of a, 23

successor, 1

surjective, 13

Functor, 107

Graph, 12

Group, 108

Hausdorff

space, 76

axiom, 76

Heine-Borel theorem, 167

Hilbert space, 66

Homeomorphism, 90

Homomorphism, 108

Homotopic

paths, 141

functions, 150

Homotopy, 141

class of paths, 145

Identification, 101

Image, 13

Inclusion mapping, 24

Indexed family, 7

product of an, 26

Interior, 84

Intermediate value theorem, 122

Intersection, 4, 7

Interval, 119, 121

closed, 3

open, 3

Inverse image, 13

Isolated point, 55

Isometric, 60

Isomorphic, 150, 155

Isomorphism, 155

Klein bottle, 192

Lebesgue number, 175

limit

of a sequence, 47

point, 54

Lindelöf’s theorem, 184

Loop, 135

Mapping, 13

open, 89

(see also Function)

Metrically equivalent, 60

Metric space, 30

complete, 178

Möbius strip, 189

Natural numbers, 1–2

Neighborhood, 41, 73

space, 77

relative, 93

Non-orientable, 190

n-sphere, 59, 127

n-tuple, 11

One-point compactification, 172

Open ball, 40

Ordered pair, 9

Orientable, 190

Oriented, 190

Partition, 131

Path, 134

closed, 135

component, 138

constant, 145

initial point of a, 134

product of, 145

terminal point of a, 134

Path-connected, 134

Peano’s axioms, 1

Positive integers, 1

Product of sets, 26

Product of spaces, 98

Projection, 16, 26

first, 15

second, 15

Projective plane, 195–197

Quotient, 16

Rational density theorem, 87

Real line, 72

Refinement, 185

Relation, 15

equivalence, 16

reflexive, 16

symmetric, 16

transitive, 16

Relatively closed, 95

Relatively open, 92

Relative neighborhood, 93

Relative topology, 93

Retract, 150

deformation, 150

Schwarz’s lemma, 66

Second axiom of countability, 184

Separable, 185

Set, 2

bounded, 49, 164

closed, 54, 74

empty, 3

null, 3

open, 52, 71, 77

Simply connected, 151

Subcovering, 158

Subset, 3

improper, 3

proper, 3

Subspace, 58, 92

Surface symbol, 198

Suspension, 106

Topologically equivalent, 62

Topological property, 115

Topological space, 71

associated with a metric space, 71

Topology, 71

discrete, 72

identification, 102

Torus, 106, 190

Transformation, 13

Uncountable, 181

Uniformly continuous, 177

Unit n-cube, 59, 171

Union

of two sets, 4, 7

of an indexed family of sets, 7

Venn diagrams, 5

Weaker, 95

Table of Contents

Title Page

Copyright Page

Contents

Preface

1 Theory of Sets

1 Introduction

2 Sets and subsets

3 Set operations: union, intersection, and complement

4 Indexed families of sets

5 Products of sets

6 Functions

7 Relations

8 Composition of functions and diagrams

9 Inverse functions, extensions, and restrictions

10 Arbitrary products

2 Metric Spaces

1 Introduction

2 Metric spaces

3 Continuity

4 Open balls and neighborhoods

5 Limits

6 Open sets and closed sets

7 Subspaces and equivalence of metric spaces

8 An infinite dimensional Euclidean space

3 Topological Spaces

1 Introduction

2 Topological spaces

3 Neighborhoods and neighborhood spaces

4 Closure, interior, boundary

5 Functions, continuity, homeomorphism

6 Subspaces

7 Products

8 Identification topologies

9 Categories and functors

4 Connectedness

1 Introduction

2 Connectedness

3 Connectedness on the real line

4 Some applications of connectedness

5 Components and local connectedness

6 Path-connected topological spaces

7 Homotopic paths and the fundamental group

8 Simple connectedness

5 Compactness

1 Introduction

2 Compact topological spaces

3 Compact subsets of the real line

4 Products of compact spaces

5 Compact metric spaces

6 Compactness and the Bolzano-Weierstrass property

7 Surfaces by identification

Bibliography

Index