JaBuka Topology

8/12/2019 JaBuka Topology

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Topology

Stanislav Jabuka

Notes for Math 440/640University of Nevada, Reno


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Contents

Chapter 1. Continuity and convergence in Euclidean spaces 51.1. Continuity in Euclidean space 51.2. Open and closed subsets of Euclidean space 81.3. Continuity and convergence in terms of open and closed sets 91.4. Some properties of open and closed sets in Euclidean space 121.5. Exercises 13

Chapter 2. Topological spaces 152.1. Definition of a topological space 152.2. Examples of topological spaces 162.3. Properties of open and closed sets 252.4. Bases and subbases of a topology 282.5. Exercises 33

Chapter 3. Continuous functions and convergent sequences 373.1. Continuous functions 373.2. Convergent sequences 45

3.3. Uniform convergence of functions 503.4. Space filling curves 513.5. Exercises 56

Chapter 4. Separation axioms 614.1. Degrees of separation 614.2. Examples 634.3. Topological invariants 654.4. A first application of topological invariants 674.5. The Urysohn Lemma 704.6. Exercises 76

Chapter 5. Product spaces 795.1. Finite products 795.2. Infinite products 855.3. Exercises 88

Chapter 6. Compactness 916.1. Definition and first examples 91

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4 CONTENTS

6.2. Compactness for metric spaces 946.3. Properties of compact spaces 986.4. The one point compactification 101

6.5. Flavors of compactness 1046.6. Exercises 105

Chapter 7. Connectedness and Path-Connectedness 1097.1. Connected topological spaces 1097.2. Path-connectedness 1147.3. Local connectivity 1197.4. Exercises 121

Chapter 8. Quotient spaces 1238.1. Quotient spaces: Definitions, properties and first examples 1238.2. Surfaces as quotient spaces of polygons 131

8.3. Lens spaces 1348.4. Seifert fibered spaces 1368.5. Real and complex projective spaces 1368.6. Cylinders, cones and suspensions 1368.7. Exercises 136

Chapter 9. Foundational theorems 1399.1. The Tietze extension theorem 1399.2. Metrization theorems 1429.3. The Borsuk-Ulam theorem 1429.4. The Tychonoff compactness theorem 142

9.5. Exercises 142Chapter 10. Spaces of functions 143


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CHAPTER 1

Continuity and convergence in Euclidean spaces

his chapter is motivational in nature and provides impetus for the definitionsof a topological space X and a continuous function f : X Y betweentwo topological spaces X and Y , given in Chapters 2 and 3 respectively.These definitions are straightforward but upon first encounter seem ad hoc

and arbitrary. To convey a sense of where they are coming from, in Section 1.1 wefirst examine the familiar setting of continuity of functions f : Rn

Rm on Euclidean

spaces in terms of its customary definition in the , language from analysis. In Section1.2we introduce two families of special subsets ofRn, theopenand closedsubsets. Thediscussion of this chapter culminates in Section1.3where we recast continuity in termsof open and closed subsets ofRn. The main results of this chapter are Theorems1.3.1and1.3.2. The remaining Section1.4scrutinizes some properties of open and closedsubsets ofRn.

1.1. Continuity in Euclidean space

Let R denote the set of real numbers. We define intervalsin R as any of the subsetsofR of the form

[a, b] ={x R | axb},[a, b={x R | ax < b},a, b] ={x R | a < xb},a, b={x R | a < x < b}.

The first of these are referred to as segmentsor closed intervals, the last in turn arecalledopen intervals.

The n-dimensional Euclidean spaceRn is defined as the set of ordered n-tuples ofreal numbers:

Rn ={(x1,...,xn) | xi R },where, as usual, the symbol R denotes the set of real numbers. We will adopt theconvention of denoting the n-tuple (x1,...,xn) simply by x, the n-tuple (y1,...,yn) byy, etc. Two elementsx = (x1,...,xn) andy = (y1,...,yn) from R

n can be added to eachother and each can be multiplied by a real number R in the familiar ways:

x+y = (x1+y1,...,xn+yn) and x= ( x1,..., xn),endowing Rn with the structure of a realn-dimensional vector space.

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6 1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES

Definition1.1.1. For a real number p1 and for p =, we define the functionsdp : R

n Rn [0, as

(1.1) dp(x, y) = (|x1 y1|p +...+

|xn

yn

|p)

1p ; ifp

=

max {|x1 y1|, ..., |xn yn|} ; ifp =.

Whenp = 2 we call the resulting function d2 theEuclidean distance function. We willrefer to the functions dp as metrics onR

n, something we will justify in Exercise 2.5.6.

For example

d1(x, y) =|x1y1|+ +|xnyn| and d 34

(x, y) =|x1 y1| 34 + + |xn yn| 34

43

,

whiled2 takes on the familiar form

d2(x, y) = (x1 y1)2 +

+ (xn

yn)2.

Definition 1.1.2. Let r > 0 be a real number and let x Rn be a point inEuclidean space. We define theopen (Euclidean) ballBx(r) of radiusr and with centerx as the subset ofRn given by

Bx(r) ={y Rn | d2(x, y)< r}.Note that this definition of Bx(r) used the Euclidean distance function d2. We

could equally well have used any of the distance functions dp (including the case ofp=) in this definition. Thus, well let Bpx(r) denote the open ball with center x andradiusr but taken with respect to the metric dp:

Bpx(r) ={y Rn | dp(x, y)< r}.

Figure1 illustrates the shapes ofBp

(0,0)(1) R2

for the choices ofp= 1, 2, .

B1(0,0)(1) B2(0,0)(1) =B(0,0)(1) B

(0,0)(1)

Figure 1. Examples of the open balls Bp(0,0)(1) R2 for p = 1, 2, .The dotted lines indicate that the boundaries of these shapes are notpart of the balls.


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1.1. CONTINUITY IN EUCLIDEAN SPACE 7

Definition 1.1.3. A function f : Rn Rm is continuous at the pointx Rn iffor every >0 there exists a >0 such that d2(x, y)< implies d2(f(x), f(y)) < .Said differently, f is continuous at xRn if for every >0 there exists a >0 suchthat

f(Bx())Bf(x)().The function f is said to be continuousif it is continuous at every point xRn. SeeFigure2 for an illustration.

x

f(x)

f

Rn Rm

Bx()

Bf(x)()

Figure 2. The function f is continuous at x Rn if for every > 0there is a >0 such that the ball of radius centered at x (shaded diskon the left) maps into the ball of radius >0 with center f(x) (shadeddisk on the right). The amoeba-like shape inside of Bf(x)() indicatesthe image ofBx() under f.

Continuity of functions is closely tied to convergence of sequences, a connection webriefly review next. Recall that a sequencex inRn is merely a function x : N R.We denoted x(k) by xk and we write{xk}k=1 or simply{xk}k to denote the entiresequence (rather than writing x, the name of the function, so as to not confuse it withour notation x for points in Rn). To indicate that the sequence{xk}k belongs to Rn,we write{xk}k Rn.

Definition1.1.4. Let{xk}k be a sequence in Rn .(a) We shall say that{xk}k converges to x Rn if for every >0 there exists a

positive integer k0 such that for all k k0 the relation xk Bx() holds. Inthis case we write limk xk =x or simply lim xk = x.

(b) The sequence{xk}k is called a Cauchy sequence if for every >0 there is aninteger k0 such that whenever k, mk0 thend2(xk, xm)< .


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The following two theorems are standard results discussed in any introductoryanalysis course, their proofs are omitted.

Theorem 1.1.5. Letf : Rn

Rm be a function andx

Rn be any point. Thenf

is continuous atx if and only if for every sequence{xk}k Rn that converges to x inRn, the sequence{yk}k withyk =f(xk), converges toy = f(x) inRm. Said differently,f is continuous atx if and only if it commutes with the limit symbol lim when appliedto any sequence{xk}k with limitx:

f

limk

xk

= limk

f(xk).

Theorem 1.1.6. A sequence{xk}k Rn is convergent if and only if it is Cauchy.1.2. Open and closed subsets of Euclidean space

Given a setXand a subsetA

X, we define thecomplement ofA inXas the set

X A={xX| x /A}.Definition1.2.1. A subsetA Rn is calledclosedif for every convergent sequence

{ai}iA the limit limi ai also lies in A. A subset B Rn is calledopenifRn Bis closed.

Example 1.2.2. When n = 1, all closed intervals [a, b] are closed sets while openintervalsa, b are open sets in R. The sets R and are both closed and open whilethe intervals [a, b are neither closed nor open. Any finite subset ofR is closed.

Example 1.2.3. When n = 2 examples of closed sets are the closed rectangle[a, b]

[c, d], the closed circles

{(x, y)

R2

|x2 + y2

r

}, the upper half-plane

{(x, y)

R2 | y0}, the graph f={(x, f(x)) | x R}of a continous functionf : R R. Themain examples of open sets are the open balls Bx(r).

The following proposition gives a characterization of open sets independent of thedefinition of closed sets.

Proposition1.2.4. A subsetU Rn is open if and only if for everyxU thereexists a >0 such thatBx()U.

Proof. = Let U be an open subset of Rn and let x U be an arbitraryelement. Suppose there were no > 0 for which the inclusion Bx() U were true.Then, for any positive integer i, the ballBx(

1i) would have to intersect the complement

V = Rn

U ofU. Pick an arbitrary element biV Bx(1i ), one for each i N, thus

creating a sequence{bi}iV. The property d2(x, bi)< 1i implies that bi converges tox. However, sinceU is open, Vmust be closed and so x = lim bi must lie in V. Butclearlyx /V, creating a contradiction. Therefore, an with the property Bx()Umust exist.

= To prove thatUis open we need to prove that V = RnUis closed. Supposedthat this fails. Then there must exists a convergent sequence{bi}i V whose limit


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1.3. CONTINUITY AND CONVERGENCE IN TERMS OF OPEN AND CLOSED SETS 9

x= lim bi lies inU. Pick an >0 so thatBx()U. Since none of thebi lie inU, wesee that d2(bi, x). This however is impossible for a convergent sequence, creatinga contradiction. We are then forced to conclude that V is a closed subset ofRn and

hence that U is open. Example 1.2.5. The open balls Bx(r) are indeed open set. This follows easily

from the preceding proposition. For ifyBx(r), let = min{d2(x, y), r d2(x, y)}. Ify=x then is positive and By()Bx(r). Ify = x we may take = r and the sameconclusion holds.

1.3. Continuity and convergence in terms of open and closed sets

The purpose of this section is to prove that the continuity of a function f : Rn Rmeither at a point or globally, can be expressed entirely in terms of open or closed sets.

Recall that the preimagef1(V) of a function f :X

Ybetween two setsXand

Y, and of a subset VY, is defined asf1(V) ={xX| f(x)V}.

In particular, f1(V) is a subset ofX. We remark that the notationf1 used abovedoes not suggest that f has an inverse function. For example, if f : R R is thefunctionf(x) =x2, thenf1([0, 1]) = [1, 1].

Theorem 1.3.1. Letf : Rn Rm be a function.(a) The functionfis continous atx Rn if and only if for every open setV Rm

containingf(x), there exists an open setU Rn containingx with the propertythatf(U)V.

(b) The function f is continuous if and only if for every open set V Rm

thepreimageU=f1(V) ofV underf, is an open subset ofRn.

Proof. (a) = Assume that f is continuous at x Rn and let V Rm beany open set containing f(x). Since V is open, by Proposition1.2.4, there exists an >0 such that Bf(x)() is contained in V. But continuity off at x then implies theexistence of a > 0 such that f(Bx()) Bf(x)(). Since Bx() is an open set (seeexample1.2.5) and sinceBf(x)() is contained in V, by taking U=Bx(), we see thatf(U)V , as claimed. See Figure3for an illustration.

= Assume that fhas the property that for every open subset V Rm thatcontains f(x) there is an open subset U Rn containing x such that f(U) V.Wed like to show that then f is continuous at x. Pick an arbitrary > 0 and let

V = Bf(x)(). Note that V is an open set containing f(x) and so we are guaranteedthe existence of an open set Ucontainingx such that f(U)Bf(x)(). But since U isopen, there must exist a >0 such that Bx()U (by Proposition1.2.4). Thus wefound a such that f(Bx())Bf(x)() and so f is continuous at x, see Figure4.

(b) = Suppose f is continuous and let V Rm be any open set. Wed liketo show that U = f1(V) is also an open set. Let x U be any point. Since f is


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x

f

Rn Rm

Bx()

Bf(x)()

V

Figure 3. This picture illustrates the portion (a) = of the proof ofTheorem 1.3.1. V is an arbitrary open set containing f(x) and > 0is chosen so that Bf(x)() (shaded disk on the right) is contained in V.By continuity of f at x, there is a > 0 so that Bx() (shaded diskson the left) maps into B(f(x)() under f. The image ofBx() under f isindicated by the smaller amoeba-like shape inside ofBf(x)().

continuous at x, part (a) of the theorem shows that there exists an open set Ux withf(Ux)V . In particular, UxU. But since Ux is open and xUx, there must exista > 0 so that Bx() Ux and so Bx() U. According to proposition1.2.4, thisshows that U is open.

= Suppose thatfhas the property thatf1(V) is an open set whenever V is anopen set. Wed then like to show thatfmust be continuous. By definition, this meansthat we must show that f is continuous at each point x Rn. Using part (a) of thetheorem, demonstrating the continuity off atx is equivalent to showing that for eachopenV Rm containingf(x), there exists an open set U Rn containingx and suchthat f(U) V. But our working assumption allows us to simply take U = f1(V),thus completing the proof of the theorem.

The importance of the preceding theorem is that it provides a definition of continu-

ity that only relies on the notion of open sets. This observation will serve as the basisfor the definition of a topological space (in Chapter2) and the generalization of conti-nuity to such settings. The next theorem testifies that one can also define continuityin terms of closed sets only.

Theorem 1.3.2. A functionf : Rn Rm is continuous if and only iff1(B) is aclosed subset ofRn wheneverB is a closed subset ofRm.


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1.3. CONTINUITY AND CONVERGENCE IN TERMS OF OPEN AND CLOSED SETS 11

x

f

Rn Rm

Bx()

V =Bf(x)()

U

Figure 4. This picture illustrates the portion (a)= of the proof ofTheorem1.3.1. Given any >0 we setV =Bf(x)() rendering it an openset containingf(x). By assumption, there is an open setU(amoeba-likeset on the left) containing x and such that f(U) V. But since U isopen, there is a >0 so thatBx() (shaded disk on the left) is containedinU. The images ofU andBx() are indicated inside ofV.

Proof. = Suppose thatf is continuous and let V Rm be a closed set. Wedlike to show that A = f1(B) is a closed subset ofRn. To see this, note that

f1(Rm B) = Rn f1(B) = Rn A.Since Rm B is open and since according to Theorem 1.3.1 f1(Rm B) then alsomust be open, the above equality of sets shows that Rn Ais open. By definition, thismeans that A is closed.

= Suppose that f1(B) is closed whenever B is closed. LetV Rm be anyopen set, let U =f1(V) and set B =Rm V (note that B is closed). Then, on onehand, A = f1(B) is closed while on the other,

f1(B) =f1(Rm V) = Rn f1(V) = Rn U.Thus Umust be open and so, according to Theorem1.3.1,fmust be continuous.

We next turn to sequences where, as with continuity, convergence can be expressedsolely in terms of open sets.

Theorem 1.3.3. A sequence{xk}k Rn converges to x Rn if and only if forevery open setU Rn containingx, there exists a natural numberk0 such that for allkk0 we obtainxkU.


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Proof. = Suppose that lim xk = x and that U is an open set containing x.Then there exists an >0 such that Bx()U. Find a natural number k0 such thatfor all kk0 the pointsxk lie inBx(). Clearly then xkUalso for all kk0.

= Suppose now that xk is a sequence in Rn

with the property that for everyopen set Ucontaining x, there is a natural number k0 such that k k0 implies thatxk U. Given any > 0, we need to demonstrate that there is a k0 so that k k0implies thatxkBx(). Of course, choosing U=Bx() finishes the proof.

1.4. Some properties of open and closed sets in Euclidean space

In the proof of Theorem 1.4.1 below, we shall use DeMorgans laws (1.2) below.To state them, let Xbe any set and let Ui Rn be a family of subsets ofX with irunning through some indexing setI. Then the following, easy to prove equalities ofsets hold:

X (iIUi) =iI(X Ui)X (iIUi) =iI(X Ui)(1.2)

Said differently, these relations imply that

The complement of the intersection is the union of the complements.The complement of the union is the intersection of the complements.

With these preliminaries in place, we are ready to turn to the main result of thissection.

Theorem 1.4.1. The following are properties of open subsets of Euclidean space:

(a) The setsRn and are both open and closed sets.(b) The union of any number of open sets is an open set.(c) The intersection of any finite number of open sets is an open set.

Proof. (a)Follows directly from the definitions of open and closed subsets ofRn.(b)LetIbe an arbitrary indexing set and for each i I, let Ui Rn be an open

set. LetU=iIUi. We need to show that Uis an open set. Let xU. ThenxUifor some i I. By Proposition1.2.4, there must exist a >0 such that Bx()Ui.But then Bx()U since Ui U. This shows that for everyxUthere is a >0such thatBx()U. According to Proposition1.2.4this means thatUis an open set.

(c) LetU1,...,Um be a finite family of open sets and let V =

mj=1Uj. To see that

V is open, pick an arbitraryxV. ThenxUifor everyj {1, 2,...,m} and so thereexist numbersj >0 with the property that Bx(j)Uj . Let = min{1,...,m}andnote that > 0. Clearly thenBx()Uj for every j {1,...,m} so that Bx()V.This shows that V is an open set.

Corollary 1.4.2. The intersection of any number of closed subsets ofRn is aclosed set. The union of any finite number of closed subsets ofRn is a closed set.


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1.5. EXERCISES 13

Proof. The corollary is a direct consequence of Theorem 1.4.1and DeMorganslaws (1.2). Namely, given a family Vi Rn of closed sets, indexed by an indexing set

I, setUi= Rn Vi. Clearly each Ui is then open and so by part (b) of Theorem1.4.1,so is U =iIUi. But thenV = R

n

U is closed, whereas by DeMorgans laws, VequalsV = Rn U= Rn (iIUi) =iI(Rn Ui) =iI(Rn (Rn Vi)) =iIVi.

The case of finite unions of closed sets follows similarly and is left as an easy exercise.

There are many examples of infinite families of open sets whose intersection is notopen, and infinite families of closed sets whose union is not closed.

Example 1.4.3. For each i N let Ui =1

i , 1i

. Then each set Ui is an open

subset ofR, but the intersectioni=1Ui={0} is not open.Example 1.4.4. For i N let Vi= 1i , 3 1i . EachVi is a closed subset ofRbut

their unioni=1Vi=0, 3is not closed.1.5. Exercises

1.5.1. Determine if the following subsets ofRare open, closed or neither:

(a) [0, 1 1, 2].(b) Q.(c)nZ[2n, 2n+ 1].(d) TheCantor setC. (The Cantor setCis obtained from [0, 1] by dividing it into 3

segments of equal length, and removing the middle open interval13

, 23. In the

remaining union of two segments [0,

1

3 ]2

3 , 1], each is divided into 3 segments ofequal length, and from each the middle open interval is again removed so as toobtain[0, 1

9] [ 2

9, 1

3] [ 2

3, 7

9] [ 7

9, 1]. The Cantor set is obtained by continuing this

process ad infinitum. Thus at each stage, one divides each remaining segmentinto 3 segments of equal length, and removes the open middle interval. SeeFigure5 for the first several iteration of this construction.)

C1= [0, 1]

C2= [0,13 ] [ 23 , 1]

C3= [0,19

] [ 29 ,

13

] [ 23 ,

79

] [ 89 , 1]

C4= [0, 127 ] [ 227 , 19 ] [ 89 , 2527 ] [ 2627 , 1]C5= [0,

181 ] [ 281 , 127 ] [ 2627 , 7981 ] [ 8081 , 1]

Figure 5. The first five iterations Cn in the construction of the Cantor set.

1.5.2. Proof the claims made in


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(a) Example1.2.2,(b) Example1.2.3.

1.5.3. Let Abe a subset ofR that is both open and closed. Show that A is either

the empty set or else A = R. Is the same claim true for subsets ofRn?

1.5.4. For anyp[1, {}once can define the p-open subsets ofRn as the setsU Rn such that for every xUthere exists an >0 with Bpx()U. Show thatthe 1-open subsets and the-open subsets ofRn are the same as the open subsets ofRn. More generally, show that the set ofp-open subsets is independent of the choiceofp[1, {}.

1.5.5. A point xRn is said to be an accumulation pointof the subset ARn ifevery open set U Rn that contains x, intersects A nontrivially. Show that ARnis a closed set if and only if it contains all of its accumulation points.


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CHAPTER 2

Topological spaces

ere we formally introduce the notion of a topological spaceXas a nonemptyset Xequipped with a topologyT. A topology onX is merely a collectionof subsets ofX, subject to axioms motivated by the results of Theorem1.4.1for the case of X = Rn. We give many examples of topological spaces in

Section2.2, examples which we rely on in later chapters to examine various proper-ties of topological spaces. In Section2.3 we introduce the important notions of the

interior, closureand boundaryof a subset of a topological space, and verify some oftheir properties. The final section of this chapter introduces basesand subbasesof atopology, and defines notions such as first and second countabilityand separabilityof atopological space.

2.1. Definition of a topological space

Definition2.1.1. AtopologyT on a set Xis a collection of subsets ofX subjectto the following three rules, called the Axioms of a topology:

1. The empty setand all ofXbelong toT.2. LetIbe an arbitrary indexing set and for each i I, pick a setUi T. Thenwe require thatiIUi also belong toT. Said differently,T must be closed

under takingarbitraryunions.3. For elements U1,...,Un T, the set U1 ... Un must also belong toT. Said

differently,Tmust be closed under taking finiteintersections.A topological space is a pair (X, T) consisting of a set Xand a topologyT on X. Asubset U ofX is called an open set ifU T and a subset V X is called closed ifX V is open. A neighborhood of a pointxX is any open set UXthat containsx.

This definition is central to the remainder of the book and so, before moving on

to consider examples, we first pause to elucidate its various aspects. The choice of thethree axioms of a topology should not be too surprising given the results from Chapter1. Specifically, they are modeled on the three properties of open subsets of Euclideanspace proved in Theorem 1.4.1. Keeping in mind that open subsets ofRn were usedto recast the definition of continuity (Theorem 1.3.1), the attentive reader will havelittle difficulty in guessing what the definition of a continuous function between twotopological spaces should be (for an answer, see Definition 3.1.1).

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16 2. TOPOLOGICAL SPACES

Given a topological space (X, T) we shall often simply write Xwhen the topologyT is understood from context, and refer to X as a topological space. On the otherhand, when several topological spaces are involved in a discussion, we may label the

topologyT byTX to indicate that it belongs with X. For example, we shall write(X, TX) or (Y, TY) to label topological spaces.The notion of open and closed subset of a topological space X shall be crucial to

all subsequent chapters. Whether or not a given subset AXis open or not, dependson the choice of a topologyT on X. As we shall see in the examples below, a setXadmits many different topologies and a subset A Xmay be open with respect tosome but not with respect to other topologies. The most common misconception aboutopen and closed sets among novices, is the notion that they form a dichotomy.

Remark 2.1.2. In a topological space X, the notions of open subsetand closedsubsetdo not form a dichotomy. That is, the failure of a subset A X to be opendoes not typically imply that A is closed. Conversely, the failure ofA to be closed does

not typically render it open. Typically, a topological space has numerous subsets thatare neither open nor closed, but can also have subsets that are both open and closed.The empty set and all ofXare examples of the latter.

Definition2.1.3. Given a set Xand two topologiesT1 andT2 on X, we say thatT1 is finer thanT2 or, equivalently, thatT2 is coarser thanT1, ifT2 T1. We shallwriteT2 T1 to denote this relation between the two topologies. As usual, we shallwriteT20 such that Bx(r)U}.


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2.2. EXAMPLES OF TOPOLOGICAL SPACES 17

Recall that Bx(r) denotes the open ball{y Rn | d2(x, y) < r} from Definition1.1.1.With this definition ofTEu, the verification of the three axioms of topology is providedcourtesy of Theorem1.4.1.

Example 2.2.2. The relative or subspace topology. Let (X, T) be a giventopological space and let A be a subset of X. Then A automatically inherits thestructure of a topological space from X by equipping it with the relative topologyorsubspace topologyTA defined as:

TA={U A | U T }We call the topological space (A, TA) asubspace ofX. Saying that Ais a subspace ofXmeans that we have given the subset A ofXthe subspace topology.

It is quite straightforward to verify thatTAsatisfies the axioms of a topology onA:1. Since, X T then A=andX A= A belong toTA.2. Let Ui

TA, i

Ibe given and set U =

i

IUi. For eachUi there exists a

Vi T such that Ui = Vi A. But thenU = V A where V =iIVi Tshowing that U TA.

3. LetU1,...,Un TA and find setsV1,...,Vn T such thatUi= Vi A. Then thesetU=U1 ... Un equalsV AwhereV =V1 ... Vn T and is thereforecontained inTA.

The subspace topology gives us immediately a myriad of examples of topologicalspaces by applying it to various subsets of (Rn, TEu) from the previous example. Forinstance, each of

then-sphere Sn ={x Rn+1 | d2(x, 0) = 1} Rn+1,the graph off : Rn Rm f ={(x, f(x)) Rn Rm | x Rn} Rn+m,

the 2-dimensional torus T2

={x R4

| x2

1+x2

2 = 1 and x2

3+x2

4= 1} R4

,becomes a topological space with the relative Euclidean topology. In the definition ofT2, the symbol x was used to denote the ordered quadruple (x1, x2, x3, x4).

Example 2.2.3. Metric spaces. A metric space is a pair (X, d) consisting of anon-empty set Xand a function d: X X[0, , referred to as the metric onX,that is subject to the next three axioms of a metric:

1. d(x, y) = 0 if and only ifx = y.2. Symmetry: d(x, y) =d(y, x) for all x, yX.3. Triangle inequality: d(x, z)d(x, y) +d(y, z) for all x, y,zX.

When the metric d is understood from context, we will call X itself a metric space. In

analogy to the case of the Euclidean metric d2 on Rn

, here too we can define what weshall again call the open ball with centerxUand radiusr >0 as

Bx(r) ={yX| d(x, y)< r}.Every metric space (X, d) comes equipped with a natural choice of topologyTd, calledthe metric topology, defined by

Td={UX| pUr >0 such that Bp(r)U}.


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The reader will have noticed that this definition agrees with the definition of the Eu-clidean topology from Example2.2.1. Indeed, the premier examples of metric spacesare (Rn, dp) withp[1, {}(withdp as given in Definition1.1.1). The verifica-tion of the axioms of a metric for dp is deferred to Exercise 2.5.6. The fact thatTd isindeed a topology on Xfollows exactly as in proof of Theorem 1.4.1for the Euclideanspace. Indeed, the proof of the latter never uses the fact that it deals with (Rn, d2)explicitly.

We shall encounter more about metric spaces in later chapters and so for nowwe limit ourselves to only one additional example. Consider the segment [a, b] Rand let X =C0([a, b],R) be the set of continuous functions f : [a, b] R. Defined: X X[0, as

d(f, g) =

ba

|f(t) g(t)| dt.Then (X, d) is a metric space and thus becomes a topological space. The second

and third axiom of a metric are readily verified (and their verification does not usecontinuity of the functions involved):

d(f, g) =

ba

|f(t) g(t)| dt= ba

|g(t) f(t)| dt= d(g, f)

d(f, h) =

ba

|f(t) h(t)| dt

=

ba

|(f(t) g(t)) + (g(t) h(t))| dt

ba

|f(t) g(t)| + |g(t) h(t)| dt

=

ba

|f(t) g(t)| dt+ ba

|g(t) h(t)| dt=d(f, g) +d(g, h)

The demonstration of axiom 1 of a metric is left for Exercise2.5.1.

Example 2.2.4. The discrete and indiscrete topologies. Every set Xalwaysadmits topologies. Namely, every set can be made into a topological space by choosingeither the indiscrete topologyTindis (also referred to as the trivial topology) or thediscrete topologyTdis defined as

Tindis ={, X},Tdis ={A | AX}.

ThusTindis only contains the empty set and all of X and is therefore the smallestpossible topology onX(according to the first axiom of a topology) while Tdisequals theentire power set ofXand is consequently the largest topology onX. In the notation of


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Definition2.1.3,any topologyT onXsatisfies the double inequalityTindis T Tdis.The axioms of a topology are trivially true forTindis andTdis.

These two extreme topologies onXdo not lead to interesting topological properties.

For example, as we shall see in Chapter 3, every function on (X, Tdis) is continuous.Fertile ground for topological exploration lies with those topologies that live betweenthese two extremes.

Example 2.2.5. Topologies on finite sets. On finite sets, topologies are bynecessity also finite and can be listed by simply listing their elements. For instance, ifX={1, 2, 3, 4, 5}, then each of the following is a topology onX(Exercise2.5.2):

(a)T1 ={, X, {1, 2}, {3, 4, 5}}(b)T2 ={, X, {1}, {2}, {1, 2}}(c)T3 ={, X, {1, 2, 3}, {2, 3, 4}, {2, 3}, {1, 2, 3, 4}}(d)T4 ={, X, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}Example 2.2.6. Included point topology. LetXbe any non-empty set and let

pXbe an arbitrary point. We define the included point topologyon X asTp={UX| U is the empty set, or pU}.

To see that (X, Tp) is a topological space, we need to verify the axioms of topology forTp from Definition2.1.1.

1. Clearly Tp be definition. Also, X Tp sincepX.2. LetUi Tp withi IwhereIis any indexing set and let U=iIUi. If each

Ui is the empty set then so is Uand is therefore contained inTp. If at least onesetUi is not empty, then pUi and thuspUshowing again that U Tp. Soin either case Umust belong to

Tp.

3. Let U1,...,Un Tp and let V =ni=1Ui. If even one ofU1,...,Un is empty thenV is empty as well and thus a member ofTp. If none ofU1,...,Unis empty thenthey each must contain p and therefore V must contain p as well. So in thiscase V is also inTp.

Example2.2.7. The excluded point topology. LetXagain be any non-emptyset and, as in the previous example, pick an arbitrary point pX. Theexcluded pointtopology Tp onX is then defined to be

Tp ={UX| U equalsX, or p(X U)}.Lets verify the axioms of a topology:

1. Xbelongs toTp

be definition andbelongs toTp

since p / .2. LetUi Tp,i Iand setU=iIUi. If at least oneUiequalsXthenU=X

and so U Tp. If none of the Ui equals X then no Ui can contain p and so pcannot be contained in Ueither. Thus, in this case too, we get U Tp.

3. Take U1,...,Un Tp and let V = U1 ... Un. If all of the setUi happen toequalXthen so doesVand is thus automatically contained inTp. Conversely,if there is at least oneUi not equal toXthen that particularUi cannot contain


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pand consequently neither can V. Thus, in this case too, V is again an elementofTp.

See Exercise2.5.3 for a generalization of the included/excluded point topology.

Definition2.2.8. LetXbe any non-empty set. ApartitionPonXis a collectionof subsets ofX such that

1. IfA, B Pare two distinct elements ofP thenA B=.2. The elements ofPcover all ofX:APA= X.

Thus a partition Pis a way of dividing all ofXinto mutually disjoint set. The partitionP={X} consisting of only Xshall be referred to as the trivial partition.

Examples of partitions abound. For instance, if we take X= R, then

P1={[a, a+ 1 | a Z},

P2=

{Q,R

Q

},

P3={x+ Z | x , + 1]},(2.1)are all examples of partitions.

Example 2.2.9. Partition topology Let X be a non-empty set and letP ={Ui | i I} be a partition on X. We then define the partition topologyTP as

TP={jJUj | J I}.Thus elements ofTPare obtained by taking unions of set fromP. To see that this isa topology we check the three axioms of a topology.

1. ChoosingJ =andJ =Irenders the setjJUj equal to the empty set andall ofX respectively.

2. LetJk, k K be a family of subsets ofIgiving rise to the sets Vk =jJkUjfromTPand let V =kKVk. Rewriting this definition ofVwe see that

V =jLUj with L=kKJk IThus, of course, V belongs toTP.

3. This case proceeds in complete analogy with the previous point. Let V1 =jJ1Uj ,...,Vn=jJnUj and set W =nk=1Vk. But then

W =jLUj with L=nk=1Jk Iwe see again that W lies inP.

For instance, let us pick the partion

P1 from (2.1) on X = R. Examples of open

sets in the associated partition topologyTP1 are intervals of the form [a, b as well as[a, and, b with a, b Z. However0, 1 is not an open set in this topology(verify this!).

Example 2.2.10. Finite complement topology. On a non-empty set X wedefine the finite complement topologyTfc as

Tfc={UX| U= or X Uis a finite set}.


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IfX is itself a finite set then the finite complement topology agrees with the discretetopology (from Example 2.2.4) but if X is infinite,Tfc andTdis are rather differenttopologies. We leave the verification of the axioms of topology forTfc for Exercise2.5.8.

Example 2.2.11. The countable complement topology. The countable com-plement topologyTcc on a set X is define as

Tcc={UX| U= or X Uis a countable set}.Note that ifX is itself countable thenTcc agrees with the discrete topologyTdis onX.However, for example on X = R, the two topologies are different. The axioms of atopology for this example are addressed in Exercise 2.5.8.

Example 2.2.12. The Fort topology. This example is obtained by combiningthe excluded point topology with the finite complement topology from above. Assume

that X is an infinite set and let pXbe an arbitrary point. We then define theForttopologyTF,p asTF,p ={UX| EitherX U is finite or p /U}.

We turn to the verification of the axioms of a topology.

1. Since p / and since X X is a finite set, we find that, X TF,p.2. Let Ui TF,p, i Ibe a family of sets in this topology and set U =iIUi.

To see that Utoo must belong toTF,p we must consider two cases separately.Firstly, suppose that none of the sets Ui contains p. In this case U doesntcontainp either and soU TF,p. In the second case, suppose that at least oneUi contains p (and therefore Umust also contain p). This particular Ui then

has to have finite complement XUi. Since XU XUi (this followsfor example from DeMorgans laws (1.2)) we see that X U is also finite andtherefore that U TF,p.

3. Left as an exercise (Exercise 2.5.9).

For instance, choosing X=R and p= 0, the closed sets of the Fort topologyTF,0are all finite subsets ofRand all subsets ofRthat contain 0.

Example 2.2.13. The order topology In this example we suppose that the setX is equipped with an ordering , a notion which we briefly review before definingthe associated order topologyT.

Recall than arelationr on Xis simply a subset ofX X. It is customary to writexry if (x, y) r. A typical choice for the name of a relation is a relation symbol,such as, , , etc. For example, if we called a relation then we would writexy to indicate that (x, y) belongs to this relation.

A (total) orderingon X is a relation onXsubject to the conditions1. is reflexive: xx for anyxX.2. is antisymmetric: Ifxy and yx then x = y.3. is transitive: Ifxy and yzthen xz.


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4. satisfies the trichotomy law: For all x, y X, either x = y or x < y ory < x.

We shall write x < y to mean that x

y but x

=y.

Given an ordering on a set X with at least two elements, let us consider thefollowing special subset ofX

Lx={yX| y < x} and Rx ={zX| x < z}.In terms of these we define the order topologyT associated to the ordering as

T={UX| Uis obtained by taking arbitrary unions offinite intersections of the sets Lx and Ry withx, yX}

Examples of elements fromT are Lx Ry which we shall denote byx, y and referto as the open intervalsof the order topology. Note that

x, y

is an empty set unless

there exists an element zXwith x < z < y.Since the set Tis closed under finite intersection and arbitrary unions (by definition

ofT), axioms (2) and (3) of a topology (definition 2.1.1) are trivially true. To seethat the empty set and X belong toT we proceed as follows. Given two distinctelementsx, y(recall that we assumed thatXhas at least two elements), we have eitherx < y or y < x (according to trichotomy axiom above). Suppose that y < x, thenx, y=showing that T. On the other hand, for these same x, yXwe obtainX=Lx Ry (Exercise2.5.11) and so X T.

As an illustration of the order topology, we consider the lexicographic ordering onRn. Let denote the standard ordering of the real numbers R and extend it to anordering on Rn by the following rule: (x1,...,xn)< (y1,...,yn) if

x1 < y1 or,x1 = y1 andx2 < y2 or,

x1= y1, x2 = y2 andx3 < y3 or,...

x1= y1, x2 = y2,...,xn2= yn2 andxn1 < yn1 or,x1 = y1, x2= y2,...,xn2 = yn2, xn1 = yn1 and xn< yn.

The thus obtained ordering is called the lexicographic ordering onRn. Whenn = 1,this topology equals the Euclidean topology on R (from Example 2.2.1). However,when n 2 the resulting order topology on Rn is quite different from its Euclideancounterpart. For example, when n = 2, consider the interval(0, 0), (1, 0) R

2

givenby

(0, 0), (1, 0)={(x, y) R2 | 0< x 0} {(1, y) R2 | y


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1

Figure 1. The shaded region in R2 represents the interval (0, 0), (1, 0)in the order topology on R2 associated to the lexicographic ordering. Thedotted lines are not part of the region while the full lines are. The regionextends infinitely vertically in both directions.

Example2.2.14. The lower and upper limit topologies on R. Thelower-limittopologyTll on R is defined as

Tll ={UX| Uis obtained by taking unions of finite intersections ofsets [a, b witha, b R}.

The definition ofTll shows that it is closed under finite intersections and arbitraryunions while the empty set and R belong toTll because, for example,= [0, 1 [3, 5and R =aZ[a, a+ 1. Thus,Tll is indeed a topology.

Theupper limit topologyTul on R is defined analogously by replacing the sets [a, bin the definition ofTll by the setsa, b].

Notice that the open intervals a, b belong both to Tll and to Tul since, for example,

a, b=

n1

a+

1

n, b

Tll.

The starting value ofn in the above union is chosen large enough so that a + 1/n < b.From this observation it is not hard to show that

TEu

Tll and

TEu

Tul. However,

the set [0, 1belong toTll but not toTEu showing thatTEu=Tll. A similar observationapplies to the upper limit topology.

Example 2.2.15. The topologists sine curve. This example and the two sub-sequent ones, do not define new types of topologies, but rather look at subspaces of(R2, TEu) (see Examples2.2.1and2.2.2) with certain special properties that shall beexplored in later chapters.


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We define thetopologists sine curveto be the subspace X R2 given byX={(x, sin(1/x)) | x 0, 1]} ({0} [0, 1]).

Thus, the topologists sine curve is the union of the graph of sin(1/x) over

0, 1] unionthe closed segment [1, 1] on the y-axis. This space is illustrated in Figure2a.

(a)

1

1

1

(b)

...

(c)

...

Figure 2. (a) The topologists sine curve. (b) The infinite broom. (c)

The Hawaiian earrings.

Example 2.2.16. The infinite broom. Let In R2 be the closed straight-linesegment joining the origin (0, 0) to the point (1, 1/n) for n N. The infinite broomisthe subspace Xof (Rn, TEu) defined by (see Figure2b)

X= (n=1In) ([0, 1] {0}).


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2.3. PROPERTIES OF OPEN AND CLOSED SETS 25

Example 2.2.17. Hawaiian earrings. For nN let CnR2 be the circle withcenter (1/n, 0) and with radius rn=

1n

. Note that all off these circles pass through theorigin (0, 0).TheHawaiian earringsis the subspace X of (R2, TEu) given as the unionof these circles (see Figure2c):

X=n=1Cn =n=1{(x, y) R2 | (x 1n)2 +y2 = ( 1n)2}.2.3. Properties of open and closed sets

This section discusses some general properties of open and closed sets of a topolog-ical space (X, T). Recall that a subset UXis called open ifU T, a subsetAXis called closed ifX A T. While typically a random subset YXis neither opennor closed, we shall see presently it can be approximatedby both kinds of sets.

Definition2.3.1. LetAXbe an arbitrary subset ofX. Define the interior A(or I nt(A)) and the closureA(or C l(A)) ofA as

A= union of all open sets contained inA,

A= intersection of all closed sets containingA.

Theboundary or frontierA ofAis defined as A= A A.Lemma 2.3.2. LetA be a subset of the topological spaceX. Then

(a) The interiorA ofA is an open set and it is the largest open set contained in

A. The equalityA= A holds if and only ifA is open.(b) The closureA ofA is a closed set and it is the smallest closed set containing

A. The equalityA= A holds if and only ifA is closed.(c) A pointx

Xbelongs to Aif and only if every neighborhood ofx intersectsA.

Proof. These claims follow readily from the definition of interior and closure of aset.

(a) Since A is obtained as a union of open sets (those contained in A) it is anopen set. IfU is any other open set containing A, then U is one of the sets from theunion which defines Aand is thus contained in A, showing that the interior ofA is thelargest open set contained in A. IfA= A then clearlyA must be open since Ais open.Conversely, ifA is open then it is clearly the largest open set containing A and thusby necessity equal to A.

(b) By definition, A is an intersection of closed set and must therefore be closed.IfB is any closed set containing A, then B occurs in the intersection of sets defining

A and hence A B. This shows that A is the smallest closed set containing A. IfA = A then A is closed, since A is. IfA is closed then A itself is the smallest closedset containing A and is thereby equal to A.

(c) Suppose firstly that x A and let U be a neighborhood of x. If we hadAU = then AUwould be a closed set containing A and would properly becontained in A, an impossibility according to part (b) of the lemma. Thus we musthave A U=.


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Conversely, suppose that x X is a point such that U A= for every neigh-borhood U of x. If we had x / A, we could take U = X A which would give animmediate contradiction. Thus we must have x A.

Definition2.3.1shows that the inclusions

AA A,hold for every subset A ofXwhile Lemma2.3.2shows that A is always an open setand A is always a closed set. It is in this sense thatA can be approximated by bothan open set (its interior) and a closed set (its closure). How good an approximation of

Ais given by the sets Aand Adepends heavily on the topology on X.

Example 2.3.3. Consider the set R and its subset A =0, 1. Find the interiorA, the closure A and the boundary A ofA with respect to the following choices oftopologies on R:

1. The Euclidean topology (Example 2.2.1). With this toplogy A itself is openand so, by Lemma2.3.2, the interior ofA equals A itself. The closure ofA isa closed set cointaining A. We guess that A = [0, 1]. Indeed, [0, 1] is closedand containsA and the only smaller subsets cotaining A are [0, 1,0, 1] andAitself, neither of which is closed. Thus

A= A =0, 1, A= [0, 1], A ={0, 1},a result which confirms our Euclidean intuition.

2. The included point topology with p = 0 (Example2.2.6). Sincep / A we seethat A is in fact closed so that A = A. On the other hand, the only open setfrom

Tp that is contained in Ais the empty set, thus A=

. Therefore,

A=, A= A =0, 1, A = A =0, 1.3. The included point topology with p = 1/2 (Example2.2.6). SincepA we see

that A= A. However, since pA, the only closed set containing A is all ofRshowing that A= R. We arrive at

A= A =0, 1, A= R, A =, 0] [1, .4. The finite complement topology (Example2.2.10). In this topology, closed set

are finite subsets ofRand all ofR. This makes is clear that A= R. No subsetofAhas finite complement showing that A=. In summary

A=

, A= R, A = R.

The next lemma provides an alternative definition of the boundary A.

Lemma 2.3.4. LetXbe a topological space and letA be a subset ofX. Then

(a) X A= XA.(b) Int(X A) =X A.(c) A = A X A.


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2.3. PROPERTIES OF OPEN AND CLOSED SETS 27

Proof. (a)By definition, the closure ofX Ais the intersection of all closed setscontainingX A:

X

A=

B

BB with

B=

{B

X

|B is closed andX

A

B

}.

Let Cbe the collection of subset ofXgotten from Bby taking complements of elementsfromB:

C ={CX| X C B}.The key observation now is that C C if and only ifC is open (since X Cmust beclosed) andCis contained in A (sinceXCcontainsXA). Therefore, by definition

A=CCC.An applicatioin of DeMorgans laws1.2 yields the desired result:

X

A=

BBB=

CC(X

C) =X

CCC=X

A.

(b)This part follows in much the same way as part (a). Namely, let now Bbe thefamily of closed set containing A andCbe the family of open sets contained in X A.As before, there is a bijective correspondenceB C by sending B to X B. Usingagain DeMorgans law finishes the proof:

Int(X A) =CCC=BB(X B) =XBBB=X A.(c)This follows easily from part (a) of the present theorem:

A = A A= A (XA) = A X A.

Definition2.3.5. Let X be a topological space. A subset A ofX is called denseinX ifA= X. The spaceX is calledseparableif there exists a countable dense subsetAofX.

As many topological spaces Xare uncountable as sets, the notion of separabilityprovides a measure of how big X is as a topological space. A dense subsetA Xhas the property that it intersects every open set ofX(Corollary2.3.7). Thus, if onecan find a countable dense subset of X (i.e. if X is separable), one should think ofX as having relatively few open sets and accordingly as being a relatively smalltopological space.

Example 2.3.6. The subset A =

0, 1

ofR is dense with respect to either theparticular point topologyTp with p = 1/2 or with respect to the finite complementtopologyTfc (Example2.3.3).

Corollary 2.3.7. A subsetA of the topological space(X, T) is dense if and onlyifA U= for allU T other thanU=.

Proof. This is a direct consequence of part (c) of Lemma 2.3.2.


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Example2.3.8. The set Q of rational number is dense in (R, TEu) since it intersectsevery open interval a, b and every open set is a union of open intervals. More generally,by the same principle, Qn is dense in (Rn, TEu). Consequently, since Qn is a countableset for each n0, (R

n

, TEu) is a separable topological space for all n0.Example 2.3.9. Consider the discrete topologyTdis on R (Example2.2.4). In this

topology every subset A R is open and closed so that A = A for every A R.Therefore the only dense subset of (R, Tdis) is R itself. We infer that (R, Tdis) is notseparable.

2.4. Bases and subbases of a topology

The reader familiar with linear algebra will recall that a basis for a real finitedimensional vector space V is a set of vectors{e1,...,en} V such that every vectorv

V can be written uniquely as a linear combination v = 1e1 +.. + nen with

i R. Thus, while V is typically an infinite set, we can capture the totality ofits vectors with the finite set{e1,...,en} by relying on the vector space operations ofvector additionand scalar multiplication.

As a topologyTon a set Xis a collection of subsets ofX, it comes equipped withtwo operations among its elements, namely those of taking unions and taking inter-sections. It is thus conceivable, in analogy with the vector space basis, that there aresubsets ofT which generateall ofT by means of taking unions and/or intersectionsof its elements. This is indeed that case and we shall consider both subsetsB Twhich generateall ofT by means of only taking unions of elements from B, andsubsetsS T which will generateTby relying on both unions and intersection. Thefirst of these cases will lead the notion of a basis for (X,

T), the closest analogy to a

vector space basis. The second will lead to the notion of a subbasis, one that is withoutanalogue in the world of vector spaces.

Definition 2.4.1. Let (X, T) be a topological space and letB andSbe subsetsofT such that

(a) Every set U T is a union of sets fromB.(b) Every set U T is a union of finite intersections of sets fromS.

ThenBis called a basis for the topologyT whileSis called a subbasis for the topologyT. We also say thatT is generated byB(by mean of taking unions of elements fromB) or thatT is generated byS (by means of taking unions of finite intersections ofelements from

S.

IfBandSare given byB={Ui T | i I} andS ={Vj T |j J } withIandJ being two indexing sets, then properties (a) and (b) from Definition2.4.1meanthat every set U Tcan we written as

(a) U=iIUUi, for some subsetIU ofI.(b) U =LU(jJVj), for some family of indicesLUand for finite families of

subsetsJ ofJ with LU.


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2.4. BASES AND SUBBASES OF A TOPOLOGY 29

The indexing setLUfrom (b) above, is of course allowed to be infinite. While we willtypically start out with a topological space (X, T) and then seek bases and subbasesforT, it is meaningful to ask about going the other way. That is, given a setX andtwo collectionsB, Sof subsets ofX, under what conditions areBandSa basis and asubbasis for some topologyT. The next two lemmas address this question.

Lemma 2.4.2. A collectionBof subsets of a setXis a basis for some topologyTif and only ifBhas the properties

(a) X=BBB.(b) For every B1, B2 B and every point x B1B2, there exists an element

B3 Bsuch thatxB3B1 B2.If these two conditions are met, the topologyT generated byBconsists of all possibleunions of elements fromB:

T =

{U

X

|U is a union of sets from

B}.

Proof. One implication of the lemma is immediate. Namely, ifBis a basis for thetopologyT then, since X T, it must be that X=BBB. Likewise, ifB1, B2 B,thenB1 B2 is an open set and therefore a union of elements fromB.

Lets turn to the other implication. We assume thatB is a collection of subsets ofXsubject to conditions (a) and (b) from the lemma and letTbe the set

T ={UX| U is a union of elements fromB}.It is then clear that X T(by condition (a)) and T(the empty set is the emptyunion of any sets). By definition ofT, it is obviously closed under unions since unions ofunions are again just unions. To see thatT is closed under finite intersection it sufficesto show that it is closed under twofold intersections. Thus, let U1, U2

T, we need to

show that U1 U2 is a union of elements fromB. Let xU1 U2 be any point. Bydefinition ofT, there must exist elements B1, B2 B with xBiUi, i= 1, 2. Butthen by property (b), there is an elementBx Bsuch thatxBxB1B2U1U2.But then clearly

U1 U2 =

xU1U2Bx,

and so U1U2 T. This shows thatT is indeed a topology on X. Moreover, thedefinition ofT shows thatBis a basis forT, as claimed.

Lemma 2.4.3. A collectionSof subsets of a setX is a subbasis for a topologyTonX, if and only ifX=

SSS. In this case, the topology

T is obtained as

T ={UX| Uis a union of finite intersections of sets fromS}.Proof. IfSis a subbasis for a topology on X, then the condition X=SSS is

clearly satisfied.Conversely, suppose that X=SSSand letT be defined as in the statement of

the lemma. Then T and X Tby the stated condition onS. The fact thatT isclosed under finite intersections and arbitrary unions, follows from the definition ofT,


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showing thatTis a topology. Likewise, the factSis a subbasis ofT also follows fromthe definition.

Lemmas 2.4.2 and 2.4.3 give us ways to define topologies on a set X by either

picking a basis or a subbasis first and letting them generate the topology. Here are afew examples.

Example 2.4.4. The sets

B1={a, b | a, b R, a < b},S1={, b, a, | a, b R},

are a basis and subbasis for (R, TEu) (recall thatTEu denotes the Euclidean topologyon R). Similarly, by relying on the density of the rational numbers Q in R, one canshow that

B2 =

{a, b

|a, b

Q, a < b

},

S2 ={, b, a, | a, b Q},are also a basis and a subbasis for (R, TEu). The key difference between the twoexamples is that the setsB2 andS2 are countable sets while both ofB1 andS1 areuncountable.

Example 2.4.5. Consider the included point topologyTp on R (Example2.2.6).The sets{p}and{x, p}, for anyx R, are open sets. However, neither{p}nor{x, p}can be obtained as a union of other nonempty open sets and must therefore be part ofevery basisB. Accordingly, every basisBfor (R, Tp) has uncountably many elements.The smallest possible basis for (R, Tp) is

B={{p}, {x, p} | x R {p}}.Example 2.4.6. Let (X, ) be an ordered set and consider the order topologyT

on X(Example2.2.13). Then a basis and a subbasis for (X, T) are given byB={a, b | a, bX,a < b} and S={La, Rb | a, bX}.

In the case ofX=R and withbeing the usual ordering of real numbers, these twosets agree withB1 andS1 from Example2.4.4.

Example 2.4.7. Consider the set R and letS be the collection of subsets ofRgiven by

S=

{x+ Q+, y+ Q

|x, y

R

},

where Q+ and Q are the sets of the positive and of the negative rational numbersrespectively. According to Lemma2.4.3,S is a subbasis for a topologyTS on R. Anexample of an open set in this topology is1, 1 Q.

Just as vector spaces are divided into finite dimensional and infinite dimensionalexamples according to whether or not they possess a finite basis or not, so too topolog-ical spaces can be group into two distinct categories. The first attempt to define the


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2.4. BASES AND SUBBASES OF A TOPOLOGY 31

analogue of a finite dimensional vector space for topological spaces, might be to de-mand the existence of finite basis for the topology. However, since a topology generatedby a finite basis is by necessity finite, this definition would exclude most interesting

examples (for instance, no Euclidean space (Rn

, TEu) withn1 admits a finite basis).Instead, we will divide topological spaces into two categories according to whether ornot they possess a countable basis or not.

Definition 2.4.8. A topological space (X, T) is called second countable if it pos-sesses a countable basisB.

We should think of a second countable topological space as being small and ofone that isnt second countable, as being large. Another measure of size for atopological space we encountered previously was that of being separable (Definition2.3.5). The next lemma explicates the relation between the two.

Lemma 2.4.9. A second countable topological space is separable. The converse isfalse in general (Example2.4.10).

Proof. LetB={Ui | iN} be a countable basis for the second countable topo-logical space (X, T). Without loss of generality we can assume thatUi= for anyi N. LetaiUi be any element and let A ={ai | i N} X. ThenA is a countableset and we claim that A= X. For ifUXis any nonempty open set then there existssomej N withUjU. But thenA Uis nonempty (as it contains aj) showing thatAis dense according to Corollary2.3.7.

Example2.4.4shows that the Euclidean space (R, TEu) is second countable whileExample2.4.5shows that (R, Tp) is not second countable.

Example 2.4.10. We just saw that the set of real numbers R with the includedpoint topologyTp isnt a second countable space. On the other hand, let A ={p} Rand notice that A= R(since closed sets in this topology are sets either not containing

p or else all ofR). Thus (R, Tp) is separable.We finish this section by considering a local version of the notion of second count-

ability.

Definition2.4.11. Let (X, T) be a topological space and let xXbe a point inX. A neighborhood basis aroundx is a collection

Bx of neighborhoods ofx such that

for every neighborhood U ofx there is an element V Bx with xV U. We saythat (X, T) isfirst countable if every point xXpossesses a countable neighborhoodbasis.

It should be clear that a second countable space is automatically first countable.The converse is false as the next example demonstrates.


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Example2.4.12. Consider the space (R, Tp) where Tpis the included point topology(example2.2.6) and letx Rbe any point. A neighborhood basisBx for x is given by

Bx = {{p}} ; ifx = p{{x, p}} ; ifx=pThus (R, Tp) is first countable but it isnt second countable according to Example2.4.5.

Example 2.4.13. The real number line R with the finite complement topologyTfc (Example2.2.10) is not first countable. For suppose thatBp={Ui | i N}were acountable neighborhood basis at some point p R, withUi = R{ai1, . . . , aini} for someai1, . . . , a

ini R {p} and some ni N. Additionally, let A =iN {ai1, . . . , aini} and

note thatA is a countable set, and hence that R Ais infinite. For any neighborhoodV = R {b1,...,bn} of p there would have to be some i N with Ui V , that iswith

{b1, . . . , bn

} {ai1, . . . , a

ini

} A. A contradiction is obtained by simply choosing

elements bj from R (A {p}), showing that phas no countable neighborhood basis.Compare to Exercise2.5.22.

Example 2.4.14. A metric space (X, d) equipped with the metric topologyTd(Example2.2.3) is always first countable. Namely, given a point xXwe can defineBx as

Bx ={Bx(r) | rQ+}.where, as before, Q+ is the set of positive rational number. ClearlyBx is a countableset and ifU is any neighborhood ofx, then there must exist some real number r >0such that Bx(r) U. Taking anyr 0, r Q gives an element Bx(r) Bx withx

Bx(r

)

Bx(r).

We saw that second countability and separability are generally two distinct mea-sures for the size of a topology (Lemma2.4.9). However, for metric spaces, these twonotions agree. The reason for this lies in the first countability.

Proposition 2.4.15. A metric space (X, d) equipped with the metric topologyTdis separable precisely when it is second countable.

Proof. LetA ={ai | i N}be a countable dense subset ofXand for eachaiAletBi ={Bai(r) | rQ+} be a countable neighborhood basis for ai. LetB=i=1Bi.SinceBis a countable union of countable sets, it is itself a countable set. To see thatBis a basis for (X, T), letUXbe an open set and letxUbe any point. There has toexist a rational numberr >0 such thatBx(r)U(be definition ofTd, Example2.2.3).Consider the open set Bx(r/2). By Corollary2.3.7the intersectionA Bx(r/2) has tobe nonempty and so without loss of generality we can suppose that a1A Bx(r/2).But then x Ba1(r/2) since a1 Bx(r/2), and additionally Ba1(r/2) Bx(r) for ify Ba1(r/2) then d(y, x)d(y, a1) +d(a1, x) < r/2 + r/2 =r. Since Ba1(r/2) B,we have shown that for every point x U there is a set Ux B with x Ux U.ThereforeU=xUUx showing thatBis a basis.


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2.5. EXERCISES 33

Remark 2.4.16. It is not true in general that a separable and first countable spaceis second countable (Exercise2.5.24).

Definition2.4.17. We say that a topological space (X,T

) ismetrizableif there isa metric onXwhose associated metric topology agrees withT.

Proposition 2.4.15and the observation from Example 2.4.14can be used to findfirst examples of non-metrizable topologies (see Section9.2 for more on metrizability).For instance, the space (R, Tp) is not metrizable as according to Example 2.4.10it isseparable but not second countable (note however that it is a first countable spaceaccording to Example 2.4.12). Similarly, (R, Tfc) is not metrizable as it is not firstcountable according to Example2.4.13.

Theorem 2.4.18. Let(X, TX) be a topological space and letYXbe a subspaceofX. IfXis either second countable or first countable, then so isY. Separability of

Xdoes not in general imply separability ofY.Proof. Suppose that X is second countable and letB={Ui | i N} be a basis

for the topology on X. ThenBY ={Ui Y| i N}is a countable basis for the relativetopology onY. To see this, letVY be an open set inYand letUbe an open set inXsuch that V =U Y . SinceBis a basis for the topology on X, then U =iMUifor some subset M N. Thus V =iM(Ui Y) proving the claim.

The case of first countability follows analogously. For a topological space showingthat separability is not necessarily inherited by a subspace, see Example 2.4.19.

Example 2.4.19. Consider the included point topologyTp on X = R. This is aseparable space with a dense subset given by

{p

}. Let Y = R

{p

} be given the

subspace topology. Since the closed sets in Xare those not containing p and X itself,it follows that the closed subsets ofY are all subsets ofY. Accordingly, A = A foranyAY. Thus the only dense subset ofY is Y itself. SinceY is not countable, itfollows that it is not separable.

2.5. Exercises

2.5.1. Verify axiom 1 for the metricd(f, g) =ba|f(t)g(t)| dtfrom Example2.2.3.

2.5.2. Verify that each ofT1,T2,T3,T4 from Example2.2.5is a topology on X ={1, 2, 3, 4, 5}.

2.5.3. This exercise generalizes Examples2.2.6and2.2.7. Let X be a topologicalspace andAXa subset ofX.

(a) DefineTA as the collection of subset ofXgiven byTA={UX| U is the empty set, or AU}.

Show thatTA is a topology on X (called the Included subset topology).


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(b) DefineTA as the collection of subset ofXgiven byTA ={UX| U equalsX, or A(X U)}.

Show thatTA

is a topology on X (called the Excluded subset topology).2.5.4. Let Xbe a set and p1, p2Xtwo distinct points. Define the collectionTp2p1

of subsets ofX as

Tp2p1 ={UX|p1U, or p2(X U)}.Show thatTp2p1 defines a topology on X (called theIncluded-excluded points topology).

2.5.5. On Rdefine the setTof subsets ofRasT ={UX| U is the empty set, or R Uis a finite union of closed intervals}.

Show thatTdefines a topology on R (called the Compact complement topology).

2.5.6. Show that the functions dp from Definition1.1.1are each a metric for anychoice ofp[1, {}.2.5.7. Two metrics d andd on a set Xare called equivalentif there exist positive

real numbers c1, c2 such that

c1 d(x, y)d(x, y)c2 d(x, y), for allx, yX.(a) Show that the notion of equivalence between metrics on a setXis an equivalence

relation.(b) Show that equivalent metrics induce the same metric topology, that is show

that ifd and d are equivalent metrics, thenTd=Td (Example2.2.3).(c) Show that for any pair p, p

[1,

{}, the two metrics dp and dp (Defi-

nition1.1.1) are equivalent metrics on Rn. Conclude that all of the metrics dp,p [1, {} induce the Euclidean topology on Rn (Hint: Prove that foranyp[1, , the double inequality

1p

n dp(x, y)d(x, y)dp(x, y)

holds for all x, y Rn, and use parts (a) and (b) of the exercise.).2.5.8. Show that the finite complement topologyTfc and the countable comple-

ment topologyTcc from Examples2.2.10and2.2.11,satisfy the axioms of a topology(Definition2.1.1).

2.5.9. Verify axiom 3 of a topology (Definition 2.1.1) for the Fort topologyTF,pfrom Example2.2.12.

2.5.10. Show that Xbelongs to the order topologyT from Example 2.2.13, byshowing that X=Lx Ry for a pair of elements x, yX with y < x.

2.5.11. For the topological space (X, T), determine the interior, closure and bound-ary of the given subset AX.


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2.5. EXERCISES 35

(a) X= Rwith the Euclidean topologyTEu and A = Q.(b) X= R with the Fort topologyTF,p (Example2.2.12) withp = 0 andA = [0, 1].(c) X= R2 with the lexicographic order topology T(Example2.2.13) andAbeing

the unit square A = [0, 1] [0, 1].(d) X= Rwith the lower limit topologyTll (Example2.2.14) and withA =0, 1.2.5.12. Are the irrational numbers dense in Requipped with the

(a) Euclidean topologyTEu?(b) Countable complement topologyTcc (Example2.2.11)?(c) Fort topologyTF,p (Example2.2.12)? Does the choice ofp Rmatter?2.5.13. Let X be a set and letT1, T2 be two topologies on X and assume that

T1 T2 (Definition2.1.3). For a subset A ofX, write Inti(A), Cli(A) and i(A) forthe interior, closure and boundary ofA with respect toTi. Show that

(a) Int1(A)

I nt2(A).(b) Cl1(A)Cl2(A).2.5.14. For a topological space Xand subsets A, B X, show the following rela-

tions: (a) Cl(A B) =Cl(A) Cl(B). (b) Int(A) Int(B)I nt(A B). Cl(A B)Cl(A) Cl(B). Int(A) Int(B) =I nt(A B). Cl(Cl(A)) = Cl(A). Int(Int(A)) =I nt(A).

Show by example that the inclusions from the second point of (a) and first pointof (b) may be proper inclusions.

2.5.15. Let Xbe a topological space and let Z Y X be subsets. LetTYXandTZXbe the relative topologies on Y andZrespectively, induced by the topologyon X. Furthermore, letTZY be the relative topology on Z induced by the topologyTYX on Y. Show thatTZY =TZX.

2.5.16. Let X be a topological space and YXa subspace. Show that a subsetAY is closed in Y if and only if there exists a closed subset BX ofXsuch thatA= B Y.

2.5.17. LetXbe a topological space and let A, BXbe two subspaces ofXwithAB.

(a) Show that ifA is open inB andB is open in X then A is also open in X.(b) Show that ifA is closed in B andB is closed in X then A is also closed inX.(c) Find an example ofA, B and X with A open in B but not in X. Similarly,

find an example with A closed inB but not in X.

2.5.18. Let Xbe a topological space and Y X a subspace. WriteClX(A) andClY(A) for the closure ofA in X and Y respectively. Similarly, write IntX(A) andIntY(A) for the interior ofA in XandY respectively. Show that for a subset A ofY,the following inclusions hold:

(a) ClY(A)ClX(A).


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(b) IntX(A)I ntY(A).Show by example that both inclusions may be proper.

2.5.19. Let Xbe a set and letT1

,T2

be two topologies on X withT1 T2

. Showthat if (X, T1) is separable, then so is (X, T2). Conclude that (R, Tll) and (R, Tul)(Example2.2.14) are separable.

2.5.20. Show that Q with the discrete topology (Example2.2.4) is second countable.Show on the other hand that Rwith the discrete topology is not second countable.

2.5.21. Let Xbe a set andP ={Ui X| i I} a partition ofX. Show that Xequipped with the partition topologyTP (Example2.2.9) is second countable if andonly if the indexing setI is countable.

2.5.22. Show that (R, Tcc) from Example2.2.11is neither first nor second countable.2.5.23. Show that (R,

Tfc) from Example2.2.10is separable.

2.5.24. Show that (R, Tll) from Example2.2.14, is separable and first countable butnot second countable.

2.5.25. Show that any open subset U Rwith respect to the lower limit topologyTll (Example2.2.14), is a countable union of intervals [a, b witha, b R, a < b.


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CHAPTER 3

Continuous functions and convergent sequences

ontinuous functions and convergent sequences are introduced, following theintuition from Euclidean spaces introduced in Chapter1(see specifically The-orems1.3.1and1.3.3). Many examples of continuous functions are presented,and some of their basic properties are illuminated in Section 3.1. Section3.2

focuses on convergent sequences in topological spaces, and heeds special attention tothe question of uniqueness of limit. The connection between continuous functions

and convergent sequences in general topological spaces holds in weakened form, andTheorem3.2.11presents the generalization of Theorem 1.1.5from the Euclidean case.Section 3.3 presents only one result - the Uniform convergence theorem(Theorem3.3.2) - which speaks to when a sequence of continues functions that converges pointwise, defines a continuous limit function. This theorem is used several times in theremainder of the text, the first instance of which occurs in Section 3.4. The latterdiscusses space filling curves, that is continuous surjections from the closed interval[0, 1] to the closed cube [0, 1]n for anyn N.

3.1. Continuous functions

Definition3.1.1. Let (X,

TX) and (Y,

TY) be topological spaces and letf :X

Y

be a function.

(a) We say that f iscontinuous atxX if for every neighborhoodV off(x) thereexists a neighborhood U ofx such that f(U)V.

(b) We say that f is continuousiff1(V) TX for everyV TY.We shall use the termmap as synonymous with continuous function.

This definition of both local continuity (part (a)) and global continuity (part (b))are directly motivated by Theorem1.3.1. In particular, we see that with the choicesof X = Rn and Y = Rm and with bothTX andTY being the Euclidean topologies,Definition 3.1.1 agrees with the usual definition of continuity on Euclidean spacesfamiliar from analysis.

Global continuity of a function between Euclidean spaces (first encountered in Def-inition1.1.3) is defined to simply mean local continuity at each point of the domain. Incontrast, Definition3.1.1gives separate meaning to both local and global continuity.Nevertheless, the relation between the two remains the same.

Theorem 3.1.2. A functionf :XYbetween two topological spaces is continu-ous if and only if it is continuous at every pointxX.

37


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38 3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES

Proof. = Suppose that f : X Y is continuous globally and let x be apoint in X. Pick an arbitrary neighborhood V off(x) in Y, then U = f1(V) is aneighborhood ofx in Xwithf(U)V.

= Suppose thatf :XYis continuous at every point xX, letVbe an opensubset ofY and set U=f1(V). Wed like to show that U is open. For that purpose,let xU be an arbitrary point (ifU is the empty set then it is automatically open)and note that V is a neighborhood off(x). By continuity off at x, there must exista neighborhood Ux ofx with f(Ux)V . This latter relation shows that UxU andconsequently we obtain U=xUUx. Being a union of open sets, U is itself open.

Example 3.1.3. Let (X, TX) = (R, Tfc) and let f : X X be the functionf(x) = x2. Let U Tfc be any nonempty open set and assume that U = R{x1, . . . , xk, y1, . . . y} with x1, . . . , xk < 0 and y1, . . . , y 0. Then f1(U) = R{y1, y1,..., y, y} which has finite complement in R. Since in addition f1() =,we see that f is continuous.

Example 3.1.4. Let (X, TX) = (R, Tfc), let (Y, TY) = (R, Tp) with p = 0 and letf : X Y be again the function f(x) = x2. Then{0, 1} Tp but f1({0, 1}) ={1, 0, 1} / Tfc showing that f is not continuous.

Example3.1.5. LetXbe a non-empty set and letP1 andP2 be two partitions onXand letT1 andT2 be the two associated partition topologies on X(Example2.2.9).Let f :X Xbe the identity function f(x) = x whose domain is equipped withT1and codomain withT2. Thenf is continuous if and only if every element inP2 is aunion of elements fromP1.

Example3.1.6

.The constant function f :XY, given by f(x) =pY for allxX, is always continuous since f1(U) is either the empty set ifp /Uor all ofX

ifpU.Example 3.1.7. LetXbe equipped with the discrete topology, then any function

f : X Y is continuous. Conversely, ifX is given the indiscrete topology, then afunctionf :XY to a Hausdorff space Y is continuous if and only if it is constant.For iffwere not constant then we could find two points a, bXwithf(a)= (b). TheHausdorff property guarantees the existence of two open and disjoint sets U, V Ywithf(a)Uand f(b)V. But thenf1(U)= (sincea /f1(U)) andf1(U)=X(sinceb /f1(U)), showing that f is not continuous.

Example3.1.8. Let (X, dX) and (Y, dY) be two metric spaces and letTdX andTdYbe the associated metric topologies. Then a functionf :XYis continuous atxXif and only if for every >0 there exists a >0 such that xX anddX(x, x) < implies that dY(f(x

), f(x)) < . Thus continuous functions between metric spacessatisfy the familiar (, )-rule for continuity from analysis. We leave the verificationof this claim as an exercise (Exercise 3.5.7), it follows along the lines of the proof ofTheorem1.2.1.


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3.1. CONTINUOUS FUNCTIONS 39

The next theorem provides alternative definitions of continuity. Part (a) is a gen-eralization of Theorem1.3.2from the Euclidean case to general topological spaces.

Theorem 3.1.9. Letf : (X, TX) (Y, TY) be a function between two topologicalspaces. Thenf is continuous if and only if any of the mutually equivalent conditionsbelow is met:

(a) f1(B) is a closed subset ofXfor any closed subsetB ofY.(b) For all subsetsBY, the inclusionf1(B)f1(B) holds.(c) For all subsetsAX the inclusionf(A)f(A) holds.(d) For all subsetsBY the inclusionf1(Int(B))I nt(f1(B)) holds.(e) Given a basisB={ViY| i I} forTY, f1(Vi) is open for every i I.Proof. We will show that properties (a) and (e) are each equivalent to f being

continuous. We will then prove the implications (a)=

(b)=

(c)=

(a). Showing that

(d) is equivalent to the continuity offis left as an exercise (Exercise 3.5.6).(a) Suppose that f is continuous and let B Y be a closed set. Then f1(B) =

Xf1(YB) is also closed sinceYBis open and continuity offforcesf1(YB)to be open.

Conversely, suppose thatfhas property (a) and letVY be any open set. Thenf1(V) =X f1(Y V) is open since Y V and f1(Y V) are both closed, thelatter by property (a).

(a)=(b) Let B be any subset ofY. Since B B, we obtain f1(B)f1(B).By property (a) off, the set f1(B) is closed but since the set f1(B) is the smallestclosed set containing f1(B) (see Lemma 2.3.2), the inclusion f1(B) f1(B) isimmediate.

(b)=(c) Take A X to be any subset of X and apply property (b) of f toB = f(A) to obtain f1(f(A)) f1(f(A)). Since A f1(f(A)) we also getAf1(f(A)). Applyingfto the inclusion Af1(f(A)) yields the desired result.

(c)=(a) Let B be a closed subset ofY and set A = f1(B). Pick a point x A.Then, according to property (c) we must have f(x) f(A) = B = B showing thatxA. This implies that A= A and thus that A is closed.

(e) The necessity of property (e) for a continuous function is obvious. Suppose thenthat fpossesses property (e) and let VY be an open set. LetJ Ibe such thatV =jJVj . Thenf1(V) = f1(jJVj) =jJf1(Vj) showing that f1(V) is aunion of open sets and therefore open.

The reader may have noticed that parts (b), (c) and (d) of Theorem 3.1.9expresscontinuity in terms of an image/preimage under f, combined with a choice of taking theinterior/closure of a set, see Table3.1. Of the four possible combinations resulting inthis manner, one is noticeably absent, namely the one involving taking images underfcombined with taking interiors of set. The next example demonstrates that this fourthpossibility cannot be added to Theorem3.1.9.


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Preimage Image

Closure (b) (c)Interior (d) ?

Table 1. Conditions (b), (c) and (d) from Theorem3.1.9each involvea combination of taking an image/preimage underfand taking the clo-sure/interior of a set.

Example 3.1.10. Consider the function f : R2 R given by f(x, y) = x whereeach ofR2 and Rare equipped with the Euclidean topology. Then f is clearly contin-uous, but, taking A ={(x, 0) R2 | x R}, we find that

Int(f(A) =I nt(R) = R while f(Int(A)) =f() =.Thus the inclusion I nt(f(A)

f(Int(A)) fails in general for continuous functions.

On the other hand, consider the function g : R Rgiven byg(x) = 0 and assumethat both copies ofRcome with the Euclidean topology. Pick B = R, then

g(Int(B)) =g(R) ={0} while Int(g(B)) = I nt({0}) =.We see that the inclusion g(Int(B)) Int(g(B)) also fails in general for continuousfunctions (see however part (a) of Proposition3.1.22below).

We next single out some simple functions that are always continuous. The readerwill no doubt recognize familiar properties of continuous functions from the Euclideancase.

Proposition 3.1.11. Let X, Y, Z be topological spaces and let f : X

Y and

g: YZ be continuous functions. Additionally, letAXbe any subspace ofX.(a) The inclusion function : A X is continuous. In particular, the identity

function id:XX is always continuous.(b) The composition functiong f :XZis continuous.(c) The restriction functionf|A: AY is continuous.(d) LetUiX, i I, be a collection of open subsets ofX such thatX=iIUi

and leth: XYbe a function withh|Ui :UiYcontinuous for eachi I.Thenh is continuous.

Proof. (a) For an open subset U X, the preimage 1(U) equals U A and istherefore open in A with respect to its relative topology.

(b) Let Wbe an open subset ofZ. ThenV =g1

(W) is an open subset ofY andthus U = f1(V) must be open in X. Since U = f1(g1(W) = (gf)1(W), theclaim follows.

(c) This follows from parts (a) and (b) since f|A = f where : A X is theinclusion map.

(d) Set hi = h|Ui and let VYbe an open set. Then h1(V) =iIh1i (V) and,since each h1i (V) must be open in Ui (in its relative topology), there must be open


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subsetsWiXwithh1i (V) =Ui Wi. As both Ui andWi are open in Xthen so isUi Wi showing that h1(V) is a union of open sets and therefore open.

The following simple lemma will prove a useful tool in subsequent sections.

Lemma3.1.12. Let(X, TX)be a topological space andA, BX two subspaces withX = A B. Assume that either bothA andB are open or that both are closed. Let(Y, TY) be another topological space and letf :AY andg : B Y be continuous

functions which agree onA B, that is assume thatf(x) =g(x) for everyxA B.Then the functionh: XY defined by

h(x) =

f(x) ; xA,g(x) ; xB,

is continuous.

Proof. LetV

Ybe any subset ofY. Then

h1(V) = (h1(V) A) (h1(V) B) =f1(V) g1(V).IfA,B are both open, choose Vto be open also. Continuity off andg shows the setsf1(V) and g1(V) to be open subsets ofAand B respectively. Accordingly they arealso open inXsinceA and B are open subsets ofX(part (a) of Exercise2.5.17). Thush1(V) is a union of two open sets and hence open, demonstrating the continuity ofh.

If A, B are both closed, pick V Y also closed. The same reasoning as in theprevious case (and by relying on part (b) of Exercise 2.5.17) shows that h1(V) is aclosed set. Continuity ofh now follows from part (a) of Theorem3.1.9.

Definition 3.1.13. A function f : X

Y between topological spaces is called

a homeomorphism if f is a continuous bijection with a continuous inverse functionf1 : Y X. We say that two topological spacesX and Y are homeomorphic, andwriteX=Y, if there exists at least one homeomorphism f :XY.

A functionf :XY is called a local homeomorphismif it is surjective and everypoint x X has a neighborhood U such that f|U : U f(U) is a homeomorphism,whereU and f(U) come equipped with their relative topologies inherited from XandY respectively. Two spaces are calledlocally homeomorphicif there exists at least onelocal homeomorphism between them.

Remark 3.1.14. The relation of being homeomorphic to between topologicalspaces is an equivalence relation:

1. Reflexivity: X=X. A homeomorphism from X to Xis given by the identitymap (see part (a) of Proposition3.1.11).

2. Symmetry: X= Y implies Y= X. Iff : X Y is a homeomorphism, thenf1 :YXis also a homeomorphism.

3. Transitivity: X= Y and Y=Z imply X=Z. Iff :X Y and g :YZare homeomorphisms, theng f :X Z is also a homeomorphism (part (b)of Proposition3.1.11).


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In a similar vein one can also prove that the relation of being locally homeomorphicisan equivalence relations among topological spaces (Exercise ??).

From an abstract point of view, that is, from a point of view where we disregardthe labels attached to the points of a set, two homeomorphic spaces (X, TX) and(Y, TY) are indistinguishable. For all intents and purposes, two homeomorphic spacescan be considered as being identical, a practice that we shall adopt from hereon out,sometimes without specifically saying so. To imprint this point further, we turn to asimple example.

Example 3.1.15. Let the sets X ={1, 2, 3} and Y ={a,b,c} be given the twotopologies

TX={, X, {1, 2}} and TY ={, Y, {b, c}}.The reader will have no trouble recognizing that the two spaces (X, TX) and (Y, TY)are homeomorphic with f : X Y given by f(1) = c, f(2) = b and f(3) = a beingone possible homeomorphism. Thus, if we disregard that the elements inXare labeledby 1, 2, 3 and those in Y by a,b,c, the spaces (X, TX) and (Y, TY) become identical.We can think of the homeomorphism f : X Y as a relabeling tool, one whichtakes an element x from Xand gives it a new label f(x). This relabeling has to bedone with care so as to respect the topologies on X and Y, the relabeling of an opensubset fromXhas to result in an open subset ofY.

Example 3.1.16. The function f : (R, T) (R, T) given by f(x) = x3 is ahomeomorphism for every choice ofT {TEu, Tp, Tp, Tfc, Tcc, TF,p}, with the choice of

p being either 0 or1. Proving this amounts to showing the f and its inverse arecontinuous function with respect to the listed topologies. For instance, when

T =

Tp,

and V R is a nonempty open set, then pV and so pf1(V) since f(p) =p forp {1, 0, 1}. Thusf1(V) is open and hence fcontinuous. The verification of theother claims is left as an exercise (Exercise3.5.8).

Example 3.1.17. The function f : (R, TEu) (R, TEu) given by f(t) = sinh t =12

(et et) is a homeomorphism with inverse function f1(t) = sinh1 t = ln(x+x2 + 1).

Example 3.1.18. Let (X, dX) and (Y, dY) be two metric spaces. A function f :X Y is called an isometry if dY(f(a), f(b)) = dX(a,b, ) for any pair of pointsa, bX. Considering X and Yequipped with their associated metric topologiesTdXandTdY respectively, any surjective isometryf :XY becomes an homeomorphism.

To see this we first verify that any isometry is continuous. Let x X be anypoint and let V be any neighborhood off(x). Then there exists an r > 0 such thatBf(x)(r) V. But then Bx(r) is a neighborhood of x with f(Bx(r)) V for ifaBx(r) then dX(x, a) =r = dY(f(x), f(a)) showing that f(x)Bf(x)(r).

It remains to see thatfis a bijection and that its inverse function is also continuous.These are deferred to Exercise3.5.11.


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Example 3.1.19 (Stereographic projections). Consider Rn and Rn+1 (withn N)both equipped with their Euclidean topologies and let Sn Rn+1 be the subspace

Sn =

{(x1,...xn+1)

Rn+1

|x21+x

22+...+x

2n+1= 1

}.

We shall refer to Sn as the n-dimensional sphere or n-sphere for short. The goal ofthis example is to show that Sn {p} is homeomorphic to Rn where p Sn is anarbitrarily chosen point. For convenience, we choose p = N = (0, 0,..., 0, 1) to be thenorth poleofSn.

Let N : (Sn {N})Rn be the function, referred to as stereographic projection

from the north poleN, given by

N(x1,...,xn+1) =(x1,...,xn)

1 xn+1 .

Note the N is well defined since the only point (x1,...,xn+1) Sn with xn+1 = 1is the north pole N. It is easy to see that N is a bijection with inverse function1N : Rn (Sn {N}) given by

1N (y1,...,yn) =(2y1, 2y2,..., 2yn, |y|2 1)

|y|2 + 1 ,

where|y|2 stands for y 21+y22+...+y2n.Leta = (a1,...,an+1)Sn {N}be any point and let >0 be chosen at will. Let

M= 1 an+1 and define as any positive real number subject to the inequality

M/2. Thus

(d2(N(a), N(a)))2 =

ni=1

ai1 an+1 ai

1 an+1

2=

ni=1

|ai(an+1 an+1) + (ai ai)(1 an+1)|2|1 an+1|2 |1 an+1|2


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N

x

N(x)

Figure 1. The geometric interpretation of the stereographic projection

N : (Sn

{N}) Rn

. Given a point xSn

{N},N(x) is obtainedas the intersection point of the line throughNandx, with the equatorialplane Rn {0} Rn+1.

Similar stereographic projection can be obtained from any point p Sn to Rnwhere Rn is characterized as those points in Rn+1 that have dot product zero with p,i.e. Rn={x Rn+1 | x p = 0}. In Exercise3.5.9you will be asked to find formulasdescribing the stereographic pro

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