Topological Spaces - David R Wilkins

8/14/2019 Topological Spaces - David R Wilkins

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Course 221: Michaelmas Term 2006

Section 4: Topological Spaces

David R. Wilkins

Copyright c David R. Wilkins 19972006

Contents

4 Topological Spaces 2

4.1 Topological Spaces: Definitions and Examples . . . . . . . . . 24.2 Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 44.3 Subspace Topologies . . . . . . . . . . . . . . . . . . . . . . . 44.4 Continuous Functions between Topological Spaces . . . . . . . 64.5 Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 8

4.6 Sequences and Convergence . . . . . . . . . . . . . . . . . . . 84.7 Neighbourhoods, Closures and Interiors . . . . . . . . . . . . . 94.8 Product Topologies . . . . . . . . . . . . . . . . . . . . . . . . 114.9 Cut and Paste Constructions . . . . . . . . . . . . . . . . . . . 144.10 Identification Maps and Quotient Topologies . . . . . . . . . . 164.11 Connected Topological Spaces . . . . . . . . . . . . . . . . . . 19

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4 Topological Spaces

The theory of topological spaces provides a setting for the notions of continu-ity and convergence which is more general than that provided by the theoryof metric spaces. In the theory of metric spaces one can find necessary andsufficient conditions for convergence and continuity that do not refer explic-itly to the distance function on a metric space but instead are expressed interms of open sets. Thus a sequence of points in a metric space Xconvergesto a point p ofX if and only if every open set which contains the point palso contains all but finitely many members of the sequence. Also a functionf: X Y between metric spaces X and Y is continuous if and only if thepreimage f

1

(V) of every open set V in Y is an open set in X. It followsfrom this that we can generalize the notions of convergence and continuityby introducing the concept of atopological space: a topological space consistsof a set together with a collection of subsets termed open sets that satisfyappropriate axioms. The axioms for open sets in a topological space aresatisfied by the open sets in any metric space.

4.1 Topological Spaces: Definitions and Examples

Definition Atopological spaceXconsists of a set Xtogether with a collec-tion of subsets, referred to as open sets, such that the following conditionsare satisfied:

(i) the empty setand the whole set Xare open sets,(ii) the union of any collection of open sets is itself an open set,

(iii) the intersection of anyfinitecollection of open sets is itself an open set.

The collection consisting of all the open sets in a topological space X isreferred to as a topologyon the set X.

Remark If it is necessary to specify explicitly the topology on a topologicalspace then one denotes by (X, ) the topological space whose underlying setis Xand whose topology is . However if no confusion will arise then it iscustomary to denote this topological space simply by X.

Any metric space may be regarded as a topological space. Indeed let Xbe a metric space with distance function d. We recall that a subset V ofX is an open set if and only if, given any point v ofV , there exists some > 0 such that{x X : d(x, v) < } V. The empty set and thewhole space Xare open sets. Also any union of open sets in a metric space

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is an open set, and any finite intersection of open sets in a metric space is an

open set. Thus the topological space axioms are satisfied by the collectionof open sets in any metric space. We refer to this collection of open sets asthe topologygenerated by the distance function d on X.

Any subsetXofn-dimensional Euclidean space Rn is a topological space:a subset V ofX is open in X if and only if, given any point v ofV, thereexists some >0 such that

{xX :|x v|< } V .

In particular Rn is itself a topological space whose topology is generated by

the Euclidean distance function onRn

. This topology onRn

is referred toas theusual topologyon Rn. One defines the usual topologies on Rand Cinan analogous fashion.

Example Given any set X, one can define a topology on X where everysubset of X is an open set. This topology is referred to as the discretetopologyon X.

Example Given any set X, one can define a topology on X in which theonly open sets are the empty setand the whole set X.

Definition Let Xbe a topological space. A subset F ofX is said to be aclosed set if and only if its complement X\ F is an open set.

We recall that the complement of the union of some collection of subsetsof some set Xis the intersection of the complements of those sets, and thecomplement of the intersection of some collection of subsets of X is theunion of the complements of those sets. The following result therefore followsdirectly from the definition of a topological space.

Proposition 4.1 LetXbe a topological space. Then the collection of closedsets ofXhas the following properties:

(i) the empty set and the whole setXare closed sets,(ii) the intersection of any collection of closed sets is itself a closed set,

(iii) the union of any finite collection of closed sets is itself a closed set.

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4.2 Hausdorff Spaces

Definition A topological spaceXis said to be a Hausdorff spaceif and onlyif it satisfies the following Hausdorff Axiom:

ifxand y are distinct points ofXthen there exist open sets U and Vsuch that xU,yV and U V =.

Lemma 4.2 All metric spaces are Hausdorff spaces.

Proof Let Xbe a metric space with distance function d, and let x and y bepoints ofX, where x= y. Let = 1

2d(x, y). Then the open balls BX(x, )

and BX(y, ) of radius centred on the points x and y are open sets. IfBX(x, ) BX(y, ) were non-empty then there would exist zX satisfyingd(x, z)< andd(z, y)< . But this is impossible, since it would then followfrom the Triangle Inequality that d(x, y) < 2, contrary to the choice of.Thus xBX(x, ), y BX(y, ), BX(x, ) BX(y, ) =. This shows thatthe metric space X is a Hausdorff space.

We now give an example of a topological space which is not a Hausdorffspace.

Example The Zariski topologyon the set R of real numbers is defined as

follows: a subset UofR

is open (with respect to the Zariski topology) if andonly if either U=or else R \ U is finite. It is a straightforward exercise toverify that the topological space axioms are satisfied, so that the set R of realnumbers is a topological space with respect to this Zariski topology. Nowthe intersection of any two non-empty open sets in this topology is alwaysnon-empty. (Indeed ifU and Vare non-empty open sets then U = R \ F1and V = R \ F2, where F1 and F2 are finite sets of real numbers. But thenU V = R \ (F1 F2), which is non-empty, since F1 F2 is finite and R isinfinite.) It follows immediately from this that R, with the Zariski topology,is not a Hausdorff space.

4.3 Subspace Topologies

Let Xbe a topological space with topology , and let A be a subset ofX.Let A be the collection of all subsets ofA that are of the form V A forV. Then A is a topology on the set A. (It is a straightforward exerciseto verify that the topological space axioms are satisfied.) The topology Aon Ais referred to as the subspace topology on A.

Any subset of a Hausdorff space is itself a Hausdorff space (with respectto the subspace topology).

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Lemma 4.3 LetXbe a metric space with distance functiond, and letAbe

a subset ofX. A subsetW ofA is open with respect to the subspace topologyonA if and only if, given any pointw ofW, there exists some > 0 suchthat

{aA : d(a, w)< } W.Thus the subspace topology onA coincides with the topology onA obtainedon regardingA as a metric space (with respect to the distance functiond).

ProofSuppose thatWis open with respect to the subspace topology on A.Then there exists some open set U in Xsuch that W =U A. Let w be apoint ofW. Then there exists some >0 such that

{xX :d(x, w)< } U.But then

{aA : d(a, w)< } U A= W.Conversely, suppose that W is a subset ofA with the property that, for

anywW, there exists some w >0 such that{aA : d(a, w)< w} W.

Define Uto be the union of the open balls BX(w, w) as w ranges over all

points ofW, where

BX(w, w) ={xX :d(x, w)< w}.The setUis an open set in X, since each open ball BX(w, w) is an open setin X, and any union of open sets is itself an open set. Moreover

BX(w, w) A={aA : d(a, w)< w} Wfor any wW. Therefore U AW. However W U A, since,W Aand{w} BX(w, w) U for any w W. ThusW = U A, where U isan open set in X. We deduce that W is open with respect to the subspacetopology on A.

Example Let Xbe any subset ofn-dimensional Euclidean space Rn. Thenthe subspace topology on Xcoincides with the topology on Xgenerated bythe Euclidean distance function on X. We refer to this topology as theusualtopologyon X.

LetXbe a topological space, and let Abe a subset ofX. One can readilyverify the following:

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a subsetB ofA is closed in A (relative to the subspace topology onA)

if and only ifB = A F for some closed subset F ofX; ifA is itself open in Xthen a subset B ofA is open in A if and only

if it is open in X;

ifA is itself closed in Xthen a subsetB ofA is closed in A if and onlyif it is closed in X.

4.4 Continuous Functions between Topological Spaces

Definition A functionf: X

Y from a topological spaceXto a topological

spaceYis said to becontinuous iff1(V) is an open set in Xfor every openset V in Y, where

f1(V) {xX :f(x)V}.A continuous function from X to Y is often referred to as a map from Xto Y.

Lemma 4.4 LetX, Y andZ be topological spaces, and letf: X Y andg: Y Z be continuous functions. Then the compositiong f: X Z ofthe functionsf andg is continuous.

Proof Let V be an open set in Z. Then g1(V) is open in Y (since g iscontinuous), and hence f1(g1(V)) is open in X (since f is continuous).But f1(g1(V)) = (gf)1(V). Thus the composition function gf iscontinuous.

Lemma 4.5 Let X and Y be topological spaces, and let f: X Y be afunction fromX to Y. The functionf is continuous if and only iff1(G)is closed inXfor every closed subsetG ofY.

Proof If G is any subset of Y then X\ f1(G) = f1(Y\G) (i.e., thecomplement of the preimage ofG is the preimage of the complement ofG).The result therefore follows immediately from the definitions of continuityand closed sets.

We now show that, if a topological space X is the union of a finite col-lection of closed sets, and if a function from X to some topological space iscontinuous on each of these closed sets, then that function is continuous onX.

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Lemma 4.6 LetX andYbe topological spaces, letf: X

Y be a function

fromX toY, and letX=A1A2 Ak, whereA1, A2, . . . , Ak are closedsets inX. Suppose that the restriction offto the closed setAi is continuousfori= 1, 2, . . . , k. Thenf: XY is continuous.Proof Let Vbe an open set in Y. We must show that f1(V) is open inX. Now the preimage of the open set V under the restriction f|Ai off toAi isf

1(V) Ai. It follows from the continuity off|Ai that f1(V) Ai isrelatively open inAifor eachi, and hence there exist open setsU1, U2, . . . , U kinXsuch thatf1(V)Ai=UiAifor i = 1, 2, . . . , k. LetWi=Ui(X\Ai)for i = 1, 2, . . . , k. ThenWi is an open set in X (as it is the union of the

open sets Ui and X\ Ai), and Wi Ai =Ui Ai =f1

(V) Ai for each i.We claim that f1(V) =W1 W2 Wk.Let W =W1 W2 Wk. Thenf1(V)W, sincef1(V)Wi for

eachi. Also

W =ki=1

(W Ai)ki=1

(Wi Ai) =ki=1

(f1(V) Ai)f1(V),

sinceX=A1A2 AkandWiAi=f1(V) Aifor eachi. Thereforef1(V) = W. But W is open in X, since it is the intersection of a finitecollection of open sets. We have thus shown that f1(V) is open inX for

any open set V in Y. Thus f: XYis continuous, as required.Alternative ProofA functionf: XYis continuous if and only iff1(G)is closed in X for every closed set G in Y (Lemma 4.5). Let G be an closedset inY. Thenf1(G) Ai is relatively closed in Ai for i = 1, 2, . . . , k, sincethe restriction off toAi is continuous for eachi. ButAi is closed in X, andtherefore a subset ofAi is relatively closed in Ai if and only if it is closed inX. Thereforef1(G) Ai is closed in X for i= 1, 2, . . . , k. Now f1(G) isthe union of the sets f1(G) Ai for i= 1, 2, . . . , k. It follows that f1(G),being a finite union of closed sets, is itself closed in X. It now follows fromLemma 4.5 that f: X

Y is continuous.

Example LetYbe a topological space, and let : [0, 1]Y and: [0, 1]Ybe continuous functions defined on the interval [0, 1], where (1) =(0).Let : [0, 1]Ybe defined by

(t) =

(2t) if 0t 1

2;

(2t 1) if 12t1.

Now|[0, 12

] = where: [0, 12

][0, 1] is the continuous function definedby(t) = 2tfor allt[0, 1

2]. Thus|[0, 1

2] is continuous, being a composition

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of two continuous functions. Similarly

|[12

, 1] is continuous. The subinter-

vals [0, 12 ] and [12 , 1] are closed in [0, 1], and [0, 1] is the union of these twosubintervals. It follows from Lemma 4.6 that : [0, 1]Y is continuous.

4.5 Homeomorphisms

Definition LetXandYbe topological spaces. A functionh: XYis saidto be a homeomorphismif and only if the following conditions are satisfied:

the function h: X Y is both injective and surjective (so that thefunction h: XYhas a well-defined inverse h1: YX),

the functionh: XYand its inverseh1: YXare both continuous.Two topological spacesXandYare said to be homeomorphicif there existsa homeomorphismh: XY fromX to Y.

Ifh: X Y is a homeomorphism between topological spaces X and Ythen h induces a one-to-one correspondence between the open sets ofXandthe open sets ofY. Thus the topological spacesX and Y can be regardedas being identical as topological spaces.

4.6 Sequences and ConvergenceDefinition A sequencex1, x2, x3, . . .of points in a topological spaceXis saidtoconvergeto a pointp ofXif, given any open setUcontaining the point p,there exists some natural number N such that xj U for all j N. If thesequence (xj) converges to pthen we refer to pas a limitof the sequence.

This definition of convergence generalizes the definition of convergencefor a sequence of points in a metric space.

It can happen that a sequence of points in a topological space can havemore than one limit. For example, consider the set R of real numbers with the

Zariski topology. (The open sets ofR in the Zariski topology are the emptyset and those subsets ofR whose complements are finite.) Let x1, x2, x3, . . .be the sequence in R defined by xj = j for all natural numbers j. Onecan readily check that this sequence converges to every real number p(withrespect to the Zariski topology on R).

Lemma 4.7 A sequencex1, x2, x3, . . .of points in a Hausdorff spaceX con-verges to at most one limit.

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Proof Suppose that p and qwere limits of the sequence (xj), where p

=q.

Then there would exist open sets U and V such that p U, q V andU V =, sinceXis a Hausdorff space. But then there would exist naturalnumbersN1 and N2 such that xjU for allj satisfying jN1 and xjVfor all j satisfying j N2. But then xj U V for all j satisfying j N1and jN2, which is impossible, since U V =. This contradiction showsthat the sequence (xj) has at most one limit.

Lemma 4.8 Let X be a topological space, and letF be a closed set in X.Let(xj :j N) be a sequence of points inF. Suppose that the sequence(xj)converges to some pointp ofX. ThenpF.

ProofSuppose thatp were a point belonging to the complement X\FofF.Now X\F is open (since F is closed). Therefore there would exist somenatural number Nsuch that xjX\ F for all values ofj satisfying jN,contradicting the fact that xjF for all j . This contradiction shows that pmust belong to F, as required.

Lemma 4.9 Let f: X Y be a continuous function between topologicalspaces X and Y, and let x1, x2, x3, . . . be a sequence of points in X whichconverges to some point p of X. Then the sequence f(x1), f(x2), f(x3), . . .converges to f(p).

Proof Let V be an open set in Y which contains the point f(p). Thenf1(V ) is an open set in X which contains the point p. It follows thatthere exists some natural number N such that xj f1(V) whenever jN. But then f(xj) V whenever j N. We deduce that the sequencef(x1), f(x2), f(x3), . . .converges to f(p), as required.

4.7 Neighbourhoods, Closures and Interiors

Definition Let X be a topological space, and let x be a point ofX. Let

N be a subset of Xwhich contains the point x. Then N is said to be aneighbourhood of the point x if and only if there exists an open set U forwhichxU and UN.

One can readily verify that this definition of neighbourhoods in topolog-ical spaces is consistent with that for neighbourhoods in metric spaces.

Lemma 4.10 LetXbe a topological space. A subsetV ofX is open inXif and only ifVis a neighbourhood of each point belonging to V.

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Proof It follows directly from the definition of neighbourhoods that an open

set Vis a neighbourhood of any point belonging to V. Conversely, supposethatVis a subset ofXwhich is a neighbourhood of eachvV. Then, givenany point v ofV, there exists an open set Uv such that vUv and UvV.Thus V is an open set, since it is the union of the open sets Uv as v rangesover all points ofV.

Definition Let Xbe a topological space and let A be a subset ofX. Theclosure A of A in X is defined to be the intersection of all of the closedsubsets ofXthat contain A. The interior A0 ofAinXis defined to be theunion of all of the open subsets ofXthat are contained in A.

LetXbe a topological space and letAbe a subset ofX. It follows directlyfrom the definition ofA that the closure A ofA is uniquely characterized bythe following two properties:

(i) the closure AofAis a closed set containing A,

(ii) ifFis any closed set containing Athen F contains A.

Similarly the interiorA0 ofAis uniquely characterized by the following twoproperties:

(i) the interior A0 ofAis an open set contained in A,

(ii) ifU is any open set contained in Athen U is contained in A0.

Moreover a point x ofA belongs to the interior A0 ofAif and only ifAis aneighbourhood ofx.

Lemma 4.11 Let X be a topological space, and let A be a subset of X.Suppose that a sequencex1, x2, x3, . . .of points ofA converges to some pointpofX. Thenp belongs to the closureA ofA.

Proof IfFis any closed set containingAthenxjFfor allj , and thereforepF, by Lemma 4.8. Therefore pA by definition ofA.

Definition Let Xbe a topological space, and let A be a subset ofX. Wesay that A is dense inX ifA= X.

Example The set of all rational numbers is dense in R.

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4.8 Product Topologies

The Cartesian product X1 X2 Xn of sets X1, X2, . . . , X n is definedto be the set of all ordered n-tuples (x1, x2, . . . , xn), where xi Xi for i =1, 2, . . . , n.

The sets R2 and R3 are the Cartesian products R R and R R Rrespectively.

Cartesian products of sets are employed as the domains of functions ofseveral variables. For example, ifX, Y and Z are sets, and if an elementf(x, y) of Z is determined for each choice of an element x of X and anelementy ofY, then we have a function f: X YZwhose domain is theCartesian product X

Y ofX and Y: this function sends the ordered pair

(x, y) to f(x, y) for all xXand yY.

Definition Let X1, X2, . . . , X n be topological spaces. A subset U of theCartesian productX1 X2 Xn is said to beopen(with respect to theproduct topology) if, given any point p ofU, there exist open sets Vi in Xifor i = 1, 2, . . . , nsuch that{p} V1 V2 VnU.

Lemma 4.12 LetX1, X2, . . . , X n be topological spaces. Then the collectionof open sets inX1 X2 Xn is a topology onX1 X2 Xn.

Proof LetX=X1X2 Xn. The definition of open sets ensures thatthe empty set and the whole set Xare open in X. We must prove that anyunion or finite intersection of open sets in Xis an open set.

Let Ebe a union of a collection of open sets in Xand letp be a point ofE. Thenp D for some open set D in the collection. It follows from thisthat there exist open sets Vi in Xi for i= 1, 2, . . . , nsuch that

{p} V1 V2 VnDE.

Thus Eis open in X.LetU=U1

U2

Um, whereU1, U2, . . . , U mare open sets inX, and let

pbe a point ofU. Then there exist open sets Vkiin Xifor k = 1, 2, . . . , mandi= 1, 2, . . . , nsuch that{p} Vk1Vk2 VknUk fork = 1, 2, . . . , m.Let Vi= V1i V2i Vmi for i= 1, 2, . . . , n. Then

{p} V1 V2 VnVk1 Vk2 VknUkfor k = 1, 2, . . . , m, and hence{p} V1 V2 VnU. It follows thatU is open in X, as required.

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Lemma 4.13 LetX1, X2, . . . , X n andZbe topological spaces. Then a func-

tionf: X1 X2 XnZis continuous if and only if, given any pointp of X1 X2 Xn, and given any open setU inZ containing f(p),there exist open setsVi inXi fori = 1, 2, . . . , nsuch thatpV1V2 Vnandf(V1 V2 Vn)U.Proof LetVibe an open set in Xifor i= 1, 2, . . . , n, and letUbe an open setinZ. ThenV1V2 Vnf1(U) if and only iff(V1V2 Vn)U.It follows thatf1(U) is open in the product topology onX1X2 Xnifand only if, given any pointp ofX1X2 Xnsatisfyingf(p)U, thereexist open setsViin Xifori = 1, 2, . . . , nsuch thatf(V1V2 Vn)U.The required result now follows from the definition of continuity.

Let X1, X2, . . . , X n be topological spaces, and let Vi be an open set inXi for i = 1, 2, . . . , n. It follows directly from the definition of the producttopology that V1 V2 Vn is open in X1 X2 Xn.Theorem 4.14 Let X = X1X2 Xn, where X1, X2, . . . , X n aretopological spaces and X is given the product topology, and for each i, letpi: X Xi denote the projection function which sends (x1, x2, . . . , xn)Xto xi. Then the functionsp1, p2, . . . , pn are continuous. Moreover a functionf: ZXmapping a topological spaceZ into X is continuous if and only ifpi

f: Z

Xi is continuous fori= 1, 2, . . . , n.

Proof Let Vbe an open set in Xi. Then

p1i (V) =X1 Xi1 V Xi+1 Xn,and therefore p1i (V) is open inX. Thuspi: XXi is continuous for all i.

Let f: Z X be continuous. Then, for each i, pi f: Z Xi is acomposition of continuous functions, and is thus itself continuous.

Conversely suppose that f: Z Xis a function with the property thatpi fis continuous for all i. LetUbe an open set in X. We must show thatf1(U) is open in Z.

Letzbe a point off1

(U), and letf(z) = (u1, u2, . . . , un). NowUis openin X, and therefore there exist open sets V1, V2, . . . , V n in X1, X2, . . . , X nrespectively such that uiVi for alli and V1 V2 VnU. Let

Nz =f1

1 (V1) f12 (V2) f1n (Vn),

where fi =pi f for i= 1, 2, . . . , n. Now f1i (Vi) is an open subset ofZ fori= 1, 2, . . . , n, sinceViis open in Xi andfi: ZXi is continuous. ThusNz,being a finite intersection of open sets, is itself open in Z. Moreover

f(Nz)V1 V2 VnU,

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so thatNz

f1(U). It follows thatf1(U) is the union of the open sets Nz

as zranges over all points off1(U). Therefore f1(U) is open in Z. Thisshows that f: ZXis continuous, as required.Proposition 4.15 The usual topology on Rn coincides with the producttopology onRn obtained on regardingRn as the Cartesian productR R R ofn copies of the real lineR.ProofWe must show that a subset UofRn is open with respect to the usualtopology if and only if it is open with respect to the product topology.

Let Ube a subset ofRn that is open with respect to the usual topology,and let u

U. Then there exists some >0 such that B(u, )

U, where

B(u, ) ={xRn :|x u|< }.Let I1, I2, . . . , I n be the open intervals in Rdefined by

Ii={tR :ui n

< t < ui+

n}

for i = 1, 2, . . . , n. Then I1, I2, . . . , I n are open sets in R. Moreover

{u} I1 I2 InB(u, )U,

since

|x u|2 =ni=1

(xi ui)2 < n

n

2=2

for all x I1 I2 In. This shows that any subset U ofRn that isopen with respect to the usual topology on Rn is also open with respect tothe product topology on Rn.

Conversely suppose that U is a subset ofRn that is open with respectto the product topology on Rn, and let u U. Then there exist opensets V1, V2, . . . , V n in R containing u1, u2, . . . , un respectively such that V1V2

Vn

U. Now we can find 1, 2, . . . , n such that i > 0 and

(uii, ui + i)Vifor alli. Let >0 be the minimum of1, 2, . . . , . . . , n.Then

B(u, )V1 V2 VnU,for ifxB(u, ) then|xi ui|< i for i= 1, 2, . . . , n. This shows that anysubset U ofRn that is open with respect to the product topology on Rn isalso open with respect to the usual topology on Rn.

The following result is now an immediate corollary of Proposition 4.15and Theorem 4.14.

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Corollary 4.16 LetXbe a topological space and letf: X

Rn be a function

fromX to Rn. Let us write

f(x) = (f1(x), f2(x), . . . , f n(x))

for allxX, where the componentsf1, f2, . . . , f n offare functions fromXto R. The functionfis continuous if and only if its componentsf1, f2, . . . , f nare all continuous.

Letf: X R andg: X R be continuous real-valued functions on sometopological spaceX. We claim thatf+ g,fgand f.gare continuous. Nowit is a straightforward exercise to verify that the sum and product functions

s:R2 R and p:R2 R defined by s(x, y) = x+ y and p(x, y) = xyare continuous, and f+g = sh and f.g = ph, where h: X R2 isdefined byh(x) = (f(x), g(x)). Moreover it follows from Corollary 4.16 thatthe function h is continuous, and compositions of continuous functions arecontinuous. Therefore f+g and f.g are continuous, as claimed. Alsogis continuous, and f g = f+ (g), and therefore f g is continuous. Ifin addition the continuous function g is non-zero everywhere on Xthen 1/gis continuous (since 1/g is the composition ofg with the reciprocal functiont1/t), and therefore f /g is continuous.

Lemma 4.17 The Cartesian productX1 X2 . . . X n of Hausdorff spacesX1, X2, . . . , X n is Hausdorff.

Proof Let X=X1 X2 . . . , X n, and let u andv be distinct points ofX,where u = (x1, x2, . . . , xn) and v = (y1, y2, . . . , yn). Then xi= yi for someinteger i between 1 and n. But then there exist open setsU and V in Xisuch that xi U, yi V and UV = (since Xi is a Hausdorff space).Letpi: XXidenote the projection function. Thenp1i (U) andp1i (V) areopen sets in X, since pi is continuous. Moreoveru p1i (U), v p1i (V),and p1i (U) p1i (V) =. Thus Xis Hausdorff, as required.

4.9 Cut and Paste Constructions

Suppose we start out with a square of paper. If we join together two oppositeedges of this square we obtain a cylinder. The boundary of the cylinderconsists of two circles. If we join together the two boundary circles we obtaina torus (which corresponds to the surface of a doughnut).

Let the square be represented by the set [0, 1][0, 1] consisting of allordered pairs (s, t) where s and t are real numbers between 0 and 1. Thereis an equivalence relation on the square [0, 1] [0, 1], where points (s, t) and

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(u, v) of the square are related if and only if at least one of the following

conditions is satisfied:

s= u and t= v; s= 0,u= 1 and t = v; s= 1,u= 0 and t = v; t= 0, v= 1 and s = u; t= 1, v= 0 and s = u;

(s, t) and (u, v) both belong to{(0, 0), (0, 1), (1, 0), (1, 1)}.Note that if 0 < s < 1 and 0 < t < 1 then the equivalence class of thepoint (s, t) is the set{(s, t)} consisting of that point. If s = 0 or 1 andif 0 < t < 1 then the equivalence class of (s, t) is the set{(0, t), (1, t)}.Similarly ift= 0 or 1 and if 0 < s


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Another well-known identification space obtained from the square is the

Klein bottle (Kleinsche Flasche). The Klein bottle K2 is obtained from thesquare [0, 1] [0, 1] by identifying (0, t) with (1, 1 t) for all t [0, 1] andidentifying (s, 0) with (s, 1) for alls[0, 1]. These identifications correspondto an equivalence relation on the square, where points (s, t) and (u, v) of thesquare are equivalent if and only if one of the following conditions is satisfied:

s= u and t= v; s= 0,u= 1 and t = 1 v;

s= 1,u= 0 and t = 1

v;

t= 0, v= 1 and s = u; t= 1, v= 0 and s = u; (s, t) and (u, v) both belong to{(0, 0), (0, 1), (1, 0), (1, 1)}.

The corresponding set of equivalence classes is the Klein bottle K2. Thuseach point of the Klein bottle K2 represents an equivalence class consistingof either one point in the interior of the square, or two points (0, t) and(1, 1 t) with 0< t


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It follows directly from the definition that any identification map is con-

tinuous. Moreover, in order to show that a continuous surjectionq: X Yis an identification map, it suffices to prove that ifV is a subset ofY withthe property that q1(V) is open inX then V is open in Y.

Example LetS1 denote the unit circle {(x, y) R2 :x2+y2 = 1} in R2, andlet q: [0, 1] S1 be the continuous map defined by q(t) = (cos2t, sin2t)for all t[0, 1]. We show thatq: [0, 1]S1 is an identification map. Thismap is continuous and surjective. It remains to show that ifVis a subset ofS1 with the property that q1(V) is open in [0, 1] then V is open in S1.

Note that|q(s)q(t)| = 2| sin (st)| for all s, t [0, 1] satisfying|s t|

1

2 . Let Vbe a subset ofS1

with the property that q1

(V) is openin [0, 1], and let v be an element of V. We show that there exists > 0such that all points uofS1 satisfying|u v|< belong to V. We considerseparately the cases when v= (1, 0) and when v= (1, 0).

Suppose that v = (1, 0). Then (1, 0) V, and hence 0 q1(V) and1 q1(V). But q1(V) is open in [0, 1]. It follows that there exists a realnumber satisfying 0 < < 1

2 such that [0, ) q1(V) and (1, 1]

q1(V). Let = 2 sin . Now if then the Euclidean distancebetween the points (1, 0) and (cos , sin ) is 2 sin 1

2||. Moreover, this distance

increases monotonically as|| increases from 0 to . Thus any point on theunit circle S1 whose distance from (1, 0) is less than must be of the form(cos , sin ), where|| < 2. Thus if u S1 satisfies|uv| < thenu= q(s) for some s[0, 1] satisfying either 0s < or 1 < s1. Butthen sq1(V), and hence uV.

Next suppose that v= (1, 0). Then v = q(t) for some real number tsatisfying 0 < t < 1. But q1(V ) is open in [0, 1], and t q1(V). Itfollows that (t , t+)q1(V) for some real number satisfying >0.Let = 2sin . If u S1 satisfies|uv| < then u = q(s) for somes(t , t+). But then sq1(V), and hence uV.

We have thus shown that ifV is a subset ofS1 with the property thatq1(V ) is open in [0, 1] then there exists > 0 such that u

V for all

elements u ofS1 satisfying|u v| < . It follows from this thatV is openinS1. Thus the continuous surjection q: [0, 1]S1 is an identification map.

Lemma 4.18 LetXbe a topological space, letYbe a set, and letq: XYbe a surjection. Then there is a unique topology onY for which the functionq: XY is an identification map.

Proof Let be the collection consisting of all subsets U of Y for whichq1(U) is open in X. Now q1() =, and q1(Y) = X, so that and

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Y

. If

{V :

A

} is any collection of subsets ofY indexed by a set A,

then it is a straightforward exercise to verify thatA

q1(V) =q1

A

V

,

A

q1(V) =q1

A

V

(i.e., given any collection of subsets ofY, the union of the preimages of thesets is the preimage of the union of those sets, and the intersection of thepreimages of the sets is the preimage of the intersection of those sets). Itfollows easily from this that unions and finite intersections of sets belongingto must themselves belong to . Thus is a topology on Y , and thefunction q: X Yis an identification map with respect to the topology .Clearlyis the unique topology on Yfor which the functionq: XY is anidentification map.

Let X be a topological space, let Y be a set, and let q: X Y be asurjection. The unique topology on Yfor which the function qis an identifi-cation map is referred to as the quotient topology (or identification topology)on Y.

Let be an equivalence relation on a topological space X. IfY is thecorresponding set of equivalence classes of elements of X then there is asurjection q: X Ythat sends each element ofX to its equivalence class.Lemma 4.18 ensures that there is a well-defined quotient topology on Y,

where a subset U of Y is open in Y if and only if q1(U) is open in X.(Appropriate equivalence relations on the square yield the torus and theKlein bottle, as discussed above.)

Lemma 4.19 Let X and Y be topological spaces and let q: X Y be anidentification map. Let Z be a topological space, and let f: Y Z be afunction fromY to Z. Then the functionf is continuous if and only if thecomposition functionf q: XZ is continuous.ProofSuppose thatfis continuous. Then the composition function f qis

a composition of continuous functions and hence is itself continuous.Conversely suppose that f q is continuous. Let Ube an open set in Z.Thenq1(f1(U)) is open inX(sincefqis continuous), and hencef1(U)is open in Y (since the function q is an identification map). Therefore thefunction fis continuous, as required.

Example Let S1 be the unit circle in R2, and let q: [0, 1] S1 be themap that sends t [0, 1] to (cos2t, sin2t). Then q: [0, 1] S1 is anidentification map, and therefore a function f: S1 Z from S1 to sometopological spaceZis continuous if and only iffq: [0, 1]Zis continuous.

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Example The Klein bottle K2 is the identification space obtained from the

square [0, 1] [0, 1] by identifying (0, t) with (1, 1 t) for all t [0, 1] andidentifying (s, 0) with (s, 1) for all s [0, 1]. Let q: [0, 1][0, 1] K2be the identification map determined by these identifications. Let Z be atopological space. A functiong: [0, 1] [0, 1]Zmapping the square into Zwhich satisfiesg(0, t) =g(1, 1 t) for allt[0, 1] andg(s, 0) =g(s, 1) for alls[0, 1], determines a corresponding function f: K2 Z, where g=f q.It follows from Lemma 4.19 that the function f: K2 Zis continuous if andonly ifg: [0, 1] [0, 1]Z is continuous.

Example Let Sn be the n-sphere, consisting of all points x in Rn+1 satisfy-

ing |x|= 1. Let RPn

be the set of all lines in Rn+1

passing through the origin(i.e., RPn is the set of all one-dimensional vector subspaces ofRn+1). Letq: Sn RPn denote the function which sends a point x ofSn to the elementofRPn represented by the line in Rn+1 that passes through both x and theorigin. Note that each element of RPn is the image (under q) of exactlytwo antipodal points xandxofSn. The function qinduces a correspond-ing quotient topology on RPn such that q: Sn RPn is an identificationmap. The set RPn, with this topology, is referred to as real projective n-space. In particular RP2 is referred to as the real projective plane. It followsfrom Lemma 4.19 that a function f:RPn Z from RPn to any topologicalspace Z is continuous if and only if the composition function f

q: Sn

Z

is continuous.

4.11 Connected Topological Spaces

Definition A topological space Xis said to be connectedif the empty setand the whole space X are the only subsets of X that are both open andclosed.

Lemma 4.20 A topological space X is connected if and only if it has thefollowing property: if U and V are non-empty open sets in X such that

X=U V, thenU V is non-empty.Proof IfUis a subset ofXthat is both open and closed, and ifV =X\ U,then U and V are both open, U V = X and U V =. Conversely ifUand V are open subsets ofX satisfying U V = X and U V =, thenU=X\V, and henceUis both open and closed. Thus a topological space Xis connected if and only if there do not exist non-empty open sets U and Vsuch that U V =X and U V =. The result follows.

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Let Z be the set of integers with the usual topology (i.e., the subspace

topology on Z induced by the usual topology on R). Then{n} is open foralln Z, since

{n}= Z {t R :|t n|< 12}.

It follows that every subset ofZis open (since it is a union of sets consistingof a single element, and any union of open sets is open). It follows thata function f: X Z on a topological space X is continuous if and only iff1(V) is open in X for any subset V ofZ. We use this fact in the proof ofthe next theorem.

Proposition 4.21 A topological space X is connected if and only if every

continuous functionf: X Z fromXto the setZ of integers is constant.ProofSuppose thatXis connected. Letf: X Z be a continuous function.Choosenf(X), and let

U={xX :f(x) =n}, V ={xX :f(x)=n}.Then U and V are the preimages of the open subsets{n} and Z \ {n} ofZ, and therefore both U and V are open in X. Moreover U V =, andX=U V . It follows thatV =X\ U, and thusUis both open and closed.MoreoverUis non-empty, sincenf(X). It follows from the connectednessofX that U=X, so that f: X

Zis constant, with value n.

Conversely suppose that every continuous functionf: X Z is constant.Let S be a subset ofXwhich is both open and closed. Letf: X Z bedefined by

f(x) =

1 ifxS;0 ifxS.

Now the preimage of any subset of Z under f is one of the open sets,S, X\ S and X. Therefore the function f is continuous. But then thefunction fis constant, so that either S=or S=X. This shows that X isconnected.

Lemma 4.22 The closed interval [a, b] is connected, for all real numbersaandb satisfyingab.Proof Letf: [a, b] Z be a continuous integer-valued function on [a, b]. Weshow that f is constant on [a, b]. Indeed suppose that fwere not constant.Then f()=f(a) for some [a, b]. But the Intermediate Value Theoremwould then ensure that, given any real numbercbetweenf(a) andf(), therewould exist some t[a, ] for which f(t) =c, and this is clearly impossible,sincef is integer-valued. Thus fmust be constant on [a, b]. We now deducefrom Proposition 4.21 that [a, b] is connected.

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A

U =

. Thus ifU is an open set in Xand ifA

U is non-empty then

A Umust also be non-empty.Now let U and V be open sets in X such that AU and AV are

non-empty and A U V. Then AU and AV are non-empty, andA U V. ButA is connected. Therefore A U V is non-empty, andthusA U V is non-empty. This shows that Ais connected.

Lemma 4.25 Let f: X Y be a continuous function between topologicalspacesX andY, and letA be a connected subset ofX. Thenf(A) is con-nected.

Proof Let g: f(A) Zbe any continuous integer-valued function on f(A).Then g f: A Z is a continuous integer-valued function on A. It followsfrom Proposition 4.21 that g f is constant on A. Therefore g is constanton f(A). We deduce from Proposition 4.21 that f(A) is connected.

Lemma 4.26 The Cartesian productX Y of connected topological spacesX andY is itself connected.

Proof Letf: XY Z be a continuous integer-valued function fromXYto Z. Choose x0 X and y0 Y . The function xf(x, y0) is continuousonX, and is thus constant. Thereforef(x, y0) =f(x0, y0) for allx

X. Now

fix x. The function y f(x, y) is continuous on Y , and is thus constant.Therefore

f(x, y) =f(x, y0) =f(x0, y0)

for all x X and y Y. We deduce from Proposition 4.21 thatX Y isconnected.

We deduce immediately that a finite Cartesian product of connected topo-logical spaces is connected.

Proposition 4.27 LetX be a topological space. For eachx

X, letSx be

the union of all connected subsets ofX that containx. Then

(i) Sx is connected,

(ii) Sx is closed,

(iii) ifx, yX, then eitherSx=Sy, or elseSx Sy =.

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Proof Let f: Sx

Z be a continuous integer-valued function on Sx, for

somexX. Lety be any point ofSx. Then, by definition ofSx, there existssome connected set A containing bothx and y . But thenfis constant on A,and thus f(x) = f(y). This shows that the function f is constant on Sx.We deduce that Sx is connected. This proves (i). Moreover the closure Sx isconnected, by Lemma 4.24. ThereforeSxSx. This shows thatSxis closed,proving (ii).

Finally, suppose that x andy are points ofXfor which Sx Sy=. Letf: Sx Sy Zbe any continuous integer-valued function on Sx Sy. Thenf is constant on both Sx and Sy. Moreover the value off on Sx must agreewith that on Sy, since Sx Sy is non-empty. We deduce that f is constanton Sx Sy. ThusSx Sy is a connected set containing both x and y, andthusSx SySx andSxSySy, by definition ofSx andSy. We concludethat Sx=Sy. This proves (iii).

Given any topological space X, the connected subsets Sx ofXdefined asin the statement of Proposition 4.27 are referred to as the connected com-ponents ofX. We see from Proposition 4.27, part (iii) that the topologicalspace Xis the disjoint union of its connected components.

Example The connected components of{(x, y) R2 :x= 0} are

{(x, y) R2 :x >0} and{(x, y)R2 :x

Topological Spaces - David R Wilkins

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