23 Topology Course.pdf

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Topology Course Lecture Notes

Aisling McCluskey and Brian McMaster

August 1997


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Chapter 1

Fundamental Concepts

In the study of metric spaces, we observed that:

(i) many of the concepts can be described purely in terms of open sets,

(ii) open-set descriptions are sometimes simpler than metric descriptions, e.g.continuity,

(iii) many results about these concepts can be proved using only the basicproperties of open sets (namely, that both the empty set and the under-lying set Xare open, that the intersection of any two open sets is againopen and that the union of arbitrarily many open sets is open).

This prompts the question: How far would we get if we started with a collection

of subsets possessing these above-mentioned properties and proceeded to defineeverything in terms of them?

1.1 Describing Topological Spaces

We noted above that many important results in metric spaces can be provedusing only the basic properties of open sets that

the empty set and underlying set Xare both open, the intersection of any two open sets is open, and

unions of arbitrarily many open sets are open.

We will call any collection of sets on X satisfying these properties a topology.In the following section, we also seek to give alternative ways of describing thisimportant collection of sets.

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1.1.1 Defining Topological Spaces

Definition 1.1 A topological space is a pair (X, T) consisting of a set Xand a familyT of subsets ofX satisfying the following conditions:(T1) T andX T(T2)T is closed under arbitrary union(T3)T is closed under finite intersection.The setXis called a space, the elements ofXare calledpointsof the space andthe subsets ofXbelonging toT are called open in the space; the familyT ofopen subsets ofX is also called a topologyfor X.

Examples

(i) Any metric space (X, d) is a topological space whereTd, the topology forXinduced by the metric d, is defined by agreeing that G shall be declaredas open whenever each x in G is contained in an open ball entirely in G,i.e.

GX is open in (X, Td)xG,rx> 0 such that xBrx(x)G.

(ii) The following is a special case of (i), above. Let R be the set of realnumbers and letI be the usual (metric) topology defined by agreeingthat

GXis open in (R,I) (alternatively,I-open) xG,rx> 0 such that (x rx, x + rx)G.

(iii) Define T0={, X} for any set X known as the trivialor anti-discrete topology.(iv) DefineD={GX :GX} known as the discrete topology.(v) For any non-empty set X, the familyC ={G X : G = or X\ G is

finite} is a topology for X called the cofinite topology.(vi) For any non-empty setX, the familyL ={G X : G = or X\ G is

countable} is a topology for Xcalled the cocountable topology.

1.1.2 NeighbourhoodsOccasionally, arguments can be simplified when the sets involved are not over-described. In particular, it is sometimes suffices to use sets which contain opensets but are not necessarily open. We call such sets neighborhoods.

Definition 1.2 Given a topological space (X, T) with x X, then N X issaid to be a (T)-neighbourhoodofx open setG withxGN.

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It follows then that a set U X is open iff for every x U, there exists aneighbourhood Nx ofx contained in U. (Check this!)Lemma 1.1 Let (X, T) be a topological space and, for each x X, letN(x)be the family of neighbourhoods ofx. Then

(i) U N(x)xU.(ii)N(x) is closed under finite intersections.

(iii) U N(x) andUVV N(x).(iv) U N(x) W N(x) such that W U and W N(y) for each

yW.Proof Exercise!

Examples

(i) LetxX, and defineTx={, {x}, X}. ThenTx is a topology for XandVXis a neighbourhood ofx iffxV. However, the only nhd ofyXwherey=x isX itself

(ii) Let xXand define a topologyI(x) for Xas follows:I(x) ={GX :xG}{}.

Note here that everynhd of a point in X is open.

(iii) Let x

Xand define a topologyE

(x) for Xas follows:

E(x) ={GX :xG} {X}.Note here that{y} is open for every y=x in X, that{x, y} is notopen,is not a nhd ofx yet isa nhd ofy .

In fact, the only nhd ofx is X.

1.1.3 Bases and Subbases

It often happens that the open sets of a space can be very complicated andyet they can all be described using a selection of fairly simple special ones.When this happens, the set of simple open sets is called a base or subbase

(depending on how the description is to be done). In addition, it is fortunatethat many topological concepts can be characterized in terms of these simplerbase or subbase elements.

Definition 1.3 Let (X, T) be a topological space. A familyB T is calleda base for (X, T) if and only if every non-empty open subset of X can berepresented as a union of a subfamily ofB.

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It is easily verified thatB T is a base for (X, T) if and only if wheneverxG T, B Bsuch that xBG.Clearly, a topological space can have many bases.Lemma 1.2 IfB is a family of subsets of a setXsuch that(B1) for anyB1, B2 Band every pointxB1 B2, there existsB3 Bwith

xB3B1B2, and(B2) for everyxX, there existsB B such thatxB,thenB is a base for a unique topology onX.Conversely, any baseB for a topological space(X, T) satisfies (B1) and (B2).Proof (Exercise!)

Definition 1.4 Let (X, T) be a topological space. A familyS T is called asubbase for (X, T) if and only if the family of all finite intersectionski=1Ui,whereUi S fori= 1, 2, . . . , k is a base for(X, T).

Examples

(i) In any metric space (X, d),{Br(x) : xX, r > 0} forms a base for theinduced metric topologyTd on X.

(ii) For the real lineR with its usual (Euclidean) topology, the family{(a, b) :a, bQ, a < b} is a base.

(iii) For an arbitrary setX, the family

{{x

}: x

X

}is a base for (X,

D).

(iv) The family of all semi-infinite open intervals (a, ) and (, b) in R isa subbase for (R,I).

1.1.4 Generating Topologies

From the above examples, it follows that for a set X one can select in manydifferent ways a familyT such that (X, T) is a topological space. IfT1 andT2are two topologies forXandT2 T1, then we say that the topologyT1 is finerthan the topologyT2, or thatT2 is coarser than the topologyT1. The discretetopology forXis the finest one; the trivial topology is the coarsest.IfXis an arbitrary infinite set with distinct points x andy, then one can readilyverify that the topologies

I(x) and

I(y) are incomparable i.e. neither is finer

than the other.By generating a topology for X, we mean selecting a familyT of subsets ofXwhich satisfies conditions (T1)(T3). Often it is more convenient notto describethe familyT of open sets directly. The concept of a base offers an alternativemethod of generating topologies.

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Examples

[Sorgenfrey line] Given the real numbers R, letB be the family of allintervals [x, r) where x,r R, x < r and r is rational. One can readilycheck thatB has properties (B1)(B2). The space Rs, generated byB,is called the Sorgenfrey line and has many interesting properties. Notethat the Sorgenfrey topology is finer than the Euclidean topology on R.(Check!)

[Niemytzki plane] Let L denote the closed upper half-plane. We define atopology forLby declaring the basic open sets to be the following:

(I) the (Euclidean) open discs in the upper half-plane;

(II) the (Euclidean) open discs tangent to the edge of the L, togetherwith the point of tangency.

NoteIfyny in L, then(i) y not on edge: same as Euclidean convergence.

(ii) y on the edge: same as Euclidean, but yn must approach y frominside. Thus, for example,yn= (

1n

, 0)(0, 0)!

1.1.5 New Spaces from Old

A subset of a topological space inherits a topology of its own, in an obviousway:

Definition 1.5 Given a topological space (X,

T) withA

X, then the family

TA={A G: G T } is a topology forA, called thesubspace(orrelative orinduced) topology forA. (A, TA) is called a subspace of(X, T).ExampleThe interval I = [0, 1] with its natural (Euclidean) topology is a (closed) sub-space of (R,I).Warning: Although this definition, and several of the results which flow fromit, may suggest that subspaces in general topology are going to be easy inthe sense that a lot of the structure just gets traced onto the subset, there isunfortunately a rich source of mistakes here also: because we are handling twotopologies at once. When we inspect a subsetB ofA, and refer to it as open(or closed, or a neighbourhood of some point p . . . . ) we must be exceedinglycareful as towhichtopology is intended. For instance, in the previous example,[0, 1] itself is open in the subspace topology on I but, of course, not in thebackground topology ofR. In such circumstances, it is advisable tospecifythetopology being used each time by sayingT-open,TA-open, and so on.

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1.2 Closed sets and Closure

Just as many concepts in metric spaces were described in terms of basic opensets, yet others were characterized in terms of closed sets. In this section we

define closed sets in a general topological space and examine the related notion of the closure of a given set.

1.2.1 Closed Sets

Definition 1.6 Given a topological space (X, T) with F X, then F is saidto beT-closed iff its complementX\ F isT-open.From De Morgans Laws and properties (T1)(T3) of open sets, we infer that

the familyFof closed sets of a space has the following properties:(F1) X F and F(F2)Fis closed under finite union(F3)F is closed under arbitrary intersection.Sets which are simultaneously closed andopen in a topological space are some-times referred to asclopensets. For example, members of the base B={[x, r) :x, rR, x < r, r rational} for the Sorgenfrey line are clopen with respect tothe topology generated byB. Indeed, for the discrete space (X, D),everysubsetis clopen.

1.2.2 Closure of Sets

Definition 1.7 If(X, T) is a topological space andAX, then

AT

={F X :AF andF is closed}is called theT-closure ofA.

Evidently,AT

(orAwhen there is no danger of ambiguity) is the smallest closedsubset ofXwhich contains A. Note thatA is closedA = A.Lemma 1.3 If(X, T) is a topological space withA, BX, then

(i)

=(ii) A A

(iii) A= A

(iv) A B= A B.Proof Exercise!

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Theorem 1.1 Given a topological space withAX, then x A iff foreachnhdU ofx, U A=.Proof

: Let x A and let U be a nhd of x; then there exists open G with xGU. IfU A=, then G A=and so AX\ G AX\ GwhencexX\G, thereby contradicting the assumption that UA=.

: If x X\ A, then X\ A is an open nhd of x so that, by hypothesis,(X\ A) A=, which is a contradiction (i.e., a false statement).

Examples

(i) For an arbitrary infinite setXwith the cofinite topology

C, the closed sets

are just the finite ones together with X. So for any AX,

A=

A ifA is finiteX ifA is infinite

Note that any two non-empty open subsets of X have non-empty inter-section.

(ii) For an arbitrary uncountable setXwith the cocountable topologyL, theclosed sets are the countable ones andXitself. Note that if we letX=R,then [0, 1] = R! (In the usual Euclidean topology, [0, 1] = [0, 1].)

(iii) For the space (X, Tx) defined earlier, if AX, then

A= X ifxA

X\ {x} ifxA

(iv) For (X,I(x)) with AX,

A=

A ifxAX ifxA

(v) For (X, E(x)) with AX,

A=

A ifxAA {x} ifxA

(vi) In (X, D), every subset equals its own closure.

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1.3 Continuity and Homeomorphism

The central notion of continuity of functions is extended in this section to generaltopological spaces. The useful characterization of continuous functions in metricspaces as those functions where the inverse image of every open set is open isused as a definition in the general setting.Because many properties of spaces are preserved by continuous functions, spacesrelated by a bijection (one-to-one and onto function) which is continuous in bothdirections will have many properties in common. These properties are identifiedas topological properties. Spaces so related are called homeomorphic.

1.3.1 Continuity

The primitive intuition of a continuous process is that of one in which small

changes in the input produce small, non-catastrophic changes in the corre-sponding output. This idea formalizes easily and naturally for mappings fromone metric space to another: f is continuous at a point p in such a settingwhenever we can force the distance between f(x) and f(p) to be as small asis desired, merely by taking the distance between x and p to be small enough.That form of definition is useless in the absence of a properly defined distancefunction but, fortunately, it is equivalent to the demand that the preimage ofeach open subset of the target metric space shall be open in the domain. Thusexpressed, the idea is immediately transferrable to general topology:

Definition 1.8 Let(X, T)and(Y, S)be topological spaces; a mappingf :XY is calledcontinuous ifff1(U)

T for eachU

S i.e. the inverse image

of any open subset ofY is open inX.

Examples

(i) If (X, D) is discrete and (Y, S) is an arbitrary topological space, thenanyfunctionf :XY is continuous!Again, if (X, T) is an arbitrary topological space and (Y, T0) is trivial,anymappingg : XY is continuous.

(ii) If (X, T), (Y, S) are arbitrary topological spaces and f : X Y is aconstant map, then f is continuous.

(iii) Let Xbe an arbitrary set having more than two elements, with x X.LetT =I(x),S =Tx in the definition of continuity; then the identitymapidX :XX is continuous. However, if we interchangeT withS sothatT =Tx andS =I(x), then idX : X X is notcontinuous! Notethat idX : (X, T1) (X, T2) is continuous if and only ifT1 is finer thanT2.

Theorem 1.2 If (X1, T1), (X2, T2) and (X3, T3) are topological spaces andh:X1X2 andg : X2X3 are continuous, theng h: X1X3 is continuous.

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Proof Immediate.

There are several different ways to recognise continuity for a mapping betweentopological spaces, of which the next theorem indicates two of the most usefulapart from the definition itself:

Theorem 1.3 Letfbe a mapping from a topological space(X1, T1) to a topo-logical space(X2, T2). The following statements are equivalent:

(i) f is continuous,

(ii) the preimage underfof each closed subset ofX2 is closed inX1,

(iii) for every subsetA ofX1, f(A)f(A).Proof It is easy to see that (i) implies (ii). Assuming that (ii) holds, apply it tothe closed set f(A) and (iii) readily follows. Now if (iii) is assumed andG is agiven open subset ofX2, use (iii) on the set A = X1 \ f1(G) and verify that itfollows thatf1(G) must be open.

1.3.2 Homeomorphism

Definition 1.9 Let (X, T), (Y, S) be topological spaces and let h : X Ybe bijective. Then h is a homeomorphism iff h is continuous and h1 iscontinuous. If such a map exists,(X, T)and(Y, S)are calledhomeomorphic.

Such a map has the property that

G T f(G) S.It follows that any statement about a topological space which is ultimatelyexpressible solely in terms of the open sets (together with set-theoretic relationsand operations) will be true for both (X, T) and (Y, S) if it is true for either.In other words, (X, T) and (Y, S) are indistinguishable as topological spaces.The reader who has had abstract algebra will note that homeomorphism is theanalogy in the setting of topological spaces and continuous functions to thenotion of isomorphism in the setting of groups (or rings) and homomorphisms,and to that of linear isomorphism in the context of vector spaces and linearmaps.

ExampleFor every space (X, T), the identity mapping idX :XXis a homeomorphism.A property of topological spaces which when possessed by a space is also pos-sessed by every space homeomorphic to it is called a topological invariant.We shall meet some examples of such properties later.

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One can readily verify that iff is a homeomorphism, then the inverse mapping

f

1

is also a homeomorphism and that the compositiong fof two homeomor-phisms f and g is again a homeomorphism. Thus, the relation X and Y arehomeomorphic is an equivalence relation.In general, it may be quite difficult to demonstrate that two spaces are homeo-morphic (unless a homeomorphism is obvious or can easily be discovered). Forexample, to verify that (R,I) is homeomorphic to (0, 1) with its induced metrictopology, it is necessary to demonstrate, for instance, that h : (0, 1)R whereh(x) = 2x1

x(x1)is a homeomorphism.

It is often easier to show that two spaces are not homeomorphic: simply exhibitan invariant which is possessed by one space and not the other.ExampleThe spaces (X,I(x)) and (X, E(x)) are nothomeomorphic since, for example,(X,

I(x)) has the topological invariant each nhd is open while (X,

E(x)) does

not.

1.4 Additional Observations

Definition 1.10 A sequence (xn) in a topological space(X, T) is said to con-verge to a pointxX iff (xn) eventually belongs to every nhd ofx i.e. iff forevery nhdU ofx, there existsn0Nsuch thatxnU for allnn0.CautionWe learnt that, for metric spaces, sequential convergence was adequate to de-scribe the topology of such spaces (in the sense that the basic primitives ofopen set, neighbourhood, closure etc. could be fully characterised in terms

of sequential convergence). However, for general topological spaces, sequentialconvergence fails. We illustrate:

(i) Limits arenotalways unique. For example, in (X, T0),eachsequence (xn)converges to everyxX.

(ii) In R with the cocountable topologyL, [0, 1] is not closed and so G =(, 0) (1, ) is not open yet ifxnx wherexG, then Assign-ment 1 shows thatxnG for all sufficiently largen. Further, 2[0, 1]L,yet no sequence in [0, 1] can approach 2. So another characterisation failsto carry over from metric space theory.

Finally, everyL-convergent sequence of points in [0, 1] must have its limitin[0, 1] but [0, 1] isnotclosed (in

L)!

Hence, to discuss topological convergence thoroughly, we need to develop a newbasic set-theoretic tool which generalises the notion of sequence. It is called anet we shall return to this later.

Definition 1.11 A topological space(X, T)is calledmetrizableiff there existsa metricdonXsuch that the topologyTdinduced bydcoincides with the originaltopologyT onX.

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The investigations above show that (X, T0) and (R, L) are examples of non-metrizable spaces. However, the discrete space (X, D) is metrizable, beinginduced by the discrete metric

d(x, y) =

1 ifx=y0 ifx = y

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Chapter 2

Topological Properties

We explained in the previous chapter what a topological property (homeomor-phic invariant) is but gave few good examples. We now explore some of themost important ones. Recurring themes will be:

When do subspaces inherit the property? How do continuous maps relate to the property? Does the property behave specially in metric spaces?

2.1 Compactness

We all recall the important and useful theorem from calculus, that functionswhich are continuous on a closed and bounded interval take on a maximum andminimum value on that interval. The classic theorem of Heine-Borel-Lebesgueasserts that every covering of such an interval by open sets has a finite subcover.In this section, we use this feature of closed and bounded subsets to define thecorresponding notion, compactness, in a general topological space. In addition,we consider important variants of this notion: sequential compactness and localcompactness.

2.1.1 Compactness Defined

Given a set X with AX, a cover for A is a family of subsetsU ={Ui :I}ofXsuch that A

iIUi. A subcoverof a given cover

U forA is a subfamily

V Uwhich still forms a cover for A.IfA is a subspace of a space (X, T),U is an open cover for A iffU is a coverforA such that each member ofUis open in X.The classic theorem of Heine-Borel-Lebesgue asserts that, in R, every open coverof a closed bounded subset has a finite subcover. This theorem has extraordi-narily profound consequences and like most good theorems, its conclusion hasbecome a definition.

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Definition 2.1 (X, T) is said to becompact iff every open cover ofXhas afinite subcover.Theorem 2.1 (Alexanders Subbase Theorem) LetS be any subbase for(X, T). If every open cover of X by members ofShas a finite subcover, thenX is compact.

The proof of this deep result is an application of Zorns lemma, and is not anexercise for the faint-hearted!

Examples

(i) (R,I) is not compact, for considerU ={(n, n) : n N}. Similarly,(C,

Tusual) is not compact.

(ii) (0, 1) isnot compact, for considerU={( 1n

, 1) : n2}.

(iii) (X, C)iscompact, for any X.

(iv) GivenxX, (X, E(x)) is compact; (X,I(x)) is not compact unless X isfinite.

(v)T finite for any X(X, T) compact.

(vi) Xfinite,T any topology for X(X, T) compact.

(vii) X infinite(X, D) not compact.

(viii) Given (X, T), if (xn) is a sequence in Xconvergent to x, then{xn :nN} {x} is compact.

2.1.2 Compactness for Subspaces

We call a subset A of (X,

T) a compact subsetwhen the subspace (A,

TA) is a

compact space. Its a nuisance to have to look atTA in order to decide on this.It would be easier to use the originalT. Thankfully, we can!Lemma 2.1 A is a compact subset of (X, T) iff everyT-open cover ofA hasa finite subcover.

Proof Exercise.

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Lemma 2.2 Compactness is closed-hereditary and preserved by continuous maps.

Proof Exercise.ExampleThe unit circle in R2 is compact; indeed, paths in any space are compact.

2.1.3 Compactness in Metric Spaces

In any metric space (M, d), every compact subset Kis closed and bounded:(bounded, since given any x0M,

K B(x0, 1) B(x0, 2) B(x0, 3) K ji=1B(x0, ni)

where we can arrange n1 < n2 < . . . < nj. Thus K

B(x0, nj) and so anytwo points ofK lie within nj ofx0 and hence within 2nj of each other i.e. K isbounded.Kis closed, since ifx Kand xK, then for eachyK,dy = 12d(x, y)> 0 sowe may form the (open) cover ofKas follows:{B(y, dy) : yK} which reducesto a finite subcover{B(yi, dyi) : yi K, i = 1, . . . , n}. The correspondingneighbourhoods ofx, namely B(x, dyi),i= 1, . . . , n, may be intersected givinga neighbourhood ofx which misses Kcontradiction!)Neither half is valid in al l topological spaces;

compact bounded doesnt even make sense since bounded dependson the metric.

compact closed makes sense but is not always true. For example,in (R, C), (0, 1) is not closed yet it is compact (since its topology is thecofinite topology!)

Further, in a metric space, a closed bounded subset neednt be compact (e.g.consider M with the discrete metric and let A M be infinite; then A isclosed, bounded (since AB(x, 2) = M for any xM), yet it is certainly notcompact! Alternatively, the subspace (0, 1) is closed (in itself), bounded, butnot compact.)However, the Heine-Borel theorem asserts that such is the case for R and Rn;the following is a special case of the theorem:

Theorem 2.2 Every closed, bounded interval [a, b] inR is compact.

Proof LetUbe any open cover of [a, b] and let K={x[a, b] : [a, x] is coveredby a finite subfamily ofU}. Note that ifx K and a y x, then y K.Clearly, K= since aK. Moreover, givenxK, there exists x > 0 suchthat [x, x+x) K (since x some open U chosen finite subcover ofU).SinceKis bounded, k = sup Kexists.

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(i) k K: Choose U U such that k U; then there exists > 0 such that(k

, k]

U. Since there exists x

K such that k

< x < k,

k K.(ii) k =b : Ifk < b, choose U U with k Uand note that [k, k + )U

for some >0 contradiction!NoteAn alternative proof [Willard, Page 116] is to invoke the connected natureof [a, b] by showing Kis clopen in [a, b].

Theorem 2.3 Any continuous map from a compact space into a metric spaceis bounded.

Proof Immediate.

Corollary 2.1 If (X, T) is compact and f : X R is continuous, then f isbounded and attains its bounds.

Proof Clearly,fis bounded. Letm= sup f(X) andl = inff(X); we must provethat m f(X) and l f(X). Suppose thatm f(X). Sincef(X) = f(X),then there exists >0 such that (m , m+) f(X) = i.e. for all xX,f(x)m . . .contra!Similarly, if l f(X), then there exists > 0 such that [l, l+ )f(X) =whencel + is a lower bound for f(X)!

2.1.4 Sequential Compactness

Definition 2.2 A topological space(X, T)is said to besequentially compactif and only if every sequence inXhas a convergent subsequence.

Recall from Chapter 1 the definition of convergence of sequences in topologicalspaces and the cautionary remarks accompanying it. There we noted that,contrary to the metric space situation, sequences in topology can have severaldifferentlimits! Consider, for example, (X, T0) and (R, L). In the latter space,if xn l, then xn = l for all n some n0. Thus the sequence 1, 12 , 14 , 18 , . . .does notconverge in (R, L)!Lemma 2.3 Sequential compactness is closed-hereditary and preserved by con-tinuous maps.

Proof Exercise.We shall prove in the next section that in metric spaces, sequential compactnessand compactness are equivalent!

Definition 2.3 Given a topological space (X, T), a subsetA ofX andxX,x is said to be an accumulation point of A iff every neighbourhood of xcontains infinitely many points ofA.

Lemma 2.4 Given a compact space(X, T)with an infinite subsetA ofX, thenA has an accumulation point.

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Proof Suppose not; then for each xX, there exists a neighbourhood Nx ofxsuch that NxA is (at most) finite; the family{Nx :xX} is an open coverofX and so has a finite subcover{Nxi : i = 1, . . . , n}. But A X and A isinfinite, whence

A= A X=A (Nxi) =ni=1(A Nxi)is finite!

Lemma 2.5 Given a sequentially compactmetricspace(M, d)and >0, thereis afinite number of open balls, radius, which coverM.

Proof Suppose not and that for some > 0, there exists no finite family ofopen balls, radius , covering M. We derive a contradiction by constructing asequence (xn) inductively such that d(xm, xn)

for alln,m (n

=m), whence

no subsequence is even Cauchy!Letx1Mand suppose inductively thatx1, . . . , xkhave been chosen inMsuchthatd(xi, xj) for alli, jk, i=j .By hypothesis,{B(xi, ) : i = 1, . . . , k}isnot an (open) cover ofMand so there existsxk+1Msuch thatd(xk+1, xi)for 1ik. We thus construct the required sequence (xn), which clearly hasno convergent subsequence.

Theorem 2.4 A metric space is compact iff it is sequentially compact.

Proof

: Suppose (M, d) is compact. Given any sequence (xn) in M, either A =

{x1, x2, . . .

} is finite or it is infinite. IfA is finite, there must be at least

one point l in A which occurs infinitely often in the sequence and itsoccurrences form a subsequence converging to l. IfA is infinite, then bythe previous lemma there exists xXsuch that every neighbourhood ofx contains infinitely many points ofA.

For each k , B(x, 1k

) contains infinitely many xns: select one, call itxnk , making sure that nk > nk1 > nk2 . . .. We have a subsequence(xn1 , xn2 , . . . , xnk , . . .) so that d(x, xnk) 0, > 0 such thatd1(x, y)< for x, yX1d2(f(x), f(y))< .Ordinary continuity offis a local property, while uniform continuity is a globalproperty since it says something about the behaviour offover the whole spaceX1. Since compactness allows us to pass from the local to the global, the nextresult is not surprising:

Theorem 2.5 If(X, d)is a compact metric space andf :XR is continuous,thenf is uniformly continuous onX.NoteResult holds for any metric space codomain.Proof Let > 0; since f is continuous, for each x X,x > 0 such thatd(x, y) < 2x |f(x) f(y)| < 2 . The family{Bx(x) : x X} is an opencover of X and so has a finite subcover{Bxi (xi) : i = 1, . . . , n} of X. Let = min{xi : i = 1, . . . , n}; then, given x, y X such that d(x, y) < , itfollows that|f(x) f(y)|< (for xBxi (xi) for some i, whence d(x, xi) < xi and so d(y, xi)d(y, x) +d(x, xi)< + xi2xi |f(y) f(xi)|< 2 .Thus|f(x) f(y)| |f(x) f(xi)| + |f(xi) f(y)|< 2 + 2 =).NoteCompactness is not a necessary condition on the domain for uniform con-

tinuity. For example, foranymetric space (X, d), letf :XXbe the identitymap. Then fis easily seen to be uniformly continuous on X.

2.1.6 Local Compactness

Definition 2.4 A topological space (X, T) is locally compact iff each pointofXhas a compact neighbourhood.

Clearly, every compact space is locally compact. However, the converse is nottrue.Examples

(i) With Xinfinite, the discrete space (X, D) is clearly locally compact (foreachxX,{x} is a compact neighbourhood ofx!) butnotcompact.

(ii) With X infinite and x X, (X,I(x)) is locally compact (but not com-pact).

(iii) (R,I) is locally compact (xR[x 1, x + 1] is a compact neighbour-hood ofx).

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(iv) The set of rational numbersQ with its usual topology is nota locally com-

pact space, for suppose otherwise; then 0 has a compact neighbourhoodC in Q so we can choose >0 such that J =Q [, ]C. Now J isclosed in (compact) C and is therefore compact in R. Thus, J must beclosed in Rbut JR = [, ]!

Lemma 2.6 (i) Local compactness is closed-hereditary.

(ii) Local compactness is preserved bycontinuous open maps it is not pre-served by continuous maps in general. Consider idQ : (Q, D) (Q,IQ)which is continuous and onto; (Q, D)is locally compact while(Q,IQ)isnt!

Proof Exercise.

2.2 Other Covering ConditionsDefinition 2.5 A topological space(X, T) is said to be

(i) Lindelofiff every open cover ofXhas a countable subcover

(ii) countably compact iff every countable open cover of X has a finitesubcover.

Thus, a space is compact precisely when it is both Lindel of and countably com-pact. Further, every sequentially compact space is countably compact, althoughthe converse is not true. Moreover, sequential compactness neither implies noris implied by compactness.However, for metric spaces, or more generally, metrizable spaces, the conditionscompact, countably compact and sequentially compact are equivalent.Note Second countable separable; separable + metrizable second count-able . . . and so in metrizable spaces, second countability and separability areequivalent.

2.3 Connectedness

It is not terribly hard to know when a set on the real line is connected, or ofjust one piece. This notion is extended to general topological spaces in thissection and alternative characterizations of the notion are given. In additionthe relationship between continuous maps and and connectedness is given. Thisprovides an elegant restatement of the familiar Intermediate Value Theoremfrom first term calculus.

2.3.1 Definition of Connectedness

Apartitionof (X, T) means a pair of disjoint, non-empty,T-open subsets whoseunion is X. Notice that, since these sets are complements of one another, theyare both closed as well as both open. Indeed, the definition of partition is notaffected by replacing the term open by closed.

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Definition 2.6 Aconnectedspace(X, T)is one which hasnopartition. (Oth-erwise, (X, T) is said to bedisconnected.)If = A (X, T), we call A a connected set in X whenever (A, TA) is aconnected space.

Lemma 2.7 (X, T) is connected iff X and are the only subsets which areclopen.

Examples

(i) (X, T0) is connected.

(ii) (X, D) cannot be connected unless|X| = 1. (Indeed the only connectedsubsets are the singletons!)

(iii) The Sorgenfrey lineRs is disconnected (for [x, ) is clopen!).

(iv) The subspaceQ of (R,I) is not connected because

Q [

2,

2] closed in Q

=Q (

2,

2) open in Q

is clopen and is neither universal nor empty.

(v) (X, C) is connected except when Xis finite; indeed, every infinite subsetofX is connected.

(vi) (X, L) is connected except when X is countable; indeed, every uncount-able subset ofX is connected.

(vii) In (R,I), A R is connected iff A is an interval. (Thus, subspaces ofconnected spaces arenotusually connected examples abound in (R,I).)

2.3.2 Characterizations of Connectedness

Lemma 2.8 A(X, T) isnot connected iff there existT-open setsG, Hsuch thatAG H, A G=, A H=andA G H =. (Again, wecan replace open by closed here.)

Proof Exercise.NoteBy an interval in R, we mean any subset I such that whenever a < b < cand whenever aI and cI then bI. It is routine to check that the only

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ones are (a, b), [a, b], [a, b), (a, b], [a, ), (a, ), (, b), (, b], (, ) = Rand{a} for real a, b, a < bwhere appropriate.It turns out that these are exactly the connected subsets of (R,I):-Lemma 2.9 InR, if [a, b] =F1 F2 whereF1, F2 are both closed andaF1,bF2 thenF1F2=.Proof Exercise.

Theorem 2.6 Let I(R,I). ThenI is connected iffI is an interval.Proof

: IfI is notan interval, then there exist a < b < c with aI, b I andc I. Take A = I(, b) and B = I(b, ). Then AB = I,A

B=

, A

=

, B

=

, A

I, B

I and A, B are both open in I i.e.

Aand B partition Iand so I is not connected.

: Suppose I is not connected and that I is an interval. By the closedversion of Lemma 2.8, there exist closed subsets K1, K2 ofR such thatI K1K2, IK1=, IK2= and IK1K2 =. SelectaI K1, bI K2; without loss of generality, a < b. Then [a, b]Iso that [a, b] = ([a, b]K1)([a, b]K2), whence by Lemma 2.9, =[a, b] K1K2I K1 K2=!

2.3.3 Connectedness and Continuous Maps

Lemma 2.10 Connectedness is preserved by continuous maps.

Proof Exercise.

Corollary 2.2 (Intermediate Value Theorem) If f : [a, b]R is contin-uous andf(a)< y < f(b), theny must be a value off.

Proof Exercise.

Corollary 2.3 (Fixed point theorem for [0, 1]) Iff : [0, 1] [0, 1] is con-tinuous, then it has a fixed point i.e. there exists some x [0, 1] such thatf(x) = x.

Proof Consider g(x) = f(x) x. Theng : [0, 1] R is continuous. Further,g(0) =f(0)0 and g(1) =f(1) 10 so that 0 is intermediate between g (0)andg(1). Thus, by the Intermediate Value Theorem, there existsx

[0, 1] such

that 0 =g(x) = f(x) x i.e. such that f(x) = x.NoteGiven continuoush : [a, b][a, b], it follows thath has a fixed point since[a, b]=[0, 1] and every continuous function has a fixed point is a homeomorphicinvariant.

Lemma 2.11 Let (X, T) be disconnected with YX, Y clopen. If A isany connected subset ofX, thenAY orAX\ Y.

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Proof IfA Y= =A (X\ Y), then A YA and A Y is clopen inA. Thus,A is notconnected! It follows that either A Y =or A X\ Y =i.e. either AX\ Y orAY.Lemma 2.12 If the family{Ai :iI} of connected subsets of a space(X, T)has a non-empty intersection, then its unioniIAi is connected.Proof Suppose not and that there exists a non-empty proper clopen subsetY ofiIAi. Then for each iI, either AiY orAi iIAi\ Y. However if forsome j, Aj Y, then AiY for each iI (sinceiIAi=) which impliesthatiIAiY!Similarly, if for some kI, Ak iIAi\Y, theniIAi iIAi\Y!Corollary 2.4 Given a family{Ci : i I} of connected subsets of a space(X,

T), ifB

Xis also connected andB

Ci

=

for al li

I, thenB

(

iICi)

is connected.

Proof Take Ai= B Ci in Lemma 2.12.Lemma 2.13 IfA is a connected subset of a space (X, T) and AB AT,thenB is a connected subset.

Proof If B is notconnected, then there exists Y B which is clopen inB. By Lemma 2.11, eitherA Y or A B\ Y. SupposeA Y (a similarargument suffices for AB \ Y); then AT YT and soB \ Y =B (B \ Y) =ATB (B\ Y)YTB (B\Y) = Y (B\ Y) = a contradiction!Definition 2.7 Let (X, T) be a topological space with x X; we define thecomponent ofx, Cx, in (X, T) to be the union of all connected subsets ofXwhich containx i.e.

Cx={AX :xA andA is connected}.For eachxX, it follows from Lemma 2.12 thatCxis themaximumconnectedsubset ofXwhich contains x. Also it is clear that ifx, yX, either Cx = CyorCx Cy =(for ifzCx Cy, thenCx CyCzCx Cy whenceCx=Cy(=Cz)). Thus we may speak ofthecomponents of a space (X, T) (withoutreference to specific points ofX): they partition the space into connected closedsubsets (by Lemma 2.13) and are precisely the maximalconnected subsets ofX.Examples

(i) If (X, T) is connected, (X, T) has only one component, namely X!(ii) For any discrete space, the components are the singletons.(iii) In Q (with its usual topology), the components are the singletons. (Thus,components need not be open.)

Definition 2.8 A space(X, T)istotally disconnectediff the only connectedsubsets ofX are the singletons (equivalently, the components of(X, T) are thesingletons).

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Thus, by the previous examples, we see that the space Q of rationals, the space

R \Qof irrationals and any discrete space are all totally disconnected. Further,the Sorgenfrey line Rs is totally disconnected.

2.3.4 Pathwise Connectedness

Definition 2.9 A topological space(X, T)ispathwise connected iff for anyx, yX, there exists a continuous functionf : [0, 1]Xsuch thatf(0) =xandf(1) =y. Such a functionf is called apath fromx to y.

Theorem 2.7 Every pathwise connected space is connected.

Proof Let (X, T) be pathwise connected and let aX; for every xX, thereexists a path px : [0, 1] X from a to x. Then, for each x X, px([0, 1])is connected; moreover, pa(0) = a

xXpx([0, 1]) so that by Lemma 2.12,

X=xXpx([0, 1]) is connected.Note wellThe converse is false. Consider the following example,the topologistssine curve:

V ={(x, 0) : x0} {(x, sin1x

) : x >0}is a connected space, but no path can be found from (0, 0) to any point (x, sin 1

x)

withx >0(for suppose, w.l.o.g., there exists a path p: [0, 1]X with p(0) = ( 1

, 0) and

p(1) = (0, 0). Then 1p, being continuous, must take all values between 0and 1

, in particular 1

(2n+12)

for each n i.e. there exists tn [0, 1] such that1 p(tn) = 1(2n+1

2)

for each n. Thus, p(tn) = ( 1

(2n+ 12)

, 1)(0, 1) asn .

Now tn [0, 1] for all n which implies that there exists a subsequence (tnk)in [0, 1] with tnk . Then p(tnk) p() and so 1p(tnk) 0. Thusp() = (0, y) for some y , whence y = 0 (sincep()X)!)

2.4 Separability

Definition 2.10 A topological space is said to be

(i) separable iff it has a countable dense subset.

(ii) completely separable(equivalently,second countable) iff it has a count-able base.

Examples(i) (R,I) is separable (since Q= R).

(ii) (X, C) is separable for any X.(iii) (R, L) is notseparable.

Theorem 2.8 (i) Complete separability implies separability.

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(ii) The converse is true in metric spaces.

Proof We prove only (ii). In metric space (M, d), letD ={x1, x2, . . .}be dense.ConsiderB ={B(xi, q) : i , q Q, q > 0}, a countable collection of opensets. One can show thatBis a base forTd . . . over to you!Theorem 2.9 (i) Complete separability is hereditary.

(ii) Separability is not hereditary. (Consider the included point topologyI(0)onR; then(R,I(0)) is separable, since{0}= R. However, R \ {0} isnotseparable because it is discrete.)

ExampleSeparability does notimply complete separability since, for example, (R,I(0))is separable butnotcompletely separable.(Suppose there exists a countable base

B for its topology. Given x= 0,{0, x} is an open neighbourhood ofx and sothere exists Bx B such that x Bx {0, x}.Thus Bx ={0, x} i.e. B isuncountable . . . contradiction!

Theorem 2.10 Separability is preserved by continuous maps.

Proof Exercise.NoteComplete separability is notpreserved by continuous maps.

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Chapter 3

Convergence

In Chapter 1, we defined limits of sequences in a topological space (X, T) so asto assimilate the metric definition. We noted, however, that not everything weknew about this idea in metric spaces is valid in topological spaces.We will examine two main ways around this difficulty:

develop a kind of super-sequence or netwhich does for general topologywhat ordinary sequences do for metric spaces.

identify the class of topological spaces in which the old idea of sequentiallimit is good enough.

3.1 The Failure of SequencesThe following important results are probably familiar to us in the context ofmetric spaces, or at least in the setting of the real line, R.

Theorem 3.1 Given (X, T), AX, pX: if there exists some sequence ofpoints ofA tending to p, thenp A.Theorem 3.2 Given (X, T), AX: ifA is closed, thenA includes the limitof every convergent sequence of points ofA.

Theorem 3.3 Givenf : (X, T)(Y, T): iffis continuous, thenf preserveslimits of sequences i.e. wheneverxnl inX, thenf(xn)f(l) inY.In each case above, it is routine to prove the statement true in a general topo-logical space as asserted. We illustrate by proving Theorem 3.3:Let f be continuous and xn l in X. We must show that f(xn) f(l).Given a neighbourhood N off(l), there exists openG such that f(l)GN.Then l f1(G) f1(N) i.e. f1(N) is a neighbourhood of l and so xnf1(N)nn0 say. Thusf(xn)Nnn0, whence f(xn)f(l).

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In metric spaces, the converses of these results are also true but our main point

here is that in general topology, the converses are not valid.ExampleIn (R, L), (0, 1) = R. So, for example, 5(0, 1) and yet the only way a sequence(xn) converges to a limit l is for xn= l from some stage on. So no sequence in(0, 1) can converge to 5 proving that the converse of Theorem 3.1 is false.Continuing, the limit of any convergent sequence in (0, 1) must belong to (0, 1)for the same reason and yet (0, 1) is not closed. Thus, Theorem 3.2s converseis false.Further,idR : (R, L)(R,I) is notcontinuous and yet it does preserve limitsof sequences.Now this is a great nuisance! Sequences are of immense usefulness in real analysisand in metric spaces and elsewhere and their failure to describe generaltopology adequately is a technical handicap. What to do?

3.2 Nets - A Kind of Super-Sequence

Recall that a sequence is just a function having the positive integers as do-main. The set of positive integers, of course, possesses a particularly simpleordering; there is a first member, second member, third member, etc. Not allsets are naturally endowed with so simple an ordering. For example, dictionary(lexographical) ordering of words is more complex (though still relative nice asorderings go). By replacing the domain of positive integers with a set having amore complicated ordering we will:

define a net (in analogy with sequence),

identify an associated notion of convergence, show that net convergence is sufficient to characterize closure of sets, and that compactness can be characterized in terms of convergence of

subnets.

Note that these last two items generalize the role of sequences in a metric space.

3.2.1 Definition of Net

Definition 3.1 A binary relation on a setP is said to be apre-order iff

(i) pp pP(ii) pq andqr implypr p, q, rP.

We often refer to Pas being a pre-ordered setwhen it is understood that isthe pre-order in question.If it is also true that for p, qP,

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(iii) p q and q p implyp = q, P is said to be a partially ordered set(or poset).

Definition 3.2 A pre-ordered setP is said to bedirected (or updirected) iffeach pair of members ofPhas an upperbound.

(i.e. ifp, qP, then there exists sP such that ps, qs.)Definition 3.3 Let(P, ) be a poset. Then ifx, yP withxy andyx,we writexy and say thatx andy areincomparable;If E P, then E is said to betotally unordered (or diverse) iff x, y Eimpliesx= y orxy.IfCP, thenCis said to belinear(or achainor atotal order) iffx, yCimpliesx < y , x= y ory < x.(P,

) is said to be alattice iff each pair of members ofPhas a greatest lower

bound and a least upper bound.A lattice (P, ) is said to becomplete iff every non-empty subsetY ofP hasa greatest lower bound (Y) and a least upper bound (Y).An element v of a poset (P, ) is said to be maximal ( minimal) iff v x(xv), xP v = x.Definition 3.4 A net in a (non-empty set) X is any function x : A Xwhose domainA is a directed set.

In imitation of the familiar notation in sequences, we usually write the net valuex() as x. A typical net x : A Xwill usually appear as (x, A) or(x)A or some such notation.

Examples of Nets

(i) N,Z,NNare all directed sets, where suitable pre-orders are respectivelythe usual magnitude ordering for N and Z, and (i, j)(m, n) iffimand j n, in N N. Thus, for example, a sequence is an example of anet.

(ii) The real function f : R \ {0} R given by f(x) = 3 1x

is a net, sinceits domain is a chain. Anyreal function is a net.

(iii) Given x(X, T), select in any fashion an element xN from each neigh-bourhoodNofx; then (xN)NNx is a net inX( since it defines a mappingfrom (

Nx,

) into X). Recall that

Nx is ordered byinverse set inclusion!

3.2.2 Net Convergence

Definition 3.5 A net (x)A in (X, T) converges to a limit l if for eachneighbourhood N of l, there exists some N A such that x N for allN.

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In such a case, we sometimes say that the net ( x)A

eventuates N. Clearly, this definition incorporates the old definition of limit ofa sequence. The limit of the net fdescribed in (ii) above is 3. In (iii), the netdescribed converges to x no matter how the values xNare chosen . . . prove!

3.2.3 Net Convergence and Closure

Our claim is that nets fully describe the structure of a topological space. Ourfirst piece of evidence to support this is that with nets, instead of sequences,Theorems 3.1, 3.2 and 3.3 have workable converses:

Theorem 3.4 Given(X, T), AX, pX: p A iff there exists a net inAconverging to p.

Proof If some net of points of A converges to p, then every neighbourhoodof p contains points of A (namely, values of the net) and so we get p A.Conversely, ifp is a closure point ofA then, for each neighbourhood N ofp, itwill be possible to choose an element aN ofA that belongs also to N. The netwhich these choices constitute converges to p, as required.

Theorem 3.5 Given(X, T), AX, A is closed iff it contains every limit ofevery (convergent) net of its own points.

Proof This is really just a corollary of the preceding theorem.

Theorem 3.6 Givenf : (X, T)(Y, T), f is continuous ifff preserves netconvergence.

Proof Exercise.

3.2.4 Nets and Compactness

Definition 3.6 Let (x)A be any net and let 0 A. The th0 tail of thenet is the set{x: 0}= x([0, )). We denote it byx(0).Definition 3.7 Let(x)A and(y)B be any two nets. We call(y)B asubnet of (x)A provided that every tail of (x) contains a tail of (y) i.e.provided:

0A 0B such thatx(0)y(0).We expected a definition like subsequence to turn up here and we are disap-pointed that it has to be so complicated.Net theory ceases to be a straightforward generalisation of sequence theoryprecisely when we have to take a subnet . . . so well try to avoid this wheneverpossible! There is however one result certainly worth knowing:

Theorem 3.7 (X, T) is compact iff inX, every net has (at least one) conver-gent subnet.

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(So, for example, (n) is a net in R with no convergent subnet.)

Proof Not required.Corollary 3.1 Compactness is closed-hereditary

Proof (for if (x) is a net in a closed setF X, then it has a convergent subnet(y) inX. Thus there exists a subnet (z) of (y) inFwhich converges in X,whence its limit is in F).

Corollary 3.2 Compactness is preserved by continuous maps

Proof (for if X is compact and f continuous, let (y)A be a net in f(X).Then for each A, y = f(x) for some x X. The net (x)A has aconvergent subnet (z)B, sayzl, whencef(z)f(l). Then (f(z))Bis a convergent subnet of (y)A).ExampleIf (xnk) is a subsequence of a sequence (xn), then it is a subnet of it also; because

theith0 tail of the sequence (xn) is

{xi0 , xi0+1, xi0+2, . . .} ()

while the ith0 tail of the subsequence (xnk) is:

{xni0 , xni0+1, xin0+2, . . .} ()and we see that ()() merely becauseni0i0.Lemma 3.1 If a net(x) converges to a limit l, then so do all its subnets.

Proof Let (y) be a subnet of (x); let Nbe a neighbourhood ofl . Then thereexists 0 such that x N for all 0. Further, there exists0 such that{y :0} {x : 0}and so yN for all 0.

3.3 First Countable Spaces - Where SequencesSuffice

Why do sequences suffice to describe structure in R,Cand other metric spacesbut not in many other topological spaces? The key here is recognizing that manyproofs regarding convergence in metric spaces involve constructing sequences

of nested open sets about a point. Sometimes these describe the topologicalstructure near the point and other times not. In what follows we

identify the local characteristic of topological space that makes theseproofs work,

and prove that sequences suffice to describe the topological structure ofspaces with this characteristic.

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3.3.1 First Countable Spaces

So what characteristic common to R,Cand other metric spaces makes sequencesso good at describing their structure?

Definition 3.8 Let x (X, T). A countable neighbourhood base at xmeans: a sequenceN1, N2, N3, . . . of particular neighbourhoods ofx such thatevery neighbourhood ofx shall contain one of theNis.

Note that we may assume that N1 N2 N3 because, if not, then wecan work with N1, N1N2, N1N2 N3, . . .Definition 3.9 We call (X, T) first-countable when every point inX has acountable neighbourhood base.

ExampleThe classic example of a first-countable space is any metric (or metriz-able) space because ifx(M, d), then B (x, 1),B (x, 12),B (x, 13), . . . is a count-able neighbourhood base at x.

Theorem 3.8 First-countability is hereditary and preserved by continuous openonto maps.

Proof Left to the reader.

Theorem 3.9 (i) Complete separability implies first countability.

(ii) Converse not always true.

(iii) Converse valid on a countable underlying set.

Proof

(i) IfB is a countable base for (X, T) and pX, consider{B B :pB}which is a countable family of neighbourhoods ofp. Moreover, they forma neighbourhood base at p.

(ii) An uncountable discrete space is first countable ( since metrizable), yet isnot completely separable.

(iii) SupposeXcountable and (X, T) first countable. For eachxX, choosea countable neighbourhood base: N(x, 1),N(x, 2),N(x, 3), .... Each is aneighbourhood ofx and so contains an open neighbourhood ofx: G(x, 1),G(x, 2),G(x, 3), . . . .

ThenB ={G(x, n) : n N, x X} is a countable family of open setsand is a base for (X, T). Thus, (X, T) is completely separable.

ExampleThe Arens-Fort space (see, for example, Steen and Seebach, Counterexamplesin Topology is not first-countable because otherwise it would be completelyseparable which is false!

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3.3.2 Power of Sequences in First Countable Spaces

The following three results illustrate that sequences suffice for first-countablespaces in the sense that we dont need to use nets to describe their structure.This is why sequences are sufficiently general to describe, fully, metric andmetrizable spaces.

Theorem 3.10 Given a first-countable space(X, T)(i) p X, A X, then p A iff there exists a sequence of points of A

converging to p.

(ii) AXis closed iffA contains every limit of every convergent sequence ofits own points.

(iii) f : (X,T

)

(Y,T

) is continuous iff it preserves limits of (convergent)sequences.

Proof

(i) Theorem 3.1 said that if there exists a sequence in A converging to somepX, then p A.Conversely, ifp A, thenp has a countable base of neighbourhoods N1N2 N3 , each of which must intersect A. So choosexj Nj Afor all j1. Then (xj) is a sequence in A and, given any neighbourhoodH ofp, H must contain one of the Nj s i.e. H Nj0 Nj0+1 sothatxjH for all jj0. That is, xjp.

(ii) Corollary of (i).(iii) fcontinuous implies that it must preserve limits of sequences (by Theo-

rem 3.3). Conversely, iff is not continuous, there existsAXsuch thatf(A)f(A). Thus, there existspf(A)\f(A) so p = f(x), somex A.So there exists a sequence (xn) in A withxnx.Yet, iff(xn)f(x)(=p),p would be the limit of a sequence in f(A) sothat p f(A) contradiction! Thus f fails to preserve convergence ofthis sequence.

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Chapter 4

Product Spaces

A common task in topology is to construct new topological spaces from otherspaces. One way of doing this is by taking products. All are familiar withidentifying the plane or 3-dimensional Euclidean space with ordered pairs ortriples of numbers each of which is a member of the real line. Fewer are probablyfamilar with realizing the torus as ordered pairs of complex numbers of modulusone. In this chaper we answer two questions:

How do the above product constructions generalize to topological spaces? What topological properties are preserved by this construction?

4.1 Constructing ProductsThe process of constructing a product falls naturally into two stages.

The first stage, which is entirely set-theoretic, consists in describing anelement of the underlying set of the product. This task is primarily oneof generalizing the notion of ordered pair or triple.

The second stage is describing what open sets look like. This will be doneby describing a subbasis for the topology. The guiding goal is to providejust enough opens sets to guarantee the continuity of certain importantfunctions.

4.1.1 Set-Theoretic ConstructionSuppose throughout that we are given a family of topological spaces{(Xi, Ti) :iI} whereIis some non-empty labelling or index set.Our first task is to get a clear mental picture of what we mean by the productof the setsXi. Look again at the finite case where I={1, 2, . . . , n}. Here, the

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product set

X=X1X2 X3 . . . Xn =ni=1

Xi={(p1, p2, . . . , pn) : piXi, iI}.

i.e. the elements ofXare the functions x : I ni=1Xi such that x(1)X1,x(2)X2, . . . , x(n)Xn i.e. x(i)Xii where, for convenience, we usuallywritexi instead ofx(i). In this form, the definition extends immediately to anyI, finite or infinite i.e. if{Xi: iI}is anyfamily of sets, then their product is

{x: I iIXi for which x(i)Xi iI}except that we normally write xi rather than x(i).Then a typical element of X =

Xi will look like: (xi)iI or just (xi). We

will still call xi theith coordinate of (xi)iI. ( Note that the Axiom of Choiceassures us that

Xi is non-empty provided none of the Xis are empty.)

4.1.2 Topologizing the Product

Of the many possible topologies that could be imposed on X =

Xi, wedescribe the most useful. This topology is just right in the sense that it is barelyfine enough to guarantee the continuity of the coordinate projection functionswhile being just course enough allow the important result of Theorem 4.1.

Definition 4.1 For eachiI, theith projectionis the map i :

XiXiwhich selects theith coordinate i.e. i((xi)iI) =xi.

An open cylindermeans the inverse projection of some non-emptyTi-open seti.e. 1i (Gi) whereiI, Gi=, Gi Ti.An open box is the intersection of finitely many open cylindersnj=11ij (Gij ).The only drawable case I={1, 2} may help explain:(Here will be, eventually, a picture!)We use these open cylinders and boxes to generate a topology with just enoughopen sets to guarantee that projection maps will be continuous. Note that theopen cylinders form a subbase for a certain topologyT on X =Xi andtherefore the open boxes form a base forT;T is called the [Tychonoff]producttopology and (X, T) is the product of the given family of spaces. We write(X, T) = {(Xi, Ti) :iI}= iI(Xi, Ti) or evenT = iITi.Notice that ifnj=11ij (Gij ) is any open box, then without loss of generality wecan assume i1, i2, . . . in all different because if there were repetitions like

. . . 1ik (G) 1ik (H) . . .we can replace each by

. . . 1ik (G H) . . .and thus eliminate all repetitions.

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It is routine to check that ifTn is the usual topology on Rn, andT the usualtopology onR, then

(R, T) (R, T) . . . (R, T) = (Rn, Tn)as one would hope!

Lemma 4.1 In a product space(X, T),Nis a neighbourhood ofpXiff thereexists some open boxB such thatpBN.Lemma 4.2 For eachiI,

(i) i is continuous

(ii) i is an open mapping.

Proof

(i) Immediate.

(ii) Given open G X, then G is a union of basic open sets{Bk : k K}in X, whence i(G) is a union of open subsets{Bik :kK}ofXi and istherefore open. (The notation here is intended to convey thatBik is thecomponent along the i-th coordinate axis of the open box Bk.)

Theorem 4.1 A map into a product space is continuous iff its composite witheach projection is continuous.

Proof Since the projections are continuous, so must be their composites withany continuous map. To establish the converse, first show that ifSis a subbasefor the codomain (target) of a mapping f, then f will be continuous provided

that the preimage of every member ofSis open; now use the fact that the opencylinders constitute a subbase for the product topology.Worked exampleShow that (X, T) (Y, S) is homeomorphic to (Y, S) (X, T).Solution

Define f :X YY Xg: Y XX Y

by f(x, y) = (y, x)

g(y, x) = (x, y).Clearly these are one-one,

onto and mutually inverse. It will suffice to show that both are continuous.1f= 2;2f=

1. Now

i is continuous fori = 1, 2 and sofis continuous!Similarly, g is continuous.Worked example Show that the product of infinitely many copies of (N, D) isnot locally compact.Solution

We claim thatno point has a compact neighbourhood. Suppose otherwise; thenthere existspX,CXand GXwithCcompact,G open andpGC.Pick an open box B such that p B G C. B looks likenj=11ij (Gij ).Choose in+1I\ {i1, i2, . . . , in}; thenin+1(C) is compact (since compactnessis preserved by continuous maps).Thus, pin+1 in+1(B) = Xin+1 in+1(C) Xin+1 = (N, D). Thus,in+1(C) = (N, D) . . . which is notcompact!

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4.2 Products and Topological Properties

The topological properties possessed by a product depends, of course, on theproperties possessed by the individual factors. There are several theorems whichassert that certain topological properties are productive i.e. are possessed bythe product if enjoyed by each factor. Several of these theorems are given below.

4.2.1 Products and Connectedness

Theorem 4.2 Any product of connected spaces must be connected.

Proof is left to the reader.

4.2.2 Products and Compactness

Theorem 4.3 (Tychonoffs theorem) Any product of compact spaces is com-pact i.e. compactness is productive.

Proof It suffices to prove that any covering ofXby open cylinders has a finitesubcover. Suppose not and letC be a family of open cylinders which covers Xbut for which no finite subcover exists. For each iI, consider

{Gij :GijXi and1i (Gij ) C}.This cannot cover Xi (otherwise, Xi, being compact, would be covered byfinitely many, say Xi= Gi1 Gi2 . . . Gin , whence

X=1i (Xi) = 1i (Gi1 . . . 1i (Gin).

all inC, contrary to the choice ofCSelect, therefore, zi Xi\ {thoseGij s}; consider z = (zi)iI X. SinceC covered X, z someC C. Now C = 1k (Gk) for some k I and sok(z) = zkGk, contradicting the choice of the zis.To prove the above without Alexanders Subbase Theorem is very difficult ingeneral, but it is fairly simple in the special case where Iis finite. Several furtherresults show that various topological properties are finitely productive in thissense.

Theorem 4.4 If(X1, T1),(X2, T2), . . . ,(Xn, Tn)are finitely many sequentiallycompact spaces, then their product is sequentially compact.

Proof

Take any sequence (xn) X. The sequence (1(xn))n1 in sequentially com-pact X1 has a convergent subsequence 1(xnk) l1 X1. The sequence(2(xnk))k1in sequentially compact X2has a convergent subsequence (2(xnkj ))j1l2X2 and 1(xnkj )l1 also.Do this n times! We get a subsequence (yp)p1 of the original sequence suchthat i(yp)li for i = 1, 2, . . . , n. Its easy to check that yp (l1, l2, . . . , ln)so thatX is sequentially compact, as required.

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Lemma 4.3 The product of subspaces is a subspace of the product.

ProofLet (X, T) = iI(Xi, Ti); let YiXi for each iI. There appear to betwo different ways to topologise

Yi:

either(i) give it the subspace topology induced by Ti

or(ii) give it the product of all the individual subspace topologies (Ti)Yi .The point is that these topologies coincideif Gi0 is open in (Ti0)Yi0 wherei0 I i.e. Gi0 = Yi0 Gi0 for some Gi0 Ti0 , a typical subbasic open set for(ii) is

{(yi)

Yi: yi0Gi0}which equals

Yi {(xi)Xi: xi0Gi0 Ti0 , i0I}

=

Yi{a typical open cylinder in

Xi}which is a typical subbasic open set in (i). Hence, (i) = (ii).

Theorem 4.5 Local compactness isfinitely productive.

ProofGiven x = (x1, x2, . . . , xn) (X, T) =

ni=1(Xi, Ti), we must show that x

has a compact neighbourhood. Now, for all i = 1, . . . , n, xi has a compactneighbourhood Ciin (Xi, Ti) so we choose Ti-open set Gisuch that xiGiCi.Then

xG1G2 . . . Gn n11i (Gi)

C1C2 . . . Cn compact subset of

Xi

i.e. x has C1C2. . .Cn as a compact neighbourhood. (Note that theprevious lemma is used here, to allow us to apply Tychonoffs theorem to theproduct of the compact subspacesCi, and then to view this object as a subspaceof the full product!) Thus, Xis locally compact.

Lemma 4.4

YiT

=

YiTi ( in notation of previous lemma).

Proof Do it yourself! (The closure of a product is a product of the closures.)

4.2.3 Products and Separability

Theorem 4.6 Separability isfinitely productive.

Proof

For 1 i n, choose countable Di Xi where DiTi = Xi. Consider D =D1 D2 . . . Dn =

n1Di, again countable. Then

D=

DiT

=

DiTi =

Xi = X.Notice that the converses of all such theorems are easily true. For example,

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Theorem 4.7 If(X, T) =

iI(Xi, Ti) is(i) compact

(ii) sequentially compact

(iii) locally compact

(iv) connected

(v) separable

(vi) completely separable

then so is every factor space(Xi, Ti).Proof For each i

I, the projection mapping i :X

Xi is continuous, open

and onto. Thus, by previous results, the result follows.

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Chapter 5

Separation Axioms

We have observed instances of topological statements which, although true forall metric (and metrizable) spaces, fail for some other topological spaces. Fre-quently, the cause of failure can be traced to there being not enough opensets (in senses to be made precise). For instance, in any metric space, com-pact subsets are always closed; but not in every topological space, for the proofultimately depends on the observation

given x=y , it is possible to find disjoint open sets G and H withxG andyH

which is true in a metric space (e.g. put G = B(x, ), H = B(y, ) where= 1

2d(x, y)) but fails in, for example, a trivial space (X, T0).

What we do now is to see how demanding certain minimum levels-of-supplyof open sets gradually eliminates the more pathological topologies, leaving uswith those which behave like metric spaces to a greater or lesser extent.

5.1 T1 Spaces

Definition 5.1 A topological space (X, T) is T1 if, for each x in X,{x} isclosed.

Comment 5.1 (i) Every metrizable space isT1

(ii) (X, T0) isntT1 unless|X|= 1

Theorem 5.1 (i) T1 is hereditary

(ii) T1 is productive

(iii) T1every finite set is closed. More precisely, (X, T)isT1 iffT C, i.e.C is the weakest of all theT1 topologies that can be defined onX.

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Proof is left to the reader.

The respects in which T1-spaces are nicer than others are mostly concernedwith cluster point of a set (an idea we have avoided!). We show the equivalence,in T1 spaces, of the two forms of its definition used in analysis.

Theorem 5.2 Given aT1 space (X, T), pX andAX, the following areequivalent:

(i) Every neighbourhood ofp contains infinitely many points ofA

(ii) Every neighbourhood of p contains at least one point of A different fromp.

Proof Obviously, (i)(ii); conversely, suppose (i) fails; so there exists a neigh-bourhoodN ofp such thatN Ais finite. ConsiderH= [X\ (NA)]{p}; itis cofinite and is thus an (open) neighbourhood ofp. Hence N H is a neigh-bourhood ofp which contains no points ofA, except possiblyp itself. Thus, (ii)fails also.Hence, (i)(ii).

5.2 T2 (Hausdorff) Spaces

Definition 5.2 A topological space(X, T)isT2(orHausdorff) iff givenx=yinX, disjoint neighbourhoods ofx andy.Comment 5.2 (i) Every metrizable space isT2

(ii) T2

T1 (i.e. any T2 space is T1, for if x, y

T2X and y

{x

}, then

every neighbourhood ofy containsx, whencex= y.)

(iii) (X, C), withX infinite, cannot beT2Theorem 5.3 (i) T2 is hereditary

(ii) T2 is productive.

Proof

i The proof is left to the reader.

ii Let (X, T) = iI(Xi, Ti) be any product ofT2 spaces. Letx = (xi)iIand y = (yi)iIbe distinct elements ofX. Then there exists i0I suchthatxi0=yi0 inXi0 . Choose disjoint open sets G,H in (Xi0 , Ti0) so thatxi0 G, yi0 H. Then x 1i0 (G) T, y 1i0 (H) T and sinceG H=, 1i0 (G) 1i0 (H) =. Hence result.

TheT2 axiom is particularly valuable when exploring compactness. Part of thereason is thatT2 implies that points and compact sets can be separated off byopen sets and even implies that compact sets can be separated off from othercompact sets in the same way.

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Theorem 5.4 In a T2-space (X, T), if C is a compact set and x C, thenthere existT-open setsG andHso thatxG, CH andG H=.Proof A valuable exercise: separate each point of C from x using disjoint opensets, note that the open neighbourhoods of the various elements of C, thusobtained, make up an open covering of C, reduce it to a finite subcover byappealing to compactness . . .

Corollary 5.1 In aT2-space, any compact set is closed.

Corollary 5.2 In aT2-space, ifC andKare non-empty compact and disjoint,then there exist openG, H such thatCG, KH andG H=.A basic formal distinction between algebra and topology is that although theinverse of a one-one, onto group homomorphism [etc!] is automatically a ho-momorphism again, the inverse of a one-one, onto continuous map can fail to

be continuous. It is a consequence of Corollary 5.2 that, amongst compact T2spaces, this cannot happen.

Theorem 5.5 Let f : (X1, T1) (X2, T2) be one-one, onto and continuous,whereX1 is compact andX2 isT2. Thenf is a homeomorphism.

Proof It suffices to prove that f is closed. Given closed K X1, then K iscompact whence f(K) is compact and so f(K) is closed. Thus f is a closedmap.

Theorem 5.6 (X, T) isT2 iff no net inXhas more than one limit.Proof

(i) (ii): Letx= y in X; by hypothesis, there exist disjoint neighbourhoods U ofx, V of y. Since a net cannot eventually belong to each of two disjointsets, it is clear that no net in Xcan converge to bothx and y.

(ii) (i): Suppose that (X, T) is not Hausdorff and that x= y are points in Xfor which every neighbourhood ofx intersects every neighbourhood ofy.LetNx (Ny) be the neighbourhood systems at x (y) respectively. ThenbothNxandNy are directed by reverse inclusion. We order the CartesianproductNxNy by agreeing that

(Ux, Uy)(Vx, Vy)UxVx and UyVy.Evidently, this order is directed. For each (Ux, Uy) Nx Ny, Ux Uy= and hence we may select a point z(Ux,Uy) UxUy. If Wx is anyneighbourhood ofx,Wyanyneighbourhood ofyand (Ux, Uy)(Wx, Wy),then

z(Ux,Uy)Ux UyWxWy.That is, the net{z(Ux,Uy), (Ux, Uy) NxNy}eventually belongs to bothWx and Wy and consequently converges to bothx and y !

Corollary 5.3 Letf : (X1, T1)(X2, T2), g: (X1, T1)(X2, T2) be continu-ous whereX2 isT2. Then their agreement set is closed i.e. A={x : f(x) =g(x)} is closed.

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5.3 T3 Spaces

Definition 5.3 A space(X, T) is calledT3 orregular provided :-(i) it isT1, and

(ii) givenx closedF, there exist disjoint open setsG andHso thatxG,FH.

Comment 5.3 (i) Every metrizable space is T3; for it is certainly T1 andgivenx closedF, we havex openX\ F so there exists >0 so thatx B(x, ) X\ F. PutG = B(x,

2) and H ={y : d(x, y) >

2}; the

result now follows.

(ii) ObviouslyT3T2.(iii) One can devise examples ofT2 spaces which are notT3.

(iv) Its fairly routine to check thatT3 is productive and hereditary.

(v) Warning: Some books takeT3to mean Definition 5.3(ii) alone, and regularto mean Definition 5.3(i) and (ii); others do exactly the opposite!

5.4 T312

Spaces

Definition 5.4 A space(X, T) isT3 12

or completely regularor Tychonoffiff

(i) it isT1, and

(ii) given x X, closed non-empty F X such that x F, there existscontinuousf :X[0, 1] such thatf(F) ={0} andf(x) = 1.

Comment 5.4 (i) Every metrizable space isT3 12

(ii) Every T3 12

space is T3 ( such a space is certainly T1 and given x closedF, choose f as in the definition; define G = f1([0, 1

3)), H =

f1((23

, 1]) and observe thatT3 follows.)

(iii) Examples are known ofT3 spaces which fail to be Tychonoff

(iv) T3 12

is productive and hereditary.

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5.5 T4 Spaces

Definition 5.5 A space(X, T) isT4 ornormal if(i) it isT1, and

(ii) given disjoint non-empty closed subsets A, B of X, there exist disjointopen setsG, Hsuch thatAG, BH.

Theorem 5.7 Every metrizable space(X, T) isT4.Proof Certainly,Xis T1; choose a metricdonXsuch that T is Td. The distanceof a point p from a non-empty set Acan be defined thus:

d(p, A) = inf{d(p, a) : aA}Given disjoint non-empty closed sets A, B , let

G={x: d(x, A)< d(x, B)}H={x: d(x, B)< d(x, A)}.

Clearly,G H=. Also, each is open (ifxG and = 12{d(x, B) d(x, A)},

then B(x, )G, by the triangle inequality.) Now, ifd(p, A) = 0, then for allnN, there existsxnA such thatd(p, xn)< 1n . Sod(p, xn)0 i.e. xnp,whencep A. Thus for each xA, xB = B so that d(x, B)> 0 =d(x, A)i.e.xG. Hence AG. Similarly BH.Its true that T4 T3 1

2but not very obvious. First note that if G0, G1 are

open in a T4 space with G0

G1, then there exists open G 12

with G0

G 12

and G 12

G1 (because the given G0 and X\G1 are disjoint closed sets sothat there exist disjoint open sets G 1

2, H such that G0G 1

2, X\ G1H i.e.

G1 (closed)X\ HG 12

).

Lemma 5.1 (Urysohns Lemma) LetF1, F2 be disjoint non-empty closed sub-sets of aT4 space; then there exists a continuous function f : X [0, 1] suchthatf(F1) ={0}, f(F2) ={1}.Proof Given disjoint closed F1 andF2, choose disjoint open G0 and H0 so thatF1G0, F2H0. DefineG1= X\F2 (open). SinceG0 (closed) X\H0X\ F2= G1, we have G0G1.By the previous remark, we can now construct:

(i) G 12

T: G0G 12

, G 12

G1.

(ii) G 14

, G 34

T: G0G 14

, G 14

G 12

, G 12

G 34

, G 34

G1.(iii) ...and so on!

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Thus we get an indexed family of open sets

{Gr :r = m2n

, 0m2n, n1}such that r1r2 Gr1Gr2 .Observe that the index set is dense in [0, 1]: ifs < t in [0, 1], there exists somem2n

such thats < m2n

< t. Define

f(x) =

inf{r: xGr} xF21 xF2.

Certainlyf : X [0, 1], f(F2) ={1}, f(F1) ={0}. To show f continuous, itsuffices to show thatf1([0, )) and f1((, 1]) are open for 0 < < 1.Well, f(x)< iff there exists some r = m

2n such that f(x)< r < . It follows

thatf1([0, )) =

r iff there exist r1, r2 such that < r1 < r2 < f(x), implyingthat x Gr2 whence x Gr1 . It follows that f1((, 1]) =r1>(X\ Gr1),which is again open.

Corollary 5.4 EveryT4 space isT3 12

.

Proof Immediate from Lemma 5.1. (Note that there exist spaces which areT3 12

but notT4.)

Theorem 5.8 Any compactT2 space isT4.

Proof Use Corollary 5.2 to Theorem 5.3.Note Unlike the previous axioms, T4 is neither hereditary nor productive. Theglobal view of the hierarchy can now be filled in as an exercise from data supplied

above:-Metrizable Hereditary? Productive?T4T3 1

2

T3T2T1

The following is presented as an indication of how close we are to having comefull circle.

Theorem 5.9 Any completely separableT4 space is metrizable!

Sketch ProofChoose a countable base; list as

{(Gn, Hn) : n

1

}those pairs of elements of

the base for which Gn Hn. For each n, use Lemma 5.1 to get continuousfn: X[0, 1] such that fn( Gn) ={0}, fn(X\ Hn) ={1}. Define

d(x, y) =

n1

{fn(x) fn(y)2n

}2.

One confirms thatd isa metric, and induces the original topology.

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