**
**
## Page 6

Denition 5. Let (X;) be a topological space. Let x2X:A subset

NofXis said to be a neighborhood of the point xif there exists

G2such thatx2GN:Note that in general, Nneed not belong

to:It is called open neighborhood if it belongs to :Collection of

all neighborhoods of xis denoted by N(x):It is easy to see that a

nonempty subset GofXis open if and only if it is a neighborhood of

each of its point. Let Nxbe an open set containing xand contained in

G:ThenG=[x2GNxandGis open, being an union of open sets. On

the other hand if Gis open then Gitself is the neighborhood of each

of its point contained in G:

Examples:

LetX=fa;b;cgand let=f;;fag;Xgthen we have N(a) =

ffag;fa;bg;fa;cg;XgandN(b) =N(c) =X:

LetX=Nand let=f;;J1;J2;J3;:::;Jk;:::;gwhereJk=

fk;k+ 1;k+ 2;:::g:ThenN(1) =fJ1g=N; N(2) =fJ1;J2g;

N(3) =fJ1;J2;J3[f1gg;N(4) =fJ1;J2;J3[f1g;J4[f1g;J4[

f2g;J4[f1;2gg:

Denition 6. LetAbe a nonempty subset of ( X;):A pointx2X

is said to be a limit point or an accumulation point of Aif for every

neighborhood Nofxthe setA\(Nnfxg) is non empty. The set of

limit points of Ais called the derived set of Aand is denoted by A0:

Note that a limit point of a set need not be an element of the set. For

example consider the topological structure dened in example 2 above.

LetA=f2g:The point 1 is a limit point of Abut it is not in A:

Denition 7. For any set Ain a topological space ( X;);the closure

ofAinXdenoted by c(A) is dened as c(A) =A[A0:

Check Your Progress LetX=fa;b;cgand let=f;;fag;fbg;fa;bg;Xg:

Show thatfag0=fcgandfcg0=;:Find the derived sets of other sub-

sets ofX:

Following results are similar to those in a metric space.

Lemma 1.5.1. (i)(;)0=;:

(ii) IfABthenA0B0:

(iii) (A[B)0=A0[B0:

Proof. (i) For any neighborhood Nof any point xofXwe have

;\N=;and hence;\Nnfxg=;:This proves ( i):

(ii) Letx2A0:Then for any neighborhood Nofxwe haveA\Nn

fxg6=;:ButAis a subset of B:HenceA\NnfxgB\Nnfxg

which gives B\Nnfxg6=;i.e.x2B0:This proves A0B0:

6munotes.in

## Page 7

(iii) We have AA[Bwhich give A0(A[B)0SimilarlyB

A[B; which gives B0(A[B)0:These two together give us

A0[B0(A[B)0:

To get the reverse inclusion suppose x2(A[B)0but does not

belong toA0andB0:Then there exist neighborhoods N1andN2

ofxwithA\N1nfxg=;andB\N2nfxg=;:LetN=N1\N2:

ThenNis a neighborhood of xsatisfyingA\Nnfxg=;and

B\Nnfxg=;which given ( A[B)\Nnfxg= (A\Nnfxg)[

(B\Nnfxg) =;[; =;:This shows that xdoes not belong

to (A[B)0This proves ( A[B)0A0[B0which completes the

proof of (iii):

Denition 8. A setAin a topological space ( X;) is said to be closed

ifA0A:

There is a more simple characterization of closed sets.

Theorem 1.5.1. A setAin(X;)is closed if and only if XnAis an

open set in (X;):

Proof. LetAbe a closed set, which means AA:To proveG=XnA

is open in ( X;):Lety2G:Thenyis not inAand hence not in A

becauseAA:Therefore there exists a neighborhood Nofysuch that

N\Anfyg=;:Butyis not inAhence we get, A\N=;:

Now, by denition of neighbourhood there exists Gy2withy2

GyN:NowA\N=;implies,A\Gy=;; equivalently put, we

havey2GyXnA:

This proves that for any y2XnAthere exists an open set Gy

satisfyingy2GyXnA:HenceXnA=[fGy:y2XnAgwhich

givesXnAis an open set in ( X;):

Conversely, suppose XnAis open. Then, no point of XnAis a

limit point of A:Because ifyis a point of XnAthen by openness of

XnA;we get that XnAis a neighborhood of ywhich is disjoint from

A:Thereforeyis not a limit point of A:This proves A0A:That is

Ais closed.

Next theorem also is familiar to you.

Theorem 1.5.2. ;andXare closed sets in (X;):

Any intersection of closed sets in (X;)is a closed set in (X;):

Finite union of closed sets in (X;)is a closed set in (X;):

Proof. We use the above theorem to prove this.

;2implies that Xn;is closed and X2implies that;=XnX

is closed.

7munotes.in

## Page 8

LetfC:2gbe a family of closed sets. Then for each

2 the setG=XnCis an open set in ( X;):Hence

[fG:2g2T:But[fG:2g=Xn\fG:2g:

This shows that\fC:2gis closed.

LetfCk:k= 1;2;3;:::;ngbe a family of closed sets. Let Gk=

XnCk;1kn:Then eachGkis open and therefore \fGk: 1

kngis an open set in ( X;):But\1knGk=Xn[1knGk

which shows that \1knGkis closed. This completes the proof.

Following is another familiar result. We have proved it earlier for

metric spaces.

Theorem 1.5.3. Closure of a set Ain(X;)is the smallest closed

subset of (X;)containing A:

Proof. LetAbe the smallest closed set containing Aand letc(A) =

A[A0:We will prove:

(i)Ac(A);

(ii)c(A)A:

These put together will prove A=c(A):

Supposex2Xnc(A) thenxdoes not belong to Aandxdoes not

belong toA0:Therefore there exists an open set Gxsuch thatx2Gx

andA\Gxnfxg=;:Butxis not inAand therefore this set equation

implies,A\Gx=;:Note that this also implies that no point of Gx

is a limit point of A:In other words GxXnA0:On the other hand

A\Gx=;impliesGxXnA0, we getGxXnA[A0=Xnc(A):

Thus each point xofXnc(A) has a neighborhood Gxcontained entirely

inXnc(A):ThereforeXnc(A) is an open subset of ( X;):This means

c(A) is a closed subset of ( X;) containing A:NowAbeing the smallest

closed set containing AimpliesAc(A):This proves (i). Also the set

Ais closed and hence ( A)0A:ButAAimpliesA0A0:Thus we

havec(A)A[A0A:This proves ( ii) completing the proof.

1.6 Interior

This concept from metric spaces is extended to topological spaced

as follows.

LetAbe a non empty set in ( X;):We dene Interior of Aas the

union of all open subsets of A:We denote it by i(A) orA: i(A) is an

open set by denition. i(A) is in fact the largest open set contained in

A:

8munotes.in

## Page 9

Theorem 1.6.1. (1.)i(X) =X

(2.)i(A)A;for all subsets AofX:

(3.)i(i(A)) =i(A);for all subsets AofX:

(4.) IfABtheni(A)i(B):

(5.)i(A\B) =i(A)\i(B);for all subsets AandBofX:

Proof. (1.)Xis open. Hence Xis the largest open set contained in

X:This proves (1) :

(2.)i(A) is a union of subsets of Aand hence is a subset of A:This

proves (2):

(3.)i(A) is open. Hence i(i(A)) =i(A):

(4.)i(A) is an open set contained in A:Hencei(A) is an open set

contained in B:This means i(A) is contained in largest open set

contained in Bwhich isi(B):

(5.)A\BAwhich gives i(A\B)i(A):AlsoA\BBwhich

givesi(A\B)i(B):Together these two imply i(A\B)i(A)\

i(B):Now lety2i(A)\i(B):By denition of interior there exist

open setsGandHin (X;) satisfying y2GAandy2HB:

G\His an open subset of A\BhenceG\Hi(A\B):This

provesy2i(A\B) and hence i(A)\i(B)i(A\B):This

completes the proof.

Denition 9. For any set Ain the topological space the set c(A)ni(A)

is dened as the boundary of Adenoted by b(A):It is also called frontier

ofA:

Check Your Progress Show that a point xofXis a boundary

point of a subset AofXif and only if N\A6=;andN\(XnA)6=;

for allN2N(x):

1.7 Continuous functions

Now we generalize the concept of continuous functions to topological

spaces.

Denition 10. Let (X;);(Y;0) be any two topological spaces and

letf:X!Ybe a map. We say fis continuous at a point aofX;if

given any neighborhood eNoff(a) there exists a neighborhood Nofa

such thatf(N)eN: f is said to be continuous on a subset AofXif

fis continuous at every point aofA:

9munotes.in

## Page 10

Remark 1.7.1. f(N)eNimpliesNf1(eN) and therefore f1(eN)

becomes a neighborhood of a:Therefore the denition can be restated

as:fis continuous at aif for every neighborhood eNoff(a) the set

f1(eN) is a neighborhood of a:Since neighborhoods are supersets open

sets containing that point continuity at a point can be rephrased as :

fis continuous at aif given open set Hof (Y;0) withf(a)2Hthere

exists open subset Gof (X;) containing asuch thatf(G)H:

We prove some theorems on continuous functions.

Theorem 1.7.1. Let(X1;1);(X2;2);(X3;3)be topological spaces and

letf: (X1;1)!(X2;2)andg: (X2;2)!(X3;3)be maps such that

fis continuous on X1andgis continuous on X2thenh=gof :

(X1;1)!(X3;3)is continuous on X1:

Proof. Leta2X1:Letf(a) =b:LeteNbe a neighborhood of g(b) =

g(f(a)):By continuity of gatf(a) =bwe haveg1(eN) is a neigh-

borhood of b:We haveg1(eN) is a neighborhood of b:Also by con-

tinuity offatawe havef1(g1(eN)) is a neighborhood of b:But

f1(g1(eN)) = (gf)1(eN) which is a neighborhood of a:Thus we have

veried that, for any neighborhood eNof (gf)(a) the set (gof)1(eN)

is a neighborhood of a:Therefore by the denition it follows that gf

is continuous at a:

Theorem 1.7.2. Letf: (X;)!(Y;0)be any map. The following

conditions on fare equivalent:

(a)fis continuous on X:(i.e. at every point of X).

(b) IfH20thenf1(H)2:

(c) IfCis a closed subset of (Y;0)thenf1(C)is a closed subset of

(X;):

(d) For any subset AofX; f (c(A))is a subset of c(f(A)):

Proof. We will prove the implications cyclically.

(a) =)(b):Letfbe continuous on X:LetHbe an open subset

of (Y;0):Letx2f1(H):Thenf(x)2H20:This means His a

neighborhood of f(x):By continuity of fatx; f1(H) is a neighbor-

hood ofx:This proves that f1(H) is a neighborhood of each of its

point. Hence f1(H) is an open set of ( X;):

(b) =)(c):LetCbe a closed subset of ( Y;0):ThenH=YnC

is an open subset of ( Y;0):By (b) this gives f1(H) is an open subset

of (X;):Butf1(H) =f1(YnC) =Xnf1(C):Hencef1(C) is

a closed subset of ( X;):

10munotes.in

## Page 11

(c) =)(d):LetAbe any subset of X:LetC=c(f(A)): Cis a

closed subset of ( Y;0):Therefore by ( c); f1(C) is a closed subset of

(X;):But we have Af1(f(A))f1(c(f(A))) =f1(C):

NowAis a subset of the closed set f1(C):Hencec(A)f1(C)

which gives f(c(A))f(f1(C))C:Thusf(c(A))c(f(A)) is

proved.

(d) =)(c):LetCbe a closed subset of ( Y;0):LetA=f1(C):

We havef(c(A))c(f(A))c(f(f1(C))c(C) =C:Now apply

f1tof(c(A))Cwe getf1(f(c(A)))f1(C) =A:Butc(A)

f1(f(c(A))):Therefore we get c(A)A:That is,Ais a closed subset

of (X;):

(c) =)(a):Letxbe any point of Xand letHbe any open subset

of (Y;0) containing f(x):LetC=YnHthen we have f1(H) =

Xnf1(YnH) =Xnf1(C):By the result ( c); f1(C) is closed and

hencef1(H) =Xnf1(C) is open. This proves that f is continuous

at every point xinX:

Remark 1.7.2. If (X;)!(Y;0) is continuous and xis a limit

point ofx2AXthen it is not necessary that f(x) is a limit

point off(A):For example take the constant map from the usual

(or indiscrete) topology to the discrete topology on R:Show that

it is continuous. Now any point in the domain is a limit point of

Rbut its image is not a limit point of f(R):

Any function dened on a discrete topological space is continuous.

Iff: (X;)!(Y;0) is continuous then A2does not mean

f(A)20:

1.8 Glossary

In this chapter, you have learnt the following:

Denition of a topological space.

Base of a topological space.

Product and subspace topology.

Open and Closed sets.

Interior of a set.

Continuity in topological spaces.

11munotes.in

## Page 12

1.9 Bibliography

James Munkres: Topology, Pearson

George Simmons: Topology and Modern Analysis, TataMcgraw-

Hill.

M. A. Armstrong: Basic Topology, Springer UTM

Wolfgang Thorn: Topological Structures; Holt, Rinehart and

Winston, New York, Chicago.

1.10 Let Us Sum Up

A topological structure on Xis a family of subsets of Xwhich

contains;;Xand is closed under the operations of unions over arbitrary

subfamilies of and over intersections of nite subfamilies of :Any

non empty family of subsets of Xgenerates a unique topology on X:A

subfamily Bofis called a base of if every non-empty set Gincan

be expressed as a union of a subfamily of B:A subsetNofXis called

a neighborhood of x2Xif there exists Ginsatisfyingx2GN:

Associated with any set Ain a topological space ( X;) there is

another set Acalled derived set of A:The setA0consists of all limit

points ofA; where a limit point of Ais a pointxofXhaving the

property that every neighborhood of it contains at least one point of

Awhich is dierent from x:A subsetAof a topological space Xis

said to be closed if A0is a subset of A:A set is closed if and only if

its complement is open. The set A[A0is called closure of Aand is

denoted by c(A):Closure of a set satises following properties.

c(;) =;:

Ac(A);for everyAX:

c(c(A)) =c(A);for everyAX:

c(A\B) =c(A)\c(B) for allA;BX:

c(A) is the smallest closed subset of XcontainingA:

The denition of a continuous function from one topological space to

another is a generalization of the classical "denition of continuity

of a real valued function of a real variable. A map f: (X;)!(Y;0)

12munotes.in

## Page 13

is continuous at a point aofXif for any neighborhood eNoff(a) in

(Y;) there corresponds a neighborhood Nofain (X;) such that

f(N)eN:We say that fis continuous on a subset Aof (X;) if it

is continuous at every point aofA:Continuity of fonXhas several

equivalent forms. Some are listed below.

f1(H)2for everyH20:

f1(C) is a closed subset of ( X;) whenever Cis a closed subset

of (Y;0):

f(c(A))c(f(A)) for every subset AofX:

1.11 References for further reading

1. W.J. Pervin: Foundations of General Topology, Academic press ,

New York, London.

2. Kelley J.L, General Topology, Van Nostrand Reinhold Co., New

York, 1955.

3. Dugundji J.: Topology, Allyn and Bacon, Boston, 1966.

1.12 Chapter End Exercises

1. There are 26 topologies on X=fa;b;cg:List all of them.

2. LetYbe a non-empty subset of a topological space ( X;) and

lety2Y:Prove that a subset MofYis a neighborhood of yin

the subspace ( Y;0) if and only if there exists a neighborhood N

ofyin (X;) such that M=N\Y:

3. LetIbe the set of all bounded open intervals and let Ibe the

subfamily of Iconsisting of all open intervals of Ihaving rational

end points. Prove that the topology on Rgenerated by Iis the

same as that generated by I:(Note that the topology generated

byIis the usual topology of R:)

4. Prove that two bases BandBgenerate the same topology if

and only if for each x2B2Bthere exists B2Bsuch that

x2BBand to each y2B02Bthere exists B002Bsuch

thatx2B00B0:

13munotes.in

## Page 14

5. Let Bbe a base to a topological structure onXand letYbe a

nonempty subset of X:LetBy=fB\Y:B2Bg:IsBya base

to the subspace topology yon Y? Justify.

6. Give an example of a set Xand two topologies and0onX

such that[0is not a topology on X:

7. IfAis a subset of ( X;) with the property that B0ABfor

some subset BofX;prove thatAis closed.

8. LetX=N:LetIn=f1;2;3;:::;ngandJn=fn;n+1;n+2;:::g:

Let=f;;I1;I2;I3;:::;Ngand0=f;;J1=N;J2;J3;:::g:

Then check that and0both are topological structures on N:

Findf1g0with respect to both and0:

9. Letf: (X;)!(Y;0) be a constant map. Show that fis

continuous.

10. Describe the closure of an innite set in the co-nite topology.

14munotes.in

## Page 15

Chapter 2

Countability and Separation

Axioms

Chapter Structure

2.1 Introduction

2.2 Objectives

2.3 Homeomorphism and Heredity

2.4 Cardinality

2.5 Separation Axioms

1.6 Hausdro topological Spaces

2.7 Regular and Normal Topological Spaces

1.8 Glossary

2.9 Bibliography

2.10 Let Us Sum Up

2.11 References for Further Reading

2.12 Chapter End Exercises

2.1 Introduction

The basic denition of a topological structure on a space Xis in

terms of a family of subsets of Xsatisfying only three conditions,

namely the three topological axioms and yet it manages to capture in

it the most basic geometric ingredient \neighbourhood" of the real num-

ber systems R;R2etc. But most of the other mathematically important

properties of these number systems are lost in such a generalization.

We get them back one by one by imposing additional conditions on a

topological structure.

15munotes.in

## Page 16

2.2 Objectives

After going through this chapter you will know:

Countability axioms and their consequences

Separation axioms and examples of metric spaces that satisfy them.

Denition, examples and properties of Hausdor spaces.

Regular and normal topological spaces and their properties.

2.3 Homeomorphism and Heredity

The main aim of studying topological spaces is the study of proper-

ties that are invariant under homeomorphism. Homeomorphism is thus

a concept of central importance in topology.

Denition 11. A mapf: (X;)!(Y;0) is said to be a homoemor-

phism iffis bijective, fis continuous and f1is also continuous.

For example, the identity map from ( X;)!(X;) is clearly a

homeomorphism and composite of two homeomorphisms is a homeo-

morphism. This makes homeomorphism an equivalence relation on the

collection of topological spaces. Also inverse of a homeomorphism is a

homeomorphism.

Denition 12. A property Pof topological spaces is said to be a

topological property if Pis satised by a topological space implies that

it is satised by another topological space homeomorphic to it. Also a

propertyPis called hereditary if Pis satised by a topological space

implies that it is satised by every subspace of it.

Remark 2.3.1. Note that a bijective continuous map need not be a

homeomorphism. Take for example a continuous map ffrom [;)

to the unit circle S1inR2dened byf(x) = (cosx;sinx ):fis bijective

and continuous but the inverse of fis not continuous.

2.4 Cardinality

Though cardinality is not our main topic of discussion we need to

revise some results in cardinality to understand the countability axioms.

Cardinality of N, the set of natural numbers and any set which is in

bijection with Nis denoted by@0:These sets are called denumerable.

16munotes.in

## Page 17

It can be proved that the set of rational numbers is denumerable. A

set which is either denumerable or nite is called as countable. The set

of real numbers Rand any interval in Rare uncountable sets. It can

be proved that cardinality of R,R2and any interval in Rare same. We

denote it by C. Continuum hypothesis states that there is no innite

set having cardinality between @0andC:An important property of

countable sets is that all its elements can be written in the form of a

sequence.

2.4.1 Countability Axioms

Recall, ifNis a neighborhood of xand ifMis a subset of Xwith

NMthenMalso becomes a neighborhood of x:This property of

neighborhoods suggests that we need not know the entire family N(x);

it is enough to single out a subfamily of it which is such that any

neighborhood of xis a superset of a set belonging to the subfamily.

Such a subfamily, say B(x) ofN(x), which is capable of describing all

the neighborhoods of xas the supersets of sets belonging to B(x) is

called the base to the neighborhood systems of x:Thus we have two

kinds of bases in our discussion:

A base to the topological structure onX:

A base to the complete neighborhood system N(x) of a point x

in (X;):

It turns out that the two concepts are not independent; the rst kind

of base, that is a base to a topological structure determines a base to

the complete neighborhood system N(x) of a point xofXwhich the

topological structure assigns. Countability axioms are auxiliary axioms

on the topological structure of a topological space which demand that

the two bases be countable.

Denition 13. A base of the neighborhood system N(x) ofxis any

subfamily B(x) ofN(x) having the following property: If N2N(x)

then, there exists B2B(x) such that BN:If all the neighborhoods

inB(x) consist of open sets then we say that B(x) is an open base

of the neighborhood system on x:Note that, if B(x) is a base of the

neighborhood system N(x) ofxthen so isfi(B) :B2B(x)gwhich is

an open base.

Here are some illustrative examples of neighborhood bases.

In case of discrete topology P(X) on a setX;B(x) =ffxggis a

base toN(x):

Inthe usual topology on R,B(x) =f(x1=n;x +1=n) :n2Ng

is a countable base to N(x):

17munotes.in

## Page 18

Let (X;d) be any metric space. Then B(x) =fB(x;1=n) :n2

Ngis a countable base to the complete neighborhood system of

eachxinX:

Denition 14. We say that the topological space ( X;) is rst count-

able space if for each xinXthe complete neighborhood system N(x)

has a countable base. This countable base is called the fundamental

system of neighborhoods of x:We also say that such a space is a C1-

space.

Topological spaces discussed in examples 2 and 3 discussed above

areC1- spaces. In addition the discrete space ( X;P (X)) isC1be-

causeffxggis a fundamental system of neighborhoods for any xin

Xand the indiscrete space ( X;f;;Xg) isC1- spaces because Xis

the only neighborhood of every point xinX:IffNn:n2Ngis

a fundamental system of neighborhoods of a point xthen putting

fN1=N1\N2\N3we get thatffNn:n2Ngis another count-

able neighborhood base of xin which the neighborhoods decrease with

increasingn:fN1fN2fN3fNn Thus in aC1- space we

can always choose a countable, monotonically decreasing fundamental

neighborhood system of neighborhoods of each of its points. Now we

verify that rst countability is a topological property.

Letf: (X;)!(Y;0) be a homeomorphism. Let ( X;) be a

C1-space. We verify that each y2Yhas a countable fundamental

neighborhood system. Let x2Xsatisfyf(x) =y:IfMn=f(Nn)

then for each n; Mnare neighborhoods of y:In factfMn:n2Ngis a

fundamental neighborhood system of y:This proves that ( Y;0) isC1-

space. Thus homeomorphic image of a C1-space (X;) is aC1- space.

We will verify that rst countability is also a hereditary property.

Let (X;) be aC1-space and let Ybe a non-empty subset of X:Let

y2Y:Theny2X:By theC1- property of ( X;),yhas a fundamental

neighborhood system fNn:n2Ngin (X;):ThenfN

n:=Nn\Y:n2

Ngis a fundamental neighborhood system of yin (Y;y):This proves

that the subspace ( Y;y) is alsoC1- space. Hence it is a hereditary

property.

Denition 15. A topological space ( X;) is second countable (or sat-

ises the second countability axiom) if the topology has a countable

base.

A second countable space is rst countable, for if Bis a countable

basis for (X;), then B(x) =fB2Bfor whichx2BgBand

hence is countable. Thus, B(x) is a countable base to the complete

neighborhood system N(x) and hence C1- axiom is satised.

The converse is not true. A space may be rst countable without

being second countable. Consider the discrete topological structure

onR:This topological space ( R;P(R)) has no countable base. But

18munotes.in

## Page 19

it isC1- space because for every point x2Xthe singleton family

ffxggis a fundamental neighborhood system of x:This shows that

second countability is stronger than rst countability. Some examples

of second countable spaces are given below.

LetX=N:LetIn=f1;2;3;:::;ngandJn=fn;n+1;n+2;:::g:

Let=f;;I1;I2;I3;:::;Xgand0=f;;J1=N;J2;J3;:::g:

Thenand0are both topological structures on N:Both are

countable families and hence are second countable structures.

Let (R;d) be the usual metric space. B=fB(x;1=n) :n2

N;x2Qgis a countable base to the metric space. Hence the

usual metric space is second countable.

2.5 Separation Axioms

Denition 16. A topological space ( X;) is aT0-space if for any two

distinct points xandyinXthere exists an open subset of ( X;) which

contains only one of xandybut not the other.

Here are some examples:

IfXcontains at least two points then the indiscrete topologi-

cal spacefX;f;;Xg]gis notT0because an open set will either

contain both the points or contain no points.

LetX=fa;b;cg:Then=f;;fa;bg;Xgis notT0because the

condition fails for the pair a;b:

Any metric space satises the Hausdor property and hence is

T0:

Ifis the co-nite topology on any set having more than element

then it isT0:

Cocountable topology on Ris also aT0space.

Theorem 2.5.1. BeingT0is a topological property

BeingT0is a hereditary property.

Proof.Letf: (X;)!(Y;0) be a homeomorphism. Suppose

(X;) is aT0space. To prove ( Y;0) is alsoT0;consider two

distinct points y1andy2inY:There exist two distinct points x1

andx2inXsatisfyingf(x1) =y1andf(x2) =y2:

19munotes.in

## Page 20

By theT0property of ( X;) there exists an open set Gof (X;)

which contains one of the points x1;x2avoiding the other. With-

out loss of generality suppose x12Gandx262G:Letf(G) =H:

Then by dening property of the homeomorphism H20:y12H

buty262H:This proves ( Y;0) is aT0space.

Let (X;) be aT0space letYbe a non empty subset of X:We

want to prove that ( Y;y) is aT0space. Lety1;y2be two distinct

points ofY:They are also distinct points of the T0space (X;):

Therefore by the T0property of ( X;) there is an open subset G

of it, which contains only one of these points. Suppose y12Gand

y262G:LetH=G\Y:ThenHis an open subset of ( Y;y) which

containsy1but noty2:This proves that ( Y;y) is aT0space.

Theorem 2.5.2. (X;)is aT0space if and only if distinct one point

subsets of it have distinct closures.

Proof. First suppose that ( X;) is aT0space. Letxandybe distinct

points ofX:We prove that c(fxg)6=c(fyg):Sincex6=y;there exists an

open subset Gof (X;) with sayx2Gbuty62G:Nowy62Gimplies

that the closed set XnGcontainsybut does not contain x:Clearly

c(fyg)XnGand as such x62c(fyg):Butc(fxg) must contain point

x:This proves c(fyg)6=c(fxg):

Conversely, suppose for every pair of points x;y2Xwithx6=ywe

have the inequality of sets c(fxg)6=c(fyg):Then either x62c(fyg) or

y62c(fxg):For ifx2c(fyg) thenfxgc(fyg) which gives c(fxg)

c(fyg):

Similarlyy2c(fxg) impliesc(fyg)c(fxg):These two together

givec(fxg) =c(fyg);which contradict the assumption. Suppose x62

c(fyg):LetG=Xnc(fyg):ThenGis an open subset of ( X;):X2G

andy62G:This proves the T0axiom for ( X;):

We now move on to the next separation axiom.

Denition 17. A topological space ( X;) isT1space if for any two

pointsx;y2Xthere exit two open sets G;H of (X;) such that x2G;

y62Gandy2H; x62H:

Clearly every T1space is aT0space. But there are spaces which are

T0but notT1:

Here are some examples

LetX=N:LetIn=f1;2;3;:::;ng:Let=f;;I1;I2;:::;Ng:

(X;) isT0but notT1since there exists an open set that con-

tains 2 but not 1, but there is no open set containing 1 and not

containing 2 :This proves that the above space is not T1:

20munotes.in

## Page 21

Discrete topological space ( X;P (X)) and metric spaces are ex-

amples ofT1spaces.

OnX=fa;b;cgwe dene=f;;fag;fa;bg;fa;cg;Xg:(X;)

isT0but notT1:The paira;bdoes not satisfy the requirements

of aT1space.

Theorem 2.5.3. A topological space is T1space if and only if

every one point subset of it is a closed subset.

In aT1space (X;)ifxis a limit point of a subset AofXthen

every neighborhood of xcontains innitely many points of A:

Proof.First we assume that ( X;) is aT1space and prove that

for anyx2X;fxgis a closed set. Equivalently we prove that

G=Xnfxgis an open set of ( X;):

Lety2Gbe arbitrary. Then y6=x:Therefore by the T1property

there exists Gy2such thaty2Gyandx62Gy:Thusy2Gy

Xnfxg:Consequently we have Xnfxg=[fGy:y2Xnfxgg:

This shows that Xnfxgis an open set and hence fxgis closed.

Conversely suppose that fxgis closed subset of ( X;) for every

x2X:Now letxandybe two distinct points of X;thenx2

Xnfyg=Gx2and similarly we have y2Xnfxg=Gy2

Thus for the distinct points xandyinXwe have the open sets

Gx;Gywithx2Gx; y62Gxandy2Gy; x62Gy:This proves

that (X;) is aT1space.

Supposex2Xis such that a neighborhood Nofxcontains only

nitely many points of A:Say,A\Nnfxg=fx1;x2;x3;:::;xng:

Now being one point subsets of the T1space each of the sets

fx1g;fx2g;fx3g;:::;fxngare closed subsets of ( X;):Consequently

their unionfx1;x2;x3;:::;xngis also a closed subset of ( X;):

This further implies, the set Xnfx1;x2;x3;:::;xngis an open

neighborhood of X:The intersection of this neighborhood with

Nwhich isNnfx1;x2;x3;:::;xngis also a neighborhood of x:

A\(Nnfx1;x2;x3;:::;xng)nfxg=;:

This is a contradiction to the assumption that xis a limit point

ofA:Therefore the assumption that A\Nis nite is wrong.

This proves that every neighborhood of xcontains innitely many

points ofA:

LikeT0;beingT1also is topological and hereditary property. You

can prove this as an exercise. We continue our discussion of separation

axioms in the next section where T2-spaces are discussed with the name

Hausdor spaces.

21munotes.in

## Page 22

2.6 Hausdor Topological Spaces

Denition 18. A topological space ( X;) is said to be Hausdor space

or aT2space if for any two distinct points x;yofXthere exist open

subsetsG;H of (X;) such that x2G;y2HandG\H=;:

Remark 2.6.1. AxiomT2is a topological property as well as

hereditary property.

AxiomT2is stronger than T1:If a topological space satises axiom

T2then it satises axiom T1:

The converse is not true. On an innite set the conite topology

satisesT1axiom but does not satisfy T2axiom.

OnX=fa;b;cgwe can not dene a topology which is T1but

notT2:Because if the topology is T1then singletons are closed

sets. This means that all the sets having 2 elements are open sets.

This forces the topology to be P(X) which isT2:

We have proved earlier that every metric space is a Hausdor

space i.e. satises T2axiom.

Denition 19. A sequence in a metric space ( X;) is a mapa:N!

X:We denote it by the notation ( an):The sequence ( an) is said to be

convergent to an element linXif for any neighborhood Noflthere

existsn02Nsuch thatan2Nfor alln > n 0:We use the notation

liman=l:

Following theorem proves a connection between limit of a sequence

and limit point of a set.

Theorem 2.6.1. If(an)is a sequence of distinct points of Awhich

converges to lthenlis a limit point of A:

Proof. LetNbe any neighborhood of l:Since (an) converges to l;there

existsn02Nsuch thatan2Nfor alln>n 0:But since the sequence

is inAwe getan2A\N;for alln>n 0:Moreover all an's are distinct

impliesan6=lfor at least one n > n 0:In other words there exists

n > n 0such thatan2A\Nnflg:This proves A\Nnflg6=;for

every neighborhood Nofli.e.lis a limit point of A:

In general in a topological space limit of a sequence is not unique and

hence limit is not a well dened concept. For example in the indiscrete

topology (X;P (X)) any sequence converges to each point in the metric

22munotes.in

## Page 23

space. Thus if the set Xhas more than one point then each sequence

has more than one limit. Another such example is X=fa;b;cgand

=f;;fcg;fa;bg;Xg:Now dene the sequence fan:n2Ngbya2k=

aanda2k+1=bfor allk2N:This sequence converges to both bandc:

Following is an example of a sequence that converges to innitely

many points.

LetX=N:LetJn=fn;n+ 1;n+ 2;:::gand=f;;J1=

N;J2;J3;:::g:Dene the sequence ( an) byan=nfor everyn:The

sequence converges to every m2N:

We cannot nd such examples in Hausdor spaces because of the

following

Theorem 2.6.2. In a Hausdor topological space a convergent se-

quence has a unique limit.

Proof. Let (X;) be a Hausdor space. If possible suppose the sequence

fan:n2Ngconverge to two dierent points landlofX:By the

Hausdor property there exist open sets GandHof (X;) such thatl2

G;l2HandG\H=;:Now since ( an) converges to lthere exists n12

Nsuch thatan2G;for alln>n 1:Also since ( an) converges to lthere

existsn22Nsuch thatan2H;for alln>n 2:Letm= maxfn1;n2g:

Now for all n>m;n>n 1andn>n 2and soan2G\H=;:This is

a contradiction which proves uniqueness of limits.

We now dene an important property of a topological space.

Denition 20. We call a topological space ( X;) separable if there

exists a countable dense subset of it. That is, there exists A=fan:

n2Ngsatisfyingc(A) =X:

The real line is a separable metric space because the set of rational

numbers is a countable dense subset. But Rwith the co-countable

topology is not a separable topological space. Because if you take a

countable set AthenRnAis a co-countable set. This means RnAis

open andAis a closed set which gives c(A) =Ais a proper subset of

RasRis uncountable.

Lemma 2.6.1. Being separable is a topological property.

Proof. Supposef: (X;)!(Y;0) is a homeomorphism. Suppose

A= (an:n2N) is a countable dense set in X:We want to show that

the countable set B=f(A) is dense in Y:Ifc(B) is a proper subset of

YthenH=Ync(B) is a nonempty subset of ( X;):LetG=f1(H):

ThenGis a non empty subset of ( X;):Nowf1(c(B)) =c(f1(B)) =

c(A) =X:This contradicts that G=Xnf1(c(B))6=;:This proves

c(B) =Ywhich means that the countable subset BofYis dense in

Y:

23munotes.in

## Page 24

Remark 2.6.2. Being separable is not hereditary.

In fact we can prove that any non-separable topological space is a

subspace of a separable space. Let ( Y;0) be a non separable topological

space. Take any object wwhich is not a member of Y:LetX=Y[fwg:

Introduce a topology onXas follows: =f;;G[fwgjG20g:

(X;) is a topological space and ( Y;0) is a subspace of it. fwgis dense

in (X;):This proves that ( X;) is separable but subspace ( Y;0) is

not separable.

Theorem 2.6.3. Every second countable topological space is separable.

Proof. Let (X;) be a topological space. Let fBn:n2Ngbe a

base to the topology :We choose a point bnfrom each of the non-

empty sets Bnand form a countable set B=fbn:n2Ng:This set

must be a dense subset of ( X;):Because if it is not dense then G=

Xnc(B) is a non-empty open set of ( X;):ButfBn:n2Ngis a base.

Hence a subfamily fBnk:k2NggeneratesG:HenceG=[k2NBnk:

Hencefbnk:k2NgG:This contradicts the construction of G:

Thereforefbn:n2Ngis a countable dense subset of X:Hence (X;)

is separable.

2.7 Regular and normal topological spaces

Denition 21. Suppose one point sets are closed in X:ThenXis

said to be regular if each pair consisting of one point xand a closed

setBdisjoint from x;there exist disjoint open sets containing xand

Brespectively. The space is called normal if for each pair A;B of

disjoint closed sets of Xthere exist disjoint open sets containing Aand

Brespectively.

Following theorem gives us a characterization of normal and regular

spaces.

Theorem 2.7.1. Let(X;)be a topological space in which one point

sets are closed. Then

Xis regular if and only if given a point xofXand a neighborhood

Uofx;there is a neighborhood Vofxsuch thatc(V)U:

Xis normal if and only if given a closed set Aand an open set

Ucontaining A;there is an open set Vcontaining Asuch that

c(V)U:

24munotes.in

## Page 25

Proof.Suppose that Xis regular and suppose that the point x

and the neighborhood Uofxare given. Let B=XnU; then

Bis a closed set. By hypothesis, there exist disjoint open sets V

andWcontainingxandBrespectively. The set c(V) is disjoint

fromBsince ify2Bthe setWis a neighborhood of ydisjoint

fromV:Thereforec(V)Uas desired.

To prove the converse, suppose the point xand the closed set B

not containing xare given. Let U=XnB:By hypothesis there is

a neighborhood Vofxsuch thatc(V)U:The open sets Vand

Xnc(V) are disjoint open sets containing xandBrespectively.

ThusXis regular.

This proof is similar. Just replace the point xby the set A

throughout.

2.8 Glossary

In this chapter, you have learnt the following:

Homeomorphism between topological spaces.

Hereditary property of a topological space.

First and second countable spaces.

Hausdor topological space.

Regular and normal topological spaces.

Characterization of regular and normal topological spaces.

2.9 Bibliography

James Munkres: Topology, Pearson

George Simmons: Topology and Modern Analysis, TataMcgraw-

Hill.

M. A. Armstrong: Basic Topology, Springer UTM

Wolfgang Thorn : Topological Structures; Holt, Rinehart and

Winston, New York, Chicago.

25munotes.in

## Page 26

2.10 Let Us Sum Up

A topological system is a generalization of the classical number sys-

temR:But in such a wild generalization many important properties of

the number system are lost. We regain these properties by introducing

a number of additional axioms. There are many types of such axioms.

One such type of axioms are separation axioms. They are denoted by

T0;T1;T2;:::: They are about enclosing a pair of points in disjoint open

subsets, disjoint closed subsets etc. Of these we study T0;T1andT2:

These properties are both topological and hereditary.

AT0space separates a pair of points from the space by a single

open set. A space ( X;) isT0if and only if distinct singleton sets have

distinct closures. A space is T1if given any pair of distinct points there

are open sets each one of them containing one point but not the other.

A topological space is T1if and only if singleton sets are closed. In a

T1space ifpis a limit point of a set Athen every neighborhood of p

contains innitely many points of A:

AT2space is also called a Hausdor space. It separates distinct

points by disjoint open sets. In a Hausdor space a convergent sequence

has unique limit. Regular and normal spaces satisfy stronger separation

axioms. Countability axioms are other type of auxiliary axioms. We

study rst and second countability axioms. Both these axioms are

topological and hereditary. A metric space is called separable if it

has a countable dense subset. A second countable topological space is

separable.

2.11 References for further reading

1. W.J. Pervin: Foundations of General Topology, Academic press ,

New York, London.

2. Kelley J.L, General Topology, Van Nostrand Reinhold Co., New

York, 1955.

3. Dugundji J.: Topology, Allyn and Bacon, Boston, 1966.

2.12 Chapter End Exercises

26munotes.in

## Page 27

1. Show that being T1is a topological and hereditary property.

2. Show that being T2is a topological and hereditary property.

3. Letbe a topology on Rgenerated by the collection f[a;b);a;b2

R;aable space.

4. Show that in a separable topological space every collection of

non-empty pairwise disjoint open sets is countable.

5. Prove that ( X;) is aT1space if and only if for each x2X;

fxg=\fG:G2;x2Gg:

6. Show that subspace of a regular space is a regular space.

7. Show that if Xis regular, every pair of points of Xhave neigh-

borhoods whose closures are disjoint.

8. Show that if Xis normal, every pair of disjoint closed sets have

neighborhoods whose closures are disjoint.

9. Show that a closed subspace of a normal space is normal.

10. LetXbe the set of all irrational numbers with usual metric d:Is

(X;d) separable?

27munotes.in

## Page 29

Chapter 3

Compactness

Chapter Structure

3.1 Introduction

3.2 Objectives

3.3 General Denition of Compactness

3.4 Continuity and Compactness

3.5 Finite Products and Compactsness

3.6 Local Compactness and one-point Compactication

3.7 Lindelof Topological Spaces

3.8 Glossary

3.9 Bibliography

3.10 Let Us Sum Up

3.11 References for Further Reading

3.12 Chapter End Exercises

3.1 Introduction

Compactness is a word we use daily in our life, to indicate that

objects occupy less space. Even in mathematics the word has the same

sense in Rnnamely that compact objects are those which are closed

and bounded. However, in an arbitrary topological space such a nice

formulation may fail, as we shall see from examples.

In an arbitrary topological space the correct notion of compactness

is dened in terms of open covers. Using this dention, we prove that

compactness is preserved under continuous functions and under nite

products. We then move on to the notion of local compactness and also

study spaces which are not too far away from being compact: namely

spaces which admit a one-point compactication.

We also study a notion similar to compactness: namely the notion

of a Lindel of space.

29munotes.in

## Page 30

3.2 Objectives

After going through this chapter you will know:

General denition of compactness.

Continuity preserves compactness.

A nite product of compact spaces is compact.

Denition of local compactness.

Construction of one point compactication.

Lindel of Spaces.

3.3 General denition of compactness

Before going to the notion of compactness in topological spaces,

we begin by recalling an important property of real numbers: the

Archimedean property. Let x;y2Rwithx > 0:Then there exists

a natural number nsuch thatnx>y: Stated in words, this just means

that by taking enough number of units of x;we can get past any given

real number y:

In fact we will often use the following version of this property: if

y2R;there exists ny2Nsuch thatny>y: (Use the above statement

withx= 1:) Before going into the denition of compactness, let us

understand what an open cover of a topological space is.

Denition 22. (Open cover) Let Xbe a topological space and fUg2

be a family of open subsets of X:ThenfUg2is said to be an open

cover ofXifX[2U:(Note that this condition is same as saying

X=[2U;asUX;for all2:)

30munotes.in

## Page 31

From the above denition it is easy to check that fUg2is an

open cover of Xif and only if X=[2U;for open subsets UofX:

Let us see some examples of open covers:

1. LetX=f1;2;3gwith the discrete topology. Then f1g;f2;3gis

an open cover of X:

2. LetX= (0;1):Letn1 andUn= (0;1

n):ThenfUngis an open

cover ofX:

3. LetX=R:Letn1 andUn= (n;n):Then,fUngis an open

cover ofX:

4. LetX=R2:Forn1;letUn=f(x;y)jx2+y2=n2g:Then

fUngis an open cover of X:

Check Your Progress Find other open covers of Xin the exam-

ples above, distinct from the given ones.

We now study the denition of compactness for arbitrary topo-

logical spaces: this abstract dention was rst formulated in 1906 by

Maurice Fr echet. (See the website: www :==httphistory:mcs:st

andrews:ac:uk=Biographies=Frechet:html andhttp:==en:wikipedia:org=wiki=Compactness

for further details.)

Denition 23. LetXbe a topological space. Xis said to compact

if every open cover of Xadmits a nite subcover i.e. if there exists a

familyfUg2of open subsets of Xsuch thatX[2U;then there

exist nitely many 1;:::;n2 such that XU1[U2[[Un:

(In this case we say that the family fUg2admits a nite subcover.)

Remark 3.3.1. The main stress in this denition is on every open

cover . If you are able to prove that a specic open cover of Xadmits a

nite subcover that does not mean the space is compact. For example,

takeX= (0;1) and for n1; Un= (0;1

n):Then, clearly U1is the

nite subcover of X:

On the other hand, if one takes another subcover of X;dened

forn1;byVn= (0;11

n);then this does not admit a nite

subcover: for if there is a nite subcover say Vn1;Vn2;:::;Vnk;let

nj= maxfn1;:::;nkg:Then, one can check that (0 ;1)[niVni

Vnj= (0;11

nj);which is impossible by the Archimedean property.

In other words, in order to say that a topological space Xis not

compact, it is enough to produce an open cover of Xwhich admits no

nite subcover. For example, the real numbers with the usual metric

topology is not compact. Take the open cover fUn:= (n;n)gn1of

R:This cannot admit a nite subcover, for if it is does, then we would

getR(k;k);for somek2N;which is a contradiction.

Here are easy examples of compact sets:

31munotes.in

## Page 32

Example 1. Every nite set is a compact set. (Since any open cover

of this nite set will require atmost nitely many open sets to cover it.)

Example 2. LetXbe a non-empty topological space with indiscrete

topology. Then, Xis compact.

Check Your Progress

1. Show that a nite union of compact sets is compact.

2. Show that every subset of the real line Rwith the nite comple-

ment topology is compact.

3.4 Continuity and compactness

One of the main themes in mathematics is to study objects with

structures and maps between these objects. In topology, the objects

under study are topological spaces and one is more interested in maps

between such spaces that are continuous. This is because continuous

maps preserve many topological properties. For example, we will prove

here that compactness is a property that behaves well with respect to

continuity i.e., continuous image of a compact set is always compact.

This property helps us to get more examples of compact spaces.

Before proving that continuity preserves compactness, let us prove

the following useful result for compactness for subspaces:

Lemma 3.4.1. LetYbe a subspace of X(i.e.,Yis a subset of Xwith

the induced topology.) Then Yis compact if and only if every open

covering of Yby sets open in Xcontains a nite subcollection covering

ofY:

Proof. We rst recall that a subset V;YofYis open inYif and only

there exists an open set V;XofXsuch thatV;Y=V;X\Y:(This is

precisely the subspace topology on Y:)

(a) Suppose Yis compact and let fV;Xg2be an open cover of Y

by sets open in X:Then,fV;X\Yg2be an open cover of Y

by sets open in Y:AsYis compact, we get nitely many 's

that cover Y:This proves the existence of a nite subcollection

covering of Y:

(b) For the other way, start with an open covering of Y;sayfV;Yg2:

Hence there exists a family of open sets fV;Xg2ofXsuch that

V;Y=V;X\Y;i.e.,V;YV;X:Then clearly we get:

Y[2V;Y=[2(V;X\Y)[2V;X:

32munotes.in

## Page 33

Thus, we get a covering of Yby open subsets of X:The hypoth-

esis now gives a nite subcollection covering Y;i.e. there exist

1;:::;nsuch thatYV1;X[V2;X[Vn;X:Taking inter-

section of both sides with Y;we get that

Y(V1;X[V2;X[Vn;X)\Y=[n

i=1(Vi;X\Y) =[n

i=1Vi;Y:

Thus we get a nite subcover of Y;proving that Yis compact.

We will now prove the following:

Lemma 3.4.2. LetX;Y be topological spaces, with Xcompact. If

f:X!Yis continuous, then f(X)is compact in Y:

Proof. LetfV;Yg2be an open cover of f(X):Then, it is easy to check

thatff1(V;Y)g2is an open cover of X:AsXis compact, we get

1;:::;nsuch thatXf1(V1;Y)[f1(V2;Y)[[f1(Vn;Y):By

noting that ( ff1)Vi;YVi;Y;we get that f(X)V1;Y[[Vn;Y:

This proves that the continuous image of a compact set is compact.

Example 3. Using Lemma 3.4.2 above, we will prove that the unit

circleS1is compact. (Recall: S1:=f(x;y)2R2jx2+y2= 1g:)

Letf:R!R2be given by f() = (cos();sin()):Then, clearly fis

continuous and image of fisS1:In fact by observing that f([0;2]) =

S1and noting that [0 ;2] is a compact subset of R, we have that S1is

a compact subset of R2:

Lemma 3.4.3. Every closed subset of a compact space is compact.

Proof. LetXbe a compact space and let Fbe a closed subset of X:Let

fVg2be an open cover of F:Note that by the denition of subspace

topology on F;there exist open subsets UofXfor each2 with

the property that V=U\F:

SinceFis closed,Fcis open andfUg2[Fcis an open cover,

sayRofX:AsXis compact,Radmits a nite subcover R0:

IfFc2R0, then it is easy to check that fU\F:=VjU2R0nFcg

gives an open cover of F:IfFc62R0;thenfU\F:=VjU2Rg is

an open cover of F:

Lemma 3.4.4. Every compact subset of a Hausdor space is closed.

Proof. LetCbe a compact subset of a Hausdor space X:We will

prove thatXnCis open, which will prove Cis closed in X:

Letxbe a point of XnC:For each point y2C;using the Haus-

dor property, choose disjoint neighbourhoods Ux;VyinXofxandy

respectively. Then R=fVyjy2Cgis an open covering of Cand

asCis compact, this open cover admits a nite subcover of C;say

33munotes.in

## Page 34

Vy1;:::;Vyn:Hence the open set V:=Vy1[[VyncontainsC:Let

U:=Uy1\\Uyn:ClearlyUbeing a nite intersection of open sets

is open inX:Also,U\V=;:(For if there exists r2U\V;then

r2Vyifor some 1inandr2Uyifor all 1inimplies that

r2Uyi\Vyi;which is not possible as Uyi\Vyi=;:)

Check Your Progress

1. Show from rst principles (i.e., using the denition of compact-

ness) that Rwith the usual topology is not compact. (Hint: It is

enough to produce an open cover which does not admit a nite

subcover.)

2. Show that closed subsets of compact sets are compact. (Hint:

Imitate the proof in Lemma 3.4.3.)

3.5 Finite products and compactness

Compactness is preserved under taking nite products i.e., product

of nitely many compact topological spaces is compact. This imme-

diately implies for example that the torus i.e., S1S1is compact in

R2and so is the unit cube in R3;being the nite product of the unit

interval [0;1] with itself.

In order to prove this result, we require a technical lemma, called

the Tube Lemma. Before proving this lemma, we dene a few terms.

Denition 24. LetX;Y be topological spaces. If x02X;the set

x0Yis called a slice of XY:IfWis an open subset of Xcontaining

x0;the setWYis called a tube about x0Y:

Here is a picture to illustrate the above concepts:

34munotes.in

## Page 35

Lemma 3.5.1. LetX;Y be topological spaces. Then the projection

mapsp1:XY!Xandp2:XY!Yare continuous in the

product topology. Let x02X:Then the slice x0Yis homeomorphic

toY:In particular, if Yis compact, then the slice x0Yis a compact

subset ofXY;whereXYis given the product topology.

Proof. Consider the projection map p2:x0Y!Ybyp2(x0;y) =y:

We will check that p2is continuous by checking that the inverse image of

an open set VofYis open inx0Y. For this note that p1

2(V) =x0V;

which is open in x0Y:Clearlyp2is injective as well as surjective and

hence it is a bijection. We now need to prove that the direct image

of a basic open subset in x0Yis open in Y:This will then prove

thatx0Yis homeomorphic to YandYcompact implies that x0Y

is compact as well, since compactness is preserved under continuous

maps.

Note that a set is open in x0Yif and only if it is the intersection

of an open set in XYwithx0Y:Hence, it can be checked (recall

basic open sets) that Wx0Yis open if and only if there exist open

subsetsViofYsuch thatW=[i(x0Vi):Then, clearly p2(W) =[iVi;

is an open subset of Y;proving that p2is a homeomorphism.

Lemma 3.5.2. (Tube Lemma) Let X;Y be topological spaces with Y

compact. If Nis an open subset of XYcontaining the slice x0Y;

thenNcontains a tube WYaboutx0Y;whereWis a neighbourhood

ofx0inX:

Proof. Cover the slice x0Yby basis elements UV;withUVlying

inN:(HereU;V are open subsets of XandYrespectively and XYis

given the product topology. This can be achieved, as x0YNand

Nis an open set in the product topology.) By Lemma 3.5.1, x0Y

being compact admits a nite subcover by these basic elements, say

U1V1;:::;UnVn:AsUiare open in X;so is their nite intersection

U1\\Un;sayW:Asx02Ui;for all 1in;we get that

x02\n

i=1U:=W:

We now claim that the tube WYN;by proving that UiVi

coverWY;i.e., we prove that WY[n

i=1(UiVi):

For this, let ( w;y)2WY:Consider the point ( x0;y);i.e. the

point with the same yco-ordinate as the point we started with. Then,

(x0;y)2UjVjfor somej;asUiViis a cover of x0Y:Then,

(w;y)2UjVj;asw2Wimplies that w2Ujtoo. This proves that

WY[n

i=1(UiVi)N:

Here is a picture of what tube lemma achieves:

35munotes.in

## Page 36

Remark 3.5.1. One has a more general theorem which says that an

arbitrary product of compact spaces is compact too. This theorem is

known as Tychono's theorem. The proof of this is beyond the scope

of these notes, but it is a very useful theorem.

We use the Tube lemma now to prove that the product of two

compact spaces is again compact. The result for nitely many compact

spaces then follows by induction.

Theorem 3.5.1. LetX;Y be compact topological spaces. Then their

Cartesian product XY(with the product topology) is also compact.

Proof. LetRbe an open cover of XY:Givenx02X;the slicex0Y

is compact by Lemma 3.5.1. Hence, there exist nitely many open

subsets, say U1;:::;UnofRwhich cover x0Y:Then,N:=U1[[Un

is an open set containing the slice x0Y:Hence, by Tube Lemma, there

exists a tube say Wx0Yaroundx0Y;contained in N:Clearly,Wx0Y

has a nite subcover, as Wx0YN=U1[[Un:

Now for each x2X;repeat the procedure above to get a tube

WxYcontaining the slice xY:Now observe that fWxjx2Xgis

an open cover of X;asWxare open in Xby construction. As Xis

compact, there exist x1;:::;xn2Xsuch thatXWx1[[Wxn:

Hence,

XY(Wx1Y)[[ (WxnY):

As each ofWxiYhas a nite cover, so does XY:Hence,XYis

compact.

Check Your Progress

1. Letfbe a continuous real-valued function on [ a;b]:Prove that

the graph of f, i.e., the setf(x;f(x))jx2[a;b]gis a compact

subset of R2:

2. State true or false with correct justication (if false, give a counter-

example): An arbitrary union of compact sets is compact.

36munotes.in

## Page 37

3.6 Local compactness and one-point

compactication

There is another more useful notion of compactness called local

compactness. This is a weaker notion than compactness. We shall see

examples of spaces which are locally compact but not compact.

Denition 25. (Local Compactness) A topological space Xis said to

be locally compact at x2Xif there is some compact subset CofX

that contains a neighbourhood of x:IfXis locally compact at each of

its points, then Xis said to be locally compact.

Every compact space is locally compact, because the compact subset

Ccontaining a neighbourhood of xinXcan be taken to be the whole

spaceXitself. The converse need not hold: for example, Rwith the

usual topology is locally compact but not compact.

Check Your Progress

1. Verify that Ris locally compact but not compact.

2. Prove that the space Qof rationals is not locally compact.

3. The innite dimensional product space RNis not locally compact.

Denition 26. LetXbe a topological space and x2X:We sayx

has a local base of compact neighbourhoods, if xhas arbitrarily small

compact neighbourhoods in X:

Lemma 3.6.1. LetXbe a regular space and suppose Xis locally com-

pact atx:Then,xhas a local base of compact neighbourhoods in X:

Proof. LetXbe regular and locally compact at x:Thenxhas a com-

pact neighbourhood CinX:LetUCbe a neighbourhood of xin

X:By regularity, there exists a closed neighbourhood Fofxsuch that

FU:Note thatF\CUis a closed neighbourhood of xinF;since

Cis closed in X:

SinceFis compact, so is F\C:Thus, we get a chain of decreasing

compact neighbourhoods x2F\CC:Repeat the argument above

withF=F\C;to get a local base of compact neighbourhoods for

x2X:

Sometimes your space need not be compact, but adding just one

point to it makes it compact. For example, consider the open interval

(0;1) as a string of thread. If you are able to fuse the two ends of it, by

joining 0 and 1 together, then you would get a circle, which is now a

37munotes.in

## Page 38

compact space! Thus by adding one point suitably you are sometimes

able to get a compact space out of your old space. Such a process is

called as one point compactication.

More precisely, the mathematical denition goes as follows:

Denition 27. (One point compactication) Let Xbe a locally com-

pact Hausdor space. Choose ynot inX:(It is a convention to denote

such ayby the symbol1:) Consider the set Y=X[f1g:Declare

the following subsets of Yto be open:

(a)U, withUopen inX:(We call these open sets of Type I.)

(b)YnC;withCcompact subset of X:(We call these open sets of

Type II.)

We will check below that Ybecomes a topological space when the

above sets are declared as open sets. Such a space Yis called a one-

point compactication of X:

Proposition 3.6.1. LetYbe the set dened above with the open sets

of the two types. Then, Yis a topological space.

Proof. We need to check the following axioms for open sets:

1 Finite intersection of open sets in Yis again an open set in Y:

2 Arbitrary union of open sets in Yis again an open set in Y:

Before we check the rst axiom, we make the following observations:

1(a) LetU1;U2be both open sets of Type I. Then clearly, U1\U2is

again an open set of Type I.

1(b) LetU1be an open set of Type I and U2be an open set of Type

II. We will prove that U1\U2is again an open set in Y:Since

U2is of Type II, there exists a compact subset C2ofXsuch that

U2=YnC2:Hence,U1[U2=U1[(YnC2) =U1[(XnC2):Note

that asC2is compact subset of a Hausdor space, hence closed

inX:Thus,XnC2is open inXand henceU1[(XnC2) is open

of Type I.

38munotes.in

## Page 39

1(c) LetU1;U2be both open sets of Type II i.e, there exist compact

setsC1;C22Xsuch thatU1=YnC1andU2=YnC2:Hence,

U1[U2=Yn(C1\C2):As a nite intersection of compact sets

is again compact, we observe that U1[U2is an open set of Type

II.

Now letU1;U2;:::;Unbe open subsets of Y:By rearranging the

indices that occur namely, 1 ;2;:::;n; we may assume that U1;:::;Un1

are of Type I; Un1+1;Un1+2;:::;Unare of Type II, for some 1 n1n:

Using the observation above, we see that U1\\Un1is again a subset

of Type I and Un1+1\\Unis again a subset of Type II.

Hence,U1\\Unis again an open subset of Y;proving that the

rst axiom for open sets holds in Y:

We now check that an arbitrary union of open subsets of Yis open

inYby making the following observations:

1(a) LetfUg2be a family of open sets of Type I. Then clearly,

[2Uis again an open set of Type I.

1(b) LetfUg2be a family of open sets of Type I and fUg20be

a family of open subsets of Type II.

We will prove that ( [2U)[([20U) is again an open set in Y:

For this observes that ( [2U)[([20U) equalsU[(YnC) =

Yn(Cn(U\C)):Now note that U\Cis open inCand hence,

Cn(U\C) is closed in Cand hence compact, as Cis compact.

Thus, ([2U)[([20U) is a set of Type II and hence open.

1(c) LetfUg20be a family of open sets of Type II. Then [20U

is again an open set of Type II as this equals [20YnC;for

compact subsets CofX:This union now equals Yn\20C:

Now note each Cis a compact subset of the Hausdor space X

and hence is closed in X:Hence,\20Cis a closed subset of X:

Moreover, for a xed 02;\20CC0and hence we have

a closed subset of a compact set and hence \20Cis a compact

subset ofX:This implies that we get that [20Uis an open set

of Type II.

Using these observations, it easily follows that an arbitrary union

of open sets is again open.

Remark 3.6.1. Every space admits a unique one point compactica-

tion upto homeomorphism.

Here is an example to keep in mind:

39munotes.in

## Page 40

Example 4. One-point compactication of (0 ;2) is homeomorphic

with the circle S1:

Dene a map f: (0;2)[f1g!S1by:f() = (cos();sin());

for allx2(0;2) andf(1) = (1;0):Clearly,fis a bijection.

We now prove that fis continuous. Let VS1be open. Then,

either (1;0)62Vor (1;0)2V:If (1;0)62V;thenVis an arc of S1

not containing (1 ;0) and hence f1(V) is of the form ( 1;2) for some

1< 22(0;2):In thisf1(V) is an open subset of (0 ;1):

If (1;0)2V;then there exists an arc around (1 ;0) contained in V:

Thus, this arc consists of some points in the rst quadrant and some

points in the fourth quadrant, which are determined by angles 1< 2

with1>0 and2<2:Thus, in this case f1(V) =f1g[ (0;1)[

(2;2):Clearly, then f1(V) =YnC;whereCis the compact subset

[1;2] of (0;1):This nishes the proof that fis continuous.

LetY= (0;2)[f1g:We will now prove that the direct image of

an open subset of Yunderfis again open. Similar to the case above,

it is easy to check that if Uis an open subset of (0 ;2);thenf(U) is

open inS1:LetUbe an open subset of (0 ;2)[f1g containing1:

Then,Uis of the form YnC;for some compact subset CofX;then

f(U) =f(Y)nf(C):AsCis compact and we have proved that fis

continuous, we get that f(C) is compact and hence closed in S1:Hence,

f(U) =S1nf(C) is an open subset of S1;proving that fis an open

map. This completes the proof that fis a homeomorphism.

We now study the basic properties of one-point compactication.

Theorem 3.6.1. LetXbe a locally compact Hausdor space which is

not compact. Let Ybe the one-point compactication of X:Then,Yis

a compact, Hausdor space; Xis a subspace of Y;the setYnXconsists

of a single point and X=Y:

Proof. We will prove rst that Xis a (topological) subspace of Yi.e.,

we will prove a set is open in Xif and only if it is the intersection with

Xof some open set in Y:LetUXbe open. Then, clearly U=Y\U;

andUis open inYby the topology on Y:Conversely, let U=V\X;

for an open set VinY:Since open sets in Yare of two types, suppose

Vis of the rst type: i.e., let V=U1;for some open subset U1ofX:

Then clearly Vis open inX:

SupposeVis of the second type i.e., V= (YnC)\X;for someC

compact in X:Then,V= (Y\X)n(C\X) =XnC=Cc;asCX:

SinceCis a compact subset of a Hausdor space, it is closed in Xand

henceCcis open inX;proving that Vis open inX:

We will now prove that 1is a limit point of Xand this will imply

X=Y:For this we need to show that every open subset around 1

intersectsXin a point dierent from 1:For this observe that every

open subset around 1is the complement in Yof a compact subset C

40munotes.in

## Page 41

ofX:AsXis not compact, Cis a proper subset of Xand hence the

complement XnCcontains at least one point. This then implies that

every open subset around 1intersectsXin a point dierent from 1;

proving that1is a limit point of X:

Clearly,YnXis the single point innity and it now remains to

prove that Yis compact and Hausdor. First we prove compactness

ofY. LetRbe an open covering of Y:As1should belong to this

collection,Rcontains an open subset of Type II, say of the form YnC;

withCcompact in X:Now take all members of Rother than YnC

and intersect them with X:They form a collection of open sets in X

coveringC:Compactness of Cimplies that there exist nitely many of

them covering C:Take the corresponding nitely many elements of R

alongwithYnCto get an open cover of Y:

Let us now check that Yis Hausdor: let x;y2Y:If bothx;ylie

inX;thenXHausdor implies that there exist disjoint open sets U;V

inXsuch thatx2Uandy2V:Ifx2Xandy=1;then local

compactness of Xallows us to choose a compact set CinXcontaining

a neighbourhood Uofx:Then,UandYnCare disjoint neighbourhoods

ofxand1respectively in Y:

Check Your Progress

1. Prove that the composition of two homeomorphisms is again a

homeomorphism.

2. Show that the one-point compactication of (0 ;1) is also homeo-

morphic to S1:(Hint: Use that (0 ;1) and (0;2) are homeomor-

phic and use the exercise above.)

3. Check that the open sets in Xdescribed in Theorem 3.6.1 above

do give a topology on X:

3.7 Lindel of Topological Spaces

Denition 28. A topological space Xis said to be a Lindel of space,

if every open cover of Xadmits a countable subcover.

Lemma 3.7.1. The Lindel o property is preserved under continuous

functions i.e., if Xis a Lindel of topological space and f:X!Yis

continuous, then f(X)is Lindel of.

Proof. LetXbe a Lindel of space and f:X!Ybe continuous. We

will prove that f(X) is Lindel of. Let fVg2be an open cover of

f(X):Then, it is easy to check that ff1(V)g2is an open cover

41munotes.in

## Page 42

ofX:AsXis Lindel of,ff1(V)g2admits a countable subcover,

sayff1(Vi)gi2;i2N:Then it can be checked that ( Vi)i2;i2Nis a

countable subcover of fVg2forf(X):This proves that f(X) is Lin-

del of.

Check Your Progress

Show that if Ais a closed subspace of a Lindel of topological space

X;thenAis Lindel of.

Show that if Xis compact and Yis Lindel of, then XYis

Lindel of.

Theorem 3.7.1. For a metric space (X;d)the following are equivalent:

(a)(X;d)is Lindel of.

(b)(X;d)is separable.

(c)(X;d)is second countable.

Proof. (a) =)(b) Let (X;d) be a Lindel of metric space. We need

to show that Xhas a countable dense subset. For each n2N;let

Cn=fB(x;1

n)jx2Xg:Then, clearlyCnis an open cover of X:As

Xis Lindel of, there exists a countable subcover Dn=fB(xni;1

n)ji

1;i2NgofCn;for a xedn:LetD=fxn;ijn2N; i2Ng:Clearly

Dis a countable subset of X, asDis a countable union of countable

sets. We will now show that Dis dense in X;which will prove that

(X;d) is separable.

Lety2X:We will prove that for every r >0; B(y;r)\D6 =;;

which will prove that Dis dense in X:To see this, choose m2Nsuch

that1

m< r: Then,B(y;1

m)B(y;1

r):Since for this m;Dmis a cover

ofXthere exists k2N;such thaty2B(xm;k;1

m)2Dmwhich means

thatd(y;xm;k)<1

mi.e.,d(xm;k;y)<1

mi.e.,xm;k2B(y;1

m)B(y;r):

Thusxm;k2B(y;r)\D;proving that B(y;r)\D6=;;as required.

(b) =)(c) We will prove that if ( X;d) is a separable metric

space, then it is second countable. Let ( X;d) be a separable metric

space i.e., there exists a subset AofXsuch thatAis countable and

c(A) :=A=X:ConsiderB:=fB(x;r)jx2A;r2Q;r>0g:Clearly,

Bis a countable collection of open sets in X:We will show that Bis a

basis for (X;d);which will prove that ( X;d) is second countable.

For this, let x2Xand letUbe an open set in Xsuch thatx2U:

We will prove that there exists B2Bsuch thatx2BU:Sincex2U

andUis open, there exists r >0;r2Qsuch thatx2B(x;r)U:

AsA=X;there exists a2Asuch thata2A\B(x;r)nfxg:Let

B=B(a;r

2):Then,Bbelongs toBand using the triangle inequality,

it is easy to check that x2BU:This prove thatBis a countable

basis for (X;d);proving that ( X;d) is second countable.

42munotes.in

## Page 43

(c) =)(a) Let (X;d) be a second countable space. Let fUigi2I

be an open cover of X:SinceXhas a countable basis, denote this by

fVngn2N;whereVnare open subsets of X:LetS=fn2NjVn

Uifor some i2Ig:Clearly,Sis countable as a subset of a countable

set is countable. Using the fact that fVngn2Sis a basis of X;it is easy

to check thatfVngn2Sis an open cover of X:For eachn2S;choose

i(n)2Isuch thatVnUi(n);for somei(n)2I:Then,fUi(n)gn2S

gives the required countable subcover of X;proving that Xis Lindel of.

(Note here that we did not use the fact that Xwas a metric space.

Thus this result is more general, but here we will restrict ourselves to

metric spaces.)

In general, Lindel o property is not hereditary. Here is a counter-

example. Let Xbe an uncountable set and let x02X:Let=fA

Xjx062Ag:It can be checked that is indeed a topology on X:It is

also easy to check that Xis Lindel of, in fact Xis compact. It can now

be checked that ( Y;Y) withY=Xnfx0gwith the subspace topology

Yis not Lindel of.

In general, even a nite product of Lindel o spaces is not Lindel o.

Other results regarding these spaces are very technical. We shall not

go into further details about this property, but the interested reader

can look up the references for further reading.

3.8 Glossary

In this chapter, you have learnt the following:

Open covers and nite subcover of a given cover.

Denition of Compactness: Every open cover has a nite sub-

cover.

Continuous image of a compact set is compact.

Closed subset of a compact topological space in compact.

Tube lemma.

Products of nitely many compact spaces is compact.

Local compactness: every point xis contained in a compact set

containing a neighbourhood of x:

A regular space, locally compact at x;admits a local base of

compact neighbourhoods of x:

One-point compactication.

43munotes.in

## Page 44

Basic properties of one-point compactication.

3.9 Bibliography

Bartle and Sherbert, Introduction to Real Analysis, Wiley.

James Munkres: Topology, Pearson.

Rudin, W.: Principles of Mathematical Analysis. Third Edition,

Mc-Graw Hill International Editions.

Singer, I. M., Thorpe, J. A.: Lecture Notes on Elementary Topol-

ogy and Geometry.

3.10 Let Us Sum Up

In this chapter, we rst learn the abstract denition of compactness.

A topological space is said to be compact, if every open cover of it ad-

mits a nite subcover. Compactness is a nice property of a topological

space, as it is preserved under continuity and nite products. These

properties help us to get more examples of compact spaces, like the

unit circleS1and the torus, S1S1:One important ingredient in the

proof of the second property (namely, nite product of compact spaces

is compact) is the Tube Lemma.

Sometimes, a topological space may fail to be compact, but it may

be locally compact i.e., every point has a compact neighbourhood. For

example, the real line with the usual topology is locally compact, but

not compact. Out of such locally compact spaces, some spaces are not

very far away from being compact. These spaces can then be made

compact by adding one point and dening a suitable topology on this

new space, such that it becomes compact. This process is called the

Alexandro one-point compactication. We have studied the properties

of one-point compactication and seen one explicit example of such a

compactication.

3.11 References for further reading

1. George Simmons: Topology and Modern Analysis, TataMcgraw-

Hill.

44munotes.in

## Page 45

2. M. A. Armstrong: Basic Topology, Springer UTM

3. W.J. Pervin: Foundations of General Topology, Academic press ,

New York, London.

4. Kelley J.L, General Topology, Van Nostrand Reinhold Co., New

York, 1955.

5. Wolfgang Thorn: Topological Structures; Holt, Rinehart and

Winston, New York, Chicago.

3.12 Chapter End Exercises

1. Show that an arbitrary intersection of compact sets is again a

compact set.

2. This is a generalization of the above exercise: Let f:X!Y;

withYcompact. Then fis continuous if and only if the graph

off; Gf:=f(x;f(x))jx2Xgis closed in XY:

3. Show that the one-point compactication of Ris homeomorphic

toS1:

4. A relation Con a setXis said to be a simple order if it has the

following properties:

(a) (Comparibility) For every x;y2Xfor whichx6=y;either

xCy oryCx holds.

(b) (Non-re
exivity) For no x2Adoes the relation xCx hold.

(c) (Transitivity) If xCy andyCz; thenxCz:

A setXwith a simple order is said to have the least upper bound

property if every non-empty subset X0ofXthat is bounded above

has a least upper bound.

Show that every simply ordered set with the least upper bound

property is locally compact.

5. Consider the set Qof all rational numbers as a metric space with

the usual metric given by d(p;q) =jpqj:LetEbe the set of all

p2Qsuch that 2 inQ;but not compact.

6. Suppose that Mis compact and f:M!Nis continuous, one-

one and onto. Prove that fis a homeomorphism.

45munotes.in

## Page 46

7. Show that if Yis compact, then the projection 1:XY!X

is a closed map i.e., 1carries closed sets to closed sets.

8. Prove that local compactness is preserved under continuous, open

functions.

9. LetXbe an innite set with a distinguished point x0:LetT

consist of the empty set and all subsets of Xcontainingx0:Prove

that (X;T) is a locally compact space.

10. LetKRconsist of 0 and the numbers1

n;for alln1:Prove

thatKis a compact subset of R:

46munotes.in

## Page 47

Chapter 4

Compact and Complete

Metric Spaces

Chapter Structure

4.1 Introduction

4.2 Objectives

4.3 Equivalent formulations of Compactness for Metric Spaces

4.4 Compact in Rni Closed and Bounded

4.5 Completeness and Completion in Metric Spaces

4.5.1 Compete Metric Spaces

4.5.2 Completion of a Metric Space

4.6 Lebesgue Covering Lemma

4.7 Uniform Continuity Theorem

4.8 Glossary

4.9 Bibliography

4.10 Let Us Sum Up

4.11 References for Further Reading

4.12 Chapter End Exercises

4.1 Introduction

The aim of this unit is to give for metric spaces other equivalent

formulations of compactness: sequential compactness and limit point

compactness. The main theorem here is that a subset EofRnis com-

pact if and only if Eis closed and bounded in Rn:We then study

complete metric spaces and characterize them in terms of compactness

and total boundedness. We brie
y study the notion of completion of a

metric space. We then try to understand compactness better using the

notion of Lebesgue covering. This helps us to prove a very important

theorem: every continuous map from a compact metric space to any

47munotes.in

## Page 48

other metric space is uniformly continuous.

4.2 Objectives

After going through this chapter you will be able to:

Give various equivalent denitions of compactness for metric spaces.

ShowERnis compact if and only if it is closed and bounded.

Dene complete metric spaces and completion of a metric space.

State Lebesgue covering lemma and nd Lebesgue number of a cov-

ering.

Prove uniform continuity theorem.

4.3 Equivalent formulations of

compactness for metric spaces

In the previous chapter, we have studied the notion of compactness

using open covers. However, checking whether a space is compact via

this denition may not be always be easy and so, one tries to see if there

are other equivalent denitions which might work. In this section we

study these other notions of compactness: sequential and limit point

compactness and prove that all three denitions of compactness are

equivalent for metric spaces.

Denition 29. (Sequentially compact) Let Ybe a topological space.

Yis said to be sequentially compact, if every sequence in Yhas a

convergent subsequence.

Denition 30. (Limit point compact) Let Ybe a topological space.

Yis said to be limit point compact if every innite subset of Yhas a

limit point.

We state a helpful lemma which is a consequence of sequential com-

pactness.

Lemma 4.3.1. LetXbe sequentially compact. Then for every >0;

there exists a nite covering of Xby"-balls.

Proof. We prove the contrapositive of this statement: if there exists

>0 such that Xcannot be covered by nitely many -balls, then X

is not sequentially compact.

SupposeXcannot be covered by nitely many -balls. We shall

construct an innite sequence of points ( xn) such that ( xn) has no

convergent subsequence. Start with any x12X:Then, there exists x22

48munotes.in

## Page 49

XnB(x1;):(This is so as by assumption, Xcannot be covered by a

single ballB(x1;):) Having chosen x1;x2;:::;xncontinue by induction

to getxn+162B(x1;)[B(x2;)[B(xn;):

The choice of xn's implies that d(xn+1;xi);for all 1in:

Hence the sequence ( xn) can have no convergent subsequence.

The above three notions of compactness are in general not equiv-

alent. The class of metric spaces is a good class of topological spaces

where the three notions of compactness agree. In fact, the three no-

tions of compactness are equivalent for a slightly bigger class, namely

for the class of metrizable spaces. (See the book by James Munkres for

the denition and further details.) We prove some implications of this

equivalence below:

Theorem 4.3.1. Let(X;d)be a metric space. Then one has the fol-

lowing implications:

(1)Xis compact implies Xis limit point compact.

(2)Xis limit point compact implies Xis sequentially compact.

Proof.We will rst prove Xcompact implies Xlimit point com-

pact.

LetXbe compact and let Abe an innite subset of X:We have to

prove thatAhas a limit point in X:We prove the contrapositive

of this statement: if Ahas no limit point in X;thenAis nite.

AssumeAhas no limit point. Then Acontains all its limit points

and is hence closed. Thus Ais a closed subset of a compact space

and hence itself compact. Since no a2Ais a limit point of A;

there exists a neighbourhood Uaofadisjoint from Anfag:In

fact,Ua=fag;for if it contains any other b6=a;then it would

not be disjoint from Anfag:Clearly,R:=fUaja2Agis an

open covering of A:SinceAis compact,Rhas a nite subcover,

sayUa1;:::;Uan:Hence,AUa1[[Uan=fa1;:::;ang:Since

every subset of a nite set is again nite, we get that Ais nite,

as required.

We now prove Xlimit point compact implies Xis sequentially

compact.

We will prove that every sequence in Yhas a convergent sub-

sequence. Let ( xn)n2Nbe a sequence in Y:Consider the set

A=fxnjn2Ng:We make two cases: Ais nite or Ais

innite.

{SupposeAbe nite. We claim that there exists xsuch that

x=xn;for innitely many n2N:To see this, dene f:N!

49munotes.in

## Page 50

Abyf(n) =xn:Then, N=[x2Af1(x):If the number of

elements in f1(x) had been nite for every x2A;then

we would get that the set of natural numbers is a nite

set, which is a contradiction. Hence, there exists x2A

such thatf1(x) is an innite subset of Ni.e., there exist

innitely many nsuch thatf(n) :=xn=x:Now it is obvious

that (xn)n2f1(x);being the constant sequence is a convergent

subsequence of ( xn)n2N:

{SupposeAis innite. As Xis limit point compact by as-

sumption, we have that Ahas a limit point, say x:We will

now dene a subsequence of ( xn) converging to xas follows:

sincexis a limit point of A; B (x;1);the ball around xof

radius 1 contains a point of Aother than x:Thus, one can

choosen1such thatxn12B(x;1):Applying inductively the

same argument again, given a positive integer ni1we can

choose an index ni> ni1such thatxni2B(x;1=i):(The

guarantee that ni>ni1is due to the fact that Ais innite.)

The choice of xninow implies that the subsequence xn1;xn2;:::

converges to xinA:

Having come this far, it is natural to ask if a metric space ( X;d)

is sequentially compact then is it compact too? The answer is yes,

but requires the notion of Lebesgue number of a covering. This will

be developed in the last section of this chapter and we will then prove

that sequential compactness implies compactness for metric spaces.

In general, the above implications need not hold. We will now see

some examples of these.

Example 5. LetXbe a two-point space in the indiscrete topology.

Then,XNis limit point compact but not compact.

Lemma 4.3.2. LetXbe a sequentially compact topological space. Then,

Xis limit point compact.

Proof. LetAbe an innite subset of X:Then,Acontains a sequence in

X:SinceXis sequentailly compact, this sequence in Ahas a convergent

subsequence. The limit of this convergent subsequence is a limit point

ofA:

4.4 Compact in Rnif and only if closed

and bounded

50munotes.in

## Page 51

In this section, we are going to characterize compact sets in Rn:

Before proving that, let us recall some basic notions related to metrics

onRn:First recall the denition of a metric:

Denition 31. A metric on a set Xis a function d:XX!R

having the following properties:

For allx;y2X; d(x;y)0 where equality holds if and only if

x=y:

For allx;y2X; d(x;y) =d(y;x):

For allx;y;z2X; d(x;y)d(x;z) +d(z;y):(This inequality is

called the triangle inequality.)

Check Your Progress

1. Forx= (x1;x2;:::;xn)2Rnandy= (y1;y2;:::;yn)2Rn;check

thatd(x;y) = (P

i(xiyi)2)1

2is a metric. (This is called the

Euclidean metric on Rn:)

2. Forx= (x1;x2;:::;xn)2Rnandy= (y1;y2;:::;yn)2Rn;check

that(x;y) = maxfjxiyijgis also a metric on Rn:

3. With notations as above, show that (x;y)d(x;y)pn(x;y):

4. Use the fact above to prove that the topologies induced by the

Euclidean metric and the square metric are the same.

5. Also show that the topologies induced by the Euclidean metric

and square metric are same as the product topology on Rn:

We will prove an important theorem in this section. The ideas in

here are also used once again in the later sections. This theorem also

has an important application: it helps us to characterize completely

compact subsets of Rn:

We x up some notation:

Denition 32. Ifa**is called a closed interval.**

Lemma 4.4.1. LetfIngbe a sequence of closed intervals in Rsuch

that for all n2N; InIn+1:Then,\1

i=1In6=;:

Proof. The proof is based on the least upper bound property (lub prop-

erty) of real numbers. If In= [an;bn];letEbe the set of all an:Then

Eis non-empty and bounded above by b1;for example. Let xbe the

least upper bound of E, which exists as Eis a non-empty bounded

subset of R:Now observe that for all m;n2Nanan+mbn+mbm

which then implies that xbmfor eachm; x being the least among

51munotes.in

## Page 52

upper bounds of E:By denition, amx;for allm2N:Thus, for all

m2N; x2[am;bm] :=Im;proving that x2\1

i=1In. This proves that

\1

i=1In6=;:

Theorem 4.4.1. Every closed and bounded interval in Ris compact.

Proof. LetI= [a;b] be a closed and bounded interval in R:Then, for

allx;y2I;jxyjba:Letba=:We will prove Iis compact.

Suppose not, i.e., suppose that there exists an open cover fGg2of

Iwhich contains no nite subcover of I:Letcbe the mid-point of

[a;b] i.e.,c=a+b

2:Then, clearly I= [a;c][[c;b]:Then atleast one of

the intervals [ a;c] or [c;b] cannot be covered by any nite collection of

fGg2;(otherwise so would be I:) Denote the interval which cannot

be covered by any nite collection by [ a1;b1] and apply the above ar-

gument again to this interval I1:= [a1;b1]:Next subdivide I1as above

to get anI2I1which cannot be covered by any nite collection of

fGg2:Thus continuing further, we get a sequence of closed intervals

Insatisfying the requirements of Lemma 4.4.1 and also having for all

x;y2In;jxyj2n:By Lemma 4.4.1, there exists a point xlying

inInfor alln2N:SincefGg2is an open cover of I;there exists

ansuch thatx2GwithG2fGg2:SinceGis open, there

existsr>0 such thatjyxjproperty, there exists n2Nlarge such that 2nthatInG;which contradicts the choice of In:

Hence,Iis compact.

Theorem 4.4.2. A subsetAofRnis compact if and only if it is closed

and bounded (in the euclidean metric dor the square metric :)

Proof. It is enough to prove that Ais bounded under d:This follows

from the inequalities (x;y)d(x;y)pn(x;y) asAis bounded

underdif and only if it is bounded under :

LetAbe a compact subset of Rn:By Lemma3.4.4, it is closed in Rn:

Consider the collection of open sets fB(0;m)jm2Ng:Clearly this

is a nested open cover of Rnand its intersection with Agives a nested

open cover of A:Compactness of Aimplies that there exists a nite

subcover of Aindexed by m1;m2;:::;mn. Lets= maxfm1;:::;mng:

Then, it is clear that AB(0;s);proving that Ais bounded under :

Conversely, suppose that Ais closed and bounded in i.e., suppose

(x;y)Nfor every pair of points x;y2A:Choosex02Aand let

(x0;0) =b:Then,(x;0)(x;x 0) +(x0;0)N+b;for every

x2A:LetP=N+b:ThenA[P;P]n:Since [P;P]nis a cube

inRn;it is compact, being a nite product compact subsets of R:(See

Theorem 3.5.1.) As Ais a closed subset of this compact set, Ais also

closed in Rn:(See Lemma 3.4.3.)

52munotes.in

## Page 53

4.5 Completeness and completion in

metric spaces

We will study another very helpful notion in this section: complete-

ness. The word \completeness" is used in mathematics in the same

sense as in English: absence of gaps. We will then study the relation

between compactness and completeness by introducing a geometric con-

cept of total boundedness.

In the earlier chapters, we have studied the notion of one-point

compactication, which helps us to get a compact space out of a non-

compact one. Here too, we will \complete" spaces that are not com-

plete: for example the set of rational numbers is not a complete metric

space and its completion gives us the real line. Though the formal def-

inition may look very intimidating, the underlying idea is simple: one

is just trying to ll up the gaps in a space which is not complete.

4.5.1 Complete metric spaces

Before we go to complete metric spaces, we look at the notion of total

boundedness.

Denition 33. Let (X;d) be a metric space. A subset AofXis said to

be totally bounded if given >0 there exist a nite number of subsets

A1;:::;AnofXsuch that diam Ak< (for all 1kn) and such

thatA[n

k=1Ak:

Theorem 4.5.1. If a subset Aof a metric space (X;d)is totally

bounded, then Ais bounded.

Proof. IfAis totally bounded, then there exist nonempty subsets A1;A2;:::;An

ofXsuch that diam Ak<1;for all 1knandA[n

k=1Ak:For each

kbetween 1 and nletak2Akbe any point. Let D=Pn1

i=1d(ai;ai+1):

Now letx;y2A:Then, without loss of generality there exist

1lmnsuch thatx2Alandy2Am:Then,d(x;y)d(x;al)+Pm1

t=ld(at;at+1)+d(am;y):Since diamAt<1;we haved(x;al);d(am;y)<

1:Hence,d(x;y)<1 +D+ 1 =D+ 2;for allx;y2A:This proves

thatAis bounded.

We state without proof an important equivalent condition for total

boundedness in metric spaces. (For details see the book by Goldberg,

Methods of Real Analysis.)

Theorem 4.5.2. Let(X;d)be a metric space. A subset AofXis

totally bounded if and only if every sequence of points of Acontains a

Cauchy subsequence.

53munotes.in

## Page 54

Using Theorem 4.5.2, we show that bounded need not imply totally

bounded.

Example 6. Consider the metric space l2;consisting of all sequences

(xn) such thatPx2

n<1:Then,l2is a metric with d(x;y) :=jjxyjj2;

where for a sequence s= (sn)2l2;jjsjj2= (P1

i=1s2

n)1

2:For eachi2N;

leteibe the sequence all whose terms are zero, except the i-th term

which is one. Let Ebe the set of all eias above. Then one can

easily check that for j6=k; d(ej;ek) =jjejekjj2=p

2:Thus,E

is bounded with diameterp

2:HoweverEis not totally bounded as

e1;e2;:::cannot have any Cauchy subsequence, since d(ej;ek) =p

2;

which remains quite large.

4.5.2 Completion of a metric space

Denition 34. (Completion of a metric space) Let ( X;d 1) and (Y;d 2)

be two metric spaces.

An isometric imbedding of XintoYis a mapf:X!Ysuch

thatd2(f(x);f(x0)) =d1(x;x0);for allx;x02X:

If there exists an isometric imbedding fof (X;d 1) into a complete

metric space ( Y;d 2) such that f(X) is dense in Y;then (Y;d 2) is

called a completion of ( X;d 1):

A trivial examples of an isometric imbeddings is that of the identity

map from Rto itself.

The main theorem of this section is to prove the existence of a

completion of a metric space ( X;d 1):Before that we outline the con-

structions and state certain results required in the construction. The

completion of a metric space is a certain quotient space of the space

of all Cauchy sequences in X:(The material in this section depends

heavily on Section 4 of the book by K. D. Joshi, Introduction to Gen-

eral Topology. Readers are requested to look up the section there for

further details.)

Let us recall the denition of a Cauchy sequence.

Denition 35. Let (X;d) be a metric space. A sequence ( xn) inXis

said to be Cauchy, if for every >0 there exists an integer N2Nsuch

thatd(xn;xm)<for alln;mN:

We now introduce a relation on the set of Cauchy sequences. The

equivalence classes under this relation will help us to get a complete

metric space out of our given space.

Denition 36. Let (X;d) be a metric space. A sequence ( xn) inXis

said to be Cauchy, if for every >0 there exists an integer N2Nsuch

thatd(xn;xm)<for alln;mN:

54munotes.in

## Page 55

Denition 37. Two Cauchy sequences ( xn);(yn) in (X;d) are said to

be equivalent, if d(xn;yn)!0 asn!1:

It is easy to check that the above relation is an equivalence relation

and letSdenote the set of equivalence classes of Cauchy sequences in

(X;d):Given a Cauchy sequence x= (xn);let ^xdenote the image of x

inS:We are now in a position to state the main theorem:

Theorem 4.5.3. LetSbe the set of Cauchy sequences in a metric

space (X;d):Let^Xdenote the set of all equivalence classes of Sunder

the relation above. We can make ^Xinto a metric space by dening

e([^x];[^y]) = lim n!1d(xn;yn):Leth:X!^Xbe given by h(x) = [^x]:

Then, ^Xis the completion of the metric space (X;d):

Proof. To prove the theorem, we need to prove several things. We will

only list them here rst and ask to readers to supply details themselves

one by one.

Proving that the relation on Sis an equivalence relation.

Proving the function eis well-dened.

Proving that the function eis a metric.

Proving that his an isometric imbedding.

Proving that h(X) is dense in ^X:

Proving that ^Xis a complete metric space.

Remark 4.5.1. The property of completion is slightly dierent from its

sister property of compactness: compactness is a topological property,

but completion is not a topological concept i.e., it is not invariant

under homeomorphism. For example, there is a homeomorphism f:

(

2;

2)!RwithRis complete but (

2;

2) not complete.

4.6 Lebesgue Covering Lemma

In this section, we seek a measure of how big sets can be. This

quantication is made more precise using the notions of diameter of a

set and associating a number to every covering of a compact topolog-

ical space. (See http :==en:wikipedia:org=wiki=Henri Lebesgue for

more details on the mathematician Lebesgue, after whom this lemma

is named.)

We begin with the denition of the diameter of a bounded set:

55munotes.in

## Page 56

Denition 38. LetAbe a bounded subset of a metric space ( X;d).

Then, diameter of A;denoted by diam( A) is dened to be lub fd(a1;a2)ja1;a22

Ag:

Examples:

(1) LetX=Rwith the absolute metric. Let A= [0;1]:Then,

diam(A) := lubfjxyjjx;y2Ag:Note that for all x;y2A;

jxyj1:It is also easy to check that one is the least upper

bound ofA:Hence, diam( A) = 1:

(2) LetX=R2with the usual distance metric. Let A=S1:

Then, diam( S1) := lubfd(x;y)jx;y2S1g:By denition of

S1;Ifx= (x1;x2) andy= (y1;y2) then we have d(x;y) =p

(x1y1)2+ (x2y2)22:Also, note that the pair of point

(1;0) and (1;0) are at a distance 2 ;proving that 2 is the diame-

ter. (This matches with our usual notion of diameter from school

days.)

Denition 39. LetRbe an open covering of the metric space ( X;d):

If there exists >0 such that for each subset of Xhaving diameter

less than;there exists an element of Rcontaining it, then is called

a Lebesgue number for the covering R:

The main aim of this section is to prove the existence of a Lebesgue

number for compact metric spaces. Before we do this, we prove that

sequential compactness guarantees the existence of a Lebesgue number

for every open covering of X:

Lemma 4.6.1. LetXbe a sequentially compact space and let Rbe any

open cover of X:ThenRhas a Lebesgue number :

Proof. We shall prove the contrapositive of this statement: suppose

there is no >0 such that every set of diameter less than lies in at

least one element of RimpliesXisnotsequentially compact.

Now suppose there is no such ;i.e., for each >0;there exists a

subset ofXhaving diameter less than which does not lie inside any

element ofR:In particular, for each n2Nwe can choose a set Cn

having diameter less than1

nwhich is not contained in any element of

R:Choose for each na pointxn2Cn:

We claim that such a chosen sequence ( xn) has no convergent subse-

quence. For suppose ( xn) had a subsequence ( xni) converging to x:Now

xlies in some element AR:AsAis open, there exists >0 such that

B(x;)A:Chooseilarge enough so that d(xni;x)<

2and1

ni<

2:

NowCnilies in the1

nineighbourhood of xniand henceCniB(x;):

Then,CniA;contradicting the choice of the sets Cn:

56munotes.in

## Page 57

Check Your Progress The proof above is a nice illustration of

one of the ways of attacking a problem in mathematics: the existen-

tial approach. The above proof asserts the existence of a without

actually telling how one can nd or construct it. Such a method of

proof was rst developed by one of the in
uential and universal math-

ematicians of the nineteenth and twentieth centuries, David Hilbert.

(Seehttps :==en:wikipedia:org=wiki=David Hilbert ). During those

days, Hilbert was criticized for such a method of proof and in fact he

was told by Gordan (another great mathematician) that this was not

mathematics, but it was theology. Later, however Hilbert's method of

thought became a very important way for going about a proof, specially

in pure mathematics. Many of the proof in mathematics today are of

this kind and one then often needs to write down algorithms to make

the computations explicit.

Can you nd at least two more examples of such existential proofs?

(Hint: Linear Algebra).

Let us now prove that for a metric space ( X;d) sequential compact-

ness implies compactness. (See Theorem 4.3.1.)

Theorem 4.6.1. Let(X;d)be a metric space. If (X;d)is sequentially

compact, then (X;d)is compact.

Proof. LetRbe an open covering of X:SinceXis sequentially com-

pact, by Lemma 4.6.1 Rhas a Lebesgue number :Apply Lemma 4.3.1

with==3;to get a nite covering of Xby balls of radius =3:Each

of these balls has diameter atmost 2 =3<so we can choose for each

of these balls an element of Rcontaining it. Thus, we get a nite

subcollection ofRthat covers X;proving that Xis compact.

Remark 4.6.1. The property of completion is slightly dierent from its

sister property of compactness: compactness is a topological property,

but completion is not a topological concept i.e., it is not invariant

under homeomorphism. For example, there is a homeomorphism f:

(

2;

2)!RwithRis complete but (

2;

2) not complete.

We can now record the Lebesgue covering lemma for compact spaces.

The proof of it follows immediately from the facts proved above.

Lemma 4.6.2. LetRbe an open covering of the metric space (X;d):

IfXis compact, there is a >0such that for each subset of Xhaving

diameter less than ;there exists an element of Rcontaining it. Such

a numberis called a Lebesgue number for the covering R:

Here is an important consequence of completeness:

Theorem 4.6.2. Let(X;d)be a complete metric space. For each n2N

letFnbe a closed and bounded subset of Xsuch thatF1F2

FnFn+1 anddiam(Fn)!0asn!1:Then\1

n=1Fncontains

precisely one point.

57munotes.in

## Page 58

Proof. For eachn2Nletanbe any arbitrary point of Fn:Then as

theFn's form a nested sequence, we get an;an+1;an+2;:::2Fn;i.e.,

an+k2Fnfor allk0:(Call this property P:)

Since the diameter of Fn's tends to zero as ntends to innity, given

> 0 there exists an integer N2Nsuch that diam( Fn)< and

aN;aN+1;aN+2;:::all lie inFN:Thus form;nNwe haved(an;am)

diam(FN)< : This proves thatfang1

n=1is a Cauchy sequence. Since

Xis complete this Cauchy sequence converges to a point, say ainX:

By property Pand the fact that a2X;we get that ais a limit point of

Fn;for everyn1 and for every closed subset Fn:Since for all n1;

Fnis closed, we get that a2Fnfor alln1:This shows that\1

n=1Fn

is non-empty.

We now prove uniqueness: if there exist a;b2Xsuch that both

a;b2\1

n=1Fn;then there exists a Ksuch thatd(a;b)>diam(FK) for

Ksuciently large. Thus, bcannot lie in\1

n=1Fn, a contradiction.

Here is an important theorem which relates the various concepts

that you have studied till now:

Theorem 4.6.3. A metric space (X;d)is compact if and only if (X;d)

is complete and totally bounded.

Proof.Let (X;d) be compact. By Theorem 4.6.1 it is enough to

prove that every sequence ( xn) has a subsequence which converges

to a point of X:

Suppose that for each point x2X;there exists an open ball Bx

which contains xfor only nitely many values of n:The family

of all suchBxwould then be an open cover of XBy hypothesis,

Xwould then be covered by a nite number of Bx;which is

impossible as the union of nitely many Bxwould contain only

nitely many xn:

Hence there exists x2Xsuch that every open ball around x

containsxnfor innitely many n:Hence there exists n1such that

xn12B(x;1);there exists n2>n 1such thatxn22B(x;1

2);and

continuiung for any k;there exists nk> nk1such thatxnk2

B(x;1

k):This subsequence ( xnk) of (xn) converges to the point x

inX, thus proving that Xis complete.

Let (X;d) be complete and totally bounded. Suppose Xis not

compact, i.e. there exists an open cover RofXsuch that no

nite number of sets of Rform a cover of X:

AsXis totally bounded, it can be written as a union of a nite

number of bounded subsets each of whose diameter is less than

one. Then, one of these subsets, say A1cannot be covered by

nitely many subsets of R:(ElseXwould be covered by nitely

58munotes.in

## Page 59

many subsets ofR:) As diamA1= diamA1;A1is a closed subset

ofXwith diameter less than one and which cannot be covered

by a nite number of sets from R:

SinceA1is itself totally bounded, the same reasoning shows that

there exists A2A1such that diam A2<1

2andA2cannot be

covered by nitely many elements of R:Proceed inductively to

get a nested family of closed subsets of Xsuch thatA1A2

AnAn+1with diamAn<1

nand such that no nite

number of sets in Rform a covering of any An:By Theorem

4.6.2,\1

n=1Ancontains precisely one point, say x:SinceRis a

covering of X;there is a set GR such thatx2G:SinceGis

open, there exists r >0 such that B(x;r)G:Now ifN2N

is such that1

N< r; then diamAN<1

N< r: Sincex2AN;we

haveANB(x;r)G:Thus,Galone covers AN:This is a

contradiction to the fact that nitely many subsets of Rcannot

coverAN:This contradiction completes the proof of the theorem.

4.7 Uniform continuity theorem

Using the notion of Lebesgue covering that we have developed, we

prove that every continuous function from a compact metric space to

another metric space is uniformly continuous. We begin by recalling

the denitions of continuity and uniform continuity.

Denition 40. (Continuity) Let ( X;d 1) and (Y;d 2) be metric spaces.

We sayf:X!Yis continuous at x02X;if for every >0 there

exists >0 such that d1(x;x 0)< impliesd2(f(x);f(x0))< : The

functionfis said to be continuous if it is continuous at every point of

X:

Note that in this denition, depends in general on as well asx:

Denition 41. (Uniform continuity) Let ( X;d 1) and (Y;d 2) be metric

spaces. We say f:X!Yis uniformly continuous on X;if for every

>0 there exists >0 such that for all x;y2X; d 1(x;y)<implies

d2(f(x);f(y))<:

Note that in the denition of uniform continuity the chosen de-

pends only on : ie., it is independent of the choice of x2X. Thus

in this case, the same works impartially for every x2X;hence the

name uniform continuity.

Example 7. We give two examples here: one of a uniformly continuous

function and another of a function which is not uniformly continuous.

59munotes.in

## Page 60

(Uniformly continuous) Let f:R!Rbe given by f(x) = 2x:

Then, for every >0;the choice==2 is such thatjxyj<=2

implies thatjf(x)f(y)j= 2jxyj<(2)=2< : Sinceis

independent of the choice of the point xchosen, we conclude that

fis continuous.

(Not uniformly continuous) Let S=fx2Rjx > 0g:Dene

g:S!Rbyg(x) =1

x:Then, we wantjg(x)g(y)j=jyx

xyj=

jyxj

jxyj< ; wheneverjxyj< : Since this should work for all

x;y; we may choose < x2:This equation tells that depends

on bothx;and cannot be made independent of x:For ifwere

independent of x;then= 0;a contradiction. Hence the function gis not uniformly

continuous. A simple pictorial illustration below explains how

the choice of depends on x2R:

60munotes.in

## Page 61

Check Your Progress

1. Prove that every uniformly continuous function is continuous.

The converse need not hold.

2. Show that the function f(x) =1

1+x2forx2Ris uniformly con-

tinuous on R:

3. Show that the function h(x) =1

x2is uniformly continuous on the

setK= [1;1) but not uniformly continuous on (0 ;1):

Theorem 4.7.1. (Uniform continuity theorem) Let (X;dX);(Y;dY)be

metric spaces with Xcompact. Let f:X!Ybe a continuous map.

Thenfis uniformly continuous.

Proof. To provefis uniformly continuous, we will prove that given

>0;there exists >0 such that for any x1;x22X; dX(x1;x2)<

impliesdY(f(x1);f(x2))<:

Given>0;consider the open covering of YbyB(y;=2);balls of

radius=2 aroundy:Hence,f(X)Y=[y2YB(y;=2):Hence,

Xf1([y2YB(y;=2))[y2Yf1(B(y;=2)):

Asfis continuous, we have f1(B(y;=2)) is an open subset X:This

proves thatR=ff1(B(y;=2))jy2Ygis an open cover of X:

Letbe the Lebesgue covering of this open covering R:Then if

x1;x22Xare such that dX(x1;x2)< ; then the setfx1;x2ghas

diameter less than :Hence its imageff(x1);f(x2)glies in some ball

B(y;=2):Then,dY(f(x1);f(x2))<: This proves that fis uniformly

continuous.

61munotes.in

## Page 62

4.8 Glossary

In this chapter, you have learnt the following:

Sequential compactness and limit point compactness.

Equivalence of compactness, sequential compactness and limit

point compactness in metric spaces.

A subset of Rnis compact if and only if it is closed and bounded.

Complete metric space and completion of a metric space.

Lebesgue number and Lebesgue covering lemma.

Uniform continuity and the fact that every continuous function

from a compact metric space to another metric space is uniformly

continuous.

4.9 Bibliography

Goldberg, R. R, Methods of Real Analysis, Oxford's Indian Edi-

tion.

James Munkres, Topology, Pearson.

Joshi, K. D., Introduction to General Topology, New Age Inter-

national (P) Limited.

4.10 Let Us Sum Up

In this chapter we rst learnt various equivalent notions of compact-

ness for metric spaces. Compactness was also characterized in terms

of completeness and total boundedness. We then dened the notion

of a Lebesgue covering and this gave rise to the notion of Lebesgue

number of covering. The main theorem was that every sequentially

compact space has a Lebesgue number. Since the notions of sequential

compactness and compactness are equivalent in metric spaces, this also

shows that compact metric spaces have a Lebesgue number. Existence

of this Lebesgue number then helps us to prove an important theorem

62munotes.in

## Page 63

about continuous functions from a compact metric space to another

metric space: these functions are also uniformly continuous.

4.11 References for further reading

1. Armstrong, M. A., Basic Topology, Springer UTM

2. Rudin, W., Principles of Mathematical Analysis. Third Edition,

Mc-Graw Hill International Editions.

3. Sutherland, W. A., Introduction to metric and topological spaces.

Oxford University Press.

4. Willard, S., General Topology, Dover Books on Mathematics.

5. Kumaresan, S., Topology of Metric Spaces, Morgan and Claypool.

4.12 Chapter End Exercises

1. Show that if f;gare uniformly continuous on Rthen so isf+g:

2. Iff(x) =xandg(x) = sin(x) then show that both f;gare

uniformly continuous on Rbut their product fgis not uniformly

continuous on R:

3. Show that if fandgare uniformly continuous on Rand if they are

bothbounded on R;then their product fgis uniformly continuous

onR:

4. Prove that if f;gare uniformly continuous on R;then so is their

composition fg:

5. Prove that the completion of a totally bounded metric space is

compact.

6. Let (X;d) be a metric space and let C(X) be the space of all

bounded real-valued functions on Xwith a metric edened by

e(f;g) = supfjf(x)g(x)jjx2Xg:Fixa2X:Forx2X;

denehx:X!Rbyhx(u) =d(x;u)d(a;u):Prove that

hx2C(X):

7. With notations as in the above exercise, dene h0:X!C(X)

byh0(x) =hx:Prove that h0is an isometric imbedding of Xinto

C(X):

63munotes.in

## Page 64

8. Prove that a metric space Xis complete, if it contains a dense

subsetDsuch that every Cauchy sequence in Dhas a limit point

inX:

9. Prove the converse of the above statement.

10. LetX= (0;1) with metric dgiven byd(x;y) =j1

x1

yj:Show

that the sequence f1

n;n2Ngis not a Cauchy sequence in ( X;d):

(Recall that the sequence is a Cauchy sequence in the usual met-

ric.)

64munotes.in