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Chapter 1
Topological Spaces
Chapter Structure
1.1 Introduction
1.2 Objectives
1.3 Topological Spaces
1.3.1 Base
1.4 Product Topology and Subspace Topology
1.5 Open and Closed Sets
1.6 Interior
1.7 Continuous Functions
1.8 Glossary
1.9 Bibliography
1.10 Let Us Sum Up
1.11 References for Further Reading
1.12 Chapter End Exercises
1.1 Introduction
Metric spaces are generalization of real line and the usual distance.
We take the process of generalization one more step ahead when we
eliminate the concept of distance and work with the collection of open
sets. In the chapter on metric spaces we saw several concepts like
closure, interior, boundary, continuity etc. which involve concept of
distance. These concepts can be described equivalently using only open
sets and closed sets. Thus a lot of analysis can be carried out only
knowing the open sets (and the complements of open sets) without
using the notion of distance.
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1.2 Objectives
After going through this chapter you will know:
De nition and examples of a topological space.
De nition and examples of basis and subbasis of a topological space.
De nition and properties of closure of a set.
De nition and properties of interior of a set.
Properties of continuous functions.
1.3 Topological spaces
De nition 1. A topology or a topological structure on Xis a family
of subsets of Xsatisfying the following axioms:
;2 andX2 :
IffG:2gis a subfamily of , then[2Gbelongs to :
This means is closed under the union of arbitrary subfamilies
of it.
IfG1;:::;Gnbelong to then\1knGkbelongs to :This means
the family is closed under intersection over nite subfamilies of
it.
If is a topological structure on Xthen the ordered pair ( X; ) is
called a topological space. Sets belonging to are called open sets of
the topological space.
Remark 1.3.1. By de nition, a topological space is closed under nite
intersection and arbitrary union. Every metric space is a topological
space. It is easy to check that the collection of usual open sets (i.e.
open balls) forms a topology.
Remark 1.3.2. It is important to note that not every topological space
is a metric space. Let Xbe a set such that jXj>1:Then =f;;Xg
forms a topological space: the rst axiom is obvious. The other two
follow by observing that ;\X=;and;[X=X:
IfXhas more than one element then Xcannot be a metric space for
the following reason: recall rst that every metric space is Hausdor .
We will prove here that the topological space ( X; ) is not Hausdor
which will imply that it is not a metric space. Let x;ybe two distinct
elements of X:IfXwere Hausdor , we would get disjoint open sets
U1;U2inXsuch thatx2U1andy2U2:However, the only open sets
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in (X; ) are the empty set and Xitself. Hence, we cannot get sets U1
andU2as claimed above.
Examples:
We will consider here several examples of topologies on X=
fa;b;cg:You should verify yourself that in each of these examples
the above three axioms are satis ed.
{f;;Xg:This is called the indiscrete topology onX:
{f;;fag;Xg:
{f;;fag;fbg;fcg;Xg:
{f;;fa;bg;Xg:
{f;;fag;fbg;fa;bg;Xg:
{f;;fag;fa;bg;Xg:
{P(X):This is called the discrete topology onX:In this
topology every singleton set is open and hence, every subset
ofXopen too.
LetXbe a non-empty set. Let consist of;;Xand all subsets
GofXsuch thatXnGis nite. We call this as co nite topology
onX:Verify that it is a topology.
LetXbe any in nite set. Let consist of;;Xand all subsets G
ofXsuch thatXnGis countable. We call this as cocountable
topology on X:Verify that it is a topology.
LetX=N:LetIn=f1;2;3;:::;ngandJn=fn;n+1;n+2;:::g:
Let =f;;I1;I2;I3;:::;Ng:Let 1=f;;J1=N;J2;J3;:::g:
Then ; 1are both topologies on X:
Check Your Progress
Prove that the co nite topology and cocountable topology are
actually topologies on X:
Check that ; 1de ned above are actually topologies on X:
All these examples are useful to understand the concept of a topo-
logical space and also to understand the concepts that will follow. We
will now study families of subsets of Xthat generate a given topology.
This is done via the notion of a base for a topological space.
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1.3.1 Base
De nition 2. Let (X; ) be a topological space. A base to the topology
is a subfamilyBof having the following property: given any G2
there exists a family fB:2gB such thatG=[2B:
For example the family ffxg:x2Xgis a base to the discrete
topology on X:The familyf;;fag;fbg;Xgis a base to the topology
f;;fag;fbg;fa;bg;XgonX=fa;b;cg:
Let us prove a theorem that will help us to check if a given family
is a base to a given topology.
Proposition 1.3.1. A familyBof subsets of Xis a base to a topology
onXif and only if
(i)X=[B2BB:
(ii) IfB1andB2are inBand ifx2B1\B2then there exists B2B
such thatx2BB1\B2:
Proof. SupposeBis a base to a topology onX:Now,X2 and
therefore by the de ning property of the base Bin ;there exists a
subfamilyfB:2gsuch thatX=[2B=[B2BB:This proves
(i):Next, letB1;B2be any two elements of BthenB1;B2are open
subsets of ( X; ) and therefore B1\B22 :Again by de ning property
of a base, there exists a family fB:2gB such thatB1\B2=
[2B:Consequently, if x2B1\B2then there exists 2 such that
x2B(:=B):This completes the proof of ( ii):
Conversely, suppose the family Bsatis es both the properties above.
Let denote the family of subsets of Xconsisting of the empty set ;
together with subsets of Xof the type G=[2BwherefB:2g
is a subfamily ofB:By the property ( i) ofB; X2 and;2 :
Clearly the family is closed under arbitrary unions. Now let G1
andG2be any two non empty subsets in the family :Ifx2G1\G2
thenx2G1and therefore, there exists B12Bsuch thatx2B1G1:
Similarly if x2G1\G2thenx2G2and therefore, there exists B22B
such thatx2B2G2which implies x2B1\B2G1\G2:By
property (ii) ofB;there exists Bx2Bsuch thatx2BxB1\B2
G1\G2:This implies G1\G2=[fBx:x2G1\G2gwhich means
G1\G22 :
This proves that is a topology on Xand the way in which is
obtained fromBimplies thatBis a base of :
De nition 3. A collectionCof subsets of Xis called a subbasis for the
topological space ( X; ) if every set in is a union of nite intersections
of sets inC:
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We do not use subbases extensively in our further discussion. We
move on to our next important concept: the product topology. This is
one of the ways to build new topological spaces out of old ones.
1.4 Product Topology and Subspace
Topology
Let (X; ) and (Y; 0) be topological spaces. The product topology
onXYis the topology having as basis the collection Bof all sets
of the form UV;whereUis an open subset of XandVis an open
subset ofY:
Theorem 1.4.1. IfBis a basis for the topology of X;andCis a basis
for the topology Y;then the collection D=fBCjB2BandC2Cg
is a basis for the topology of XY:
Proof. Given any open set WofXYand a point xyofW;there
is a basis element UVsuch thatxy2UV2W:Because Band
Care bases for XandYrespectively, we can choose B2BandC2C
such thatx2BUandy2CV:Thenxy2BCW:Thus
Dis a basis for XY:
We also study another important notion: that of subspace topol-
ogy. A topological structure on a set Xinduces a topological structure
on any subset YofX:
De nition 4. Let (X; ) be any topological space and let Ybe a
nonempty subset of X:We consider Y=fG\Y:G2 g:Then Y
is a topological structure on Y:The topology Yis said to be induced
by topology onX:We also say that ( Y; Y) is a subspace of the
topological space ( X; ):
Check Your Progress
Show that the countable collection f(a;b)(c;d)jad;anda;b;c;d2Qgis a basis for the usual topology on R2i.e.,
the one induced by the usual metric on R2:
Show that the set of all half open intervals [ a;b) witha;b2Ris
a basis for the usual (metric) topology on R:
1.5 Open and Closed sets
You have studied neighbourhoods in metric spaces. We extend this
concept to topological spaces.
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De nition 5. Let (X; ) be a topological space. Let x2X:A subset
NofXis said to be a neighborhood of the point xif there exists
G2 such 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:
De nition 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 de ned in example 2 above.
LetA=f2g:The point 1 is a limit point of Abut it is not in A:
De nition 7. For any set Ain a topological space ( X; );the closure
ofAinXdenoted by c(A) is de ned 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:
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(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):
De nition 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 de nition of neighbourhood there exists Gy2 withy2
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.
;2 implies that Xn;is closed and X2 implies that;=XnX
is closed.
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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 de ne 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 de nition. i(A) is in fact the largest open set contained in
A:
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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 de nition 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.
De nition 9. For any set Ain the topological space the set c(A)ni(A)
is de ned 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.
De nition 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:
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Remark 1.7.1. f(N)eNimpliesNf 1(eN) and therefore f 1(eN)
becomes a neighborhood of a:Therefore the de nition can be restated
as:fis continuous at aif for every neighborhood eNoff(a) the set
f 1(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 haveg 1(eN) is a neigh-
borhood of b:We haveg 1(eN) is a neighborhood of b:Also by con-
tinuity offatawe havef 1(g 1(eN)) is a neighborhood of b:But
f 1(g 1(eN)) = (gf) 1(eN) which is a neighborhood of a:Thus we have
veri ed that, for any neighborhood eNof (gf)(a) the set (gof) 1(eN)
is a neighborhood of a:Therefore by the de nition 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) IfH2 0thenf 1(H)2 :
(c) IfCis a closed subset of (Y; 0)thenf 1(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):Letx2f 1(H):Thenf(x)2H2 0:This means His a
neighborhood of f(x):By continuity of fatx; f 1(H) is a neighbor-
hood ofx:This proves that f 1(H) is a neighborhood of each of its
point. Hence f 1(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 f 1(H) is an open subset
of (X; ):Butf 1(H) =f 1(YnC) =Xnf 1(C):Hencef 1(C) is
a closed subset of ( X; ):
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(c) =)(d):LetAbe any subset of X:LetC=c(f(A)): Cis a
closed subset of ( Y; 0):Therefore by ( c); f 1(C) is a closed subset of
(X; ):But we have Af 1(f(A))f 1(c(f(A))) =f 1(C):
NowAis a subset of the closed set f 1(C):Hencec(A)f 1(C)
which gives f(c(A))f(f 1(C))C:Thusf(c(A))c(f(A)) is
proved.
(d) =)(c):LetCbe a closed subset of ( Y; 0):LetA=f 1(C):
We havef(c(A))c(f(A))c(f(f 1(C))c(C) =C:Now apply
f 1tof(c(A))Cwe getf 1(f(c(A)))f 1(C) =A:Butc(A)
f 1(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 f 1(H) =
Xnf 1(YnH) =Xnf 1(C):By the result ( c); f 1(C) is closed and
hencef 1(H) =Xnf 1(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 de ned on a discrete topological space is continuous.
Iff: (X; )!(Y; 0) is continuous then A2 does not mean
f(A)2 0:
1.8 Glossary
In this chapter, you have learnt the following:
De nition 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.
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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 Bof is called a base of if every non-empty set Gin can
be expressed as a union of a subfamily of B:A subsetNofXis called
a neighborhood of x2Xif there exists Gin satisfyingx2GN:
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 di erent 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 satis es 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 de nition of a continuous function from one topological space to
another is a generalization of the classical " de nition of continuity
of a real valued function of a real variable. A map f: (X; )!(Y; 0)
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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.
f 1(H)2 for everyH2 0:
f 1(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:
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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 and 0onX
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;:::;Ngand 0=f;;J1=N;J2;J3;:::g:
Then check that and 0both are topological structures on N:
Findf1g0with respect to both and 0:
9. Letf: (X; )!(Y; 0) be a constant map. Show that fis
continuous.
10. Describe the closure of an in nite set in the co- nite topology.
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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 de nition 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.
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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.
De nition, 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.
De nition 11. A mapf: (X; )!(Y; 0) is said to be a homoemor-
phism iffis bijective, fis continuous and f 1is 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.
De nition 12. A property Pof topological spaces is said to be a
topological property if Pis satis ed by a topological space implies that
it is satis ed by another topological space homeomorphic to it. Also a
propertyPis called hereditary if Pis satis ed by a topological space
implies that it is satis ed 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 S1inR2de ned 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.
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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 in nite
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.
De nition 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):
In the usual topology on R,B(x) =f(x 1=n;x +1=n) :n2Ng
is a countable base to N(x):
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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:
De nition 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.
De nition 15. A topological space ( X; ) is second countable (or sat-
is es 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 satis ed.
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
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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;:::;Xgand 0=f;;J1=N;J2;J3;:::g:
Then and 0are 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
De nition 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 satis es the Hausdor property and hence is
T0:
If is 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:
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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 de ning property of the homeomorphism H2 0: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.
De nition 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:
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Discrete topological space ( X;P (X)) and metric spaces are ex-
amples ofT1spaces.
OnX=fa;b;cgwe de ne =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 in nitely 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 Gy2 such 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=Gx2 and 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 in nitely 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.
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2.6 Hausdor Topological Spaces
De nition 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 satis es axiom
T2then it satis es axiom T1:
The converse is not true. On an in nite set the co nite topology
satis esT1axiom but does not satisfy T2axiom.
OnX=fa;b;cgwe can not de ne 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. satis es T2axiom.
De nition 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 de ned concept. For example in the indiscrete
topology (X;P (X)) any sequence converges to each point in the metric
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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 de ne the sequence fan:n2Ngbya2k=
aanda2k+1=bfor allk2N:This sequence converges to both bandc:
Following is an example of a sequence that converges to in nitely
many points.
LetX=N:LetJn=fn;n+ 1;n+ 2;:::gand =f;;J1=
N;J2;J3;:::g:De ne 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 di erent 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 de ne an important property of a topological space.
De nition 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=f 1(H):
ThenGis a non empty subset of ( X; ):Nowf 1(c(B)) =c(f 1(B)) =
c(A) =X:This contradicts that G=Xnf 1(c(B))6=;:This proves
c(B) =Ywhich means that the countable subset BofYis dense in
Y:
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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[fwgjG2 0g:
(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
De nition 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:
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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.
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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 in nitely 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
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1. Show that being T1is a topological and hereditary property.
2. Show that being T2is a topological and hereditary property.
3. Let be 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?
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Chapter 3
Compactness
Chapter Structure
3.1 Introduction
3.2 Objectives
3.3 General De nition of Compactness
3.4 Continuity and Compactness
3.5 Finite Products and Compactsness
3.6 Local Compactness and one-point Compacti cation
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 de ned in terms of open covers. Using this de ntion, 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 compacti cation.
We also study a notion similar to compactness: namely the notion
of a Lindel of space.
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3.2 Objectives
After going through this chapter you will know:
General de nition of compactness.
Continuity preserves compactness.
A nite product of compact spaces is compact.
De nition of local compactness.
Construction of one point compacti cation.
Lindel of Spaces.
3.3 General de nition 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 de nition of compactness, let us
understand what an open cover of a topological space is.
De nition 22. (Open cover) Let Xbe a topological space and fU g 2
be a family of open subsets of X:ThenfU g 2is said to be an open
cover ofXifX[ 2U :(Note that this condition is same as saying
X=[ 2U ;asU X;for all 2:)
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From the above de nition it is easy to check that fU g 2is an
open cover of Xif and only if X=[ 2U ;for open subsets U ofX:
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 de nition of compactness for arbitrary topo-
logical spaces: this abstract de ntion was rst formulated in 1906 by
Maurice Fr echet. (See the website: www :==http history:mcs:st
andrews:ac:uk=Biographies=Frechet:html andhttp:==en:wikipedia:org=wiki=Compactness
for further details.)
De nition 23. LetXbe a topological space. Xis said to compact
if every open cover of Xadmits a nite subcover i.e. if there exists a
familyfU g 2of open subsets of Xsuch thatX[ 2U ;then there
exist nitely many 1;:::; n2 such that XU 1[U 2[[U n:
(In this case we say that the family fU g 2admits a nite subcover.)
Remark 3.3.1. The main stress in this de nition is on every open
cover . If you are able to prove that a speci c 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;de ned
forn1;byVn= (0;1 1
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;1 1
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:
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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 ;Xg 2be an open cover of Y
by sets open in X:Then,fV ;X\Yg 2be 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 ;Yg 2:
Hence there exists a family of open sets fV ;Xg 2ofXsuch that
V ;Y=V ;X\Y;i.e.,V ;YV ;X:Then clearly we get:
Y[ 2V ;Y=[ 2(V ;X\Y)[ 2V ;X:
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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 thatYV 1;X[V 2;X[V n;X:Taking inter-
section of both sides with Y;we get that
Y(V 1;X[V 2;X[V n;X)\Y=[n
i=1(V i;X\Y) =[n
i=1V i;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 ;Yg 2be an open cover of f(X):Then, it is easy to check
thatff 1(V ;Y)g 2is an open cover of X:AsXis compact, we get
1;:::; nsuch thatXf 1(V 1;Y)[f 1(V 2;Y)[[f 1(V n;Y):By
noting that ( ff 1)V i;YV i;Y;we get that f(X)V 1;Y[[V n;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
fV g 2be an open cover of F:Note that by the de nition of subspace
topology on F;there exist open subsets U ofXfor each 2 with
the property that V =U \F:
SinceFis closed,Fcis open andfU g 2[Fcis an open cover,
sayRofX:AsXis compact,Radmits a nite subcover R0:
IfFc2R0, then it is easy to check that fU \F:=V jU 2R0nFcg
gives an open cover of F:IfFc62R0;thenfU \F:=V jU 2Rg 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
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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 de nition 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 de ne a few terms.
De nition 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:
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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 p 1
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:
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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 justi cation (if false, give a counter-
example): An arbitrary union of compact sets is compact.
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3.6 Local compactness and one-point
compacti cation
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.
De nition 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 in nite dimensional product space RNis not locally compact.
De nition 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
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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 compacti cation.
More precisely, the mathematical de nition goes as follows:
De nition 27. (One point compacti cation) 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 compacti cation of X:
Proposition 3.6.1. LetYbe the set de ned 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.
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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) LetfU g 2be a family of open sets of Type I. Then clearly,
[ 2U is again an open set of Type I.
1(b) LetfU g 2be a family of open sets of Type I and fU g 20be
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) LetfU g 20be 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 C ofX:This union now equals Yn\ 20C :
Now note each C is a compact subset of the Hausdor space X
and hence is closed in X:Hence,\ 20C is a closed subset of X:
Moreover, for a xed 02;\ 20C C 0and hence we have
a closed subset of a compact set and hence \ 20C is a compact
subset ofX:This implies that we get that [ 20U is 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 compacti ca-
tion upto homeomorphism.
Here is an example to keep in mind:
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Example 4. One-point compacti cation of (0 ;2) is homeomorphic
with the circle S1:
De ne 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 f 1(V) is of the form ( 1;2) for some
1< 22(0;2):In thisf 1(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 f 1(V) =f1g[ (0;1)[
(2;2):Clearly, then f 1(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 compacti cation.
Theorem 3.6.1. LetXbe a locally compact Hausdor space which is
not compact. Let Ybe the one-point compacti cation 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 di erent from 1:For this observe that every
open subset around 1is the complement in Yof a compact subset C
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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 di erent from 1;
proving that1is a limit point of X:
Clearly,YnXis the single point in nity 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 compacti cation 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
De nition 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 fV g 2be an open cover of
f(X):Then, it is easy to check that ff 1(V )g 2is an open cover
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ofX:AsXis Lindel of,ff 1(V )g 2admits a countable subcover,
sayff 1(V i)g i2;i2N:Then it can be checked that ( V i) i2;i2Nis a
countable subcover of fV g 2forf(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.
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(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.
De nition 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 compacti cation.
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Basic properties of one-point compacti cation.
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 de nition 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 de ning a suitable topology on this
new space, such that it becomes compact. This process is called the
Alexandro one-point compacti cation. We have studied the properties
of one-point compacti cation and seen one explicit example of such a
compacti cation.
3.11 References for further reading
1. George Simmons: Topology and Modern Analysis, TataMcgraw-
Hill.
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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 compacti cation 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) =jp qj: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.
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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 in nite 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:
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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
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other metric space is uniformly continuous.
4.2 Objectives
After going through this chapter you will be able to:
Give various equivalent de nitions of compactness for metric spaces.
ShowERnis compact if and only if it is closed and bounded.
De ne 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 de nition may not be always be easy and so, one tries to see if there
are other equivalent de nitions which might work. In this section we
study these other notions of compactness: sequential and limit point
compactness and prove that all three de nitions of compactness are
equivalent for metric spaces.
De nition 29. (Sequentially compact) Let Ybe a topological space.
Yis said to be sequentially compact, if every sequence in Yhas a
convergent subsequence.
De nition 30. (Limit point compact) Let Ybe a topological space.
Yis said to be limit point compact if every in nite 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 in nite sequence of points ( xn) such that ( xn) has no
convergent subsequence. Start with any x12X:Then, there exists x22
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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 de nition 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 in nite 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
in nite.
{SupposeAbe nite. We claim that there exists xsuch that
x=xn;for in nitely many n2N:To see this, de ne f:N!
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Abyf(n) =xn:Then, N=[x2Af 1(x):If the number of
elements in f 1(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 thatf 1(x) is an in nite subset of Ni.e., there exist
in nitely many nsuch thatf(n) :=xn=x:Now it is obvious
that (xn)n2f 1(x);being the constant sequence is a convergent
subsequence of ( xn)n2N:
{SupposeAis in nite. As Xis limit point compact by as-
sumption, we have that Ahas a limit point, say x:We will
now de ne 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 ni 1we can
choose an index ni> ni 1such thatxni2B(x;1=i):(The
guarantee that ni>ni 1is due to the fact that Ais in nite.)
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 in nite 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
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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 de nition of a metric:
De nition 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(xi yi)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) = maxfjxi yijgis 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:
De nition 32. Ifais 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
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upper bounds of E:By de nition, 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;jx yjb a:Letb a=: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;jx yj2 n:By Lemma 4.4.1, there exists a point xlying
inInfor alln2N:SincefGg2is an open cover of I;there exists
an such thatx2G withG 2fGg2:SinceG is open, there
existsr>0 such thatjy xjproperty, there exists n2Nlarge such that 2 nthatInG ;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.)
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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
compacti cation, 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.
De nition 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=Pn 1
i=1d(ai;ai+1):
Now letx;y2A:Then, without loss of generality there exist
1lmnsuch thatx2Alandy2Am:Then,d(x;y)d(x;al)+Pm 1
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.
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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) :=jjx yjj2;
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) =jjej ekjj2=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
De nition 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 de nition of a Cauchy sequence.
De nition 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.
De nition 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:
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De nition 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 de ning
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-de ned.
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 di erent 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
quanti cation 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 de nition of the diameter of a bounded set:
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De nition 38. LetAbe a bounded subset of a metric space ( X;d).
Then, diameter of A;denoted by diam( A) is de ned to be lub fd(a1;a2)ja1;a22
Ag:
Examples:
(1) LetX=Rwith the absolute metric. Let A= [0;1]:Then,
diam(A) := lubfjx yjjx;y2Ag:Note that for all x;y2A;
jx yj1: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 de nition of
S1;Ifx= (x1;x2) andy= (y1;y2) then we have d(x;y) =p
(x1 y1)2+ (x2 y2)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.)
De nition 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:
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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 di erent 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.
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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 in nity, 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 in nitely 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> nk 1such 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
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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 de nitions of continuity and uniform continuity.
De nition 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 de nition, depends in general on as well asx:
De nition 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 de nition 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.
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(Uniformly continuous) Let f:R!Rbe given by f(x) = 2x:
Then, for every >0;the choice==2 is such thatjx yj<=2
implies thatjf(x) f(y)j= 2jx yj<(2)=2< : Sinceis
independent of the choice of the point xchosen, we conclude that
fis continuous.
(Not uniformly continuous) Let S=fx2Rjx > 0g:De ne
g:S!Rbyg(x) =1
x:Then, we wantjg(x) g(y)j=jy x
xyj=
jy xj
jxyj< ; wheneverjx yj< : 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:
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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,
Xf 1([y2YB(y;=2))[y2Yf 1(B(y;=2)):
Asfis continuous, we have f 1(B(y;=2)) is an open subset X:This
proves thatR=ff 1(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.
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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 de ned 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
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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 ede ned by
e(f;g) = supfjf(x) g(x)jjx2Xg:Fixa2X:Forx2X;
de nehx:X!Rbyhx(u) =d(x;u) d(a;u):Prove that
hx2C(X):
7. With notations as in the above exercise, de ne h0:X!C(X)
byh0(x) =hx:Prove that h0is an isometric imbedding of Xinto
C(X):
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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
x 1
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.)
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