project

1. GOVERNMENTCOLLEGE
AMBALAPUZHA
DEPARTMENTOFMATHEMATICS
SEMESTER-VI
PROJECT
i

2. AN INTRODUCTION TO TOPOLOGY
PROJECT
Submitted to University of Kerala in partial fulllment of the
requirements of for the award of the Degree of Bachelor of Science
in
Mathematics
By
VISHNU V
Candidate Code:
Exam Code: Project Code:
Under the Guidance of
Sri Asokan C.K,
Associate Professor.
Department of Mathematics
Government College, Ambalapuzha
2014
ii

3. GOVERNMENT COLLEGE
AMBALAPUZHA
DEPARTMENT OF MATHEMATICS
CERTIFICATE
This is to certify that the project work entitled AN INTRODUCTION TO
TOPOLOGYis a bonade work done by NAME (candidate code) in partial
fullment of the requirement for the award of Bachelor of Science in Mathe-
matics by the University of Kerala and this report has not been submitted by
any other university for the award of any degree to the best of my knowledge
and belief.
Ambalapuzha, Asokan C.K,
June 6, 2016 Department of Mathematics,
Govt. College, Ambalapuzha.
iii

4. DECLARATION
I hereby declare that this is a bonade record of the work done by me in
partial fullment of the requirements for the award of the degree of Bachelor
of Science in Mathematics by the University of Kerala and this report has
not been submitted to any other university for the award of any degree to
the best of my knowledge and belief.
Name
Candidate code
iv

5. ACKNOWLEDGEMENT
It is our pleasure to express our sincere thanks to god almighty showing
his choicest blessing on as for the successful completion of the project.Our
sincere thanks to our principle Prof .Geethakumari for the constant support
and encouragement. We express our sincere thanks to our guide Prof. C.K
Asokan, Associatie Professor, Departmaent Mathematics, for the valuable
guidance, observation and timely advice during the preparation of the project
report. Our sincere thanks to teaching and non teaching sta of the college
for the expert counsel in completing our project.
We express our sincere thanks to Donald Knuth, who developed the
TEXengine, to typeset Mathematics .
Last but not least, Our sincere thanks are also due to my beloved parents
and a friend whose love and encourage has helped us in completing this
project. Once again, I take an opportunity to thanks each and every persons
helped us directly and indirectly for the successful completion of the project.
v

6. INTRODUCTION
The word Topology is derived from two Greek words,toposmeaning 'sur-
face' andlogosmeaning 'discourse' or 'study'. Topology thus literally means
the study of surfaces.It sometimes referred to as the mathematics of con-
tinuity, or rubber sheet geometry, or the theory of abstract topological
spaces,is all of these,but above all,it is a language used by mathematicians
in practically all branches of our science. In this chapter, we will learn the
basic words and expression of this language as well as its grammer, i.e,
the most general notations, methods and basic results of topology. We will
also start building the library of examples, both 'nice and natural' such as
manifolds or the cantor set,other more complicated and even pathological.
Those examples often possess other structures in addition to topology and
this provides the key link between topology and other branches of geome-
try.They will serve us illustrations and the testing ground for the notations
and methods developed in later sessions.
Acircleis topologically equivalent to anellipseand asphereis equivalent
to anellipsoid. Similarly, the set of all possible positions of the hour hands,
minute hands and second hands of a clock are topologically equivalent to a
circle, a torus and a three-dimensional object.
Topological concepts like compactness, connectedness etc. are a base
to mathematicians of today as sets and functions were to those of last cen-
tury .Topology has several dierent branches- general topology(also known as
point-set topology),algebraic topology, dierential topology and topological
algebra-the rst,general topology,being the door to the study of the others.I
aim in this book to provided a through grounding in general topology.
vi

7. Contents
1 Preliminaries 1
1.1 Sets and Functions . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Metric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Topological Spaces 9
2.1 Topological Spaces and Examples . . . . . . . . . . . . . . . . 9
2.2 Bases and Sub-bases . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 Closure of a Set, Continuous Functions And Related Concepts 17
3 Topological Properties 30
3.1 Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.2 Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.3 Some Applications of Connectedness . . . . . . . . . . . . . . 44
vii

8. Chapter 1
Preliminaries
In this initial chapter we will present the background needed for the the study
of Topology. It consists of a brief survey of set operations and functions,
two vital tools for all of mathematics. In it we establish the notation and
state the basic denitions and properties that will be used throughout the
report. We will regard the word "set" as synonymous with the words "class",
"collections" and "family" and we will not dene these terms or give a list
of axioms for set theory. this approach, often referred to as " native " set
theory is quite adequate for working with sets in the context of topology.
1.1 Sets and Functions
In this section we give a brief review of terminology and notation that will
be used in this report.
if an elementxis in a setA, we write
x2A
and say thatxis aMemberofA, or thatxbelongstoA:IfxisnotonA,
1

9. we write
x =2A:
Denition 1.Two setsAandBare said to be equal and we writeA=B,
if they contain the same elements.
Thus ,to prove that the setsAandBare equal, we must show that
AB & BA:
Set Operations
Note that the set operations are based on the meaning of the words or,
and and not.
Denition 2.The following are some set operations:
1. Theunionof setAandBis the set
A[B=fx:x2A or x2Bg:
2. Theintersectionof the setsAandBis the set
A\B=fx:x2A and x2Bg:
3. Thecomplement ofBrelative toAis the set
AB=fx:x2A and x =2Bg:
The set that has no element is calledempty setand it is denoted by
. Two setsAandBare said to bedisjointif they have no elements in
common and it is expressed byA\B=:
Functions:
2

10. Denition 3.Afunctionffrom setXto setY, denotedf:X!Yis a
rule which assigns to each memberxofXa unique membery=f(x)ofY.
Ify=f(x)thenyis called theimageofxandxis called apre-imageof
y. The setXis thedomainoffandYis theco-domainorrangeoff.
Note that for a functionf:X!Yeach elementxinXhas a unique image
f(x)inY. However, the number of pre-image may be zero, one, or more
than one.
Denition 4.A functionf:X!Yisone-to-oneorinjectivemeans that
for distinct elementsx1; x22X; f(x1)6=f(x2). In other words,fis one-to-
one provided that no two distinct points in the domain have the same image.
In contrapositive form this can be stated as :f(x1) =f(x2) =)x1=x2.
A functionffor whichf(X) =Yi.e, for which the imagef(X)equals
the co-domain , is said to be mapXontoYor to be surjective.
A one-to-one function fromXontoYis called aone-to-one corre-
spondenceor abijection. Thusf:X!Yis a one-to-one correspondence
provided that each member ofYis the image underfof exactly one member
ofX. In the case there is aninverse functionf
1
:Y!Xwhich assigns
to eachyinYits unique pre-imagex=f
1
(y)inX.
Example 1.LetX=fa; b; c; d; eg; Y=f1;2;3;4;5gand the functionf:
X!Ydened by
f(a) = 1; f(b) = 2; f(c) = 3; f(d) = 4; f(e) = 5
is a bijection with inverse functionf
1
:Y!Xdened by
f
1
(1) =a; f
1
(2) =b; f
1
(3) =c; f
1
(4) =d; f
1
(5) =e
.
3

11. Denition 5.Iff:X!Yandg:Y!Zare functions on the sets, then
thecomposite functiongf:X!Zis dened by
gf(x) =g(f(x)); x2X
The composite functiongfis some times denoted simplygf.
Example 2.Consider the functionf:R!Randg:R!Rdened by
f(x) =x
2
; g(x) =x+ 1. Then the composite functiongfandfgare
both dened
gf(x) =g(f(x)) =g(x
2
) =x
2
+1; fg(x) =f(g(x)) =f(x+1) = (x+1)
2
1.2 Metric Space
In this section, we will introduce the idea of metric space and discuss the con-
cepts of neighbourhood of a point, open and closed closed sets, convergence
of sequences, and continuity of functions.
Denition 6.Ametric spaceis a setXwhere we have a notation of dis-
tance. That is, ifx; y2X, thend(x; y)is the distance betweenxand
y.The particular distance functions must satisfy the following conditions.
1.d(x; y)0for allx; y2X
2.d(x; y) = 0ix=y
3.d(x; y) =d(y; x)
4.d(x; z)d(x; y) +d(y; z)
Open ball :
4

12. Denition 7.Letx02Xandrbe a positive real number .Then theopen
ballwithcentrex0andradiusris dened to be the set
fx2X:d(x; x0)< rg:
It is denoted either byBr(x0)or byB(x0;r). It is also called theopen r-ball
aroundx0.
Open set :
Denition 8.A subsetAXis said to beopenif for everyx02A,9some
open ball aroundx02A:If their exist somer >0such thatB(x0;r)2A:
Remark 1.Before doing anything with open balls and open sets it would be
nice to know that open balls are indeed open sets.This follows trivially from
the denitions and the triangle inequality.
Note 1.Letfxngbe a sequence in metric space. Thenfxngconverges toy
inXi for every open sety2U9N2Z:
Theorem 1.Let(X;d)be a metric space. Then
1.andXare open.
2. The union of collection of open set is open.
3. The intersection of nite number of open set is open.
4.x; y2X9open setsU; Vsuch thatx2U; y2VandU\V=:
Proof :
(i) Since there are no pointse2the statementx2wheneverd(x; e)<1,
holds for alle2. Since every pointx2X, the statementx2Xwhenever
d(x; e)<1, holds8e2X
5

13. (ii) Ife2
[
2A
U, then we can nd a particular12Awithe2U1. Since
U1is open, we can nd a >0such that
x2U1; d(x; e)<
. SinceU1
[
2A
U,
x2
[
2A
U; d(x; e)<
Thus
[
2A
Uis open.
(iii) Ife2
n
\
j=1
Uj, thene2Ujfor each1jn. SinceUjis open, we can
nd aj>0such that
x2Uj; d(x; e)< j
. Setting=minfjg1jnwe have >0and
x2Uj; d(x; e)<
forall1jn. Thus
x2
n
\
j=1
Uj; d(x; e)<
and we have shown that
n
\
j=1
Ujis open.
Example 3.Thendimensional Euclidean spaceR
n
is a metric space with
the respected to the functiond:R
n
R
n
!R, dened by
d(x; y) =
(
n
X
i=1
(xiyi)
2
)
1=2
wherex= (x1; x2; : : : ; xn)andy= (y1; y2; : : : ; yn)2R
n
wherexi; yi2R.
6

14. Clearly,d(x; y)08x; y2R;
d(x; y) = 0i
(
n
X
i=1
(xiyi)
2
)
1=2
= 0
i.e, ixi=yi8i= 1;2; : : : ; n
Hencex=yid(x; y) = 0
Now let
x= (x1; x2; : : : ; xn)
y= (y1; y2; : : : ; yn)
z= (z1; z2; : : : ; zn)
be three arbitrary elements ofR
n
.
Sincexi; yi; zi2R8i= 1;2; : : : ; nand
Pi=xiyiand
Qi=yizi2R
Clearly,Pi+Qi=xiziwherei= 1;2; : : : ; n
By the corollary we just proved
(
n
X
i=1
(Pi+Qi)
2
)
1=2

(
n
X
i=1
(Pi)
2
)
1=2
+
(
n
X
i=1
(Qi)
2
)
1=2
i.e,
(
n
X
i=1
(xi+zi)
2
)
1=2

(
n
X
i=1
(xiyi)
2
)
1=2
+
(
n
X
i=1
(yizi)
2
)
1=2
i.e,d(x; y)d(x; y) +d(y; z)(Triangle inequality)
Finally,d(x; y) =
(
n
X
i=1
(xiyi)
2
)
1=2
=
(
n
X
i=1
(yixi)
2
)
1=2
=d(x; y)
7

15. All these prove thatdis a metric known asEuclidean metricorUsual
metric.
Example 4.LetRbe a set of real numbers, show that the functiond:R!R
dened byd: (a; b) =jabj;8a; b2R, is a metric onR.
Here,
1.jabj 0 =)d(a; b)0;8a; b2R
2.jabj= 0; iff ab= 0; iff a=bso thatd(a; b) = 0,a=b
3.jabj=jbaj=)d(a; b) =d(b; a);8a; b2R
4.jabj=j(ac) + (cb)j jacj+jcbj
=)d(a; b)d(a; b)d(a; c) +d(c; b); a; b; c2R
From thisdis a metric onR.
8

16. Chapter 2
Topological Spaces
2.1 Topological Spaces and Examples
In this chapter we give the much-delayed denition of a topological spaces.
We develop it from properties of a metrices space. In the second section we
give a few examples like nite spaces, discrete spaces, indiscrete spaces of
topological spaces.
Denition 9.Atopological spaceis a pair(X;T)whereXis a set and
Tis a collection of subsets ofXsatisfying:
1.; X2T.
2.Tis clodsed under arbitrary unions.
3.Tis closed under nite intersections.
The collectionTis said to be atopologyon the setX. Members ofT
are called open sets ofX. The elements ofXare called itspoints.
9

17. Example 5.LetXbe a non-empty set, and let the topology be the class
of all subsets ofX. This is called thediscrete topologyonX, and any
topological space whose topology is the discrete topology is called adiscrete
space.
Suppose thatX6=;T=}(X), cleralyXand2T. Let
A=fu=2gthenuX8
1
[
i=0
uX2}(x) =T
soTis closed under arbitrary union.
Now we are going to showTis closed under nite intersection. Take
u1; u2; : : : ; unbe a nite elements ofTthen,
n
\
i=1
uX2}(x) =T
i.e;Tis closed under arbitrary intersection. So(X;T)is a topological space.
Example 6.LetXbe a non-empty set, and let the topology consist only the
empty setand full setX. This is called thein-discrete topologyonX
and any topological space whose topology is the in-discrete topology is called
ain-discrete space.
HereT=f; Xg)Tis closed under arbitrary union, be-
causeTconsist onlyandX, so the union is eitherXor. SimilarlyTis
closed under nite intersections.
Example 7.Everymetric spaceis a topological space.
Consider a metric space(X; d). LetTbe the collection of all
open subsets ofX.
*andXare open inX,(; X)2T.
Now the union of a number of open sets inXis open. i.e;Tis closed under
arbitrary union. Similary,Tis closed under nite intersection.
10

18. )Tis a topology.
Thus every metric space is a topological space.
Remark 2.OnRd(x; y) =jxyjis a metric. Hence(R; d)is a metric
space and soRis a topological space. This topology is said beUsual topology
onR
Example 8.LetX=fa; bgand letT=f; X; ag. CleralyTis a topology
onXcalledSierpinski topology. Suppose thatd:XX!Ris a metric or
pseudometric onXthend(a; a) = 0 =d(b; b)andd(a; b) =d(b; a) =k(say).
Ifdis metric thenk >0.
Sr(a) =fagsimilarlySr(b) =fbg
)T=f; X;fag;fbggis a Discrete topology
Ifk= 0thendwill be a pseudometric. ThenSk(x) =fa; bg=X. Then
the only open sphere inXisX)T=f; Xgis In-discrete topology. From
this we can say thatnot every topological space is metrizable space.
2.2 Bases and Sub-bases
In the section we showed that any collection of subsets of a set generates
a topology on that set. In this section we shall see how the topology so
generated can be described intrinsically in terms of the original collection of
subsets. we begin with an important denition.
Denition 10.Let(X;T)be a topological spaces. A subcollectionBofT
is said to bebaseforTif every member ofTcan be expressed as the union
of some members ofB.
It is often useful to dene a topology in terms of a base. for example,
in a metric space every open set can be expressed as a union of open balls
11

19. and consequently the collection of all open balls is a base for the topology
induced by the metric.
Lets us look some examples :
Example 9.The collection of all open intervals(a; b)witha; b2Ris a base
for the standard topology onR. The collection of all open intervals(a; b)R
with rational end pointsa; b2Qis a base for the standard topology onR
Example 10.IfXis topology space with the discrete topology, then the
collection
B=fx=x2Xg
is base of the discrete topology .
Denition 11.A space is said to satisfy thesecond axiom of countability
or is said to besecond countableif its topology has a countable base.
Theorem 2.LetXbe second countable space .If a non-empty opensetG
inXis represented as the union of a classGiof open sets, thenGcan be
represented as a countable union ofGi's.
Proof:
SupposeXbe a second countable space . AndG=
[
2
Gwhere eachGis
a open set.
we want to prove thatG=
[
i2K
GiwhereKis a countable subset of.
Letx2Gthusx2Gfor some.
*fBngis a countable open base ,9Bnsuch thatx2BnGG. Clearly,
Gis the union of suchBn's and alsoGis the union of suchGn's.
Denition 12.A collectionUof sets is said to be acoverof set A if A is
contained in the union of members ofU. Asub-coverofUis subcollection
12

20. VofUwhich itself is a cover of A. If we are in a topological space then a
cover is said to beopenif all its members are open.
Theorem 3.If a space is second countable then every open cover of it has
a countable sub-cover.
Proof:
Let(X;T)be a topological space with countable baseBand letUbe the
open cover ofX.
SupposeB=fB1; B2; : : :g.Now letS=fn2N:BnUg.
For eachn2SxUn2Usuch thatBnUn. Now letC=fBn:n2Sg
andV=fUn:n2Sg;clearlyVis a countable sub-collection ofUand
coversXifCdoes.
Now we have to prove thatCis a cover ofX. For this letx2Xthenx2U
for someU2U9k2Nsuch thatx2BkandBkUclearly,k2Sand
soBk2C.SoCand consequentlyVis a cover ofX.
Note 2.The same topology may have more than one distinct bases but two
distinct topologies can never have the same collection of subsets as a base for
both of them.
Corollory 1.IfBis a cover ofXandBis closed under nite intersections
thenBis a base for a topologyTonX. Moreover,Tconsists precisely of
those subsets ofXwhich can be expressed as unions of subfamilies ofB.
Proof:
supposeSX. LetBbe the collection of all nite intersections of
elements ofX. by taking the intersections of zero sets inS. i.e,X2B.
SoBis closed under nite intersections and thatSB.
13

21. Denition 13.A collectionSof subsetsXis said to beSub-basesfor a
topologyTonXif the collection of all nite intersections of members ofS
is a base forT.
For example, for the usual topology onR, the collection of all open in-
tervals of the form(a;1)or(1; b)fora; b2Ris a sub base.
Note 3.Any base for a topology is also a sub base for the same and the sub
base can be chosen to be much smaller than a base.
Theorem 4.LetXbe a set ,Ta topology onXandSa collection of
subsets ofX. ThenSis a sub base forTiSgeneratesT.
Proof:
LetBbe the collection of nite intersections of members ofS. Suppose that
Sis a sub base forT. We want to show thatTis the smallest topology on
X S. SinceS BandBTthenT S. SupposeUis some other
topology onXsuch thatS U. Now we have to show thatTU.
SinceUis closed under nite intersections andS U,BU. i.e,U
is closed under arbitrary unions and each member ofTcan be written as
union of some members ofB. i.e,TU.
Conversely prove thatTis the smallest topology containingSandBis
a base forT. Clearly,BTsinceTis closed under nite intersections
andS T. There is a topologyUonXsuchthatBis a base forUand
BT.This meansUTand consequentlyU=T.
SinceTis the smallest topology containingSandBis a base forT.
14

22. 2.3 Subspaces
Denition 14.LetXbe a topological space with topologyT. IfYis a
subset ofXthe collection,
U=fY\U:U2Tg
is a topology onY. The topologyUis also denoted byT=Y. It is called
therelative or the subspace topologyonYinduced byT. The space
(Y;T=Y)is called asubspaceof the space(X;T).
Its open sets consists of all intersections of open sets ofXwithY. it is
trival to verify thatUis a topology onY;
T=Ycontains empty setand full setY,(=Y\&Y=Y\
X ; Y2T=Y)
And alsoT=Yis closed under nite intersection and also closed under the
arbitrary union.
Denition 15.A property of topological spaces is said to behereditaryif
whenever a space has that property, then so does every subspace of it.
A trival example of a hereditary property of being either an in-discrete
or a discrete space. We have not yet dened any properties which are not
hereditary, but there will be many examples to come, e.g. compactness and
connectedness.
Lemma 1.LetBbe a base for topologyTon a setXand letYX. Let
B=Y=fB\Y:B2Bg. ThenB=Yis a base for the topologyT=Yon
Y.
Proof:
15

23. Lety2YandGbe an open set inYcontainingy. ThenG=H\Yfor
some open setHinX. Clearlyy2Hand so there existsB2Bsuch that
y2BandBH.Theny2B\Y,B\YGandB\Y2B=Y. This
proves thatB=Yis a base forT=Y.
Corollory 2.Second countability is a hereditary property.
Proof:
Let(X;T)be a space with a countable baseBandYX. ThenB=Yas
B=Y=fB\Y:B2Bg
is countable and is a base forT=Y.
Note 4.A topological spaceXis metrizable provided that the topology ofX
is generated by a metric.
Theorem 5.Metrisability is a hereditary property.
Proof:
Assume(X;T)is metrizable. Letdbe a metric onXwhich induces
the topologyT. LetYXandebe the restriction ofdtoYY. We
claim thateinduces the topologyT=YonY. Ify2Yandr >0then
Be(Y; r) =Bd(y; r)\YandBe(y; r)2T=Y. LetUbe metric topology on
Yinduced bye. LetBbe the collections of open balls inY.
Letu2UsinceBis a base forUi.e; the elements ofUcan be written
as the union of some elements ofB. But the elements ofBare the elements
ofT=Ya topology and henceuwill be an element ofT=Y.
)UT=Y
16

24. Conversely, letG2T=YthenG=H\YwhenH2T, the topology
induced byd. Hence for eachyinG(hence inH)9Bd(y; ry)H. Then
Be(y; ry) =Bd(y; ry)\YH\Y=G. HenceG=[y2GBe(y; ry)
)G2Uthen
T=YU
Therefore the two topologies are(U=T=Y)are same, or in other words
(Y;T=Y)is metrisable.
2.4 Closure of a Set, Continuous Functions And
Related Concepts
In this section, we include the concept of closure,convergence,interior and
boundary which are geometrically in spirit.Continuity is a way of going one
space into another and we dene homeomorphism which means a basic equiv-
alence relation and we also mention some general problems in topology.
Closed sets and Closure
Denition 16.A subsetAof a tpological spaceXis said to beclosedif the
setXAis open.
Consider a simple example:
Example 11.The subset[a; b]onRis closed because its complement
R[a; b] = (1; a)[(b;+1)
is open. Similarly,[a;+1)is closed because its complement(1; a)is open.
Note 5.The set of rationals is neither open nor closed in the usual topology
on the real line. A set which is both open and closed is sometimes called a
clopenset.
17

25. It is immediate that a set is open i its complement is closed.
Theorem 6.LetXbe a topological space, then
1.X; are closed sets.
2. Any intersection of closed setsXis closed.
3. Any nite union of closed set inXis closed.
Hint :
De-morgan's law
A(B[C) = (AB)\(AC)
A(B\C) = (AB)[(AC)
proof:
LetfA:2gis the collection of closed sets, then
X
\
2
A=
[
2
(XA)
(De-morgan's law )
*(XA)is open.
[
2
is an arbitrary union of open set.
)
T
Ais closed .
Similarly, ifAiwherei= 1; ; nis closed, then
X
n
[
i=1
Ai=
n
\
i=1
(XAi)
which is a nite intersection of open sets ,
)its open, and[Aiis closed .
andXare closed because they are the complements of the open setXand
resp.
18

26. Denition 17.Theclosureof a subset of a topological space is dened as
the intersection of all closed subsets containing it.
In symbols, ifAis a subset of a space(X;T), then its closure is the set
CX:Cis closed inX,CAg. It is denoted by

A.
Some important properties of closure are given below:
LetXbe a topological space andA; Bsubsets ofXthen,
1.

Ais a closed subset ofX. IfCis closed inXandACthen

AC.
2.

=
3.Ais closed inXif

A=A
4.

A=

A
5.A[B=

A[

B
Example 12.LetX=fa; b; c; d; egandT=fX; ;fag;fc; dg;fa; c; dg;fb; c; d; egg.Show
thatf

bg=fb; eg;fa; cg=Xandfb; dg=fb; c; d; eg:
Proof:
fHint :To nd the closure of a particular set, we shall nd all the closed
sets containing that set and then select the smallest. We)begin by writing
down all of the closed sets - there are simply the complements of all the open
sets.g
The closed sets are; X;fb; c; d; eg;fb; egandfag. So the smallest closed
set containingfbgisfb; eg;
i.e,f

bg=fb; eg
Similarly,fa; cg=Xand
fb; dg=fb; c; d; eg
19

27. Denition 18.LetXbe a topological space andAX. ThenAis said to
bedenseinXif

A=X.
Lemma 2.LetXbe space and letAXthenAis dense. I for every
non-empty open setB; A\B6=.
Proof:
supposeAis dense inXandBis a open set inX. IfA\B=. then
AXBand

AXB( from the denitions of dense). sinceXBis
closed(becauseBis open). But thenXBXcontradicting thatA=X.
SoA\B6=.
Conversely suppose thatA\B6=. Clearly that the only closed set con-
tainingAisXand

A=X
In the real line the usual topology the setQof all rational numbers, as
well as its complementRQare both dense. Actually they are also dense
subsets with respect to the semi-open interval topology.
Neighbourhoods, Interior and Accumulation points
Denition 19.Let(X;T)be a space,x02XandNXis said to be a
neighbourhoodofx0, if9an open setVsuch thatx02VandVN.
And thex0is said to be theinterior pointofN.
Lemma 3.A subsetAof a topological space is open, i it is a neighbourhood
of each of its points.
Proof:
LetXbe a topological space andGX. First supposeGis open. Then
Gis a nbd of each of its points. Conversely suppose thatGis a nbd of each
points. Then for eachx2G9an open setVxsuch thatx2VxandVxG.
ThenG=
[
x2G
Vx. So eachVxis open and inG.
20

28. Denition 20.Let(X;T)be a space andAX.Then theinteriorofA
is dened to be the set of all interior points ofA,
i.e the setfx2A:Ais a nbd ofxg.It is denoted byA
0
,int(A), or intT(A).
For an example, the interior of a closed disc in the plane is a open disc.The
set of rationals has empty interior with respect to the usual topology onR
and so does its complement, the set of all irrational numbers has an empty
interior.
Remark 3.The interior of the empty set is empty.
Theorem 7.LetXbe a space andAX. ThenA
0
is the union of all open
sets contained inA.
Proof:
LetUbe the collection of all open set contained inA. LetV=
[
G2U
G.
Ifx2Vthenx2G, for someG2Ui.e;Ais a nbd ofxand sox2A
0
.
Conversely, letx2A
0
. then there is an open setHsuch thatx2Hand
HA. But then,H2U,HU, thenx2Vi.e;V=A
0
A topology on a setXinduces an operatori:P(X)!P(X)is dened
byi(A) =A
0
. It is calledinterior operatorassociated withT. The
interior operator determinesTuniquely, for it is clear that a set is open i
it coincides with its interior.
LetXbe a space andx2X. Letxbe the set of all neighbourhood of
x2X, with respect to the given topology onX. The collectionsxis called
theneighbourhood systematx.
Some properties of neighbourhood system :
LetXbe a space and forx2X, letxbe the neighbourhood system atx.
Then ,
21

29. 1. IfU2x, thenx2U.
2. For anyU; V2x; U\V2x.
3. IfV2xandUVthenU2x
4. A setGis open, iG2x8x2G.
5. IfU2xthen9V2xsuch thatVUandV2y8y2V.
Denition 21.LetXbe a space and letAX,y2X. Thenyis said
to be anaccumulation pointofAif every open set containingycontains
atleast one point ofAother thany.
In a discrete space, ther is no point is an accumulation point of any set. In
indiscrete space, a pointyis said to be an accumulation point of any setA,
ifAcontains at least one point other thany. In the real line ( under usual
topology ), every real number is an accumulation point of the set of rational
numbers while the set of integers has no point of accumulation.
Denition 22.LetXbe a space andAX. Then thederived setofA
is the set of all accumulation points ofAinXand is denoted byA
0
.
A
0
not only depends onAbut also on the topology . Properties of derived
sets are elementary. With the denition of derived set of a set, we can describe
the closure of a set more closely.
Theorem 8.LetXbe a space andAX;

A=A[A
0
.
Proof:
We claim thatA[A
0
is closed. LetY2A[A
0
. SinceYis not an accumulation
point ofAthen9an open setV y2Vsuch thatVcontains no point ofA
excepty. Buty =2A, so we haveA\V=. We claimA
0
\V=. So let
22

30. z2A
0
\V. ThenVis an open set containingz, is an accumulation point
ofA. SoV\Ais non empty, which is a contradiction. SoA
0
\V=and
henceVA[A
0
. SoA[A
0
is closed and since it containsAandA
0
, i.e,

AA[A
0
.
AndA[A
0


A, we haveA

A. Lety2A
0
, ify =2

Atheny2X

A
which is a closed set , since

Ais closed set. Butyis an accumulation point
ofA. So(XA)\A6=which is a contradiction, sinceX

AXA.
Soy2

A. i.e;A[A
0


A. So

A=A[A
0
.
Theorem 9.LetXbe a space andAX,

A=fy2X: every ngbd ofy
intersectsAg.
Proof:
LetB=fy2X:U2y)U\A6=g, then we have to show that

A=B, from the above theoremA[A
0
=B, so lety2A[A
0
ify2Athen
certainly every nbd ofyintersectsAat a pointyand soy2B. Ify2A
0
then every nbd ofycontains a point ofAand soy2Bi.e,A[A
0
B.
Lety2Brst we can assume the negation that,y =2A[A
0
, theny =2

A
and soX

Ais a nbd ofy, which doesnot intersectA. This is a contradiction
thety2B. SoBA[A
0
. ThusA[A
0
=B.
Hence the result.
Continuity and Related concepts
Denition 23.Let functionf:X!Y x02XandT;Ube topologies
onX; Yrespectively. Thenfis said to becontinuousatx0, if for every
V2Usuch thatf(x0)2V;9U2Tsuch thatx02Uandf(U)V
Theorem 10.The following statements are equivalent:
23

31. 1.fis continuous atx0.
2. The inverse image of every neighbourhood off(x0)inYis a neighbour-
hood ofx0inX.
3. For every subsetAX; x02

A)f(x0)2f(A).
Proof :
(1))(2)LetNbe a neighbourhood off(x0)inY. Then9an open setV2Y
such thatf(x0)2VandVN. Sincefis continuous atx0;9an open set
U2Xsuch thatx02Uandf(U)V. Sox02Uf
1
(V)f
1
(N)
thusf
1
(N)is a neighbourhood ofx0.
(2))(3). LetAXand suppose thatx02

A. Iff(x0)=2f(A), then
there is a neighbourhoodNoff(x0)such thatf(A)\N=i.e;f
1
(f(A))\
f
1
(N) =and hence thatA\f
1
(N) =, sinceAf
1
(f(A)). But the
inverse image of every neighbourhood off(x0)inYis a neighbourhood of
x0inX,f1(N)is a neighbourhood ofx0and soA\f1(N)6=, since
x02

Awhich is a contradiction.
(3))(4). LetVbe an open set containingf(x0). LetA=Xf
1
(V) =
f
1
(YV). Thenf(A)YVand sof(A)YVasYVis closed.
Sof(x0)=2f(A); x0=2

Afrom(3). Hence there is a neighbourhoodNofx0
such thatN\A=. Clearly thenf(N)V)fis a continuous atx0.
Continuity offatx0to mean that the image of a neighbourhood ofx0
is a neighborhood off(x0).
Continuity at a point is a local concept. It depends on the particular
point, the particular function and also the topologies on the domain and the
co-domain. In the next denition we generalise the concept.
Denition 24.Let functionf:X!YandT;Ube topologies onX; Y
respectively. Thenfis said to becontinuous(TUcontinuous), if it
24

32. is continuous at each points ofX.
Theorem 11.The following statements are equivalent :
1.fis continuous
2. For each closed subsetCofY,f
1
(C)is closed inX.
3. For each subsetAofX,f(

A)f(A).
4. There is a basisBfor the topology ofYsuch thatf
1
(B)is open in
Xfor each basic open setBinB.
5. There is a sub-basisSfor the topology ofYsuch thatf
1
(S)is open
inXfor each sub-basic open setSinS.
Proof:
We use the open set formulation to describe continuity :fis continuous i
for each open setVinY,f
1
(V)is open inX. The equivalence of(1)and
(2)follows from the duality between open sets and closed sets,
(2) =)(3): Suppose that(2)holds and letAXthenf(A)is a
closed subset ofY, so its inverse imagef
1
(f(A))is closed inX. Since
Af
1
(f(A))
and the latter set is closed then

Af
1
(f(A))
so
f(

A)f(A)
and(3)holds.
25

33. (3) =)(2)Assume(3)and letCbe a closed subset ofY. then
f(f
1
(C))ff
1
(C)

C=C
so
f
1
(C)f
1
(C)
andf
1
(C)must be a closed set. Thus(3) =)(2).
We have completed the proof(1);(2)and(3)are equivalent. Since a
basisBand sub-basisSforYconsist of open sets, it should be clear that
(1) =)(4) (5), since a basis is a sub-basis,(4) =)(5).
(5) =)(4): Suppose(5)holds and consider the basisBgeneratedS
by taking nite intersection. For any basic open setB2B,
B=
n
\
i=1
Si
for some nite collections of elementsS1; : : : ; SnofS. Then
f
1
(B) =f
1

n
\
i=1
Si
!
=
n
\
i=1
f
1
(Si)
. Since each setf
1
(Si)is open inXand the intersection of any nite
collection of open sets is open, thenf
1
(B)is open inX. Thus(5) =)(4).
(4) =)(1): Assuming(4), letObe an open set inY. By the denitions
basis,
O=
[
2I
B
for some sub-collectionfB:2Igof the basisB. Then
f
1
(O) =f
1

[
2I
B
!
=
[
2I
f
1
B
.
Since each setf
1
(B)is openXand the union of any family of open
sets is open, thenf
1
(O)is open inXandfis continuous.
26

34. Theorem 12.Iff:X!Yandg:Y!Zare continuous function , then
the composite functiongf:X!Zis continuous .
Continuous functions are also calledmapsormappings. LetX1; X2; : : : ; Xn
be the sets and letX=X1X2: : :Xn. For eachi= 1;2; : : : ; nwe
denei:X!Xiby the rulei(x1; x2; : : : ; xn) =xi. Theniis called the
projectiononXior thei
th
projection. It is a surjective function expect
in the case of someXj, whereXis empty. Ifx2Xtheni(x)is called the
i
th
coordinateofx.
Denition 25.LetX; Ybe spaces. A functionf:X!Yis said to be
open(Closed) if wheneverAXis an open ( closed ), thenf(A)Y
is open ( closed ).
I order to show that a function is open, it is enough to shoe it takes all
members of a base for the domain space to open subsets of the co-domain.
Homeomorphsim :
Next we are going to discuss an important and interesting topic about home-
omorphsim. Two groups are the same for the purposes of group theory if
they are (group) isomorphic. Two vector spaces are the same for the pur-
poses of linear algebra if they are (vector space) isomorphic. When are two
topological spaces(X;T)and(Y;U)the same for the purposes of topol-
ogy? In other words, when does there exist a bijection betweenXandY
for which open sets correspond to open sets, and the grammar of topology
(things like union and inclusion)is preserved? A little reection shows that
the next denition provides the answer we want.
Denition 26.A functionf:X!Ybetween two topological spaceXand
Yis ahomemorphismortopologically equivalentif it is a one-to-one,
27

35. onto map and bothfandf
1
are continuous. When such a homemorphism
exist, then we say that,Xis said to behomeomorphictoY.
Homemorphic space are indistinguishable as topological space. For ex-
ample, iff:X!Yis a homeomorphsim, thenGis a open inXif(G)
is open inY, and a sequence(xn)converges toXi the sequence(f(xn))
converges tof(x)inY.
A one-to-one,onto mapfalways has an inversef
1
, butf
1
need not be
continuous iffis.
Denition 27.A propertyPof a topological space is atopological prop-
ertyortopological invariantprovided that if spaceXhas propertyP, then
so does every spaceYwhich is topologically equivalent toX.
The next theorem give example of topological properties .
Theorem 13.Separability is a topologically property
Proof:
LetXbe a separable space with countable dense subsetAandYa space
homeomorphic toX. Letf:X!Ybe a homeomorphism. Clearly, for a
countable dense subset ofYisf(A). To see thatf(A)is dense inY, letObe
a non empty open set inY. Thenf
1
(O)is a non empty open set inX. Since
Ais dense inX; f
1
(O)contains some memberaofA. ThenOcontains the
memberf(a)off(A), so every non-empty open set inYcontains at least
one member off(A). Thusf(A) =YandYis separable.
The relation of being homeomorphic is obviously an equivalence rela-
tion(in the technical sense: it is reexive, symmetric, and transitive). Thus
topological spaces split into equivalence classes, sometimes called homeomor-
phy classes. In this connection, the topologist is sometimes described as a
28

36. person who cannot distinguish a coee cup from a doughnut (since these two
objects are homeomorphic). In other words, two homeomorphic topological
spaces are identical or indistinguishable from the intrinsic point of view in the
same sense as isomorphic groups are indistinguishable from the point of view
of abstract group theory or two conjugatennmatrices are indistinguish-
able as linear transformations of anndimensional vector space without a
xed basis.
Theorem 14.LetX < Ybe spaces andf:X!Ya function. Then the
following statements are equivalent:
1.fis a homeomorphsim,
2.fis a continuous bijection andfis open,
3.fbijection andf
1
is a continuous,
4. there exists a functiong:Y!Xsuch thatf; gare continuous ,
gf=idXandfg=idY.
29

37. Chapter 3
Topological Properties
3.1 Compact Spaces
Denition 28.LetXbe a space andAXthenAis said to becompact
subsetofXif every cover ofAby open subsets ofXhas a nite sub-cover.
A spaceXis said to becompactifXis a compact subset of itself.
Or equivalent denition :
Denition 29.The space(X; T)is called
1. compact if every open cover of X has a nite sub-cover;
2. sequentially compact if every sequence has a convergent subsequence;
3.compact if it is the union of a countable family of compact sets.
4. locally compact if every point has an open neighborhood whose closure
is compact in the induced topology.
It is known from elementary real analysis that for subsets of aR
n
com-
pactness and sequential compactness are equivalent. This fact naturally gen-
30

38. eralizes to metric space. A topological space(X;T)is called compact ifX
itself is a compact set.
Denition 30.A space is said to beseparableif it contains a countable
dense subset .
Theorem 15.Every second countable space isLindelo.
Proof: LetTbe a second countable .Then by denition(6) , its topology
has a countable basis. LetBbe this countable basis .
LetCbe an open cover ofT.Every set inCis the union of a subset ofB.This
union of a subset ofBis a countable sub-cover ofC.i.e,Tis Lindelo.
Theorem 16.A metric space is Lindelo if and only if it is separable.
Proof:
Since we know separable metric spaces are second countable and therefore
Lindelo, we only need to show that a Lindelof metric spaceXis separable.
For eachn2N f0g, the set of all open balls of radius1=nis an open
cover ofX; letx(n; m)2Xbe points such thatfB(x(n; m);1=n)gm2Nis a
countable sub-cover. Then the setfx(n; m)jm; n2N; n= 0gis a countable
dense subset: ifB(x; r)is an open ball, letnbe large enough that1=n < r=2;
if nox(n; m)is inB(x; r), thenxis not in anyB(x(n; m);1=n)and the set
of such balls could not be a cover.
Theorem 17.Every second countable space is separable.
proof:
LetXbe a second countable space with countable basisBandAbe the
countable set choosing a member from each basic open set . If2Bthen
choose one member from each non-empty member ofB.
It follows from the denition of basis thatAis dense inX.
31

39. The real line with the semi-open interval topology is separable but not
second countable. As for compactness , it is clear that every compact space
is Lindelo.
Theorem 18.Letaandbbe real numbers satisfyinga < b. Then the closed
bounded interval[a; b]is a compact subset ofR.
Proof:
LetUbe a collection of open sets inRwith the property that each point of
the interval[a; b]belongs to at least one of these open sets. We must show
that[a; b]is covered by nitely many of these open sets.
LetSbe the set of allT2[a; b]with the property that[a;T]is covered
by some nite collection of open sets belonging toU, and lets=supS. Now
s2 Wfor some open setWbelonging toU. MoreoverWis open inR, and
)9some >0such that(s; s+) W. Moreoversis not an upper
bound for the setS, hence9someT2SsatisfyingT> s. It follows from
the denition ofSthat[a;T]is covered by some nite collectionV1; V2; : : : Vr
of open sets belonging toU.
Lett2[a; b]satisfyTt < s+. Then[a; t][a;T][(s; s+)
V1[V2[: : :[Vr[ W, and thust2 S. In particulars2 S, and moreover
s=b, since otherwiseswould not be an upper bound of the setS. Thus
b2 S, and)[a; b]is covered by a nite collection of open sets belonging to
U, as required.
Theorem 19.LetAbe a closed subset of some compact topological spaceX.
ThenAis compact.
Proof:
LetUbe any collection of open sets inXcoveringA. On adjoining the
open setXAtoU, we obtain an open cover ofX. This open cover ofX
32

40. possesses a nite sub-cover, sinceXis compact. MoreoverAis covered by
the open sets in the collectionUthat belong to this nite sub-cover. From
thisAis compact.
Theorem 20.Any closed interval is a compact subset ofRin the standard
topology.
Proof:
We prove the proposition for[0;1]and the general statement follows easily.
SupposeCis an open cover of[0;1]inR. LetS [0;1]be the set withx2 S
if and only if there is a nite sub-collection ofCcovering[0; x]. Of course,
02 S, soSis non-empty. Letybe the least upper bound ofS. Ify2 S,
there isU2Cwithy2U. There isy
0
2Uwithy
0
< yand(y
0
; y)U, and
there is a nite sub-collectionC
0
ofCcovering[0; y
0
]. But thenC
0
[ fUgis
a nite cover of[0; y], a contradiction. Ify2 Sandy <1, then there is a
nite subsetC
0
ofCand aU2C
0
withy2U. SinceUis an open set, there
isy
0
2Uwithy
0
> y, contradicting the supposition thatyis the least upper
bound ofS. So we must havey= 1, meaning thatChas a nite sub-cover
covering[0;1].
Theorem 21.Every continuous real-valued function on a compact space is
bounded and attains its extrema.
Proof: LetXbe a compact space andf:X!Ris continuous.Now we
have to show thatfis bounded for eachx2X, letjxbe the open interval
(f(x)1; f(x) + 1)and letVx=f
1
(jx). By continuity off; Vxis an open
set containing x. Now the collectionfVx:x2Xgis an open cover ofXand
by compactness , admits a nite sub-coverfVx1; Vx2; : : : ; Vxng(say).
LetM=maxf(f(x1); f(x2); : : : ; f(xn))g+ 1and
letm=minf(f(x1); f(x2); : : : ; f(xn))g 1. For anyx2Xthere is somei
33

41. such thatx2Vxi. Thenf(xi)1< f(x)< f(xi) + 1andm < f(x)< M
shows thatfis bounded.
LetL; be the supremum and inmum offoverX. i.e,f(x) =Lthen we
dene a new functiong:X!Rbyg(x) = 1=(Lf(x))8x2X.
gis unbounded and anyR >09xsuch thatf(x)> L1=Rand hence
g(x)> R. This shows thatfattainsL. Similarlyfattains the inmum.
3.2 Connected Spaces
Denition 31.A topological spaceXisdisconnectedorseparatedif it is
the union of two disjoint, non-empty open sets. Such a pairA; Bof subsets
ofXis called sseparationofX. A space isconnectedprovided that it is
not disconnected. in other words,Xis connected if there do not exists open
subsetsAandBofXsuch that
A6= B 6= A \B= A [B=X
A subspaceYofXisconnectedprovided that it is a connected space
when assigned the sub-space topology. The termsconnected setandcon-
nected subsetare some times used to mean connected space and connected
subspace, respectively.
Example 13.The real lineRwith the usual topology is connected.
Let us suppose the negation thatRis disconnected. Then
R=A[B
for some disjoint, non-empty open setsAandBofR. Since
A=RB; B =RA
34

42. thenAandBare closed as well as open. Consider two pointsaandbwith
a2A; b2B:Without loss of generality we may assumea < b. Let
A
0
=A\[a; b]
NowA
0
is a closed and bounded subset ofRand consequently contains its
least upper boundc. Not thatc6=bsinceAandBhave no point in common.
Thusc < b. SinceA\(c; b] =then
(c; b]B
and hencec2

B:ButBis closed, soc2B. Thusc2A; B, which is a
contradicting the assumption thatAandBare disjoint. This shows thatR
is connected.
Theorem 22.The continuous image of a connected space is connected.
Proof: Letf:X!Ybe an onto continuous function andXbe a
connected space. IfAis a subset ofYsuch thatAis both open and closed,
thenf
1
(A)is both open and closed. Sincefis continuous. Sincef
1
(A)
is both open, closed andXis connected,f
1
(A)must either be all ofXor
the empty set.
)Ais either the entire space or the empty set, i.e,Yis connected.
Theorem 23.LetXbe a space andA; BX. Then
1.A[B=Xand

A\

B=.
2.A[B=X; A\B=.
3.B=XAandAis clopen inX.
4.B=XAand@Ais empty.
35

43. 5.A[B=X; A\B=andA; Bare both open inX.
Proof:(1) =)(2). Clearly

A\

B=. SinceA

AandB

B.
Also

AX

BXB=Aand so

A=Aclear thatAis closed.
(2) =)(3)shows that the complement of a closed set is open.
(3) =)(4)from this it is clear that the boundary of a clopen set is empty.
(5) =)(1). Assume thatX=A[BwhereA\B=andA; Bare open.
ThenA=XBandB=XAwhereA; Bare closed. i.e,

A=A;

B=B
this shows that

A

B=.
Theorem 24.A subset ofRis connected i it is an interval.
Proof:
Suppose thatJRis not an interval. Then there arex; y2Jandz2J
withx < z < y. Then deneA= (1; z)\JandB= (z;1)\J. Clearly,
A; Bare disjoint, non-empty, relatively open, andA[B=J. SoJis not
connected.
Conversely, suppose thatJis an interval. We will show thatJis connected.
Letf:J! f0;1gbe continuous, and suppose thatfis not constant. Then
there arex1; y12Jsuch thatf(x1) = 0andf(y1) = 1. Assume that
x1< y1. Letabe the midpoint of[x1; y1]. Iff(a) = 0, then setx2= 0and
y2=y1, and otherwise,x2=x1andy2=a. Sox1x2y2;jx2y2j
2
1
jx1y1j; f(xi)6=f(yi). Iterating this procedure we nd sequencesxnand
ynwith the following properties:
x1x2: : : xn< yn: : :y1;jxnynj 2

1
jxn1yn1j 2
n1
jx1y1j
andf(xn) = 0; f(yn) = 1. SinceRis complete,fxngconverges to somez,
and sincejxnynj !0; yn!z. Clearly,z2J.
Hence0 =limnf(xn) =f(z) =limnf(yn) = 1. This is a contradiction. So
fis constant, and this implies thatJis connected.
36

44. Denition 32.Non-empty subsetAandBof a spaceXareseparated
sets, if

A\BandA\

Bare both empty
Theorem 25.The following statements are equivalent for a topological space
X:
1.Xis disconnected.
2.Xis the union of two disjoint, non-empty closed sets.
3.Xis a union of two separated sets.
4. There is a continuous function fromXonto a discrete two-point space
fa; bg.
5.Xhas a proper subsetAwhich is both open and closed .
6.Xhas a proper subsetAsuch that

A\(XA) =:
Proof:
It will be shown that (1) implies each of the other statements and that each
statement implies(1). Assume rst thatXis disconnected and letA,Bbe
disjoint, non-empty open sets whose union isX.
(1) =)(2):B=XAandA=XBare disjoint, non-empty closed
sets whose union isX.
(1) =)(3): SinceAandBare closed as well as open, then

A\B=A\B=; A\

B=A\B=:
SoXis the union of the separated setsAandB.
37

45. (1) =)(4): The functionf:X! fa; bgdened by
f(x) =
8
>
<
>
:
aifx2A
bifx2B
is continuous and mapsXonto the discrete spacefa; bg.
(1) =)(5):A6=, and
A=XB6=X
sinceB6=. ThusAis the required set. (Bwill do equally well.)
(1) =)(6): EitherAorBcan be used as required set.
(2) =)(1): IfX=C[DwhereCandDare disjoint, non-empty
closed sets, then
D=XC; C=XD
are open as well as closed.
(3) =)(1): IfXis the union of separated setsCandD, thenCandD
are both non-empty, by denition. SinceX=C[Dand

C\D=, then

CC, soCis closed. The same argument shows thatDis also closed, and
it follows as before thatCandDmust be open as well.
(4) =)(1): Iff:X! fa; bgis continuous, thenf
1
(a)andf
1
(b)
are disjoint open subsets ofXwhose union isX. Sincefis required to have
bothaandbas images, bothf
1
(a)andf
1
(b)are non-empty.
(5) =)(1): SupposeXhas a proper subsetAwhich is both open and
closed. ThenB=XAis a non-empty open set disjoint fromAfor which
X=A[B:
(6) =)(1): SupposeXhas a proper subsetAfor which

A\(XA) =:
38

46. Then

Aand(XA)are disjoint, non-empty closed sets whose union is
X, and it follows as before that

Aand(XA)are also open.
Corollory 3.The following statements are equivalent for a topological space
X:
1.Xis connected.
2.Xis not the union of two disjoint, non-empty closed sets.
3.Xis not the union of two separated sets.
4. There is no continuous function fromXonto a discrete two-point space
fa; bg:
5. The only subsets ofXwhich are both open and closed areXand:
6.Xhas no proper subsetAfor which

A\(XA) =:
Theorem 26.A subspaceYof a space is disconnected i9open setsUand
VinXsuch that
U\Y6=; V\Y6=; U\V\Y=; YU[V:
Proof:
Suppose rstYis disconnected. Then there are disjoint, non-empty , open
setsAandBin subspace topology forYsuch thatY=A[Bby the
denition of relatively open sets, there must be open setsUandVinXsuch
that
A=U\Y; B=V\Y
39

47. It is a simple matter to check thatUandVhave the required properties.
For the reverse implication, suppose thatUandVare open subsets ofX
such that
U\Y6=; V\Y6=; U\V\Y=; YU[V
. Then
A=U\Y; B=V\Y
are non empty, disjoint relatively open sets whose union isY, soYis discon-
nected
Theorem 27.IfYis a connected subspace of spaceX, then

Yis connected.
Proof:
SupposeYis connected, the connectedness of

Ywill be shown by proving
that there is no continuous function on

Yonto a discrete two point space.
Consider a continuous functionf:

Y! fa; bgfrom

Yinto such a discrete
space. We must shown thatfis not surjective. Then restrictionfYcannot
be surjective. This means thatfmapsYto only one point offa; bg, say a:
f(Y) =fag:
Sincefis continuous, previous
f(

Y)f(Y) =fag=fag
sofis not surjective. Thus, Theorem 23,

Yis conneted.
Corollory 4.LetYbe a connected subspace f a spaceXandZa subspace
ofXsuch thatYZ

Y. ThenZis connected
cleraly, it is not true in general that the union of connected sets is always
connected. The reader should also be able to give an example to show that
40

48. the intersectionof two connected sets may fail to be connected. It does seem
reasonable, however, that if a family of connected sets all have a point in
common, then their union is connected. This is true and is proved in the
next theorem.
Theorem 28.LetXbe a space andfA:2Iga family of connected
subsets ofXfor which
\
2I
Ais not empty. Then
[
2I
Ais connected.
proof:
To show thatY=
[
2I
Ais connected. Suppose thatUandVare open sets
inXfor which
U\Y6=; U\V\Y=; YU[V
. It will be shown thatV\Y=, thus proving thatYis connected. Now
u\Y6=, soUcontains some point inA, for some
0
2I. SinceAis
connected, thenAU. Ifb2
\
2I
A, thenbmust be inAsob2U. Thus
Ucontains a pointbin eachA; 2I. SinceAis connected, thenAU
for each2I.Thus
Y=
[
2I
AU
soV\Y=.
Corollory 5.LetXbe a space,fA:2Iga family of connected subsets
ofX, andBa connected subset ofXsuch that, for each2I; A\B6=.
ThenB[(
[
2I
A)is connected.
proof:
By above theorem, each setb[A; 2I, is connected and the intersection
\2I(B[A)6=
41

49. since it containsB. Thus by above theorem
B[

[
2I
A
!
=
[
2I
(B[A)
is connected.
Denition 33.Acomponentof a topological spaceXis connected subset
CofXwhich is not a proper subset any connected subset ofX.
The following properties of the component of a space are given :
1. Each pointa2Xbelongs to exactly one component. The component
cacontainingais the union of all the connected subsets ofXwhich
containaand thus may be thought of as the largest connected subset
ofXwhich containsa.
2. For pointsa; b2Xthe componentscaandcbare either identical or
disjoint.
3. Every connected subset ofXis contained in a component.
4. Each component ofXis a closed set.
5.Xis connected ofXandA; Bform a separation ofX, thenCis a
subset ofAor a subset ofB.
Example 14.In a discrete space each components contains only one point.
Example 15.For a setRof rational number with the subspace topology
determined by the real line, each component contains only one point . Note
that the topology in this case is note discrete.
Property(6)of components states that if two pointsaandbbelongs to the
same component ofX, then they must belong to the same member of any
42

50. separation ofX. This example shows that the converse is false : It is possible
for pointsa; bto be always in the same member of any separationA; BofX
yet to belong to dierent components.
Consider the subspaceXofR
2
consisting of a sequence of line segment
converging to a line segment whose mid pointcha s been deleted. Then
[a; c)is the component ofXwhich containsaand(c; b]is the component
which containsb, soaandbbelong to dierent components. However, for
any separation ofXinto disjoint non-empty open setsAandBwhose union
isXbotha; b2Aor botha; b2B.
Denition 34.A spaceXisTotally disconnectedprovided that each com-
ponents ofXconsists of a single point.
The real line and all intervals on the real line are connected. Following
theorems show that there are no other connected subsets ofR
Lemma 4.A non empty subsetAofRis an interval i for each pairc; dof
members ofA, every real number b/wcanddis inA.
Proof:
Consider,ARwhich contains every real number b/w any two of its mem-
bers. We can split this into several cases whether or notAhas a least upper
bound or greatest lower bound and whether or not these bounds, if they
exist, belong toA.
Suppose thatAhas neither a least upper bound nor a greatest lower
bound. Then forx2Rthere are memberscanddofAfor whichc < xand
d > x. Thenx2A, so it follows thatA=RandAis the interval(1;1).
SupposeAhas greatest lower boundawhich does not belong toAandA
has no least upper bound. ThenAcontains no real number such thatxa.
Ify > athenyis not an upper bound forAso there is a elementd2A
43

51. withd > y. Similarly,yis not the greatest lower bound forA, so there is an
elementc2Awithc < y. Thenc < y < d
)y2A.ThusA= (a;1)and is a interval
The remaining cases are similar.
Theorem 29.The connected subset ofRare precisely the intervals.
proof:
Since we know that every interval is connected, it remains only to be proved
that a subsetBofRwhich is not an interval must be disconnected. Let
BRthat is not an interval. Then by lemma, there are membersC; d2B
andy2Rwithc < y < dfor whichy =2Bthen the open sets
U= (1; y); V= (y;1)
have the following properties
1.c2U\B, soU\B6=; d2V\B, soV\B6=
2.U\V=, soU\V\B=
3.BU[V
By theorem24,Bis disconnected. Hence every connected subset ofRmust
be an interval.
3.3 Some Applications of Connectedness
In the present section
Theorem 30.The Intermediate Value Theorem: Letf: [a; b]!Rbe
a continuous function , andK2Rbetweenf(a)andf(b). Then9c2[a; b]
such thatf(c) =K.
44

52. Proof:
The interval[a; b]is connected. Sincefis continuous, thenf([a; b])is a
connected subset ofRand by theorem 29, it must be a interval. Thus any
numberKb/wf(a)andf(b)must be in the imagef([a; b]). This means
thatK=f(c)for some real numberc2[a; b].
Corollory 6.Letf: [a; b]!Rbe a continuous function for which one
off(a)andf(b)is positive and the other is negative. Then the equation
f(x) = 0has a root betweenaandb.
Theorem 31.Letf: [a; b]![a; b]be a continuous function, then there is a
memberc2[a; b]such thatf(c) =c
Iff(a) =aorf(b) =b, thenfhas the required property, so we assume
thatf(a)6=aandf(b)6=b. Thusa < f(a)andf(b)< b, sincef(a)and
f(b)must be in[a; b]:Deneg: [a; b]!Rby the rule
g(x) =xf(x);8x2[a; b]
Thengis continuous and
g(a) =af(a)<0; g(b) =bf(b)>0
From Intermediate value theorem the existence ofc2[a; b]for whichg(c) = 0
thenf(c) =c.
Denition 35.Fixed pointof a functionf:X!Yis a pointxfor
whichf(x) =x. A topological spaceXhas axed-point propertyif every
continuous function fromXinto itself has at least one xed point.
Then the above theorem, Theorem 31, can be restated as follows:
"Every closed and bounded interval has the xed point property."
45

53. Theorem 32.The xed point property is a topology invariant.
Proof:
LetXbe a space which has the xed-point property,Ya space homeomorphic
toXandh:X!Y;a homeomorphism. Letf:Y!Ybe continuous
function. Since the composite functionh
1
fh:X!Xis a conitiuous
function onXit has at least one xed pointx0
h
1
fh(x0) =x0
then
f(h(x0)) =hh
1
fh(x0) =h(x0):
So the point is a xed point forfthusYhas the xed-point property.
Example 16.The real line does not have the xed pint property since, for
a example, the function
f(x) =x+ 1; x2R;
has no xed point. Since each open interval is homeomorphic toR
Example 17.ThensphereS
n
; n1, does not have the xed-point prop-
erty since the function
g(x) =x; x2S
n
;
has a xed point.
46

54. References
1. George F Simmons. Topology And Modern Analysis.
McGraw-Hill Book Company, INC.
2. James R Munkress. Topology A First Course.
Prentice Hall.
3. K D Joshi. Introduction to General Topology.
Wiley Eastern Limited.
4. Lynn Arthur Steen
J.Arthur Seebach. Counter Examples in Topology.
Springer-Verlag New York.
47