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,topos meaning 'sur-
face' and logos meaning '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.
A circle is topologically equivalent to an ellipse and a sphere is equivalent
to an ellipsoid. 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
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 element x is in a set A, we write
x ∈ A
and say that x is a Member of A, or that x belongs to A. If x is not on A,
1
9. we write
x /∈ A.
Denition 1. Two sets A and B are said to be equal and we write A = B,
if they contain the same elements.
Thus ,to prove that the sets A and B are equal, we must show that
A ⊆ B B ⊆ A.
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. The union of set A and B is the set
A ∪ B = {x : x ∈ A or x ∈ B}.
2. The intersection of the sets A and B is the set
A ∩ B = {x : x ∈ A and x ∈ B}.
3. The complement of B relative to A is the set
A − B = {x : x ∈ A and x /∈ B}.
The set that has no element is called empty set and it is denoted by
φ. Two sets A and B are said to be disjoint if they have no elements in
common and it is expressed by A ∩ B = φ.
Functions:
2
10. Denition 3. A function f from set X to set Y , denoted f : X → Y is a
rule which assigns to each member x of X a unique member y = f(x) of Y .
If y = f(x) then y is called the image of x and x is called a pre-image of
y. The set X is the domain of f and Y is the co-domain or range of f.
Note that for a function f : X → Y each element x in X has a unique image
f(x) in Y . However, the number of pre-image may be zero, one, or more
than one.
Denition 4. A function f : X → Y is one-to-one or injective means that
for distinct elements x1, x2 ∈ X, f(x1) = f(x2). In other words, f is 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 function f for which f(X) = Y i.e, for which the image f(X) equals
the co-domain , is said to be map X onto Y or to be surjective.
A one-to-one function from X onto Y is called a one-to-one corre-
spondence or a bijection. Thus f : X → Y is a one-to-one correspondence
provided that each member of Y is the image under f of exactly one member
of X. In the case there is an inverse function f−1
: Y → X which assigns
to each y in Y its unique pre-image x = f−1
(y) in X.
Example 1. Let X = {a, b, c, d, e}, Y = {1, 2, 3, 4, 5} and the function f :
X → Y dened by
f(a) = 1, f(b) = 2, f(c) = 3, f(d) = 4, f(e) = 5
is a bijection with inverse function f−1
: Y → X dened 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. If f : X → Y and g : Y → Z are functions on the sets, then
the composite function g ◦ f : X → Z is dened by
g ◦ f(x) = g(f(x)), x ∈ X
The composite function g ◦ f is some times denoted simply gf.
Example 2. Consider the function f : R → R and g : R → R dened by
f(x) = x2
, g(x) = x + 1. Then the composite function g ◦ f and f ◦ g are
both dened
g◦f(x) = g(f(x)) = g(x2
) = x2
+1, f◦g(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. A metric space is a set X where we have a notation of dis-
tance. That is, if x, y ∈ X, then d(x, y) is the distance between x and
y.The particular distance functions must satisfy the following conditions.
1. d(x, y) ≥ 0 for all x, y ∈ X
2. d(x, y) = 0 i x = y
3. d(x, y) = d(y, x)
4. d(x, z) ≤ d(x, y) + d(y, z)
Open ball :
4
12. Denition 7. Let x0 ∈ X and r be a positive real number .Then the open
ball with centre x0 and radius r is dened to be the set
{x ∈ X : d(x, x0) r}.
It is denoted either by Br(x0) or by B(x0; r). It is also called the open r-ball
around x0 .
Open set :
Denition 8. A subset A ⊂ X is said to be open if for every x0 ∈ A, ∃ some
open ball around x0 ∈ A. If their exist some r 0 such that B(x0; r) ∈ A.
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. Let {xn} be a sequence in metric space. Then {xn} converges to y
in X i for every open set y ∈ U ∃ N ∈ Z.
Theorem 1. Let (X; d) be a metric space. Then
1. φ and X are open.
2. The union of collection of open set is open.
3. The intersection of nite number of open set is open.
4. x, y ∈ X ∃ open sets U, V such that x ∈ U, y ∈ V and U ∩ V = φ.
Proof :
(i) Since there are no points e ∈ φ the statement x ∈ φ whenever d(x, e) 1,
holds for all e ∈ φ. Since every point x ∈ X, the statement x ∈ X whenever
d(x, e) 1 , holds ∀ e ∈ X
5
13. (ii) If e ∈
α∈A
Uα, then we can nd a particular α1 ∈ A with e ∈ Uα1. Since
Uα1 is open, we can nd a δ 0 such that
x ∈ Uα1, d(x, e) δ
. Since Uα1 ⊆
α∈A
Uα,
x ∈
α∈A
Uα, d(x, e) δ
Thus
α∈A
Uα is open.
(iii) If e ∈
n
j=1
Uj, then e ∈ Uj for each 1 ≤ j ≤ n. Since Uj is open, we can
nd a δj 0 such that
x ∈ Uj, d(x, e) δj
. Setting δ = min{δj}1≤j≤n we have δ 0 and
x ∈ Uj, d(x, e) δ
forall 1 ≤ j ≤ n . Thus
x ∈
n
j=1
Uj, d(x, e) δ
and we have shown that
n
j=1
Uj is open.
Example 3. The n− dimensional Euclidean space Rn
is a metric space with
the respected to the function d : Rn
× Rn
→ R, dened by
d(x, y) =
n
i=1
(xi − yi)2
1/2
where x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) ∈ Rn
where xi, yi ∈ R.
6
14. Clearly, d(x, y) ≥ 0 ∀ x, y ∈ R,
d(x, y) = 0 i
n
i=1
(xi − yi)2
1/2
= 0
i.e, i xi = yi ∀ i = 1, 2, . . . , n
Hence x = y i d(x, y) = 0
Now let
x = (x1, x2, . . . , xn)
y = (y1, y2, . . . , yn)
z = (z1, z2, . . . , zn)
be three arbitrary elements of Rn
.
Since xi, yi, zi ∈ R ∀ i = 1, 2, . . . , n and
Pi = xi − yi and
Qi = yi − zi ∈ R
Clearly, Pi + Qi = xi − zi where i = 1, 2, . . . , n
By the corollary we just proved
n
i=1
(Pi + Qi)2
1/2
≤
n
i=1
(Pi)2
1/2
+
n
i=1
(Qi)2
1/2
i.e,
n
i=1
(xi + zi)2
1/2
≤
n
i=1
(xi − yi)2
1/2
+
n
i=1
(yi − zi)2
1/2
i.e, d(x, y) ≤ d(x, y) + d(y, z) (Triangle inequality)
Finally, d(x, y) =
n
i=1
(xi − yi)2
1/2
=
n
i=1
(yi − xi)2
1/2
= d(x, y)
7
15. All these prove that d is a metric known as Euclidean metric or Usual
metric.
Example 4. Let R be a set of real numbers, show that the function d : R → R
dened by d : (a, b) = |a − b|, ∀ a, b ∈ R, is a metric on R.
Here,
1. |a − b| ≥ 0 =⇒ d(a, b) ≥ 0, ∀ a, b ∈ R
2. |a − b| = 0, iff a − b = 0, iff a = b so that d(a, b) = 0 ⇔ a = b
3. |a − b| = |b − a| =⇒ d(a, b) = d(b, a), ∀ a, b ∈ R
4. |a − b| = |(a − c) + (c − b)| ≤ |a − c| + |c − b|
=⇒ d(a, b) ≤ d(a, b) ≤ d(a, c) + d(c, b), a, b, c ∈ R
From this d is a metric on R.
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. A topological space is a pair (X, T ) where X is a set and
T is a collection of subsets of X satisfying:
1. φ, X ∈ T .
2. T is clodsed under arbitrary unions.
3. T is closed under nite intersections.
The collection T is said to be a topology on the set X. Members of T
are called open sets of X. The elements of X are called its points.
9
17. Example 5. Let X be a non-empty set, and let the topology be the class
of all subsets of X. This is called the discrete topology on X , and any
topological space whose topology is the discrete topology is called a discrete
space.
Suppose that X = φ, T = ℘(X) , cleraly X and φ ∈ T . Let
A = {uα/α ∈ λ} then uα ⊂ X ∀ α
∞
i=0
uα ⊂ X ∈ ℘(x) = T
so T is closed under arbitrary union.
Now we are going to show T is closed under nite intersection. Take
u1, u2, . . . , un be a nite elements of T then,
n
i=1
uα ⊂ X ∈ ℘(x) = T
i.e; T is closed under arbitrary intersection. So (X, T ) is a topological space.
Example 6. Let X be a non-empty set, and let the topology consist only the
empty set φ and full set X. This is called the in-discrete topology on X
and any topological space whose topology is the in-discrete topology is called
a in-discrete space.
Here T = {φ, X} ∴ T is closed under arbitrary union, be-
cause T consist only φ and X, so the union is either X or φ. Similarly T is
closed under nite intersections.
Example 7. Every metric space is a topological space.
Consider a metric space (X, d). Let T be the collection of all
open subsets of X.
φ and X are open in X, (φ, X) ∈ T .
Now the union of a number of open sets in X is open. i.e; T is closed under
arbitrary union. Similary, T is closed under nite intersection.
10
18. ∴ T is a topology.
Thus every metric space is a topological space.
Remark 2. On R d(x, y) = |x − y| is a metric. Hence (R , d) is a metric
space and so R is a topological space. This topology is said be Usual topology
on R
Example 8. Let X = {a, b} and let T = {φ, X, a}. Cleraly T is a topology
on X called Sierpinski topology. Suppose that d : X×X → R is a metric or
pseudometric on X then d(a, a) = 0 = d(b, b) and d(a, b) = d(b, a) = k(say).
If d is metric then k 0.
Sr(a) = {a} similarly Sr(b) = {b}
∴ T = {φ, X, {a}, {b}} is a Discrete topology
If k = 0 then d will be a pseudometric. Then Sk(x) = {a, b} = X . Then
the only open sphere in X is X ∴ T = {φ, X} is In-discrete topology. From
this we can say that not 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 subcollection B of T
is said to be base for T if every member of T can be expressed as the union
of some members of B.
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) with a, b ∈ R is a base
for the standard topology on R. The collection of all open intervals (a, b) ⊂ R
with rational end points a, b ∈ Q is a base for the standard topology on R
Example 10. If X is topology space with the discrete topology, then the
collection
B = {x/x ∈ X}
is base of the discrete topology .
Denition 11. A space is said to satisfy the second axiom of countability
or is said to be second countable if its topology has a countable base.
Theorem 2. Let X be second countable space .If a non-empty openset G
in X is represented as the union of a class Gi of open sets, then G can be
represented as a countable union of Gi's.
Proof:
Suppose X be a second countable space . And G =
α∈λ
Gα where each Gα is
a open set.
we want to prove that G =
i∈K
Gi where K is a countable subset of λ.
Let x ∈ G thus x ∈ Gα for some α.
{Bn} is a countable open base , ∃ Bn such that x ∈ Bn ⊂ Gα ⊂ G. Clearly,
G is the union of such Bn's and also G is the union of such Gn's.
Denition 12. A collection U of sets is said to be a cover of set A if A is
contained in the union of members of U . A sub-cover of U is subcollection
12
20. V of U which itself is a cover of A. If we are in a topological space then a
cover is said to be open if 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 base B and let U be the
open cover of X.
Suppose B ={B1, B2, . . .}.Now let S = {n ∈ N : Bn ⊂ U} .
For each n ∈ S x Un ∈ U such that Bn ⊂ Un. Now let C = {Bn : n ∈ S}
and V = {Un : n ∈ S}, clearly V is a countable sub-collection of U and
covers X if C does.
Now we have to prove that C is a cover of X. For this let x ∈ X then x ∈ U
for some U ∈ U ∃ k ∈ N such that x ∈ Bk and Bk ⊂ U clearly, k ∈ S and
so Bk ∈ C .So C and consequently V is a cover of X.
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. If B is a cover of X and B is closed under nite intersections
then B is a base for a topology T on X. Moreover, T consists precisely of
those subsets of X which can be expressed as unions of subfamilies of B.
Proof:
suppose S ⊂ X . Let B be the collection of all nite intersections of
elements of X. by taking the intersections of zero sets in S . i.e, X ∈ B .
So B is closed under nite intersections and that S ⊂ B .
13
21. Denition 13. A collection S of subsets X is said to be Sub-bases for a
topology T on X if the collection of all nite intersections of members of S
is a base for T .
For example, for the usual topology on R, the collection of all open in-
tervals of the form (a, ∞) or (−∞, b) for a, b ∈ R is 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. Let X be a set , T a topology on X and S a collection of
subsets of X. Then S is a sub base for T i S generates T .
Proof:
Let B be the collection of nite intersections of members of S. Suppose that
S is a sub base for T . We want to show that T is the smallest topology on
X ⊇ S. Since S ⊂ B and B ⊂ T then T ⊇ S. Suppose U is some other
topology on X such that S ⊂ U . Now we have to show that T ⊂ U .
Since U is closed under nite intersections and S ⊂ U, B ⊂ U . i.e, U
is closed under arbitrary unions and each member of T can be written as
union of some members of B. i.e,T ⊂ U .
Conversely prove that T is the smallest topology containing S and B is
a base for T . Clearly,B ⊂ T since T is closed under nite intersections
and S ⊂ T . There is a topology U on X suchthat B is a base for U and
B ⊂ T .This means U ⊂ T and consequently U = T .
Since T is the smallest topology containing S and B is a base for T .
14
22. 2.3 Subspaces
Denition 14. Let X be a topological space with topology T . If Y is a
subset of X the collection,
U = {Y ∩ U : U ∈ T }
is a topology on Y . The topology U is also denoted by T /Y . It is called
the relative or the subspace topology on Y induced by T . The space
(Y, T /Y ) is called a subspace of the space (X, T ).
Its open sets consists of all intersections of open sets of X with Y . it is
trival to verify that U is a topology on Y ;
T /Y contains empty set φ and full set Y , (φ = Y ∩ φ Y = Y ∩
X φ, Y ∈ T /Y )
And also T /Y is closed under nite intersection and also closed under the
arbitrary union.
Denition 15. A property of topological spaces is said to be hereditary if
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. Let B be a base for topology T on a set X and let Y ⊂ X. Let
B/Y = {B ∩ Y : B ∈ B}. Then B/Y is a base for the topology T /Y on
Y .
Proof:
15
23. Let y ∈ Y and G be an open set in Y containing y. Then G = H ∩ Y for
some open set H in X. Clearly y ∈ H and so there exists B ∈ B such that
y ∈ B and B ⊂ H.Then y ∈ B ∩ Y ,B ∩ Y ⊂ G and B ∩ Y ∈ B/Y . This
proves that B/Y is a base for T /Y .
Corollory 2. Second countability is a hereditary property.
Proof:
Let (X, T ) be a space with a countable base B and Y ⊂ X. Then B/Y as
B/Y = {B ∩ Y : B ∈ B}
is countable and is a base for T /Y .
Note 4. A topological space X is metrizable provided that the topology of X
is generated by a metric.
Theorem 5. Metrisability is a hereditary property.
Proof:
Assume (X, T ) is metrizable. Let d be a metric on X which induces
the topology T . Let Y ⊂ X and e be the restriction of d to Y × Y . We
claim that e induces the topology T /Y on Y . If y ∈ Y and r 0 then
Be(Y, r) = Bd(y, r) ∩ Y and Be(y, r) ∈ T /Y . Let U be metric topology on
Y induced by e. Let B be the collections of open balls in Y .
Let u ∈ U since B is a base for U i.e; the elements of U can be written
as the union of some elements of B. But the elements of B are the elements
of T /Y a topology and hence u will be an element of T /Y .
∴ U ⊂ T /Y
16
24. Conversely, let G ∈ T /Y then G = H ∩ Y when H ∈ T , the topology
induced by d. Hence for each yinG (hence in H ) ∃Bd(y, ry) ⊂ H. Then
Be(y, ry) = Bd(y, ry) ∩ Y ⊂ H ∩ Y = G. Hence G = ∪y∈GBe(y, ry)
∴ G ∈ U then
T /Y ⊂ U
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 subsetA of a tpological space X is said to be closed if the
set X − A is open.
Consider a simple example:
Example 11. The subset [a, b] on R is closed because its complement
R − [a, b] = (−∞, a) ∪ (b, +∞)
is open. Similarly, [a, +∞) is closed because its complement (−∞, 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
clopen set.
17
25. It is immediate that a set is open i its complement is closed.
Theorem 6. Let X be a topological space, then
1. X, φ are closed sets.
2. Any intersection of closed sets X is closed.
3. Any nite union of closed set in X is closed.
Hint :
De-morgan's law
A − (B ∪ C) = (A − B) ∩ (A − C)
A − (B ∩ C) = (A − B) ∪ (A − C)
proof:
Let {Aα : α ∈ β} is the collection of closed sets, then
X −
α∈β
Aα =
α∈β
(X − Aα)
(De-morgan's law )
(X − Aα) is open.
α∈β
is an arbitrary union of open set.
∴ Aα is closed .
Similarly, if Ai where i = 1, · · · , n is closed, then
X −
n
i=1
Ai =
n
i=1
(X − Ai)
which is a nite intersection of open sets ,
∴ its open, and ∪Ai is closed .
φ and X are closed because they are the complements of the open set X and
φ resp.
18
26. Denition 17. The closure of a subset of a topological space is dened as
the intersection of all closed subsets containing it.
In symbols, if A is a subset of a space (X, T ), then its closure is the set
∩{C ⊂ X : C is closed in X,C ⊃ A}. It is denoted by ¯A.
Some important properties of closure are given below:
Let X be a topological space and A, B subsets of X then,
1. ¯A is a closed subset of X. If C is closed in X and A ⊂ C then ¯A ⊂ C.
2. ¯φ =φ
3. A is closed in X if ¯A=A
4.
¯¯A = ¯A
5. A ∪ B= ¯A ∪ ¯B
Example 12. LetX={a, b, c, d, e} and T = {X, φ, {a}, {c, d}, {a, c, d}, {b, c, d, e}}.Show
that {¯b}={b, e}, {a, c} = X and {b, d} = {b, c, d, e}.
Proof:
{ Hint :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.}
The closed sets are φ, X, {b, c, d, e}, {b, e} and {a}. So the smallest closed
set containing {b} is {b, e};
i.e, {¯b} = {b, e}
Similarly,{a, c} = X and
{b, d} = {b, c, d, e}
19
27. Denition 18. Let X be a topological space and A ⊂ X. Then A is said to
be dense in X if ¯A = X.
Lemma 2. Let X be space and let A ⊂ X then A is dense. I for every
non-empty open set B, A ∩ B = φ.
Proof:
suppose A is dense in X and B is a open set in X . If A ∩ B = φ. then
A ⊂ X − B and ¯A ⊂ X − B( from the denitions of dense). since X − B is
closed(because B is open). But then X − B ⊂ X contradicting that A = X.
So A ∩ B = φ.
Conversely suppose that A ∩ B = φ. Clearly that the only closed set con-
taining A is X and ¯A = X
In the real line the usual topology the set Q of all rational numbers, as
well as its complement R − Q are 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, x0 ∈ X and N ⊂ X is said to be a
neighbourhood of x0 , if ∃ an open set V such that x0 ∈ V and V ⊂ N.
And the x0 is said to be the interior point of N.
Lemma 3. A subset A of a topological space is open, i it is a neighbourhood
of each of its points.
Proof:
Let X be a topological space and G ⊂ X. First suppose G is open. Then
G is a nbd of each of its points. Conversely suppose that G is a nbd of each
points. Then for each x ∈ G ∃ an open set Vx such thatx ∈ Vx and Vx ⊂ G.
Then G =
x∈G
Vx. So each Vx is open and in G.
20
28. Denition 20. Let (X, T ) be a space and A ⊂ X .Then the interior of A
is dened to be the set of all interior points of A ,
i.e the set {x ∈ A : A is a nbd of x} .It is denoted by A0
,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 on R
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. Let X be a space and A ⊂ X. Then A0
is the union of all open
sets contained in A.
Proof:
Let U be the collection of all open set contained in A. Let V =
G∈U
G.
If x ∈ V then x ∈ G, for some G ∈ U i.e; A is a nbd of x and so x ∈ A0
.
Conversely, let x ∈ A0
. then there is an open set H such that x ∈ H and
H ⊂ A. But then, H ∈ U ,H ⊂ U , then x ∈ V i.e; V = A0
A topology on a set X induces an operator i : P(X) → P(X) is dened
by i(A) = A0
. It is called interior operator associated with T . The
interior operator determines T uniquely, for it is clear that a set is open i
it coincides with its interior.
Let X be a space and x ∈ X. Let ηx be the set of all neighbourhood of
x ∈ X, with respect to the given topology on X. The collections ηx is called
the neighbourhood system at x.
Some properties of neighbourhood system :
Let X be a space and for x ∈ X, let ηx be the neighbourhood system at x.
Then ,
21
29. 1. If U ∈ ηx, then x ∈ U.
2. For any U, V ∈ ηx, U ∩ V ∈ ηx.
3. If V ∈ ηx and U ⊃ V then U ∈ ηx
4. A set G is open, i G ∈ ηx∀ x ∈ G.
5. If U ∈ ηx then ∃ V ∈ ηx such that V ⊂ U and V ∈ ηy∀ y ∈ V .
Denition 21. Let X be a space and let A ⊂ X , y ∈ X. Then y is said
to be an accumulation point of A if every open set containing y contains
atleast one point of A other than y.
In a discrete space, ther is no point is an accumulation point of any set. In
indiscrete space, a point y is said to be an accumulation point of any set A,
if A contains at least one point other than y. 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. Let X be a space and A ⊂ X. Then the derived set of A
is the set of all accumulation points of A in X and is denoted by A .
A not only depends on A but 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. Let X be a space and A ⊂ X, ¯A = A ∪ A .
Proof:
We claim that A∪A is closed. Let Y ∈ A∪A . Since Y is not an accumulation
point of A then ∃ an open set V y ∈ V such that V contains no point of A
except y. But y /∈ A, so we have A ∩ V = φ. We claim A ∩ V = φ . So let
22
30. z ∈ A ∩ V . Then V is an open set containing z , is an accumulation point
of A. So V ∩ A is non empty, which is a contradiction. So A ∩ V = φ and
hence V ⊂ A ∪ A . So A ∪ A is closed and since it contains A and A , i.e,
¯A ⊂ A ∪ A .
And A ∪ A ⊂ ¯A, we have A ⊂ ¯A. Let y ∈ A , if y /∈ ¯A then y ∈ X − ¯A
which is a closed set , since ¯A is closed set. But y is an accumulation point
of A. So (X − A) ∩ A = φ which is a contradiction, since X − ¯A ⊂ X − A.
So y ∈ ¯A. i.e; A ∪ A ⊂ ¯A. So ¯A = A ∪ A .
Theorem 9. Let X be a space and A ⊂ X, ¯A = {y ∈ X: every ngbd of y
intersects A}.
Proof:
Let B = {y ∈ X : U ∈ ηy ⇒ U ∩ A = φ}, then we have to show that
¯A = B, from the above theorem A ∪ A = B, so let y ∈ A ∪ A if y ∈ A then
certainly every nbd of y intersects A at a point y and so y ∈ B. If y ∈ A
then every nbd of y contains a point of A and so y ∈ B i.e, A ∪ A ⊂ B.
Let y ∈ B rst we can assume the negation that, y /∈ A ∪ A , then y /∈ ¯A
and so X− ¯A is a nbd of y , which doesnot intersect A. This is a contradiction
thet y ∈ B. So B ⊂ A ∪ A . Thus A ∪ A = B.
Hence the result.
Continuity and Related concepts
Denition 23. Let function f : X → Y x0 ∈ X and T , U be topologies
on X, Y respectively. Then f is said to be continuous at x0 , if for every
V ∈ U such that f(x0) ∈ V, ∃U ∈ T such that x0 ∈ U and f(U) ⊂ V
Theorem 10. The following statements are equivalent:
23
31. 1. f is continuous at x0.
2. The inverse image of every neighbourhood of f(x0) in Y is a neighbour-
hood of x0 in X.
3. For every subset A ⊂ X, x0 ∈ ¯A ⇒ f(x0) ∈ f(A).
Proof :
(1) ⇒ (2) Let N be a neighbourhood of f(x0) in Y . Then ∃ an open set V ∈ Y
such that f(x0) ∈ V and V ⊂ N. Since f is continuous at x0, ∃ an open set
U ∈ X such that x0 ∈ U and f(U) ⊂ V . So x0 ∈ U ⊂ f−1
(V ) ⊂ f−1
(N)
thus f−1
(N) is a neighbourhood of x0.
(2) ⇒ (3). Let A ⊂ X and suppose that x0 ∈ ¯A . If f(x0) /∈ f(A) , then
there is a neighbourhood N of f(x0) such that f(A)∩N = φ i.e; f−1
(f(A))∩
f−1
(N) = φ and hence that A ∩ f−1
(N) = φ, since A ⊂ f−1
(f(A)). But the
inverse image of every neighbourhood of f(x0) in Y is a neighbourhood of
x0 in X, f−1(N) is a neighbourhood of x0 and so A ∩ f−1(N) = φ, since
x0 ∈ ¯A which is a contradiction.
(3) ⇒ (4). Let V be an open set containing f(x0). Let A = X−f−1
(V ) =
f−1
(Y − V ). Then f(A) ⊂ Y − V and so f(A) ⊂ Y − V as Y − V is closed.
So f(x0) /∈ f(A), x0 /∈ ¯A from (3). Hence there is a neighbourhood N of x0
such that N ∩ A = φ. Clearly then f(N) ⊂ V ∴ f is a continuous at x0.
Continuity of f at x0 to mean that the image of a neighbourhood of x0
is a neighborhood of f(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 function f : X → Y and T , U be topologies on X, Y
respectively. Then f is said to be continuous( T − U continuous ), if it
24
32. is continuous at each points of X.
Theorem 11. The following statements are equivalent :
1. f is continuous
2. For each closed subset C of Y , f−1
(C) is closed in X.
3. For each subset A of X, f( ¯A) ⊂ f(A).
4. There is a basis B for the topology of Y such that f−1
(B) is open in
X for each basic open set B in B.
5. There is a sub-basis S for the topology of Y such that f−1
(S) is open
in X for each sub-basic open set S in S.
Proof:
We use the open set formulation to describe continuity : f is continuous i
for each open set V in Y , f−1
(V ) is open in X. The equivalence of (1) and
(2) follows from the duality between open sets and closed sets,
(2) =⇒ (3) : Suppose that (2) holds and let A ⊂ X then f(A) is a
closed subset of Y , so its inverse image f−1
(f(A)) is closed in X. Since
A ⊂ f−1
(f(A))
and the latter set is closed then
¯A ⊂ f−1
(f(A))
so
f( ¯A) ⊂ f(A)
and (3) holds.
25
33. (3) =⇒ (2) Assume (3) and let C be a closed subset of Y . then
f(f−1(C)) ⊂ ff−1(C) ⊂ ¯C = C
so
f−1(C) ⊂ f−1
(C)
and f−1
(C) must be a closed set. Thus (3) =⇒ (2).
We have completed the proof (1), (2) and (3) are equivalent. Since a
basis B and sub-basis S for Y consist 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 basis B generated S
by taking nite intersection. For any basic open set B ∈ B ,
B =
n
i=1
Si
for some nite collections of elements S1, . . . , Sn of S. Then
f−1
(B) = f−1
n
i=1
Si =
n
i=1
f−1
(Si)
. Since each set f−1
(Si) is open in X and the intersection of any nite
collection of open sets is open, then f−1
(B) is open in X. Thus (5) =⇒ (4).
(4) =⇒ (1) : Assuming (4), let O be an open set in Y . By the denitions
basis,
O =
α∈I
Bα
for some sub-collection {Bα : α ∈ I} of the basis B. Then
f−1
(O) = f−1
α∈I
Bα =
α∈I
f−1
Bα
.
Since each set f−1
(Bα) is open X and the union of any family of open
sets is open, then f−1
(O) is open in X and f is continuous.
26
34. Theorem 12. If f : X → Y and g : Y → Z are continuous function , then
the composite function g ◦ f : X → Z is continuous .
Continuous functions are also called maps or mappings. Let X1, X2, . . . , Xn
be the sets and let X = X1 × X2 × . . . × Xn. For each i = 1, 2, . . . , n we
dene πi : X → Xi by the rule πi(x1, x2, . . . , xn) = xi. Then πi is called the
projection on Xi or the ith
projection. It is a surjective function expect
in the case of some Xj, where X is empty. If x ∈ X then πi(x) is called the
ith
coordinate of x.
Denition 25. Let X, Y be spaces. A function f : X → Y is said to be
open ( Closed ) if whenever A ⊂ X is an open ( closed ), then f(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 between X and Y
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 function f : X → Y between two topological space X and
Y is a homemorphism or topologically equivalent if it is a one-to-one,
27
35. onto map and both f and f−1
are continuous. When such a homemorphism
exist, then we say that,X is said to be homeomorphic to Y .
Homemorphic space are indistinguishable as topological space. For ex-
ample, if f : X → Y is a homeomorphsim, then G is a open in X i f(G)
is open in Y , and a sequence (xn) converges to X i the sequence (f(xn))
converges to f(x) in Y .
A one-to-one,onto map f always has an inverse f−1
, but f−1
need not be
continuous if f is.
Denition 27. A property P of a topological space is a topological prop-
erty or topological invariant provided that if space X has property P, then
so does every space Y which is topologically equivalent to X.
The next theorem give example of topological properties .
Theorem 13. Separability is a topologically property
Proof :
Let X be a separable space with countable dense subset A and Y a space
homeomorphic to X. Let f : X → Y be a homeomorphism. Clearly, for a
countable dense subset of Y is f(A). To see that f(A) is dense in Y , let O be
a non empty open set in Y . Then f−1
(O) is a non empty open set in X. Since
A is dense in X, f−1
(O) contains some member a of A. Then O contains the
member f(a) of f(A), so every non-empty open set in Y contains at least
one member of f(A). Thus f(A) = Y and Y is 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 conjugate n × n matrices are indistinguish-
able as linear transformations of an n−dimensional vector space without a
xed basis.
Theorem 14. Let X Y be spaces and f : X → Y a function. Then the
following statements are equivalent:
1. f is a homeomorphsim,
2. f is a continuous bijection and f is open,
3. f bijection and f−1
is a continuous,
4. there exists a function g : Y → X such that f, g are continuous ,
g ◦ f = idX and f ◦ g = idY .
29
37. Chapter 3
Topological Properties
3.1 Compact Spaces
Denition 28. Let X be a space and A ⊂ X then A is said to be compact
subset of X if every cover of A by open subsets of X has a nite sub-cover.
A space X is said to be compact if X is 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. σ− 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 a Rn
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 if X
itself is a compact set.
Denition 30. A space is said to be separable if it contains a countable
dense subset .
Theorem 15. Every second countable space is Lindelo.
Proof: Let T be a second countable .Then by denition(6) , its topology
has a countable basis. Let B be this countable basis .
Let C be an open cover of T .Every set in C is the union of a subset of B.This
union of a subset of B is a countable sub-cover of C .i.e, T is 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 space Xis separable.
For each n ∈ N − {0}, the set of all open balls of radius1/n is an open
cover of X; let x(n, m) ∈ X be points such that {B(x(n, m), 1/n)}m∈Nis a
countable sub-cover. Then the set {x(n, m)|m, n ∈ N, n = 0} is a countable
dense subset: if B(x, r) is an open ball, let n be large enough that 1/n r/2;
if no x(n, m) is in B(x, r), then x is not in any B(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:
Let X be a second countable space with countable basis B and A be the
countable set choosing a member from each basic open set . If φ ∈ B then
choose one member from each non-empty member of B.
It follows from the denition of basis that A is dense in X.
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. Let a and b be real numbers satisfying a b. Then the closed
bounded interval [a, b] is a compact subset of R.
Proof:
Let U be a collection of open sets in R with 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.
Let S be the set of all T ∈ [a, b] with the property that [a, T ] is covered
by some nite collection of open sets belonging to U, and let s = supS. Now
s ∈ W for some open set W belonging to U. Moreover W is open in R, and
∴ ∃ some δ 0 such that (s−δ, s+δ) ⊂ W. Moreover s−δ is not an upper
bound for the set S, hence ∃ some T ∈ S satisfying T sδ. It follows from
the denition of S that [a, T ] is covered by some nite collection V1, V2, . . . Vr
of open sets belonging to U.
Let t ∈ [a, b] satisfy T ≤ t s + δ . Then [a, t] ⊂ [a, T ] ∪ (s − δ, s + δ) ⊂
V1 ∪ V2 ∪ . . . ∪ Vr ∪ W, and thus t ∈ S . In particular s ∈ S, and moreover
s = b, since otherwise s would not be an upper bound of the set S . Thus
b ∈ S , and ∴ [a, b] is covered by a nite collection of open sets belonging to
U, as required.
Theorem 19. Let A be a closed subset of some compact topological space X.
Then A is compact.
Proof:
Let U be any collection of open sets in X covering A. On adjoining the
open set X − A to U , we obtain an open cover of X. This open cover of X
32
40. possesses a nite sub-cover, since X is compact. Moreover A is covered by
the open sets in the collection U that belong to this nite sub-cover. From
this A is compact.
Theorem 20. Any closed interval is a compact subset of R in the standard
topology.
Proof:
We prove the proposition for [0, 1] and the general statement follows easily.
Suppose C is an open cover of [0, 1] in R. Let S ⊂ [0, 1] be the set with x ∈ S
if and only if there is a nite sub-collection of C covering [0, x] . Of course,
0 ∈ S , so S is non-empty. Let y be the least upper bound of S. If y ∈ S,
there is U ∈ C with y ∈ U . There is y ∈ U with y y and (y , y) ⊂ U, and
there is a nite sub-collection C of C covering [0, y ] . But then C ∪ {U} is
a nite cover of [0, y] , a contradiction. If y ∈ S and y 1 , then there is a
nite subset C of C and a U ∈ C with y ∈ U. Since U is an open set, there
is y ∈ U with y y, contradicting the supposition that y is the least upper
bound of S. So we must have y = 1 , meaning that C has 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: Let X be a compact space and f : X → R is continuous.Now we
have to show that f is bounded for each x ∈ X , let jx be the open interval
(f(x) − 1, f(x) + 1) and let Vx = f−1
(jx). By continuity of f, Vx is an open
set containing x. Now the collection {Vx : x ∈ X} is an open cover of X and
by compactness , admits a nite sub-cover {Vx1, Vx2, . . . , Vxn} (say).
Let M = max{(f(x1), f(x2), . . . , f(xn))} + 1 and
let m = min{(f(x1), f(x2), . . . , f(xn))} − 1. For any x ∈ X there is some i
33
41. such that x ∈ Vxi. Then f(xi) − 1 f(x) f(xi) + 1 and m f(x) M
shows that f is bounded.
Let L, λ be the supremum and inmum of f over X. i.e, f(x) = L then we
dene a new function g : X → R by g(x) = 1/(L − f(x)) ∀ x ∈ X.
g is unbounded and any R 0 ∃ x such that f(x) L − 1/R and hence
g(x) R. This shows that f attains L. Similarly f attains the inmum λ.
3.2 Connected Spaces
Denition 31. A topological space X is disconnected or separated if it is
the union of two disjoint, non-empty open sets. Such a pair A, B of subsets
of X is called s separation of X. A space is connected provided that it is
not disconnected. in other words, X is connected if there do not exists open
subsets A and B of X such that
A = φ B = φ A ∩ B = φ A ∪ B = X
A subspace Y of X is connected provided that it is a connected space
when assigned the sub-space topology. The terms connected set and con-
nected subset are some times used to mean connected space and connected
subspace, respectively.
Example 13. The real line R with the usual topology is connected.
Let us suppose the negation that R is disconnected. Then
R = A ∪ B
for some disjoint, non-empty open sets A and B of R. Since
A = R − B, B = R − A
34
42. then A and B are closed as well as open. Consider two points a and b with
a ∈ A, b ∈ B. Without loss of generality we may assume a b. Let
A = A ∩ [a, b]
Now A is a closed and bounded subset of R and consequently contains its
least upper bound c. Not that c = b since A and B have no point in common.
Thus c b. Since A ∩ (c, b] = φ then
(c, b] ⊂ B
and hence c ∈ ¯B. But B is closed, so c ∈ B. Thus c ∈ A, B, which is a
contradicting the assumption that A and B are disjoint. This shows that R
is connected.
Theorem 22. The continuous image of a connected space is connected.
Proof: Let f : X → Y be an onto continuous function and X be a
connected space. If A is a subset of Y such that A is both open and closed,
then f−1
(A) is both open and closed. Since f is continuous. Since f−1
(A)
is both open, closed and X is connected, f−1
(A) must either be all of X or
the empty set.
∴ A is either the entire space or the empty set, i.e, Y is connected.
Theorem 23. Let X be a space and A, B ⊂ X. Then
1. A ∪ B = X and ¯A ∩ ¯B = φ.
2. A ∪ B = X, A ∩ B = φ.
3. B = X − A and A is clopen in X.
4. B = X − A and ∂A is empty.
35
43. 5. A ∪ B = X, A ∩ B = φ and A, B are both open in X.
Proof: (1) =⇒ (2). Clearly ¯A ∩ ¯B = φ . Since A ⊂ ¯A and B ⊂ ¯B.
Also ¯A ⊂ X − ¯B ⊂ X − B = A and so ¯A = A clear that A is 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 that X = A ∪ B where A ∩ B = φ and A, B are open.
Then A = X − B and B = X − A where A, B are closed. i.e, ¯A = A, ¯B = B
this shows that ¯A ¯B = φ.
Theorem 24. A subset of R is connected i it is an interval.
Proof:
Suppose that J ⊂ R is not an interval. Then there are x, y ∈ J and z ∈ J
with x z y. Then dene A = (∞, z) ∩ J and B = (z, ∞) ∩ J. Clearly,
A, B are disjoint, non-empty, relatively open, and A ∪ B = J. So J is not
connected.
Conversely, suppose that J is an interval. We will show that J is connected.
Let f : J → {0, 1} be continuous, and suppose that f is not constant. Then
there are x1, y1 ∈ J such that f(x1) = 0 and f(y1) = 1. Assume that
x1 y1. Let a be the midpoint of [x1, y1]. If f(a) = 0, then set x2 = 0 and
y2 = y1 , and otherwise, x2 = x1 and y2 = a. So x1 ≤ x2 ≤ y2, |x2 − y2| ≤
2−1
|x1 −y1|, f(xi) = f(yi). Iterating this procedure we nd sequences xn and
yn with the following properties:
x1 ≤ x2 ≤ . . . xn yn ≤ . . . ≤ y1, |xn −yn| ≤ 2
−1
|xn−1 −yn−1| ≤ 2n−1
|x1 −y1|
and f(xn) = 0, f(yn) = 1. Since R is complete, {xn} converges to some z,
and since |xn − yn| → 0, yn → z. Clearly, z ∈ J.
Hence 0 = limnf(xn) = f(z) = limnf(yn) = 1. This is a contradiction. So
f is constant, and this implies that J is connected.
36
44. Denition 32. Non-empty subset A and B of a space X are separated
sets, if ¯A ∩ B and A ∩ ¯B are both empty
Theorem 25. The following statements are equivalent for a topological space
X :
1. X is disconnected.
2. X is the union of two disjoint, non-empty closed sets.
3. X is a union of two separated sets.
4. There is a continuous function from X onto a discrete two-point space
{a, b}.
5. X has a proper subset A which is both open and closed .
6. X has a proper subset A such that
¯A ∩ (X − A) = φ.
Proof:
It will be shown that (1) implies each of the other statements and that each
statement implies(1). Assume rst that X is disconnected and let A, B be
disjoint, non-empty open sets whose union is X.
(1) =⇒ (2): B = X − A and A = X − B are disjoint, non-empty closed
sets whose union is X.
(1) =⇒ (3): Since A and B are closed as well as open, then
¯A ∩ B = A ∩ B = φ, A ∩ ¯B = A ∩ B = φ.
So X is the union of the separated sets A and B.
37
45. (1) =⇒ (4): The function f : X → {a, b} dened by
f(x) =
a if x ∈ A
b if x ∈ B
is continuous and maps X onto the discrete space {a, b}.
(1) =⇒ (5): A = φ, and
A = X − B = X
since B = φ. Thus A is the required set. ( B will do equally well.)
(1) =⇒ (6): Either A or B can be used as required set.
(2) =⇒ (1): If X = C ∪ D where C and D are disjoint, non-empty
closed sets, then
D = X − C, C = X − D
are open as well as closed.
(3) =⇒ (1): If X is the union of separated sets C and D, then C and D
are both non-empty, by denition. Since X = C ∪ D and ¯C ∩ D = φ, then
¯C ⊂ C, so C is closed. The same argument shows that D is also closed, and
it follows as before that C and D must be open as well.
(4) =⇒ (1): If f : X → {a, b} is continuous, then f−1
(a) and f−1
(b)
are disjoint open subsets of X whose union is X. Since f is required to have
both a and b as images, both f−1
(a) and f−1
(b) are non-empty.
(5) =⇒ (1): Suppose X has a proper subset A which is both open and
closed. Then B = X − A is a non-empty open set disjoint from A for which
X = A ∪ B.
(6) =⇒ (1): Suppose X has a proper subset A for which
¯A ∩ (X − A) = φ.
38
46. Then ¯A and (X − A) are disjoint, non-empty closed sets whose union is
X, and it follows as before that ¯A and (X − A) are also open.
Corollory 3. The following statements are equivalent for a topological space
X:
1. X is connected.
2. X is not the union of two disjoint, non-empty closed sets.
3. X is not the union of two separated sets.
4. There is no continuous function from X onto a discrete two-point space
{a, b}.
5. The only subsets of X which are both open and closed are X and φ.
6. X has no proper subset A for which
¯A ∩ (X − A) = φ.
Theorem 26. A subspace Y of a space is disconnected i ∃ open sets U and
V in X such that
U ∩ Y = φ, V ∩ Y = φ, U ∩ V ∩ Y = φ, Y ⊂ U ∪ V.
Proof:
Suppose rst Y is disconnected. Then there are disjoint, non-empty , open
sets A and B in subspace topology for Y such that Y = A ∪ B by the
denition of relatively open sets, there must be open sets U and V in X such
that
A = U ∩ Y, B = V ∩ Y
39
47. It is a simple matter to check that U and V have the required properties.
For the reverse implication, suppose that U and V are open subsets of X
such that
U ∩ Y = φ, V ∩ Y = φ, U ∩ V ∩ Y = φ, Y ⊂ U ∪ V
. Then
A = U ∩ Y, B = V ∩ Y
are non empty, disjoint relatively open sets whose union is Y , so Y is discon-
nected
Theorem 27. If Y is a connected subspace of space X, then ¯Y is connected.
Proof:
Suppose Y is connected, the connectedness of ¯Y will be shown by proving
that there is no continuous function on ¯Y onto a discrete two point space.
Consider a continuous function f : ¯Y → {a, b} from ¯Y into such a discrete
space. We must shown that f is not surjective. Then restriction fY cannot
be surjective. This means that f maps Y to only one point of {a, b}, say a:
f(Y ) = {a}.
Since f is continuous, previous
f(¯Y ) ⊂ f(Y ) = {¯a} = {a}
so f is not surjective. Thus, Theorem 23, ¯Y is conneted.
Corollory 4. Let Y be a connected subspace f a space X and Z a subspace
of X such that Y ⊂ Z ⊂ ¯Y . Then Z is 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. Let X be a space and {Aα : α ∈ I} a family of connected
subsets of X for which
α∈I
Aα is not empty. Then
α∈I
Aα is connected.
proof:
To show that Y =
α∈I
Aα is connected. Suppose that U and V are open sets
inX for which
U ∩ Y = φ, U ∩ V ∩ Y = φ, Y ⊂ U ∪ V
. It will be shown that V ∩ Y = φ, thus proving that Y is connected. Now
u ∩ Y = φ, so U contains some point in Aα, for some α ∈ I. Since Aα is
connected, then Aα ⊂ U. If b ∈
α∈I
Aα, then b must be in Aα so b ∈ U. Thus
U contains a point b in each Aα, α ∈ I. Since Aα is connected, then Aα ⊂ U
for each α ∈ I.Thus
Y =
α∈I
Aα ⊂ U
so V ∩ Y = φ.
Corollory 5. Let X be a space, {Aα : α ∈ I} a family of connected subsets
of X, and B a connected subset of X such that, for each α ∈ I, Aα ∩ B = φ.
Then B ∪ (
α∈I
Aα) is connected.
proof:
By above theorem, each set b ∪ Aα, α ∈ I, is connected and the intersection
∩α∈I(B ∪ Aα) = φ
41
49. since it contains B . Thus by above theorem
B ∪
α∈I
Aα =
α∈I
(B ∪ Aα)
is connected.
Denition 33. A component of a topological space X is connected subset
C of X which is not a proper subset any connected subset of X.
The following properties of the component of a space are given :
1. Each point a ∈ X belongs to exactly one component. The component
ca containing a is the union of all the connected subsets of X which
contain a and thus may be thought of as the largest connected subset
of X which contains a.
2. For points a, b ∈ X the components ca and cb are either identical or
disjoint.
3. Every connected subset of X is contained in a component.
4. Each component of X is a closed set.
5. X is connected of X and A, B form a separation of X, then C is a
subset of A or a subset of B.
Example 14. In a discrete space each components contains only one point.
Example 15. For a set R of 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 points a and b belongs to the
same component of X, then they must belong to the same member of any
42
50. separation of X. This example shows that the converse is false : It is possible
for points a, b to be always in the same member of any separation A, B of X
yet to belong to dierent components.
Consider the subspace X of R2
consisting of a sequence of line segment
converging to a line segment whose mid point c ha s been deleted. Then
[a, c) is the component of X which contains a and (c, b] is the component
which contains b, so a and b belong to dierent components. However, for
any separation of X into disjoint non-empty open sets A and B whose union
is X both a, b ∈ A or both a, b ∈ B.
Denition 34. A space X is Totally disconnected provided that each com-
ponents of X consists 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 of R
Lemma 4. A non empty subset A of R is an interval i for each pair c, d of
members of A, every real number b/w c and d is in A.
Proof:
Consider, A ⊂ R which contains every real number b/w any two of its mem-
bers. We can split this into several cases whether or not A has a least upper
bound or greatest lower bound and whether or not these bounds, if they
exist, belong to A.
Suppose that A has neither a least upper bound nor a greatest lower
bound. Then for x ∈ R there are members c and d of A for which c x and
d x. Then x ∈ A, so it follows that A = R and A is the interval (−∞, ∞).
Suppose A has greatest lower bound a which does not belong to A and A
has no least upper bound. Then A contains no real number such that x ≤ a.
If y a then y is not an upper bound for A so there is a element d ∈ A
43
51. with d y. Similarly, y is not the greatest lower bound for A, so there is an
element c ∈ A with c y. Then c y d
∴ y ∈ A.Thus A = (a, ∞) and is a interval
The remaining cases are similar.
Theorem 29. The connected subset of R are precisely the intervals.
proof:
Since we know that every interval is connected, it remains only to be proved
that a subset B of R which is not an interval must be disconnected. Let
B ⊂ R that is not an interval. Then by lemma, there are members C, d ∈ B
and y ∈ R with c y d for which y /∈ B then the open sets
U = (−∞, y), V = (y, ∞)
have the following properties
1. c ∈ U ∩ B, so U ∩ B = φ, d ∈ V ∩ B, so V ∩ B = φ
2. U ∩ V = φ, so U ∩ V ∩ B = φ
3. B ⊂ U ∪ V
By theorem 24, B is disconnected. Hence every connected subset of R must
be an interval.
3.3 Some Applications of Connectedness
In the present section
Theorem 30. The Intermediate Value Theorem: Let f : [a, b] → R be
a continuous function , and K ∈ R between f(a) and f(b). Then ∃c ∈ [a, b]
such that f(c) = K.
44
52. Proof:
The interval [a, b] is connected. Since f is continuous, then f([a, b]) is a
connected subset of R and by theorem 29, it must be a interval. Thus any
number K b/w f(a) and f(b) must be in the image f([a, b]). This means
that K = f(c) for some real number c ∈ [a, b].
Corollory 6. Let f : [a, b] → R be a continuous function for which one
of f(a) and f(b) is positive and the other is negative. Then the equation
f(x) = 0 has a root between a and b.
Theorem 31. Let f : [a, b] → [a, b] be a continuous function, then there is a
member c ∈ [a, b] such that f(c) = c
If f(a) = a or f(b) = b, then f has the required property, so we assume
that f(a) = a and f(b) = b. Thus a f(a) and f(b) b, since f(a) and
f(b) must be in [a, b]. Dene g : [a, b] → R by the rule
g(x) = x − f(x), ∀ x ∈ [a, b]
Then g is continuous and
g(a) = a − f(a) 0, g(b) = b − f(b) 0
From Intermediate value theorem the existence of c ∈ [a, b] for which g(c) = 0
then f(c) = c.
Denition 35. Fixed point of a function f : X → Y is a point x for
which f(x) = x. A topological space X has a xed-point property if every
continuous function from X into 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:
Let X be a space which has the xed-point property, Y a space homeomorphic
to X and h : X → Y, a homeomorphism. Let f : Y → Y be continuous
function. Since the composite function h−1
fh : X → X is a conitiuous
function on X it has at least one xed point x0
h−1
fh(x0) = x0
then
f(h(x0)) = hh−1
fh(x0) = h(x0).
So the point is a xed point for f thus Y has 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, x ∈ R,
has no xed point. Since each open interval is homeomorphic to R
Example 17. The n− sphere Sn
, n ≥ 1, does not have the xed-point prop-
erty since the function
g(x) = −x, x ∈ Sn
,
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.
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