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1 DISTANCE EDUCATION SELF LEARNING MATERIAL MADHYA PRADESH BHOJ (OPEN) UNIVERSITY BHOPAL (M.P.) PROGRAMME : M. Sc MATHEMATICS (PREVIOUS) YEAR : FIRST PAPER : III TITLE OF PAPER : TOPOLOGY
Transcript of Paper IV Complex Analysis - bhojvirtualuniversity.com · 1.1 Countable Sets:- Countable sets are of...
1
DISTANCE EDUCATION
SELF LEARNING MATERIAL
MADHYA PRADESH BHOJ (OPEN) UNIVERSITY BHOPAL (M.P.)
PROGRAMME : M. Sc MATHEMATICS (PREVIOUS)
YEAR : FIRST
PAPER : III
TITLE OF PAPER : TOPOLOGY
Course Name :- M.Sc. Mathematics (Previous)
PAPER – III
DISTANCE EDUCATION SELF LEARNING MATERIAL
3
MADHYA PRADESH BHOJ (OPEN) UNIVERSITY BHOPAL (M.P.)
FIRST EDITION -
UNIVERSITY - M.P. Bhoj (Open) University, Bhopal
PROGRAMME - M. Sc Mathematics (Previous)
TITLE OF PAPER - Topologly
BLOCK NO - 1
UNIT WRITER - Dr. Smita Nair
Assistant Professor Mathematics
Sri Sathya Sai College for Women ,Bhopal
EDITOR - Dr. Anupam Jain
Professor and Head of Mathematics
Holkar Autonomus Science College ,Indore
COORDINATION - Dr. (Mrs.) Abha Swarup,
Director (Printing & Translation)
COMMITTEE - Maj. Pradeep Khare (Rtd)
Consultant, M.P. Bhoj (open) University,
Bhopal
M.P. Bhoj (Open) University ALL RIGHT RESERVED No part of this publication may be reproduced in any form, by mimeograph or any other means, without permission in writing from M.P. Bhoj (Open) University. The views expressed in this SLM are that of the author (s) & not that of the MPBOU. The cost of preparation and printing of Self-Learning Materials is met out of DEC grant. Further information on the MPBOU courses may be obtained from the University’s office at Raja Bhoj Marg, Kolar Road, Bhopal (M.P.) 462016 Publisher : Registrar, M.P. Bhoj (Open) University, Bhopal (M.P.) Phone : 0755-2492093 Website: www.bhojvirtualuniversity.com.
Course Name :- M.Sc. Mathematics (Previous)
DISTANCE EDUCATION SELF LEARNING MATERIAL
BLOCK: 1
Unit – 1 Countable & Uncountable sets & Topological Spaces
Unit – 2 Continuity in Topological Spaces Unit – 3 Separation & Compactness
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Paper III Topology Block -1
Topological spaces
Introduction:
The concept of topological spaces. Came into existence with study of real line
and Euclidean spaces and study of continuous functions on these spaces.
The block is divided into three units the first unit gives preliminary concepts
required to understand the concepts of topological spaces, the second unit
deals with concept of topological spaces and continuous functions. The third
unit gives classifications of topological spaces on the basis of separation
axioms & Compactness.
Unit – I
Countable and uncountable sets. Infinite sets and the Axiom of Choice.
Cardinal numbers and its arithmetic. Schroeder-Bernstein theorem. Cantor‟s
theorem and the continum hypothesis. Zorn‟s lemma. Well-ordering theorem.
Definition and examples of topological spaces. Closed sets. Closure. Dense
subsets. Neighbourhoods. Interior, exterior and boundary. Accumulation
points and derived sets. Bases and Sub-bases. Subspaces and relative
topology.
Unit – II
Alternate methods of defining a topology in terms of Kuratowski Closure
Operator and Neighbourhood Systems.
Continuous functions and homeomorphism.
First and Second Countable spaces. Lindelof‟s theorems. Separable spaces.
Second Countability and Separability.
Unit – III
Separation axioms T0, T1, T2, T3 1/2 , T4 their Characterizations and basic
properties. Urysohn‟s lemma. Tietze extension theorem.
Compactness, Continuous functions and compact sets. Basic properties of
compactness. Compactness and finite intersection property. Sequentially and
countably compact sets. Local compactness and one point compactification.
Stone-vech compactification. Compactness and sequential compactness in
metric spaces.
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UNIT – I
Countable & Uncountable sets & Topological spaces
Introduction:-
Topology like geometry, deals with certain objects (sets), classifies
them according to some equivalence relation and then studies those
properties of the objects which are invariant under this classification.
The unit deals with the preliminary ideas involved in a topological
space and provides an insight for further application of these ideas to
further units.
Objectives:-
After studying the unit, you would be
able to understand and identity countable sets.
able to correlate sets with different relations defined on it.
able to define various types of topologies
topological definition of various concepts such as closure, limit
point, derived set etc.
Structure:-
1.1 Countable sets
1.2 Infinite sets
1.3 Uncountable sets and theorems based on countable sets
1.4 Cardinal numbers
1.5 Continum Hypothesis
1.6 Equipotent sets
1.6.1 Schroeder Bernstein theorem
1.6.2 Cantors theorem
1.7 Axiom of choice
1.7.1 Zorn‟s Lemma
1.7.2 Well ordering theorem
1.8 Topology
1.8.1 Types of Topology
1.8.2 Open & closed sets
1.8.3 Closure
1.8.4 Interior
1.8.5 Neighbourhood
1.8.6 Limit point
1.8.7 Derived set
1.8.8 Exterior
1.9 Theorems based on derived set and closure
1.91 Base for topology
1.92 Sub base for topology
1.93 Subspace & relative topology.
1.10 Summary
1.11 Assignment/ check your progress.
1.12 Point for discussion
1.13 References
1.1 Countable Sets:-
Countable sets are of two types
1.10 Finite set
A set „A‟ is said to be finite if there exists a bijective correspondence
(one-one onto mapping) of „A‟ with some section of positive integers
i.e. „A‟ is finite if it is empty or if there is a bijection
f: A {1,2,3,______,n} for some positive integer n
Example 1 :-
A = {1,2,3,5,6,8}
B = {a,b,c,d,e,}
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1.12 Countably Infinite Set:-
A set „A‟ is said to be countably infinite set if there exists a bijective
correspondence between set „A‟ and set of natural numbers Z+ (or N)
Example 1 :-
Set of integers Z, is countably infinite as there is a bijection
f : Z Z+ defined as
( ) 2
i.e Z Z+
Example 2 :- Set of natural numbers Z+ we define the bijection
f : Z+ Z+ as
f (n) = n V n Z+
Z+ Z+
f
Countable sets are sometimes also called enumerable or
denumerable
-2
-1
0
1
2
1
2
3
4
5
-1
0
1
2
1
2
3
4
.
.
.
0
1
2
1
2
3
4
.
.
.
0
1
2
1.2 Infinite Set:-
A set „A‟ is said to be infinite set if it is not finite
Example 1 :-
A = {…………….,-5,-4,-3,-2,-1}
1.3 Uncountable Set:-
A set `A‟ is said to be uncountable or non-denumerable if it is neither
finite nor countably infinite
Example 1 :- Set of real numbers R
Set of irrational numbers
Theorem- 1 Galileo’s Paradox
Any denumerable set can be put into a one-one correspondence with a
proper subset of itself.
Proof:- Let A be a denumerable set & its elements be written as
A = {a1, a2, a3, ……………}
Let B = A - { a1} = { a2, a3, a4,………..}
we define a mapping f : A B as
f (a1) = ai+1
Then, clearly f is one-one mapping between A and its proper subset B
Diagrammatic representation is as follows
A B
f
a1
a2
a3
.
.
.
.
3
4
.
.
.
0
1
a2
a3
a4
.
.
.
.
2
3
4
.
.
.
0
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Theorem -2
A subset of a countable set is countable
Proof :- Let `A‟ be a countable set, B be a subset of `A‟ As A is
countable set by definition of countable set a bijective
correspondence
f : A Z+ then the restriction mapping
Z+ is a injective (one-one) mapping between B & Z+
therefore B is countable.
Theorem -3
The set Z+ X Z+ (or N X N) is countably infinite
Proof : - Z+ X Z+ will be countable if there exists a one-one
correspondence between Z+ X Z+ and Z+ , we define
f : Z+ X Z+ Z+ as
f ((n, m)) = 2n.3m
We shall now prove that f is one-one let us suppose
2n 3m = 2p 3q (1)
If n<p then 3m = 2 p-n 3q
This is a contradiction since L.H.S. of above equality is odd for all m while
R.H.S. of the above equality is even. Similar result holds for p<n, So we
must have p = n (2), So equation (1) now becomes 3m = 3q, if m ≠
q then we have 1 = 3q-m which is again a contradiction as only 30 = 1.
Hence we must have
m = q (3)
from (2) & (3) (p, q) = (n, m) So, by definition of f, f((p,
q)) = f ((n, m)) . This implies f is one-one, f is injective map
between Z+ X Z+ & Z+. Hence Z+ X Z+ is countable.
Theorem- 4
A countable union of countable sets is countable.
Proof let {An} be a countable family of countable sets. We enumerate the
elements of each An, n = 1, 2, 3, ………… in array as follows
A1 = {a11 a12 a13 ……… a1n ………}
A2 = {a21 a22 a23 ……… a2n ………}
A3 = {a31 a32 a33 ……… a3n ………}
.
.
.
.
.
An = {an1 an2 an3 ……… ann ………}
The elements of nUAn are now written as
a11
a21, a12
a31, a22, a13
……… ……… ………
……… ……… ………
an1, a(n-1)2, a(n-2)3, ……… a1n
……… ……… ……… ……… ………
We observe that apq is the qth element of (p+q-1)th row. The elements of
can be arranged in the infinite sequence as { a11, a21, a12, a31, a22, a13,
…………………}
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now we define a map f :
N as
f (apq) = ( )( )
The mapping so defined is a bijection therefore nUAn is countable.
Theorem -5
The set of rational numbers is denumerable.
Proof : Let Q be the set for rational numbers Then Q can be written as.
Q = nUAn where n Z+ ie An =*
+
We shall prove that An is countable, for this it is sufficient to prove that a
bijective mapping between An & Z+ Let f : Z+ An be defined as
( )
{
Then f is one-one & onto. Therefore An is countable. Q being countable
union of countable sets is therefore countable as we knew that “The union
of a countable family of countable sets is countable.”
1.4 Cardinal Numbers : -
Cardinal numbers are of two types.
1.41 Finite cardinal number
Let A be a finite set, then number of elements in
A is called the finite cardinal number of A
Example 1 :-
A = {1, 2, 3, 4, 5, 6, 7, 8,}
Cardinal number (cardinality) of A = 6
Example 2 :-
B = {a, b, c, d, e}
Cardinal number (cardinality) of B = 5
1.42 Transfinite cardinal number
Cardinal number of infinite set is called transfinite cardinal number cardinal
number of a set A is denoted by
The Cardinal number of Z+ (N) is denoted by a or by x0 (alpha nought). The
cardinal number of set of all real numbers R is denoted by c.
1.5 Continum Hypothesis :-
There exists no cardinal number λ such that a < λ <c
1.6 Equipotent Sets:-
Let A, B be any two sets. We say A & B are equivalent (or equipotent or of
same cardinality) if there exists a bijective map and is denoted
by A B.
Theorem 1.61
Schroeder-Bernstein theorem
Let A and B be two sets such that A is equipotent to a subset of B. B is
equipotent to a subset of A. Then A is equipotent to B.
Proof: - Without the loss of generality we assume that A & B are disjoint.
As A is equipotent to a subset of B by definition of equipotency a one-one
mapping f of A into B. Similarly a one-one mapping g of B into A.
We shall now produce a mapping. which is bijective. We may
assume neither f nor g is onto for if f is onto we may define to be f and if g
is onto we may take to be g-1 since f, g are both one-one so, f-1 g-1 are
defined on ( ) and g(B)
Let „a‟ be any arbitrary element of A then if ( ) exist in B, we call it first
ancestor of „a‟ if ( ) exits the is called second ancestor of „a‟. If we
continue this process we have following possibilities
1. `a‟ has last ancestor in A. In this case `a‟ has even
number of ancestors.
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2. „a‟ has last ancestor in B. In this case `a‟ has odd number
of ancestors.
3. `a‟ has infinitely many ancestors.
The set A is now partioned into three subsets, A0 containing elements having
odd number of ancestors, Ae containing elements having even number of
ancestors, Ai containing elements having infinitely many ancestors. Similarly,
the set can also be partioned into subsets. We observe that in this
process of finding ancestors f maps Ai onto Bi, Ae onto Bo, g-1 maps A0 onto
Be. We now define as.
( ) { ( )
Then is a bijection between A and B Therefore A is equipotent to B.
Theorem 1.62
No set X is equipotent to its power set P (X)
OR
For any set X, |X| < |P(X)| i.e cardinality of X is less than cardinality of power
set P (X).
Proof :- In order to prove the theorem we shall prove that if is a subset of
P(X) which is equivalent to X then proper subset P(X).
As is equivalent to X by definition of equivalent set a mapping
f : X which is one-one & onto.
We define
* ( )+
By definition of S, S is a subset of X. As f is one-one & onto for every
a unique element such that ( ) . We have following cases.
Case 1. If x є f (x) = A then by definition of S, x є S so, A ≠ S.
Case 2. If x є f (x) = A then by definition of S, x є S so A ≠ S.
In either case A ≠ S i.e. an element S of P (X) which does not belongs to
. Thus is a proper subset of P(X). And as X is equipotent to it cannot
be equipotent to its power set P(X) i.e. |x| < |P (X)|
1.7 Axiom Of Choice :-
Given a collection of disjoint non-empty sets, there exists a set C consisting
of exactly one element of i.e. a set C is such that C is contained in the
union of elements of and for each A є , the set C A contains single
element.
= { {a, b}, {c, d}, {e, f}}
C = {a, d, e}
Then {a, b} C = {a}
{c, d} C = {d}
{e, f+ C = {d}
Also {a, d, e,} {a, b} {c, d} {e, f}
i.e. {a, d, e,} { a, b, c, d, e, f}
1.71 Zorn’s Lemma:-
Every non –void (non-empty) partially ordered set in which every chain has an
upper bound has a maximal element.
1.72 Well Ordered:-
A set with an order relation < is said to be well ordered if every non-empty
set A has a smallest element.
Example: Set of natural number with relation ≤
1.73 Well Ordering Theorem:-
Proof :- Let X be any set. We now form a collection of the set of pairs ( )
where and R is a well-ordering on A. We now define a relation on A.
We now define a relation as follows.
( ) ( ) iff * + for an , with an ordering on
induced by that on . Then by definition of the relation above we infer it
is an partially ordered relation on {( ) }. We observe that every chain
*( )+ is bonded by where Therefore by Zorn‟s
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lemma every chain will have a maximal element. We claim that the maximal
element ) satisfies. , for if the condition does not holds it would
imply such that and * + be would be well ordered by well
ordering on A and the condition so the pair * * + + would be
greater than ( ) contradicting the fact that ( ) is maximal element.
Therefore we have and as A is well ordered it will imply X is well
ordered.
1.8 Topology:-
A topology on set X is a collection (or T) of subsets of X having the following
properties
(a) Ø and X are in
(b) The union of elements of any sub collection of is in
i.e if Gi є then Ui Gi є .
(c) The intersection of elements of any finite subcollection of
is in i.e. if
Gi, i = 1, 2……….n є then ⋂ i є
Thus a topology onX is a class of subsets of X which is closed under arbitrary
unions & finite intersections.
The ordered pair (X, ) consisting of a set X and a topology on X is called a
topological space.
Example 1.
X = {a, b, c, d,}
= {X, ø, {a}, {b}, {a, b}}
Example 2.
X = {1, 2, 3,}
= {Ø, X, {1, 2}, {2, 3}, {2}}
1.81 Types of Topology:-
(a) Indiscrete topology
Let X be a non empty set the collection
= {X, Ø} is also called trivial topology.
(b) Diserete topology
Let X be a non-empty set. Then the collection of all
subsets of X is called discrete topology
Example:
X = {a, b, c}
= { Ø, X, {a},{b},{c},{a, b},{b, c}, {c,a}}
(c) Co-finite topology
Let X be an infinite set. be a family consisting of ø, X
and complements of finite subsets of X. This topology is
called co-finite topology.
(d) Co-countable topology
Let X be a countable set. be a family consisting of ø, X
and complements of countable subsets of X. This topology
is called co-countable topology.
(e) Usual topology:
Let consistsx of Ø and all subsets G of R having the
property that to each Є G there exist Є > 0 such that (x-
, x+ ) G Then is a topology called usual topology.
1.82 Open & Closed Sets:-
Let (X, ) be a topological space. Let G Є then G is said to be an open set.
Example:
X = {a, b, c}
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= { Ø, X, {a}, {b}, {c}, {a, b}}
Open sets – {a}, {a, b}, {b}, X, Ø
Let (X, ) be a topological space. A subset F of X is said to be a
closed set if complement of F is Open set,
Example:
X = {a, b, c}
= { Ø, X, {a}, {b}, {c}, {a, b}}
F1 = {b, c} then F1c = {a} which is open set
F1 closed set
F2 = {a,c} then F2c = {b} Є F2
c is open & F2 is closed sets
Theorem: Arbitrary intersection of closed set is closed.
Proof:- Let (X, ) be a topology space.
Let Fi be a closed subset of X Then, by definition of closed set Fic
is open.
i.e. Fi is closed Fic is open i
Fic Є [By definition of open set]
Fic Є [arbitrary union of open set is open i.e. by definition
of topology]
= [⋂
Fic]c Є [Generalised De-morgan Law
Fic =[⋂
Fi
c]c]
= [⋂
Fic] c is open
= ⋂
Fi is closed
Thus, arbitrary intersection of closed set is closed.
Theorem
Finite union of closed set is closed
Proof:- Let (X, ) be a topological space. Fi be a closed subset of
X.
Then Fi is closed= Fic is open [By definition of closed set]
Fic Є [By definition of open set]
⋂
Fic Є [finite intersection open set is open]
[
Fi]c Є [Generalized de-morgan law]
[
Fi]c is open [By definition of open set]
Fi is closed [By definition of closed set]
Thus finite intersection of closed set is closed.
1.83 Closure:-
Let (X, ) be a topological space and The closure of A is defined as
intersection of all closed sets which contain A and is denoted by
Symbolically = i { Fi X: Fi is closed, F A }
Example :
X = {a, b, c, d}
= { Ø, X, {a}, {b}, {a, b}, {b, c} {a,b, c}}
Closed sets = { Ø, X, {b, c, d}, {a, c, d}, {c, d},
{a, d}, {d} }
A = {b, d}
Closed sets containing A = {(b, c, d), X}
= {b, c, d} X
= {b, c, d}
Evidently:
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(i) is closed, as arbitrary intersection of closed set is
closed.
(ii) .
(iii) is the smallest closed set containing A.
1.84 Interior:-
Let (X, ) be a topological space.
. A point x is called an interior point of A if G with s.t.
i.e x is called interior point of A if some open set G containing x and
contained in A and is denoted by A0 or int (A) Symbolically, A0 = U {G :
such that G A}
Evidently
(i) A0 is an open set, as arbitrary union of open set is
open.
(ii) A0
(iii) A0 is the largest open set of A.
Example:
X = {a, b, c, d}
= { Ø, X, {a}, {b}, {a, b},{b, c},{a, b, c}}
A = {a, b, d}
Then open sets contained in A are {a}, {b}, {a, b}
So, A0 = {a} U {b} U {a, b}
= {a, b}
1.85 Neighbourdhood:-
Let (X, ) be a topological space is called a neighbourhood of if
some open set G such that .
Example:
X = {a, b, c, d}
= { Ø, X, {a}, {b}, {a, b},{b, c},{a, b, c}}
A = {a, c}
Then {a} {a, c}
Thus {a, c} is neighbourdhood of a
1.86 Limit Point:-
Let (X, ) be a topological space and A point is called
a limit point of A if every neighbourhood of x contains a point of A
other than x, symbolically
[Nx – {x}] Nx
Example :
X = {a, b, c,}
J = { Ø, X, {a, b},{a, c}}
A = {b, c}
Limit point of A is c
Limit point is also called clusterpoint or accumulation
point.
1.87 Derived Set:-
Set of all limit points of a set A is called derived set of A and is denoted as D
(A)
1.88 Exterior:-
Let (X, ) be a topological space. A be any subset of X then exterior of A is
defined as interior of A complement. It is denoted by ext (A)
23
zSymbolically
Ext (A) = int (X - A) = (X - A)0
1.89 Boundary:-
Let (X, ) be a topological space. A be a subset of X then boundary of A is set
of all those points of X which belong neither to the interior of A nor to the
interior of X – A. It is denoted by (A) or b(A). Elements of b(A) are called
the boundary points of A. Boundary points are sometimes called frontier
points.
1.9 Therems Based On Derived Sets & Closure Theorem:-
Let X be a topological space and let A be a subset of X, then A is closed iff
D(A) A
Proof: Let X be a topological space and A be a subset of X. In first part of the
proof let A be a closed set. Then, by definition of closed set A‟ is open so, for
each ‟ there exists a nbd Nx of x such that Nx and, as .
This implies Nx , that is every nbd Nx of x has empty intersection
with A, so by definition of limit point x is not a limit point of A. Thus no point
of A‟ is limit point of A. A contains all its limit point, D (A) A
In next part of the proof let D (A) A and we shall prove that A is closed.
let then , since D (A) A, ( ). So by definition of limit point
there exists a nbd Nx of x such that This implies Nx A‟ Thus A‟
contains a nbd of each of its points. So A‟ is open that is A is closed.
Theorem: Let A, B be subsets of a topological space. Then
(i) D ( ) =
(ii) ( ) ( ) (iii) ( ) ( ) ( ) (iv) ( ) ( ) ( )
Proof
(i) As is closed D ( ) since we know that A is
closed ( ) Also is a subset of every set so
D ( ).
Hence D ( ) = .
(ii) Let ( ), then by definition of derived set is limit
point of A. So, by definition of limit point for every nbd Nx
of x we have
[Nx – {x} ] .
As so, [Nx – {x} ] [Nx – {x} ] Nx – {x}
.
Thus every nbd Nx of x contains point of B other than
x.So, x is a limit point of B that is ( )
(iii) As ( ) ( ) [from ii]
( ) ( ) [from ii]
( ) ( ) ( )
(iv) As ( ) ( ) [from ii]
( ) ( ) [from ii]
( ) ( ) ( ) --------- (A)
To prove the other way inclusion we prove that
( ) ( ) ( )
As ( ) ( ) ( ) ( )
x is not limit point of A nor of B.
Then by definition of limit point nbd‟s N1 and N2 of x such that
[N1 – {x} ] and [N2 – {x} ]
Now, let N = N1 N2
Then [N – {x} ] and [N – {x} ]
25
x is not limit point of
( )
So, ( ) ( ) ( ) ………………….(B)
From (A) & (B) we have
( ) ( ) ( )
Theorem A is closed iff
Proof: In first part of the proof let A is closed, then by definition of closure
. But A is closed set containing A. We know that closure of a set is
smallest closed set containing A
conversely, let then as is a closed set so A is closed.
Theorem ( )
Proof: Let A be any subset of a topological space. We have to prove
( )
In first part of the proof we shall prove that ( ) is closed for this. It is
sufficient to show that [ ( )- is open i.e. [ ( )-‟ = ( ) is
open Let ( ). Then and ( ) so, and ( ).
This implies x is not a limit point of A so, by definition of limit point there
exists an open nbd Nx which contains no point of A other that x. Thus Nx A‟
and Nx ( ) So, we have Nx A‟
Thus A‟ ( ) is nbd of each of its points and consequently A‟ ( ) is
open as been know that set is open iff it is neighbourhood of each of its
points which implies A ( ) is closed.
In next part of the proof we shall prove that closed set A ( ) is equal to
closed set . As A ( ) is a closed set containing A and is the smallest
closed set containing A, so by definition of closure A ( ) . Again as
is closed set if contains all its limit point and in particular it contains all limit
point of A. So, we have ( ) , also ( )
Thus it follows thus = A ( )
Theorem
Let A, B be subsets of a topological space then
(i)
(ii)
(iii) =
(iv) ( )
Proof
(i) is closed set as we know that A is
closed
(ii) By definition of closure , also This
implies . But is a closed set containing
A. Since is the smallest closed set containing A
we must have Hence
(iii) ,
,
…………….(A)
Again
Thus is a closed set
containing , But is the smallest closed
set containing ,
So, we must have ….…… (B)
From (A) & (B) we conclude.
(iv) ( ) , ( )
( ) , ( ) [from ii]
27
1.91 Dense Set:-
It A, B are subsets of topological space X then A is said to be dense in B iff
. In particular A is said to be dense in X iff .
Example:- Set of rational numbers is dense in set of real numbers.
1.92 Base For A Topology:-
Let X be topological space . An open base for X is a class of open sets with
the property that every open set is a union of sets in this class.
The property can also be expressed as, if G is any arbitrary open set and
, then exist an element B of such that
Example:-
X = {a, b, c,}
= { Ø, X,{a}, {b}, {a, b}, {b, c}, {c, a}}
= { Ø, {a}, {b}, {c}}
Theorem
Let (X, ) be a topological space. be a base of . Then for every B1, B2
and every B1 B2 there exists such that x B1 B2
Proof:- Let B1, B2 then by definition of open base B1, B2 and as is a
topology B1 B2 i.e. B1 B2 is an open set such that B1 B2
and as is a base by definition of base there exists a basic open set B such
that B1 B2
1.93 Open Subbase For A Topology:-
Let (X, ) be a topological space. A collection of subsets of X is called a
sub base for topology. iff and finite intersection of members of
form a base for .
Example:-
X = {a, b, c, d}
J = { Ø, X, {a},{a, c}, {a, d}, {b, c},{a, c, d}}
B* = {X,{a, c}, {a, d}}
Then, the class of finite intersection of members of B*is B = { (a), (a, c), (a,
d), X}
Then B is a base for the topology
Hence B* is a sub base.
1.94 Sub Space & Relative Topology:-
Let (X, ) be a topological space Y be a subset of X. Then Y is a topological
space with topology defined as
= {Y G : G }
The topology is called relative topology and (Y, ) subspace.
Example:-
Let X = {1, 2, 3, 4, 5}
= {X, Ø, {1}, {3, 4}, {1, 3, 4},{2,3,4,5}}
Let Y = {1, 4, 5}
={X Y,Ø Y, {1} Y, {2,3} Y,{1,3,} Y, {1, 2,
3, 4, 5} Y}
= {Y, Ø , {1}, {4}, {1, 4}, {4, 5}}
(Y, ) is the subspace and is the relative
topology
Theorem
Let (X, ) be a topological space, Y X,
Then the collection = {Y G : G } is a topology on Y
29
Proof:-
In order to prove is a topology on Y, we need to prove
satisfies following properties.
(i) Since , so by definition of y
Ø Y : Ø
Similarly X X Y
So, we have , Y
(ii) Let H1, H2 , then by definition of
open sets G1, G2 such that H1 = G1 Y
& H2 = G2 Y
So, H1 H2 = (G1 Y) (G2 Y)
As G1 , G2 G1 G2 . Hence by
definition of H1 H2 that is is
closed under finite intersection of open
sets.
(iii) Let Hi , then Gi such that
Hi = Gi Y
Now
* +
⋃* +
*⋃ +
As is closed under arbitrary union of open sets
so, Ui Gi . Hence by definition of , Ui Hi
, that is is closed under arbitrary union.
So, from (i) (ii) (iii) we conclude is a topology
on Y.
1.10 Summary:
1. A set is said to be countable if there exists a bijection between
set A and set of natural numbers.
2. Set N, Q are countable sets.
3. Set of real numbers is uncountable.
4. Countable union of countable sets is countable
5. In a topological space arbitrary union of open sets is open
6. Finite intersection of open sets is open.
7. In a topological space arbitrary intersection of closed sets is
closed and finite union of closed sets is closed.
8. Closure of a set A is the smallest closed set containing A.
9. Interior of a set A is the largest open set contained in A.
10. A set is said to be closed iff it contains all its limit point.
11. A set A is closed
12. A set A is open
1.11 Assignment / Check Your Progress:-
(i) Prove that if B is a countable subset of an uncountable
set then A-B is uncountable.
(ii) Show that set of all irrational numbers is uncountable.
(iii) Prove that if , are topologies defined on a non-empty
set X then is also a topology on X.
31
(iv) Prove that if A is an arbitrary subset of a topological
space X then = {x: each neighbourhood of x intersects
A}
(v) Prove that if A is a subset of a topological space X then
a. A0 is the largest open set contained in A
b. A is open A0 = A
(vi) Prove that if A is a subset of a topological space X then
X = A0 ext (A) b (A)
1.12 Points for discussion. ________________________________________
________________________________________
________________________________________
________________________________________
________________________________________
1.13 References:
1. George F. Simmons, Introduction to Topology and Modern
Analysis, MC Grow Hill book company, 1963.
2. K.D. Joshi, Introduction to General Topology. Wiley Eastern Ltd.
1983.
3. K.K.Jha, Advanced General Topology. Nav Bharat Prakashan,
Delhi.
4. James R, Munkres, Topology, Pearson Education Pvt. Ltd. Delhi
2000.
UNIT – II
Continuity in Topological spaces Introduction:
This unit deals with defining topology in terms of neighbourhood and
closure. The unit provides a base for extending the topological
properties from one space to another through continuous mapping and
homeomorphism. The unit also deals with special type of spaces which
are categorized on the basis of local base and base for the topology
Objectives : After studying this unit the students will be able
to define topology in terms of closure operator and Neighourhood
system.
to define first and second countable space
to understand the link between countability and separability.
Structure 2.1 Kurtowski closure axioms
2.11 Topology in terms of kurtowski closure operator
2.12 Topology in terms of Neighbourhood system
2.2 Continuity in topological space
2.21 Theorems based on continuous function
2.3 Homeomorphism
2.4 Local base at a point
2.41 First countable space
2.42 Second countable space
2.43 Separable spaces
2.44 Theorems on countability and separability
2.5 Summary
2.6 Assignment
2.7 Points for discussion
2.8 References
33
2.1 Kuratowski closure axioms:
Let (X, ) be a topological space. A closure operator on X is a function
( ) ( ) satisfy in the following four conditions known as
kuratowsi closure axioms.
(1) ( )
(2) ( )
(3) ( ) ( ) ( )
(4) ( ( )) ( )
where A and B are subsets of X.
2.11 Topology in terms of kuratowski closure operator:
Theorem: Let X be any set and let c be kuratowski closure operator on
X, that is a function
( ) ( )
such that
(1) ( )
(2) ( )
(3) ( ) ( ) ( )
(4) ( ( )) ( )
Then there exists a unique topology on X such that for each ( )
coincides with -closure of A.
Proof: - Let be the family of all subsets F of X such that ( )
that is
* ( ) +
We shall show that family consisting of complements of members of
i.e.
* + is a topology on X.
(A) By (1) ( ) so, by definition of ,
By (2) ( ) ( ) always therefore
So,
(B) Let
( ) and ( ) [by definition of ]
( ) ( )
( ) [By 3]
( )
( )
That is
(C) Let
Then ( ) by definition of we shall prove that
⋂* +
For this we have to show that
(⋂* +
) ⋂* +
from (2) we have
⋂* +
(⋂* +
)
We shall now proe that
(⋂* +
) (⋂* +
)
35
for this we first prove that
( ) ( )
now
( ) ( )
( ) ( ) ( ) , -
( ) ( )
⋂* +
(⋂* +
) ( ) , -
(⋂* +
) ⋂* ( ) +
(⋂* +
) ⋂* + , ( ) -
from I & II we conclude
(⋂* +
) ⋂* +
⋂* +
[⋂* +
]
⋃* +
that is
⋃* +
So, from A, B, C we conclude is a topology on X.
We shall now prove that ( ) closure of .
By (4) c (c (A)) = c (A) by definition of that is ( ) is -closed.
By (2) ( ) Thus ( ) is -closed set containing A. Now, let B be
any closed set containing A. Then so that ( ) Also since
we have ( ) ( )
It follows that ( ) Thus ( ) is contained in any -closed set
containing and so ( ) is the smallest -closed set containing
Hence ( ) closure of
2.12 Topology in terms of Neighbourhood system:
Theorem: Let X be any non-empty set and with each , let there
be associated a family ( ) of subsets of called neighbourhoods
satisfying the following conditions.
(1) ( )
(2) ( )
(3) ( ) ( )
(4) ( ) ( ) ( )
(5) ( ) ( )
and ( )
Then there exists a unique topology in such a way that if ( )
is the collection of neighbourhoods of defined by the topology
then
( ) ( )
Proof:- We begin the proof by defining as follows
A set iff ( )
We shall prove is a topology on .
37
(A) Since contains no points so that the statement
( ) is trivally true. We now prove that
By(1) ( ) Therefore there exists
some set ( ) Since it
follows from (3) that ( ) .
( )
(B) Let then by definition of J we have
( ) , ( )
( ) ( )
( ) , -
That is,
(C) Let then by definition of J
( ) and
( ) ⋃* +
⋃* + ( ) ⋃* +
⋃* +
that is
⋃* +
So, from A, B, C we conclude is a topology on X.
In next part of theorem we shall prove
( ) ( ) i.e. ( ) N is a -nbd of .
Let ( ) then by (5) ( ) s.t and ( ) y
This by definition of implies Thus M is a -open set such that
But this implies N is a -nbd of x i.e. ( )
So ( ) ( ) ( ) ( )
Again let ( ) then by definition of ( ), N is a -nbd of x.
Then there exists a - open set G such that now
( )
But ( ) and ( ) , -
So ( ) ( ) Hence we conclude ( ) ( )
2.2 Continuity in topological space:
Let (X, ) and (Y, ‟) be topological spaces. A mapping is said
to be continuous at iff every ‟-nbd M of ( ) there exists a
-nbd N of such that ( )
Also f is said to be continuous if it is continuous at each point of X.
2.21 Theorems based on continuous function:
(a) Theorem: Let X and Y be topological spaces. A mapping
is continuous iff the inverse image under of every open set
in Y is open in X
Proof:- In first part of the proof let f be a continuous mapping and
let H be an open set in Y we shall prove that ( ) is open in X. We
have following two possibilities.
(i) ( ) then as is open set so the theorem is true.
(ii) ( ) Let ( ) so that ( ) As is
continuous function by definition of continuous function there
exist open set in X such that and , - that is
( ) This shows that ( ) is neighbourhood of
each of its points and so it is open in X.
Conversely, let ( ) is open in X for every open set H in Y,
we shall prove that f is continuous at
39
Let H be any open set in Y such that ( ) , so that
( ) . By hypothesis ( ) is open in X. If ( )
then G is an open set in X containing such that
( ) , ( )-
Hence f is continuous.
(b) Theorem: Let X and Y be topological spaces. A mapping
is continuous iff inverse image under f of every closed set in Y
is closed in X.
Proof the theorem is similar to the proof of theorem (a)
(c) Theorem: Let X and Y be a topological spaces. Then a mapping
is continuous iff inverse image of every member of base for Y
is open in X.
Proof:- In first part of the proof let be continuous
function be a base for Y. Let we have to prove ( ) is open
in X. As B is a basic element so, by definition of open base B is an open
set in Y and as is continuous ( ) is open in X by theorem (a)
above.
Conversely,
Let inverse image of every basic open set in Y is open in X. We shall
prove that f is continuous. For this it is sufficient to prove that inverse
image of every open set in Y is open in X. Let H be any open set in Y.
Then, by definition of open base
* +
( ) , * +-
, ( ) -
By assumption inverse image of every basic open set in Y is open in
X so, ( ) is open in X and as arbitrary union of open set is open
open ( ) is open in X. Therefore f is continuous.
(d) Theorem: A mapping is continuous iff ( ) ( )
for every .
Proof:- Let be continuous function. Let A be a subset of X.
since ( ) is closed in Y and is continuous so , ( )- is closed
in X by the theorem b Therefore
, ( )- , ( )- ( ) , -
Now
( ) ( ) , ( )- , -
, ( )- , ( )- ( ), -
( ) ( )
Conversely, let ( ) ( ) for every Let F be any closed
set in Y so that . Now ( ) is a subset of X so by
hypothesis
, ( )- , ( )-
Therefore
( ) ( )
But
( ) ( )
( ) ( )
so ( ) is closed in X. Hence is continuous by theorem b.
41
2.3 Homeomorphism:
Let (X, ), (Y, ) be two topological space. A mapping is
said to be a homeomorphism if
(i) is one-one
(ii) is onto
(iii) is continuous mapping
(iv) is open mapping i.e is continuous.
A mapping is said to be an open mapping it image of
every open set in X is open in Y.
2.4 Local base at a point:
Let (X, ) be a topological space.
A non-empty collection B (x) of -neighbourhood of x is called a local
base of x iff for every -neighbourhood N of x there is a ( )
such that
Example let * +
* * + * + * + * + * ++
Then a local base at each of the points a, b, c, d, e is.
( ) {* +} ( ) {* +} ( ) {* +}
( ) {* +} ( ) ** ++
2.41 First countable space: A topological space (X, J) is said to
satisfy the first axiom of countability if each point of X possesses a
countable local base. Such a space is said to first countable space.
Example :- The topological space (R, u) is a first countable space.
2.42 Second countable spaces: A topological space(X, ) is
said to be second countable (or to satisfy second axiom of countability)
iff there exists a countable base for J
Example: Topological space (R, u) is a second countable space.
2.43 Separable space: A space X is said to be separable if it
contains a countable dense subset.
2.44 Theorems based on countability and separability
Theorem: Every second countable space is first countable.
Proof:- Let X be a second countable space and As X is second
countable there exists a countable base say for X. let *
+ Then as L is a subclass of and is countable, therefore L is
countable. We shall now prove that L is a local base for x For this let N
be a nbd of x. Then, by definition of neighbourhood there exists an
open set V such that . Since B is a base so by definition of
base there exists such that , but this implies .
This proves L is a local base at X. Thus there exists a countable local
base for each . Hence X is first countable.
Lindelof’s theorem: Let X be a second countable space.
If a non-empty openset in X is represented as the union of a class
* + of open sets, then can be represented as a countable union of
s
Proof:- Let X be a second countable space then by definition there
exists a countable open base for X. Let { } be the countable open
base. Let be a non-empty set. Then , by assumption
⋃
43
This implies there exist, some such that , as is open set, by
definition of open base there exist a basic open set such that
It we do this for each point x in we obtain a subclass of
our countable open base whose union is , and this class is countable.
Further for each basic open set in this subclass we can select a
which contains it. The class of which arises in this way is clearly
countable and its union is
Theorem: A metric space is second countable iff it is separable.
Proof :- In first part of the proof let (X, d) be a metric space with
metric topology defined on it such that (X, ) is second countable.
We shall prove that X is separable since X is second countable by
definition of second countable space there exists a countable base B
for . We choose a point b from each member B of B . Let D be the
set so obtained. The set D obtained is countable. We shall now prove
D is dense in X. Let x be any arbitrary point of X and let G be any open
neighbourhood of x, since B is a base, there exists atleast one
such that
By definition of D, such that Thus contains a point
of D. Hence x is an adherent point of D. Since x is arbitrary we have
. It implies D is a countable dense subset of X. Hence X is
separable.
Conversely.
In this part of the proof let X be a separable metric space. We shall
prove that X is second countable for this we have to prove that there
exists a countable base for X. As X is separable there exists a
countable dense subset A of X. We define * ⁄( ) +
Then B is countable we shall prove B is a base for X.
Let and G be an open set containing x. Then by definition of
open set there exists such that ( ) . Choose .
Since X is separable therefore that is every point of X is
adherent point of A. Hence every open set centered at must contain
a point of A. in particular ( ) contains a point of A, then
( ) ⁄ ( ).
We shall prove ⁄ ( ) ( ). Let
⁄( ) ( )
( )
⁄
If follows ( ) that is
⁄( ) ( )
Thus for every and open set G. containing there exists a
member ( ⁄ ) of containing and contained in G. Hence is
a countable base for X. therefore X is second countable.
2.5 Summary:
1. A mapping is continuous iff inverse image of every
open set in Y is open in X.
2. Second countability first countability but converse need not be
true.
3. Second countability is hereditary.
4. Every metric space is second countable iff it is separable
2.6 Assignment /Check your progress:
1. Prove that if is a mapping then is continuous
inverse image of every sub basic open set in Y is open in X.
2. A metric space is first countable.
3. Prove that the property of a space being second countable is
hereditary.
45
2.7 Points for discussion:
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
2.8 References:
1. George F. Simmons, Introduction to Topology and Modern
Analysis, MC Grow Hill book company, 1963.
2. K.D. Joshi, Introduction to General Topology. Wiley Eastern
Ltd. 1983.
3. K.K.Jha, Advanced General Topology. Nav Bharat Prakashan,
Delhi.
4. James R, Munkres, Topology, Pearson Education Pvt. Ltd.
Delhi, 2000.
Unit- III
Separation & Compactness
Introduction:
Since every concept of topology is defined in terms of open sets, in
order to make non-trivial and interesting statements about a space it is
necessary that the space possess a fairly rich collection of open sets. In
this unit we shall study various conditions which assert the existence of
open sets, these conditions are various separation axioms.
The notion of compactness came for proving theorems such as mean
value theorem and uniform continuity theorem. Earlier, the concept of
limit point for [a,b] was thought to be crucial, but this was a weak
formulation, so a stronger formulation, in terms of open coverings of
the space came into existence which we now call compactness.
Objectives:
After studying this unit you should be able to appreciate.
Importance of hierarchy of various separation axioms.
Importance of concept of compact space.
The interrelated concepts of separation and compact spaces.
Structure: 3.1 Hierarchy of Separation axioms. [T0,T1,T2,T31/2,T4]
3.2 Theorems
3.2.1 Urysohn‟s Lemma
3.2.2 Tietze Extension theorem
3.3 Compactness.
3.3.1 Open cover
3.3.2 Sub-cover
3.3.3 Compact space
3.3.4 Theorem
47
3.4 Compactness in metric space.
3.4.1 Bolzano Weirstrass Property (BWP)
3.4.2 Sequential Compactness
3.4.3 Theorems
3.4.7 Lebesgue Covering Lemma.
3.5 Countably compact space.
3.6 Locally compact space
3.7 Compactification
3.8 Stone –Vech compactification
3.9 Finite intersection property (FIP)
3.9.2 Heine Borel Theorem
3.10 Unit Summary
3.11 Assignment
3.12 Point for discussion
3.13 References
3.1 Hierachy of Separation axioms:
The separation axioms are of various degrees of strengths and
they are called To, T1, T2, T3, T4 axioms in ascending order of
strength, To being the weakest separation axiom.
3.1.1 To Space:
A topological space X is said to satisfy the To axiom, or is
said to be To space if given any two distinct points in X,
there exists an open set which contains one of them but
not the other.
Example: Every discrete space is a To space.
3.1.2 T1 Space:
A topological space X is said to satisfy the T1 axiom or is
said to be T1 space for every two distinct points and
an open set containing but not .
Example : All metric spaces are T1 space.
3.1.3 T2 Space
A topological space X is said to satisfy T2 axiom (or the
Hausdorff property) or is said to be a T2 (or Hausdorff)
space if for every distinct point there exist disjoint
open sets U,V in X such that and
Example; All metric spaces are T2 space.
3.1.4 Regular space:
A topological space X is said to be regular at a point
if for every closed subset C of X not containing
disjoints open sets U,V such that X is said
to be regular space if it is regular at each of its points.
3.1.5 T3 Space:
A topological space is said to satisfy T3 axiom or is said
to be T3 space if it is regular and T1.
3.1.6 Normal Space:
A space X is said to be normal if for every two disjoint
closed subsets C and there exist two disjoint open sets
U and V such that C U and V.
3.1.7 T4 Space:
A topological space is said to satisfy T4 axiom or is said
to be T4 space if it is normal and T1.
Example: All metric space are T4.
Thus we conclude.
3.1.8 Completely regular space (
):
A T1 space X is said to be completely regular if is any
point in X and F is any closed subspace of X which does
49
not contain . Then there exists a function in C (X,R) all
of whose values lie in closed unit interval [0,1] such that
( ) and ( ) .
Theorem:
A topological space (X, ) is a T1 space if every
singleton subset * + of X is closed.
Proof: Let X be a topological space, be any
arbitrary point such that
* + is closed its complement i.e. * + is open
* +
* + is neighbourhood of y
each point y different from has
a neighbourhood which does
not contains
X is a T1 Space
Theorem:
Each singleton subset of a T2 space is closed.
Proof: Let X be a Hausdorff space and let
we have show that * + is closed. Let be any
arbitrary point of X distinct from . Since the space
is Hausdorff, there exists a N of such that
It implies that is not a limit point of * +
and consequently (* +) * + * + * + is
closed set.
Theorem:
In a Hausdorff space, limits of sequences are
unique.
Proof: Let * + be a sequence in Hausdorff
space , X and suppose and as
. We have to show that Suppose on
the contrary . Then by definition of Hausdorff
space open sets U &V in X such that , V
and U V= Again, by definition of limit N1,
N2 N Such that for all for all
. Let be an integer greater than both N1
and N2. Then i.e.
contradicting that U V= . So, and thus
limits of sequence in X arc unique.
3.2.1 Theorem:
(a) Urysohn’s lemma
Let X be a normal space and Let A and B be disjoint closed subspaces
of X. Then there exists a continuous real. Function defined on X, all
of whose value lie in the closed unit interval [0,1]such that ( )
and ( )
Proof: The proof of the theorem will consist of following steps;
1. Construction of a real valued function
2. Showing is continous.
Let A and B be closed subset of X. Then as B is closed it implies Bc is
open set containing A i.e. Bc is of closed set A. By normality of X
and the resutl “X is normal iff each of a closed set F contians the
closure ofsome of F,” so
A has a say U1/2 such that
⁄
⁄ , -
As U1/2 is of so by repeated application of above result
say U1/4 such that
51
⁄
⁄
⁄
⁄
If we continue this process for each dyadic rational number of the form
( )
We have an open set of the form Ut such that
We now define our function by.
( ) {
* +
then if ( ) i.e. ( )
so ( ) Also by definition we conclude that is a real valued
function all of whose value lies in [0,1] again so, by
definition of ( ) Now we shall prove that is continuous for
this we shall prove that inverse image of every open set in [0,1] is
open in X. As we know that is continuous inverse image
of every open set in Y is open in X.” Let us consider all intervals of the
form [0,a) and (a,1] where 0<a<1. Then we observe that ( )
is in some Ut for t<a (, )) * ( ) +
and as arbitrary union of open sets is open (, )) is open.
Similarly ( ) is outside for some
This implies (( -) * ( ) + which being
arbitrary union of open set is open .This proves that is continuous.
Corollary. 3.2.1(b) Let X be a normal space let A & B be disjoint
closed subspace of X if [a,b] is any closed interval on the real line then
there exists a contiuouns real function defined on X all of whose
values lie in [a,b] such that ( ) and ( ) .
The corollary is a direct consequence of Urysohn‟s lemma.
3.2.2 Theorem (Tietze Extension theorem):
Let X be a normal space. F a closed subspace and a continuous real
function defined on F whose values lie in the closed interval [a,b].
Then has a continuous extension defined on all X whose values
also lie in [a,b].
We have following two cases.
Case-I a=b then the theorem is obvious.
Case-II a<b then using corollary 3.2.1 (b).
We prove the theorem for a=-1 & b=1 i.e. for the closed interval [-
1,1], we begin by defining f0 to be defined on F. let A0 and B0 are
disjoint nonempty closed subset in F defined as follows.
{ ( ) ⁄ } { ( )
⁄ }
Then A0 and B0 are disjoint nonempty closed subset in F and as F is
closed in X the y are closed in X. thus A0 and B0 are a pair of disjoint
nonempty closed subsets of X so by corollary3.2.1(b), a continuous
function
[ ⁄ ⁄ ] such that ( )
⁄ ( ) ⁄
We next define then
( ) ( ) ( ) ( ) ( ( )) ( ) ( )
⁄
⁄ ⁄
Again if
{ ( ) ( ⁄ )(
⁄ )} { ( ) ( ⁄ ) (
⁄ )}
then by above argument a continuous function.
[( ⁄ )(
⁄ ) ( ⁄ )(
⁄ )] such that
( ) ( ⁄ )(
⁄ ) ( ) ( ⁄ )(
⁄ )
Next we define ( )
53
Such that ( ) ( ⁄ ) By continuing in this manner we get a
sequence * + defined on F such that ( ) .
/
and a
sequence * + defined on X such that ( )
( ⁄ ) ( ⁄ ) with the property that on F we have
( )
We now define Sn by Sn = Then Sn Can be
regarded as partial sums of infinite series of functions. Then ( )
.
/ .
/
and the fact that
∑ ( ⁄ )(
⁄ )
We conclude Sn converges uniformly on X by comparison test to a
bounded continuous real function such that ( ) Since
( ) .
/
, Sn converges uniformly on F to and therefore
equals on F and is a continuous extension of on X.
3.3 Compactness:
3.3.1 Open Cover:
Let X be a topological space A class { + of open subsets of X is said
to be an open cover of X if each point in X belongs to atleast one
that is if
⋃
Example X={a,b,c,d,e}
Topology on X T=* * + * + * ++
Open Cover C= {{a,b,c} {c,d,e}
As X= {a,b,c,} {c,d,e}
3.3.2 Sub cover:
A Sub class of an open cover which is ifself an open cover is
called a sub cover.
Example: X={a,b,c,}
T=* * + * + * +* +* +* ++
Open Cover C= {{a,b} {a},{b}, {a,c}}
Sub Cover C‟= {{a,b} {a,c}}
3.3.3 Compact Space:
A topological space X is said to be compact if ever open cover of
X has a finite sub cover.
3.3.4 Theorem
Any closed subspace of a compact space is compact.
Proof: Let X be a topological space. Y be a closed subspace
of X. In order to prove Y is compact we shall prove that every
open cover in Y has a finite subcover. So, let { } be an open
cover of Y. Then by definition of open cover we have
⋃
( )
As Y is a subspace so, by definition of relative topology for each
T open set Hi in X such that
( ) ( )
, - .
, - [Distribution Property]
and as Y is closed, Yc is open.
Again X =YU Yc
* +
55
Thus the class of all Hi‟s along with Yc will form an open cover
for X and as X is compact by definition of compact space there
exist a finite subcover . In case this finite sub-cover contains Yc,
we shall remove it since it covers no part of Y. Let the remaining
finite collection of sets be * + such that
⋃
*⋃
+
⋃*
+
⋃ , -
Thus { + is a finite subcover for Y. Hence Y is
compact.
Theorem
Any continuous image of a compact space is compact.
Proof; Let X, Y be two topological space where X is compact . is a
continuous function. We have to prove ( ) is compact. Clearly ( ) is a
subspace of Y. Let * + be an open cover for X. Then by definition of open
cover.
( ) ⋃
( )
By definition of relative topology for each an open set in X.
Such that ( ) ( )
From (1) we have
( ) ⋃( ( ) )
( ) ( ) [⋃
]
( ) ⋃
(⋃
)
OR
⋃ (
) ( )
is continuous ( ) is open in X as we know that “ is
continuous inverse image of every open set in Y is open in X” from (3) we
conclude that * ( )+ is a cover for X and as X is compact a finite
subcover for it i.e.
⋃ ( )
( ) ⋃ ( )
( ) ( )⋂[⋃ ( )
]
⋃0 ( )⋂ 1
⋃ , -
{ + is a finite subcover.
Hence ( ) is compact,
57
3.4 Compactness in metric space:
This section deals with equivalence of compactness, Bolzano Weirstrass
property and sequential compactness.
3.4.1 Bolzano Weirstrass Property:
A metric space X is said to have Bolzano Weirstrass Property or
simply BWP if every infinite subset of X has a limit point.
3.4.2 Sequentially Compact:
A metric space X is said to be sequentially compact if every
sequence in X has a, convergent subsequence.
3.4.3 Theorem:
A metric space is sequentially compact it has BWP
Proof: In first part of the proof let X be a sequentially compact
metric space and let A be an infinite subset of X. Since A is
infinite set a sequence * + of distinct points of A can be
formed. And as X is sequentially compact. This sequence has a
convergent subsequence * +. Then limit of this subsequence
will be the limit point of A as we know that ”if a convergent
sequence in a metric space has infinitely many distinct points,
then its limit is a limit point of the set of points of the
sequence.” Hence X has BWP Conversely,
Let X has BWP we shall prove X is sequentially compact.
Let * + be an arbitrary sequence in X. If * + has a point
which is infinitely repeated then it has constant subsequence
and being constant subsequence it is convergent. If no point of
* + is infinitely repeated then the set A of points of this
sequence is infinite and as X has BWP A has a limit point say
and we shall extract from * + a subsequence which converges
to . Thus X is sequentially compact.
Theorem:
Every compact metric space has Bolzano Weirstrass property.
Proof: Let X be a compact metric space we shall prove X has
BWP, on the contrary let X be without BWP. Then by definition
of BWP an infinite subset A of X which has no limit point.
Let be any arbitrary element. Then as is not a limit
point so, by definition of limit point an open sphere
( ) 2
Then C={ ( ) } will clearly form an open cover for X and
as X is compact this open cover will have a finite subcover i.e.
⋃
( )
{⋃
( )}
⋃*
( )+ ⋃
i.e. * +
A is finite which is a contradiction to the fact that A
is infinite. Hence X has BWP.
3.4.4 Lebsque Number:
Let X be a metric space { } an open cover for X. A real
number a>0 is called a lebesque number if each subset
of X whose diameter is less than „a‟ is contained in at
least one
59
3.4.5 :
Let X be a metric space. If is given, a subset A of X
is called an if A is finite and ( )
3.4.6 Totally bounded:
A metric space X is said to be totally bounded of for every
there exist
Theorem: Lebesque’s Covering Lemma
In a sequentially compact metric space, every open
cover a Lebesque number.
Proof: Let X be a sequentially compact metric space. { } be
an open cover for X. We being the proof by defining “big
set”. A subset of X is said to be big if it is not contained in
any . We have following two cases.
Case -1 X has no big set
In this case any positive real number will become
a Lebesque number „a‟ since any set having diameter less
than a, not being a big set will be contained in some
Case-2 X has big set
We define „a‟ to be the greatest lower bound of diameter
of big set in X. Then clearly . Again if
then any real number a will serve as lebesque number
and if a‟ is real then we can take a to be a‟. we shall ow
prove that a‟>0 for this let a‟=0. Since every big set must
have atleast two points, we infer from a‟=0 that for each
positive integer n there exists a big set Bn such that
( )
…….(1), we now choose a point from
each Bn and form a sequence * + As X is sequentially
compact by definition there exists a convergent
subsequence of * + which converges to some point in
X. As * + is an open cover for X by definition of open
cover some open set such that
. And as
is open in metric space X, by definition of open set there
exists open sphere ( ) such that ( ) . Let
⁄( ) be the concentric sphere with radius ⁄ . Since
the subsequence of * + converges to , by definition of
limit there are infinitely many positive integers n, for
which ⁄( ) .
Let n0 be one of these positive integers. which is so large
that 1/n0 <r/2 ……(2) Also form (1) ( )
⁄
⁄
i.e. diameter of is less than . So, by definition of
open sphere
⁄( ) ( ) i.e.
, a big set is
contained in which is a contradiction to the definition
of . Hence we must have a‟>0. which will be the
required lebesgue number.
3.5 Countably Compact Space:
A topological space X is said to be countably compact iff every
countable open cover of X has a finite subcover.
3.5.1 Theorem:
A countably compact topological space has BWP.
Proof: Let X be a countably compact space and let X does not have
BWP. Then by definition of BWP and our assumption there exists
an infinite set having no limit point. Let A be countably infinite
subset of S, then A will have no limit point. Also for each is
not a limit point of A so, there exists an open set and
* + Then the collection * + is a countable
open cover of X. This cover has no finite subcover . For if we
remove a single it will not be a cover of X since then will not
61
be a covered Hence X is not countably compact which is a
contradiction. Therefore X must have BWP.
3.6 Locally compact:
A topological space X is said to be locally compact iff every point in X
has atleast one neighbourhood whose closure is compact.
3.6.1 Theorem
Every compact topological space X is locally compact.
Proof: Let X be compact space since X is both open and closed. It
is neighbourhood of each of its points which implies
Thus for each point is a neighbourbhood whose closure
is compact. Hence X is locally compact.
Theorem:
Every closed subspace of a locally compact space is locally
compact.
Proof: Let Y is a closed subspace of X, and let be arbitrary
point . Then and X being locally compact a
neighbourhood N of y such that is compact. Now Y,N both
are neighbourhoods of is also a neighbourhood of y
as we know that “N1, N2 are neighbourhood of is
also a neighbourhood of ”.
Also , -. Thus is
closed subset of a compact space it is compact we know
that a closed subset of compact space is compact. Thus we
have shown that every point in Y has a neighbourhood in Y
whose closure is compact in Y. Hence Y is locally compact.
3.7 Compactification:
If Y is a compact Hausdorff space and X is a proper subspace of Y
whose closure equals Y, then Y is said to be a compactification of X. If
Y-X equals a single point then Y is called one point compactification of
X.
3.8 Stone- ech compactification.
Let X be an arbitrary completely regular space. Then there exists a
compact Hausdorff space ( ) with the following properties;
1. X is dense subspace of ( )
2. Every bounded continuous real function defined on X has a
unique extension to a bounded continuous real function
defined on ( )
Then ( ) is called stone- ech compactification of given
completely regular space.
3.9 Finite intersection Property (FIP)
A collection of subsets of X is said to have finite intersection
property if for every finite sub collection
* + of the intersection is non
empty.
3.9.1 Theorem:
A topological space is compact every class of closed sets with
finite intersection property has non empty intersection.
Proof: In first part of the proof let X be a compact space.
* + be a class of closed subsets of X with FIP. We
have to prove ⋂ * + on the contrary
⋂* +
[⋂* +
]
⋃* +
, -
63
* + is a class of open sets whose unions is X it is an open
cover for X and as X is compact a finite subcover for it
⋃
[⋃
]
⋂
, -
* is a finite subcollection of , with empty intersection
which is a contradiction to the fact that has FIP therefore
⋂* +
Conversely,
Let every class of closed subsets of X with FIP has non-empty
intersection. We have to prove X is compact. Let { } be an open
cover for X then by definition of open cover
⋃
[⋃
]
⋃ , -
Thus { + is a class of closed sets having empty intersection. Then
by assumption * + does not have FIP i.e. T a finite subcollection
*
+ such that
⋂
[⋂
]
, -
⋃
Thus *
+ is a finite subcover X is compact.
3.9.2 Theorem: Heine-Borel Theorem
Every closed and bounded subspace of the real line is
compact.
Proof: A closed and bounded subspace of real line is a
closed subspace of some closed interval [a,b] so for the
proof of the theorem it is sufficient to prove that [a,b] is
compact as we now that “ a closed subspace of a compact
space is compact.”
If a=b then the theorem is trivial so, let a<b. Now the
class of all intervals of the form [a,d) and (c,b], where c and
d are any real numbers, such that a<c<b and a<d<b is an
open subbase for [a,b] the class of all [a,c]‟s and all [d,b]‟s
is a closed subbase.
Consider S={[a,ci], [dj,b]} be a class of these subbasic closed
sets with FIP, we shall prove intersection of all sets in S is
non empty which will imply that [a,b] is compact as we now
that “X is compact every class of closed sets with FIP has
non-empty intersection”.
We assume that S is non-empty. If S contain only intervals of
the type. [a,ci]‟s or [dj,b]‟s there intersection clearly contains
a or b respectively. Thus we assume S contains intervals of
both type. We define d=sup {dj} and shall prove that d ci
for every i. Suppose on the contrary let for some
65
Then by definition of there exists a djo such that .
Then [ ] [ ] which contradicts the fact that S
has F.I.P. Hence for every i which implies S has non-
empty intersection. Therefore [a,b] is compact.
3.9.3 Theorem
In a Hausdorff space, any point and disjoint compact
subspace can be separated by open sets in the sense that
they have disjoint neihbourhoods
Proof: Let X be a Hausdorff space. be arbitrary, be
a compact subspace. Such that we shall prove there
exist disjoint openset , H such that
Let Since X is a Hausdorff space i.e.
and are distinct points, so by definition of Hausdorff space
there exists disjoint open sets and such that and
if we vary over we obtain a class
of whose union contains i.e. the class of form an
open cover for and as is compact by definition of
compact set, this open cover has a finite subcover i.e.
⋃
If
+ are the neighbourhoods of which
corresponds to , we take
⋂
Then are disjoint open sets s.t and
Theorem
Every compact subspace of a Hausdorff space is closed.
Proof: Let X be a Hausdorff space be a compact subspace. We
have to prove is closed .for this it is sufficient to show
that c is open . If c is empty then it is open, so we
assume c is non-empty. Let then . Thus is
point in Hausdorff space ad is a disjoint compact
subspace so by theorem 3.93 has neighbourhood such
that . If we vary over then we obtain a
class of open sets * + whose union is i.e.
⋃
And as arbitrary union of open set is open is open
which implies is closed.
3.10 Unit Summary/Things to remember;
1. The axioms To, T1, T2, T3, and T4 form a hierarchy of
progressively stronger conditions.
2. Weaest axiom is To axiom & strongest axiom is T4 .
3. All metric spaces are T4.
4. Any closed subspace of a compat space is compact.
5. Continuous image of a compact space is compact.
6. The following statement are equivalent.
ii. X is compact
iii. X is sequentially compact
iv. X has BWP
v. X is totally bounded
7. Every closed and bounded subspace of real line is compact.
8. Every compact subspace of real line is compact.
3.11 Assignment/ Activity check your progress;
1. Prove that one-to one continuous mapping of a compact space
onto a Hausdorff space is a homeomorphism.
2. Prove that every compact Hausdorff space is normal.
67
3. Let X be a T1 space then show that X is normal each
neighbourhood of a closed set F contains the closure of some
neighbouhood of F.
4. Prove that every sequentially compact metric space is totally
bounded.
5. Prove that every sequentially compact metric space is
compact.
3.12 Points for Discussion/Clarification
____________________________________________________________________________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________________________________________________________________________________________________
3.13 References.
1. George F. Simmons, Introduction to Topology and Modern
Analysis, MC Grow Hill book company, 1963.
2. K.D. Joshi, Introduction to General Topology. Wiley Eastern Ltd.
1983.
3. K.K.Jha, Advanced General Topology. Nav Bharat Prakashan,
Delhi.
4. James R, Munkres, Topology, Pearson Education Pvt. Ltd. Delhi
2000.
DISTANCE EDUCATION
SELF LEARNING MATERIAL
MADHYA PRADESH BHOJ (OPEN) UNIVERSITY
BHOPAL (M.P.) Course Name :- M.Sc. Mathematics (Previous)
PAPER – III
DISTANCE EDUCATION SELF LEARNING MATERIAL
PROGRAMME : M. Sc MATHEMATICS (PREVIOUS)
YEAR : FIRST
PAPER : III
TITLE OF PAPER : TOPOLOGY
69
MADHYA PRADESH BHOJ (OPEN) UNIVERSITY BHOPAL (M.P.)
FIRST EDITION -
UNIVERSITY - M.P. Bhoj (Open) University, Bhopal
PROGRAMME - M. Sc Mathematics (Previous)
TITLE OF PAPER - Topology
BLOCK NO - II
UNIT WRITER - Dr. SmitaNair
Assistant Professor Mathematics
Sri Sathya Sai College for Women ,Bhopal
EDITOR - Dr. Anupam Jain
Professor and Head of Mathematics
Holkar Autonomus Science College ,Indore
COORDINATION - Dr. (Mrs.) Abha Swarup,
Director (Printing & Translation)
COMMITTEE - Maj. Pradeep Khare (Rtd)
Consultant, M.P. Bhoj (open) University,
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Course Name :- M.Sc. Mathematics (Previous)
DISTANCE EDUCATION SELF LEARNING MATERIAL
BLOCK: II Unit -4 Connectedness, Product Spaces
Unit -5 Paracompactness, Net & Filters, Fundamental Groups & Covering Spaces
Paper – III Topology Block : II
Connectedness, Net & Filters &
Homotopy
Introduction:
The reader gets an inside of basic concepts of topological spaces
in Block – I the Block – II takes a step further to some of the
properties of topological spaces such as connectedness, product
spaces. The Block also gives description of a new type of mapping
a long with concept of net & filters.
Unit- IV
71
Connected spaces. Connectedness on the real line. Components.
Locally connected spaces.
Tychonoff product topology in terms of standard sub-base and its
characterizations. Projection maps. Separation axioms and product
spaces. Connectedness and product spaces. Compactness and
product spaces (Tychonoff‟s theorem). Countability and product
spaces.
Embedding and metrization. Embedding lemma and Tychonoff
embedding. The Urysohn metrization theorem.
Unit – V
Nets and filters. Topology and convergence of nets. Hausdorffness
and nets. Compactness and nets. Filters and their convergence.
Canonical way of converting nets to filters and vice-versa. Ultra-
filters and Compactness.
Metrization theorems and Paracompactness-Local finiteness. The
nagata Smirnov metrization theorem. Paracompactness. The
Smirnov metrization theorem.
The fundamental group and covering spaces – Homotopy of paths.
The fundamental group Covering spaces. The fundamental group
of the circle and the fundamental theorem of algebra.
Unit-IV
Connectedness, Product Spaces Introduction:
The unit is divided into three sections. The first section deals with an
important topological property named connectedness and gives some
necessary conditions for a space to be connected.
The second section is extension of topological space to Cartesian
product of topological space. The section deals with generalization of
basic topological properties such as connectedness, Compactness,
Separation, Countability to Cartesian product of topological spaces.
The third section deals with the concept of evaluation map and
embedding.
Objectives: After completion of this unit the students will be able to
Understand an important concept of connectedness.
Extend and generalize topological properties to the Cartesian
product of topological spaces.
Define embedding and its linkage with projection mapping.
Structure:
4.1 Connected Space.
4.2 Theorems on connectedness.
4.3 Connectedness on real line.
4.4 Definitions
4.42 Locally connected Space.
4.43 Theorems an locally connected space and components.
4.5 Tychonoff Product topology
4.53 Separation axioms and product topology
73
4.54 Connectedness and product topology
4.55 Tychonoff Theorem
4.56 Countability and product space
4.6 Definitions
4.61 Evaluation function
4.64 Embedding
4.7 Prepositions
4.71 Embedding Lemma
4.73 Urysohn‟s Metization theorem
4.8 Summary
4.9 Assignment
4.10 Point for discussion
4.11 References.
4.1 Definitions:
4.12 Separation: Let X be a topological space. A separation of X
is a pair of non-empty disjoint open subsets of X whose union is X
i.e.
and
Example: * +
{ * + * +}
* + * +
* + * + * +
Also * + * +
* + * + is a separation of
4.12 Connected Space: Let X be a topological space. Then X is
said to be connected if there does not exists a separation for X.
Example: , - with usual topology is a connected space.
4.13 Disconnected Space: Let X be a topological space. Then X
is said to be disconnected if there exists a separation for X.
Example: * +
{ * + * +}
* + * +
such that
is disconnected space.
4.2 Theorems s on Connectedness:
Theorem(A): If the sets C and D form a separation of X, and if Y
is a connected subspace of X, then Y lies entirely within either C or D.
Proof: Let X be a topological space and C and D be separation of X,
then by definition of separation
________________(1)
Again and are open in X so, by definition of relative topology
and are open in Y.
Then
( ) ( ) ( )
( )
Also, ( ) ( ) ( )
( )
Thus we have two open disjoint sets and whose union is
Y. If both of these are non-empty sets then they form a separation for
Y but this is a contradiction to the fact that Y is connected. Hence
75
either which implies or which implies
Theorem (B): The union of a collection of connected subspaces of
X that have a point in common is connected.
Proof: Let * + be a collection of connected subspaces of
X such that ⋂ . We have to prove
is connected. On, the contrary let is disconnected , then
by definition of disconnected space there exists a separation for
that is there exist two non-empty disjoint open sets in X
such that and .
As ⋂ this implies
So,
or
Suppose then as and arc connected sets so
by theorem a. This implies H is an empty set as .
But this is a contradiction to our assumption that H is . Hence
is connected.
Theorem (c): The image of a connected space under a continuous
map is connected.
Proof: Let X and Y be two topological spaces. be
continuous mapping. We have to prove ( ) is connected. On the
contrary let ( ) is disconnected then by definition of disconnected set
there exists a separation for ( ) that is there exists non-empty
disjoint set in ( ) such that ( ) and
__________(1)
As is continuous mapping so ( ) and ( ) arc open sets in X
since we Know that “A mapping is continuous inverse
image of every open set in Y is open in X”.
Also ( ) ( ) and ( ) ( ) Form (1)
Thus ( ) ( ) arc non-empty disjoint open set whose union is X.
This implies X is disconnected but this a contradiction since it is given
that X is connected. Hence ( ) is connected.
4.3 Connectedness on real line:
Theorem: A subspace of real line R is connected it is an interval.
Proof: Let A is be a subspace of real line. In first part of the proof let
A is connected. Suppose, if possible, A is not an interval then real
numbers x,y,z such that with but we define
( ) and ( )
Clearly and arc non-empty disjoint open sets in A as ( )
and ( ) Also,
, ( )- , ( )-
,( ) ( )-
, * +-
And ⋂ , ( )- , ( )-
,( ) ( )-
Thus and form a separation for A which implies A is disconnected
but this is a contradiction to the fact that A is connected. Therefore A
is on interval.
Conversely, Suppose A is an interval let it be possible A is
disconnected. Then there exists nonempty disjoint open sets say
such that
Let and . Since so . Without loss of generality
let , Then , - . Since A is an interval and a point
in , - is either in or in we define.
77
* +
Obviously , so But P being closed in A by definition of
y, it follows Since P it follows and
this shows . Again, by definition of y wherever
which means that every neighbouhood of y contains atleast one point
of Q other than y. Hence y is limit point of Q and as Q is closed .
But implies which is a contradiction to the
fact that . Hence A is connected.
Corollary The real line R is connected.
Proof: Since ( ) is an interval so, by above theorem R is
connected.
4.4 Definition:
4.41 Component: Let (X, ) be a topological space. A maximal
connected subspace which is not properly contained in any larger
connected subspace is called a component of the space and is denoted
by .
4.42 Locally connected: Let (X, ) be a topological space. Then
X is said to be locally connected if is any point in it and any
neighourhood of , then contains a connected neighbourhood of .
4.43 Theorems of components and locally connected space:
Theorem: If X is an arbitrary topological space ,then we have the
following;
1. Each point in X is contained in exactly one component of X.
2. Each connected subspace of X is contained in a component
of X.
3. A connected subspace of X which is both open and closed is
a component of X.
4. Each component of X is closed.
Proof:
1. Let be a point X. * + be a class of all connected subspace
of X which contain . This class is non-empty since * } itself
is connected. Also is connected subspace of X
which contains x As we know that if * + is a class of
connected sets such that ⋂ then is
connected. Clearly is maximal and therefore X is
component of X because any connected subspace of X which
contains is one of the and is thus contained in . We
shall now prove that, is the only component of X which
contains x. for if is another, by definition of * + it is
clearly among the and is therefore contained in and
since is maximal as a connected subspace of X, we must
have *= .
2. Let A be a connected subspace then let by part 1 x is
contained in exactly one component say C of X and as A is
connected & it will entirely lie in C.
3. Let A be connected subspace of X which is both open and
closed. By (2) A is contained in some component . If A is a
proper subset of then ( ) ( ) is a
disconnection of . this contradicts the fact that , being a
component is connected and we conclude A= .
4. If a component is not closed then its closure is a
connected subspace of X which properly contains and this
contradicts the maximality of as a connected subspace of
X.
Theorem: A space X is locally connected iff for every open set U of
X, each component of U is open in X.
Proof: Suppose X is locally connected U be an open set in X. Let C be
a component of U if is a point of , we can choose a neighbourhood
79
V of such that . Since V is connected, it must lie entirely in the
component C of U. Therefore C is open in X.
Conversely, suppose that components of open sets in X are open.
Given a point and X and a neighbourhood U of let C be the
component of U containing Now C is connected since it is open in X
by assumption, X is locally connected at .
4.5 Tychonoff Product topology:
Let *( ) + be an indexed family of topological spaces and let
be the cartesian product
is called product topology.
4.51 Projection Map:
Let * + be an indexed family of sets and let
. For
projection is the function denoted by ( ) ( )
that is ( ) denotes the co-ordinate in ( )
4.52 Standard sub base for product topology:
Let be the product topology on the set
where * + is
indexed collection of topological spaces. Then the family of all subsets
of the form
( ) for is a subbase of known as standard
sub base for the topology.
One thing to be kept in mind is product topology is the smallest
topology on X which makes each projection continuous. That is
product topology is the weakest topology determined by family of
projections +
4.53 Separation axioms and Product Spaces:
Theorem: The product space
is iff each co-
ordinate space is .
Proof: In first part of the proof let X be a space then we observe
that the projection is a one to one continuous, open
mapping .Thus each co-ordinate space is homeomorphic to a
subspace of X. And as being is a hereditary property every
subspace of X is also so coordinate space homeomorphic to this
subspace is also .
Conversely,
Let each co-ordinate space be and let * + be any element
of X. then for every . Since is * } is closed in for
every Since projection map is continuous it follows
,* +- is
closed in X for every . Hence their intersection⋂( ,* +-) is also
closed in X But evidentl⋂ ,* +- * + for if
⋂( ,* +-)
,* +- .
Therefore it follows that * + is closed in X. Thus we have proved that
every singleton subset of X is closed .Hence X is a .
Theorem: The product space
is Hausdorff iff each co-
ordinate space is Hausdorff.
Proof: Let X be a Hausdorff space. We shall prove that the co-
ordinate space is Hausdorff for arbitrary . Let and be any
two distinct points of . We choose and in X such that and
differ in the coordinate such that and . Since X is
Hausdorff by definition of Hausdorff space there exist open sets and
in such that and and ⋂ . There exist basic open
81
sets
and
such that and and
. It follows that and are open sets in such that
and and ⋂ . Hence is
Hausdorff.
Conversely, let each co-ordinate space be Hausdorff. Let
* + and * + be two distinct points of the product space
X, then for some where and . Hausdorff
there exists open sets and in such that
( )
Since ( ) ( ) the above relation (1) implies that
, -
, -
[ ] , -
, -
, -
But
[ ]
, - arc open in X being subbasic members of
product topology. Thus to each pair of distinct points X, there exist
two disjoint open sets one containing and the other containing .
Hence X is Hausdorff.
4.54 Connectedness and product space:
The product space
is connected iff each co-ordinate space
is connected.
Proof: Let
be a connected space. Since each projection
is continuous and is the image of X under it follows that
continuous image of connected space is connected. Thus each co-
ordinate space is connected.
Conversely, Let each co-ordinate space be connected and *
} be a fixed point of X. let C be a component of X to which a
belongs. We shall prove every point of X belongs to C.
Let
be any arbitrary basic open set containing a point *
} of X, where is open in X, and if
.Then
is the set of all points
* + such that if is homeomorphic to
But
is connected hence its
homeomorphic image
is connected. As C is
maximal connected subset of X it follows A but the set A contains
the points * + for which if and
for . This point * +lies in
which was an
arbitrary basic open set containing * + It follows that the point
* + is in closure of C, thus we have shown that each basic
neighbourhood of x contains a point of A and hence a point of C. But C
being a component is closed C . Hence * + C Thus every
point of X belongs to C so that C But C Hence C But C is
connected set. Hence X is also connected.
4.55 Compactness and Product space (Tychonoff’s theorem):
The product of any non-empty class of compact spaces is compact.
Proof: Let * + be a non-empty class of compact spaces and
be the product space let { } be a non-empty subclass of closed
subbase for the product topology. This implies that each is a product
of the form
where is a closed subset of for all i‟s but
one. We assume that the class { } has finite intersection property and
83
shall show that ⋂
is non-empty which will imply that X is compact as
we know that “A topological space is compact iff every class of closed
sets with has non-empty intersection.” For a given fixed * + is a
class of closed subsets of with finite intersection property and as
are compact there exists a point in which belongs to ⋂ . If we
do this for each i, we obtain a point * + in X which is in ⋂ .
Hence X is compact.
4.56 Countability and Product spaces:
Theorem: The product of two second countable spaces is second
countable.
Proof: Let X and Y be two second countable spaces. We have to
show X x Y is also second countable space by definition of second
countable spaces let and be countable bases for X and Y
respectively. We define a collection * +
Then as product of two countable sets is countable therefore D is
countable base for X x Y. Hence X x Y second countable.
4.6 Definition:
4.61 Evaluation Function: Let * + be an indexed family of
sets. Suppose X is a set and let for each is a function.
Then the function
defined by ( )( ) ( ) is
called evaluation function of the indexed family* + of functions.
Hence ( )( ) implies co-ordinate of ( ).
4.62 Distinguish Points:
An indexed family * + of functions all defined on the same domain
X is said to distinguish points if for distinct X there exists
such that ( ) ( ).
4.63 Distinguish Points form closed sets:
An indexed family * + where X and arc topological
spaces is said to distinguish points form closed sets in X if for any
and any closed subset C of X not containing there exists such
that ( ) ( ) in
4.64 Embedding (or Imbedding):
Let (X, ), (Y,U) be topological spaces. An embedding of X into Y is a
function which is a homeomorphism when regarded as a
function form (X, ) onto . ( ) ( )⁄ /.
4.65 Cube:
A cube is a space of the form , - , where I is some set. If the set I is
denumerable the cube is called a Hilbert cube
4.66 Preposition :
Let * + be a family of sets, X a set and for each . a
function. Then evaluation is the only function from X to
whose
composition with projection equals to for all .
Proof: Let be the evaluation map defined by ( )( )
( ) and be the projection map we have prove the
only function for which
.
Firstly we shall prove that , Then we shall prove uniqueness
of e.
Consider
( )( ) , ( )-
85
( )( ), +
( ), )
Thus
Let there exists another function satisfying for
all . Let by any arbitrary element then,
( )( ) ( ( )), -
( ) , -
( )( ), -
4.67 Preposition :
The evaluation function of a family of functions is one to one iff it
distinguishes points.
Proof: Let * + be family of sets, X be any set such that
be a mapping for each .
be evaluation map.
Let be distinct points. Then ( ) ( ) iff such that
( )( ) ( )( )
But
( )( ) ( )
( )( ) ( ), -
So this implies ( ) ( ) thus we conclude ( ) ( ) implies that
some such that ( ) ( )
4.68 Preposition :
The evaluation function is continuous iff each is continuous.
Proof: We know that from preposition 4.66 for all .
Again we have if “X is a topological product of an indexed family of
topological spaces * ): + and let Y be any topological space.
Then is continuous iff for each , the composition
is continuous.” from the above two results we have e is
continuous is continuous
4.69 Preposition:
A topological space is completely regular iff the family of all continuous
real valued function on it distinguishes points form closed sets.
Proof: Let X be a topological space. F be a family of all continuous
real valued function. In first part of the proof let X be completely
regular, then by definition of completely regular space for any arbitrary
point and a closed subset C of X not containing a continuous real
valued function , - such that ( ) and ( ) * +. Then
can be regardes as a function from X to R. Then F and ( )
( ) since {1} is a closed subset of R. so distinguishes points form
closed sets.
Conversely, let F distinguish point form closed sets and shall prove that
X is completely regular.
Let and C be a closed subset of X not containing . Then there
exists some F such that ( ) ( ) . Now * ( )+ { ( ) } are closed,
disjoint subsets of R. and R being normal by Urysohn‟s lemma some
continuous function , -such that ( ( )) ( ( ) )
……… (a)
Let , - be the composite function gof then ( ) ( )
, ( )-
for each
( ) ( ) , ( ))- [from a]
Thus is a real valued function such that ( ) and ( ) . Thus
X is completely regular space.
4.70 Preposition:
87
Let * + be a family of functions which distinguishes points
from closed sets in X. Then the corresponding evaluation function
is open when regarded as a function from X on ( )
Proof: Let X be any topological space * + be family of
functions which distinguishes points form closed sets in X.
be evaluation map. We have to prove e is an open mapping form X to
e( )
We know that “ is said to be open iff - image of every open
set in X is open inY.” So for the proof of the preposition it is sufficient
to show that if V is an open set in X then ( ) is open in e( )
Let e( ) be any arbitrary point of e( ), then corresponding to e( )
Now as V is open X-V is closed set not containing . As family
of function distinguishes points from closed set such that
( ) ( ) Let ( ) . As closure of an x set is
closed set therefore ( ) is closed set therefore its complement G
is open in Y .So,
( ) is open subset of
we know that
projection map is continuous.” We claim that
( ) ( ) ( )
Let ( ) ( )
( ) ( )
( ) and some z such that ( )
So, ( ) ( ( ))
( )
( ) [By preposition 4.66]
From this it follows because if
( ) ( ) ( ) which is a contradiction as we
have proved above that ( ) Hence ( ) ( ). Thus
( ) ( ) ( ) The set
( ) ( ) is open in relative
topology on e(X) containing ( ) Thus e(V) is a neighbourhood of ( )
in e(X). Thus e(V) is neighbourhood of each of its points Hence e(V) is
open as we know that “a set G is open it is neighbourhood of each
of its points.”
4.71 Embedding Lemma:
Let * + be a family of continuous functions which distinguishes
points and also distinguishes points from closed sets. Then the
corresponding evaluation map is an embedding of X into the product
space
Proof: In order to prove e is an embedding of X in
we have to
prove X is homeomorphic to e(X). For this we have to prove
e is one to one in e(X)
e is continuous in e(X)
e is open in e(X)
The continuity of e follows from preposition “evaluation map is
continuous iff each f is continuous.” Here arc continuous therefore
e is continuous. As. * + distinguishes points therefore by
preposition 4.67 e is one to one.
Again * + distinguishes points from closed sets. Hence e is
open when regarded as a mapping form X on to e(X) by preposition
4.69.
Thus e is an embedding of X into
4.72 Theorem:
89
A space is embeddable in the Hilbert cube iff it is second countable
and T3.
Proof : Let X be a Hilbert cube. In first part of the proof let X is
embeddable in Hilbert cube, we shall prove that X is second countable
and T3
As we know the “ Every subspace of Hilbert cube is second countable
metric space. “ Therefore embedding of X is second countable and
metrizable .Also as “ all metric space arc T4(hence T3)” The embedding
of X is second countable and T3.
Conversely, let X be second countable and T3. We shall prove the
theorem by applying embedding lemma. So for the proof of the
theorem we shall construct a family of functions that distinguishes
points and also distinguishes point from closed set .
As X is second countable and T3 therefore X is normal because “every
regular and second countable space is normal.”
Let B be countable base for X. Enumerate B as {B1, B2, B3………..}, if B
is finite we repeat some members of B infinitely. Now let
*( ) + I is then countable set. For each say
( ) we apply Urysohn‟s Lemma to the disjoint closed sets
and of X and obtain a map , - such that ( )=0
and ( ) We shall prove that * + distinguishes points
from closed set in X. Let and C be a closed set of X not containing
then which is open as C is closed. As B is a base so for
some such that . As X is regular and open set
G in X such that and . But as G is open such that
therefore ( ) Then the corresponding
evaluation function vanishes on and takes value 1 on and
in particular on C as , So ( ) ( ) . Thus * +
distinguishes points form closed sets.
Now as X is a T1 space all singleton sets are closed so * +
distinguishes points, as well.
So, by embedding lemma corresponding evaluation map , -
is an embedding . If I is countably infinite, this completes the proof of
the theorem because , - is then homeomorphic to , - which is the
Hilbert cube.
4.73 Theorem- Urysohn’s Metrization Theorem:
A second countable space is metrizable iff it is
Proof: Let X be a second countable space. In first part of the proof let
X be metrizable then X will be as we know that “Every metrizable
space is ”.
Conversely let X be a second countable space we shall prove X is
metrizable. As we know that “ A space is embeddable in Hilbert cube iff
it is second countable and Thus every second countable space is
homeomorphic to a subspace of Hilbert cube since Hilbert cube is
metrizable and mertizability is hereditary property therefore its
subspace is metrizable and hence the inverse image of embedding that
is X is metrizable.
4.8 Summary:
1. Continuous image of connected set is connected.
2. Union of connected sets is connected if their intersection is non-
empty.
3. Real line is connected.
4. Each point of space in a component.
5. Each connected subspace lies in a component.
6. components are closed sets.
7. Product topology is smallest topology.
8. Product of compact space is compact.
9. Product of connected space is connected.
91
10. A space is embeddable in a Hilbert cube it is second countable
and .
11. A second countable space is metrizable it is
4.9 Assignment /Check your progress:
1. Prove that if E be a connected subset of X, F is a subset of X such
that then F is connected in particular is connected.
2. A topological space X is disconnected iff there exists a continuous
mapping of X onto the discrete two point space [0,1].
3. Show that every discrete space is locally connected.
4. Prove that projection map is open.
5. Show that product space of two T0 space is T0- space.
6. Show that product space of two completely regular space is
completely regular.
7. Show that a topological space is a Tychonoff space iff it is
embeddable into a cube.
4.10 Points for Discussion:
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
4.11 References:
1. George F. Simmons, Introduction to Topology and Modern
Analysis, MC Grow Hill book company, 1963.
2. K.D. Joshi, Introduction to General Topology. Wiley Eastern
Ltd. 1983.
3. K.K.Jha, Advanced General Topology. Nav Bharat Prakashan,
Delhi.
4. James R, Munkres, Topology, Pearson Education Pvt. Ltd.
Delhi 2000.
UNIT – V
Paracompactness, Net & Filters, Fundamental Groups & Covering Spaces
Introduction: The unit deals with three concepts the first part of the unit
deals with the extension of continuous functions in forms of nets and gives a
base of another algebraic structure called filter and at the same time provides
a conversion technique of nets into filter & vice-versa . The second part deals
with other type of continuous mapping called homotopy. While the third
section deals with concepts of paracompactness and locally finite spaces.
Objectives: After completing this unit you should be able to
Understand the concepts of net & filter
Appreciate the transformation of mappings through homotopy
Define locally finite spaces and its related concepts
93
Structure:
5.1 Definition Based on nets
5.18 Theorems Based on nets
5.2 Filter
5.21 Types of filters
5.3 Definition Based on filter
5.4 Theorem Based on filter
5.5 Definition Related to Homotopy
5.6 Theorems Based on Homotopy
5.64 Fundamental Theorem of Algebra
5.7 Definition Based on Paracompactness
5.8 Theorems and Lemma
5.84 Nagata Smirnov Metrization theorem
5.87 Smirnov Metrization theorem
5.88 Summary
5.89 Check your Progress/Assignment
5.9 Point for Discussion
5.91 References
5.1 Definitions Based on Nets:
5.11 Directed Set:
A directed set is a pair ( ) where D is a non-empty set and a
binary relation on D satisfying
(i) For all m and
(ii) For all
(iii) For all there exists such that and
Example-1: Set of natural number N with relation is usual
ordering on N is a directed set.
Example-2: Let X be a topological space . be the
neighbourhood system of We define for , ,
Then ( ) is directed set.
5.12 Net:
A net in a set X is a function where D is a directed set.
Example - defined by ( ) is a net
Example - defined by ( ) is a net.
5.13 Convergence of Net:
Let ( ) be a topological space. be a net. Then is said to
converge to a point if given any open set U containing , there
exists such that for all , imples (or ( )) .
In this case is a limit of .
5.14 Eventual Subset:
A subset of a directed set is said to be eventual if there exists
such that for all implies that . A net
is said to be eventually in a subset of A of if the set ( ) is an
eventual subset of .
5.15 Co-Final Subset
Let ( ) be a directed set. A subset F of is said to be a cofinal
subset of if for every , there exists such that . A net
is said to be frequently in a subset A of if ( ) is a co-
final subset of .
5.16 Subnet:
Let , be nets. Then is said to be a subnet of if
there exists a function such that
(i)
95
(ii) For any , there exists such that for all
implies ( ) in .
5.17 Cluster Point:
Let be a net. A point is said to be a cluster point of
if for every neighbourhood U of in and there exist in
such that and .
5.18 Theorems Based on Nets:
Theorem: A topological space is Hausdorff iff limits of all nets in it
are unique.
Proof:- In first part of the proof let be a Hausdorff space.
be a net in we have to prove S converges to a unique point on the
contrary, let S converges to two distinct points say . Then as
and is a Hausdorff space by definition of Hausdorff space there
exists open sets U and V such that , and . As is
limit of S, by definition of convergence there exists such that
for all , implies similarly y is limit of S, so there
exist such that for all implies . Now as
is a directed set implies there exists such that
and and . This implies
which is a contradiction as . Hence that is limit of nets
are unique. Conversely, let limits of nets in are unique we shall prove
that is Hausdorff space. On the contrary let is not Hausdorff then
there exist two distinct points in which do not have mutually
disjoint neighbourhoods in . Let
and for ( ) , ( ) we define ( )
( ) iff and . Then is a directed set. We define a
net as follows for any and as by
assumption, we define ( ) to be any point in . We shall
prove that S converges to . Let G be an open neighbourhood of .
Then ( ) . Now if ( ) ( ) in then so ( )
⋂ Thus converges to . Similarly converges to . But
this is a contradiction as nets in have unique limit. Therefore our
assumption that is not Hausdorff is false.
Preposition: Suppose is a net and is a cofinal subset
of If converges to a point in , then is a cluster
point of .
Proof: Let be a neighourhood of in . As F is a cofinal subset of S
there exists such that for any , implies that .
Now, let be given. As D is directed set by definition of directed set
such that and . Corresponding to choose
such that . Then . So is frequently in and since was
arbitrary, is a cluster point of .
Preposition:
Let be a subset of a space and let Then iff there exists a
net in (that is a net which takes values in the set ) which converges
to in .
Proof: Let is a net which when regarded as a net in
converges to let U be any neighbourhood. Then by definition of
convergence there exists such that for all implies
. Since therefore for all So ⋂ .
Thus every neighbourhood of intersects and so by definition of
limit point is limit point of therefore ( ) and as ( ) .
This implies .
Conversely, suppose Then every neighboushood of intersects
“As we know that ( ) .” Let be the neighbourhood system
of in . Then ( ) where is a directed set. We
define a net by ( ) ⋂ . We shall prove
that converges to in . Let be any open set in containing .
97
Then for any implies and hence ( ) So,
converges to in .
Theorem: Any topological space is compact iff every net in has a
cluster point in .
Proof: Let be a compact space and be a net in .
Suppose has no cluster point in . Then each is not a cluster
point of X which implies there exists a neighbourhood of x and an
element such that for all implies .
We cover by such neighbourhoods. As is compact there exists
a finite subcover for that is
Let the corresponding
elements in be Because is a directed set there
exists such that But by
assumption ( )
which is a
contradiction . So, has atleast one cluster point in .
Conversely, let every net in has a cluster point .Let be a family of
closed sets of having finite intersection property. Let be the family
of all finite intersections of members of . For define
. Then is a directed set because (i) for all D, E, F we have
this implies that is (ii)
for all that is (iii) for all D, E, F , ,
similarly Also each member of is
non-empty because C is having finite intersection property. We define
a net by ( ) By assumption has a
cluster point say in . We shall prove ⋂ For, if not there
exists such that . Then is a neighbourhood of . Also
So, by definition of a cluster point, there exists such that
and ( ) But then and so
contradicting that ( ) So ⋂ Thus X is compact since we
know that “X is compact every class of closed sets having has
nonempty intersection.
5.2 Filter:
A filter on a set is a non empty family of subsets of such that
1.
2. is closed under finite intersections
3. if and then
5.21 TYPES OF FILTER
(1) The singleton family * + is a filter on .
(2) Let A non-empty subset of X. Then the collection of all supersets of
( ) is a filter on X. Such a filter is known as an atomic filter,
the set A is called the atom of the filter.
(3) Let X is infinite, the family of all co finite subsets of X is a filter on
X. Such a filter is called a co finite filter.
(4) Suppose is a topology on X. Then for any , the
neighbourhood system at x is a filter. It is called the -
neighbourhood filter at x. It depends both on x and
Filter Associated With Net
(5) Let be a net. For each , let * ( )
+ Let * for some + In other words,
is the collection of all supersets of sets of the form for
then is a filter on X, it depends on the net S and is called the
filter associated with the net S.
5.3 Definition:
5.31 Base Let be a filter on a set X. Then a sub-family of is
said to be a base for (or a filter base) if for any there exists
such that .
Preposition: Let be a family of non-empty subsets of a set X.
Then there exists a filter on X having as a base iff has the
property that for any , there exists such that
99
Proof: Let there exists a filter on X having as a base. Then by
definition of base and hence . Also let
Then as and so as is closed under
finite intersections. So by the definition of a base, there exists
such that Conversely suppose satisfies the given
condition. We shall prove that a filter can be formed from the sets of
as follows. Let be the family of all supersets of members of . We
shall prove that is a filter (i) as the empty set cannot be a superset
of any other set. Hence it follows that . (ii) We now prove that
is closed under finite intersections. For this it is sufficient to show
that the intersection of any two members of is again in let
. Then by definition of there exist such that
and We are given that there exist such that
. But then is a superset of and so
by definition of . (iii) As is the family of all supersets of
members of so by its very construction the third property of filter is
satisfied. Thus is a filter on X
5.32 Sub Base: Let be a filter on a set X. Then a subfamily of
is said to be a sub-base for if the family of all finite intersections
of members of is a base for . We also say generates
5.33 Limits & Cluster Points: Let ( )be a topological space
and let be a filter on the set X. A point x of X is said to be a limit of
w.r.t. if every neighbourhood of x belongs to , i.e. if . Also
a point is said to be a cluster point of if every neighbourhood
of y intersects every member of .
5.34 Nets Associated with filter
There is a canonical way of converting nets to fillers and vice versa. Let
be a net given a filter on we define a net associated with
it as follows. Let *( ) + For ( ) ( ) we
define ( ) ( ) if .Then directs because is closed
under finite intersections. Now we define ( )
Then is a net in It is called the net associated with .
5.4 Ultra filter:
Let X be any set a filter is called an ultra filter if it is maximal filter
that is it is not contained in any other filter
Theorem: Every filter is contained in an ultra filter.
Proof: Let be a filter on a set . let be a collection of all filters
on containing . Then as therefore so is nonempty,
We partially order by inclusion. Let * + be a non-empty chain in
. Let We shall prove is a filter on . will be a filter if it
satisfies following properties Clearly because for all
We shall show that is closed under finite intersection it is
sufficient to show that intersection of two members of is again in .
Let ,then by definition of there exists such that
and . Again as the collection { + is a chain under inclusion
relation therefore it implies either or . Let this
implies so as is a filter. Similarly in second
case . In either case as now let
and D is a superset of in X we shall prove that As
it implies for some and as is a filter by
definition of filter . This implies Thus by definition of
we conclude is a filter in X clearly contains as each contains
. So and by its construction it has an upper bound for the
chain * + Thus we have shown that every chain in has an
upper bound in . So, by Zorn‟s lemma contains a maximal element
say H. We shall now prove that H is an ultrafilter. For this we have to
show that H is maximal element in the set of all filters in X that is H is
not contained in any other filter on X Suppose K is a filter on X such
that . Then as so K by definition of . But H
101
is maximal in so K this implies . Thus H is an ultra filter
containing .
Theorem for a filter on a set X the following statements are
equivalent
1. is an ultra filter
2. For any either or
3. For any , either
Proof: We shall first prove that is (1) equivalent to (2). Let is an
ultra-filter on and is a subset of . If then contains
members of , or equivalently every member of intersects
— . Then the family * + has the finite intersection
property and so it generates a filter “if S is a family of subsets of X
than there exists a filter on X having as subbase iff S has fip”. Since
is maximal, no filter on can properly contain . So we must have
which implies that Conversely we assume (2) holds. If
is not an ultrafilter then by definition of ultrafilter there exists a
filter which properly contains . Then there exists — . Since
— by (2). Hence — . So contains as well
as — which contradicts the finite intersection property of a filter.
Thus is an ultraffiter.
Let us assume (3) holds. Let as (
) so by (3) or ( ) Conversely we assume (2)
holds. Let Since is a superset of as well as , one
way implication in (3) is immediate from the very definition of a filter.
For the other way, suppose but neither nor
Then by (2), — and — and So
( — ) ⋂( — ) But ( — ) ⋂ ( — ) — ( ) So
contains as well as its complement, which is a contradiction. So
(3) holds. Thus we have shown that (2) is equivalent to (3).
Preposition: An ultra filter converges to a point iff that
point is a cluster of it.
Proof: Let be an ultra filter which converges to a point x
then x is a cluster point by definition of cluster point for filter
Conversely, suppose is a space and is a cluster
point of an ultrafilter on . If does not converge to ,
then by definition of convergence of filter there exists a
neighbourhood of such that By the theorem
proved above, we have that — But since x is a
cluster point of every of intersects every member
of , whereas ⋂ ( — ) This is a contradiction
hence converges to in
Theorem: A topological space is compact if every ultrafilter
in it is convergent.
Proof: We know that a space is compact iff every filter in it
has a cluster point In particular every ultrafilter has a cluster
point and hence is convergent by the above proposition.
Conversely suppose is a space with the property that every
ultralilter on it is convergent. To show is compact we shall
show that every filter on has a cluster point. Suppose is
a filter on . Then there exists an ultrafilter containing .
By assumption converges to a point say on . Then x is
also a cluster point of . So every of meets every
member of and in particular every member of since
. So is also a cluster point of Thus every filter on
has a cluster point in . This proves that is compact..
103
Preposition: Let be a net and be the filter associated
with it Let Then converges to as a net iff converges to as
a filter.
Proof: Let be a net such that converges to . Let be any
neighourhood of in . then by definition of convergence of net there
exists such that where * ( ) + But this
implies , by definition of filter associated with net . So i.e.
converges to
Conversely suppose converges to . Let be an open
neighbourhood of . Then by definition of filter associated with
net there exists such that . This implies ( ) for all
Thus converges to in .
Theorem: A topological space is Hausdorff iff no filter can
converge to more than one point in it.
Proof: Let be a Hausdorff space be a filter convergingto say two
points and in . Then the net associated with also converges
to and in . As we know that “if is a filter and net associated
with it then converges to a point iff the net associated with it
converges to . “ But this a contradiction since we know that” is
Hausdorff iff every net in has unique limit.” Therefore converges
to unique limit. Conversely, let every filter in converge to unique
limit. Then no net in converges to more than one point i.e. limit of
nets are unique then the space is Hausdorff, as stated in the statement
above.
5.5 Definitions
5.1 Homotopy
If f and f ‟ are continuous maps of the space X into the space Y, then f is said
to be homotopic to f if there is a continuous map such that
( ) ( ) and ( ) ( ) for each x here , -. The map f is
called a homotopy between and . If is homotopic to we write .
If and is a constant map then the homotopy is called nullhomotopy
Path homotopy
Let f and f‟ be two paths between I=[0, 1] into X. Then f and f‟ are said to be
path homotopic if they have same intial point and same final point and
if there is a continuous map such that
( ) ( ) and ( ) ( )
( ) and ( )
For each and each then F is called path homotopy between f and f‟
and is denoted by
Example:-
Let and be any two maps of a space X into we define
( ) ( ) ( ) ( )
Then ( ) ( ) ( ) ( )
is homotopy between f and g
5.52 Product of paths
It f is a path in X from and if g is a path in X from we define
the product f g of f and g to be the path h given by.
( ) ( ) { ( ) ,
-
( ) ,
-
The function h is well defined and continuous, it is a path in X from .
105
The product operation on paths induces a well defined operation on path
homotopy classes, defined by equation
, - , - , -
5.53 Reverse path
If f is a path in X from Let be the path defined by ( )
( ) Then is said to be reverse of f
5.54 Loop
Let X be a space and be any arbitrary point. A path in x that
begins and ends at is called a loop based at .
5.55 Fundamental group.
The set of path homotopy classes of loops based at with the
operation * is called the fundamental group of X relative to the base
point . It is denoted by ( ).
5.56 let be a path in X from . We define a
mapping
( ) ( )
as follows
(, -) , - , - , -
5.57 Simply connected
A space X is said to be simply connected if it is a path connected space
that is between any two points in X there exists a path and if
( ) is a trivial (one-element) group for some
5.58 Homomorphism induced
Let h : ( ) ( ) be continuous map. We define
( ) ( ) by equation
(, -) , - then is called Homomorphism induced by h
relative to base point
5.59 Evenly covered: Let be a continuous surjective
(onto) map. The open set of is said to be evenly covered by if
the inverse image ( ) can be written as the union of disjoint open
sets in such that for each the restriction of p to is a
homeomorphism of onto U, the collection * + is called a partition of
( ) into slices.
5.60 Covering Space: Let be continuous and surjective
map. If every point of has a neighbourhood that is evenly
covered by , then p is called a covering map and is said to be a
covering space of .
Example The mapping given by equation
( ) ( ) a covering map.
5.61 Lifting: Let be a map. If is a continuous mapping
of some space X into B, a lifting of is a map such that
E
X B
107
5.62 Lifting correspondence: Let be a covering map
. Choose in so that ( ) Given an element , - of
( ) Let be lifting off to a path in E that beings at . Let
(, -) denote the end point ( ) of . Then is a well defined set
map
( ) ( )
is called lifting correspondence derived from the covering map p.
5.63 Theorem
The map is a group isomorphism
Proof:- Let X be any space be a path in X from . We define
( ) ( ) as
(, -) , - , - , -
We have to prove is a isomorphism for this we have to prove
(i) is one-one
(ii) is onto
(iii) is homomorphism
Now will be one-one and onto if there exists an inverse of
. Let denote the reverse of i.e. . We shall prove is the
inverse of So, consider
(, -) [ ] , - , - , -
, - , - , - , -
Now
. (, -)/ [, - , - , -] , , -
, - , - , - , - , -
, - [ , - , - [ ] ]
This implies is inverse of therefore is one-one onto. Now, we
shall prove that is homomorphism. So, consider
(, -) (, -) (, - , - , -) (, - , - , -)
, - , - , - , -
(, -)
This implies is a homomorphism Thus we have proved is a
isomorphism.
Lemma: In a simply connected space , any two paths having some initial
and final points are path homotopic.
Proof: Let be simply connected space be paths in such that ( )
( ) and ( ) ( ) we have to prove and are path
homotopic for his it is sufficient to show that , - , -
As and are paths in from is a path from Also ( )
( )
Therefore ( ) is defined as
Then ( ) { ( ) , -
( ) , -
Then ( ) ( )
( ) ( )
Thus is a loop at and as X is simply connected, this loop is path
homotopic to constant loop Then,
[ ] , - [ ] , - [ ]
, -, -
But
[ ] , - , - [ ]
, -
Thus we conclude , - , -.
109
This implies and are path homotopic.
Theorem: are covering maps then
is a covering map.
Proof: Let be and U and U‟ be neighbouhoods of b and b‟
respectively, that are evenly conversed by p and p‟ respectively. As are
covering maps by definition of covering map * + and * + be partitions of
( ) and ( ) ( ) respectively into slices. Then the inverse image under
p X p‟ of the open set U x U‟ is the union of all the set . These are
disjoint open sets of exe, and each is mapped homeomorphically on UxU‟ by
therefore satisfies all conditions of covering map.
Theorem: Let be covering map, let ( ) If E is path
connected, then the lifting correspondence.
( ) ( ) is surjective. If E is simply connected, it is bijective.
Proof: Let E be path connected and let ( )
( ) so, by
definition of path connected space. There is a path in E. form to .
Then because .
( ) [ ( )] ( ) ( )
( ) [ ( )] ( ) ( )
Thus is a loop based at in B and as is lifting correspondence so, by
definition (, -) ( ) Thus, we have proved for every
( ) , - ( ) such that (, -) is surjective.
Now, let E be simply connected, then by definition of simply connected space
E is path connected, then from the result proved above we infer is surjetive
so, for the proof of the theorem we only have to prove is one-one. Let
, - , - ( ) such that (, -) (, -). Then be the liftings of
and respectively to paths in E that begins at , then by definition of we
have (, -) ( ) (, -) ( ) as (, -) (, -), this implies ( )
( ) Thus are paths in E such that ( ) ( ) ( ) ( ) and as
the space is simply connected a path homotopy between and so, let F
be the path homotopy between and . Then is path homotopy
between and because “ If is a continuous map and if
and are paths in X with F as path homotopy then and are path
homotopic with path homotopy in ” therefore [ ] , -, but
and , - , - Thus we have proved (, -) (, -)
, - , -
This implies is injective (one-one)
Theorem: The fundamental group is isomorphic to the additive group of
integers.
Proof: is the fundamental group of circles and we have to prove is
isomorphic to the additive group of integers. We begin the proof by defining a
mapping as follows
( ) ( )
Then p is a covering map. Let and ( ). Then ( ) is a set of
intergers Z, as R is simply connected, the lifting correspondence.
( ) is a bijection as we know that “ Let be a covering
map, let ( ) If E is simply connected then , the lifling correspondence
( ) ( ) is bijection.
Thus is the mapping between fundamental group of circles and additine
group of integers and from the above quoted theorem is bijective.
Therefore to prove the theorem it is sufficient to show that is a
homomorphism.
Let , - and , - be any two arbitrary element of ( ). Then , be their
respective lifting to paths in R beginning at 0. Let ( ) and ( )
Then by definition of , we have (, -) and (, -) .
We define a path ( ) ( )
Then ( ) , ( )-
, ( )- , -
, ( )- [ ( )]
( )
( ) , -
Thus is a lifting of g. Also we have ( ) ( )
111
Also ( ) ( )
As ( ) ( ) is defined and is given by
( ) { ( ) , -
( ) , -
Also [ ] ( ) ( ) , -
Thus is a lifting of that begins at 0 as ( ) ( )
( ) ( ) ( )
Therefore
(, - , -) (, -)
( )( ) ( )
( ) ( ) ( )
(, - , -)
Thus is a homomorphism. We have proved that a mapping
( ) which is a bijection and homomorphism. Therefore is a
isomorphism.
5.64 Theorem (Fundamental theorem of Algebra) A polynomial equation
of degree n>0
with real or complex co-efficient has atleast one (real or complex) root.
Proof: The theorem will be proved in following steps.
(i) We shall define a mapping and prove that
corresponding induced homomophism is injective.
(ii) Using the result proved in first step we shall prove that the
mapping is not null homotopic.
(iii) We shall prove the result for a particular case.
(iv) Generalize a result.
We define a map defined as ( ) where z is a complex
number Now, be the standard loop in defined as ( )
( )……………… (1)
Then, the induced homomophism of fundamental group is defined by
, ( )- , ( )-, we shall prove that f is injective
Now, ( ) , ( )- ( )
( )
( )………………………. (2)
From (1) & (2) we observe that the loop lifts the path to in the covering
space R. Thus is multiplication by n in fundamental group in particular
injective.
We define another map as ( ) and shall prove that g is
not null homotopic. We observe that equals map define in step 1 through
the inclusion map . Therefore,
( ) , ( )-
, - , - , - ( ) , - Then is injective because “If is a retract (subset) of then the
homomorphism of fundamental groups, induced by inclusion is
injective. “Here is a retract of Therefore is injective,
which implies that cannot be null homotopic
Now, we shall prove a special case of the theorem Given a polynomial
equation
( )
Such that and shall prove that equation
has a root lying in the unit ball on the contrary let it has no such root. We
define a map by equation ( ) let
h be the restriction of to as h extends to a map of unit ball into ,
the map h is null homotopic because “Let be a continuous map.
Then following conditions are equivalent
1. is null homotopic
2. extends to a continuous map
We now define a homotopy F between h and the map defined as
given by ( ) (
) Then ( ) ( ) ( ) (
) ( )
113
Thus is homotopic to which is null homotopic, which implies is also null
homotopic but this is a contradiction, as we have already proved that is not
null homotopic hence our assumption (A) has no roots in unit ball is false.
Now we prove the general case, we have the polynomial equation
Let us choose a real number and substitute we have ( )
( )
( )
We choose c so large such that
|
| |
| |
| |
|
Then from step (3) this equation will have root say the
original equation will have root Hence the theorem is
proved
5.7 Definition:
5.71 Locally finite: Let X be a topological space. A collection of
subsets of X is said to be locally finite in X if every point of X has a
neighbourhood that intersects only finitely many elements of .
Example I. The collection of intervals
* +
is locally finite in the topological space ,
5.72 Countably Locally finite A collection of subsets of is said to
be countably locally finite if can be written as the countable union of
collection , each of which is locally finite.
5.73 Refinement. Let be a collection of subsets of the space A
collection of subsets of is said to be a refinement of (or is said to
refine ) if for each element B of , there is an element A of containing
B. If the elements of are open sets, we call an open refinement of if
they are closed sets, we call a closed refinement.
5.74 Set. A subset of a space is called a set in if it equals the
intersection of a countable collection of open subsets of
5.75 Paracompact: A space is paracompact if every open covering
of has a locally finite open refinement that covers
5.8 Theorem and Lemma
5.81 Let be a locally finite collection of subsets of X. Then:
(a) Any sub collection of is locally finite.
(b) The collection * + of the elements of is locally finite.
(c) .
Proof
(a) Let A be a locally finite collection of subsets of X. B be a
subcollection of A i.e . Then as A is locally finite it by
definition of locally finiteness every point in X will have a
neighbourhood that will intersects only finitely many elements of A
or in particular only finitely many elements of B as .
Therefore B will be locally finite.
(b) Let A be a locally finite collection of subsets of X. Let
* + we have to prove that B is also locally finite for this we will
have to prove that for every point in X a neighbourhood that will
intersects only finitely many elements of B so, let and U be
any neighbourhood of that intersects only finitely many elements
of A. Let where Ai A . Thus U will
intersect finitely many elements of B. showing that B is locally
finite.
(c) We have to prove
⋃
⋃
Let
115
⋃
So we have to prove the result
⋃
By definition of we have
So for each since we know that
Thus
⋃
( )
Now, let and U be neighbourhood of that intersects finitely
many elements of A say Then will belong to
some and hence will belong to because if
does not belong to any and hence will belongto
any then will be a
neighbourhood of that will intersect with no elements of A as we
have assume U intersects and will therefore not
intersect . Thus there exists neighbourhood of which does not
intersect this implies is not limit point of and hence
which is a contradiction to our assumption that
This implies
⋃ ( )
From (1) & (2) we infer
⋃
⋃
⋃
Lemma 5.82. Let be a regular space with a basis that is countably
locally finite Then is normal, and every closed set in is a set in
In the proof of the theorem we shall first prove that for any open
set in there exists a countable collection { } of open sets of
such that
⋃ ⋃
As B is a countably locally finite basis of by definition of countably locally
finite we have B = where each collection is locally finite. We now
form a subclass of B, consisting of those basis elements B for which
and Thus be a subclass of locally finite class B it will also
be locally finite as we know that “ Any subcollection of a locally finite class is
locally finite”. We now define
⋃
As B being basis elements are open and arbitrary union of open sets being
open will imply that is open set
By theorem 5.81 we have
⋃
⋃
And as it implies
i.e. but
so we have ( )
Now let as is a regular space by definition there exists a basis
element such that and . As B is countably locally finite
for some n as .Then by definition of , then
by definition of so we have proved , This implies
( )
From (1) and (2) we have
117
( )
We shall now prove the theorem using the above proved result. Let C Be any
closed set in X we shall show that C is a set in X. Then is an
open set by the result proved in eq. (3)
there exists open sets in such that
Then
, -
, -
[Generalized De-Morgans law]
, -
Thus C is countable intersection of open sets of X and hence is a set.
We shall now prove that X is normal. Let C and D be two closed disjoint
subsets of X. X will be normal if two disjoint neighbourhoods for C and D.
As D is closed set it implies is open so by the result proved in (3) there
exist a countable collection of open sets {Un} such that .
As C and D are disjoint sets .
Thus C is covered by {Un} where each is disjoint from D. Following similar
argument for C we obtain a class of open sets {Vn} that cover D and whose
closures are disjoint from C. We now define
⋃
and
⋃
Then by its very construction
are complements of closed sets and
hence are open and at the same time disjoint. Then the sets
U ‟ =
, V ‟ =
Are arbitrary union of open sets hence are open sets containing C and D
respectively also U ‟ V ‟ = . Thus, X is normal.
Lemma 5.83 Let be a metrizable space. If is an open covering of ,
then there is an open covering of refining that is countably locally
finite.
Theorem 5.84 (Nagata-Smirnov metrization). space is
metrizable if and only if is regular and has a basis that is countably locally
finite.
Proof:- In first part of the proof let X be a metrizable space we shall prove
that X is regular and has a countably locally finite basis since all metric spaces
are regular therefore X is also a regular space. We shall now prove a
countably locally finite basis. As X is a metrizable space let d be the metric
defined on X. Let we cover X by open balls of radius 1/m. Let this
covering be . Let this covering be . Then by lemma 5.88 there is an open
covering of X refining such that is countably locally finite. Let B be
the union of the collections ,
Then as is countably locally finite therefore B is also countably
locally finite. We shall prove that B is a basis. Let , then there exists an
open ball ( ) centred at x and let As is a basis
so there exists an element B of such that ( ). Thus B is a
basis which is countably locally finite.
In next part of the proof let X be regular and it has a countably locally finite
basis we shall prove that X is metrizable for this we shall prove that X can be
imbedded in metric space (( )) for some I. Let be countably locally
finite basis let . Now for each positive integer n and each basis
element we define a function , - such that
( ) for and ( )
for . Then by definition of the collection * + distinguishes points
from closed sets in X because for a given point a basis element B such
that . Then as , ( ) and vanishes outside u. let I
be subset of
We define a mapping , -
as ( ) ( ( ))( )
119
Then by Urysohn‟s imbedding lemma is an imbedding when consider as a
mapping between X and product space [0, 1]. We shall prove that is an
imbedding with respect to topology induced by uniform metric.
Now, as uniform topology is larger than product topology relative to uniform
metric is bijective and onto when considered as a mapping between X and
(X) by Urysohns lemma will be an imbedding if is continuous.
Let n be an positive integer. and be the neighbourhood of that
intersects only finitely many element of . This implies that all but finitely
many of the functions are identically equal to zero as each is
continuous we choose a neighbourhood of contained in on which
each of the remaining function for varies by at most .
Let N be a positive integer such that
and Then W
is the neighbourhood of . Let then for we have.
| ( ) ( )| .
as the function either vanishes or varies at most in w. Also for
| ( ) ( )|
as is a mapping between X and [0, 1/n]
Therefore by definition of uniform metric
( ( ) ( )) * ( ) ( ) +
* ( ) ( )
Therefore is continuous and is an imbedding of X into , - and X is
metrizable.
5.85 Theorem: Every closed subspace of a paracompact space is
paracompact.
Let X be a para compact space. Y be a closed subspace of . We have to
prove is also para compact for this we have to prove that every open
covering of Y has a locally finite open refinement that covers . So, let A be
an open cover of by sets open in Y. Then by definition of open cover
As is a subspace, so by definition of relative topology, for each there
exist an open set A‟ in X such that We observe that the sets
along with form an open cover for X and as X is paracompact there
exists a locally finite open refinement say B of this covering of X. Then the
collection
* +
will be the locally finite refinement of . Therefore is also paracompact.
5.86 Locally Metrizable: A space is locally metrizable if every point
of has a neighborhood that is metrizable in the subspace topology.
5.87 Theorem: (Smirnov metrization theorem). A space is
metrizable if and only if it is a paracompact Hausdorff space that is locally
metrizable.
In first part of the proof let X be a metrizable space then X will be locally
metrizable. Also as metrizable space are, Hausdorff space. X will be a
Hausdorff space. Again as every metrizable space is paracompact X will be a
paracompact Hausdorff space that is locally metrizable.
Conversely let X be a paracompact Hausdorff space that is locally metrizable
we shall prove that X is metrizable. In order to prove X is metrizable. We shall
prove that X is a regular space with a countably locally finite basis. As we
know that from Nagata Smirnov Metrization theorem “A space is metrizable if
and only if it is regular and has a countably locally finite basis.”
Now, as every paracompact space is regular we only have to prove that there
exists a countably locally finite basis. As X is locally metrizable so, by
definition of locally metrizable each point has neighbourhood U that is
metrizable in subspace topology. Let us cover X by such open sets. Then for
this cover we choose a locally finite reginement say that covers X. Then
clearly each is metrizable in subspace topology. Let be
the metric that gives the topology of . Now for we define ( ) to
be set of all points y in such that ( ) Then ( )are open balls in
121
C and hence are open sets in X. we shall now cover X by all such open balls
of radius 1/m, m Z+ . Let the covering be Am. So, we have.
* (
) +
As X is para compact by definition of paracompact spaces there is a locally
finite open refinement of that covers X. Let it be . Let D be the union
of collections then clearly D is countably locally finite.
We shall prove that D is a basis for X. For this we have to show for and
any neighbourhood U of element such that . So, let
and U be any neighbourhood of . As is a locally finite refinement
that covers will belong to only finite many elements of say 1, 2, 3
…………………. n.
Then being finite intersection of ofen set is an open set containing .
Therefore So, by definition of open set in metric space there exist a
ball ( ) such that
( )
We choose m so that
min * + As covers X by
definition of open cover such that As is a locally finite
refinement of there must exists an element ( ) of for some
and some that contains Thus we have .
/
As so
Since diameter of ( ) is at most
we have.
( )
( )
i.e.
Therefore is countably locally finite basis. So X is metrizable.
5.88 Summary A space is Hausdorff iff nets in it converge to unique limit.
A space is Hausdorff iff filter in it converge to unique limit.
There is a canonical way of converting nets into filters and vice-versa.
Every filter is contained in an ultra filter.
Every paracompact Hausdorff space is regular.
5.89 Check Your Progress/ Assignment Prove that a subset A of space X is closed iff limits of nets in A are in A.
A subset B of space X is open iff no net in the complement X – B can
converge to a point in B prove.
Prove that relation of homotopy is an equivalence relation.
Every Paracompact Hausdorff space is normal.
5.90 Points for Discussion: ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… 5.91 References:
5. George F. Simmons, Introduction to Topology and Modern Analysis, MC
Grow Hill book company, 1963.
6. K.D. Joshi, Introduction to General Topology. Wiley Eastern Ltd. 1983.
7. K.K.Jha, Advanced General Topology. Nav Bharat Prakashan, Delhi.
8. James R, Munkres, Topology, Pearson Education Pvt. Ltd. Delhi.
