Download Set Topology-MTH251-Lecture notes-11

Set Topology
MTH 251
• Lecture # 11 to 32
• Lecture Notes
• Introduction:
• One way to describe the subject of Topology is to say that
it is qualitative geometry.
• The idea is that if one geometric object can be
continuously transformed into another, then the two
objects are to be viewed as being topologically the same
• For example, a circle and a square are topologically
equivalent.
• Physically, a rubber band can be stretched into the form of
either a circle or a square.
• A figure eight curve formed by two circles touching at a
point is to be regarded as topologically distinct from a
circle or square
• A qualitative property that distinguishes the circle from
the figure eight is the number of connected pieces that
remain when a single point is removed
1
• When a point is removed from a circle what remains is still
connected, a single arc, whereas for a figure eight if one
removes the point of contact of its two circles, what
remains is two separate arcs, two separate pieces.
• The term used to describe two geometric objects that are
topologically equivalent is homeomorphic.
• Thus a circle and a square are homeomorphic.
• if we place a circle C inside a square S with the same
center point, then projecting the circle radially outward to
the square defines a function f :C→S , and this function is
continuous: small changes in x produce small changes in f
(x). The function f has an inverse
: S→C obtained by
projecting the square radially inward to the circle, and this
is continuous as well. One says that f is a homeomorphism
between C and S .
• One of the basic problems of Topology is to determine
when two given geometric objects are homeomorphic.
• Our first goal will be to define exactly what the ‘geometric
objects’ are that one studies in Topology.
• These are called topological spaces.
2
• Definition #1: (Topological)
• Let X be a non-empty set. A collection of subset of X is
called a topology on X, if it meets the following
requirements:
( )
and X belong to .
( ) the union of any number of sets in
belongs to ;
( ) the intersection of any two sets (and hence of any
finite number of sets) in belongs to ;
• Definition #2: (Topological Space)
• A pair (X, ), where X is a non-empty set, and is a
topology on X is called a topological space.
• The set X is called its ground set and the elements of X are
called its points.
• Members of a topology
subsets of X.
on a set X are called open
• If ∈ then is open subset of X and
closed subset of X.
−
is known as
3
• Example #1.
• Let X={a,b}. Consider the following four collection of
subsets of X:
={ , { , } }
= { , { }, { , } }
= { , { }, { , } }
= { , { }, {b }, { , } }
Then each
• Remark :
(for = 1,2,3,4) is a topology on X.
• From the above example it is clear:
• There may be more than one topologies on a set X
• therefore there exist more than one topological spaces of
which X is the ground set.
• Example #2:
• Let X={a,b,c,d}. Consider the following collection of subsets
of X.
= { }, { , }, { , , , }
=
=
, { }, { , }
, { }, { , }, { , , }, { , , , }
4
(i)
=
, { }, { }, { , , , }
is not a topology on X, because
to
does not belong
(ii)
is not a topology, because X does not belong to
(iii)
is not a topology, because { } = { , } ∩ { , , }
does
not belong to
; and so ( ) is not
satisfied.
(iv)
is not a topology, because { } ∪ { } = { , } does
not belong to ; and so ( ) is not satisfied.
• Remark:
• The above example shows that not every collection of
subsets of X is a topology on X.
• Example #3: (Co-finite topology)
• Let X be a non-empty set and let be the collection of the
null set and all subsets of X, whose complements are
finite. Then is a topology on X.
• This topology is called co-finite topology or
(The significance of
-topology
will appear in later Lectures).
5
Example #4:
• Let X be a non-empty set
• let = P(X) where P(X) is power set of X, i,e., the set of all
subsets of X.
• Then is a topology on X. This topology is called discrete
topology .
• (X, ) is called discrete topological space.
• Example # 5:
• The Co-finite topology on X is discrete if X is a finite set.
• Example # 6: (Indiscrete Space)
• For any non-empty set X,
•
= { , } is a topology on X. This topology is called
Indiscrete topology and (X, ) is called indiscrete space.
• Note:
If
1.
2.
and
∪
∩
are two topologies on X. Then:
may not be a topology on X.
is always a topology on X.
6
Example #7: (metric topology)
 Let ( , ) be a metric space. Let
subsets
of X such that, for each
consist of all those
∈ , there is an open
sphere
is a topology on X. This
( ) ⊆ . Then
topology is called the metric topology on X.
• Example # 8:
• Let
.
= {1,2,3, … } and
. Then = { ,
• Example # 9:
,
={ ,
, …,
+ 1,
+ 2, … } for
, … } is a topology on
• Let = and a collection of subsets of
which are
unions of open intervals in . Then is a topology on .
This topology is called usual topology on .
• Example # 10:
• Let =
and a collection of subsets of
which are
unions of open discs in . Then is a topology on . This
topology is called usual topology on .
• Example # 11:
• Let = { , } and = { , { }, }. Then is a topology
on X. This topology is call Sierpinski topology and (X, ) is
called Sierpinski space.
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Lecture # 12
• In this Lecture we will define
• Neighbourhood of a point in
• Coarser and Finer topologies
• And discuss results on open sets and
• Give characterizations of neighbourhood of a point
• Remember:
•
If is a topology on X, then:
• Union of any collection of open sets is again open set in X
• Intersection of finite number of open sets is again open set
in X
• If
is open set in X then
−
is closed set in X.
• Union of finite number of closed sets is a closed set in X.
• Intersection of any number of closed sets is a closed set in
X.
• Problem # 1: Let
:
→ ( ,
) be a function from a
non-empty set X into a topological space ( , ) . Further
let
be the class of inverses of open subsets of Y:
8
={
( ):
Show that is a topology on X.
• Proof.
∈
}
Open sets and Neighbourhoods
• Definition # 1. Let (X, ) be a topological space. A subset of
X is said to be open if and only if it belongs to .
The conditions ( ) − ( ) imply that
(A ) the union of any number of open sets is open;
(A ) the intersection of any two open sets (and hence of
any finite number of open sets) is open;
(A ) ∅ and X are open.
Definition # 2 (Neighbourhood of a point)
• Let (X, ) be a topological space and let
be a point of X.
• A subset N of X is called a neighbourhood of
there exists an open set
for short, x ∈
⊆ N).
such that
∈
if and only if
and U⊆ N (or,
• In other words, N is a Neighbourhood of , if and only if it
contains some open set to which belongs.
• Example # 1: Let X= { , , } and = {∅, { }, { , }, }. Find
neighbourhoods of the points of X.
9
•
Example # 2: In an indiscrete topological space, each point
has a single neighbourhood which is the ground itself.
• The following example illustrate that a neighbourhood of a
point may not be an open set.
• Example # 3: Let X = { , , , } and = { , { }, }.
Clearly the set { , } is a neighbourhood of , but it is not
open because it is not a member of .
• Example#4: Let
[ − ,
∈ . Then each closed interval
+ ] with centre at
is a neighbourhood of .
• Theorem # 1. If (X, ) is a topological space, then a subset A
of X is open, if and only if A is a neighbourhood of each of
its points.
Proof. Suppose that A is open.
• We shall show that A is a neighbourhood of each of
its points.
• Let
•
∈
⊆
, where A is open.
• It follows that A is neighbourhood of each of its
points.
• Conversely,
10
• Let A be a neighbourhood of every point belonging to
it,
• than for each x ∈ A there exists an open set
•
•
•
that x ∈
⊆ A
A = U {x: x∈ A} ⊆ U{
A ⊆ U{
A = U{
, such
:x ∈ A} ⊆ A.
:x ∈ A} and U {
:x ∈ A}
: x∈ A} ⊆ A
Since the union of open sets is open it follows that A is open.
This completes the proof.
The most important properties of neighbourhoods in a
topological space are established in the following Theorem.
Theorem # 2: Let (X, ) be a topological space. Then
(
) each x ∈ X has a neighbourhood,
(
) If A is a neighbourhood of x and A⊆B, then B is also a
neighbourhood of x,
(
) If A and B are neighbourhoods of x; then A ∩ B is also a
neighbourhood of
( ) If A is a neighbourhood of x ∈ X , then there exists a set B
such that B is also a neighbourhood of x and A is a
neighbourhood of each point of B.
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•
. ( ): Since X is open, it follows from theorem 1,
that X is a neighbourhood of each of its points.
• In other words, each point of X has at least one
neighbourhood, i.e., X itself.
(
•
): since A is neighbourhood of x,
there exists an open set
• Since B is a neighbourhood of x,
•
such that x ∈ U ⊆ A.
there exists an open set V such that x ∈
•
x∈
∩
⊆ B). Hence x ∈
and U ∩ V ⊆ A ∩ B (because
∩
⊆A∩B.
⊆ .
⊆ A and V
But U ∩ V , being the intersection of two open sets is open.
Therefore A ∩ B is neighbourhood of x.
• (
•
): Since A is a neighbourhood of x,
there exists an open set U, such that x∈ U ⊆A.
• But A⊆B so x∈ U ⊆ B.
• Hence B is a neighbourhood of x.
(
•
) Since A is neighbourhood of x,
there exists an open set B such that x∈ B⊆ A.
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• Since x is a point of B and B is open, by theorem 1, B is a
neighbourhood of x.
• If y ∈ B, then because B is open. (Theorem 1). B is a
neighbourhood of y .
• Since B⊂A therefore A is also a neighbourhood of y by
( ).
• A is therefore a neighbourhood of each point of B.
This completes the proof.
• Definition # 3.
• If
and
• then
are two topologies on X, and
⊆
is said to be coarser (or weaker) than
• alternatively
•
is said to be finer or stronger than
,
:
.
• Definition# 4. Two topologies on X are
incomparable if neither is coarser than the other.
called
• Note:
• The indiscrete topology is contained in any topology on X
and is, therefore, the coarsest topology on X.
13
• At the other extreme is the discrete topology which
contains any topology on X and is , therefore, the finest
topology on X.
• Example # 5. The topologies
and
in example below are not comparable:
on X = { , } given
={ , { , } }
= { , { }, { , } }
= { , { }, { , } }
= { , { }, {b }, { , } }
• Example # 6. Let X= { , , } and let
= {∅, { , }, }
and
= {∅, { , }, { , }, { }, } then τ and τ are
topologies on X. Since every member of
is also a
member of
.
τ is coarser than τ
⊆
τ is finer than τ
P1=page 44. Bsc book
• Theorem#3.Let
and
∩
be any two topologies on a non empty set X. Then
is also a topology on X.
14
Proof. (O ) : Let {X ∶ ∈ I} be any subcollection
of ∩ . Then X ∈
and X ∈
, for each ∈
∪
∪
I and so
X ∈
and
X ∈
because
∈ I
∈ I
∪
and are topologies on X. Hence
X ∈
∩ .
∈ I
Thus
∩
satisfies (O ).
(O ) : Let X and X be any two sets of
and X both belong to
and
.
and
∩
Then X
both being
topologies on X, satisfy (O ). Hence X ∩ X ∈
X ∩ X ∈
so X ∩ X ∈
∩ . Therefore
satisfies (O ).
(O ) : Condition (O ) for
belong to both
∩
Hence
and
∩
implies that X and ∅
. Hence X and ∅ belong to
and (O ) is satisfied.
∩
and
and
is a topology on X.
• Example # 7. Let X= {a,b,c,d} . Then each of the collections:
•
on X.
•
∪
= {∅, { , }, } and
= {∅, { }, } is a topology
is not topology on X.
15
Lecture#13.
Lecture Plan
• Neighbourhood systems
• Properties of nbd system
• Closed sets in
• Characterizations of closed sets
• Exercises
• Definition: Neighbourhood of a point
• Let
be a point in a topological space (X, ). Then the set
of all neighbourhoods of
is called the neighbourhood
system of the point and it is denoted by ( ) .
• Example #1.
• Let X= { , , } and
of ,
,
• Solution.
= {∅, { }, { , }, } find nbd systems
• Let X= { , , } and = {∅, { }, { , }, }
•
•
•
Then
( )=
( )=
( ) ={
{ }, { , }, { , },
,{ , }
}
16
• The main properties of the neighbourhood systems are
summed up in the following.
Theorem #3. Let ( , ) be a topological space and let
be the neighbourhood system of the point
∈ X. Then:
(1).
is not empty
(2) the intersection of any two members of
to
.
(3) If a subset of X contains a member of
subset also belongs to
.
•
. ( ) From ( ) it follows that
neighbourhood, namely X, Therefore
• (2) From (
belongs
then this
has at least one
is non-empty.
) it follows that the intersection of any two
neighbourhoods of x is a neighbourhood of x . Hence the
intersection of any two members of ( ) is a member of
(
).
• (3) From ( ) , it follows that if a subset A of X contains a
neighbourhood of x . Then A must also be a nbd of x . In
other words if a subset of X contain a member of ( )
then it must also a member of
(
).
17
• Definition: Closed sets
• A subset of X is said to be closed in a topological space
(X, ) if its complement w.r.t. X is open or equivalently. If it
is the complement of an open set.
• Example #3
• Let X={ , , , } and let =
closed sets in (X, ) are X, { ,
the complements of ∅,{ },{ ,
{∅, { }, { , }, }. Then the
, }, { , } and ∅, which are
} and X respectively.
• The following examples indicate that unlike doors, sets in a
given topological space may be both open and closed, or
neither open nor closed.
Example #4. In the topological space whose ground set is
X={ , , } whose topology is {∅, { }, { , }, },
• the set { } is open as well as closed , being the
complement of { , }.
•
But { } and { } are neither open nor closed.
• Example #5. In any topological space, the empty set and
the ground set are both open and closed, because both
belong to the topology and are complements of each
other.
18
• Example #6. In a discrete topological space (X, P(X)) every
set is both open and closed. Because, if A is a subset of X,
then A ∈ P(X). Also A is the complement of the set X-A,
which being a subset of X, is also open.
• Hence A is both open and closed.
• Example #7. In an indiscrete topological space, any non
empty set different from the ground set is neither open
nor closed,
• It can be easily verified.
• The collection G of all closed sets in any topological space
has three basic properties,
• These are summed up in the following theorem.
•
Theorem #5. Let (X, )
a topological space. Then the
collection G of closed sets has the following properties:
•
•
•
• Let {
( ) The intersection of any number of
closed sets is closed.
( ) The union of any two closed sets (and
hence of any finite number of closed sets is closed.
(
) X and ∅ are closed sets.
.(
:
):
∈ 1} be any number of closed sets in G
19
• We shall show that ∩
is closed.
• By proving that for each
• By De Morgan's Law (∩
• Since each
,
and
• Now
∪
• By condition (
• (
∈
is open
is closed
• Then
• Hence
,
is open.
• Hence ∩
): Let
) =∪
is closed, therefore each
• Hence ∪
• (
Complement is open.
∈
∪
be any two closed sets
are open.
=
)
∩
.
∩
is open.
is closed , as required.
): The complement of X is the open set ∅
• the complement of ∅ is the open set X.
• Hence X and ∅ are closed sets.
i.
ii.
• Q#2. Explain why each of the following collection is not a
topology on X = { , , , , , }
= {∅, { , }, { , }, X}
= {∅, { , , }, { , , }, X}
20
iii.
iv.
= { }, { , }, { , } , { , , } , X
= ∅, { }
Q#3. Explain why each of the following collections is not a
topology on the set Z of all integers:
i.
The collection of all finite subsets of Z,
ii.
The collection of all infinite subsets of Z,
iii.
The collection of all subsets of Z whose complements are
finite,
iv.
The collection of all subsets of Z of which 0 is an element,
v.
The collection of all subsets of Z of which 0 is not an
element.
Q#4. Show that following collections of subsets of X are
topologies on X.
X = { , , , } are topology on X,
= {∅, { }, { , }, X}
= {∅, { }, { }, { , }, X}
= {∅, { }, { , }, { , , }, X}
Q#5. Write down the discrete topological space of which
X= { , , } is the ground set.
Q#6. Are the topologies comparable?
21
X={ , , , }
= {∅, { }, { , }, }
= {∅, { }, { }, { , }, }
= {∅, { }, { , }, { , , }, }
Q#7. Find the neighbourhoods and the neighbourhood
systems of , , ,
for each of the topologies .
X = { , , , } are topology on X,
= {∅, { }, { , }, X}
= {∅, { }, { }, { , }, X}
= {∅, { }, { , }, { , , }, X}
Q#8. List the closed sets in each of the three topological
spaces:
={ , , , }
= {∅, { }, { , }, X}
= {∅, { }, { }, { , }, X}
= {∅, { }, { , }, { , , }, X}
Q#9. Prove that there are twenty-nine different
topologies on the set { , , }.
22
Lecture# 14.
Lecture Plan
• Interior Points
• Theorems
• Worked Examples
Definition # 1.
Let A be a subset of a given topological
space X. Then a point
•
∈
is called an interior point of
if and only if there exists a neighbourhood N of
that
∈ N ⊆ A.
• 1. Every neighbourhood N of
containing ,
such
, contains an open set
• 2. Since every open set V in X is a neighbourhood of each
of its points, and x ∈ V⊆ V, ∀ x ∈ V,
It follows that every point of an open set V is an interior
point of V.
23
•
. (Interior of a set)
• The set of all interior points of a subset A of a topological
space X is called the interior of
Int(A)
and shall be denoted by
• Example 1.
• Let R be the topological space with the usual metric
topology. Let A =[0,1[ be a subset of R.
• Then every point of A except 0 is an interior point of A.
Hence Int (A)= ] 0,1 [ which is an open interval of R.
•
.
• Let
be the topological space with its usual metric
topology. Let
• A= {( , ): ( , ) ∈
,
• Let ( , ) ∈ A such that
• Then ( , ) ∈
+
+
≤ 1} be a subset of
.
=
( , ) ⊆ A, Where r=1− ,
• This implies that ( , ) ∈ Int(A).
• Example # 3:
• Let
be the topological space with its usual metric
topology. Let
24
• A= {( , ): ( , ) ∈
,
+
• Let ( , ) ∈ A be such that
+
≤ 1} be a subset of
.
=1.
• Then it is easy to see that there is no sphere centered at
( , ), and contained in A.
• Hence, it follows that
• Int(A) ={( , ): ( , ) ∈
•
# .
,
+
• Let X={ , , } and the topology
= { , }. Find Int((A)
< 1}
=
• Example # 5. Let X = { , , , } and
For
•
= { , }. Find
X. Then
( )
, ∅, { }, { , } , Let
=
, ,{ , } .
# . Let A be any subset of a topological space
(1) Int (A) ⊆ A;
(2) Int (A) is the largest open set of X which is contained in
A
•
. ( ) Let x ∈ Int (A).
• It follows from definition, that A contains a neighbourhood
of x and hence,
∈
. Therefor we conclude that
25
Int(A) ⊆ A.
(2). To prove that Int(A) is the largest open set contained
in A, we first establish that every open set contained in
A, is contained in Int(A). (This means: Int(A) is largest open
set containing A)
 Let x∈ , where U ⊆ is an open set.
 Now being an open set, is a nbd of x and x∈
It follows that x is an interior point of A.
Thus every point of
⊆ A.
is an interior point of A
and hence U⊆ Int(A).
• Now we show that Int(A) is open,
• we shall prove that Int(A) is union of open sets of X
• Let
∈ Int (A).
• It follows from definition that there exists a nbd N( ),
(say) of such that
∈N( ) ⊆ A
……………..(1)
It follows from the definition of neighbourhood that there
exists an open set U( ). (say), such that
∈ ( ) ⊆ N(y)
……………..(2)
• From (1) and (2) we get
26
∈ U( ) ⊆ N ( ) ⊆ A
Since U( ) is an open set contained in A, it follows that
U( ) ⊆ Int(A)
Let =U ( ) ⊆A. It follows that
∈ Int(A)
Since U is open. Therefore U ⊆Int(A). Also,
Int(A)=U{ } ⊆ U ( )= . Hence we conclude that
∈ Int A
IntA=
and consequently Int(A) is open.
• Corollary 1. A subset A of a topological space X is open if
and only if Int(A) =A
. Suppose A is open.
Then A is a neighbourhood of each of its points and as a
consequence every point of A is an interior point of A.
Hence
Int(A)=A
Conversely,
If Int(A)=A, then A is open
• Corollary 2. Int(A) is the union of all the open sets
contained in A.
•
• Then
. Let {
} be family of all open sets contained in A.
⊆ Int(A) and so ∪
⊆ Int(A)
27
But Int(A) is open and Int A ⊆ A . Hence Int (A) ⊆ U
,
hence it follows that
U
=Int(A)-
•
: .
• Let X= {1,2,3,4} and = , ∅, {1}, {3}, {1,3} Let A ={1,2,3}
and B={2,4}. Then Int{A}={1,3} and Int (B)=∅
•
topology,
. Let R be the topological space with the usual
Let A=[1,2]. Then
Int (A)=]1,2[ Which is clearly open and is the
largest open set contained in A.
•
. Let A and B be any two subsets of a
topological space. Then
1. Int (Int (A)=Int (A),
2. Int (AUB) ⊇ Int (A)UInt (B),
3. Int (A∩B)= Int (A)∩Int (B),
. . Since Int (A) is open, therefore
Int (Int (A))=Int (A)
28
Proof 2.
• Let x ∈ Int (A) U Int (B).
• Then either x ∈Int (A) or x ∈ Int (B).
• For the sake of definiteness, we assume that x ∈ Int (A).
• It follows from the definition of an interior point that there
exists a neighbourhood N of x such that
•
x ∈ N ⊆ A.
• Since A ⊆ AUB, therefore, we have
•
x ∈ N ⊆ AUB
•
⇒ x ∈ Int (AUB)
•
Hence Int (AUB) ⊇ Int(A) U Int (B)
• Proof 3.
• Let x ∈ Int (A∩B).
• Then there exists a neighbourhood N of x such that
•
•
•
•
x ∈ N ⊆ A∩B
⇒ x ∈ N ⊆ A and x ∈ N ⊆ B
⇒ x ∈ Int (A) and x ∈ Int (B)
⇒ x ∈ Int (A) ∩ Int (B)
29
• Therefore Int (A ∩B) ⊆ Int (A) ∩ Int (B)
(1)
• Now let x ∈ Int (A) ∩ Int (B). Hence
• x ∈ Int (A) and x ∈ Int (B)
• ⇒ x ∈ N ⊆ A and x ∈ N ⊆ B, where
• N and N are neighbourhoods of x
• ⇒ x ∈ N ∩ N ⊆ A∩B
• ⇒ x ∈ Int (A ∩B)
• ⇒ Int (A) ∩ Int (B) ⊆ Int (A∩B)
(2)
From (1) and (2) the required result follows
. To illustrate the fact that Int (AUB) is not, in
general, equal to Int(A) U Int(B), we cite the following
example.
Let X= { , , , } and = X, ∅, { , }, { , }
Let A = { , } and B = { , }, Then it is evident that
Int (A) = ∅, Int (B) = ∅,
Int (AUB)= Int (X)=X. And Int (A) U Int (B)= ∅.
Hence Int (AUB)≠Int (A) U Int (B)
30
Lecture#15.
Lecture Plan
• Exterior Points
• Closure of a set
• Theorems
• Worked Examples
•
. (Exterior Point)
• Let A be a subset of a topological space X. A point x is
called an exterior point of if there exists neighbourhood
N of x such that
x ∈ N ⊆ X−A
. A point x ∈ X is an exterior point of A if
i.
x ∈ X−A and
ii.
There exists at least One neighbourhood of x which does
not intersect A
•
. Let X= { , , } and = X, ∅, { , }
If A= { , }, then A has no exterior point because is the
only point which belongs to X−A but there is no
neighbourhood of contained in X−A.
31
•
. The set of all exterior points of A is called
the Exterior of and we shall denote it by Ext (A)
•
. Let X be any non-empty set and = {X, ∅} be
the trivial topology on X. Then for every non-empty subset
A of X, Ext(A)= ∅
• because there is no point of X for which there exists a
neighbourhood contained in X−A
•
. Let X={ , , } be the discrete topological space
Let A = { , } Then Ext (A)= { }=X−A.
This result is true for all subsets of X (why)?
•
Let
X=
= X, ∅, { }, { , }, { , , }
.
Let
•
and
= { , }, Then verify that Ext (A)= { }
. Let R be the topological space with its usual
metric topology. Let A= ] , ]
Then Ext (A)={x: x∈ R. x<
•
{ , , , }
or x>b}
. Let R be the topological space with its usual
metric topology. Let
= ( , ): ( , ) ∈ R ,
+
≤ 4 Then
32
Ext
•
= ( , ): ( , ) ∈ R ,
.
+
>4
• Let A be a subset of a topological space X. Then the
intersection of all closed subsets of X containing A, i,e., the
smallest closed set of X containing A, is called the closure
of A, and is denoted by CI (A)
•
. Let X={ , , } and = X, ∅, { , }
Let A={ , }, The A =X because the smallest closed subset
containing A is X itself.
•
topology.
• Let
•
. Let R be the topological space with the usual
A= ( , ): ( , ) ∈ R ,
+
≤ 1 . Then A = Cl(A)
. Let X= {1,2,3,4} and let
= X, ∅, {1,2}
and
two topologies on X.
= X, ∅, {1}, {1,2}, {1,2,3}
Let A={4}. Find Cl(A) w.r.t
and
be
33
•
. Let X be a topological space. Then
1. A subset A of X is closed if and only if A=A;
2. A=A;
3. A ∪ B = A ∪ B for any two subsets A and B of X;
4. x =X, ∅ = ∅
Proof 1. Suppose A is closed. Then A itself is a member of
the family of closed sets containing A. Hence A being the
intersection of this family, is contained in A. But A ⊆ A.
Hence A = A . Conversely,, if A = A then A is closed
because A is closed.
2. This follows from (1) since A is closed.
3. By definition of the closure, A is the smallest
closed set containing A. Since
A ⊆ A∪B ⊆ A ∪ B,
It follows that A ⊆ A ∪ B because A ∪ B is a closed set
which contains A. Similarly B ⊆ A ∪ B. Hence
A∪B ⊆ A ∪ B
(i)
Since A ⊆ A , B ⊆ B it follows that
A ∪ B ⊆ A ∪ B. which is a closed set. Hence
A ∪ B ⊆A∪B
(ii)
From (i) and (ii) we get
A ∪ B = A ∪ B.
34
•
. Let A and B be any two subsets of a topological
space X such that A⊆B. Prove that A ⊆ B
. Since A ⊆B ⊆ B and B is closed, it follows that
•
A ⊆ B . Cl(B) is any closed set containing A and Cl(A) is
smallest closed set containing A. Thus Cl(A) ⊆ Cl(B)
. We give an example to show that, in general,
A∩B ≠A∩B
• Let
• Let
= { , , } and = { , , { }, { }, { , }}.
= { } and
= { , }. Then A ∩ B ≠ A ∩ B
. A point x ∈ A if and only if every open set
containing x contains a point of A.
Proof.
Let x ∈ A and be an open set containing x. Since A = A
∪
. It follows that either x ∈ A or x ∈ .
If x ∈ A, then ∩ A is non-empty because it contains x. If x
∈
, then , being an open set containing x is a
neighbourhood of x, and hence by the definition of
,
U ∩ A ≠ ∅. Therefore, U contains a point of A.
Conversely, we suppose that every open set
U containing x satisfies A ∩ U ≠ ∅. If x ∈ A, then clearly x
∈ A ⊇ A. if x ∉ A, then every open set containing x
35
•
contains a point of X−A, viz., x itself. That is to say, every
open set containing x contains a point of A and a point of
X−A. Hence x ∈
A ⊆ A. Therefore x ∈ A.
•
A point x ∈ X is called a limit point of if every open set
containing x contains a point of A other than x .
. Let A be a subset of a topological space X.
. The set of all limit points of A is called the
derived set of
•
and is usually denoted by A .
. Let X={ , , } and = {X, ∅}
Let A={ , } Then it is clear that
•
•
A =X
. Let
= X , ∅, { }, { } be a topology on X=
{ , }. Let A={ }. Then A =∅ because A has no limit
points.
. Let R be the topological space with its usual
metric topology
• let A={x: x∈ R, x = 1 +
Then clearly A ={1}
1
,
∈ N}
36
Lecture#16
Lecture Plan
• Limit point of a set
• Derived set
• Examples
• Theorem on
• Limit point
• Exercise problems
•
•
. Let A be a subset of a topological space X.
A point x ∈ X is called a limit point of if every open set
containing x contains a point of A other than x .
. The set of all limit points of A is called the
derived set of
•
and is usually denoted by A .
. Let X={ , , } and = {X, ∅}
Let A={ , } Then it is clear that
A =X
37
. Let
•
= X , ∅, { }, { } be a topology on X=
{ , }. Let A={ }. Then A =∅ because A has no limit
points.
. Let R be the topological space with its usual
metric topology
• let A={x: x∈ R, x = 1 +
•
Then clearly A ={1}
• For any
•
•
,
∈ N}
. A point x ∈ A if and only if every open set
containing x contains a point of A.
• Proof. Since
•
1
∈
∈ ,
( )
( )=
∪
∈
∪
∈
∈
for each open set
• This completes the proof.
containing ;
∩
≠
38
• Q#1.
Let
X={1,2,3,4,5,6}
and
= X, ∅, {1}, {1,2,3,4}, {2,3,4}, {1,5,6}, {5,6}, {2,3,4,5,6}
Let A = {1,2,3}, B={1,3,4},C={5,6,2} and D={1,2,4,5}. Find
the interior, exterior, closure and the derived sets for
each of A,B,C, D.
• Q#2. Let A be a subset of X. Prove that
i.
ii.
Int (X−A)= X −A;
X − A = X − Int (A).
Q#3. If A and B are any subsets of a topological space X
then prove that
A∩B ⊆A∩B
• Q#4. Let R be the topological space with its usual
topology
Let A={( , ): (
) ∈ R , x ≥ 0,
Find Int (A), Ext (A) and A .
≥ 0, +
≤ 1}
39
Lecture#17
Lecture Plan
• Bases
• subbases
•
•
•
base for
.
. A subcollection of a topology is called a
if every member of is a union of members of
.
Let
X={ , , , , }
and
let
= {∅, { , }, { , }, { , , , }, X} be a topology on X,
= { , }, { , }, ∅, X is a subcollection of , which
meets the requirement for a base .
. Let
= ∅, { }, { , }, { , , }, { , , , }, { , }, { , , }, X
be a topology on X={ , , , , }. Consider the following
two subcollections
of :
= ∅, { }, { , }, { , , , }, { , }
= ∅, { }, { , }, { , }
Each
•
form base for .
. If X is a non-empty set, then the collection of
all singletons of X is a base for discrete topology on X.
40
The following theorem gives a condition under which a
collection of sets is a base for given topology.
•
. A subcollection
of a topology is a base for
• if and only if
• for any points
exists B ∈
Proof.
belonging to an open set U∈ , there
such that x ∈ B ⊆ .
Proof. Let be a base for and let x ∈ . Where
Then = ∪ B , where each B ∈
Since x ∈
=∪B
x ∈ B , for some
Hence x ∈ B
∈
.
=
⊆∪B =
Conversely, if for each x ∈
such that x ∈ B ⊆ , then
, there exists a set B ∈
= ∪ {{x} : x ∈ } ⊆ ∪ {B : x ∈ }
But ∪ { B : x ∈ } ⊆
(1)
(2)
Hence from (1) and (2), if follows that
= ∪ {B ∶ x ∈
}
41
Hence every open set
∈ is expressile as a union of
members of . Thus is a base for .
• Note. Bases for the usual topology on the line and the
plane.
I.
Consider the usual topology on the real line R. Then a base
for this topology is the set of all open intervals.
II. Consider the usual topology on the plane R . Each of the
following collections is a base for the usual topology of the
plane:
i.
The collection
of all open discs,
ii.
The collection
of all open rectangles, bounded by the
sides, which are parallel to the co-ordinate axes.
Note: An alternate definition for Base of a topology
. A collection of subsets of a non-empty set X
is a base for some topology on X if meets the following
requirements:
(B ) if , ∈
∈ such that
∈
⊆
and
∩
∈
∩
, then there exists a set
;
(B ) ∪ = , where ∪
members of
denotes the union of all
42
•
• Let
. (Base at a point)
be any point of a topological space X
• A collection
the point
of open sets containing
is called a base at
• (or a local base at ), i.e.,
• if for every open set V to which
set B ∈ such that ∈ B ⊆ V
•
. A collection
space X is a base,
belongs, there exists a
of open sets in a topological
• if and only if
•
contains a base at each point of X.
Proof.
Let be a base for a topology on X. We shall show contains a
base at each point of x. Let x be any arbitrary point of X and let
V be any open set containing x.
∪
B ,B ∈
∈I
∪
Hence x ∈ V ⇒ x ∈
B
∈I
Since
is a basis, V =
⇒ x ∈ B for some
for some index set I.
∈I
43
But B ⊆ V. Thus there exists an open set B ∈
B ⊆ V.
Hence
contains a base at each point x ∈ X.
Conversely suppose that
such that x ∈
contains a base at each point of X.
We shall prove that is a base for the topology on X. Let be
any open set and let x ∈ . Since contains a base at the point
x, there exists some B ∈ such that x ∈ B ⊆ .
Hence
=∪{{x} : x ∈ } ⊆
But each B ⊆
∪
∈ B
Hence ∪ B ⊆
(1)
(2)
From (1) and (2), it follows that
Therefore
Thus
∪
∈ B
=
is expressible as a union of members of .
is a base and the theorem is proved.
44
•
. Let X= { , , , , } and let
= {∅, { }, { }, { }, { , }, { , }, { , }, { , , }, { , , }, { , , , }, { , , , }, X}
be a topology on X,then β , a base at point a, is { },
•
because
the
open
sets
containing
are
{ },{ , }, { , }, { , , }, { , , , } and X, { } is a subset
of each.
. The set of all open discs with Centre is a base
at w.r.t. the usual topology on the plane, because if is
any open set containing , then we can find an open disc
D with Centre at
such that x ∈ D ⊆ .
. In a discrete topological space X, the base at
each point is {{ }}. For if ∈ X, then the singleton set { },
which is open serves as a base at , because { } is a subset
of every open set which contains . Hence
={{ }}.
• Subbases:
•
. If is a topology on X, a subcollection
called a subbase for ,
of is
• if and only if
• finite intersections of members of
•
form a base for .
X={ , , , , }
and
let
= ∅, { }, { , }, { , , }, { , , , }, { , }, { , , }, X
.
Let
45
• Be a topology on X. A subbase
collection:
•
for
is given by the sub-
= { }, { , , }, { , , , } , { , , }
• Q#1. Let X={ , , , }, show that the collection
= { , }, { , } cannot be a base for any topology.
• Q#2. Find a subbase (with as few members as possible) for
each of the following topologies on X ={ , , }.
• {∅, { }, X},
• {∅, { }, { , }, X},
• {∅, { }, { , }, { , }, X}
• Q#3. Let X= { , , , , }. Find the topologies on X
generated by the following collections of subsets of X.
•
•
= { , }, { , }, X ,
= { , }, { , , }, { , } , X
46
Lecture#18
Lecture Plan
• Continuity
• Homeomorphism
•
. (continuous function)
• Let f : X→Y be a function from topological space X to a
topological space Y,
• Let x ∈ X. Then f is said to be continuous at x if and only
if for every open set V containing f(x ) in Y, there exists an
open set
•
Y={1,2} ,
.
in X such that x ∈
Let
X={ , , },
⊆f
(V).
= X, ∅, { }, { , }
and
= Y, ∅, {1}, {2} . Let f : X →Y be defined by
f(a)=1, f(b)=1, f(c)=2. Then f is continuous at x = a, x = b
but not continuous at x =c.
•
. Let X be an indiscrete topological space and let
Y be a discrete topological space. Then every function
• f : X →Y which is not a constant function, is discontinuous
at all points of X.
•
. Let X be a arbitrary topological space and let Y
be an indiscrete topological space. The every function
47
• f : X →Y is continuous at all points of X (prove)!.
•
. A function f: X → Y from a topological space X
to topological space Y is called a
(or
) if it is continuous at all points of X.
•
. Let f : (X,
) → (Y,
Then f is continuous if and only if f
∈ .
Proof.
) be a given function.
(V) ∈
Let us assume that f is continuous. Let V ∈
for every V
and let A
= f (V). Let x be any point of A. Since f is continuous and
f(x) ∈ V ∈
, it follows that there exists an open set
(say), such that
Now
f
f
x∈
⊆ f
(V)
f (V) =∪ {{ } ∶ ∈
∪{ ∶ ∈
( )} ⊆
( )} ⊆ f (V), and therefore, f (V)={ ∶ x ∈
(V) which is open because each
is open. Hence
(V) ∈ .
Conversely, we assume that f (V) ∈ for every V ∈ .
Let x be any point of X and let V be an open set in Y which
contains f(x). Since x ∈ f (V) and f (V), being an open
set by the hypothesis, is a neighbourhood of x, it follows
that there exists an open set in X such that
48
x∈
(V)
⊆ f
This implies that f is continuous at x. Since x is an arbitrary
point of X, it follows that f is continuous from X to Y.
•
Let
.
Y={ , , , },
X={1,2,3},
•
and
let
= Y, ∅, { }, { }, { , }
let f : X → Y be defined by
continuous.
•
= X, ∅, {1,2}
f(1) = f(2) = f(3) = a. Then f is
. Let X={1,2},
= X, ∅, {1}
and Y={ , },
= Y, ∅, { }, { } . Let f be given by f(1) =
Then f is not continuous.
, f(2) =
.
. Let X be any topological space and let Y be an
indiscrete topological space. Then every function from X to
Y is continuous because there are only two members of
namely, Y, ∅ whose inverse images under any f are
•
respectively X, ∅ which belongs to
•
x ∈ X. Then i is continuous.
. Let i : (X,
)→ (X,
. Let f : (X,
) →(Y,
be a base of
∈
for every
.
) be given by i (x) = x, ∀
) be a function and let
. Then f is continuous if and only if f
B∈
.
(B)
49
Proof.
Suppose f is continuous. Then f
(V) ∈
for every V ∈
Since
⊆ , it follows, in particular that f
every B ∈ .
Conversely we assume that f
Let V ∈
. Since
(B) ∈
is a base of
(B) ∈
for every B ∈
.
for
.
∪
B ,
∈
for all
∈ I.
, we have V =
for some index set I, where B ∈
∪
Therefore, it follows that f (V)=
f (B ) which is
∈
open by (O ). Hence f (Y) ∈
for every V ∈
and
consequently f is continuous.
. Let f : X →Y and g : Y →Z be two continuous
functions. Then g of : X →Z is a continuous function.
Proof. Let h= gof. To prove that h is continuous we have
to establish that for every open set W in Z, h (W) is open
in X. This fact may be established as follows:
Let W be an open set in Z. Since g: Y →Z is continuous
(given), it follows that g (W) is open in Y. Now g (W) is
open in Y and f: X →Y is continuous (given), therefore
f (g (W)) is open in X.
Since h (A)= (gof) (A)= f
(g
(A)) for all A ⊆ Z,
50
f
(g
(W))= (gof) (W) = h (W)
Therefore h (W) is open in X for every open set W in Z
and hence h=gof is continuous.
•
.
(Homeomorphism)
• Let f : X→Y be a function from a topological space X to a
topological space Y which meets the following
requirements.
(H ) f is continuous,
(H ) f is bijective (i.e f is both one-one and onto),
(H ) f is continuous , where f
function of f.
denotes the inverse
Then f is called a homeomorphism
•
•
•
. Let X= {1,2},
= X, ∅, {1}
and Y={ , },
= Y, ∅, { } . Let f : X →Y be given by f(1) = , f(2) = .
Then f is a homeomorphism between X and Y.
. Let i : X →X be the identity function on a
topological space X. Then i is homeomorphism.
.
Y={ , , },
Let
X={ , , }, i = X, ∅, { }, { , } and
= Y, ∅, { } . Let f : X →Y be given by f( ) =
51
, f( ) = , f( )= . Then it can be seen that f is continuous
and bijective.
•
. Let X and Y be two discrete topological spaces
and let f : X → Y be a bijective function. Prove that f is a
52
Lecture#19
Lecture Plan
• Subspaces
• Relative Topologies
• Note: Let X be a topological space with topology . Let A
be a non-empty subset of X. Although it is possible to
assign many different topologies to A without making any
reference to the topology , We would like to assign to A a
definite topology, which it inherits from .
• In this Lecture we shall construct such topologies and
study some of the properties of the resulting topological
space.
• Relative Topology:
• Definition. Let be a subset of a space ( , ). Then a
topology
= { ∩ : ∈ } is called a relative topology
on
and ( ,
• Example.
) is called a subspace of ( , ).
• Let = { , , , , } with
= { , { }, { , }, { , , }, { , , , }, }.
• Let
= { , , }. Then
53
•
. Let X be a topological space with a given
topology and let A be a given non-empty subset of X.
Then prove that the collection
of subsets of A, defined
as follows:
= A∩ :
Proof. In order to prove that
the following
(
∶
) : Let {A ∩
∪
. Now
(A ∩
∈I
being a topology
Hence
Thus (
∪
(A∩
∈I
∈ I } be any subcollection of
∪
)=A ∩
by distributive law
∈I
∪
=
∈
∈I
∈
)A∩ ∈
and C= A ∩
( ) : let B = A ∩
where
and
belong to .
Since
=(A ∩ (
and
(
) : Since X, ∅ ∈
A∩X=A∈
) ∩ (A ∩
∩
∈
be any two members of
.
)
)
belong to ,
Hence B ∩ C = A ∩
is a topology on A.
is a topology on A, we verify
) is satisfied.
Then B ∩ C = (A ∩
,
∈
=
∩
∈
and the condition (
and A ∩ ∅ = ∅ ∈
) is satisfied.
54
Hence A, ∅ ∈
Thus
•
and (
) is satisfied.
is a topology on A.
. Let (X, ) be a topological space and let A be
a non-empty subset of X. Then the topology
defined by
= A∩ : ∈
is called the relative topology or the
induced topology on A.
•
. A non-empty subset A of a topological space
(X, ) is called a subspace of X if and only if it has been
assigned the relative topology .
•
. Let X be a non-empty set and let be the
indiscrete topology on X. Let A be a non-empty subset of
X. Since ={∅, X},
={A ∩ ∅, A ∩ X}={∅, A}
• Thus the relative topology on A is the indiscrete topology.
• Note.
• A subspace of a discrete space is discrete
• A subspace of indiscrete space is indiscrete
•
.
• Let X={ , , , , } and let
•
= {∅, { , }, { , }, { }, { , , }, X}
55
• Find the relative topology on A={ , , } and the resulting
subspace.
•
.
• Let A be a straight line
• where ,
+
+ = 0,
are real constants
• one of the constants
is not zero.
• Let be the usual topology on the plane R of which A is a
subset. Describe the relative topology .
•
. Let be the usual topology on a real line R.
Determine whether or not the following subsets of the
subspace [1,2] are open w.r.t the relative topology on
[1,2]:
• [1, 4/3[ ,
• ]11/6, 2],
iii. ]1, 3/2].
•
. If (A, ) is a subspace of a topological space
(X, ), then a subset B of A is closed in the subspace, if and
only if B= A∩F, where F is some closed set in (X, ).
56
Proof. Suppose that B ⊆ A is closed in the subspace. Hence
B is the complement relative to A of some open set in A.
i.e., A-B=A ∩ , where ∈ we have
B =A −(A ∩
Where F=X−
II.
•
)=A− =A ∩ (X− )=A ∩ F
is a closed set of X
Conversely, let B= A ∩ F, where F is a closed set of
(X, ).
Now F=X− where U is some open set in X.
Hence = ∩ ( − )
= − = −( ∩ )
Since ∈ , ∩ ∈
Hence B, being the component of an open set of the
subspace is closed in the subspace.
. Let X={ , , , , }
• let
={∅, { , }, { , }, { }, { , , , }, { , , }, X}
topology on X.
be
a
• Find the closed sets of the subspace whose ground set is
A={ , , }.
•
.
• Let a< < < and let A=[ , ] ∪ ] , [ be considered
as a subspace of the real line R with the usual topology .
57
• Show that [ , ] and ] , [ are both open and closed in the
subspace.
58
Lecture#20
Lecture Plan
• Topological property
• Subspaces
Continuity w.r.t. spaces and subspaces
•
. Let
be a base for a topological space ( , )
• If A is a subspace of X, then prove that the collection
= {A ∩
: B ∈ } is a base for A.
Proof. To show that
= {A ∩ : B ∈ } is a base for A,
we need to prove that every open set of A can be
expressed as union of members of
.
Let
=
be any open set of the subspace A. Then
∩ for some open set U in X.
Since B is a base for X, therefore
∈ for each
.
Hence
Since
=
∈ ,
∩ (∪
∩
:
∈
Hence every open set
as union of members of
=∪
) = ∪ {( ∩
for each
.
):
:
, where
}
of the subspace A is expressible
.
59
Hence
is a base for relative topology on A. This
completes the proof.
Example.
Let
={ , , , , }
and
let
τ = {ϕ, {a, b, c}, {c, d}, {d, e}, {c}, {d}, {a, b, c, d}, {c, d, e}, X}
and let = { , { }, { }, { , }, { , }, { , , }} be a base
for . Find the base for the subspace whose ground set is
={ , , }
• Q#1. Let X be a non-empty set and let be the discrete
topology on X. Let A be a non-empty subset of X. show
that the relative topology on A is also discrete.
• Q#2.
Let
X={ , , , , }
and
let
= X, ∅, { }, { , }, { , , }, { , , , }, { , , }
be
a
topology on X. Find the relative topology on each of the
following subsets of X:
i.
ii.
iii.
iv.
A={ , , }
B={ , , , }
C={ , , , }
D={ , }
Indicate the sets which are open (closed) in the subspace so
obtained, but are not open (closed) in the given topological
space.
60
• Q#3. if A is a subspace of X and B is a subspace of A, prove
that B is subspace of X.
• Q#4. Let
be the usual topology on the real line R.
Determine whether or not the following subsets of the
subspace [1,3] are open in the relative topology:
ii.
3
[1, 2 [,
4
iii.
]1, 5/2].
i.
]1 5/7, 3],
• Q#5.
Let
X={ , , , , , }
and
let
=
{∅, { , }, { , , }, { }, { , }, { }, { , , }, { , , , }, { , , , }, { , , , , }, { , , , , }X}
be
a
topology
on
X.
Let
= { }, { }, { , }, { , , }, { , , , , } be a base for . Find a
base for each of the subspaces whose ground sets are :
i.
ii.
iii.
A= { , , }
B= { , , , }
C= { , , , }
Hence find the relative topologies
,
.
61
TOPOLOGICAL Property
• Recall that if X, Y are homeomorphic, then open sets of Y
are the images of the open sets of X and open sets of X
are the inverse images of the open sets of Y. Therefore any
property of X expressed entirely in terms of set operations
and open sets is also possessed by each space
homeomorphic to X.
•
length angle, boundedness,
Cauchy
sequences,
straightness and being triangular or circular are not
topological properties.
• Whereas limit point, interior, nbd, boundary and
first/second countability are topological properties.
Definition (Topological property)
• A topological property is a property which, if
possessed by a topological space, is also possessed by
all topological spaces homeomorphic to that space.
Example
• Let X =(-1,1) and f: X→R defined by f(x)=tan x/2.
• Then f is a homeomorphism and therefore (-1,1)≃R
• Note that (-1,1), R have different lengths,
• therefore ‘length’ is not a topological property.
62
• Also X is bounded and R is not bounded,
• therefore ‘boundedness’ is not a topological property.
Example.
• Let f : (0, ∞) → (0, ∞) defined by f(x) =1/x, then f is
homeomorphism. Consider the sequences
• (x ) = (1, 1/2,1/3, … ). Then (f(x )) = (1,2,3, … ) in (0, ∞).
(x ) is a Cauchy sequence, where (f(x )) is not.
Therefore ‘being a Cauchy sequence is not a topological
property.
• Example. Consider f : D → D defined by f(r, )=(2r, )
• Then f is a homeomorphism. Note that the area of D is
half of the area of D . Therefore ‘area’ is not a topological
property.
• Example. Being ‘triangular’ is not topological property:
• since a triangle can be continuously deformed into a circle
and conversely.
• Whereas limit point, interior point, boundary, nbd and
first/second countability are topological properties
Theorem#1. If A is a subspace of a space X, then G is open in A
if and only if G= A ∩ O, O ∈ .
63
Theorem#2. Let A be a subspace of a space X, then H is closed
in A if and only if H= A ∩ C, C is closed in X.
Proof. Let H be closed in A. Then H = A−W, where W is open in
A. Since W = A ∩ O, O ∈ , we have
H= A −(A ∩ O) = A−O = A ∩ (X –O) = A ∩ C, where C is closed in
X.
Conversely, H= A ∩ C, where C is closed in X.
Now C = X−O, O ∈ . Then H= A ∩ (X−O) = A−O = A−(A ∩O).
Since O ∈ , A ∩ O ∈
and hence H is closed in A.
Theorem#3. Let A be a subspace of a space X, then N is a nbd
of x in A if and only if N =A ∩ U , U
is a nbd of x in X.
Proof.
• Suppose
is a nbd of
• We show that
• Suppose
∩
=
is a nbd of
• Then ∃ an open set
• Also
in X.
is a nbd of
∈
∩
in
in
such that
⊆
in X s.t.
∩
in .
∈
in X,
• There exists an open set
• This gives
is a nbd of
∈
⊆
⊆
64
• Since
• Implies
∩
is open in
∩
is a nbd
(say) of
in
and conversely.
Theorem#4. Let A be a subspace of a space X, if x ∈ A and
is
a nbd base at x in X, then {A∩B: B ∈ } is a nbd base at x in A.
Proof.
• By theorem 3,
• If
•
∩
is any nbd of
=
∩ ,
• Then,
• So
• Or
=
∩
is a nbd of
in , then
is a nbd of
⊆V, for some
∩
⊆U
⊇
• This gives { ∩ :
∩
∈
∈
in X
∈
} forms a nbd at
in
Theorem#5. Let A be a subspace of a space X, and is a base
for the topology on X, then {A∩B: B ∈ } is a base for the
relative topology for A.
Proof.
• We show that each open set in can be expressed as a
union of members of { ∩ : ∈ }
• Let
∈
65
• Then
=
• Since
where
∩
∈
is a base for
• Therefore
• Thus
=∪ , for some
=
∩
=
∈
∩ (∪ ) =∪ ( ∩ ),
• This completes the proof
∈
Theorem#6. Let A be a subspace of a space X, and E ⊆ A, then
CI (E) = A ∩ CI (E), where CI (E) denotes the closure of E in
A.
Proof.
• Let
∈
∩
• Then for all
• Since
⊆ ,
• Therefore
• Or (
,
Theorem#7. Let
∩
≠
∩( ∩ )≠
∩ )∩
• This implies
( )
∈
≠
( ) and conversely.
f : X → Y
be continuous and
A ⊆ X.
Then f : A → Y is continuous, where f denote the restriction
of f to A.
66
Proof. Since f
(G) = A ∩ f (G). G ⊆ Y, we have if G ∈ ,
then f (G) ∈ . Therefore A ∩ f (G) ∈
, gives f is
continuous. This completes the proof.
67
Lecture#21
Lecture Plan
• First Countable Space
• Second Countable Space
• Definition. First Countable Space
• A space X is said to be first countable (or to satisfy the first
axiom of countability) if, there exists a countable base at
every point ∈ .
• Example.
• Let X be a discrete space. Each { } is open in X and is
contained in every open set containing
every discrete space is first countable.
. This implies that
(second Countable Space)
• A space X is said to be second countable If it has countable
base for .
Remark:
• Every second countable space is first countable.
• For if
• if
is a countable base for a space X, and
consists of members of
which contain x ∈ X,
68
• then
is a countable base at x.
• on the other hand,
• any uncountable discrete space is first countable without
being second countable.
Example
• The usual line R is second countable,
• since it has countable base consisting of open intervals
(a,b), a,b ∈ Q.
Example
• Let R have the discrete topology.
• Then any base for this topology contains singletons
• and hence cannot be a second countable space.
Example.
with the usual topology is second countable.
Theorem. The continuous open surjective image of a second
countable space is second countable.
Proof. Let f: X →Y be an onto continuous open map. We shall
prove that f( ) = {f(B): B ∈ } is a base for , where is a base
for . Let V ∈
and b ∈ V. then f (V) is open in X and if we
pick a ∈ f (b) ⊆ f (V), then for some basic open set B ∈ , a
∈ B ⊆ f (V). it follows that b = f(a) ∈ f(B) ⊆ ff (V) ⊆ V or b ∈
69
f(B) ⊆ V and thus sets f(B), B∈
completes the proof.
, form a base for
. This
• Corollary. Second countability is a topological property
Theorem. Prove that first countability is a hereditary property.
Proof Let A ⊆ X be a subspace of a space X. Then x ∈ A gives
x ∈ X. By hypothesis, X is first countable, therefore there
exists a countable base
={B : n ∈ N} at x in X. Then , =
{A ∩ B : n ∈ N} is a base at x in A. which is countable. This
completes the proof.
70
Lecture#22
Lecture Plan
• First Countable Space
• Hereditary Property
• Separation Axioms
Theorem. First countability is preserved under open continuous
surjection
OR
The continuous open surjective image of a first countable space
is first countable.
Proof.
• Definition. A property of a space X is said to be hereditary
if every subspace of X also possesses the same property.
• Definition ( −Space). If for x, y ∈ X, x ≠ y, there exists
open set U of X such that x ∈ U, y∉ U. Then X is said to
satisfy T – axiom.
• X with
− axiom is called a
− Space.
Example. Let X = { , , } and ={∅, { }, { }, { , }, X} Then
, :
, :
∈ { },
∈ { },
∉{ }
∉{ }
71
, :
∈ { }, ∉ { }
This shows that X is a T − Space.
• Definition (
−Space). If for x, y ∈ X, x≠y,
• there exist open sets U,V of X such that x ∈ U, y ∉ U and
x ∉ V, y ∈ V , then X is said to satisfy T − axiom.
• X with
− axiom is called
− space.
• Remark.
− space is a
− space but not conversely
as the following example shows:
• If X= {1,2} and ={∅, {1}, X}, then X is a T − space
• But not T − space.
Example. The cofinite space is T − space. (verify)
• Definition (
− space). if for x, y ∈ X, x ≠ y, there exist
open sets U, V of X such that x ∈ U, y ∈ V and U∩V=∅,
• then X is said to satisfy T − axiom or Hausdorff property.
• X with T − axiom is called a T − space or Hausdorff
space.
• Remark. Every
conversely
− space is a
• as the following example shows:
− space but not
72
Example. Let R be with the cofinite topology. Then R is a
T − space but not T .
• Example.
Let
X=
{ , , }
={∅, X, { }, { }, { }, { , }, { , }, { , }, X}.
and
Then X is a T − space as well as T − space.
• Remark. Note that:
•
→
→
• reverse implications do not hold in general
Theorem (Characterization of
− Space). A space X is a
T −space if and only if each singleton in X is closed.
Proof. Necessity. Let X be a T −space. We show that each
singleton {x} is closed, that is, X−{x}, simply we write X−x, is
open.
Let y ∈ X−x, x ≠ y. Then there exists an open set U such
that y ∈ U , and x ≠ Y. Then y ∈ U ⊆ X−x and hence ∪ {U ∶ y
∈ X−x} = X−x. This gives X−x is open.
Sufficiency. Suppose each {x} is closed. We show that X is a
T −space. Let x, y ∈ X, x ≠y. then X−x, X−y are open in X. Thus
x ∈ X−y, y ∉ X−y and x ∉ X−x, y ∈ X−x imply X is a T −space.
This completes the proof.
73
Theorem. The property if being a T − space,
topological property.
= 0,1, is a
Proof. Let f: X → Y be a homeomorphism and X, a T −space.
We show that Y is a T −space. For this, we show that each
singleton in Y is closed. Let y ∈ Y. Then there exists x ∈ X
such that f(x) = y or f({x}) = {y}. Since each singleton {x} in X
is closed, therefore f is closed implies {y} is closed in Y. This
gives that Y isT . This completes the proof.
74
Lecture#23
Lecture Plan
• We will continue our study on Separation Axioms
• Hereditary Property
Theorem. The property if being a T − space, = 0,1, is a
1.hereditary property.
2.Productive property.
Proof. (1)Let Y ⊆ X be a subspace of a T −space X. We show
that each singleton {y} is closed in Y or X −{y} is open in Y.
Since y ∈ X, and X is T , therefore X−{y} is open in X, Thus
y ∈ Y ⊆ X ⇒ Y ∩ (X−{y}) = Y – {y}.
Hence Y−{y} is open in Y or {y} is closed in Y. This proves
that Y is a T −space.
(2) Let X , X be T −spaces, we show that X × X is T . It
suffices to show that each {(x , x )} is closed in X × X .
since each X is T ,therefore by theorem 32.1, each {x } is
closed in X . Therefore {(x , x )} = {x } × {x } gives
CI({(x , x )})= CI({x })× CI({x }) = {x } × {x } ={(x , x )} or
CI({(x , x )}) = {(x , x )} implies {(x , x )} is closed in X ×
X . Hence by theorem 32.1, X × X is T . This completes
the proof.
75
Theorem.
(1)
A bijective closed or open image of a Hausdorff space
is Hausdorff. Hence the property of being a Hausdorff is
topological.
(2)
Hausdorff property is hereditary.
Proof. (1). Since each bijective closed map is open, therefore
the image of disjoint open nbds are disjoint open nbds and
this proves (1).
(2). Let ⊆ be a subspace of X and ,
, ≠ . Since X
is Hausdorff, there exist disjoint open nbds U, V respectively.
Then ∩ , ∩ are also disjoint open nbds of x and y
respectively. This proves that A is Hausdorff.
Theorem. If f : X → Y is a continuous injection and Y is T , Then
X is T .
Proof. Since f: X→ Y is a continuous injection, therefore
f ∶ f(X) →X is bijective and open. Since f(X) is T , by the
result “bijective open image of Hausdorff space is
Hausdorff” we get, X is T . This completes the proof.
76
Lecture#24
Lecture Plan
• Regular spaces
•
spaces
Definition. (Regular Space)
A space X is said to be a regular space if it satisfies the axiom:
(R) for any closed set A and x ∉ A, there exist disjoint open sets
U, V such that x ∈U, A⊆ V.
• Example. Let X= { , , } and ={∅, { }, { , }, X}
• then X is regular space.
• Remark. Note that X is not a
not closed.
− space. Since each
Example. An indiscrete space is regular, it is neither T nor T
is
.
• Definition ( − Space). A space is said to be a T −
Space, if it is regular and T .
• thus we have the following :
77
Theorem. Every
−space X is
.
Proof. Let a, b ∈ X, a ≠ b. Since each {x} is closed in X,
therefore {a} is closed and b ∉ {a}. X is regular implies there
exist disjoint open sets U, V such that b ∈ U, {a} ⊆ V or b ∈ U, a
∈ V. This gives that X isT . This completed the proof.
• Theorem. The following are equivalent for a space X:
1. X is regular.
2. If U is open in X and ∈ U, then there exists an open set V
such that ∈ V ⊆ CI(V) ⊆ U. Each ∈ has a nbd base
consisting of closed sets.
3. Each
∈ X has a nbd base consisting of closed sets.
• Proof. (1)
• (2)
(2)
(3).
• (2) gives that every open set containing contains (V)
which is closed nbd of . Therefore, closed nbds of forms
the nbd base.
• (3)
(1).
78
• Suppose each
sets.
has a nbd base consisting of closed
∈
• Let be a closed set s.t.
of .
∉ . Then
• By (3), there exists a closed nbd
or
•
⊆
−
• This implies
• Thus
•
∈
( )⊆
( ) and
( ) and
−
⊆
of
( − )=
−
s.t.
is an open nbd
∈
⊆
−
−
are disjoint open sets s.t.
−
• This implies that X is a regular space.
Theorem. Every subspace of a regular space (T −space) is
regular (T −space) .
Proof. Suppose X is regular and A is a subspace of X. We show
that A is regular. Suppose Y is closed in A. Then = ∩ , Z is
closed in X. Now if ∈ and ∉ , then ∉ . Then X is
regular implies there exist disjoint open sets U, V such that
, ⊆ . This gives ∩ , ∩ are disjoint open sets in A
such that ∈ ∩ , ⊆ ∩ . This proves that A is regular.
79
Lecture#25
Lecture Plan
• Characterization of Normal Spaces
• Urysohn Lemma
Definition (Normal Space). A space X is said to be normal if it
satisfies the axiom:
•
(N) for any disjoint closed sets U, V of X, there exist
disjoint open sets G, H such that U ⊆ G and V ⊆ H.
• Example. Let X= { , , } and = {∅, { }, { , }, X}.
• Then X is normal.
• Note that X is neither
nor regular.
• All discrete spaces are normal.
Theorem. T −space (normal withT ) is a regular space.
Proof. Let X be a T −space. Suppose F is a closed set of X such
that x ∉ F. By T −axiom, each {x} is closed. Therefore {x}, F are
disjoint closed sets. Since X is normal, therefore, there exist
disjoint open sets U, V such that {x} ⊆ U, F ⊆ V or x ∈ U, F ⊆ V.
This proves that X is regular. This completes the proof.
80
Theorem. The following are equivalent in a space X:
(1) X is a normal space.
(2) For each closed set F and open set H containing F, there
exists an open set G such that F ⊆ G ⊆ CI(G) ⊆ H.
Proof. (1) ⇒(2) Suppose X is normal. Also F ⊆ H, F is closed and
H is open. Then X−H is closed and F ∩ (X−H) = ∅. Since X is
normal, there exist disjoint open sets G and G such that F ⊆ G
and X−H ⊆ G . But G ∩ G = ∅ implied G ⊆ X−G . X−H ⊆ G
implies X−G ⊆ H. Where X−G is closed. Thus
F ⊆ G ⊆ CI(G) ⊆ CI(X−G ) = X−G ⊆ H or F ⊆ G ⊆ CI(G) ⊆ H.
This proves (2)
(2) ⇒ (1) Let F , F be disjoint closed sets. Then F ⊆ X−F and
X−F is open by (2), there exists an open set G such that
F ⊆ G ⊆ CI (G) ⊆ X−F
But CI(G) ⊆ X−F implies F ⊆ X−CI(G).G ⊆ CI(G) implies G ∩
(X−CI(G)) = ∅.
Thus F ⊆ G, F ⊆ X−CI(G) and G ∩ (X−CI(G)) = ∅. This proves
(1). Hence the proof.
81
Theorem. The continuous closed surjective image of a normal
space is normal.
In other words, normality is invariant under continuous
closed surjections and hence is a topological property.
Proof. Let f: X→ Y be a continuous closed surjection and X a
normal space. Let F , F be disjoint closed sets in Y. Then
f ( F ), f ( F ), are disjoint closed sets in X. Since X is
normal, there exist disjoint open sets U, V such that f ( F )
⊆ U, f ( F ), ⊆ V. Since f is closed, therefore G= Y−f(X−V),
H= Y−f(X−V) are open in Y. Then it is easy to see that G, H
are disjoint. This completes the proof.
82
Lecture#26.
Lecture Plan
• We will continue our study on normal spaces
• Hereditary of Normal Spaces
• Urysohn Lemma
Theorem. Normality is a closed hereditary property.
Proof. If C is closed in X and A, B are disjoint closed sets in C,
then, A, B are closed in X. Since X is normal, therefore there
exist disjoint open sets U, V in X such that A ⊆ U, B ⊆ V.
Then U ∩ C, V ∩ C are disjoint open sets in C such that A ⊆ U
∩ C, B ⊆ V ∩ C. This proves that C is normal.
Theorem (Urysohn Lemma). Let X be a normal space. If
F , F are any disjoint closed sets in X, then there exists a
continuous map f : X→ [0,1] with f( )=0, f(F )=1.
Proof.
Step I. F ∩ F = ∅ implies F ⊆ X−F and X−F is open.
Therefore there exists an open set U
F ⊆U
/
⊆ CI(U
/
) ⊆ X−F
/
such that
83
Step II. Now U / and X−F are open subsets of closed set F
and CI(U / ) respectively. Therefore, in the same way, there
exist open sets U
F ⊆U
/
⊆CI(U
/
and U
/
)⊆U
/
such that
⊆ CI(U
/
/
)⊆U
/
⊆ CI(U
/
) ⊆ X−F
Step III. Let d=m/2 , n=1,2,… , m=1,2,… , 2 −1. Numbers of
the form d are called dyadic rational numbers. If we continue
this process, we obtain an open set U such that
<
⇒F ⊆U
⊆CI(U ) ⊆ U
⊆CI(U ) ⊆ X− F .
Step IV. We define f : X →[0, 1] as : f(x)=0, x ∈ U and
f(x) = sup{d: x ∉ U }. It is clear that f(F ) = 0 and f(F )=1
Step V. We show that f is continuous.
Consider the family {[0, ), ( , 1],0 <
< 1},
which is a subbase for [0,1].
We simply show that f
We note that
f(x)<
⇔ x ∈ U , for some
This gives that f
Similarly, f(x) >
f
[0, ) and f
( , 1] are open .
< .
[0, ) ={x : f(x) < }= ∪ U , which is open.
⇔ x∉CI(U ), for some
( , 1] = {x : f(x) > }=
> . Then
∪
(X−CI(U )), which is open.
>
84
This proves that f is continuous . Hence the proof .
85
Lecture#27
Lecture Plan
• Compactness
• Definition
• Examples
• Theorems
Definition ( Open Cover )
•
Let X be a topological space. A class F= {F : i ∈ I} of
open subsets of X is called an open convering or cover of
X, If ∪ F =X, i ∈ I.
• A class G= {F : i ∈ I }, I ⊆ I, of open subsets of X is
called a subcover of the cover F, If ∪ F =X, i∈ I .
• It is called a finite (respt. countable ) subcover, if I is finite
(respt. countable).
• Example.
Consider the class of
=
:
∈
×
,
is the open disc in the plane with radius
1 and centre = ( , ), , ∈ . Then is a cover of ,
i.e., every point in
belongs to at least one member of
.
86
• On the other hand , the class of open discs
× } Where
cover of
. ( , )∈
={
has centre at p and radius
∗
:
∈
is not a
but not belong to any member of H.
• Definition (Compactness). A space X is said to be compact,
• if every open cover of X has a finite subcover.
• Example. The interval = (0,1) on the real line
the usual topology is not compact.
with
• Definition (Compact subset). A space X is said to be
compact, if every open cover of X has a finite subcover.
• A subset A of a space X is compact, if it is compact as a
subspace of X.
• That is , if A is compact and
⊆
Example.
∈
A⊆
∈
, Then
, where , I is finite and each F is open in X.
(1)
All finite spaces are compact and are called trivial
compact spaces.
(2)
Every cofinite space is compact. (verify)
(3)
R with the usual topology is not compact.
(4)
No infinite discrete space is compact (verify)
87
Lecture#28
Lecture Plan.
• We will continue the study of Compactness
• Union of compact subsets
• Image of compact subsets
• Hereditary property
• Finite Intersection Property
Theorem. Any finite union of compact subsets of a space X is
compact.
Proof.
Theorem. A continuous image of a compact space is compact.
Proof. Let f: X→ Y be a continuous map and X, a compact space.
We show that f(X) is compact as a subspace of Y. Let {G : i ∈ I}
be an open cover of f(X). Then each G = f(X) ∩ H , H open in Y.
Clearly F= {f (H ): i ∈ I} is an open cover of X. Since X is
compact, therefore F has a finite subcover {f (H ): i ∈ I }, I is
finite, that is to say,
f
(H ) ∪ f
(H ) ∪ …∪ f
(H ) = X.
88
Then the corresponding finite family (G , G , … , G ) is an open
cover of f(X) which is elaborated as:
H ∪ H ∪… ∪ H ⊇
ff
G
H
ff
(H )
= f(X).
∪
ff
(H )
∪… ∪
Or ∪ H ⊇ f(X). Now G = f(X) ∩ H gives
=1
∪ G
∪ … ∪ G = f(X) ∩ ( ∪ H ) ⊇ f(X) ∩ f(X) = f(X)
=1
∪ G ⊇ f(X). This gives that {G , j= 1, 2, … , n} is a finite
=1
open cover of f(X). This completes the proof.
Or
Theorem (Closed Hereditary Propety)
A closed subspace of a compact space is compact.
Proof. Let A be a closed subspace of a compact space X and {G
= i ∈ I} an open cover of A. Then G = A ∩ H , i ∈ I, where H is
open in X. Clearly {X−A, H : i ∈ I}
Is an open cover of X.
Since X is compact, therefore X has a finite open cover {X−A,
H , i ∈ I }, I is finite. It follows that the corresponding finite
class {G = i ∈ I } is a finite open cover of A. This proves that A is
compact. Hence the proof.
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Example.
(1)
All finite spaces are compact and are called trivial
compact spaces.
(2)
Every cofinite space is compact. (verify)
Example.
(1)
All finite spaces are compact and are called trivial
compact spaces.
(2)
Every cofinite space is compact. (verify)
(3)
R with the usual topology is not compact.
(4)
No infinite discrete space is compact (verify)
Theorem. Any finite union of compact subsets of a space X is
compact.
Proof.
90
Lecture#29
Theorem. Let be a subspace of . Then is compact if
and only if every covering of by sets open in contains a
finite subcollection covering .
• Note. Remember that:
• Every class of closed sets with f.i.p has a non empty
intersection
• is equivalent to
• Every class of closed sets with empty intersection has a
finite subclass with empty intersection
Definition (Finite Intersection Property).
• Let F= F : i ∈ I be a class of subsets of space X.
• Then F is said to have a finite intersection property,
if every finite subclass F , F , … , F
of {F } has a
nonempty intersection, that is, F ∩ F ∩ … ∩ F ≠ ∅.
• Example. Let X= R and
• F={…,(−∞, −2), (−∞, −1), (−∞, 0), (−∞, 1), (−∞, 2),…}
is a class of open intervals.
• Then F has a finite intersection property.
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• Example. Let X = R and F = {(0,1), (0, ½), (0,1/4), …}.
• Then F has a finite intersection property.
Theorem (Characterization of Compactness).
following are equivalent in a space :
The
(1) X is compact
(2) Every class of closed sets with empty
intersection has a finite subclass with empty intersection.
Proof: (1) ⇒ (2) Suppose {F } is a class of closed sets with ∩
F = ∅. Then by De-Morgan’s law
X = X−∅ =X− ∩ F = ∪ (X−F )
This implies {X−F } is an open cover of X. Since X is compact
therefore X has a finite open cover {X−F , X−F , … X − F },
that is, ∪ (X− F ) = X. Again by De-Morgan’s Law
i=1
This gives (2).
∩ F = X−X = ∅.
j=1
(2) ⇒ (1) is similar and hence is left for the reader.
Theorem (Characterization). The following are equivalent
in a space X :
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(1) X is compact.
(2) Every class of closed sets having finite
intersection property has a nonempty intersection.
Proof. Proof is similar as above.
Theorem A space X is compact, if every basic open cover
of X has a finite subcover.
Proof. Let {G : i ∈ I} be an open cover of X and = : j ∈ J
an open base for . Then each G is the union of some B ′s.
Then class of all such B ′s gives a basic open cover of X. By
hypothesis, this class has a finite subcover
. then for
each
∈ , B ⊆ G , for some i ∈ I. The class of all such G ′s
gives us a finite open cover of X. This proves that X is compact.
Hence the proof.
• Example . Let X= { , , , } Then each of the collections
= {∅, { , }, X} and
= {∅, { }, X} is a topology of X.
• But
∪
= {∅, { , }, { }, } is not a topology on X,
because it does not meet the requirement ( ), for { , },
• { }∈ ∪
to ∪ .
but { , } ∪ { } = { , , } does not belong
93
Lecture#30
Theorem. If X is a Hausdorff space and C, a compact subspace
of X such that ∉ then and can be separated by open
sets.
Proof Let y ∈ C. Then x ≠y and since X is Hausdorff, there exist
disjoint open sets G , H such that x ∈ G , y ∈ H . Thus for
each y ∈ C, we obtain an open set H , such that C ⊆ ∪ U , y ∈
C. since C is compact, therefore there exists a finite open cover
for C, that is
H ,H ,… H
such that C ⊆
Correspondingly, we get G , G , … G
contain x. put
∪ H .
=1
∩ G
=1
= G and ∪ H = H. Then x and C can be separated by open
=1
sets G and H. This completes the proof.
Theorem. Every compact subspace of a Hausdorff space is
closed.
Proof Let C be a compact subspace of a Hausdorff space X. We
prove that C is closed, that is, X−C is open. Let x ∈ X−C.
By theorem 37.1, x and C can be separated by open sets G and
H. Thus x ∈ G ⊆ X−C implies X−C is the union of open sets and
hence is open. This completes the proof.
94
Theorem. Let : → be a bijective continuous function . If
is compact and is Hausdorff, then is homeomorphism.
Proof Let f: X → y be a bijective continuous map and X, a
compact space and Y a Hausdorff space. It suffices to show that
f is closed. Suppose F ⊆ X is closed, then by theorem 36.5, F is
compact and by theorem 36.3, f(F) is compact. Since a compact
subspace of a Hausdorff space is closed, therefore f(F) is closed.
This proves that f is closed. Hence the proof.
Definition. (Locally Compact)
• A space X is said to be locally compact at x ∈ X
• If there is some compact subspace Y of X that contains a
nbd of x.
• If X is locally compact at each of its points, then X is called
a locally compact space.
Renark. Every compact space is locally compact but converse is
not true in general.
• Example.
• The real line
• The point
is locally compact.
lies in some interval ( − ,
+ ), which in
tern is contained in the compact subspace { − ,
+ }
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Theorem. Prove that every compact space is locally compact.
Proof.
Theorem. Let : → be an open continuous surjection. If
is locally compact then prove that is locally compact.
Proof. Let y ∈ Y. Then there is x ∈ X such that f(x) = y. Let U be
a compact nbd of x in X, since X is - compact. Also f(x) ∈ f(U )
⊆ f(U ) gives f(U ) is a nbd of f(x), since f is open implies f(U )
is open in Y. since f is continuous, therefore by theorem 36.4,
f(U ) is compact. So each y ∈ Y has a compact nbd. This proves
that Y is - compact. Hence the proof.
96
Lecture#31
Lecture Plan
• Connected Spaces
• Connected subsets
• Some spaces are in a sense ‘disconnected’, being the union
of two or more completely separate subspaces.
• For example the space X ⊂ R consisting of the two
intervals
• A = [0, 1] and B = [2, 3] should certainly be disconnected
Definition.
• A space X is connected if it is not the union of two
nonempty disjoint open sets.
• In other words,
• if X is disconnected, then there exist disjoint nonempty
open sets A, B such that
A ∪ B= X.
• The pair (A, B) is called a disconnection of X.
97
Examples
(1) (0,1) – {1/2} is disconnected.
(2) {x} ⊆ R is connected.
(3) Every indiscrete space is connected.
(4) Every discrete space with more than one point is
disconnected.
Example
• Let X= {0,1} and = ∅, {0}, X .
• Then X is connected, called Sierpinski space.
Example
• Let X ={ , , } and
• and
= {∅, { }, { , }, X}
= {∅, { }, { }, { , }, X}.
• Then (X,
) is disconnected and (X,
) is connected.
Theorem (characterization of Connected Space)
In a space X, the following are equivalent :
(1) X is connected.
(2) The only open and closed subsets of X are ∅, X.
(3) There does not exist a continuous map
98
f : X → {0,1}
from a space X onto the discrete space {0,1}.
Proof ~(2) ⇒ ~(1).Suppose A ⊆ X is both open and closed,
and A≠ ∅, A ≠X. Then X = A ∪ (X−A) gives a disconnection of X.
~(3) ⇒ ~(2). Suppose there is a continuous map f: X → {0,1}
from a space X onto the discrete space {0, 1}.Then f (0} and
f {1} are open and X = f {0} ∪ f {1}. Thus f {0}, f {1}
are the nonempty open and closed subsets of X.
~(1) ⇒ ~ (3) If X = A ∪ B, A, B are nonempty disjoin open sets.
Then define
g : X → {0,1} as :
g(x) =
0, x ∉ A
1, x ∈ A.
Then g is continuous surjective. This completes the proof.
Theorem. The continuous image of a connected space is
connected.
Proof supposes f: X → Y is continuous and X is connected. We
show that f(X) is connected. Suppose f(X) is disconnected. Then
there is continuous surjection g: f(X) → {0, 1} (discrete). Then
gof: X → {0, 1} is also continuous surjection.
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This shows that X is disconnected, a contradiction. This
completes the proof.
Definition
• A subset A of a space X is disconnected if there exist open
subsets G, H of X such that ∩ ≠ and
• (1) ( ∩ ) ∪ ( ∩ ) =
• (2) ( ∩ ) ∩ ( ∩ ) =
Theorem. If f : X →Y is continuous and A, a connected subset
of X, then f(A) is a connected subset of Y.
Proof Suppose f(A) is not connected. Then there exist
nonempty open sets U, V in Y such that f(A) ⊆ U ∪ V, U ∩ V ⊆
Y−f(A) and U ∩ f(A) ≠ ∅, V ∩ f(A) ≠ ∅. Since f is continuous,
therefore f (U), f (V) are open in X. But A ⊆ f f(A) ⊆
f (U ∪ V) = f (U) ∪ f (V) or A ⊆ f (U) ∪ f (V) … (I)
Also f (U) ∩ f (V) = f (U ∩ V) ⊆ f
f (Y)− f f(A) = X−f f(A) ⊆ X−A
Or f
(U) ∩ f
(V) ⊆ X−A
It is easy to verify that f
(Y-f(A)) =
(U) ∩ A ≠ ∅, f
…
(II)
(V) ∩ A ≠ ∅.
From (I) and (II), it follows that A is not connected, a
contradiction. Thus f(A) is connected. This completed the proof.
100
101
Lecture#32
Lecture Plan
• Components of Connected Spaces
• Totally Disconnected subsets
• Locally Connected Spaces
• Path Connected Spaces
• Definition (component)
• A component E of a topological space X is a maximal
connected subset of X;
• that is, E is connected and E is not a proper subset of any
connected subset of X.
• Example
• If X is connected, then X has only one component: X itself
• Consider the following topology on
•
={ , , , , }
= { , , { }, { , }, { , , }, { , , , }}
• The components of X are { } and { , , , }
• Any other connected subset of X, such as { , , } is a
subset of one of the components.
102
• Example
• If X is connected, then X is itself a component.
• Let X be a discrete space. Then each singleton is a
component
• Let
X={ , , , , }, = {∅, { }, { , }, { , , }, { , , , }, X}
• Then connected subsets of X are
{ }, { }, { }, { , , } , { , , , }
• Thus the component of X are :
•
{ }, { , , , } .
• Definition (Totally Disconnected)
• A space X is totally disconnected,
• if for each x, y ∈ X,
• x ≠y, there exist disjoint open sets G, H
•
such that x ∈ G, y ∈ H and X= G ∪ H.
• Example
• One point space is totally disconnected
•
Discrete space is totally disconnected
103
• R with the upper limit topology generated by open- closed
intervals (a,b] is totally disconnected. For, if x, y ∈ R, x < y,
then (−∞, x]= G, (x, ∞)= H are such that R= (−∞,x] ∪ (x,
∞) and x ∈ G, y ∈ H,
• R with the usual topology is not totally disconnected.
• Theorem. The component of a totally disconnected space
are singletons.
• Proof. suppose a component C of X is not singleton.
• Let x, y ∈ C, x ≠ y.
• Since X is totally disconnected,
• therefore there exist a disconnection of X,
• that is, X = G ∪ H, x ∈ G, y ∈ H.
• Consequently, C ∩ G and C ∩ H are open and disjoint
subsets of C such that C= (C ∩ G) ∪ (C ∩ H) and x ∈ C ∩ G,
y ∈ C ∩ H.
• This shows that C is disconnected,
• a contradiction. This proves that C is singleton.
• Hence the proof.
• We know that
104
• every totally disconnected space is Hausdorff, but the
converse is not true in general .
• Definition (Locally Connected Space)
• A space X is said to be locally connected at x ∈ X ,
• if for every open set U containing x,
• there exists a connected open set V such that x ∈ V ⊆ U.
• A space is said to be locally connected,
• if it is locally connected at each of its points.
• A subset A of a space X is called locally connected,
• if it is locally connected as subspace.
• Example. Every discrete space X is locally connected.
• Definition. Paths
• Let = [0,1], the closed unit interval.
• A path from a point to a point in a topological space X
is a continuous function : → X with
•
(0) =
and (1) = .
• A is called initial point of the path
• B is called terminal point of the path
•
105
• Example
• For any
•
∈ , the function : →
defined by
is a constant function and hence a path.
( )=
• It is called constant path.
• Let : →
• then g: →
be a path from
to .
defined by ( ) = (1 − )
• g is a path from
to .
• Definition. Path Connected Space
• A space X is path connected,
• if each pair of points can be joined by a path in X.
• Theorem. Let X be a space and ∈ . Then X is path
connected if and only if each ∈ can be joined to by a
path in X.
• Proof.
• Let X be path connected.
• By definition, each
X.
• Converselly
• Let ,
∈
∈
can be joined to
by a path in
106
• Let : →
• g: →
be a path from
a path from
• Then h: →
•
to
and
to
defined by
(2 ), 0 ≤ ≤ 1 2
(2 − 1), 1 2 ≤ ≤ 1
( )=
• Is a path from
to
in X.
• This completes the proof.
• Theorem. Every path connected space is connected.
• Proof.
• Suppose a path connected space X.
• We shall prove that X is connected.
• Assume contrary
• Suppose X is disconnected.
• Let = ∪ , where G and H are non empty open sets
form disconnection for X.
• Let : →
•
∈ ,
• Then
∈
be a path in X from a to b in X such that
( ),
( ) form a disconnection for I,
107
• a contradiction to the fact that [0,1] is connected
• This proves that X is connected.
• This completes the proof.
End.
108