Topology Ii Unit-Iii Definition

TOPOLOGY II
UNIT-III
DEFINITION:-
Let X be a metric space .If h:X->Y is an isometric imbedding of X

̅̅̅̅̅̅ of Y is a
into a complete metric space Y,then the subspace ℎ(𝑥)
complete metric space .It is called the completion of X.
SEC:46
POINTWISE AND COMPACT CONVERGENCE:-
DEFN:POINTWISE CONVERGENT TOPOLOGY
Given a point xX and an open set U of the topological space

Y .Let S(x,U)={f/f𝑌 𝑋 & f(x)U}.The sets S(x,U) form a subbasis
for a topology on 𝑌 𝑋 which is called the topology of pointwise
convergence or point open topology.
NOTE:-
Since any basic open set is a finite intersection of subbasic open
sets,any basic open set in the pointwise convergent topology is of the
form ∩𝑛𝑖=1 S(𝑥𝑖 , 𝑈𝑖 ) where each 𝑥𝑖 X and each 𝑈𝑖 is open in Y.
THEOREM:46.1
A sequence (𝑓𝑛 ) of functions converges to the function f in the

topology of pointwise convergence iff for each xX,the sequence
(𝑓𝑛 (𝑥)) of points of Y converges to the point f(x).
PROOF:-
Let (𝑓𝑛 )->f in 𝑌 𝑋 with respect to pointwise convergent topology.
To prove that (𝑓𝑛 (𝑥))->f(x) in Y.
Let U be a nbd of f(x) in Y.
Then f(x)U.
 f S(x,U) which is a subbasic open set in 𝑌 𝑋 with respect to the

pointwise convergent topology.
Since (𝑓𝑛 )->f ,there exists N𝑍+ such that 𝑓𝑛  S(x,U)  n  N.
= > 𝑓𝑛 (x) U  nN.

 (𝑓𝑛 (𝑥))->f(x) in Y.
Conversely,
Suppose that (𝑓𝑛 (𝑥))->f(x) inY.
To prove that (𝑓𝑛 )->f in 𝑌 𝑋 w.r to pointwise convergent topology.
Let S (x,U) be any subbasic open set in 𝑌 𝑋 w.r to pointwise

convergent topology containing f where xX and U is any open set
in Y.
ie) f S(x,U)=>f(x)U.
since (𝑓𝑛 (𝑥))->f(x) in Y,by defn of convergence , there exists a

positive integer N such that 𝑓𝑛 (x) U  nN.
𝑓𝑛  S(x,U)  nN.
 (𝑓𝑛 )->f in 𝑌 𝑋 w.r to pointwise convergent topology.
RESULT:-
Limit of a convergent sequence of continuous functions need not be
continuous in pointwise convergent topology.
Eg:-
Consider the space 𝑅𝐼 where I=[0,1], the sequence (𝑓𝑛 ) of continuous

functions given by 𝑓𝑛 (𝑥) =𝑥 𝑛 converges in the pointwise convergent
0 𝑓𝑜𝑟 0 𝑥 < 1
topology to the function f defined by f(x)={ but this
1 𝑓𝑜𝑟 𝑥 = 1
f is not continuous.
NOTE:-
From the above example,we know that  (I,R) is not closed in 𝑅𝐼 in

the pointwise convergent topology.
Since the limit f is not continuous ie) f(I R).
DEFNITION:-
Let (Y,d) be a metric space.Let X be a topological space ,given an

element f of 𝑌 𝑋 , a compact subspace C of X and a number  >0.Let
𝐵𝑐 (f,)={g𝑌 𝑋 /sup[(f(x),g(x)/xC]<}.The sets 𝐵𝑐 (f,) form a basis
for a topology on 𝑌 𝑋 which is called a topology of compact
convergence (or sometimes the topology of uniform convergence on
compact sets).
NOTE:-
If g𝐵𝑐 (f,) then there exists a >0 such that 𝐵𝑐 (g,) 𝐵𝑐 (f,) where
=-sup{d(f(x),g(x)/xc}.
h 𝐵𝑐 (g,) =>sup{d(h(x),g(x)/xc}<=
THEOREM:46.2
A sequence (𝑓𝑛 ) from X to Y of functions converges to a function f

in the topology of compact convergence iff for each compact
subspace C of X ,the sequence 𝑓𝑛 /C converges uniformly to f/C.
PROOF:-
Suppose that (𝑓𝑛 )->f in 𝑌 𝑋 with compact topology.
To prove that:The sequence 𝑓𝑛 /C->f/C uniformly for every compact

subspace C of X.
Consider 𝐵𝑐 (f,) which is a basic open set w.r to the compact
convergence topology.
Clearly f𝐵𝑐 (f,)
By Hypothesis,the sequence (𝑓𝑛 )->f in 𝑌 𝑋 w.r to compact

convergent topology.
 There exists N𝑍+ such that 𝑓𝑛  𝐵𝑐 (f,)  nN
= > sup (d(f(x), 𝑓𝑛 (x)) <   nN and  xC.
= > (d(f(x), 𝑓𝑛 (x)) <  nN and  xC
 The sequence 𝑓𝑛 /C ->f/C uniformly.
Conversely,
Suppose that the sequence 𝑓𝑛 /C->f/C uniformly.
T.P.T:Seq (𝑓𝑛 )->f w.r to compact convergent topology.
Let 𝐵𝑐 (f,) be any neighbourhood of f w.r to compact convergent

topology where C is a compact subspace of X and >0.
f𝐵𝑐 (f,)
By Hypothesis,then seq 𝑓𝑛 \C->f\C uniformly.
There exists N 𝑍+ such that d(f(x), 𝑓𝑛 (x)) <   xC and  nN.
Sup (d(f(x), 𝑓𝑛 (x)) <   xC and  nN.
𝑓𝑛 𝐵𝑐 (f,)  nN.
The sequence (𝑓𝑛 )->f in 𝑌 𝑋 with respect to compact convergent

topology.
DEFN:COMPACTLY GENERATED
A space X is said to be compactly generated if it satisfies the

following condition:
A set A is open in X if AC is open in C for each compact subspace

C of X .
ie) AC is open in C => A is open in X.
NOTE:-
An equivalent definition for compactly generated space.

A set B is closed in X if BC is closed in C for each compact
subspace C of X.ie) BC is closed in C =>B is closed in X.
LEMMA:46.3
If X is locally compact or if X satisfies the first countability axiom

then X is compactly generated.
[X is locally compact => X is compactly generated.
X is I countable =>X is compactly generated.]
PROOF:-
Let X be a locally compact space.
T.P.T:X is compactly generated.
Let AC be open in C for every compact subspace C in X.
T.P:A is open in X.
Let xA.
Since X is locally compact and since xX by defn of locally
compactness,there exists a neighbourhood U of x and a compact
subspace C in X such that xUC.
By Hypothesis,AC is open in C.
= > (AC)U is open in U.
= > A(CU) is open in U.
= > AU is open in U. [since UC =>CU=U]
Now AU is open in U and U is open in X.
 AU is open in X.
x AUA
A is open in X.
X is compactly generated.
Let X be a I countable space.
T.P:X is compactly generated.

Let BC be closed in C for all compact subspace C in X.
T.P:B is closed in X.
ie)T.P:B=𝐵̅.
It is enough to prove that 𝐵̅ B
Let x𝐵̅.
Since X is I countable,X has a countable basis at x.
There exists a sequence (𝑥𝑛 ) in B such that (𝑥𝑛 )->x.
Let C={x}U{𝑥𝑛 /n𝑍+ }
Claim:C is compact.
Let A be a open cover for C.
Since,xC,there exists a open set 𝑈˳  A such that x𝑈˳ .
Now,(𝑥𝑛 )->x = > there exists N𝑍+ such that 𝑥𝑛 𝑈˳  nN.
For i=1,2,3…….N-1,choose an open set 𝑈 ᵢ such that 𝑥𝑖  𝑈ᵢ .

𝑈˳ , 𝑈₁ ,………. 𝑈 𝑁̯−1 form a finite subcover of A covering C.
Thus the open cover A has a finite subcover.
Hence C is compact.
Now by hypothesis, BC is closed in C.
𝑥𝑛 BC  n.[since 𝑥𝑛 B and 𝑥𝑛 𝐶̅  n]
= > 𝑥𝑛 𝐵
̅̅̅̅̅̅̅
∩𝐶
= > x ̅̅̅̅̅̅̅
𝐵∩𝐶
= > xBC
= >xB.
x𝐵̅ = >xB
𝐵̅  B
B is closed in X.
 X is compactly generated.
BY ,
I.PAVITHRA II M.Sc Maths

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TOPOLOGY II

Let X be a metric space .If h:X->Y is an isometric imbedding of X

POINTWISE AND COMPACT CONVERGENCE:-

DEFN:POINTWISE CONVERGENT TOPOLOGY

Given a point xX and an open set U of the topological space

A sequence (𝑓𝑛 ) of functions converges to the function f in the

Let (𝑓𝑛 )->f in 𝑌 𝑋 with respect to pointwise convergent topology.

To prove that (𝑓𝑛 (𝑥))->f(x) in Y.

Let U be a nbd of f(x) in Y.

 f S(x,U) which is a subbasic open set in 𝑌 𝑋 with respect to the

Since (𝑓𝑛 )->f ,there exists N𝑍+ such that 𝑓𝑛  S(x,U)  n  N.

= > 𝑓𝑛 (x) U  nN.

Suppose that (𝑓𝑛 (𝑥))->f(x) inY.

To prove that (𝑓𝑛 )->f in 𝑌 𝑋 w.r to pointwise convergent topology.

Let S (x,U) be any subbasic open set in 𝑌 𝑋 w.r to pointwise

since (𝑓𝑛 (𝑥))->f(x) in Y,by defn of convergence , there exists a

𝑓𝑛  S(x,U)  nN.

 (𝑓𝑛 )->f in 𝑌 𝑋 w.r to pointwise convergent topology.

Consider the space 𝑅𝐼 where I=[0,1], the sequence (𝑓𝑛 ) of continuous

From the above example,we know that  (I,R) is not closed in 𝑅𝐼 in

Since the limit f is not continuous ie) f(I R).

Let (Y,d) be a metric space.Let X be a topological space ,given an

A sequence (𝑓𝑛 ) from X to Y of functions converges to a function f

Suppose that (𝑓𝑛 )->f in 𝑌 𝑋 with compact topology.

To prove that:The sequence 𝑓𝑛 /C->f/C uniformly for every compact

Clearly f𝐵𝑐 (f,)

By Hypothesis,the sequence (𝑓𝑛 )->f in 𝑌 𝑋 w.r to compact

 There exists N𝑍+ such that 𝑓𝑛  𝐵𝑐 (f,)  nN

= > sup (d(f(x), 𝑓𝑛 (x)) <   nN and  xC.

= > (d(f(x), 𝑓𝑛 (x)) <  nN and  xC

 The sequence 𝑓𝑛 /C ->f/C uniformly.

Suppose that the sequence 𝑓𝑛 /C->f/C uniformly.

T.P.T:Seq (𝑓𝑛 )->f w.r to compact convergent topology.

Let 𝐵𝑐 (f,) be any neighbourhood of f w.r to compact convergent

Sup (d(f(x), 𝑓𝑛 (x)) <   xC and  nN.

𝑓𝑛 𝐵𝑐 (f,)  nN.

The sequence (𝑓𝑛 )->f in 𝑌 𝑋 with respect to compact convergent

A space X is said to be compactly generated if it satisfies the

A set A is open in X if AC is open in C for each compact subspace

ie) AC is open in C => A is open in X.

An equivalent definition for compactly generated space.

If X is locally compact or if X satisfies the first countability axiom

[X is locally compact => X is compactly generated.

X is I countable =>X is compactly generated.]

Let X be a locally compact space.

T.P.T:X is compactly generated.

Let AC be open in C for every compact subspace C in X.

= > (AC)U is open in U.

= > A(CU) is open in U.

= > AU is open in U. [since UC =>CU=U]

Now AU is open in U and U is open in X.

Let X be a I countable space.

T.P:X is compactly generated.

It is enough to prove that 𝐵̅ B

Since X is I countable,X has a countable basis at x.

There exists a sequence (𝑥𝑛 ) in B such that (𝑥𝑛 )->x.

Let C={x}U{𝑥𝑛 /n𝑍+ }

Let A be a open cover for C.