Page 17 - DMTH503_TOPOLOGY

Unit 1: Topological Spaces

From (1) and (2), we get Notes
x  G  N  A …(3)
x x
 x  G  A
x
which is true  x  A

 G x  A …(4)
x A

Let G = G x and an arbitrary union of open sets is open and so G is an open set.
x A
 G  A …(5) [Using (4)]

for any x  A  x  G  G  x  G  A  G …(6)
x
from (5) & (6), we get
A = G
 A is an open set.

Theorem 8: Let X be a topological space. Then the intersection of two nhds of x  X is also a nhd
of x.
Proof: Let N and N be two nhds of x  X then  open sets G and G such that
1 2 1 2
x  G  N and
1 1
x  G  N
2 2
 x  G  G  N  N
1 2 1 2
 G  G is an open set containing x and contained in N  N .
1 2 1 2
This shows that N  N is also a nhd of x.
1 2
Theorem 9: Let (y, ) be a subspace of a topological space (X, T). A subset of Y is -nhd of a point
y  Y iff it is the intersection of Y with a T-nhd of the point y  Y.
Proof: Let (y, )  (X, T) and y  Y be arbitrary, then y  X.

Step I: Let N be a -nhd of y, then
1
 V  s.t. y  V  N …(1)
1
To show: N = N  Y for some T-nhd N of y.
1 2 2
y  V   G  T s.t. V = G  Y
 y  G  Y  y  G, y  Y … (2)

Let N = N G
2 1
Then N  N , G  N …(3)
1 2 2
From (2) and (3), y  G  N where G  T
2
This shows that N is a T-nhd of y.
2
N  Y = (N  G)  Y = (N  Y)  (G  Y)
2 1 1
= (N  Y)  V = N  V = N [by (1)]
1 1 1
N  Y and V  N
1 1

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