Page 102 - DMTH503_TOPOLOGY

Topology

Notes To show that d is metric on F.
(i) d (x, y) 0. For |x – y | 0  n
n n
(ii) d (x, y) = 0 x = y

 1 x  y
For d (x, y) = 0   n n n  0
n
n 1 2 1 x  y n

1 x  y n
n
 .  0  n
2 n 1  x  y n
n
0
 x  y    x  y  n
n n n n n
 x = y
(iii) d (x, y) = d (y, x)
For |x – y | = |y – x |
n n n n
(iv) d (x, y)  d (x, z) + d (z, y)

Here we use the fact that
   
 
1    1  1 
In view of this, we have

 x  y 1
d (x, y) =  n n . n
n

n 1 1 x  y n 2
 1 (x  z ) (z  y )

=  n . n n n n


n
n
n
n

n 1 2 1 (x  z ) (z  y )
 1 x  z  1 z  y
  n . n n   n . n n
n
n


n 1 2 1 x  z n n 1 2 1 z  y n
= d (x, z) + d (z, y)
Thus d is metric on F. The fair (F, d) is a metric space and this metric space is called Frechet space.
Example 7: In a metric space (X, d), prove that
F is closed  D (F) F.
Prove that a subset F of a metric space X contains all its limit points iff X – F is open.
Solution: Let (X, d) be a metric space and F X.
We know that F is closed X – F is open.
Step I: Let X – F be open so that F is closed,
Aim: D (F) F.

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