Page 99 - DMTH503_TOPOLOGY

Unit 9: The Metric Topology

Solution: Let (X, d) be a discrete metric space. Let x, y X be arbitrary. By definition of discrete Notes
metric,

 1 if x  y
d(x, y) = 
 0 if x  y
Let r be any positive real number s.t. r 1.

Then S = {y X : d (y, x) < r 1}
r (x)
= {y X : d (y, x) < 1}
= {y X : d (y, x) = 0} (by definition of d)

= {y  X : y = x} = {0}
or S = {0}
r (x)
But every open sphere is an open set.
{x} is an open set is X  x X.
If A = {x , x , ..., x } = finite set X, then
1 2 n
n
A = {x } = finite union of open sets.
r

r 1
= open set.
Hence every finite subset of X is open set. ...(1)

If B = {x , x , x , ....} X, then
1 2 3
B is an infinite subset of X.


Now B = {x }
r
r 1

= Arbitrary union of open sets
= Open set,
B is an open set. ...(2)
From (1) and (2), it follows that every subset (finite or infinite) is an open set in X.

Problem: A finite set in any metric space has no limit point.
Solution: Let A be a finite subset of a metric space (X, d). We know that “x X is a limit point of
any set B if every open sphere S contains an infinite number of points of B other than x.”
r (x)
This condition can not be satisfied here as A is finite set.
Hence A has no limit point.
Theorem 6: Let (X, d) be a metric space. A subset A of X is closed if given any x X – A, d (x, A)  0.
Proof: Let (X, d) be a metric space and A X be an arbitrary closed set.
To prove that

Given any x X – A, d (x, A) 0

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