Page 234 - DMTH503_TOPOLOGY

Topology

Notes Thus, x is the limit point of B.
 x is a limit point of A, as B  A.
Hence X has the Bolzano Weierstrass Property.
Conversely, let X has the Bolzano Weierstrass Property. Let x  be a sequence in X. Let A be the
n
range of x . If A is infinite, then there is some term of x  which is infinitely repeated and that
n n
gives us a convergent subsequence of x . If A is infinite then by our assumption the set A has a
n
limit point, say x.
Since A is infinite and x is a limit point of A, therefore there exists a subsequence x  of x 
n n
k
such that x  x.
n
k
Thus proves that X is sequentially compact.
Theorem 2: Every compact metric space has the Bolzano Weierstrass Property.
Proof: Let X be a compact metric space.
To prove: X has Bolzano Weierstrass Property.
Let A be an infinite subset of X. Suppose that A has no limit point. Then to each x  X, there exists
an open sphere S which contains no other point of A other than its centre x.
x
Thus, the class {S } of all such open spheres is an open cover of X.
x
But X is compact, therefore its open cover is reducible to a finite subcover say

{S : i  1,2,...,n}, so that
i x
n
A  S .
i x
i 1

Each S contains no point of A other than its centre x , i = 1,2,...,n
i x i
 A = {x , x ,...,x }
1 2 n
 A is finite.
This contradicts the fact that A is infinite.
Hence A must have a limit point.

Thus, the compact metric space X has BWP.
Theorem 3: A compact metric space is separable.
Proof: Let (X, d) be a compact metric space.
To prove that (X, d) is separable.
Fix a positive integer n.

Each open sphere forms an open set.
Consider the family {S : x  X}
(x, 1/n)
Clearly it is an open cover of X which is known to be compact.

Hence this cover must be reducible to a finite sub cover, say
{(S x nr ,1/n) : r  1,2,.....,K }
n

228 LOVELY PROFESSIONAL UNIVERSITY