Page 141 - DMTH503_TOPOLOGY

Unit 14: Limit Point Compactness

(Otherwise X could be covered by a single -ball), choose x to be a point of X not in B(x , ). In Notes
2 1
general, given x , ...., x , choose x to be a point in the union
1 n n+1
B(x , )  ..... B (x , )
1 n
using the fact that these balls do not cover X. Note that by construction d (x , x )  for i = 1, ..., n.
n+1 i
Therefore, the sequence (x ) can have no convergent subsequence; in fact, any ball of radius /2
n
can contain x for at most one value of n.
n
Finally, we show that if X is sequentially compact, then X is compact. Let  be an open covering of
X. Because X is sequentially compact, the open covering  has a Lebesgue number . Let = /3; use
sequential compactness of X to find a finite covering of X by open - balls. Each of these balls has
diameter at most 2/3, so it lies in an element of . Choosing one such element of  for each of
these -balls, we obtain a finite subcollection of  that covers X.

Example 2: Prove that a continuous image of a sequentially compact set is sequentially
compact.
Solution: Let (X, T) be a sequentially compact topological space so that every sequence x  in X
n
has a convergent subsequence x : K  N and let this subsequence converge to x , i.e.,
k i 0 i
x  x  X.
K i 0 i
Let f : (X, T)  (Y, U) be a continuous map.
To prove that f (X) is sequentially compact set.

f is continuous map  f is sequentially continuous
Furthermore x  x .
K i 0 i
This implies that    f x
  .
f x
K i 0 i
Showing thereby f (X) is sequentially compact.
Example 3: A finite subset of a topological space is necessarily sequentially compact.

Solution: Let (X, T) be a topological space and A  X be finite and x  be a sequene in A so that
n
x  A  n. Also x  contains infinite number of terms. It follows that at least one element of A,
n n
say x must appear infinite number of times in x . Thus x , x , x , … is a subsequence of x  and
0 n 0 0 0 n
this subsequence converges to x  A, showing thereby A is a sequentially compact.
0
Theorem 3: A metric space is sequentially compact iff it has the Bolzano Weierstrass Property.
Proof I: Let (X, d) be a sequentially compact metric space. To prove that (X, d) has Bolzano
Weierstrass Property,

Let A  X be an infinite set.
If we show that A has a limit point in X, the result will follow.
A is an infinite set  A contains an enumerable set, say {x : n  N}
n
 x  A, n  N is a sequence with infinitely many distinct points.
n
By the assumption of sequential compactness, the sequence x  has a convergent subsequence
n
x : n  N (say). Let this convergent sequence x : n  N converge to x . Then x  X and x 
in in 0 0 n
also converges to x , i.e., x  x . Consequently x is a limit point of the set {x : n  N}.
0 n 0 0 n
Evidently {x : n  N}  A.
n

LOVELY PROFESSIONAL UNIVERSITY 135