Page 140 - DMTH503_TOPOLOGY

Topology

Notes
Example 1: Let Y consist of two points; give Y the topology consisting of Y and the empty
set. Then the space X = Z × Y is limit point compact, for every non-empty subset of X has a limit
+
point. It is not compact, for the covering of X by the open sets U = {n} × Y has no finite
n
subcollection covering X.
14.1.2 Sequentially Compact

Let X be a topological space. If (x ) is a sequence of points of X, and if
n
n < n < ... < n < ....
1 2 i
is an increasing sequence of positive integers, then the sequence (y ) defined by setting y = x is
i i ni
called a subsequence of the sequence (x ). The space X is said t be sequentially compact if every
n o
sequence of points of X has a convergent subsequence.
Theorem 2: Let X be a metrizable space. Then the following are equivalent:
1. X is compact
2. X is limit point compact
3. X is sequentially compact
Proof: We have already proved that (1) (2). To show that (2) (3), assume that X is limit point
compact. Given a sequence (x ) of points of X, consider the set A = {x n  Z }. If the set A is finite,
n n +
then there is a point x such that x = x for infinitely many values of n. In this case, the sequence
n
(x ) has a subsequence that is constant, and therefore converges trivially. On the other hand, if A
n
is infinite, then A has a limit point of x. We define a subsequence of (x ) converging to x as
n
follows.
First choose n so that
1
x  B (x, 1)
n 1
Then suppose that the positive integer n is given. Because the ball B (x, 1/i) intersects A in
i – 1
infinitely many points, we an choose an index n > n such that
i i – 1
x  B (x, 1/i)
ni
Then the subsequence x , x , ..., converges to x.
n1 n2
Finally, we show that (3) (1). This is the hardest part of the proof.
First, we show that if X is sequentially compact, then the Lebesgue number lemma holds for X.
(This would follow from compactness, but compactness is what we are trying to prove.) Let  be
an open covering of X. We assume that there is no > 0 such that each set of diameter less than
 has an element of  containing it, and derive a contradiction.
Our assumption implies in particular that for each positive integer n, there exists a set of diameter
less than 1/n that is not contained in any element of ; let C be such a set. Choose a point x C ,
n n n
for each n. By hypothesis, some subsequence (x ) of the sequence (x ) converges, say to the point
ni n
a. Now a belongs to some element A of the collection ; because A is open, we may choose an
> 0 such that B (a, ) A. If i is large enough that 1/n < /2, then the set C lies in the 
i ni 2
neighborhood of x ; if i is also chosen large enough that d (x , a) < /2, then C lies in the
ni ni ni
-neighborhood of a. But this means that C  A, contrary to hypothesis.
ni
Second, we show that if X is sequentially compact, then given > 0, there exists a finite covering
of X by open -balls. Once again, we proceed by contradiction. Assume that there exists an > 0
such that X cannot be covered by finitely many -balls. Construct a sequence of points x of X as
n
follows: First, choose x to be any point of X. Noting that the ball B (x , ) is not all of X
1 1

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