Page 239 - DMTH503_TOPOLOGY

Unit 28: Compactness in Metric Spaces

In this event (1) takes the form Notes
n
X  S ...(2)
(  i x )

i 1
Let {G : i  } be an open cover of X which is known to be sequentially compact so that, by
i
theorem (lebesgue covering lemma),  a Lebesgue number, say,  for the cover {G} . Set  = 3.
i i  
The diameter of an open sphere of radius r is less than 2r.


i.e., d(S (x )) < 2 = 2.  
 i 3
 d(S (x ) < 
 i
By definition of Lebesgue number,  an open set
G  {G : i  } s.t. S (x )  G for 1  k  n.
ik i  k ik
n n
From which we get S (  k x )  G ik
k 1 k 1


n
On using (2), X  G ik ...(3)

k 1
But X is a universal set,

n
G  X ...(4)
ik
k 1

n
Combining (3) and (4), we get X = G .
ik

k 1
This implies that the family {G : 1  K  n} is an open cover of X.
ik
Thus the open cover {G : i  } of X is reducible to a finite the subcover {G : 1  K  n} showing
i ik
thereby X is compact.
Sequentially compact  compact  Countably compact

Theorem 7: A metric space (X, d) is compact iff it is complete and totally bounded.
Proof: If X is a compact metric space then X is complete. The fact that X is totally bounded is a
consequence of the fact that the covering of X by all open  - balls must contain a finite
subcovering.
Conversely, Let X be complete and totally bounded.

To prove: X is sequentially compact.
Let <x > be sequence of points of X. We shall construct a subsequence of <x > i.e. a Cauchy
n n
sequence, so that it necessarily converges.
First cover X by finitely many balls of radius 1. At least one of these balls, say B , contains x for
1 n
infinitely many values of n. Let J be the subset of Z consisting of those indices n for which
1 +
x  B .
n 1

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