Page 241 - DMTH503_TOPOLOGY

Unit 28: Compactness in Metric Spaces

Notes
  
x  E  x  S a ,  for some i, 1  i  n by (1)

 i 2 

 d(x, a ) < (1  i  n) …(3)
i
2
Hence d (y, a )  d (y, x) + d (x, a)
i i
 
< + =  by (2) and (3).
2 2
 y  S (a , ) (1  i  n)
i
Thus y  E  y  S (a , ) for some i, 1  i  n.
i
n
 E  S a ,  
 i

i 1
A = {a , a , …, a } is an -net for E
1 2 n
 E is totally bounded.
Conversely, let E be totally bounded. Then since E  E , E is totally bounded since every
subspace of a totally bounded metric space is totally bounded.

Example 5: Let A be a compact subset of a metric space (X, d). Show that for any B  X
there is a point p  A such that
d (p, B) = d (A, B).

Solution: By the definition, we have
d (A, B) = inf {d (a, b) : a  A, b  B}.
Let d (A, B) = .
  = inf {d (a, b) : a  A, b  B}  d (a, b),
a  A, b  B being arbitrary which follows that

 n  N, a  A and b  B such that
n n
1
  d (a , b ) <  + .
n n n
Since A is compact, it is also sequentially compact and so the sequence a  has a subsequence
n
a i n which converges to a point p  A.

We claim that d (p, B) = 
Let, if possible, d (p, B) > 
Let d (p, B) =  +  where  > 0

Since a i n converges to p there must exist a natural number n such that
0

d p,a n  <
2
0
1
and d a , b n  <  
 n
n 0
0 0

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