Page 240 - DMTH503_TOPOLOGY

Topology

Notes 1
Next, cover X by finitely many balls of radius . Because J is infinite, at least one of these balls,
2 1
say B , must contain x for infinitely many values of n in J . Choose J to be the set of those indices
2 n 1 2
n for which n  J and x  B . In general, given on infinite set J of positive integers, choose J
1 n 2 k k+1
1
to be an infinite subset of J such that there is a ball B of radius that contains x for all
k  1
k k+1 n
n  J .
k+1
Choose n  J . Given n , choose n  J such that n > n ; this we can do because J is an
1 1 k k+1 k+1 k+1 k k+1
infinite set. Now for i, j  k, the indices n and n both belong to J (because J  J  ... is nested
i j k 1 2
sequence of sets). Therefore, for all i, j  k, the points x and x are contained in a ball B of
i n j n k
1
radius . It follows that the sequence x  is a Cauchy sequence, as desired.
i n
k
Theorem 8: Let X be a space; let (Y, d) be a metric space. If the subset  of (X, Y) is totally bounded
under the uniform metric corresponding to d, then  is equicontinuous under d.
Proof: Assume  is totally bounded. Give 0 <  < 1, and given x , we find a nhd U of x such that
0 0
d(f(x), f(x )) <  for x  U and f  F.
0
Set  = /3 ; Cover  by finitely many open  - balls.
B(f , ), ...., B(f , ) in (X, Y). Each function f is continuous; therefore, we can choose a nhd of x
1 n i 0
such that for i = 1 ,.., n.
d(f (x), f (x )) < 
i i 0
whenever x  U.
Let f be an arbitrary element of F. Then f belongs to at least one of the above  balls say to
B(f , ). Then for x  U, we have
i
d(f(x),f (x))   ,
i
d(f (x), f (x ) < 
i i 0
d (f (x ), f(x )) < .
i 0 0
The first and third inequalities hold because p(f,f )   and the second holds because x  U.
,
i
Since  > 1, the first and third also hold if d is replaced by d; then the triangle inequality implies
that for all x  U, we have d(f(x)), f(x0) < , as desired.

Example 4: Let E be a subspace of a metric space X. Show that E is totally bounded  E
is totally bounded.
Solution: Let E be totally bounded and  > 0 be given.

Let A = {a , a , …, a } be an net for E so that
1 2 n
2
n   
E  S a ,  …(1)

i
i 1  2 

Let y be any element of E .
Then there exists x  E such that

d(x, y) < …(2)
2

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