Page 251 - DMTH503_TOPOLOGY

Unit 30: Ascoli’s Theorem

1 Notes
-
or f(x) f(y) < ¢ k, for x - y <  , f where k = .
3
This proves that  is equicontinuous.
Step II: Suppose  is uniformly bounded and equicontinuous.
To prove:  is compact.
Since C [0, 1] is complete and  is a closed subset of it and so  is complete. Hence we need only
to show that  is totally bounded.
[As we know that “A metric space is compact iff it is totally bounded and complete.”]
Given> 0,positive integer n s.t.
o
1 
-
x y <  f(x) f(y)- < f
n 5
o
for each f, we can construct a polygon arc p s.t. f - p <  and p connects points belonging
f f f
to

ì 1 2 n  ü
P = (x,y) : x = 0, , ,¼ ,1,y = ,n is an integer . ý
í
î n o n o 5 þ
Write  = {p : f  }
f
We want to show that  is finite and hence an-net for .
A is uniformly bounded.

 B is uniformly bounded.
Hence a finite number of points in  will appear in the polygonal arcs in . It means that there
can only be a finite number of arcs in , showing thereby  is an -net for  and so  is totally
bounded. Also  is complete. Consequently  is compact.
Remark: Ascoli’s theorem is also sometimes called Arzela-Ascoli’s theorem.
Theorem 2: Every compact metric space is separable.

Proof: Let (X, d) be a compact metric space.
Let m be a fixed positive number.

ì æ 1 ö ü
Let  = S x, ÷ : x X ý be a collection of open spheres.

í ç
î è m ø þ
( each open sphere forms an open set.)
Then  is clearly an open cover of X. Since X is compact and hence its open cover is reducible to
a finite sub cover say

ì æ 1 ö ü
S x ,
¢ = í ç m ÷ : i = 1, 2, ¼ , k ý
î è i m ø þ
x
Let A = { m : i = 1, 2,¼ , k } .
m i
Thus for each mN, we can construct A in above defined manner.
m

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