Page 250 - DMTH503_TOPOLOGY

Topology

Notes Theorem 1: Let f be an equicontinuous sequence of functions on (X, d). Suppose that f (x)f(x)
n n
pointwise. Then f(x) is continuous.
Proof: Let xX and > 0, choose> 0 so that d(x, y) < f (x) f (y)- <  for any n.
n n 2
Then f(x) f(y) = lt f (x) f (y)
-
-
n n
n¥
 Sup f (x) f (y)- n
n
n

 < .
2
30.1.3 Statement and Proof of Ascoli’s Theorem

Statement: Let  be a closed subset of the function space C [0, 1]. Then  is compact iff  is
uniformly bounded and equicontinuous.
Proof: Let  be closed subset of the function space C [0, 1].
Step I: Let  be compact.

To prove :  is uniformly bounded and equicontinuous.
 is compact   is totally bounded
  is bounded.

Now  is a bounded subset of C [0, 1] and each member of C [0, 1] is uniformly continuous. It
means that  is uniformly bounded as a set of functions. Remains to show that  is equicontinuous.
By definition of totally bounded,  has an -net Denote this -net by . We can take

B = {f , f , …, f } s.t. for any
1 2 m
f, f  s.t. f - f <k, where k > 0
i o o i
{
where f - f = sup f(x) f (x) : x [0,1]-  }
o i o i
-
 f(x) f (x) < kx[0, 1]. …(1)
o i
Let x, y[0, 1] and f be arbitrary.

f(x) f(y) = f(x) f (x) f (x) f (y) f (y) f(y)- + - + -
-
o i o i o i o i
-
-
< f(x) f (x)- + f (x) f (y) + f (y) f(y)
o i o i o i o i
Using (1), f(x) f(y) < k + f (x) f (y)- +k …(2)
-
o i o i
f f f is uniformly continuous on [0, 1].
i o i o i o
-
  > 0 s.t. x - y <  f (x) f (y) < k …(3)
i i i i
Take = min{ ,  , …,  }. Then, by (3), we get
1 2 n
x - y <   f (x) f (y)- <  k, Using this in (2),
o i o i

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