An Alternate Proof of Ascoli-Arzela
by greenturtle3141, Apr 13, 2024, 4:02 AM
Reading Difficulty: 4.5/5
Prerequisites: Understand the title
I make no claim that this is superior to the "Bolzano-Weierstrass + diagonalization" proof.
Theorem (Ascoli-Arzela): Let
be a compact metric space. Then a family 
is precompact iff its is pointwise bounded and equicontinuous.
Proof.
Exercise.
We approach this by showing that 
is totally bounded. Fix 
.
Note that is actually uniformly equicontinuous (why?). So pick 
such that 
for all 
with 
and for all 
.
Find a finite cover of
with 
balls 
of radius . For each tuple 
of real numbers, construct a continuous function 
satisfying 
for all 
as follows:
Where 
is some decreasing continuous "weighting function" with the values 
and 
(and 
for 
). Something like 
would work. But whatever. The whole point of this construction is that:
Now for each
find 
for which 
, let 
be the set of all multiples of 
in the interval ![$[-M_i,M_i]$](//latex.artofproblemsolving.com/d/6/2/d62a7705c75f831001b7c630102851127db10ac9.png)
, and define the family
Note that 
is a finite subset of 
. We are done if we can show that any element of is within 
of some member of .
Indeed, take and for each , find 
for which 
. We claim that 
. Well, for any 
, we can find two points among 
, 
and 
, such that
But 
, and ditto for 
. So in fact,
We're done. 
Prerequisites: Understand the title
I make no claim that this is superior to the "Bolzano-Weierstrass + diagonalization" proof.
Theorem (Ascoli-Arzela): Let
be a compact metric space. Then a family 
is precompact iff its is pointwise bounded and equicontinuous.Proof.
Exercise.
We approach this by showing that 
is totally bounded. Fix 
.Note that
is actually uniformly equicontinuous (why?). So pick 
such that 
for all 
with 
and for all 
.Find a finite cover of
with 
balls 
of radius . For each tuple 
of real numbers, construct a continuous function 
satisfying 
for all 
as follows:
Where 
is some decreasing continuous "weighting function" with the values 
and 
(and 
for 
). Something like 
would work. But whatever. The whole point of this construction is that:
is continuous
is always expressed as a weighted average of the values 
for which 
. In particular, among all , there must be two such points and such that 
. Ultimately this is the property that we'll be exploiting.
is continuous
is always expressed as a weighted average of the values 
for which 
.
.Now for each
find 
for which 
, let 
be the set of all multiples of 
in the interval ![$[-M_i,M_i]$](http://latex.artofproblemsolving.com/d/6/2/d62a7705c75f831001b7c630102851127db10ac9.png)
, and define the family
Note that 
is a finite subset of 
. We are done if we can show that any element of is within 
of some member of .Indeed, take
and for each , find 
for which 
. We claim that 
. Well, for any 
, we can find two points among 
, 
and 
, such that
But 
, and ditto for 
. So in fact,
We're done. 
March 2025
November 2024