Rudin's proof of Arzela-Ascoli teorem

[Epsilon36819]

This is little Rudin's proof 7.25: If K is compact, f_n complex and continuous functions and if {f_n} is pointwise bounded and equicontinuous on K, then {f_n} contains a uniformly convergent subsequence.

The proof requires a countable dense subset of K, called E. Then by compactness (and density?), there exists finitely many x_i elements of E s.t K is covered by U V(x_i, delta). Then the usual triangle inequality applies.

I don't see why a countable dense subset of K is needed. Why can we not directly invoke compactness and choose the x_i's elements of K, instead? U V(x, delta), x elements of K, is an infinite cover of K, so why can we not directly choose the x_i's? The only requirement in the proof is that d(x,x_i)<delta be valid for some i from 1 to m and for all x elements of K.

What am I completely missing?

Thanks in advance!

 

[mrandersdk]

You need the set because you want to use theorem 7.23, to insure the subsequences there exist to be pointwise convergent.

 

[Epsilon36819]

Sorry if I'm being a little dense (), but thm 7.23 only requires that {f_n} be pointwise bounded (which we have). Why can the countable set used in 7.23 not be the set of all x_i's taken directly from K such that U V(x_i, delta) covers K? I don't see the need for density...

 

[mrandersdk]

It is becase the convergent subsequence g_i, need to be independent of the given epsilon. You see the delta choosen depent on epsilon, and the there for the set E you want to consruct from U V(x,delta), depends on epsilon, and because that E depends on epsilon, when you use theorem 7.23 on that g_i depends on epsilon.

By doing it the way rudin does it, g_i is independent of epsilon. If g_i depends on epsilon, that is when I give you an epsilon you need to choose a new equence you are in trouble because, then you haven't proved that g_i converges, becase for every epsilon you choose a new sequence, you see?

It is a very subtle point, I missed it my self the first time I learned the proof, good catch.

 

[Epsilon36819]

HA! It's all clear now. Thanks a lot!

 

[mrandersdk]

no problem. Glad to help.