Rudin's proof of Arzela-Ascoli teorem

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Discussion Overview

The discussion revolves around Rudin's proof of the Arzelà-Ascoli theorem, specifically focusing on the necessity of a countable dense subset of a compact set K in the context of proving the existence of a uniformly convergent subsequence from a sequence of complex continuous functions that are pointwise bounded and equicontinuous.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the necessity of a countable dense subset of K, suggesting that elements of K could be used directly to cover K without invoking density.
  • Another participant asserts that the countable dense set is needed to apply theorem 7.23, which ensures the subsequences are pointwise convergent.
  • A different participant expresses confusion about the requirement for density, arguing that theorem 7.23 only requires pointwise boundedness, which they believe could be satisfied by choosing points directly from K.
  • One participant explains that the independence of the convergent subsequence from the chosen epsilon is crucial, indicating that using a set dependent on epsilon could lead to issues in proving convergence.
  • A later reply acknowledges the clarification and expresses gratitude for the explanation provided.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of a countable dense subset in the proof, with some supporting its inclusion for the application of theorem 7.23 and others questioning its relevance. The discussion remains unresolved regarding the necessity of density in this context.

Contextual Notes

The discussion highlights the subtlety of the relationship between the choice of points in the proof and the dependence on epsilon, which may not be fully resolved within the conversation.

Epsilon36819
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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!
 
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You need the set because you want to use theorem 7.23, to insure the subsequences there exist to be pointwise convergent.
 
Sorry if I'm being a little dense (:rolleyes:), 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...
 
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.
 
HA! It's all clear now. Thanks a lot!
 
no problem. Glad to help.
 

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