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Rudin's proof of Arzela-Ascoli teorem

  1. Jul 27, 2008 #1
    This is little Rudin's proof 7.25: If K is compact, f_n complex and continous 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!
     
  2. jcsd
  3. Jul 27, 2008 #2
    You need the set because you want to use theorem 7.23, to insure the subsequences there exist to be pointwise convergent.
     
  4. Jul 27, 2008 #3
    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...
     
  5. Jul 27, 2008 #4
    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.
     
  6. Jul 27, 2008 #5
    HA! It's all clear now. Thanks a lot!
     
  7. Jul 27, 2008 #6
    no problem. Glad to help.
     
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