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Compactness and space dimension question

  1. Nov 13, 2008 #1
    I am having trouble accepting two well known results of analysis as non contradictory.

    First, given a vector space equipped with norm ||.||, the unit ball is compact iff the space is finite dimensional.

    Second, the Arzela-Ascoli theorem asserts that given a compact set X, a set S contained in the space of all continuous functions f(X) equipped with the supremum norm is compact iff it is uniformly closed, uniformly bounded and equicontinuous.

    My problem is the following: let S be the set of all equicontinuous functions on X with ||f|| <= 1, where ||.|| is the supremum norm. Is the space of all equicontinuous functions not an infinite dimensional subspace? Yet S is compact.

    I realize there is something fundamental I'm not getting. Please help! :confused:
     
    Last edited: Nov 13, 2008
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  3. Nov 13, 2008 #2

    Office_Shredder

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    The limit of an equicontinuous uniformly converging sequence is continuous, but not necessarily equicontinuous if I recall correctly. Hence you don't have closure, and thus the space isn't compact.

    I've never really done anything with equicontinuous functions, so I could be wrong
     
  4. Nov 13, 2008 #3
    Can't we use an epsilon/3 argument, just as in the proof that uniform convergence of continuous functions => continuity of the limit? In this case, the given delta valid for fn will be the one used for f, so f will be part of the equicontinuous family.

    Now that I think of it, though, I don't think that the set of equicontinuous functions is closed under addition and scalar multiplication, hence it is not a subspace.

    Nonetheless, it's still foggy to me why the first thm applies only to vector spaces, and not sets.
     
  5. Nov 14, 2008 #4

    Office_Shredder

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    Well, it needs to be a vector space for norm, unit ball and finite dimensional to be defined.
     
  6. Nov 14, 2008 #5
    It can be a subset of a vector space.
     
  7. Nov 14, 2008 #6

    morphism

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    It doesn't make sense to speak of an equicontinuous function. Equicontinuity is a property of sets of functions.

    So in your original post, the following bit is nonsensical:
     
  8. Nov 14, 2008 #7

    Office_Shredder

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    In such a case there's no reason to believe it contains any unit vectors. And the set still has no dimension on it unless it's a vector space itself
     
  9. Nov 14, 2008 #8

    Hurkyl

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    One common thing to do in mathematics is to generalize a theorem by carefully studying its proof, and removing unused hypotheses, or weakening overly strong hypotheses.

    So, you could try to do that here -- study the proof to see exactly what properties of "normed vector spaces", "unit balls", and "dimension' are being used here, and which aren't.
     
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