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Any compact subset is a contained in finite set + a convex set?

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data
    So I am trying to understand this proof and at one point they state that an arbitrary compact subset of a Banach space, or a completely metrizable space is the subset of a finite set and an arbitrary convex neighborhood of 0. I've been looking around and can't find anything to support this. Where does this come from?

    2. Relevant equations



    3. The attempt at a solution
    I keep thinking that maybe it can be approached by the set being totally bounded since it is compact. So it seems for a given ε can cover the subset in open balls of radius ε. Then the centers of these balls is a finite set, then somehow choosing a corresponding convex set might let you represent any of the elements of the original compact set as the sum of elements of these new sets. But I don't know, I just don't have any basis from which to approach this.
     
  2. jcsd
  3. Feb 21, 2013 #2

    mfb

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    I would expect that you can prove it like this (just more formal):
    Assume there is a compact set which cannot be described as a subset of a convex set and a finite set. This would require an infinite number of elements "far away" (without bound). You can cover each of this element with an open set (using an infinite amount of them), such that the only covering subset is the whole set of those open sets. Therefore, there is no finite subcover.
     
  4. Feb 21, 2013 #3
    So if they weren't unbounded, then you could cover it with a ball of radius bigger than the largest element? I think I am okay with this, but in the proof where this was stated, it was for an convex neighborhood about 0 which is a subset of any arbitrary open neighborhood of 0. So in turn it seems like the convex neighborhood is really arbitrary. So then it seems like the convex set wouldn't be free to be large necessarily. For example, say it is a closed ball in a finite dimensional space, it still seems counter-intuitive that the closed ball is the subset of a finite set and an arbitrary convex neighborhood. Thanks for the help!
     
  5. Feb 21, 2013 #4

    mfb

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    Sure

    I have no idea what you mean in the rest of the post.
     
  6. Feb 21, 2013 #5
    Haha, sorry. So this all comes up in a proof for "the closed convex hull of a compact subset of a completely metrizable space is compact". It can be found on page 186 here:

    http://books.google.com/books?id=4h...epage&q=closed convex hull is compact&f=false

    Anyway, the very first part is that for any arbitrary open neighborhood U about 0 there exists a convex neighborhood of 0, V which is a subset U and such that the compact subset in question is a subset of a finite set plus this convex set. But at the end of the proof they shrink U since it was arbitrary, so then it seems even more surprising. I don't know if that made sense at all?
     
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