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More Heine-Borel

  1. Apr 29, 2007 #1
    The other thread about heine-borel theorem just reminded me of something that has been unclear to me. I understand how you can prove, that a closed and bounded subset of [tex]\mathbb{R}^n[/tex] is compact, but isn't this true also for an arbitrary metric space? The proof I've read relies on the fact that we can first put the subset in a box [tex][-R,R]^n[/tex], and then start splitting this box into smaller pieces, but how could you replace this procedure with something in an arbitrary metric space?
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  3. Apr 29, 2007 #2


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    I'm pretty sure no. Try looking at function spaces.
  4. Apr 29, 2007 #3
    Oh, well. No wonder I didn't understand how to extend this proof into general metric spaces...

    Now when you mentioned function spaces, I just remembered, that I do know the Riesz's theorem of non-compactness. Just couldn't put pieces together. :rolleyes: Ok, sorry for bothering!
  5. Apr 29, 2007 #4


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    You don't need function spaces. The Heine-Borel theorem, for the real numbers, is equivalent to the "least upper bound property". That means that it is NOT true for the set of rational numbers with the "usual" topology.

    In particular, {x| x2<= 2} is both closed and bounded as a subset of the rational numbers but is not compact. (The crucial point in the proof is that there is no rational number, x, such that x2= 2.)
  6. Apr 29, 2007 #5
    In general a metric space is compact if and only if it is complete and totally bounded.
  7. May 2, 2007 #6


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    deadwolfes remark means compactness follows if every sequence has a cauchy subsequence, and every cauchy sequence converges.

    notice R^n is already complete, so any closed subset is also complete. what property does R^n have causing every bounded subset to be totally bounded?
    Last edited: May 2, 2007
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