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Hints only please: compact iff bicompact

  1. Aug 2, 2004 #1
    I'd like hints only please. I have an analysis book and I could look up the proof myself but I'm trying to prove it myself as an exercise; so giving the full proof would be redundant as well as counterproductive to my own learning.

    X is a metric space.

    In this other book, K is compact iff every sequence in K has a convergent subsequence. I know that topologically, this is not the standard definiton of compact; some authors would call this sequentially compact.

    K is bicompact if every open cover of K admits a finite refinement that also covers K. This is how compactness is usually defined as far as I knew.

    Hints on how compact implies bicompact and bicompact implies compact would be very much appreciated.

    Right now, I'm working on compact implies bicompact so that is the current priority. I don't need hints on the other direction at this time since I haven't tried it.

    A general hint like "prove the contrapositive" or "use contradiction" might be helpful though I've tried the first of the two already. I suppose as I await a reply, I'll try contradiction...

    Thanks!
     
  2. jcsd
  3. Aug 2, 2004 #2

    plover

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    Did you mean "K is a metric space"?
     
  4. Aug 2, 2004 #3
    Sorry for the oversight. I meant that X is a metric space and that K is a subset of X.
     
  5. Aug 3, 2004 #4

    arildno

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    I'll give you 2 lemmas to prove first:

    1. If K is (sequentially) compact, then K is totally bounded.
    (A set is called totally bounded, if for any e>0, there exist a finite set [tex]\{x_{k}\}\in{K}[/tex] such that K is covered by the union of disks [tex]D(x_{k},e)[/tex])
    2.Let [tex]\{U_{i}\}[/tex] be an open cover of K.
    Then there exist r>0, such that for any [tex]y\in{K}[/tex], the open disk
    D(y,r) is contained in some [tex]U_{i}[/tex]

    (I thought 2. was shockingly strong the first time I saw it..)

    Proof by contradiction..
     
    Last edited: Aug 3, 2004
  6. Aug 3, 2004 #5
    Thank you so much. I went ahead and proved that compact --> bicompact using the two lemmas; now I'm working on proving the first lemma. I'm sort of stuck on it though I haven't tried everything yet. In the next day or two, I may ask for a hint on how to prove lemma 1. I'm now resorting to proving it with contradiction. I want to exhibit a sequence with no convergent subsequence... Still working on it. Thanks again.
     
  7. Aug 4, 2004 #6

    arildno

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    Good luck!
    I guess you've already figured out how the two lemmas together will prove your main objective..
     
  8. Aug 4, 2004 #7
    How does this look for a proof of the first lemma?

    We will show that if [tex]K[/tex] is not totally bounded then [tex]K[/tex] is not compact by constructing a sequence in [tex]K[/tex] that has no convergent subsequence. Let [tex]x_{1}[/tex] be any element of [tex]K[/tex]. Note that since [tex]K[/tex] is not totally bounded, [tex]K[/tex] is not covered by [tex]B_{1}\left( x_{1}\right) [/tex]. So let [tex]x_{2}\in K\backslash B_{1}\left( x_{1}\right) [/tex]. Likewise, [tex]K[/tex] is not covered by [tex]B_{1}\left( x_{1}\right) \cup B_{2}\left( x_{2}\right) [/tex]. For [tex]n>2[/tex] let [tex]x_{n}\in K\backslash \bigcup_{i=1}^{n-1}B_{1}\left( x_{i}\right) [/tex]. [tex]x_{n}[/tex] exists because if [tex]K[/tex] is covered by [tex]\bigcup_{i=1}^{n-1}B_{1}\left( x_{i}\right) [/tex], it would be totally bounded. The sequence [tex]\left\{ x_{i}\right\} [/tex] has no convergent subsequence because for all (unequal) [tex]i,j[/tex], [tex]d\left( x_{i},x_{j}\right) >1[/tex] and if there were an increasing function [tex]\tau :Z^{+}\rightarrow Z^{+}[/tex] such that there is an [tex]I\in Z^{+}[/tex] such that if [tex]i\geq I[/tex], then [tex]d\left( x_{\tau \left( i\right) },x\right) <1/2[/tex] for some [tex]x\in X[/tex]. That would imply that [tex]d\left( x_{\tau \left( I\right) },x_{\tau \left( I+1\right) }\right) \leq d\left( x_{\tau \left( I\right) },x\right) +d\left( x,x_{\tau \left( I+1\right) }\right) <1[/tex] when we know that [tex]d\left( x_{\tau \left( I\right) },x_{\tau \left( I+1\right) }\right) >1[/tex]. Therefore [tex]K[/tex] is not compact as we have exhibited a sequence with no convergent subsequence.
     
  9. Aug 4, 2004 #8

    arildno

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    Is [tex]B_{1}[/tex] a ball with radius 1?
    If so, your argument is (slightly) flawed:
    The negation is:
    There exist some e>0 such that K cannot be covered by finitely many balls of radius e.
    (As far as I can see, you can merely substitute e for 1 in your argument to get the proof)
     
  10. Aug 4, 2004 #9
    Cool deal. Now I'm working on the second lemma...
     
  11. Aug 4, 2004 #10
    Could I have a hint for the second lemma? Thanks in advance.
     
  12. Aug 5, 2004 #11

    arildno

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    The negation is:
    For every r>0, there exist some y in K, such that the disk D(y,r) is not (fully) contained in any Ui
     
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