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Compactness contradiction physics

  1. Feb 24, 2005 #1
    Let A be a compact subset of a metric space (X,d). Show that there exist a,b in A such that d(A) = d(a,b) where d(A) denotes the diameter of A.

    I guess...we're supposed to use the fact that a compactness of A implies that it is closed and bounded or alternately...we could assume that there are no a,b in A such that d(A)=d(a,b) and arrive at a contradiction to the fact that A is compact.
    Any suggestions??
     
  2. jcsd
  3. Feb 24, 2005 #2
    Find appropriate sequences (a_n) and (b_n) then apply Bolzano-Weierstrass.
     
  4. Feb 24, 2005 #3

    matt grime

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    THis may be exactly the same hint, but it's also known that compact and seqentially compact are equivalent for metric spaces. But then there is another way of doing it: every continuous real valued function on a compact spaces is bounded and attains its bounds.
     
  5. Feb 24, 2005 #4
    That's what I meant by Bolzano-Weierstrass. But I probably should have been more explicit, since that's really a generalization of the theorem and not the theorem iteself.
     
  6. Feb 25, 2005 #5
    thanks.... i did do the sequences thing but i feel that it's not an elegant way of doing it. why? that's because you're talking about distance in A (cross) A
    that's what d(an,bn) means...
    so i think it's definitely not the best way of doing it....
    "every continuous real valued function on a compact spaces is bounded and attains its bounds. " can you please elaborate? :confused:
     
  7. Feb 25, 2005 #6

    matt grime

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    A is compact, hence AxA is compact. d(?,?) is a continuous function on AxA hence there is a point in AxA such that d(?,?) is maximized, let that point be (a,b).
     
  8. Feb 26, 2005 #7
    thanks a lot...i figured that out finally!
     
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