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Compact implies Sequentially Compact

  1. Feb 1, 2009 #1
    The problem statement, all variables and given/known data[/b]
    I need help proving that if X is a metric space and E a subset of X is compact, then E is sequentially compact.


    I know I need to consider a sequence x_n in E, and I want to say that there is a point a in E and a radius r > 0 so that Br(a) [the ball of radius r with center a] contains x_k for infinitely many k's. If I show this, then I think I can conclude that any subsequence of x_n converges to a. Can someone please help?
     
  2. jcsd
  3. Feb 1, 2009 #2

    Hurkyl

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    So... your conjecture is that every sequence in a compact metric space is actally convergent?
     
  4. Feb 1, 2009 #3
    No, that is not one of the assumptions...my second paragraph in the problem description is a hint in the back of the textbook as to how to go about the problem.
     
  5. Feb 1, 2009 #4

    Office_Shredder

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    If any subsequence of xn converges to a, then xn converges to a too. Rather, what you can conclude is a single subsequence of xn converges to a (which is what sequential compactness is about)
     
  6. Feb 16, 2009 #5
    This is not true- consider a_n=(-1)^n=1,-1,1,-1,1,-1,...

    Then a_{2n}=1 is a subsequence that converges to 1, but a_n does not converge at all- nonetheless to 1.
     
  7. Feb 17, 2009 #6

    HallsofIvy

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    Office Shredder meant "if every subsequence". He chose the wrong word.
     
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