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Compactness with accumulation points

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Let K be a subset of R. Prove that if every sequence in K has an accumulation point, then K must be compact.

    2. Relevant equations

    I tried to proof it below. Am I on the right track?


    3. The attempt at a solution

    My intuition;

    Let x_n be sequence in K whose accumulation point is x, then there is a sub-sequence.
    x_n_k converges to x.
    Since sub-sequence x_n_k converges to x, x_n_k is a Cauchy sequence.
    We know that Cauchy sequences are bounded. Thus x_n is bounded. so K is bounded.

    Since all accumulation points are in K so K is closed.

    By Heine-Borel Thm, K is compact.
     
  2. jcsd
  3. Oct 12, 2011 #2

    PAllen

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    Not quite. I think you're better off showing the equivalent (each implies the other):

    if K is not compact, there exists a sequence without an accumulation point.
     
  4. Oct 12, 2011 #3

    Fredrik

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    Here you appear to be saying that if a sequence has a bounded subsequence, the sequence is bounded. Consider 1/2, 2, 1/3, 3, 1/4, 4, ...
     
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