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Convergent sequence in compact metric space

  1. Dec 9, 2012 #1

    In Baby Rudin, Thm 3.6 states that If p(n) is a sequence in a compact metric space X, then some subsequence of p(n) converges to a point in X.

    Why is it not the case that every subsequence of p(n) converges to a point in X? I would think a compact set would contain every sequence (finite or infinite) that originates within it.

  2. jcsd
  3. Dec 9, 2012 #2
    Take [itex]X=[-1,1][/itex] and let [itex]p(n)=(-1)^{n}[/itex]. Then there exists a convergent subsequence, namely the constant sequence [itex]p(2n)=1[/itex]. But the subsequence [itex]p(n)[/itex] (= the entire sequence) is not convergent.
  4. Dec 9, 2012 #3


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    As Micromass points out, you can shuffle several convergent sequences that converge to different points.

    Can you suffle infinitely many convergent sequences converging to different points? If so what happens?
  5. Dec 9, 2012 #4
    I think the issue was that I had assumed bounded sequences converge. This is clearly not the case.

    To answer my question using the example you gave, there are 2 subsequences of p(n) which converge: p(2n) and p(2n+1), but a number of other subsequences (infinitely many to be exact) that diverge in X = [-1,1], a compact metric space. But if these 2 subsequences converge then infinitely many subsequences converge (i.e. p(4n), p(6n), ...). Right? And clearly p(3n), p(5n), ... all diverge. So then infinitely many subsequences diverge. Is this right?

    In general at least one subsequence of p(n) in a compact metric space X converges to a point in X, but if the sequence p(n) converges to a point p [itex]\in[/itex] X, then every subsequence converges to p [itex]\in[/itex] X.

    Please correct me if I'm wrong
  6. Dec 9, 2012 #5
    That's right!

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