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Can a bounded subsequence have infinitely many convergent subsequences?

  1. Nov 29, 2011 #1
    I'm not sure if I am confusing myself or not, but a friend and I were trying to figure this out. Basically, I know that if a sequence is bounded, we are guaranteed at least one convergent subsequences. However, is it possible for a bounded sequence to have infinitely many of such subsequences?
     
  2. jcsd
  3. Nov 29, 2011 #2

    mathman

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    Yes: 1, 1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4, 1, 1/2, 1/3, 1/4, 1/5, .......

    1/n is a convergent subsequence limit for every n.
     
  4. Nov 29, 2011 #3

    jgens

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    A bounded sequence can actually have uncountably many ([itex]2^{\aleph_0}[/itex]) convergent subsequences. In particular, for each i in N, let {xi,n} be an enumeration of the rational numbers in [0,1]. Then consider the sequence:

    x1,1, x1,2, x2,1, x1,3, x2,2, x3,1, ...

    That is, enumerate all the xi,n such that i+n = 2, then all elements such that i+n = 3, then all elements such that i+n = 4, and so on. Since the rationals are dense in R, this gives us a convergent subsequence for every real number in [0,1].
     
    Last edited: Nov 29, 2011
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