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Every sequence has a convergent subsequence?

  1. Nov 2, 2014 #1
    I'm not sure if this is true or not. but from what I can gather, If the set of Natural numbers (divergent sequence) {1, 2, 3, 4, 5,...} is broken up to say {1}, is this a subsequence that converges and therefore this statement is true?
     
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  3. Nov 2, 2014 #2

    Simon Bridge

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    You are wondering if it is always possible to find at least one convergent subsequence in any sequence however divergent?
    Your specific question amounts to whether a sequence with a finite number of elements is convergent.
    To answer that, check the definition of "converge".

    Where does the question come up?
     
  4. Nov 2, 2014 #3
    if it's possible to find at least one convergent subsequence in ANY sequence.

    Definition of converge: "A sequence {a(n)} converges to a real number A iff for each epsilon>0 there is a positive integer N such that for all n >= N we have |a(n) - A| < epsilon."
     
  5. Nov 2, 2014 #4

    Simon Bridge

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  6. Nov 2, 2014 #5
    Hey, thanks a lot. Now i understand it.
     
  7. Nov 4, 2014 #6
    Only a bounded sequence has a convergent subsequence. An unbounded one, like 1,2, 3, 4...may not.
     
  8. Nov 6, 2014 #7

    WWGD

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    Does {1,2,3,..,n,.. } , i.e., a_n:=n have a convergent subsequence? Of course, this depends on your topology, but, as
    a subspace of the Reals, does this have a convergent subsequence? This is one of the characterization of compact metric spaces, as every sequence having a convergent subsequence. And, in the subspace topology of the Reals,
    {1,2,3,....} is/not compact (find a cover by open sets so that each contains a single number).
     
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