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Homework Help: Cauchy sequence proof

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data

    If {s_n} is a Cauchy sequence of real numbers which has a subsequence converging to L, prove that {s_n} itself converges to L.

    2. Relevant equations


    3. The attempt at a solution

    I know that all Cauchy sequences are convergent, and I know that any subsequences of a convergent sequence are convergent to the same limit as the sequence, but I am not sure if I can turn the second part of the statement around to say that if a subsequence is convergent to L, then the sequence converges to the same limit. Any ideas? Thanks.
     
  2. jcsd
  3. Jun 14, 2010 #2

    HallsofIvy

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    Actually, you can't just say "if a subsequence converges to L, then the sequence must converge to L". For example, the sequence [itex]\{a_n\}[/itex] with [itex]a_n= 1- 1/n[/itex] for n even and [itex]a_n= 1/n[/itex] for n odd has a subsequence ([itex]\{a_n\}[/itex] for a even) that converges to 1 but the sequence itself does not converges.

    What you can say is that if the sequence converges, then because, as you say, all subsequences must converge to that limit, yes, the sequence converges to whatever limit any subsequence converges to. The difference is that you must know the sequence does converge. Which you know here because it is a Cauchy sequence.
     
  4. Jun 14, 2010 #3
    Well, suppose that the Cauchy sequence converges to a number M different than L. Show that this leads to a contradiction by showing that there exists an epsilon such that some sequence value is always further away from M than epsilon.
     
  5. Jun 14, 2010 #4
    Thank you!
     
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