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Cauchy Sequences Proof

  1. Feb 19, 2009 #1

    I need to prove that any infinite subsequence {xnk}of a Cauchy sequence {xn}is a Cauchy sequence equivalent to {xn}.

    My problem is that it seemed way too easy, so I'm concerned that I missed something. Please see the attachment for my solution, and let me know what you think.


    Attached Files:

  2. jcsd
  3. Feb 19, 2009 #2


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    The definition of "Cauchy" sequence is that |an- am| goes to 0 as m and n go to infinity, independently. That can be stated as "Given any [itex]\epsilon> 0[/itex], there exist N such that if m> N and n> N, then [itex]|a_n- a_m|< \epsilon[/itex]. If an and am are from a subsequence, then they are also in the original sequence so that must be true.

    Showing that the two sequences are "equivalent" means showing that the converge to the same limit. Again, if {an} converges to L, then, for any [itex]\epsilon> 0[/itex], there exist N such that if n> N then [itex]|a_n- L|< \epsilon[/itex]. If an is from a subsequence, it is from the sequence and so that is still true.

    Strictly speaking, you only need to prove the second because any convergent sequence is a Cauchy sequence.
  4. Feb 19, 2009 #3
    For the the equivalence part, I want to show that the difference of the Cauchy sequence and its subsequence is a null sequence. Let xn and xnk exist in the Cauchy sequence where xnk is also an element in the subsequence. Therefore, |xn - xnk| < epsilon for n, nk < N, so Cauchy sequence - subsequence is null. Sound legitimate? Just seems too simple.

    I hear where you're coming from with the limits being the same for equivalence, but we're supposed to use the "difference is null" definition.
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