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Math Help - Cauchy sequences and subsequences

  1. #1
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    Cauchy sequences and subsequences

    I have the following theorem

    If {x_n} is Cauchy and {x_n_{k}} is a convergent subsequence, then {x_n} converges.

    n,m > N implies d(x_n,x_m) < \frac{\epsilon}{2}

    and

    d(x_n_{k},x) < \frac{\epsilon}{2}

    I want to say

    By triangle inequality, d(x_n,x) <= d(x_n,x_m) + d(x_n_{k},x) < \epsilon

    But I am not really sure if the triangle inequality applies here. Is this correct or can I make a different argument?

    Thank you very much.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Jame View Post
    I have the following theorem

    If {x_n} is Cauchy and {x_n_{k}} is a convergent subsequence, then {x_n} converges.

    n,m > N implies d(x_n,x_m) < \frac{\epsilon}{2}

    and

    d(x_n_{k},x) < \frac{\epsilon}{2}

    I want to say

    By triangle inequality, d(x_n,x) <= d(x_n,x_m) + d(x_n_{k},x) < \epsilon

    But I am not really sure if the triangle inequality applies here. Is this correct or can I make a different argument?

    Thank you very much.
    for any \varepsilon >0 there exists an $$N_{\varepsilon} such that:

    n,m > N_{\varepsilon} implies d(x_n,x_m) < \frac{\varepsilon}{2}

    So if n,n_k > N_{\varepsilon}, then:

    d(x_n,x) \le d(x_n,x_{n_k}) + d(x_{n_k},x) < \varepsilon

    etc ...

    CB
    Last edited by CaptainBlack; April 1st 2011 at 12:00 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    As (x_n) is a Cauchy sequence, for all \epsilon>0 there exists N\in \mathbb{N} such that n,m>N implies d(x_n,x_m)<\epsilon/2.

    As (x_{n_k}) is convergent, there exists N'\in \mathbb{N} such that k>N' implies d(x_{n_k},x)<\epsilon/2.

    Choose N''=\max\{N',N''\}. Then, for all n,k>N'' we have:

    d(x_n,x)\leq d(x_n,x_{n_k})+d(x_{n_k},x)<\epsilon/2+\epsilon/2=\epsilon


    Edited: Sorry, I didn't see CaptainBlack's post.
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