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Thread: Cauchy sequences and subsequences

  1. #1
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    Cauchy sequences and subsequences

    I have the following theorem

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

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

    and

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

    I want to say

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

    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 $\displaystyle {x_n}$ is Cauchy and $\displaystyle {x_n_{k}}$ is a convergent subsequence, then $\displaystyle {x_n}$ converges.

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

    and

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

    I want to say

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

    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 $\displaystyle \varepsilon >0$ there exists an $\displaystyle $$N_{\varepsilon}$ such that:

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

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

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

    etc ...

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

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

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

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


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