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  1. MHB Craftsman
    Alexmahone's Avatar
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    Prove that if a subsequence of a Cauchy sequence converges then so does the original Cauchy sequence.

    I'm assuming that we're not allowed to use the fact that every Cauchy sequence converges. Here's my attempt:

    Let $\displaystyle\{s_n\}$ be the original Cauchy sequence. Let $\displaystyle \{s_{n_k}\}$ be the convergent subsequence.

    Given $\epsilon>0$,

    $\exists N_1\in\mathbb{N}$ such that $\displaystyle|s_n-s_m|<\frac{\epsilon}{2}$ whenever $n\ge N_1$ and $m\ge N_1$.

    $\{s_{n_k}\}$ converges, say, to $L$.

    So $\exists N_2\in\mathbb{N}$ such that $\displaystyle|s_{n_k}-L|<\frac{\epsilon}{2}$ whenever $k\ge N_2$.

    $\displaystyle|s_n-L|=|s_n-s_{n_k}+s_{n_k}-L|\le |s_n-s_{n_k}|+|s_{n_k}-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ whenever $n\ge N_1$, $n_k\ge N_1$ and $k\ge N_2$.

    How do I wrap up this proof by finding the $N$ such that $|s_n-L|<\epsilon$ holds whenever $n\ge N$?

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    Hi Alexmahone,

    It'll be useful to use the fact that for all $k\in \Bbb N$, $n_k \ge k$. Let $N = \max\{N_1,N_2\}$. If $n \ge N$, then $n\ge N_1$ and $n_N \ge N \ge N_1$, which implies $\lvert s_n - s_{n_N}\rvert < \epsilon/2$. Also, $n \ge N_2$ and $n_N \ge N \ge N_2$, so that $\lvert s_{n_N} - L\rvert < \epsilon/2$. Thus $\lvert s_n - L\rvert \le \lvert s_n - s_{n_N}\rvert + \lvert s_{n_N} - L\rvert < \epsilon/2 + \epsilon/2 = \epsilon$.

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