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 $n\ge N$?

  • 4
    $\begingroup$ Set $N=\max\{N_1,N_2\}$ $\endgroup$ – user194469 Oct 9 '16 at 21:47
  • $\begingroup$ Could you please explain how that would imply $n_k\ge N_1$ and $k\ge N_2$? $\endgroup$ – Thomas Oct 9 '16 at 21:51

Let $N=\max\{N_1,N_2\}$. You want to argue that $$ |s_n-L|<\epsilon $$ whenever $n\geq N$.

If $n\geq N$, then $n\geq N_1$ and $n\geq N_2$. Now choosing $k\geq N$, you would have $$ n_k\geq k\geq N. $$

  • $\begingroup$ Thanks, @Jack! A quick question: Wouldn't the inequality hold for $n\ge N_1$ provided we choose $k=\max\{N_1,N_2\}$? $\endgroup$ – Thomas Oct 10 '16 at 23:54
  • $\begingroup$ Well, $n$ has nothing to do with $k$, but $n_k$ does. $\endgroup$ – Jack Oct 10 '16 at 23:59

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