How do I complete this convergence proof?

[alexmahone]

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 $n\ge N$?

 

[Euge]

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 $\lvert s_n - L\rvert \le \lvert s_n - s_{n_N}\rvert + \lvert s_{n_N} - L\rvert < \epsilon/2 + \epsilon/2 = \epsilon$.

 

[math771]

To complete the proof, we can choose $N = \max\{N_1, n_{N_2}\}$. Then, for any $n \geq N$, we have $n \geq N_1$ and $n_k \geq n_{N_2}$, so by the triangle inequality, we have
$$|s_n - L| \leq |s_n - s_{n_k}| + |s_{n_k} - L| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Therefore, we have shown that for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $|s_n - L| < \epsilon$ for all $n \geq N$, which proves that the original Cauchy sequence $\{s_n\}$ converges to $L$.