# Thread: How do I complete this convergence proof?

1. ## How do I complete this convergence proof?

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 $n\ge N$?

2. ## Re: How do I complete this convergence proof?

Take N equal to the larger of $\displaystyle N_1$ and $\displaystyle N_2$.

3. ## Re: How do I complete this convergence proof?

Originally Posted by HallsofIvy
Take N equal to the larger of $\displaystyle N_1$ and $\displaystyle N_2$.
Why would that imply that $n_k\ge N_1$ and $k\ge N_2$?

4. ## Re: How do I complete this convergence proof?

Originally Posted by alexmahone
Why would that imply that $n_k\ge N_1$ and $k\ge N_2$?
Here is another way. $N_1+N_2$ is an integer greater than either $N_1\text{ or }N_2$.
So let $N=N_1+N_2$.

5. ## Re: How do I complete this convergence proof?

Originally Posted by alexmahone
Why would that imply that $n_k\ge N_1$ and $k\ge N_2$?
Given that N is the larger of $\displaystyle N_1$ and $\displaystyle N_2$, $\displaystyle N\ge N_1$ and $\displaystyle N\ge N_2$. So if $\displaystyle n_k\ge N$ we have $\displaystyle n_k\ge N\ge N_1$. And if $\displaystyle k\ge N$ we have $\displaystyle k\ge N\ge N_2$.