# Cauchy sequences and subsequences

• Mar 31st 2011, 10:30 PM
Jame
Cauchy sequences and subsequences
I have the following theorem

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

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

and

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

I want to say

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

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.
• Apr 1st 2011, 12:32 AM
CaptainBlack
Quote:

Originally Posted by Jame
I have the following theorem

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

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

and

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

I want to say

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

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

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

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

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

etc ...

CB
• Apr 1st 2011, 12:49 AM
FernandoRevilla
As $(x_n)$ is a Cauchy sequence, for all $\epsilon>0$ there exists $N\in \mathbb{N}$ such that $n,m>N$ implies $d(x_n,x_m)<\epsilon/2$.

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

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

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

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