# Thread: Cauchy sequences and subsequences

1. ## Cauchy sequences and subsequences

I have the following theorem

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

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

and

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

I want to say

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

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.

2. Originally Posted by Jame
I have the following theorem

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

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

and

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

I want to say

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

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

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

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

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

etc ...

CB

3. As $\displaystyle (x_n)$ is a Cauchy sequence, for all $\displaystyle \epsilon>0$ there exists $\displaystyle N\in \mathbb{N}$ such that $\displaystyle n,m>N$ implies $\displaystyle d(x_n,x_m)<\epsilon/2$.

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

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

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

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