Show that a subsequence of a Cauchy subsequence is also Cauchy My approach for solving this problem was to first prove that the sequence is convergent and then use that to prove that the subsequence is convergent and therefore Cauchy. I'm pretty sure this process seems right but I'm not sure how to exactly prove this. If anyone can help, I'd appreciate it. Thanks
 A: You can take that approach, though to really be rigorous you would first have to prove that any metric space embeds in a complete metric space (for otherwise, if you are in an incomplete metric space, a Cauchy sequence need not converge (in case you live in $\mathbb R$ only, then this difficulty disappears, but still it is rather unnatural to proceed by the strategy you offer)). Having said that, assuming the knowledge that every metric space embeds in a complete one, you may assume the metric space you have is complete. Then certainly, your argument is already a proof. The thing is though that you don't need the heavy machinery of the completion of a space just in order to show that a subsequence of a Cauchy sequence is Cauchy. Indeed, it is quite easy to give a very direct proof: state (precisely) what it means for the given sequence to be Cauchy. Now state (precisely) what you need to do in order to show the subsequence is Cauchy. Remember well that its a subsequence and you'll realise you solved the problem. 
A: Let $(a_{n})_{n \geq 1}$ be Cauchy; let $(a_{n_{k}})_{k \geq 1}$ be a non-Cauchy subsequence of $(a_{n})_{n \geq 1}$. Then there is some $\varepsilon > 0$ such that $N \geq 1$ only if there are some $p,q \geq N$ such that
$|a_{n_{p}} - a_{n_{q}}| \geq \varepsilon$; but by assumption there is some $N' \geq 1$ such that $p,q \geq N'$ only if $|a_{p} - a_{q}| < \varepsilon$, a blatant contradiction.
A: Since, your original sequence, say (x$_{n}$) is Cauchy, $\exists$ n$_{0}$ in naturals such that $\forall$n,m$\geq$n$_{0}$, |x$_{n}$-x$_{m}$|< $\epsilon$, for any $\epsilon$>0. 
Let us say that you have a sub-sequence (x$_{n_{j}}$) of (x$_{n}$), then to be Cauchy, we have to make sure that we pass the n$_{0}$th term of the original sequence. Since, we are talking about any general sub-sequence of the given sequence, n$_{n_{0}}$ is sufficient in deciding that tail of the sequence obeys Cauchy behavior, because 
n$_{n_{0}}$$\geq$n$_{0}$
as  $\forall$n$_{j}$,n$_{k}\geq$n$_{n_{0}}$$\geq$ n$_{0}$, 
|x$_{n_{j}}$-x$_{n_{k}}$|< $\epsilon$ 
