Guidance on Proof of proposition involving subsequences I am trying to prove the following proposition: "If $x_n$ is a convergent sequence, then every subsequence of $x_n$ is convergent and converges to the same limit as $x_n$."
I am not looking for an answer - I would not like a direct answer - but rather some guidance on how to prove this.
Firstly, I think I need to show that every subsequence of $x_n$ is convergent. So let $x_{n_r}$ be a subsequence. By definition, the $n_r's$ are strictly increasing, so can I deduce from here that the subsequence $x_{n_r}$ is strictly increasing as well? 
I know as well that as $x_n$ is convergent, it is bounded, viz $|x_n|\leq M$ where $M > 0$. So as the terms in a subsequence are contained in the set of all the $x_n's$, this means that every subsequence of $x_n$ is bounded as well?
If I can deduce that the $x_{n_r}'s$ are bounded and monotone, then I know that every subsequence of a convergent sequence is convergent.
Now the hard part of showing that every subsequence converges to the same limit, of which I have no idea; I could begin though to assume the negation that there exists a subsequence such that it converges to a different limit, say $M$ while the $x_n's$ converge to $L$ instead.
 A: @ Theo Buehler Ok I've thought of something but i'm not sure if it's right. Let us say that the limit of $x_{n_r}$ is $X$. Then $|x_{n_r} - X|$ = $|x_{n_r} - x_n + x_n - X|$ $\leq$ $|x_{n_r} - x_n| + |x_n - X|$. Can I conclude that: 
$|x_{n_r} - x_n| < \frac{\epsilon}{2}$? My reason would be that $x_{n_r}$ and $x_n$ are convergent sequences.
Secondly, can I say that in order for $|x_n - X|$ to be less than $\frac{\epsilon}{2}$, it must be that $X$ is the limit of the sequence $x_n$?
"Better to answer the right question wrong than the wrong question right" - Richard Hamming
Ben
A: @Theo,
So, putting these facts together, let $X$ be lim $x_n$. Then for $n \geq N$, $|x_n - X| < \epsilon$. So as you said that for $r$ large, $n_r \geq r$ by definition of what it means for $x_{n_r}$ to be a subsequence of $x_n$. So if $n_r \geq r \geq N$, we have that
$|x_{n_r} - X| < \epsilon$??
Ben
A: For the second part, if a sub sequence converges to L1 (using x as suggested above might get a little confusing) while another sub-sequence converges to L2 you need to show L1=L2. Well if L1 does not equal L2 there is some positive distance between them. (What does this say about convergence of the original sequence?)
