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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.

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    As you said for all $\epsilon \gt 0$ there is $N = N(\epsilon)$ such that for all $n \gt N$ we have $|x-x_{n}| \lt \epsilon$. As $x_{n_r}$ is a subsequence, we have $n_{r} \lt n_{r+1}$. Show that there exists $R = R(\epsilon)$ such that for all $r \gt R$ we have $n_{r} \gt N(\epsilon)$.2011-04-09

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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?)

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    This seems a bit complicated. You already have a candidating limit floating around. Just show directly that any subsequence converges to the same limit as the mother sequence.2011-04-09
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@ 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

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    The formal way to go about it is sketched in my last comment to your question. Feel free to post your solution here, I'll certainly have a look at it.2011-04-09
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@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

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    ...for beyond $N(\frac{1}{2\cdot 10^k})$ we can't wiggle more than $\frac{1}{2\cdot 10^{k}}$, so at least $k-1$ digits of the limit are fixed. A subsequence must approach the limit at least as fast as the original sequence due to the restriction that we *must move to the right*.2011-04-10