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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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The $n_r$'s are strictly increasing but, it doesn't imply that $x_{n_r}$ is increasing as well. Given $\epsilon >0$, how we can make sure that $|x_{n_r}-L| < \epsilon$ using the fact that $x_n\to L$? –  user9077 Apr 9 '11 at 13:09
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It seems that you are confusing several things: Convergence has nothing to do with monotonicity, a priori. The fact that a bounded and monotone sequence of real numbers converges is only a sufficient condition for convergence, it is by no means necessary: Look at $x_n = \frac{(-1)^n}{n}$, for example. Suggestion: Let $x$ be the limit of $x_{n}$ and prove that $x$ is also the limit of $x_{n_r}$. Since limits are unique, you're done then. To prove this, write out what it means for $x_n$ to converge to $x$ and think about why this implies that the same holds for $x_{n_r}$. –  t.b. Apr 9 '11 at 13:10
    
@Theo Buehler So when I say that $x_n \rightarrow x$, this means that for sufficiently large $n$ all the terms of the sequence lie within a distance of $\epsilon > 0$ of the limit $x$. So does this then mean that as $x_{n_r}$ is a subsequence of $x_n$, it too will lie within some ball of radius $\epsilon$ about its limit? But then how does that tell me that its limit is the same as the original sequence? I think I'm missing something here. –  fpqc Apr 9 '11 at 13:26
    
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 $n_{r} \gt N(\epsilon)$. –  t.b. Apr 9 '11 at 13:30
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3 Answers

@ 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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First of all, you're already assuming that $x_{n_r}$ converges. You don't want to do that, because it's exactly what you want to prove. Second, you're making your life a bit too complicated. Assume that $x_n \to X$. For $r$ large you have $n_{r}$ large, so automatically $|x_{n_r} - X| \lt \epsilon$, since $x_n$ converges to $X$. –  t.b. Apr 9 '11 at 13:55
    
Ok thanks theo, I think I finish it off from here. –  fpqc Apr 9 '11 at 14:04
    
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. –  t.b. Apr 9 '11 at 14:06
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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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Yes, exactly. We have $n_{r} \geq r$ for all $r$, so for $r \geq N$ we have $n_{r} \geq N$ and hence $|x_{n_{r}} - X| \lt \epsilon$ by definition. You still seem to be unsure. Note that "$(x_{n_r})_{r=1}^{\infty}$ is a subsequence" means that each $x_{n_r}$ occurs in the sequence $(x_n)_{n=1}^{\infty}$ as you have a strictly monotonic function $r \mapsto n_{r}$ and a function $n \mapsto x_n$ (a sequence in $\mathbb{R}$ is nothing but a function $\mathbb{N} \to \mathbb{R}$. The assignment $r \mapsto x_{n_r}$ means the composite $r \mapsto n_r \mapsto x_{n_r}$. –  t.b. Apr 9 '11 at 15:33
    
By the way: The @-notification doesn't work in answers, so you should add a comment somewhere in order to notify somebody. –  t.b. Apr 9 '11 at 15:34
    
Ok thanks for your reply, perhaps after a while these ideas will sink in! –  fpqc Apr 9 '11 at 16:14
    
Ben: Maybe I should do this less formally. Think of a sequence as an ordered collection $(x_1, x_2, x_3, \ldots, x_{n-1},x_n,x_{n+1},\ldots)$ of real numbers. Now a subsequence is obtained by picking $(x_{n_1}, x_{n_2}, x_{n_3},\ldots)$ from this list. Moreover, you are only allowed to move to the right. Now convergence means that if you go far to the right, then you're necessarily close to $x$. Think for instance of the decimal expansion of $\pi: (3, 3.1, 3.14, 3.141,\ldots)$. At each step you're approaching $\pi$ a bit more closely. A sub-sequence would be, e.g. $(3.14, 3.14159,\ldots)$ –  t.b. Apr 9 '11 at 16:28
    
@TheoBuehler A specific example of yours like that was helpful, I'm trying to think of a non-trivial one that helps illuminate this idea. Oh course I could come up with $\{\frac{1}{n}\}_{n=1}^{\infty}$ and the subsequence $\frac{1}{2n}$, but that would not be very exciting. Yours though on the decimal expansion of $\pi$ helps much more. –  fpqc Apr 10 '11 at 1:11
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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. –  wildildildlife Apr 9 '11 at 14:22
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