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Suppose $(x_n)_{n\in \mathbb{N}}$ and $(y_n)_{n\in \mathbb{N}}$ are two bounded sequences. Show that there exists an increasing sequence of integers $n_k$ so that the subsequences $x_{n_k}$ and $y_{n_k}$ both converge.

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closed as off-topic by Michael Albanese, Nehorai, Claude Leibovici, SchrodingersCat, USER91500 Dec 30 '15 at 11:17

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Albanese, Nehorai, Claude Leibovici, SchrodingersCat, USER91500
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My first idea was to multiply the two bounded sequences together and then that is certainly bounded. Then I used bolzano-Weistrass to get a convergent subseqence. I was trying to show that each half of the multiplied subsequence converges. But i'm unsure if this is correct or not. Any help is appreciated. – Mathstudent May 7 '12 at 17:55
    
Please add your own work to the question, instead of adding it as a comment. – TMM May 7 '12 at 17:57
    
Ok will do next time. Thanks for the tip. – Mathstudent May 7 '12 at 18:00
up vote 5 down vote accepted

Any bounded sequence (in $\mathbb{R}^n$, presumably) lies in some compact set, so must have a convergent subsequence.

So start with $x_n$, then $x_{n_k} \rightarrow \hat{x}$ along the subsequence $n_k$. Now consider the sequence $y_{n_k}$, by the same considerations, $y_{n_{k_j}} \rightarrow \hat{y}$ along some subsequence $n_{k_j}$. Since $x_{n_k}$ converges, it also converges along the subsequence $x_{n_{k_j}}$, hence you have $x_{n_{k_j}} \rightarrow \hat{x}$, as well, which is what you are trying to show.

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This assumes that $\{x_n\}$ and $\{y_n\}$ are sequences in $\Bbb R^n$.

Use Bolzano-Weierstrass on $\{x_n\}$ to get a convergent subsequence $\{x_{n_k}\}$. Now look at $\{y_{n_k}\}$. Can you use Bolzano-Weierstrass on this?

Yes, you can, because $\{y_{n_k}\}$ is a subset of the bounded set $\{y_n\}$. Use that to get another convergent subsequence $\{y_{n_{k_l}}\}$ and look at $\{x_{n_{k_l}}\}$...

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Weierstrass wants to know what you have against him. :-) – Brian M. Scott May 7 '12 at 17:57
    
@BrianM.Scott If it weren't for him, I wouldn't have to deal with those darn differentiable-nowhere cts functions. – Alex Becker May 7 '12 at 17:59
    
@Brian Nothing at all. I have added him in. – user21436 May 7 '12 at 17:59

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