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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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    Ok will do $n$ext time. Thanks $f$or the tip.2012-05-07

2 Answers 2

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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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    @Brian Nothing at all. I have added him in.2012-05-07