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.


closed as off-topic by Michael Albanese, 3SAT, Claude Leibovici, SchrodingersCat, user91500 Dec 30 '15 at 11:17

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  • $\begingroup$ 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. $\endgroup$ – Mathstudent May 7 '12 at 17:55
  • $\begingroup$ Please add your own work to the question, instead of adding it as a comment. $\endgroup$ – TMM May 7 '12 at 17:57
  • $\begingroup$ Ok will do next time. Thanks for the tip. $\endgroup$ – Mathstudent May 7 '12 at 18:00

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.


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 $\{x_{n_{k_l}}\}$...

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

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