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The question is as follows:

Let $\{(x_n,y_n)\}$ be a sequence of points in $\mathbb R^2$ s.t. the sequences $\{x_n\}$ and $\{y_n\}$ are bounded. Prove that $\{(x_n,y_n)\}$ has a convergent subsequence.

Now, I am positive that I should tackle this problem using B-W theorem. Can anyone give an insight into this? Appreciate your help.

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Hints:

There exists a convergent subsequence $\;\{x_{n_k}\}\subset \{x_n\}\;$

Now look at $\;\{(x_{n_k}\,,\,y_{n_k})\}\;$ . This, again, is a bounded sequence, so again apply B-W to this sequence...on the second coordinate this time.

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  • $\begingroup$ would this work then? Suppose on the contrary that {(x_n,y_n)} does NOT have a convergent subsequence. Now, since both {x_n} and {y_n} are bounded, by the B-W Th'm, x_n has a convergent subsequence x_n_k and similary, y_n has a convergent subsequence y_n_k. Since {(x_n,y_n)} is a sequence of points in R^2 s.t. {x_n} and {y_n} are bounded and they both have convergent subsequences, {(x_n,y_n)}, by the B-W. th'm, must have another convergent subsequence of points {(x_n_k),(y_n_k)}, a contradiction. $\endgroup$ – user152967 May 25 '14 at 18:19
  • $\begingroup$ In your third line, it may well be that the indexes $\;n_k\;$ for $\;x_n\;$ are not the same as the indexes for $\;y_n\;$ ...and still then I can't see any sound contradiction in your argument. What isn't clear in the above answer that cannot take you to the end of the proof in 2-3 lines more? $\endgroup$ – DonAntonio May 25 '14 at 18:23

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