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Here's the statement:

A sequence $(x_n)$ is bounded iff every subsequence of $(x_n)$ has a convergent subsequence.

Okay so the $(\Rightarrow) $ part is easy to prove. Since $(x_n)$ is bounded, any of its subsequences are bounded thus by the Bolzano Weierstrass Theorem, each of those subsequences have a convergent subsequence.

Now, the fate of the statement lies on the $(\Leftarrow ) $ part. However, I neither could prove it nor provide a counterexample. Hints will be appreciated!

Here's what I tried though: Since a subsequence of a subsequence of a sequence is itself a subsequence of that sequence. Hence, we know there are some convergent subsequences of that particular sequence! I'm not sure how this helps.

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If the sequence is unbounded, then there is a $n_1\in\mathbb N$ such that $|x_{n_1}|>1$. And there is a $n_2>n_1$ such that $|x_{n_2}|>2$. And so on. Obviously, $\left(x_{n_k}\right)_{k\in\mathbb N}$ doesn't converge.

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  • $\begingroup$ Does this need to assume using $R^n$ and euclidean metric? $\endgroup$ – Clark Makmur May 30 '18 at 15:47
  • $\begingroup$ @ClarkMakmur: this direction of the equivalence holds in any metric space. The other direction requires some assumptions on the metric space. Jose: the comment isn't from the OP (and +1), $\endgroup$ – Rob Arthan May 30 '18 at 15:50
  • $\begingroup$ @RobArthan thanks $\endgroup$ – Clark Makmur May 30 '18 at 15:53
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Hint:

Prove the contrapositive: if $(x_n)$ is not bounded, then there exists a subsequence of $(x_n)$ which has no convergent subsequence.

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