# Prove or disprove a statement about subsequences

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.

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.
• Does this need to assume using $R^n$ and euclidean metric? – Clark Makmur May 30 '18 at 15:47
Prove the contrapositive: if $(x_n)$ is not bounded, then there exists a subsequence of $(x_n)$ which has no convergent subsequence.