# Prove that a set $S \subseteq \mathbb{R}$ is closed and bounded if and only if every sequence in $S$ has a sub sequence converging to a point of $S$.

$(\to)$ Assume that $S \subseteq R$ is bounded. If $(x_n)$ is a sequence in $S$ , then $(x_n)$ is bounded and thus it has a convergent subsequence.

$(\leftarrow)$ Assume that $S$ is not bounded. Then for any integer $n$ there is an $x_n$ in $S$ such that $|x_n| > n$. Since any convergent sequence is bounded, the sequence $(x_n)$ cannot have a convergent subsequence.

Can someone check to see whether my proof is missing something.

thank you.

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For the $\Rightarrow$ direction, why is the limit of the convergent sub-sequence an element of $S$? –  Arthur Fischer Nov 21 '12 at 4:04

First for this direction $(\Rightarrow)$ you must prove that the sequence converges to $S$ (i.e. the limit is in $S$).
In this direction $(\Leftarrow)$ you only proved that if any sequence in $S$ has a converging subsequence in $S$ then $S$ is bounded.
Use the closeness of $S$ in this $(\Rightarrow)$ and prove it in this $(\Leftarrow)$ to finish your proof.