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 $S$ 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 $S$.
The direction $\Rightarrow$ was easy. But I don't have an idea how to prove $\Leftarrow$. Maybe I first assume that S is unbounded. This means for all $n\in\mathbb{N}$ it is $|x_n|\ge n$. But how to continue to get a contradiction? Then I have to assume that $S$ is not closed. But I don't know how to prove this direction in detail.
 A: Well, suppose $x_n \to x$ with $x_n \in S$, then any subsequence also converges to $x$, hence we see that $x \in S$ and so $S$ is closed.
So, we only need to show that $S$ is bounded. If $S$ was unbounded, we could select a sequence $x_n \in S$ such that $|x_n| >n$. Hence no subsquence can converge, a contradiction.
Alternative proof of closure:
Suppose $x \notin S$. Then I claim that there is some $\epsilon>0$ such that
$B(x,\epsilon)$ does not intersect $S$. Then this will show that $S^c$ is open and hence $S$ is closed. If not, then by letting $x_n \in B(x,{1 \over n}) \cap S$, we have $x_n \to x$, and since every subsequence of $x_n$ converges to $x$, we see that $x \in S$, a contradiction.
A: Hint: If you have an unbounded set, then you can consider the intervals $[x,x+1]$ for integers $x$ and infinitely many of these contain a point in your unbounded set (why?). So choose one point from each such interval to get a sequence with no convergent subsequence. Then assume your set is not closed. This means there is a limit point $z$ not in the set. Consider the sets $X_n$ defined by the set of points $x$ that satisfy $1/({n+1}) \leq |z - x| \leq 1/n$. Infinitely many of these sets must contain a point in your set (why?). So choose one point from each such set. What is the limit? Is it in your set?
