# Prove that the set is closed and bounded

Prove that if set $K\subset \mathbb{R}^k$ has the property that for every sequence $\left\{ x_n \right\}\subset K$ we can choose convergent subsequence $\left\{ x_{n_j} \right\}\rightarrow x\in K$ then $K$ is closed and bounded.

Seems very hard, but very interesting. I don't know how to approach. Can anybody help?

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You already have a more detailed answer from Arturo, but the short answer is contraposition. –  Jonas Meyer May 31 '12 at 20:57
@JonasMeyer, right.. In retrospect, there was rather no other option. By contradiction there is always something we can "seize". –  xan May 31 '12 at 21:04
@xan: Contraposition and contradiction are not the same. Contraposition means using equivalence of $A\implies B$ and (not $B)\implies($not $A)$, whereas contradiction could be used for example in the form of showing that $(A$ and (not $B))$ leads to a contradiction. –  Jonas Meyer May 31 '12 at 21:08
@xan: If you have a contraposition proof, contradiction could be unnecessarily thrown in (because $A$ and (not $A)$ is always a contradiction). But for example, nowhere is the assumption of $K$ being sequentially compact (q.v.) used in Arturo's proof sketch that if $K$ is not closed and bounded (=not closed or not bounded), then $K$ is not sequentially compact. Thus, it is a proof by contraposition. –  Jonas Meyer May 31 '12 at 21:16
@JonasMeyer, thank you very much for pointing it out! Now I see the difference, but so far I thought that this is the same. Especially because in my native language those two words are pronounced identically. –  xan May 31 '12 at 21:16

If $K$ were unbounded, then for each $n\in\mathbb{N}$, we can choose some $x_n\in K$ such that $\lVert x_n\rVert>n$. No subsequence of such a sequence can converge at all, much less to a point of $K$. If $K$ were not closed, then there is some limit point $x$ of $K$ with $x\notin K$. That means every neighborhood of $x$ contains a point of $K$ (so cannot be $x$), and in particular, for each $n\in\mathbb{N}$, we can choose $x_n\in K$ with $\lVert x_n-x\rVert<\cfrac{1}{n+1}$. This sequence (and every subsequence) converges to $x$, so does not converge to a point in $K$, by uniqueness of limits.
If $K$ is not bounded, then for every $M\gt 0$ there exists $x\in K$ such that $\lVert x\rVert \gt M$. Pick $x_n$ with $\lVert x_{n+1}\rVert \gt \lVert x_n\rVert+1$ to get a sequence with no convergent subsequence.
If $K$ is not closed, pick $y\in\overline{K}\setminus K$; if a sequence of points of $K$ converges to $y$, can a subsequence converge to a point in $K$?