Real Analysis - proving compactness

Question

Let $K⊆R^n$ be a set such that every infinite subset of $K$ has a limit point in $K$.

How can we show that K is closed and bounded?

Answer 1

Suppose $K$ is not bounded. The there exists a sequence $\{x_n\}$ in $K$ such that $\|x_n\| \to \infty$. This gives an infinite subset $\{x_1,x_2,...\}$ with no limit point in $K$,a contradiction. Hence $K$ is bounded. Now suppose $\{y_n\}$ is a sequence in $K$ converging to some point $y$. If $y \notin K$ then $\{y_1,y_2,...\}$ is a sequence in $K$ which has no limit point in $K$: if there is a limit point $z$ in $K$ then there would be a subsequence $\{y_{n_k}\}$ which is convergent, but the limit of this subsequence can only be $y$. This completes the proof.