# Show that every compact subspace of a metric space is bounded in that metric and is closed [duplicate]

For the closed part, I just noted that it's a compact subspace of a Hausdorff space and therefore it's closed. For the bounded part, I know intuitively that since every open cover has a finite subcover, I just have to take the largest ball that includes all these covers, but I don't know how to write it rigorously. I know they'd all fit a large ball...

I also must find a metric space in which not every closed subspace is compact. Which is an example of a metric space in which not every closed is compact? Because once I know that, I could just take the metric $0,1$

## marked as duplicate by user228113, Claude Leibovici, user91500, E. Joseph, астон вілла олоф мэллбэргNov 17 '16 at 9:45

• Are you unable to use the general Heine-Borel theorem? – A.Riesen Nov 17 '16 at 1:14
• Well, in $\Bbb R$ there are closed subsets which are not compact. For instance $\Bbb R$ itself (since it's not bounded). – user228113 Nov 17 '16 at 1:14
• Compact imply finite subcover for every open cover. Then for the cover $\bigcup \Bbb B(x,\epsilon)$ for all $x$ in the compact set this imply that the set is bounded. – Masacroso Nov 17 '16 at 1:20
• @G.Sassatelli but I needed a bounded one :c – Guerlando OCs Nov 17 '16 at 1:23
• @GuerlandoOCs You did not mention it, though. Pick your favourite closed interval of $\Bbb Q$. – user228113 Nov 17 '16 at 1:28

Let $S$ be the compact set. Pick any point $x \in S$. Now consider the cover $\{B(x, n) \ | \ n \in \mathbb{N} \}$, where $B(x, n)$ denotes the open ball centered at $x$ of radius $n$. Notice that $S$ is contained inside the largest $B(x, k)$ in the finite subcover this cover admits.