TITLE: Efficient way of saying a subset $A$ of a metric space $X$ is compact implies A is bounded
QUESTION [6 upvotes]: Can anyone come up with a one line answer why a set $A \subset X$ being compact implies it's bounded. I figure that we could take the closure of $A$'s open subcover and every $x \in A$ would be contained in this set. Is that correct? And is there a 'nicer' way of saying it (in one sentence)?

REPLY [14 votes]: Fix $a\in A$ and consider the open cover defined by $B(a,n)=\{x\in X\mid d(a,x)<n\}$ for $n\in\mathbb N$.

REPLY [10 votes]: Take $X$-balls of radius $1$ around every point in $A$ and then take a finite subcover of that.