# Efficient way of saying a subset $A$ of a metric space $X$ is compact implies A is bounded

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)?

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Fix $a\in A$ and consider the open cover defined by $B(a,n)=\{x\in X\mid d(a,x)<n\}$ for $n\in\mathbb N$.
Is the point being that for some $m \in N$, when $n \ge m$, $A \subset B_n(a)$? Is that implied in your answer? (I am still learning what is implied and what has to be stated explicitly). – sonicboom Nov 24 '12 at 19:55
Yes. The union $B(a,n)\cup B(a,m)$ is simply $B(a,\max\{m,n\})$. Take a finite subcover of this cover, and its union is $B(a,n)$ for some $n$. – Asaf Karagila Nov 24 '12 at 19:57
Take $X$-balls of radius $1$ around every point in $A$ and then take a finite subcover of that.