Compactness implies boundedness

Compactness implies boundedness: Let $(X,d)$ be a metric space and let $K\subseteq X$ be compact . Then $K$ is bounded : there exist $x_0\in X$ and $R>0$ such that $K\subseteq B(x_0,R).$

This is one of the basic “finiteness” consequences of compactness and is used to control sequences and covers.

Proof sketch: Fix $x_0\in X$. The function $x\mapsto d(x,x_0)$ is continuous on $K$. Since $K$ is compact, it attains its maximum $R$ on $K$. Then $d(x,x_0)\le R$ for all $x\in K$, so $K\subseteq B(x_0,R)$.