Compactness implies boundedness: Let (X,d)(X,d) be a and let KXK\subseteq X be a . Then KK is a ; equivalently, diam(K)<\operatorname{diam}(K)<\infty, where diam\operatorname{diam} is the induced by dd.

Remarks

This is a first step toward “compact sets behave like finite sets” in metric settings, and it complements .