This question may seem trivial, but in topology we were taught that in a complete metric space, a subset of that space was compact if and only if it is closed and bounded. Moreover, we are told that for any topological space $X$, by definition $X$ is closed. Does it follow that if $X$ is a complete metric space, and $X$ is bounded in its metric, then $X$ is compact in that topology?
[Math] Bounded complete metric space is compact
compactnessgeneral-topologymetric-spaces
Best Answer
No. The general characterization is that a metric space is compact if and only if it is complete and totally bounded. The latter means that for any $\varepsilon > 0$ the space has a finite cover by balls of radius at most $\varepsilon$ (this is sometimes called an "$\varepsilon$-net"). This rules out, for instance, the closed unit ball in an infinite dimensional Banach space, which is a closed, bounded, noncompact metric space.