Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / E / E4 / a / iii

In any topological space a closed subset of a compact set is compact: add its open complement to an open cover and take a finite subcover of the compact set. However, compact subsets need not be closed in a general space. In the indiscrete two-point space a singleton is compact but not closed. Conversely a closed subset need not be compact: $\mathbb{R}$ itself is closed in $\mathbb{R}$ but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover.

In a Hausdorff space every compact subset $K$ is closed. Given $x\notin K$, separate $x$ and each $k\in K$ by disjoint neighborhoods; compactness selects finitely many neighborhoods covering $K$, and the intersection of their companion neighborhoods of $x$ misses $K$. Thus the complement is open. Closedness still need not imply compactness, as the same real-line example shows. It does imply compactness if the ambient space is compact.

In a metric space compactness implies boundedness: finitely many radius-one balls cover the set, so their centers have a finite maximal distance from any fixed center and the triangle inequality gives a global bound. The converse is false. An infinite set with discrete metric $d(x,y)=1$ for $x\ne y$ is bounded, but its singleton cover has no finite subcover. It is even closed and complete, so closedness plus boundedness alone does not suffice in general metric spaces.