Past exam of the options II course of the University of Oxford / 2010 / E / E4 / a / iii / Solution

In an arbitrary topological space, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb{R}$ is closed in itself but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a compact space $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.

In a Hausdorff space, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint open sets $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.

In every metric space, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the triangle inequality bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus compact implies closed and bounded in a metric space, while closed and bounded need not imply compact.