Past exam of the options I course of the University of Oxford / 2005 / E

A topology on $X$ contains $\varnothing ,X$, is closed under arbitrary unions and under finite intersections. The cofinite family has these properties: a nonempty union contains one cofinite member and therefore has finite complement; the complement of a finite intersection is a finite union of finite sets. The empty union and the sets $X,\varnothing$ are included explicitly.

Any two nonempty open sets intersect, because their intersection has finite complement in the infinite set $X$. Thus no two disjoint nonempty open sets can separate $X$, proving connectedness. For compactness, take any open cover and choose a nonempty member $U$. Only finitely many points lie outside $U$. Choose one covering member for each of those points. Together with $U$ these give a finite subcover. Thus $X$ is compact.

Continuity means that the inverse image of every open subset of $Y$ is open in $X$. If $x\in \overline{A}$ and $V$ is an open neighborhood of $f(x)$, then $f^{-1}(V)$ is an open neighborhood of $x$, so it meets $A$. Thus $V$ meets $f(A)$. The closure characterization givesIf $B$ is empty the assertion is immediate. Otherwise, if $B$ is bounded, choose a centre $p$ and $R<\infty$ with $d(p,b)\le R$ for all $b\in B$. The closed ball $\{y:d(p,y)\le R\}$ is closed because $|d(p,y)-d(p,z)|\le d(y,z)$. It therefore contains $\overline{B}$. Hence the closure of a bounded set is bounded. Empty sets cause no exception. If $f(A)$ is bounded, its closure is bounded by this result, and the displayed inclusion makes $f(\overline{A})$ bounded too.