Past exam of the topology course of the University of Oxford / 2014 / 1 / b

A proper closed subset in the cofinite topology is finite, so cannot contain the infinite subset $A$. Its only closed superset is therefore the whole space:In particular, every infinite subset is dense in this topology.

Because $f$ is surjective, $f^{-1}(B)$ is infinite whenever $B$ is infinite; a finite preimage could have only finitely many image points. By part (i), its closure is $X$. Apply part (a)(ii):The other inclusion is automatic, soBoth continuity and surjectivity enter the argument.

Any proper closed subset $F$ of $Y$ must be finite: if it were infinite, part (ii) would give $F=\overline{F}=Y$. Therefore every nonempty open subset of $Y$, being the complement of a proper closed set, is cofinite.The empty set belongs to both topologies. Inclusion need not be equality; the target may have fewer open sets.

Any two nonempty open sets of $Y$ are cofinite by part (iii). Their complements have finite union, so the two sets must intersect because $Y$ is infinite. Thus $Y$ is a hyperconnected space.

A Hausdorff space with at least two points has disjoint nonempty open neighbourhoods of two distinct points. This is impossible here. Therefore $Y$ is not Hausdorff.