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

The closure $\overline{A}$ is the intersection of every closed subset of $X$ containing $A$. Thus it is the smallest such closed subset. Equivalently,For equivalence, if an open neighbourhood misses $A$, its complement is a closed superset of $A$ missing $x$. Conversely, a closed superset missing $x$ has an open complement that is a neighbourhood of $x$ disjoint from $A$.

Suppose $f$ is continuous and $x\in \overline{A}$. Every open neighbourhood $V$ of $f(x)$ has an open preimage containing $x$, which meets $A$. Hence $V$ meets $f(A)$, proving $f(x)\in \overline{f(A)}$.

Conversely, suppose the indicated inclusion holds for every subset. To prove continuity, take any closed set $C\subseteq Y$ and put $A=f^{-1}(C)$. Since $f(A)\subseteq C$ and $C$ is closed,Thus $\overline{A}\subseteq f^{-1}(C)=A$, so $A$ is closed. Every closed set has closed preimage, which is the closed-set characterization of continuity.This is the closure characterization of continuity; it requires no Hausdorff or metric hypothesis.

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.

The quotient set $X/\mathcal{R}$ consists of the equivalence classes. Let $p(x)=[x]$. Equip that set with the quotient topologyPreimages preserve the empty set, whole set, arbitrary unions and finite intersections, so this is a topology. The canonical projection is continuous and surjective, and the displayed equivalence specifies the quotient space completely.

For a nonempty open subset $U$ of the quotient, $p^{-1}(U)$ is nonempty and open in the cofinite space. Thereforeis finite. Surjectivity implies that $(X/\mathcal{R})\setminus U$ is finite as well: it is the image of that finite preimage. Hence every quotient-open set is cofinite or empty, proving the desired inclusion.

Equality is not necessary. Give $X=\mathbb{Z}$ its cofinite topology and use two equivalence classes, the even and odd integers. Neither singleton of the quotient is open, since its preimage and its complement are both infinite. The quotient is the two-point indiscrete topology, whereas the cofinite topology on a finite two-point set is discrete.This is the quotient topology of a cofinite space phenomenon, not a failure of continuity of the projection.