On an infinite set the cofinite topology has every infinite subset dense and no two nonempty open sets disjoint

Let $X$ be an infinite set with the cofinite topology ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}$, whose open sets are $\mathrm{\varnothing }$ together with the sets of finite complement (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Finite, countably infinite, countable, uncountable). Then:

1. Closures. For $A\subseteq X$, $\bar{A}=\{\begin{array}{ll}A& A\text{ finite}\\ X& A\text{ infinite,}\end{array}\mathrm{int}(A)=\{\begin{array}{ll}A& X\setminus A\text{ finite}\\ \mathrm{\varnothing }& X\setminus A\text{ infinite.}\end{array}$
2. A subset is dense if and only if it is infinite (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets); in particular every infinite subset is dense, and no finite subset is.
3. No two nonempty open sets are disjoint. If $U,V\in {\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}$ are nonempty then $U\cap V\mathrm{\ne }\mathrm{\varnothing }$.
4. Every singleton is closed, so points are distinguishable by closed sets; nevertheless claim 3 says distinct points are never separated by disjoint open sets, so the space is as far from Hausdorff as a space with closed points can be.

• Claim 3 is the reason the cofinite topology is a standard counterexample factory. A space with at least two points in which any two nonempty open sets meet cannot be metrizable (Distinct points of a metric space have disjoint balls around them, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), since a metric separates distinct points by disjoint balls. It does not follow that sequential limits fail to be unique there: the cocountable topology on $\mathbb{R}$ also has no two disjoint nonempty open sets and its sequential limits are unique (In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$, and a sequence converges iff it is eventually constant). In the cofinite topology uniqueness does fail, but the argument needs an injective sequence and not merely the meeting of open sets (In the indiscrete topology every sequence converges to every point, and in the cofinite topology on an infinite set an injective sequence converges to every point).

• Finiteness of $X$ collapses the example. If $X$ were finite then every subset would have finite complement, the topology would be discrete, and all four claims would read differently or vacuously. The hypothesis that $X$ is infinite is used in steps 1.5 and 2.2 and is not decoration.

• The same computation with "at most countable" in place of "finite" gives the cocountable topology, whose behaviour on $\mathbb{R}$ is the subject of the next example (In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$, and a sequence converges iff it is eventually constant). The two differ in exactly one respect that matters here: on $\mathbb{R}$ the cocountable topology still has closed points and no two disjoint nonempty open sets, but its convergent sequences are far more restricted.