The cofinite topology on an infinite set, and the cocountable topology on $\mathbb{R}$, are $T_1$ with a diagonal whose closure is the whole square; on a countably infinite set the cocountable topology is discrete instead

Standard topologies are as in The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, diagonals as in The diagonal ${\mathrm{\Delta }}_X\subseteq X\times X$, the diagonal map ${\delta }_X$, and the pairing $⟨f,g⟩$ of two maps, and every square carries the product topology (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

1. Cofinite, on an infinite set. Let $X$ be infinite (Finite, countably infinite, countable, uncountable) and give it the cofinite topology ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}$. Then $(X,{\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}})$ is $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces), no two nonempty open sets are disjoint, $\overline{{\mathrm{\Delta }}_X}=X\times X\mathrm{\ne }{\mathrm{\Delta }}_X,$ so ${\mathrm{\Delta }}_X$ is not closed and the space is not Hausdorff.
2. Cocountable, on $\mathbb{R}$. Give $\mathbb{R}$ the cocountable topology ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{c}}$. The same three conclusions hold: $(\mathbb{R},{\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{c}})$ is $T_1$, no two nonempty open sets are disjoint, and $\overline{{\mathrm{\Delta }}_{\mathbb{R}}}=\mathbb{R}\times \mathbb{R}\mathrm{\ne }{\mathrm{\Delta }}_{\mathbb{R}}$.
3. "Infinite" is the wrong hypothesis for the cocountable half. If $Z$ is countably infinite then ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{c}}$ on $Z$ is the discrete topology, which is Hausdorff and whose diagonal is therefore closed. So clause 2 must be asserted of a set large enough that a cocountable set is a genuine restriction, and $\mathbb{R}$ is such a set; an arbitrary infinite set is not.

In every case the verdict on the diagonal matches A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology, as it must.

• Why clause 3 is stated rather than left implicit. The cofinite and the cocountable topologies behave alike only when the underlying set is large enough for the excluded sets to be a genuine restriction. On a countably infinite set "at most countable complement" excludes nothing, so the cocountable topology collapses to the discrete one and every conclusion of clause 2 reverses. Stating the two clauses with the same hypothesis would be a falsehood, and the falsehood is invisible unless the degenerate case is written out.

• The closure of the diagonal is as large as it can be. In both spaces of clauses 1 and 2 it is the entire square, so the diagonal is not merely non-closed: it is dense. That is the extreme opposite of the metric picture of The diagonal of $\mathbb{R}$ is closed in ${\mathbb{R}}^2$, computed from the product basis, where the diagonal is closed and its complement is open.

• $T_1$ is doing no work here. Both spaces satisfy $T_1$ and neither satisfies $T_2$, which is exactly the separation between the two axioms; the diagonal criterion detects the second and is blind to the first, since it is a statement about the square rather than about singletons.