## On the topological closure of the diagonal of a space

When we take a class in general topology and learn about $T_2$-spaces, one of the standard exercises we do is:

Prove that a space $X$ is Hausdorff if and only if the diagonal $\Delta_X = \{(x,x) : x \in X\}$ is closed in the product topology of $X \times X$.

Given any topological space $X$, the closure of the diagonal, that is $cl(\Delta_X)$, is a binary relation, which in the case of Hausdorff spaces is pretty boring. So a natural question to ask is: Which binary relations on a set $X$ arise as the closure of the diagonal for a suitable topology $\tau$ on $X$? We will call those binary relations topologically realizable.

It is easy to see that for any topology, $cl(\Delta_X)$ is reflexive and symmetric, and it turns out that a large class of reflexive and symmetric relations are topologically realizable:

Proposition. Any equivalence relation is topologically realizable.

Proof. Let $X$ be a set and $R \subseteq X\times X$ be an equivalence relation. We set $\textrm{Block}(R)=\{[x]_R : x \in X\}$ — so it is the set of equivalence classes or “blocks” that $R$ induces. Using the Axiom of Choice we pick for each block $B \in$ $\textrm{Block}(R)$ a representative $r(B) \in B$ and we define

$\tau = \{ U\subseteq X :$ for all $B \in$ $\textrm{Block}(R)$ : if $U\cap B \neq \emptyset$ then $r(B)\in U\}$.

It is not hard to verify that $\tau$ is indeed a topology. We need to show that $cl(\Delta_X) = R$.

$cl(\Delta_X) \subseteq R$: Let $(x,y) \in (X\times X)\setminus R$. So $U= [x]_R$ and $V= [y]_R$ are disjoint open sets, therefore $U\times V$ is an open neighborhood of $(x,y)$ that does not intersect $\Delta_X$, so $(x,y)$ is not in $cl(\Delta_X)$.

$cl(\Delta_X) \supseteq R$: Let $(x,y) \in R$ that is $[x]_R = [y]_R$. By construction of the open sets, every open neighborhood of $x$ and every open neighborhood of $y$ contains $z := r([x]_R) = r([y]_R)$. Therefore every open neighborhood of $(x,y)$ in the product space $X\times X$ contains $(z,z) \in \Delta_X$ which implies that  $(x,y) \in cl(\Delta_X)$.

Does the converse hold? We might wonder whether conversely, every topologically realizable relation is transitive. It turns out that this is not true; see Example 4.1. in this paper by Maria-Luisa Colasante and me.

I would like to see the following: Is there a set $X$ and a reflexive, symmetric relation $R \subseteq X\times X$ such that $R$ is not topologically realizable?