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?

About dominiczypen

I'm interested in general topology, order theory, and graph theory. This link takes you to my preprints on arXiv.
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