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

Prove that a space is Hausdorff if and only if the diagonal is closed in the product topology of .

Given any topological space , the closure of the diagonal, that is , 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** arise as the closure of the diagonal for a suitable topology on ? We will call those binary relations **topologically realizable**.

It is easy to see that for any topology, 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 be a set and be an equivalence relation. We set — so it is the set of equivalence classes or “blocks” that induces. Using the Axiom of Choice we pick for each block a representative and we define

for all : if then .

It is not hard to verify that is indeed a topology. We need to show that .

: Let . So and are disjoint open sets, therefore is an open neighborhood of that does not intersect , so is not in .

: Let that is . By construction of the open sets, every open neighborhood of and every open neighborhood of contains . Therefore every open neighborhood of in the product space contains which implies that .

**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 and a reflexive, symmetric relation such that is not topologically realizable?

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