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?