The specialisation order is a way of turning any topological space into a preordered set (with the same underlying set).
Let if and only if belongs to the closure of . One may also use the opposite convention.
is if and only if its specialisation order is a partial order. is iff its specialisation order is equality. is (like but without ) iff its specialisation order is an equivalence relation.
Given a continuous function between topological spaces, it is order-preserving relative to the specialisation order. Thus, we have a faithful functor from the category of of topological spaces to the category of preordered sets.
If we restrict to a finite underlying set, then the categories and of finite prosets and finite topological spaces are equivalent in this way. The corresponding topology can be recovered from a finite proset through its specialization topology. More generally, the category of Alexandroff spaces (spaces in which an arbitrary intersection of open sets is open) is equivalent to the category of all preordered sets in the same way.
There's an adjunction here; I should think about which way it goes and whether it's a reflection or something. —Toby
Mike: If it exists, it’s either a reflection or a coreflection, since the functor is fully faithful.