One knows that the Alexandrov topology on a preordered set is the finest topology that induces the same [specialization] preorder on the set.

Given this, one finds a one-to-one correspondence between the Alexandrov topologies on a set and the pre-orders on that set.

On the other hand, every pre-order can be characterised by a thin category.

My question is:

If there's anyway to formulate the Alexandrov topology on the pre-ordered set totally algebraically in terms of the thin category?