Classification of Finite Topologies

Question

Does there exist a classification of finite topologies?

I define a finite topology as a finite Set $T$ of Sets which respects the following properties:

• $\forall a,b \in T: a \cap b \in T$,
• $\forall a,b \in T: a \cup b \in T$,
• $ \emptyset \in T$,
• $\exists S\in T\ |\ \forall a \in T , a \subseteq S$.

This seems like a natural thing to do in the vein of classifying finite groups, so i'm curious what current research in this area looks like.

Answer 1

Finite topologies and finite preorders (reflexive & transitive relations) are equivalent:

Let $T$ be a topological space with finite topology $\mathcal{O}$. Define $\leq$ on $T$ by: $$x\leq y \Leftrightarrow \forall U\in \mathcal{O} : x\in U \Rightarrow y\in U$$

Then $\leq$ is clearly a preorder, called the specialization order of $T$.

Given a preorder $\leq$ on $T$, define the set $\mathcal{O}$ to be set of all upwards-closed sets in $(T,\leq)$, that is all sets $U$ with:

$$\forall x,y\in T : x\leq y \text{ and } x\in U \Rightarrow y\in U$$

Then $\mathcal{O}$ is a topology, called the specialization topology or Alexandroff topology of $(T,\leq)$.

The constructions are functorial and can be turned into an equivalence of categories $\mathsf{FinTop}$ and $\mathsf{FinPros}$ (I don't have time to work out the details right now, however).

Answer 2

There is a huge amount of literature about finite topologies. Actually this topic is one of the major chapters in universal algebra, under the name of distributive lattices. Namely, sets $L$ endowed with two associative, commutative and idempotent operations $\vee$ (“join”) and $\wedge$ (“meet”) which furthermore satisfy the following equations: $$ x\vee(x\wedge y) = x = x\wedge(x\vee y) $$ (absorption), and $$ x\vee(y\wedge z) = (x\vee y)\wedge (x\vee z) $$ $$ x\wedge(y\vee z) = (x\wedge y)\vee (x\wedge z) $$ (distributivity). In the case at hand, we are looking at bounded distributive lattices, i.e., having two elements $0$ and $1$ that satisfy $$ x \vee 0 = x \qquad x \vee 1 = 1 $$ for all $x\in L$. You'll check immediately that every finite topology on a set $S$ is a concrete interpretation of this axioms, since $\cup$ and $\cap$, $\emptyset$ and $S$ satisfy the defining identities. Moreover, every finite bounded distributive lattice is isomorphic to some finite topology on a finite set (considered as an algebraic structure): This follows from Priestley's representation theorem.

Just perform a web search for more on this.