TITLE: Classification of Finite Topologies
QUESTION [12 upvotes]: 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.

REPLY [8 votes]: 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.