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.

  • $\begingroup$ I seem to recall seeing somewhere that every finite CW complex is weakly homotopy equivalent to a finite topological space. This pretty much makes classification hopeless if true. $\endgroup$ Dec 20, 2015 at 2:06
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    $\begingroup$ @MattSamuel: Classifying finite spaces up to weak equivalence is hard, but classifying them up to isomorphism is "easy" (they are just posets). The hard part of classifying them up to weak equivalence is determining when two posets have weak equivalent nerves. $\endgroup$ Dec 20, 2015 at 3:26
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    $\begingroup$ One paper I've spent a lot of time reading (in an attempt to do something similar) is about estimating the number of topologies on a finite set. "On the number of finite topologies" by Kleitman (the man with the smallest Erdos-Bacon number and, I believe, Michael Saks of Rutgers' advisor) and Rothschild. $\endgroup$ Dec 20, 2015 at 3:40

2 Answers 2


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).

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    $\begingroup$ Ah this is simple so then we are simply tasked with finding all directed, acyclic graphs with a single source and single sink. And each non isomorphic graph corresponds to a different topology $\endgroup$ Dec 19, 2015 at 22:59
  • $\begingroup$ @frogeyedpeas Almost. Note, that we do not assume antisymmetry. That is: If there is a cycle, then every vertix $v$ in that cycle has an edge $e : v \to w$ for every other vertix $w$ in that cycle. Similarly for "sources" and "sinks": there are unique only up to "isomorphism", meaning there is a pair of antiparallel edges between any two candidates. $\endgroup$ Dec 19, 2015 at 23:11
  • $\begingroup$ @frogeyedpeas: There need not be a unique source and sink. It's just any finite partially ordered set (possibly with indistinguishable copies of points, i.e. distinct points $x$ and $y$ such that $x\leq y$ and $y\leq x$). $\endgroup$ Dec 20, 2015 at 3:25

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.

  • $\begingroup$ What do you mean by "every finite bounded distributive lattice is isomorphic to some finite topology on a finite set"? $\endgroup$ Dec 21, 2015 at 9:36
  • $\begingroup$ @StefanPerko I mean that, given such an $L$, there is a finite topology $T$ (as defined by the OP) on a finite set $S$ and a one-one onto mapping $f:L\to T$ such that $f(x\vee y) = f(x)\cup f(y)$, and the same with the other operations. I'll add some hint in the answer. $\endgroup$ Dec 21, 2015 at 12:12
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    $\begingroup$ I am just not sure, whether the statement you are making is: "Every finite topological space is sober.", i.e. whether finite spaces are completely determined by their lattice of open sets. This fails for example for finite sets with more then one element and the trivial topology. What I am saying is: Mapping spaces to their lattice of open sets is not an injective mapping. $\endgroup$ Dec 21, 2015 at 12:37
  • $\begingroup$ @StefanPerko You're right, but as you can check, that's not what I said. This is a characterization of finite topologies (as rings of sets), albeit not one of finite topological spaces. Thanks for your remark. $\endgroup$ Dec 21, 2015 at 14:25
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    $\begingroup$ Fair enough. More generally, every topology is a complete heyting algebra (a.k.a. locale). $\endgroup$ Dec 21, 2015 at 14:38

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