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One could ask this question for any topological property $P$ of a finite topological space:

What proportion of topological spaces with $n$ points (up to homeomorphism?) have property $P$?

Are there any familiar properties of topological spaces for which this question has an interesting answer as $n$ approaches in infinite? By interesting I suppose I mean that the proportion doesn't approach either $1$ or $0$. And I'm not sure how much the "up to homeomorphism" condition matters in this question.

Mike Pierce
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    Are you familiar with the work on classifying finite spaces up to homeo? It's incredibly complicated. – Randall Sep 09 '19 at 18:06
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    @Randall oh yeah, I know it's complicated. Are there any asymptotic-sort of results as to how many homeomorphism-classes of spaces there are on $n$ points? But anyways, I think there being an interesting answer to this question is a long shot, and I don't know whether considering the spaces up to homeomorphism or not increases the odds. :) – Mike Pierce Sep 09 '19 at 18:12
  • If you don't consider things up to homeo then it's impossible to answer. – Randall Sep 09 '19 at 18:15
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    @Randall See I was leaning towards to exact opposite conclusion: if you consider things up to homeomorphism it might be impossible, but without that consideration this is more of a purely combinatorial question. – Mike Pierce Sep 09 '19 at 18:32
  • If you don't identify homeomorphic spaces then for every $n>0$ there is a proper class of spaces with $n$ points – Alessandro Codenotti Sep 09 '19 at 20:02
  • @AlessandroCodenotti Oh you must know what I really mean though. ... Like, should I really have to say we're considering topological spaces up to set isomorphism? – Mike Pierce Sep 09 '19 at 20:06
  • I don't know what you mean with "not up to homeomorphism" then. What is a pair of spaces that you'd consider the same when considering spaces up to homeomorphism but distinct otherwise? – Alessandro Codenotti Sep 09 '19 at 20:08
  • @AlessandroCodenotti The space with two points ${x,y}$ has two nonsilly topologies. The two topologies are distinct because one has the open set ${x}$ whereas the other has the open set ${y}$, but the topologies are the same up to homeomorphism. – Mike Pierce Sep 09 '19 at 20:16
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    @AlessandroCodenotti Just fix the set $X={1,\ldots,n}$ and ask how many of the finitely many subsets of $\mathscr{P}(X)$ are topologies (we have "just" $2^{2^n}$ such candidates, really fewer, as $\emptyset$ and $X$ must be in it). And then count the equivalence classes modulo "being homeomorphic", which is probably an order of magnitude smaller. No "proper classes" are involved, just a finite (but big) counting problem. – Henno Brandsma Sep 09 '19 at 21:22
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    @alessandro you can fix n points and ask for all topologies on that set. This does strike me as an easier question than finding all topologies on that set up to homeomorphism. – Cheerful Parsnip Sep 09 '19 at 21:23

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For "compact" the answer is all, obviously. Likewise for countably compact or Lindelöf. So covering properties are not very interesting.

Neither are separation axioms, mostly: counting $T_0$ topologies on finite sets is just counting partial orders, and so open (but interesting). $T_1$ and $T_2$ implies discrete (so only one topology on $n$ points obeys it).

It matters quite a lot whether we just count all topologies on $\{1,\ldots,n\}$ or all topologies up to homeomorphism. For $n=3$ we have $9$ essentially different topologies (classes of topologies under homeomorphism), but $29$ in total, see my answer here.

There's been some research on asymptotics but I know of no definite results beyond the trivial ones I mentioned. Already $T_0$ and connected are already tricky to count.

Henno Brandsma
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