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Given a finite set $X$ with cardinality $n$. There exist finitely many distinct topologies $T$ on the set.

Every topology also has its own cardinality $k$. And I'm wondering what's the distribution of $K$ be like?

peng yu
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  • See my answer to Finite topological Hausdorff spaces, Matt Samuel's answer to Estimates on number of topologies on a finite set, and Sloane sequence A000798. – Dave L. Renfro Sep 22 '21 at 15:48
  • I believe there is a difference between your links and my question, they were meant to count the number of all possible topologies to $X$, meanwhile, the question is about, for all possible topologies to $X$ what's the distribution of their cardinality. e.g. $k=1$, it's impossible, as trivial topology has $k=2$. it's aiming to group topologies via $k\in [2, 2^n]$ @DaveL.Renfro – peng yu Sep 22 '21 at 16:08
  • If not even the number of distinct topologies on $X$ is known, what can be said about the distribution (except for small $n$)? – Paul Frost Sep 22 '21 at 22:21
  • it's a fair point, didn't expect it to be that hard. Meanwhile, maybe understanding about the distribution can help solving the total number. – peng yu Sep 22 '21 at 22:24
  • @PaulFrost that's not necessarily a good observation. For example we don't know exactly how many primes are between given two numbers, yet we know quite a bit about their distribution. – freakish Sep 23 '21 at 11:17
  • That being said, I'm not sure what "distribution" means in this context. @pengyu are you asking about the number of topologies on $X$ grouped by each cardinality $k$? – freakish Sep 23 '21 at 11:19
  • Yes, first $k$ must have some range, second there must be some pattern on how many different topology when $k=5$ for example – peng yu Sep 23 '21 at 12:32
  • @freakish You are right, even if we do not need to know the exact number of topologies on an $n$-point set $X_n$, it could theoretically be possible to get an "asymptotic estimate" of some distribution (where the meaning of "distribution" must be defined). But the problem here is that we have not enough instances to make a guess (as in the case of prime numbers). Only for very small $n$ there are data available. – Paul Frost Sep 23 '21 at 16:30
  • Look at https://cs.uwaterloo.ca/journals/JIS/VOL9/Benoumhani/benoumhani11.pdf. – Paul Frost Sep 23 '21 at 16:36
  • thanks for the paper, it is very relevant, kind of curious, if $k$ is continuous on $[2, 2^n]$, maybe it's trivial to prove? – peng yu Sep 23 '21 at 17:09

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