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I've wondered for a while whether or not most finite topologies are $T_{0}$, in the sense that as n goes to infinity, does the proportion of finite topologies of size n that are $T_{0}$ go to 1 or to 0? or does it limit to something inbetween? or not at all?

I'm thinking about the number of finite topologies up to homeomorphism, but if there are similar results about, for example, random finite topologies I would also be interested to hear them. I had at first assumed it would be obvious from the first terms of the sequences in the OEIS, but they seem to grow at similar rates.

https://oeis.org/A000112

https://oeis.org/A001930

(the number of $T_{0}$ topologies of size n is the same as the number of partial orders, due to the isomorphism of categories between Alexandrov topologies and preorders)

HallaSurvivor
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Zoe Allen
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1 Answers1

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Write

  • $T(n)$ for the number of topologies on an $n$ element set
  • $T_0(n)$ for the number of $T_0$ topologies
  • $T^c(n)$ for the number of connected topologies
  • $T^c_0(n)$ for the number of connected $T_0$ topologies

There's a nice paper Counting Finite Posets and Topologies by Marcel Erné and Kurt Stege which mentions (at the bottom of written page 261) a result by the first author that these four sequences are all asymptotically equivalent. In particular, $\frac{T_0(n)}{T(n)} \to 1$, as you expect.

Unfortunately, the only reference for this fact that I can find is the author's original paper Struktur- und anzahlformeln für topologien auf endlichen mengen which (as you may have guessed) is in German.

That said, we can still tease out the key idea, which is mentioned on written page 252 of Counting Finite Posets and Topologies. We work with preorders and partial orders instead (which, as you've mentioned, is equivalent).

We can get every preorder on $n$ elements by choosing a partial order on $k \leq n$ elements and "blowing up" these $k$ points into $n$-many total. Essentially we're undoing the quotient from a preorder onto its partial order by un-identifying those $x,y$ with $x \leq y$ and $y \leq x$. Of course, we can count the number of ways to do such un-quotienting using the stirling numbers, and then we get the claim by pushing on this idea for a while.

I'm not sure how you get the same asymptotics for connected topologies, unfortunately.

You initially asked about the number of topologies up to homeomorphism, and the claim above is for all topologies. I remember reading somewhere that we also get asymptotic equivalence in the up-to-homeomorphism case, but I can't seem to find a reference right now... The best I can do is Kleitman and Rothschild's The Number of Finite Topologies.

If we write $T'(n)$ (resp. $T'_0(n)$) for the number of (resp. $T_0$) topologies on a set of size $n$ up to homeomorphism, then these authors show $\log T$, $\log T'$, $\log T_0$, and $\log T'_0$ are all asymptotically equivalent.

If somebody happens to find a reference showing that $T'$ and $T'_0$ are asymptotically equivalent, please post it here! I don't have time to look any longer, unfortunately.

I hope this helps ^_^

HallaSurvivor
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