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Let $X$ be a finite set of $n$ ($>1$) elements and $\tau$ be a topology on $X$ having exactly $m$ elements.

Can we give any description of $m$ as it relates to $n$?


Obviously $2\le m\le 2^n$ and the extreme cases are attained: $m=2$ for the indiscrete topology and $m=2^n$ for the discrete topology.


For $n=2$, $m=2, 3, 4$ are all possible.

For $n=3$, $m=2, 3,\dotsc, 6, 8$ are possible, only $m=7$ isn't possible.

For $n=4$, $m=2, 3,\dotsc, 10, 12, 16$ are possible, $m=11, 13, 15$ aren't possible.

Is it possible to obtain a result similar to Lagrange's theorem in group theory (in a finite group, the order of a subgroup must divide the order of the group)?

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    $\begingroup$ One small comment is that another good way to study finite topological spaces is via their specialisation preorders. This gives an equivalence of categories to finite preorders, so the question is how many upwards closed sets a preorder on $n$ elements has. $\endgroup$ May 22, 2023 at 11:19
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    $\begingroup$ The paper by Kari Ragnarsson and Bridget Tenner at arxiv.org/abs/0802.2550 is relevant. $\endgroup$ May 22, 2023 at 13:14
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    $\begingroup$ OEIS doesn't know this sequence. $\endgroup$
    – LSpice
    May 25, 2023 at 11:14
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    $\begingroup$ The paper mentioned by Richard Stanley is Obtainable sizes of topologies on finite sets, published as doi.org/10.1016/j.jcta.2009.05.002 $\endgroup$
    – David Roberts
    May 25, 2023 at 12:52
  • $\begingroup$ Well, the OEIS does describe the sequence alluded to by Richard Stanley: oeis.org/A137813 $\endgroup$ May 25, 2023 at 13:29

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