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A sober space is a topological space such that every irreducible closed subset is the closure of exactly one point. Looking for examples I convinced myself that the following is true.

Every finite $T_0$ topological space is sober.

As I could not find this mentioned anywhere, can someone provide a proof to have it as a reference?

PatrickR
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2 Answers2

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This is true. It suffices to show that every finite irreducible space has a generic point, since $T_0$ implies that generic points are unique. So, let $X$ be a finite irreducible space. Then $X$ is the union of the closures of its points, but this is a finite union of closed sets, so irreducibility says that one of these closures is $X$!

diracdeltafunk
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Suppose $X$ is a finite $T_0$ space and $A\subseteq X$ is an irreducible closed subset. Let $B\subset A$ be a maximal closed proper subset (which exists by finiteness), and let $x\in A\setminus B$. By maximality of $B$, $B\cup\overline{\{x\}}$ must be all of $A$. By irreducibility of $A$, this means either $B$ or $\overline{\{x\}}$ must be all of $A$, and thus $\overline{\{x\}}=A$ and $x$ is a generic point of $A$.

Eric Wofsey
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  • Very slick! I just want to point out that the $T_0$ assumption is necessary to deduce that the generic point of $A$ is unique. – diracdeltafunk Jul 15 '21 at 03:45
  • Yes, I was assuming OP was aware that $T_0$ is equivalent to the "uniqueness" part of sobriety. – Eric Wofsey Jul 15 '21 at 03:47
  • Yes, I am aware of that. Finite spaces are quasi-sober (= every irreducible closed subset has a (not necessarily unique) generic point). – PatrickR Jul 15 '21 at 04:11