# Finite $T_0$ spaces are sober

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?

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 $$X$$!
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 $$A$$.
• 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. Jul 15 at 3:45
• Yes, I was assuming OP was aware that $T_0$ is equivalent to the "uniqueness" part of sobriety. Jul 15 at 3:47