Finite $T_0$ spaces are sober

Question

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?

score 6 · Accepted Answer · answered Jul 15 '21 at 03:40

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 $X$!

score 5 · Answer 2 · answered Jul 15 '21 at 03:37

5

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 $A$.

answered Jul 15 '21 at 03:37

Eric Wofsey

342,377

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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
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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
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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

score 3 · Answer 3 · edited Dec 30 '24 at 04:58

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Actually we can prove a little bit stronger result:

Proposition: Every locally finite space is quasi-sober.

(A quasi-sober space is a topological space such that every irreducible closed subset is the closure of at least one point)

Then the original statement follows from this proposition, “Finite $\Rightarrow$ Locally finite” (obvious) and “Quasi-sober + $T_0$ $\Rightarrow$ sober” (Different points have different closures in $T_0$).

The proof only needs a slightly modification from @Eric Wofsey's answer:

Proof: Let $A \subseteq X$ be a nonempty irreducible closed subset, then $A$ is also locally finite, i.e., each point in $A$ has a finite neighborhood. Hence, there exists a minimal (finite nonempty) open set $U \subseteq A$.

Let $V \subseteq A$ be another nonempty open set, then $U \cap V \neq \varnothing$ by irreducibility and $U \cap V$ can't be proper in $U$ by minimality.

Therefore every nonempty open set contains $U$ and therefore $A$ is the closure of every point in $U$.

edited Dec 30 '24 at 04:58

PatrickR

7,165

answered Dec 30 '24 at 04:45

Kitsune Kiriha

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Thanks for the writeup. Nice result. – PatrickR Dec 30 '24 at 04:57
the $U$ and the $V$ are open in the subspace topology of $A$, I suppose? – PatrickR Dec 30 '24 at 05:00
Yes all the contents from the second sentence are discussed in the subspace topology of $A$. – Kitsune Kiriha Dec 30 '24 at 05:17
@kaba Actually, it is not the case that every finitely generated space (= Alexandrov space) is quasi-sober. See π-Base, Search for Alexandrov+~Quasi-sober. An easy example is the right ray topology on $\omega$. – PatrickR May 17 '25 at 19:39
@PatrickR Yes, you seem to be correct. With finitely generated spaces the above proof shows only that the closure of U in A equals A. – kaba May 17 '25 at 20:07

Finite $T_0$ spaces are sober

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