Lemma 5.23.8. Let $X$ be a spectral space. The following are equivalent:

1. $X$ is profinite,

2. $X$ is Hausdorff,

3. $X$ is totally disconnected,

4. every quasi-compact open is closed,

5. there are no nontrivial specializations between points,

6. every point of $X$ is closed,

7. every point of $X$ is the generic point of an irreducible component of $X$,

8. the constructible topology equals the given topology on $X$, and

Proof. Lemma 5.22.2 shows the implication (1) $\Rightarrow$ (3). Irreducible components are closed, so if $X$ is totally disconnected, then every point is closed. So (3) implies (6). The equivalence of (6) and (5) is immediate, and (6) $\Leftrightarrow$ (7) holds because $X$ is sober. Assume (5). Then all constructible subsets of $X$ are closed (Lemma 5.23.6), in particular all quasi-compact opens are closed. So (5) implies (4). Since $X$ is sober, for any two points there is a quasi-compact open containing exactly one of them, hence (4) implies (2). Parts (4) and (8) are equivalent by the definition of the constructible topology. It remains to prove (2) implies (1). Suppose $X$ is Hausdorff. Every quasi-compact open is also closed (Lemma 5.12.4). This implies $X$ is totally disconnected. Hence it is profinite, by Lemma 5.22.2. $\square$

Comment #4225 by Laurent Moret-Bailly on

Suggestion for (8): the constructible topology is the original one. (Obviously, this is another way to state (4) but it could be worth mentioning).

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