**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.
$\square$

## Comments (2)

Comment #4225 by Laurent Moret-Bailly on

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