**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 5.23.5), 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). 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.
$\square$

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