Lemma 66.15.1. Let $S$ be a scheme. Let $X$ be a Zariski locally quasi-separated algebraic space over $S$. Then the topological space $|X|$ is sober (see Topology, Definition 5.8.6).

**Proof.**
Combining Topology, Lemma 5.8.8 and Lemma 66.6.6 we see that we may assume that there exists an affine scheme $U$ and a surjective, quasi-compact, étale morphism $U \to X$. Set $R = U \times _ X U$ with projection maps $s, t : R \to U$. Applying Lemma 66.6.7 we see that the fibres of $s, t$ are finite. It follows all the assumptions of Topology, Lemma 5.19.8 are met, and we conclude that $|X|$ is Kolmogorov^{1}.

It remains to show that every irreducible closed subset $T \subset |X|$ has a generic point. By Lemma 66.12.3 there exists a closed subspace $Z \subset X$ with $|Z| = |T|$. Note that $U \times _ X Z \to Z$ is a quasi-compact, surjective, étale morphism from an affine scheme to $Z$, hence $Z$ is Zariski locally quasi-separated by Lemma 66.6.6. By Proposition 66.13.3 we see that there exists an open dense subspace $Z' \subset Z$ which is a scheme. This means that $|Z'| \subset T$ is open dense. Hence the topological space $|Z'|$ is irreducible, which means that $Z'$ is an irreducible scheme. By Schemes, Lemma 26.11.1 we conclude that $|Z'|$ is the closure of a single point $\eta \in |Z'| \subset T$ and hence also $T = \overline{\{ \eta \} }$, and we win. $\square$

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