Proposition 66.12.4. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then the topological space $|X|$ is sober (see Topology, Definition 5.8.4).

**Proof.**
We have seen in Lemma 66.12.3 that $|X|$ is Kolmogorov. Hence it remains to show that every irreducible closed subset $T \subset |X|$ has a generic point. By Properties of Spaces, Lemma 64.12.3 there exists a closed subspace $Z \subset X$ with $|Z| = |T|$. By definition this means that $Z \to X$ is a representable morphism of algebraic spaces. Hence $Z$ is a decent algebraic space by Lemma 66.5.3. By Theorem 66.10.2 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 $\eta \in T$ and hence also $T = \overline{\{ \eta \} }$, and we win.
$\square$

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