Lemma 68.23.5. Let $S$ be a scheme. Let $X$ be a decent locally Noetherian algebraic space over $S$. Let $x \in |X|$. Then
is a Noetherian, spectral, sober, Jacobson topological space.
Proof. We may replace by any open subspace containing $x$. Thus we may assume that $X$ is quasi-compact. Then $|X|$ is a Noetherian topological space (Properties of Spaces, Lemma 66.24.2). Thus $W$ is a Noetherian topological space (Topology, Lemma 5.9.2).
Combining Lemma 68.14.1 with Properties of Spaces, Lemma 66.15.2 we see that $|X|$ is a spectral topological space. By Topology, Lemma 5.24.7 we see that $W \cup \{ x\} $ is a spectral topological space. Now $W$ is a quasi-compact open of $W \cup \{ x\} $ and hence $W$ is spectral by Topology, Lemma 5.23.5.
Let $E \subset W$ be an irreducible closed subset. Then if $Z \subset |X|$ is the closure of $E$ we see that $x \in Z$. There is a unique generic point $\eta \in Z$ by Proposition 68.12.4. Of course $\eta \in W$ and hence $\eta \in E$. We conclude that $E$ has a unique generic point, i.e., $W$ is sober.
Let $x' \in W$ be a point such that $\{ x'\} $ is locally closed in $W$. To finish the proof we have to show that $x'$ is a closed point of $W$. If not, then there exists a nontrivial specialization $x' \leadsto x'_1$ in $W$. Let $U$ be an affine scheme, $u \in U$ a point, and let $U \to X$ be an étale morphism mapping $u$ to $x$. By Lemma 68.12.2 we can choose specializations $u' \leadsto u'_1 \leadsto u$ mapping to $x' \leadsto x'_1 \leadsto x$. Let $\mathfrak p' \subset \mathcal{O}_{U, u}$ be the prime ideal corresponding to $u'$. The existence of the specializations implies that $\dim (\mathcal{O}_{U, u}/\mathfrak p') \geq 2$. Hence every nonempty open of $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u}/\mathfrak p')$ is infinite by Algebra, Lemma 10.61.1. By Lemma 68.12.1 we obtain a continuous map
Since the generic point of the LHS maps to $x'$ the image is contained in $\overline{\{ x'\} }$. We conclude the inverse image of $\{ x'\} $ under the displayed arrow is nonempty open hence infinite. However, the fibres of $U \to X$ are finite as $X$ is decent and we conclude that $\{ x'\} $ is infinite. This contradiction finishes the proof. $\square$
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