Lemma 67.23.4. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then $X$ is Jacobson if and only if $|X|$ is Jacobson.

**Proof.**
Assume $X$ is Jacobson and that $T \subset |X|$ is a closed subset. By Morphisms of Spaces, Lemma 66.25.6 we see that $T \cap X_{\text{ft-pts}}$ is dense in $T$. By Lemma 67.23.3 we see that $X_{\text{ft-pts}}$ are the closed points of $|X|$. Thus $|X|$ is indeed Jacobson.

Assume $|X|$ is Jacobson. Let $f : U \to X$ be an étale morphism with $U$ an affine scheme. We have to show that $U$ is Jacobson. If $x \in |X|$ is closed, then the fibre $F = f^{-1}(\{ x\} )$ is a finite (by definition of decent) closed (by construction of the topology on $|X|$) subset of $U$. Since there are no specializations between points of $F$ (Lemma 67.12.1) we conclude that every point of $F$ is closed in $U$. If $U$ is not Jacobson, then there exists a non-closed point $u \in U$ such that $\{ u\} $ is locally closed (Topology, Lemma 5.18.3). We will show that $f(u) \in |X|$ is closed; by the above $u$ is closed in $U$ which is a contradiction and finishes the proof. To prove this we may replace $U$ by an affine open neighbourhood of $u$. Thus we may assume that $\{ u\} $ is closed in $U$. Let $R = U \times _ X U$ with projections $s, t : R \to U$. Then $s^{-1}(\{ u\} ) = \{ r_1, \ldots , r_ m\} $ is finite (by definition of decent spaces). After replacing $U$ by a smaller affine open neighbourhood of $u$ we may assume that $t(r_ j) = u$ for $j = 1, \ldots , m$. It follows that $\{ u\} $ is an $R$-invariant closed subset of $U$. Hence $\{ f(u)\} $ is a locally closed subset of $X$ as it is closed in the open $|f|(|U|)$ of $|X|$. Since $|X|$ is Jacobson we conclude that $f(u)$ is closed in $|X|$ as desired. $\square$

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