Lemma 65.15.2. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. The topological space $|X|$ is a spectral space.

Proof. By Topology, Definition 5.23.1 we have to check that $|X|$ is sober, quasi-compact, has a basis of quasi-compact opens, and the intersection of any two quasi-compact opens is quasi-compact. By Lemma 65.15.1 we see that $|X|$ is sober. By Lemma 65.5.2 we see that $|X|$ is quasi-compact. By Lemma 65.6.3 there exists an affine scheme $U$ and a surjective étale morphism $f : U \to X$. Since $|f| : |U| \to |X|$ is open and continuous and since $|U|$ has a basis of quasi-compact opens, we conclude that $|X|$ has a basis of quasi-compact opens. Finally, suppose that $A, B \subset |X|$ are quasi-compact open. Then $A = |X'|$ and $B = |X''|$ for some open subspaces $X', X'' \subset X$ (Lemma 65.4.8) and we can choose affine schemes $V$ and $W$ and surjective étale morphisms $V \to X'$ and $W \to X''$ (Lemma 65.6.3). Then $A \cap B$ is the image of $|V \times _ X W| \to |X|$ (Lemma 65.4.3). Since $V \times _ X W$ is quasi-compact as $X$ is quasi-separated (Lemma 65.3.3) we conclude that $A \cap B$ is quasi-compact and the proof is finished. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).