Lemma 66.15.2. 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 66.15.1 we see that |X| is sober. By Lemma 66.5.2 we see that |X| is quasi-compact. By Lemma 66.6.3 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 66.4.8) and we can choose affine schemes V and W and surjective étale morphisms V \to X' and W \to X'' (Lemma 66.6.3). Then A \cap B is the image of |V \times _ X W| \to |X| (Lemma 66.4.3). Since V \times _ X W is quasi-compact as X is quasi-separated (Lemma 66.3.3) we conclude that A \cap B is quasi-compact and the proof is finished. \square
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