Lemma 28.2.4. Let $X$ be a quasi-compact and quasi-separated scheme. Then the underlying topological space of $X$ 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. This follows from Schemes, Lemma 26.11.1 and 26.11.2 and Lemma 28.2.3 above.
$\square$

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