Lemma 5.23.5. Let X be a spectral space. Let E \subset X be closed in the constructible topology (for example constructible or closed). Then E with the induced topology is a spectral space.
Proof. Let Z \subset E be a closed irreducible subset. Let \eta be the generic point of the closure \overline{Z} of Z in X. To prove that E is sober, we show that \eta \in E. If not, then since E is closed in the constructible topology, there exists a constructible subset F \subset X such that \eta \in F and F \cap E = \emptyset . By Lemma 5.15.15 this implies F \cap \overline{Z} contains a nonempty open subset of \overline{Z}. But this is impossible as \overline{Z} is the closure of Z and Z \cap F = \emptyset .
Since E is closed in the constructible topology, it is quasi-compact in the constructible topology (Lemmas 5.12.3 and 5.23.2). Hence a fortiori it is quasi-compact in the topology coming from X. If U \subset X is a quasi-compact open, then E \cap U is closed in the constructible topology, hence quasi-compact (as seen above). It follows that the quasi-compact open subsets of E are the intersections E \cap U with U quasi-compact open in X. These form a basis for the topology. Finally, given two U, U' \subset X quasi-compact opens, the intersection (E \cap U) \cap (E \cap U') = E \cap (U \cap U') and U \cap U' is quasi-compact as X is spectral. This finishes the proof. \square
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