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The Stacks project

Lemma 61.2.4. Let X be a w-local spectral space. If Y \subset X is closed, then Y is w-local.

Proof. The subset Y_0 \subset Y of closed points is closed because Y_0 = X_0 \cap Y. Since X is w-local, every y \in Y specializes to a unique point of X_0. This specialization is in Y, and hence also in Y_0, because \overline{\{ y\} }\subset Y. In conclusion, Y is w-local. \square


Comments (2)

Comment #443 by Kestutis Cesnavicius on

Proof: The subset of closed points is closed because . Since is -local, every specializes to a unique point of . This specialization is in , and hence also in , because . In conclusion, is -local.

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