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
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