Lemma 61.2.4 (096B)—The Stacks project

Table of contents
Part 3: Topics in Scheme Theory
Chapter 61: Pro-étale Cohomology
Section 61.2: Some topology
Lemma 61.2.4 (cite)

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 February 07, 2014 at 14:32

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.

Comment #444 by Johan on February 08, 2014 at 13:06

Thank you very much. Your proof was added here.

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