Lemma 60.2.4. 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 $Y_0 \subset Y$ 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.

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• 2 comment(s) on Section 60.2: Some topology

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