The Stacks project

Proof. This follows immediately from Lemma 5.12.13. $\square$

Proof. Follows immediately from Lemma 5.16.1 and the definition of a continuous map. $\square$

Proof. Assume $E$ is constructible and $Z \subset X$ irreducible closed. Then $E \cap Z$ is constructible in $Z$ by Lemma 5.16.2. Hence $E \cap Z$ is a finite union of nonempty locally closed subsets $T_ i$ of $Z$. Clearly if none of the $T_ i$ is open in $Z$, then $E \cap Z$ is not dense in $Z$. In this way we see that (1) implies (2).

Conversely, assume (2) holds. Consider the set $\mathcal{S}$ of closed subsets $Y$ of $X$ such that $E \cap Y$ is not constructible in $Y$. If $\mathcal{S} \not= \emptyset$, then it has a smallest element $Y$ as $X$ is Noetherian. Let $Y = Y_1 \cup \ldots \cup Y_ r$ be the decomposition of $Y$ into its irreducible components, see Lemma 5.9.2. If $r > 1$, then each $Y_ i \cap E$ is constructible in $Y_ i$ and hence a finite union of locally closed subsets of $Y_ i$. Thus $E \cap Y$ is a finite union of locally closed subsets of $Y$ too and we conclude that $E \cap Y$ is constructible in $Y$ by Lemma 5.16.1. This is a contradiction and so $r = 1$. If $r = 1$, then $Y$ is irreducible, and by assumption (2) we see that $E \cap Y$ either (a) contains an open $V$ of $Y$ or (b) is not dense in $Y$. In case (a) we see, by minimality of $Y$, that $E \cap (Y \setminus V)$ is a finite union of locally closed subsets of $Y \setminus V$. Thus $E \cap Y$ is a finite union of locally closed subsets of $Y$ and is constructible by Lemma 5.16.1. This is a contradiction and so we must be in case (b). In case (b) we see that $E \cap Y = E \cap Y'$ for some proper closed subset $Y' \subset Y$. By minimality of $Y$ we see that $E \cap Y'$ is a finite union of locally closed subsets of $Y'$ and we see that $E \cap Y' = E \cap Y$ is a finite union of locally closed subsets of $Y$ and is constructible by Lemma 5.16.1. This contradiction finishes the proof of the lemma. $\square$

Proof. It is clear that (1) implies (2). Assume (2). Consider the set $\mathcal{S}$ of closed subsets $Y$ of $X$ containing $x$ such that $E \cap Y$ is not a neighbourhood of $x$ in $Y$. If $\mathcal{S} \not= \emptyset$, then it has a minimal element $Y$ as $X$ is Noetherian. Suppose $Y = Y_1 \cup Y_2$ with two smaller nonempty closed subsets $Y_1$, $Y_2$. If $x \in Y_ i$ for $i = 1, 2$, then $Y_ i \cap E$ is a neighbourhood of $x$ in $Y_ i$ and we conclude $Y \cap E$ is a neighbourhood of $x$ in $Y$ which is a contradiction. If $x \in Y_1$ but $x \not\in Y_2$ (say), then $Y_1 \cap E$ is a neighbourhood of $x$ in $Y_1$ and hence also in $Y$, which is a contradiction as well. We conclude that $Y$ is irreducible closed. By assumption (2) we see that $E \cap Y$ is dense in $Y$. Thus $E \cap Y$ contains an open $V$ of $Y$, see Lemma 5.16.3. If $x \in V$ then $E \cap Y$ is a neighbourhood of $x$ in $Y$ which is a contradiction. If $x \not\in V$, then $Y' = Y \setminus V$ is a proper closed subset of $Y$ containing $x$. By minimality of $Y$ we see that $E \cap Y'$ contains an open neighbourhood $V' \subset Y'$ of $x$ in $Y'$. But then $V' \cup V$ is an open neighbourhood of $x$ in $Y$ contained in $E$, a contradiction. This contradiction finishes the proof of the lemma. $\square$

Proof. This follows formally from Lemmas 5.16.3 and 5.16.4. $\square$