Lemma 5.16.3. Let $X$ be a Noetherian topological space. Let $E \subset X$ be a subset. The following are equivalent:

1. $E$ is constructible in $X$, and

2. for every irreducible closed $Z \subset X$ the intersection $E \cap Z$ either contains a nonempty open of $Z$ or is not dense in $Z$.

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. $\square$

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