Lemma 5.15.15. Let $X$ be a topological space. Suppose that $Z \subset X$ is irreducible. Let $E \subset X$ be a finite union of locally closed subsets (e.g. $E$ is constructible). The following are equivalent

1. The intersection $E \cap Z$ contains an open dense subset of $Z$.

2. The intersection $E \cap Z$ is dense in $Z$.

If $Z$ has a generic point $\xi$, then this is also equivalent to

1. We have $\xi \in E$.

Proof. The implication (1) $\Rightarrow$ (2) is clear. Assume (2). Note that $E \cap Z$ is a finite union of locally closed subsets $Z_ i$ of $Z$. Since $Z$ is irreducible, one of the $Z_ i$ must be dense in $Z$. Then this $Z_ i$ is dense open in $Z$ as it is open in its closure. Hence (1) holds.

Suppose that $\xi \in Z$ is a generic point. If the equivalent conditions (1) and (2) hold, then $\xi \in E$. Conversely, if $\xi \in E$ then $\xi \in E \cap Z$ and hence $E \cap Z$ is dense in $Z$. $\square$

Comment #7353 by Hao Peng on

I think this lemma is true without the hypothesis that $Z$ is closed. One proof is given in https://arxiv.org/abs/1005.1423v2 Lemma2.1

Comment #7354 by Hao Peng on

Ah it seems this $Z$ can be any subset of $X$. But this lemma only makes sense when $Z$ is the closure of $E$, so it is still slightly more general.

Comment #7400 by Alex Scheffelin on

I think the end of the proof is more complicated then need be. If $\xi \in E$, then $\xi\in E\cap Z$, and we can immediately conclude that $E\cap Z$ is dense in $Z$, no need to introduce any $U_i$.

Comment #7401 by Laurent Moret-Bailly on

In addition to #7400, we can simplifiy the exposition of (1)$\Leftrightarrow$(2) as follows. First, (1)$\Rightarrow$(2) is clear. Next, $E\cap Z$ is a finite union of locally closed subsets $Z_i$ of $Z$, so we can assume $Z=X$ and forget about $X$. Then since $Z$ is irreducible, one of the $Z_i$ must be dense in $Z$, hence open because it is open in its closure.

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