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

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

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

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

We have $\xi \in E$.

**Proof.**
Write $E = \bigcup U_ i \cap Z_ i$ as the finite union of intersections of open sets $U_ i$ and closed sets $Z_ i$. Suppose that $E \cap Z$ is dense in $Z$. Note that the closure of $E \cap Z$ is the union of the closures of the intersections $U_ i \cap Z_ i \cap Z$. As $Z$ is irreducible we conclude that the closure of $U_ i \cap Z_ i \cap Z$ is $Z$ for some $i$. Fix such an $i$. It follows that $Z \subset Z_ i$ since otherwise the closed subset $Z \cap Z_ i$ of $Z$ would not be dense in $Z$. Then $U_ i \cap Z_ i \cap Z = U_ i \cap Z$ is an open nonempty subset of $Z$. Because $Z$ is irreducible, it is open dense. Hence $E \cap Z$ contains an open dense subset of $Z$. The converse is obvious.

Suppose that $\xi \in Z$ is a generic point. Of course if (1) $\Leftrightarrow $ (2) holds, then $\xi \in E$. Conversely, if $\xi \in E$, then $\xi \in U_ i \cap Z_ i$ for some $i = i_0$. Clearly this implies $Z \subset Z_{i_0}$ and hence $U_{i_0} \cap Z_{i_0} \cap Z = U_{i_0} \cap Z$ is an open not empty subset of $Z$. We conclude as before.
$\square$

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