The Stacks project

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. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 005K. Beware of the difference between the letter 'O' and the digit '0'.