Lemma 29.22.4. Let $X$ be a scheme. Let $x \in X$. Let $E \subset X$ be a locally constructible subset. If $\{ x' \mid x' \leadsto x\} \subset E$, then $E$ contains an open neighbourhood of $x$.

**Proof.**
Assume $\{ x' \mid x' \leadsto x\} \subset E$. We may assume $X$ is affine. In this case $E$ is constructible, see Properties, Lemma 28.2.1. In particular, also the complement $E^ c$ is constructible. By Algebra, Lemma 10.29.4 we can find a morphism of affine schemes $f : Y \to X$ such that $E^ c = f(Y)$. Let $Z \subset X$ be the scheme theoretic image of $f$. By Lemma 29.6.5 and the assumption $\{ x' \mid x' \leadsto x\} \subset E$ we see that $x \not\in Z$. Hence $X \setminus Z \subset E$ is an open neighbourhood of $x$ contained in $E$.
$\square$

## 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Your email address will not be published.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.

## Comments (0)