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 $E$. $\square$

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).