Lemma 37.28.4. Let $f : X \to Y$, $s : Y \to X$ be as in Situation 37.28.1. If $f$ is of finite presentation then $X^0$ is locally constructible in $X$.

**Proof.**
Let $x \in X$. We have to show that there exists an open neighbourhood $U$ of $x$ such that $X^0 \cap U$ is constructible in $U$. This reduces us to the case where $Y$ is affine. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation, endowed with a section $s_ i : \mathop{\mathrm{Spec}}(A_ i) \to X_ i$ whose base change to $Y$ recovers $f$ and the section $s$. By Lemma 37.28.2 it suffices to prove the lemma for $f_ i, s_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

Assume $Y$ is a Noetherian affine scheme. Since $f$ is of finite presentation, i.e., of finite type, we see that $X$ is a Noetherian scheme too, see Morphisms, Lemma 29.15.6. In order to prove the lemma in this case it suffices to show that for every irreducible closed subset $Z \subset X$ the intersection $Z \cap X^0$ either contains a nonempty open of $Z$ or is not dense in $Z$, see Topology, Lemma 5.16.3. Let $x \in Z$ be the generic point, and let $y = f(x)$. By Lemma 37.28.3 there exists a nonempty open subset $V \subset \overline{\{ y\} }$ such that $X^0 \cap X_ V$ is open and closed in $X_ V$. Since $f(Z) \subset \overline{\{ y\} }$ and $f(x) = y \in V$ we see that $W = f^{-1}(V) \cap Z$ is a nonempty open subset of $Z$. It follows that $X^0 \cap W$ is open and closed in $W$. Since $W$ is irreducible we see that $X^0 \cap W$ is either empty or equal to $W$. This proves the lemma. $\square$

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