Lemma 66.8.7. Let $S$ be a scheme. Let $X$ be a quasi-separated algebraic space over $S$. Let $E \subset |X|$ be a subset. Then $E$ is étale locally constructible (Properties of Spaces, Definition 64.8.2) if and only if $E$ is a locally constructible subset of the topological space $|X|$ (Topology, Definition 5.15.1).

**Proof.**
Assume $E \subset |X|$ is a locally constructible subset of the topological space $|X|$. Let $f : U \to X$ be an étale morphism where $U$ is a scheme. We have to show that $f^{-1}(E)$ is locally constructible in $U$. The question is local on $U$ and $X$, hence we may assume that $X$ is quasi-compact, $E \subset |X|$ is constructible, and $U$ is affine. In this case $U \to X$ is quasi-compact, hence $f : |U| \to |X|$ is quasi-compact. Observe that retrocompact opens of $|X|$, resp. $U$ are the same thing as quasi-compact opens of $|X|$, resp. $U$, see Topology, Lemma 5.27.1. Thus $f^{-1}(E)$ is constructible by Topology, Lemma 5.15.3.

Conversely, assume $E$ is étale locally constructible. We want to show that $E$ is locally constructible in the topological space $|X|$. The question is local on $X$, hence we may assume that $X$ is quasi-compact as well as quasi-separated. We will show that in this case $E$ is constructible in $|X|$. Choose open subspaces

and surjective étale morphisms $f_ p : V_ p \to U_ p$ inducing isomorphisms $f_ p^{-1}(T_ p) \to T_ p = U_ p \setminus U_{p + 1}$ where $V_ p$ is a quasi-compact separated scheme as in Lemma 66.8.6. By definition the inverse image $E_ p \subset V_ p$ of $E$ is locally constructible in $V_ p$. Then $E_ p$ is constructible in $V_ p$ by Properties, Lemma 28.2.5. Thus $E_ p \cap |f_ p^{-1}(T_ p)| = E \cap |T_ p|$ is constructible in $|T_ p|$ by Topology, Lemma 5.15.7 (observe that $V_ p \setminus f_ p^{-1}(T_ p)$ is quasi-compact as it is the inverse image of the quasi-compact space $U_{p + 1}$ by the quasi-compact morphism $f_ p$). Thus

is constructible by Topology, Lemma 5.15.14. Here we use that $|T_ p|$ is constructible in $|X|$ which is clear from what was said above. $\square$

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