Definition 66.5.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. We say $X$ is quasi-compact if there exists a surjective étale morphism $U \to X$ with $U$ quasi-compact.
66.5 Quasi-compact spaces
Lemma 66.5.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Then $X$ is quasi-compact if and only if $|X|$ is quasi-compact.
Proof. Choose a scheme $U$ and an étale surjective morphism $U \to X$. We will use Lemma 66.4.4. If $U$ is quasi-compact, then since $|U| \to |X|$ is surjective we conclude that $|X|$ is quasi-compact. If $|X|$ is quasi-compact, then since $|U| \to |X|$ is open we see that there exists a quasi-compact open $U' \subset U$ such that $|U'| \to |X|$ is surjective (and still étale). Hence we win. $\square$
Lemma 66.5.3. A finite disjoint union of quasi-compact algebraic spaces is a quasi-compact algebraic space.
Proof. This is clear from Lemma 66.5.2 and the corresponding topological fact. $\square$
Example 66.5.4. The space $\mathbf{A}^1_{\mathbf{Q}}/\mathbf{Z}$ is a quasi-compact algebraic space.
Lemma 66.5.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Every point of $|X|$ has a fundamental system of open quasi-compact neighbourhoods. In particular $|X|$ is locally quasi-compact in the sense of Topology, Definition 5.13.1.
Proof. This follows formally from the fact that there exists a scheme $U$ and a surjective, open, continuous map $U \to |X|$ of topological spaces. To be a bit more precise, if $u \in U$ maps to $x \in |X|$, then the images of the affine neighbourhoods of $u$ will give a fundamental system of quasi-compact open neighbourhoods of $x$. $\square$
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