Lemma 100.9.16. Let $\mathcal X$ be an algebraic stack. Let $\mathcal{X}_ i$, $i \in I$ be a set of open substacks of $\mathcal{X}$. Assume

$\mathcal{X} = \bigcup _{i \in I} \mathcal{X}_ i$, and

each $\mathcal{X}_ i$ is an algebraic space.

Then $\mathcal{X}$ is an algebraic space.

**Proof.**
Apply Stacks, Lemma 8.6.10 to the morphism $\coprod _{i \in I} \mathcal{X}_ i \to \mathcal{X}$ and the morphism $\text{id} : \mathcal{X} \to \mathcal{X}$ to see that $\mathcal{X}$ is a stack in setoids. Hence $\mathcal{X}$ is an algebraic space, see Algebraic Stacks, Proposition 94.13.3.
$\square$

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