Lemma 69.17.1. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. If $X$ is not a scheme, then there exists a closed subspace $Z \subset X$ such that $Z$ is not a scheme, but every proper closed subspace $Z' \subset Z$ is a scheme.

Proof. We prove this by Zorn's lemma. Let $\mathcal{Z}$ be the set of closed subspaces $Z$ which are not schemes ordered by inclusion. By assumption $\mathcal{Z}$ contains $X$, hence is nonempty. If $Z_\alpha$ is a totally ordered subset of $\mathcal{Z}$, then $Z = \bigcap Z_\alpha$ is in $\mathcal{Z}$. Namely,

$Z = \mathop{\mathrm{lim}}\nolimits Z_\alpha$

and the transition morphisms are affine. Thus we may apply Lemma 69.5.11 to see that if $Z$ were a scheme, then so would one of the $Z_\alpha$. (This works even if $Z = \emptyset$, but note that by Lemma 69.5.3 this cannot happen.) Thus $\mathcal{Z}$ has minimal elements by Zorn's lemma. $\square$

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