Lemma 100.10.2. Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{X}' \subset \mathcal{X}$ is a closed substack, $\mathcal{X}$ is reduced and $|\mathcal{X}'| = |\mathcal{X}|$, then $\mathcal{X}' = \mathcal{X}$.
Proof. Choose a presentation $[U/R] \to \mathcal{X}$ with $U$ a scheme. As $\mathcal{X}$ is reduced, we see that $U$ is reduced (by definition of reduced algebraic stacks). By Lemma 100.9.11 $\mathcal{X}'$ corresponds to an $R$-invariant closed subscheme $Z \subset U$. But now $|Z| \subset |U|$ is the inverse image of $|\mathcal{X}'|$, and hence $|Z| = |U|$. Hence $Z$ is a closed subscheme of $U$ whose underlying sets of points agree. By Schemes, Lemma 26.12.7 the map $\text{id}_ U : U \to U$ factors through $Z \to U$, and hence $Z = U$, i.e., $\mathcal{X}' = \mathcal{X}$. $\square$
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