Lemma 101.49.1. Let $\mathcal{X}$ be a decent algebraic stack. Then $|\mathcal{X}|$ is Kolmogorov (see Topology, Definition 5.8.6).

Proof. Let $x_1, x_2 \in |\mathcal{X}|$ with $x_1 \leadsto x_2$ and $x_2 \leadsto x_1$. We have to show that $x_1 = x_2$. Let $\mathcal{Z} \subset \mathcal{X}$ be the reduced closed substack with $|\mathcal{Z}|$ equal to $\overline{\{ x_1\} } = \overline{\{ x_2\} }$. By Lemma 101.48.3 we see that $\mathcal{Z}$ is decent. After replacing $\mathcal{X}$ by $\mathcal{Z}$ we reduce to the case discussed in the next paragraph.

Assume $|\mathcal{X}|$ is irreducible with generic points $x_1$ and $x_2$. Pick an affine scheme $U$ and $u_1, u_2 \in U$ and a smooth morphism $f : U \to \mathcal{X}$ such that $f(u_ i) = x_ i$. Then we find a third point $u_3 \in U$ which is the generic point of an irreducible component of $U$ whose image $x_3 \in |\mathcal{X}|$ is also a generic point of $|\mathcal{X}|$. Namely, we can simply choose $u_3$ any generic point of an irreducible component passing through $u_1$ (or $u_2$ if you like). In the next paragraph we will show that $x_1 = x_3$ and $x_2 = x_3$ which will prove what we want.

By symmetry it suffices to prove that $x_1 = x_3$. Since $x_1$ is a generic point of $|\mathcal{X}|$ we have a specialization $x_1 \leadsto x_3$. By Lemma 101.47.1 we can find a specialization $u'_1 \leadsto u_3$ in $U$ (!) mapping to $x_1 \leadsto x_3$. However, $u_3$ is the generic point of an irreducible component and hence $u'_1 = u_3$ as desired. $\square$

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