Lemma 67.12.3. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then $|X|$ is Kolmogorov (see Topology, Definition 5.8.6).

Proof. Let $x_1, x_2 \in |X|$ with $x_1 \leadsto x_2$ and $x_2 \leadsto x_1$. We have to show that $x_1 = x_2$. Pick a scheme $U$ and an étale morphism $U \to X$ such that $x_1, x_2$ are both in the image of $|U| \to |X|$. By Lemma 67.12.2 we can find a specialization $u_1 \leadsto u_2$ in $U$ mapping to $x_1 \leadsto x_2$. By Lemma 67.12.2 we can find $u_2' \leadsto u_1$ mapping to $x_2 \leadsto x_1$. This means that $u_2' \leadsto u_2$ is a specialization between points of $U$ mapping to the same point of $X$, namely $x_2$. This is not possible, unless $u_2' = u_2$, see Lemma 67.12.1. Hence also $u_1 = u_2$ as desired. $\square$

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