Lemma 68.12.3. 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 68.12.2 we can find a specialization u_1 \leadsto u_2 in U mapping to x_1 \leadsto x_2. By Lemma 68.12.2 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 68.12.1. Hence also u_1 = u_2 as desired. \square
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