Lemma 70.5.2. With same notation and assumptions as in Lemma 70.5.1 we have |X| = \mathop{\mathrm{lim}}\nolimits _ i |X_ i| as topological spaces.
Proof. We will use the criterion of Topology, Lemma 5.14.3. We have seen that |X| = \mathop{\mathrm{lim}}\nolimits _ i |X_ i| as sets in Lemma 70.5.1. The maps f_ i : X \to X_ i are morphisms of algebraic spaces hence determine continuous maps |X| \to |X_ i|. Thus f_ i^{-1}(U_ i) is open for each open U_ i \subset |X_ i|. Finally, let x \in |X| and let x \in V \subset |X| be an open neighbourhood. We have to find an i and an open neighbourhood W_ i \subset |X_ i| of the image x with f_ i^{-1}(W_ i) \subset V. Choose 0 \in I. Choose a scheme U_0 and a surjective étale morphism U_0 \to X_0. Set U = X \times _{X_0} U_0 and U_ i = X_ i \times _{X_0} U_0 for i \geq 0. Then U = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_ i in the category of schemes by Lemma 70.4.1. Choose u \in U mapping to x. By the result for schemes (Limits, Lemma 32.4.2) we can find an i \geq 0 and an open neighbourhood E_ i \subset U_ i of the image of u whose inverse image in U is contained in the inverse image of V in U. Then we can set W_ i \subset |X_ i| equal to the image of E_ i. This works because |U_ i| \to |X_ i| is open. \square
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