Lemma 69.5.2. With same notation and assumptions as in Lemma 69.5.1 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 69.5.1. 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 69.4.1. 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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