The Stacks project

Proof. We will use the criterion of Topology, Lemma 5.14.3. We have seen that $S_{set} = \mathop{\mathrm{lim}}\nolimits _ i S_{i, set}$ in Lemma 32.4.1. The maps $f_ i : S \to S_ i$ are morphisms of schemes hence continuous. Thus $f_ i^{-1}(U_ i)$ is open for each open $U_ i \subset S_ i$. Finally, let $s \in S$ and let $s \in V \subset S$ be an open neighbourhood. Choose $0 \in I$ and choose an affine open neighbourhood $U_0 \subset S_0$ of the image of $s$. Then $f_0^{-1}(U_0) = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}(U_0)$, see Lemma 32.2.2. Then $f_0^{-1}(U_0)$ and $f_{i0}^{-1}(U_0)$ are affine and

either by the proof of Lemma 32.2.2 or by Lemma 32.2.1. Choose $a \in \mathcal{O}_ S(f_0^{-1}(U_0))$ such that $s \in D(a) \subset V$. This is possible because the principal opens form a basis for the topology on the affine scheme $f_0^{-1}(U_0)$. Then we can pick an $i \geq 0$ and $a_ i \in \mathcal{O}_{S_ i}(f_{i0}^{-1}(U_0))$ mapping to $a$. It follows that $D(a_ i) \subset f_{i0}^{-1}(U_0) \subset S_ i$ is an open subset whose inverse image in $S$ is $D(a)$. This finishes the proof. $\square$