The Stacks project

Proof. Set $Z = \overline{\{ x\} } \subset |X|$ and $Z_ i = \overline{\{ x_ i\} } \subset |X_ i|$. Since $|X| \to |X_ i|$ is continuous we see that $Z$ maps into $Z_ i$ for each $i$. Hence we obtain an injective map $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i$ because $|X| = \mathop{\mathrm{lim}}\nolimits |X_ i|$ as sets (Lemma 70.5.1). Suppose that $x' \in |X|$ is not in $Z$. Then there is an open subset $U \subset |X|$ with $x' \in U$ and $x \not\in U$. Since $|X| = \mathop{\mathrm{lim}}\nolimits |X_ i|$ as topological spaces (Lemma 70.5.2) we can write $U = \bigcup _{j \in J} f_ j^{-1}(U_ j)$ for some subset $J \subset I$ and opens $U_ j \subset |X_ j|$, see Topology, Lemma 5.14.2. Then we see that for some $j \in J$ we have $f_ j(x') \in U_ j$ and $f_ j(x) \not\in U_ j$. In other words, we see that $f_ j(x') \not\in Z_ j$. Thus $Z = \mathop{\mathrm{lim}}\nolimits Z_ i$ as sets.

Next, endow $Z$ and $Z_ i$ with their reduced induced scheme structures, see Properties of Spaces, Definition 66.12.5. The transition morphisms $X_{i'} \to X_ i$ induce affine morphisms $Z_{i'} \to Z_ i$ and the projections $X \to X_ i$ induce compatible morphisms $Z \to Z_ i$. Hence we obtain morphisms $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i \to X$ of algebraic spaces. By Lemma 70.4.3 we see that $\mathop{\mathrm{lim}}\nolimits Z_ i \to X$ is a closed immersion. By Lemma 70.4.4 the algebraic space $\mathop{\mathrm{lim}}\nolimits Z_ i$ is reduced. By the above $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i$ is bijective on points. By uniqueness of the reduced induced closed subscheme structure we find that this morphism is an isomorphism of algebraic spaces. $\square$