Lemma 69.5.9. Notation and assumptions as in Situation 69.5.5. If $X$ is separated, then $X_ i$ is separated for some $i \in I$.

Proof. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. For $i \geq 0$ set $U_ i = U_0 \times _{X_0} X_ i$ and set $U = U_0 \times _{X_0} X$. Note that $U_ i$ and $U$ are affine schemes which come equipped with surjective étale morphisms $U_ i \to X_ i$ and $U \to X$. Set $R_ i = U_ i \times _{X_ i} U_ i$ and $R = U \times _ X U$ with projections $s_ i, t_ i : R_ i \to U_ i$ and $s, t : R \to U$. Note that $R_ i$ and $R$ are quasi-compact separated schemes (as the algebraic spaces $X_ i$ and $X$ are quasi-separated). The maps $s_ i : R_ i \to U_ i$ and $s : R \to U$ are of finite type. By definition $X_ i$ is separated if and only if $(t_ i, s_ i) : R_ i \to U_ i \times U_ i$ is a closed immersion, and since $X$ is separated by assumption, the morphism $(t, s) : R \to U \times U$ is a closed immersion. Since $R \to U$ is of finite type, there exists an $i$ such that the morphism $R \to U_ i \times U$ is a closed immersion (Limits, Lemma 32.4.16). Fix such an $i \in I$. Apply Limits, Lemma 32.8.5 to the system of morphisms $R_{i'} \to U_ i \times U_{i'}$ for $i' \geq i$ (this is permissible as indeed $R_{i'} = R_ i \times _{U_ i \times U_ i} U_ i \times U_{i'}$) to see that $R_{i'} \to U_ i \times U_{i'}$ is a closed immersion for $i'$ sufficiently large. This implies immediately that $R_{i'} \to U_{i'} \times U_{i'}$ is a closed immersion finishing the proof of the lemma. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).