Lemma 69.23.8. Notation and assumptions as in Lemma 69.23.4. If $f$ is proper, then there exists an $i_3 \geq i_0$ such that for $i \geq i_3$ we have $f_ i$ is proper.

Proof. By the discussion in Remark 69.23.5 the choice of $i_1$ and $W$ fitting into a diagram as in (69.23.2.1) is immaterial for the truth of the lemma. Thus we choose $W$ as follows. First we choose a closed immersion $X \to X'$ with $X' \to Y$ proper and of finite presentation, see Lemma 69.12.1. Then we choose an $i_3 \geq i_2$ and a proper morphism $W \to Y_{i_3}$ such that $X' = Y \times _{Y_{i_3}} W$. This is possible because $Y = \mathop{\mathrm{lim}}\nolimits _{i \geq i_2} Y_ i$ and Lemmas 69.10.2 and 69.6.13. With this choice of $W$ it is immediate from the construction that for $i \geq i_3$ the algebraic space $X_ i$ is a closed subspace of $Y_ i \times _{Y_{i_3}} W \subset B_ i \times _{B_{i_3}} W$ and hence proper over $Y_ i$. $\square$

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