The Stacks project

Proof. For each $i$ let $U_ i = \{ x \in X_ i \mid \dim _ x((X_ i)_{f_ i(x)}) \leq d\}$. This is an open subset of $X_ i$, see Morphisms, Lemma 29.28.4. Set $Z_ i = X_ i \setminus U_ i$ (with reduced induced scheme structure). We have to show that $Z_ i = \emptyset$ for some $i$. If not, then $Z = \mathop{\mathrm{lim}}\nolimits Z_ i \not= \emptyset$, see Lemma 32.4.3. Say $z \in Z$ is a point. Note that $Z \subset X$ is a closed subscheme. Set $s = f(z)$. For each $i$ let $s_ i \in S_ i$ be the image of $s$. We remark that $Z_ s$ is the limit of the schemes $(Z_ i)_{s_ i}$ and $Z_ s$ is also the limit of the schemes $(Z_ i)_{s_ i}$ base changed to $\kappa (s)$. Moreover, all the morphisms

are closed immersions by assumption (4). Hence $Z_ s$ is the scheme theoretic intersection of the closed subschemes $(Z_ i)_{s_ i} \times _{\mathop{\mathrm{Spec}}(\kappa (s_ i))} \mathop{\mathrm{Spec}}(\kappa (s))$ in $X_ s$. Since all the irreducible components of the schemes $(Z_ i)_{s_ i} \times _{\mathop{\mathrm{Spec}}(\kappa (s_ i))} \mathop{\mathrm{Spec}}(\kappa (s))$ have dimension $> d$ and contain $z$ we conclude that $Z_ s$ contains an irreducible component of dimension $> d$ passing through $z$ which contradicts the fact that $Z_ s \subset X_ s$ and $\dim (X_ s) \leq d$. $\square$