The Stacks project

Proof. The notation in the statement means this: Let $p_ i : T \to T_ i$ be the projection morphisms, let $t_ i = p_ i(t)$ and $t'_ i = p_ i(t')$. Finally $\xi _ i \in |X_{T_ i}|$ is the image of $\xi $. By Divisors on Spaces, Lemma 71.4.7 we have $\xi _ i \in \text{Ass}_{X_{T_ i}/T_ i}(\mathcal{F}_{T_ i})$. Thus the only point is to show that $t_ i \not\in f_{T_ i}(\overline{\{ \xi _ i\} })$ for some $i$.

Let $Z_ i \subset X_{T_ i}$ be the reduced induced scheme structure on $\overline{\{ \xi _ i\} } \subset |X_{T_ i}|$ and let $Z \subset X_ T$ be the reduced induced scheme structure on $\overline{\{ \xi \} } \subset |X_ T|$. Then $Z = \mathop{\mathrm{lim}}\nolimits Z_ i$ by Limits of Spaces, Lemma 70.5.4 (the lemma applies because each $X_{T_ i}$ is decent). Choose a field $k$ and a morphism $\mathop{\mathrm{Spec}}(k) \to T$ whose image is $t$. Then

because limits commute with fibred products (limits commute with limits). Each $Z_ i \times _{T_ i} \mathop{\mathrm{Spec}}(k)$ is quasi-compact because $X_{T_ i} \to T_ i$ is of finite type and hence $Z_ i \to T_ i$ is of finite type. Hence $Z_ i \times _{T_ i} \mathop{\mathrm{Spec}}(k)$ is empty for some $i$ by Limits of Spaces, Lemma 70.5.3. Since the image of the composition $\mathop{\mathrm{Spec}}(k) \to T \to T_ i$ is $t_ i$ we obtain what we want. $\square$