The Stacks project

Proof. Let $T_0 \subset T$ be the set of $t \in T$ which do not specialize to another point of $T$. If $T_0$ is infinite, then $T' = T_0$ works. Hence we may and do assume $T_0$ is finite. Inductively, for $i > 0$, consider the set $T_ i \subset T$ of $t \in T$ such that

1. $t \not\in T_{i - 1} \cup T_{i - 2} \cup \ldots \cup T_0$,

2. there exist a nontrivial specialization $t \leadsto t'$ with $t' \in T_{i - 1}$, and

3. for any nontrivial specialization $t \leadsto t'$ with $t' \in T$ we have $t' \in T_{i - 1} \cup T_{i - 2} \cup \ldots \cup T_0$.

Again, if $T_ i$ is infinite, then $T' = T_ i$ works. Let $d$ be the maximum of the dimensions of the local rings $\mathcal{O}_{S, t}$ for $t \in T_0$; then $d$ is an integer because $T_0$ is finite and the dimensions of the local rings are finite by Algebra, Proposition 10.60.9. Then $T_ i = \emptyset $ for $i > d$. Namely, if $t \in T_ i$ then we can find a sequence of nontrivial specializations $t = t_ i \leadsto t_{i - 1} \leadsto \ldots \leadsto t_0$ with $t_0 \in T_0$. As the points $t = t_ i, t_{i - 1}, \ldots , t_0$ are in $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, t_0})$ (Schemes, Lemma 26.13.2), we see that $i \leq d$. Thus $\bigcup T_ i = T_ d \cup \ldots \cup T_0$ is a finite subset of $T$.

Suppose $t \in T$ is not in $\bigcup T_ i$. Then there must be a specialization $t \leadsto t'$ with $t' \in T$ and $t' \not\in \bigcup T_ i$. (Namely, if every specialization of $t$ is in the finite set $T_ d \cup \ldots \cup T_0$, then there is a maximum $i$ such that there is some specialization $t \leadsto t'$ with $t' \in T_ i$ and then $t \in T_{i + 1}$ by construction.) Hence we get an infinite sequence

of nontrivial specializations between points of $T \setminus \bigcup T_ i$. This is impossible because the underlying topological space of $S$ is Noetherian by Lemma 28.5.4. $\square$