Lemma 28.5.13. Let $S$ be a Noetherian scheme. Let $T \subset S$ be an infinite dense subset. Then there exist a countable subset $E \subset T$ which is dense in $S$.

Proof. Let $T'$ be the set of points $s \in S$ such that $\overline{\{ s\} } \cap T$ contains a countable subset whose closure is $\overline{\{ s\} }$. Since a finite set is countable we have $T \subset T'$. For $s \in T'$ choose such a countable subset $E_ s \subset \overline{\{ s\} } \cap T$. Let $E' = \{ s_1, s_2, s_3, \ldots \} \subset T'$ be a countable subset. Then the closure of $E'$ in $S$ is the closure of the countable subset $\bigcup _ n E_{s_ n}$ of $T$. It follows that if $Z$ is an irreducible component of the closure of $E'$, then the generic point of $Z$ is in $T'$.

Denote $T'_0 \subset T'$ the subset of $t \in T'$ such that there is no nontrivial specialization $t' \leadsto t$ with $t' \in T'$ as in Lemma 28.5.12 whose results we will use without further mention. If $T'_0$ is infinite, then we choose a countable subset $E' \subset T'_0$. By the argument in the first paragraph, the generic points of the irreducible components of the closure of $E'$ are in $T'$. However, since one of these points specializes to infinitely many distinct elements of $E' \subset T'_0$ this is a contradiction. Thus $T'_0$ is finite, say $T'_0 = \{ s_1, \ldots , s_ m\}$. Then it follows that $S$, which is the closure of $T$, is contained in the closure of $\{ s_1, \ldots , s_ m\}$, which in turn is contained in the closure of the countable subset $E_{s_1} \cup \ldots \cup E_{s_ m} \subset T$ as desired. $\square$

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