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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