Does every closed set of prime ideals of a noetherian (commutative) ring contain a finite dense subset?
EDIT 1. Here is the motivation. In books like Mumford’s red book or Eisenbud-Harris, the authors describe the topology of the spectrum of some interesting noetherian rings by describing only the closures of points. Zev’s answer shows that they are right after all: this tells the whole story, because any closed set is a finite union of such closures.
EDIT 2. For the sake of completeness here is a proof of the fact that a noetherian ring $A$ contains only finitely many minimal prime ideals. Let $R$ be the set of those ideals $\mathfrak a$ of $A$ which are equal to their radical $r(\mathfrak a)$. We'll use the following facts:
(a) $r(\mathfrak a\mathfrak b)=r(\mathfrak a\cap\mathfrak b=r(\mathfrak a)\cap r(\mathfrak b)$; in particular $\mathfrak a,\mathfrak b\in R\Rightarrow$ $\mathfrak a\cap\mathfrak b\in R$,
(b) if $\mathfrak p\in R$ and if the conditions $\mathfrak a,\mathfrak b\in R$ and $\mathfrak p=\mathfrak a\cap\mathfrak b$ imply $\mathfrak p=\mathfrak a$ or $\mathfrak p=\mathfrak b$, then $\mathfrak p$ is prime.
Claim (a) is clear. Let $\mathfrak p$ satisfy the assumptions of (b) and suppose that $\mathfrak p$ is not prime. By (a) there are ideals $\mathfrak c,\mathfrak d$ such that $\mathfrak p\supset\mathfrak c\cap\mathfrak d$, $\mathfrak p\not\supset\mathfrak c$, $\mathfrak p\not\supset\mathfrak d$. Then $\mathfrak p$ contains neither $\mathfrak a:=r(\mathfrak c+\mathfrak p)$ nor $\mathfrak b:=r(\mathfrak d+\mathfrak p)$. To get a contradiction it suffices to prove $\mathfrak p=\mathfrak a\cap\mathfrak b$. The inclusion $\mathfrak p\subset\mathfrak a\cap\mathfrak b$ is obvious, and the reverse inclusion follows from (a).
Let $I\subset R$ be the set of finite intersections of primes. As the existence of a maximal element of $R\backslash I$ would contradict (b), we have $I=R$. In particular, the nilradical $\mathfrak n$ is the intersection of the primes $\mathfrak p_1,\dots,\mathfrak p_n$. Let $\mathfrak p$ be a prime. As $\mathfrak p$ contains $\mathfrak n$, it contains one the $\mathfrak p_i$.