Lemma 10.31.5. If $R$ is a Noetherian ring then $\mathop{\mathrm{Spec}}(R)$ is a Noetherian topological space, see Topology, Definition 5.9.1.

**Proof.**
This is because any closed subset of $\mathop{\mathrm{Spec}}(R)$ is uniquely of the form $V(I)$ with $I$ a radical ideal, see Lemma 10.17.2. And this correspondence is inclusion reversing. Thus the result follows from the definitions.
$\square$

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