Lemma 61.5.7. Let $A$ be a ring. Let $V(I) \subset \mathop{\mathrm{Spec}}(A)$ be a closed subset which is a profinite topological space. Then there exists an ind-Zariski ring map $A \to B$ such that $\mathop{\mathrm{Spec}}(B)$ is w-local, the set of closed points is $V(IB)$, and $A/I \cong B/IB$.

**Proof.**
Let $A \to A_ w$ and $Z \subset Y = \mathop{\mathrm{Spec}}(A_ w)$ as in Lemma 61.5.3. Let $T \subset Z$ be the inverse image of $V(I)$. Then $T \to V(I)$ is a homeomorphism by Topology, Lemma 5.17.8. Let $B = (A_ w)_ T^\sim $, see Lemma 61.5.1. It is clear that $B$ is w-local with closed points $V(IB)$. The ring map $A/I \to B/IB$ is ind-Zariski and induces a homeomorphism on underlying topological spaces. Hence it is an isomorphism by Lemma 61.3.8.
$\square$

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