Lemma 61.6.1. Let A be a ring. Let X = \mathop{\mathrm{Spec}}(A). Let T \subset \pi _0(X) be a closed subset. There exists a surjective ind-Zariski ring map A \to B such that \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A) induces a homeomorphism of \mathop{\mathrm{Spec}}(B) with the inverse image of T in X.
Proof. Let Z \subset X be the inverse image of T. Then Z is the intersection Z = \bigcap Z_\alpha of the open and closed subsets of X containing Z, see Topology, Lemma 5.12.12. For each \alpha we have Z_\alpha = \mathop{\mathrm{Spec}}(A_\alpha ) where A \to A_\alpha is a local isomorphism (a localization at an idempotent). Setting B = \mathop{\mathrm{colim}}\nolimits A_\alpha proves the lemma. \square
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