Proof.
Using Limits, Proposition 32.11.2 there is an immediate reduction to the case where X is reduced. Details omitted. In the rest of the proof we endow every closed subset of X with the induced reduced closed subscheme structure.
We argue by induction that we can find an affine open U \subset Z_1 \cup \ldots \cup Z_ r containing T \cap (Z_1 \cup \ldots \cup Z_ r). For r = 1 this holds by assumption. Say r > 1 and let U \subset Z_1 \cup \ldots \cup Z_{r - 1} be an affine open containing T \cap (Z_1 \cup \ldots \cup Z_{r - 1}). Let V \subset X_ r be an affine open containing T \cap Z_ r (exists by assumption). Then U \cap V contains T \cap ( Z_1 \cup \ldots \cup Z_{r - 1} ) \cap Z_ r. Hence
\Delta = (U \cap Z_ r) \setminus (U \cap V)
does not contain any element of T. Note that \Delta is a closed subset of U. By prime avoidance (Algebra, Lemma 10.15.2), we can find a standard open U' of U containing T \cap U and avoiding \Delta , i.e., U' \cap Z_ r \subset U \cap V. After replacing U by U' we may assume that U \cap V is closed in U.
Using that by the same arguments as above also the set \Delta ' = (U \cap (Z_1 \cup \ldots \cup Z_{r - 1})) \setminus (U \cap V) does not contain any element of T we find a h \in \mathcal{O}(V) such that D(h) \subset V contains T \cap V and such that U \cap D(h) \subset U \cap V. Using that U \cap V is closed in U we can use Lemma 33.42.5 to find an element g \in \mathcal{O}(U) whose restriction to U \cap V equals the restriction of h to U \cap V and such that T \cap U \subset D(g). Then we can replace U by D(g) and V by D(h) to reach the situation where U \cap V is closed in both U and V. In this case the scheme U \cup V is affine by Limits, Lemma 32.11.3. This proves the induction step and thereby the lemma.
\square
Comments (2)
Comment #3210 by Kollar on
Comment #3212 by Johan on