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Extra info for Forcing Idealized

Sample text

This is a nontrivial -ideal, the forcing PI is not proper though. Let M be a countable elementary submodel of a large structure and consider the Borel set B = x ∈ 2 x is Mgeneric . I will show that the set A ∩ B is Borel, so B ∈ I and PI is not proper. Suppose x ∈ B is a point. We have x ∈ A iff M x = x ∈ A (by analytic absoluteness) ˙ (by the forcing theorem) iff x ∈ C ∈ PI ∩M iff ∃C ∈ PI ∩M x ∈ C ∧C • x˙ gen ∈ A ˙ . Thus A ∩ B is Borel as desired. 4. Let X denote the Hilbert cube 0 1 . 5. Let I be the -ideal on X generated by the zerodimensional sets.

Suppose that I is a -ideal on a Polish space X such that the forcing PI is proper. Suppose that B ∈ PI is a Borel set, Y is a Polish space, f B → Y is a Borel function, and An n ∈ are analytic or coanalytic subsets of the space X. Then there is a Borel set C ⊂ B such that for every number n ∈ , the image f C ∩ An ⊂ Y is a Borel set. Proof. The expression y˙ = f˙ x˙ gen is a PI -name for a point in the space Y . Write ˙ where P adds the point y˙ and Q ˙ is the remainder forcing, adding the PI = P ∗ Q point x˙ gen .

Proof. On one hand, a review of the definitions shows that if B C D witness the ˙ x˙ and the latter set is in the ideal J by an statement I ⊥ J then B • C ∩ V ⊂ D gen absoluteness argument. On the other hand, if B • C˙ ∩ V ∈ J then there is a name ˙ ∈ J such that B • C˙ ∩ V ∈ I. 2, thinning out for a Borel set A ˙ =D ˙ x˙ . the set B if necessary, I can find a Borel set D ⊂ B × C such that B • A gen Using an absoluteness argument, thinning out the set B if necessary again, I can find the set D in such a way that its vertical sections are in the ideal J .