Projective measure without projective Baire

Transcription

1 Projective measure without projective Baire David Schrittesser Westfälische Wilhems-Universität Münster 4th European Set Theory Conference Barcelona David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

2 Regularity notions for sets of reals Let P be a forcing such that conditions are trees T 2 <ω ordered by reverse inclusion Definition X 2 ω is P-null* p P q P s.t. q p and [q] X =. P-null ideal: generated σ-ideal. Definition X 2 ω is P-regular p P [q] 2 ω \ X). q P s.t. q p and ([q] X or Similar for ω ω instead of 2 ω. David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

3 Our main examples: measure and category Meausure Equivalent ways of saying X R is measurable (LM) : X is B-regular (Random forcing) T s.t. µ([t ]) > 0 T T s.t. µ([t ]) > 0 and [T ] X is null or [T ] X X = B N where B Borel (or F σ ), N null Category Equivalent ways of saying X R has the Baire property (BP) : X is C-regular (Cohen forcing) p basic clopen q basic clopen s.t. q X is meager or q X X = B M, where B is Borel (or open), M meager Note that the 3rd formulation only works since C and B are ccc. David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

4 More examples S Sacks perfect trees on 2 <ω V Silver uniform perfect trees on 2 <ω M Miller superperfect tree on ω ω (infinite splitting) L Laver every node above stem has infinite splitting B measure (perfect) T s.t. µ([t ]) > 0 C Cohen full tree above stem Other examples: Mathias, Heckler, Matet, eventually different forcing,... David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

5 Which projective sets are regular? Independence in L there are 1 2 sets which are not P-regular for all above P After collapsing an inaccessible, all projective sets are P-regular for all above P (Solovays model) Let Γ be a pointclass (projective in most cases). Γ(P) denotes all sets of complexity Γ are P-regular. Separating notions of regularity E.g.: Can we have, from small large cardinals, consistency of: 1 3 (M) Σ1 3 (L)? (Yes, Fischer-Friedman-Khomskii) Σ 1 ω(s) 1 3 (V)? most (unlike the first one given here) are open David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

6 Separating notions of regularity (P) and Σ1 2 (P) can be related to transcendence over L in terms of P. 2 More recently, a theory has been established for 1 3, Σ1 3 and partially at Σ 1 4 and 1 4. (see Yurii Khomskii s talk at this conference) 3 Not everything is possible, there are implications (the first ones are easy, some are quite involved to show; not a complete list): Γ(L) Γ(M) Γ(S) Γ(B) Γ(S) Γ(C) Γ(M) Σ 1 2 (V) Σ1 2 (M) Σ 1 2 (B) Σ1 2 (C) David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

7 Separating Measure and Category Theorem (S. Friedman, S.) Assume a Mahlo cardinal. Consistently, Σ 1 ω(b) 1 3 (C) Σ 1 3 (B) ω 1 is inaccessible to reals (Shelah). This in turn implies Σ 1 2 (C), by the fact below. So the complexity of the irregular set is optimal. For the large cardinal assumption this is unclear. Fact Σ 1 2 (C) r R {s s is C-generic over L[r]} is comeager. David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

8 Separating Measure and Category, one direction. Theorem (Shelah) Assume that ZFC is consistent. Then it is consistent that all projective sets have the Baire property. Theorem (Shelah) If all Σ 1 3 -sets are Lebesgue-measurable, ω 1 is inaccessible in L. Corollary Consistently, Σ 1 ω(c) Σ 1 3 (B). David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

9 Precursor of the main result (other direction) Theorem (Shelah) Assume there is an inaccessible. You can force and then go to HOD( ω On {X}) (for a particular X) to get every set is measurable, X does not have the Baire-property (not projective!). Compare, again: Theorem (S. Friedman, S.) Assume a Mahlo cardinal. You can force Σ 1 ω(b) and 1 3 (C). David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

10 Some ideas of the proof The forcing is an iteration of length κ, the least Mahlo. Use a new type of amalgamation to construct Boolean algebras with automorphsims for Random sub-algebras. Immitate Solovays proof. The set X without the Baire property is added in a straightforward manner (Cohens reals added at an unbounded set of stages in the iteration, with a similar set in the complement - close of under automorphisms). X is made projective using a Easton supported variant of Jensen coding using David s trick. Develop the concept of stratified extension to show κ becomes ω 1. Components of the iteration Q ξ aren t well-behaved, but initial segments P ξ are. David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

11 A first result on other notions of regularity Theorem (Laguzzi) Assume an inaccessible. Consistently, ω 1 is inaccessible to reals, all sets are V-regular but there are sets which are not P-regular for P {M, B} This can be turned into a projective result: Theorem Assume a Mahlo. Consistently, ω 1 is inaccessible to reals and Σ 1 ω(v) 1 3 (M) 1 3 (B) David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

12 Some more ideas of the proof Look at Shelah s model where all sets are measurable and X does not have the Baire property. The reason X does not have the Baire property is that X and R \ X contains a dense set of reals which are Cohen (over any initial segment of the iteration). X must be closed under automorphisms. To have both, show X is not mixed up in unexpected ways by the automorphism. Cohen reals are unbounded. Random forcing (over which we amalgamate) cannot make these reals bounded. In Laguzzi s proof: unreachability separates Silver forcing from Miller and Random. David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13

13 Two questions: 1 How can we separate other notions of regularity? E.g. Silver from Sacks? What is the right separating property? 2 Can we force, from modest large cardinal assumptions Σ 1 7(C) 1 8 (C) In a model with just the right finite number of Woodin cardinals, 1 7 (C) Σ1 8 (C) holds. But something of the order of an inaccessible should suffice to force it. David Schrittesser (WWU Münster) Projective measure without projective Baire ESTC / 13