Forcing and generic absoluteness without choice

Transcription

1 Forcing and generic absoluteness without choice Philipp Schlicht, Universität Münster Daisuke Ikegami, Kobe University Logic Colloquium Helsinki, August 4, 2015

2 Introduction

3 Introduction Iterated forcing

4 Introduction Iterated forcing σ-closed forcing

5 Introduction Iterated forcing σ-closed forcing Random forcing

6 Introduction Iterated forcing σ-closed forcing Random forcing A switch

7 Introduction Iterated forcing σ-closed forcing Random forcing A switch Generic absoluteness

8 Introduction We force over a countable transitive model M of ZF, working in ZFC. Alternatively, we could work inside a model of ZF and construct Boolean-valued models. Question 1. What are the implications of combinatorial properties of forcings such as σ-closed and ccc in arbitrary models of ZF? Can σ-closed forcings collapse cardinals? 2. How can we ensure cardinal preservation?

9 Our motivation for studying forcing over models of ZF was to study models of ZF with more generic absoluteness than what is possible in ZFC. Definition (Hamkins) 1. A button is a statement ϕ such that we can force ϕ, and ϕ remains true in any subsequent forcing extension. 2. A switch is a statement ϕ such that in every generic extension we can force ϕ and we can force ϕ in further extensions. Example In ZFC 1. CH is a switch. 2. ω L 1 ă ω 1 is a button. Question 1. Can models of ZF satisfy more generic absoluteness than models of ZFC? 2. A there switches, provably in ZF?

10 Example Cohen constructed a symmetric extension from a sequence xx n n P ωy of Cohen reals which does not add the sequence but adds A tx n n P ωu. In this model A is Dedekind finite, i.e. there is no injection ω Ñ A. Lemma (Karagila-S) Suppose that in this model, κ is an uncountable (well-ordered) cardinal and we add a bijection f : A Ñ κ with finite conditions. Then no (well-ordered) cardinals are collapsed and A has size κ in the extension. Example Gitik constructed from a proper class of strongly compact cardinals a model of ZF in which the axiom of choice fails badly. In Gitik s model, cofpκq ω for every uncountable cardinal κ. Over this model, any forcing which well-orders the reals collapses ω 1.

11 Iterated forcing Definition A definable forcing is a formula ϕpxq which provably defines a forcing. Similarly we use definable names etc. Definition A forcing P is κ-linked if there is a linking function f : P Ñ κ such that all p, q P P with fppq fpqq are compatible. A definable forcing P is definably κ-linked if there is a definable linking function for P and this is provable in ZF.

12 A Borel code for a Borel subset A of the Cantor space ω 2 is a well-founded tree which codes how A is built from basic open sets. Example Random forcing consists of Borel codes for Borel subsets of ω 2 with positive measure. Let rps denote the Borel set coded by p. Let U t tx P ω 2 t Ď xu denote the basic open sets in the Cantor space. Lemma (Karagila-S) Every Borel subset of ω 2 has a Borel code if and only if AC ωpborelq holds. Lemma Random forcing is definably ω-linked. Proof. Suppose that ps iq ipω is a definable enumeration of ăω ω. For every condition p P P, there is some i P ω such that µprpsxus i q µpu si, by the q Lebesgue density theorem for ω 2. Let fppq denote the least such i.

13 Lemma Definably κ-linked forcings P preserve all cardinals µ ą κ. Proof. Suppose that l : P Ñ κ is a linking function for P. Suppose that λ ă µ are cardinals with κ ď λ. Let D α denote the dense set of conditions p deciding fpαq 9 for α ă λ. Let h: λ ˆ κ Ñ µ, hpα, βq γ if there is a condition p P D α with lppq β and p, P fpαq 9 γ. Then h is onto, contradicting the assumption that κ ă λ are cardinals.

14 Lemma Suppose that κ, µ are infinite cardinals and P is definably κ-linked. Then finite support iterations of P preserve all cardinals µ ą κ. Proof. Let P γ denote the finite support iteration of P of length γ. Let D denote the set of conditions p P P such that p decides the linking function for ppαq for all α P suppppq.

15 Claim D is dense. Proof. Suppose that p 0 P P with support supppp 0q s 0. We construct P n, Q n, s n, δ n. Let P 0 Q 0 tp 0u and δ 0 maxpsupppp 0qq. Let P n`1 denote the set of conditions q P P γ such that for some p P Q n with q ď p: 1. p æ pγzδ nq q æ pγzδ nq and 2. q decides the linking function for ppαq for all α P suppppqzδ nq. Let ď m lex denote the lexicographical order on rords m. Find s n`1 such that 1. s n`1 supppqq for some q P P m, 2. m s n`1 is minimal with s n`1 is ď m lex-minimal with 1. and 2. Let Q n`1 tq P P n`1 supppqq s nu. If s n Ĺ s n`1, let δ n`1 maxpδ n`1zδ nq, otherwise δ n`1 δ n. There is some n with δ n δ n`1. Then there is a sequence p 0 ě p 1 ě ě p n with p i P Q i for i ď n.

16 Suppose that λ ă µ are cardinals with κ ď λ. Suppose that P γ is a finite support iteration of P of length γ and f 9 is a P γ-name for a surjection from λ onto µ. Let X denote the set of finite partial functions g : γ Ñ κ. Let h: X ˆ λ Ñ µ denote the onto partial function with hpg, αq β if there is a condition p with 1. domp domg 2. p decides the linking function for ppγq as gpγq for all γ P domp 3. p, fpαq 9 β. We now work in Lrhs. Let X γ tα P X pα, γq P domphqu for γ ă λ. Let h γ : X γ Ñ λ, h γpαq hpα, γq. By the delta system lemma in Lrhs, X γ with reverse inclusion has the pκ`q Lrhs -c.c. in Lrhs. Let x X γ γ ă λy be a sequence such that X γ is a maximal antichain in X γ of size ď κ. Then h æ Ť γăλ X γ ˆ λ is onto µ.

17 σ-closed forcing Question Over which models of ZF can some σ-closed forcing collapse ω 1? Definition Suppose that κ, µ are cardinals and X is a set. 1. Colpκ, Xq tp: κ Ñ X Dγ ă κ domppq Ď γu. 2. Col cl pµ`, Xq tpp, fq p P Colpµ`, Xq f : domppq Ñ µ is bijectiveu. 3. Addpκ, 1q Colpκ, 2q. 4. Add cl pµ`, 1q Col cl pµ`, 2q. The conditions are ordered by reverse inclusion in the first coordinate. Definition Suppose that µ is a cardinal and X is a set. 1. DC µpxq denotes the statement: If T is a ă µ-closed tree on X of height µ, then there is a branch of length µ in T. 2. DC ďµpxq states that DC λ pxq holds for all cardinals λ ď µ. 3. DCpXq denotes DC ωpxq.

18 Definition A forcing is (weakly) κ-distributive if Ş αăκ Uα is dense (nonempty) for every sequence pu α α ă κq of dense open sets. Lemma Suppose that xp, ďy is a forcing and xg, ďy is weakly µ-distributive for every P-generic G over V. Then P is µ-distributive. Proof. Suppose that pu α α ă µq is a sequence of dense open sets in Q. Let G be Q-generic over V with p P G. Then U α is dense open in G below p for every α. Since G is a generic filter, for every q, r P G there is s ď q, r in G. So U α æ p tq P U α : q ď pu is dense in pg, ďq for every n. Hence Ş αăµ puα æ pq H.

19 Lemma Suppose that µ is an infinite cardinal and X ě 2. Let P Colpµ`, X µ q or P Col cl pµ`, X µ q. 1. If DC ďµpx µ q, then P preserves all cardinals λ ď µ`. 2. If DC ďµpx µ q fails, then P singularizes µ`. 3. If P singularizes µ`, then P collapses µ`. Proof. The first claim holds since there is a surjection from X µ onto P. Suppose that DC ďµpx µ q fails. Forcing with P wellorders px µ q V, so DCppX µ q V q holds in V rgs. So P cannot be µ-distributive. If µ` is regular in V rgs, then P is µ-distributive by the previous lemma. Suppose that µ` is singular in V rgs. Since in V rgs, there is a wellorder of Ppµq V, µ` is collapsed.

20 Lemma Suppose that µ is an infinite cardinal. Let P Addpµ`, 1q or P Add cl pµ`, 1q. 1. If DC ďµp2 µ q, then P preserves all cardinals λ ď µ`. 2. If DC ďµp2 µ q fails, then P singularizes µ`. 3. If P singularizes µ`, then P collapses µ`. Lemma The following are equivalent. 1. DC 2. There is a σ-closed forcing which collapses ω 1.

21 Random forcing Question Does the random forcing B κ on a cardinal κ preserve cardinals? Definition Suppose that κ is an infinite ordinal. 1. We say that p is a Borel code in 2 κ if p is of the form pσ, cq such that σ is a countable subset of κ and c is a Borel code in the space 2 σ, where we endow 2 σ with the product topology. 2. For a Borel code p pσ, cq in 2 κ, let µppq µ LpB cq Lrps, where B c is the decode of c in the space 2 σ and µ L is the product measure on the space 2 σ. 3. Let B κ be the set of Borel codes p in 2 κ. Given p and q in B κ, we set p ď q if µppzqq 0. We also set p q if both p ď q and q ď p hold.

22 Question Is random forcing a complete Boolean algebra? Let Fnpκ, 2, ωq denote the set of finite partial functions κ Ñ 2. Definition 1. For a p in B κ and a code t in Fnpκ, 2, ωq, let r p,t be the following real number: r p,t µpp X tq. µptq 2. For a sequence r xr t P r0, 1s t P Fnpκ, 2, ωqy of real numbers, let A r be the following Borel set in 2 κ : A r tx P 2 κ lim nñ8 r xæn 1u. 3. For a p in B κ, we define the Borel set ϕppq to be the set A rp.

23 Lemma There is an OD function r ÞÑ b r such that 1. b r is defined when r is a sequence of real numbers in r0, 1s indexed by elements of Fnpκ, 2, ωq, i.e., r : Fnpκ, 2, ωq Ñ r0, 1s, 2. b r is in B κ and b r codes the Borel set A r in 2 κ, 3. b r ď b r 1 if and only if r t ď r 1 t for all t in Fnpκ, 2, ωq, 4. for all p in B κ, b rp p, and 5. p q in B κ if and only if b rp b rq. Proof. The function is OD, since the set A r has a Π 0 3 definition with a parameter r uniformly in r. Claim 4. follows from the fact that p ϕppq A rp by Lebesgue s Density Theorem in Lrps.

24 Theorem The Boolean algebra pb κ{, ďq is complete. Proof. It suffices to show that the preorder B κ has supremums for all its subsets. Let X be any subset of B κ and let r be the pointwise supremum of r p for p P X, i.e., r t sup ppx r p,t for all t in Fnpκ, 2, ωq. It follows from the previous lemma that b r is a supremum of X in B κ.

25 Lemma Let G be a B κ-generic filter over V. Then 1. for any inner model M of ZFC, G X M is B M -generic over M, and 2. for any set of ordinals A in V rgs, there is an inner model M of ZFC in V such that A P MrG X Ms. Proof. B M is c.c.c. in M and B M is the measure algebra in M. Then any maximal antichain in M is countable in M and hence it stays a maximal antichain of B in V. So G X M is a B M -generic over M. For the second claim, suppose that A Ď γ and 9 A G A. Let M HOD t 9Au. Let 9 B be the B κ{ -name 9B tpˇα, b αq α ă γu. where b α is a representative of rˇα P 9 As Bκ{ chosen in an ODp 9 Aq way. A 9 A G 9 B G P MrG X Ms, as desired.

26 Corollary B κ preserves cardinals. Proof. Suppose G is a B κ-generic filter over V and let λ be a cardinal in V. Let γ be an ordinal less than λ and f : γ Ñ λ be any function in V rgs. We will show that f is not surjective. There is an inner model M of ZFC in V such that f P MrG X Ms. But B M is c.c.c. in M, so B M preserves cardinals.

27 A switch Lemma (Woodin) There is a switch in ZF, i.e. a sentence ϕ such that ϕ and ϕ can be forced over any model of ZF. Proof. We consider the statement: p q For any subset A of ω 1, there is a random real over LrAs. We show that this can be forced to be true and false in some generic extensions respectively, We first force it to be false. Let C ω1 be the forcing adding ω 1-many Cohen reals with finite conditions. Let G pc α α ă ω 1q be a C ω1 -generic filter over V. Let A G.

28 Claim There is an ω 1-sequence of Borel codes pb α α ă ω 1q for Lebesgue null sets in the Cantor space in LrGs such that in V rgs, ω 2 Ť αăω 1 B bα, where B bα is the decode of b α for each α. In particular, there is no random real in V rgs over LrGs. The statement can also be forced to hold. Let κ ω V 2. Suppose that G is B κ-generic filter over V. Let A be a subset of ω 1 in V rgs. Then there is an inner model M of ZFC in V such that A P MrG X Ms. Then the support of A has size ď ω1 V and hence there is a random real over LrAs in V rgs.

29 Generic absoluteness Definition Suppose that C is a class of forcings. C-absoluteness (generic absoluteness for C) is the statement V ( ϕ ðñ V rgs ( ϕ for all sentences ϕ and all generic extensions V rgs by forcings in C.

30 Lemma If κ` is singular for some infinite cardinal κ and Colpω, κq-absoluteness holds, then ω 1 is singular. Proof. Let G be Colpω, κq-generic over V. Since Colpω, κq has the κ`-c.c. and is well-ordered, ω V rgs 1 κ`. Moreover κ` is singular in V rgs. Lemma If κ is regular for some uncountable regular κ and Colpω, ă κq-absoluteness holds, then ω 1 is regular. Proof. There are no antichains in Colpω, ă κq by working in LrAs, hence Colpω, ă κq has the κ-c.c. Since Colpω, ă κq is well-ordered, this is sufficient to show that κ is not singular in V rgs, where G is Colpω, ă κq-generic over V. Hence ω V rgs 1 κ.

31 Lemma Suppose that generic absoluteness for Colpω, µq, Colpω, ă µq, and Addpµ, λq holds for all cardinals µ, λ. Then every uncountable cardinal has cofinality ω. Proof. Otherwise µ` is regular for every cardinal µ. Let X` supptα P Ord there is a surjection f : X Ñ αu. Find some cardinal κ with cofpκq ω 2 and pα ω q` ă κ for all α ă κ. Then κ ω Ť αăκ αω, since cofpκq ą ω. Then Addpω, κq does not add a surjection f : pα ω q V Ñ κ for any α ă κ. Let G be P-generic over V. Let θ p2 ω q`v rgs. If θ ą κ`, then there is a surjection f : pκ ω q V Ñ κ` in V rgs. Since pκ ω q V Ť αăκ pαω q V, this contradicts the regularity of κ` in V rgs. Hence θ κ` pκ`q V rgs. Then p2 ω q` is the successor of a cardinal of cofinality ω 2 in V rgs. Similarly, there is a generic extension where p2 ω q` is the successor of a cardinal of cofinality ω 3.

32 Proposition (Woodin) If Generic Absoluteness holds for forcings adding κ-many Cohen reals for each κ, then every uncountable cardinal is singular. Proof. A cardinal κ is called a strong limit if for each α ă κ, there is no surjection from V α to κ. It is enough to show that for each strong limit κ, the cofinality of κ is ω because for each regular cardinal µ, there is a strong limit cardinal κ with cofinality µ. Let κ be any strong limit cardinal of uncountable cofinality. Let c ω supt α there is an α-sequence of distinct realsu. Claim In V Cκ, c ω is equal to κ. Claim In V C κ`, c ω is equal to κ`.

33 Question Is C κ-absoluteness consistent? Definition Let c κ : supt γ there is an injection f : γ Ñ κ ω u. Lemma c V Cκ ω c κ. Lemma In Gitik s model, c κ κ for all cardinals κ ě ω. Hence C κ-absoluteness fails for some κ.