Forcing and generic absoluteness without choice

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

Introduction

Introduction Iterated forcing

Introduction Iterated forcing σ-closed forcing

Introduction Iterated forcing σ-closed forcing Random forcing

Introduction Iterated forcing σ-closed forcing Random forcing A switch

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

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?

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?

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.

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.

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.

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.

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.

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 1. 3. 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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`.

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.