ITERATIONS WITH MIXED SUPPORT

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Cornelius Watts
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1 ITERATIONS WITH MIXED SUPPORT VERA FISCHER Abstract. In this talk we will consider three properties of iterations with mixed (finite/countable) supports: iterations of arbitrary length preserve ω 1, iterations of length ω 2 over a model of CH have the ℵ 2 -chain condition and iterations of length < ω 2 over a model of CH do not increase the size of the continuum. Definition 1. Let P κ be an iterated forcing construction of length κ, with iterands Q α : α < κ such that for every α < κ α Q α is σ-centered or α Q α is countably closed. Then P κ is finite/countable iteration if and only if for every p P κ, support(p) = {α < κ : p(α) 1 α } is countable and Fsupport(p) = {α : Q α is σ-centered, p(α) 1 α } is finite. Remark 1. In the context of the above definition, whenever α Q α is σ-centered we will say that α is a σ-centered stage and correspondingly, whenever Q α is countably closed we will say that α is a countably closed stage. From now on P κ is a finite/countable iteration of length κ. Definition 2. Let p, q P κ. We say than p D q if and only if p q and for every σ-centered stage α < κ, p α p(α) = q(α). Similarly p C q if and only if for every countably closed stage α, p α p(α) = q(α). Claim. Both D and C are transitive relations. Lemma 1. Let p n n ω be a sequence in P κ such that for every n ω, p n+1 D p n. Then there is a condition p P κ such that for every n ω, p D p n. Date: August 2,

2 2 VERA FISCHER Proof. Define p inductively. It is sufficient to define p for successor stages α. Suppose we have defined p α so that for every n ω, p α D p n α. If α is σ-centered then p α p 0 (α) = p 1 (α) =... and so we can define p(α) = p 0 (α). If α is countably closed stage, then p α p 0 (α) p 1 (α)... and since α Q α is countably closed there is a P α -name p(α) such that p α p(α) p n (α) for every n ω. Lemma 2. Let p, q P κ be such that p q. Then there is a condition r P κ such that p C r D q. Proof. Again we will define r inductively. Suppose we have defined r α so that p α C r α D q α. Then if α is a σ-centered stage, let r(α) = q(α). If α is a countably closed stage, define r(α) to be a P α -term such that: if r p α then r α r(α) = p(α), if r is incompatible with p α then r α r(α) = q(α). To verify p C r note that if α is a σ-centered stage then p α p(α) q(α) = r(α). If α is a countably closed stage, then p α p(α) = r(α). To see that r D q note that if α is a σ-centered stage then by definition r α r(α) = q(α). If α is countably closed stage, it is sufficient to show that 1 α r(α) q(α). Let r P α. If r p α fix a common extension t. Then t p(α) = r(α) p(α) q(α) and so t r(α) q(α). If r is incompatible with p α then by definition r r(α) = q(α). Definition 3. Let α be a σ-centered stage and let ṡ α be a P α -name such that α (ṡ α : Qα ω) [ p, q Q α (ṡ α (p) = ṡ α (q) p q)]. Condition p P κ is determined if and only if for every α Fsupport(p) there is n ω such that p α ṡ α (p(α)) = ň. Claim. The set of determined conditions in P κ is dense. Proof. Proceed by induction on the length of the iteration κ. It is sufficient to consider successor stages. Let α = β + 1 and p P α. We can assume that β is a σ-centered stage. By inductive hypothesis there is a determined condition r p β such that r ṡ β (p(β)) = ň for some n ω. If r P α is such that r β = r and r(β) = p(β), then r is determined and r p. Lemma 3. Let q 1, q 2 be (determined) conditions in P κ such that Fsupport(q 1 ) = Fsupport(q 2 ) = F

3 ITERATIONS WITH MIXED SUPPORT 3 and for every α F there is n ω such that q 1 α ṡ α (q 1 (α)) = ň and q 2 α ṡ α (q 2 (α)) = ň. Furthermore, let p P κ such that q 1 C p and q 2 p. Then q 1 and q 2 are compatible. Proof. The common extension r of q 1 and q 2 will be defined inductively. Suppose we have defined r α for some α < κ such that r α q 1 α and r α q 2 α. If α is a σ-centered stage, then there is n ω such that r α ṡ α (q 1 (α)) = ṡ α (q 2 (α)) = ň and so there is a P α -name r(α) for a condition in Q α such that r α r(α) q 1 (α) r(α) q 2 (α). If α is countably closed then r α q 1 (α) = p(α) q 2 (α) p(α). Thus we can define r(α) = q 2 (α). Definition 4. An antichain q ξ : ξ < η of determined conditions is concentrated with witnesses p ξ : ξ < η if and only if ξ < η(q ξ C p ξ ) and ζ < ξ(p ξ D p ζ ). Lemma 4. There are no uncountable concentrated antichains. Proof. Suppose to the contrary that q ξ : ξ < ω 1 is a concentrated antichain with witnesses p ξ : ξ < ω 1. For every ξ < ω 1 let F ξ = Fsupport(q ξ ). Since a subset of a concentrated antichain is a concentrated antichain, we can assume that F ξ : ξ < ω 1 form a -system with root F such that for some α < κ, F α < min F ξ \F for every ξ < ω 1. Claim. q ξ α : ξ < ω 1 is a concentrated antichain in P α. Proof. Suppose to the contrary that there are ζ < ξ such that for some r P α, r q ζ α and r q ξ α. Then for every γ α, define r(γ) as follows: if γ F ζ let r(γ) = q ζ (γ), otherwise let r(γ) = q ξ (γ). Inductively we will show that r is a common extension of q ζ and q ξ. It is sufficient to show that for all countably closed stages γ if r γ q ξ γ and r γ q ζ γ, then r γ r(γ) q ζ (γ) r(γ) q ξ (γ). Note that r γ (q ζ (γ) = p ζ (γ)) (p ξ (γ) p ζ (γ)) (q ξ (γ) = p ξ (γ)) and so r γ r(γ) = q ξ (γ) q ζ (γ). For every ξ < ω 1 let f ξ : F ω be such that f ξ (γ) = n if and only if q ξ γ ṡ γ (q ξ (γ)) = ň. Since there are only countably many such functions, there are ζ < ξ such that f ζ = f ξ. Then q ζ α, q ξ α and p ζ α satisfy the hypothesis of Lemma 3 and so q ζ α and q ξ α are compatible, which is a contradiction.

4 4 VERA FISCHER Lemma 5. Let p P κ and let ẋ be a P κ -name such that p ẋ V. Then there is q D p and a ground model countable set X such that q ẋ ˇX. Proof. Inductively construct concentrated antichain q ξ : ξ < η < ω 1 with witnesses p ξ : ξ < η < ω 1 such that for all ξ < η, q ξ p, p ξ D p and x ξ V, q ξ ẋ = ˇx ξ. Furthermore we will have that if ξ ζ, then x ξ x ζ. Suppose q ξ : ξ < η and p ξ : ξ < η have been defined. Since η is countable, by Lemma 1 there is condition p such that ξ < η(p D p ξ ). Case 1. If p ẋ {x ξ : ξ < η}, then let q = p and X = {x ξ : ξ < η}. Case 2. Otherwise, there is q η p, x η V such that x η / {x ξ : ξ < η} and q η ẋ = ˇx η. By Lemma 2 there is p η such that q η C p η D p. This extends the concentrated antichain and so completes the inductive step. Since there are no uncountable concentrated antichains at some countable stage of the construction Case 1 must occur. Corollary 1. Let p be a condition in P κ and let f be a P κ -name such that p f : ω V. Then there is q D p and X V, X countable such that q f ω ˇX. Therefore P κ preserves ω 1. Proof. Inductively define a sequence of conditions p n n ω in P κ, where p 1 = p and a sequence of countable sets {X n } n ω V such that for every n, p n+1 D p n and p n f(n) ˇX n. Let q P κ be such that q D p n for every n and let X = n ω X n. Then X is a countable ground model set, q D p and q f ω ˇX. Theorem 1 (CH). Let P ω2 be a finite/countable support iteration with iterands Q α : α < ω 2 such that α < ω 2, α Q α = ℵ 1. Then P ω2 is ℵ 2 -c.c. Proof. It is sufficient to show that for every α < ω 2 there is a dense subset D α in P α of cardinality ℵ 1. Suppose q ξ : ξ < ω 2 is an antichain in P ω2 of size ℵ 2. We can assume that F ξ : ξ < ω 2 where F ξ = Fsupport(q ξ ) form a -system with root F such that for some α < ω 2, F α < min F ξ \F for every ξ < ω 2. Then q ξ α : ξ < ω 2 is an antichain in P α of size ℵ 2 which is not possible. As an additional requirement we will have that for every p P α there is d D α such that d D p. Proceed by induction. Suppose α = β + 1 and we have defined D β D -dense in P β of size ℵ 1. Let { d ξ : ξ < ω 1 } be a set of P β -terms such that β Q β = { d ξ : ξ < ω 1 }. For every countable antichain A D β and function f : A ω 1 let q(f) be a P β -term such that for

5 ITERATIONS WITH MIXED SUPPORT 5 every p P β the following holds: if there is a A such that p a then p q(f) = d f(a) ; if p is incompatible with every element of A then p q(f) = 1 β. The collection T of all such names is of size ℵ 1 and so D α = {p P α : p β D β and p(β) T } is of size ℵ 1 as well. We will show that D α is D -dense in P α. Consider arbitrary p P α and let A be a maximal antichain of conditions in D β such that for every a A there is ξ < ω 1 such that a p(β) = d ξ. Claim. There is q D p β s. t. A = {a A : a q} is countable. Proof. Fix an enumeration a ξ : ξ < ω 1 of A and let ẋ = { ˇξ, a ξ : ξ < ω 1 }. Then ẋ is a P β -name for an ordinal and so repeating the proof of Lemma 4 we can obtain X V [ω 1 ] ω and q D p such that q ẋ ˇX. Then q ẋ sup ˇX and so ξ > sup X(q a ξ ). Let f : A ω 1 be such that f(a) = ξ if and only if a p(β) = d ξ. Then q q(f) = p(β). By inductive hypothesis, we can assume that q D β and so if r P α is such that r β = q and r(β) = q(f) then r D p and r D α. Suppose α is a limit and for every β < α we have defined a D - dense subset D β of P β of size ℵ 1. Let D β be the image of D β under the canonical embedding of P β into P α. Then D = β<α Dβ is of size ℵ 1 and furthermore there is a set D P α of size ℵ 1 which contains D and such that for every sequence p n n ω D for which n(p n+1 D p n ) there is p D such that n ω(p D p n ). We will show that D is D -dense in P α. Let p P α. If sup(support(p)) = β < α then by inductive hypothesis there is d D β such that d D p. Otherwise fix an increasing and cofinal sequence α n n ω in α such that Fsupport(p) α 0. Inductively define a sequence d n n ω such that for all n, d n P αn and d n+1 D d n p α n+1. If d D is such that n(d D d n ), then n(d D p α n ) and so d D p. Lemma 6 (CH). A forcing notion which preserves ω 1 and has a dense subset of size ℵ 1 does not increase the size of the continuum. Proof. Suppose P is a forcing notion which preserves ω 1 and has a dense subset D of size ℵ 1. Let T be the collection of all pairs p, ẏ where p D, ẏ is a P-name for a subset of ω and for every n ω, there is a countable antichain of conditions in D, deciding ň ẏ which is maximal below p. Then T 2 ℵ 0 = ℵ 1. We will show that V P 2 ℵ 0 T.

6 6 VERA FISCHER Consider any p P and ẏ a P-name for a subset of ω. Then for every n ω let r n,ξ : ξ < ω 1 be a maximal antichian of conditions in D deciding ň ẏ. Let f be a P-term such that f(n) = ξ iff r n,ξ Ġ. Then p f : ω ω 1 and since P preserves ω 1 and D is dense, there is q D such that q p and q f ω ˇβ for some β < ω 1. Then q, ẏ is a pair in T. Corollary 2 (CH). Let P ω2 be a finite/countable iteration of length ω 2 with iterands Q α : α < ω 2 such that α Q α 2 ℵ 0. Then for every α < ω 2, V Pα CH. Proof. Proceed by induction on α, repeating the proof of Theorem 1 and using Lemma 6. References [1] J. Baumgartner Iterated forcing, Surveys in set theory (A.R.D. Mathias, editor), London Mathematical Society Lecture Notes Series, no. 87, Cambridge University Press, Cambridge, 1983, pp [2] P. Dordal, A Model in which the base-matrix tree cannot have cofinal branches, The Journal of Symbolic Logic, Vol. 52, No. 3, 1987, pp [3] K. Kunen Set Theory, North Holland, Amsterdam, 1980.

The Bounded Axiom A Forcing Axiom

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