SOME REMARKS ON NON-SPECIAL COHERENT ARONSZAJN TREES

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1 SOME REMARKS ON NON-SPECIAL COHERENT ARONSZAJN TREES MICHAEL HRUŠÁK AND CARLOS MARTÍNEZ RANERO Abstract. We introduce some guessing principles sufficient for the existence of non-special coherent Aronszajn trees and show how they relate to some of the standard principles in Set Theory (like MA ω1 and ). A variant of a question of I. Juhasz asks whether the principle implies the existence of a non-special Aronszajn tree. Motivated by this question, we investigate when a coherent Aronszajn tree associated with the ρ 1 function of Todorčević (see [5]) is not special. To do this, we define principles 0 and 1, and their corresponding weak versions w 0 and w 1. The principles 0 and 1 are strong enough to construct non-special coherent Aronszajn trees. All these principles are weak in the sense that are all consistent with MA σ centered and some of them are strong in the sense that they do not follow from. Our notation is mostly standard (see Kunen[4] and Jech[2] ). We will use Λ to denote the collection of all countable limit ordinals. A B will be used to denote that A is an initial segment of B, whenever A, B are subsets of ω 1. If A is a subset of ω 1, we will use ot(a) to denote the order-type of A. The symbol denotes concatenation. By a C-sequence (see [5]) we mean a sequence C α : α ω 1 with the following properties: C α+1 = {α}, C α is a cofinal subset of α of order-type ω, whenever α is a countable limit ordinal > 0. Definition 1. The principles 1, w 1, 0, w 0 are defined as follows: 0 There is a C-sequence S α : α ω 1 such that for every ϕ : Λ ω there are α, β Λ such that ϕ(α) = ϕ(β), S β α S α and α S β. w 0 There is a C-sequence S α : α ω 1 such that for every ϕ : Λ ω there are α, β Λ such that ϕ(α) = ϕ(β) and α S β. 1 There is a C-sequence S α : α ω 1 such that for every stationary set S there are α, β S such that S β α S α and α S β. w 1 There is a C-sequence S α : α ω 1 such that for every stationary set S there are α, β S such that α S β Mathematics Subject Classification. Primary 03E05, 03E35. Key words and phrases. Aronszajn tree, guessing principle. The research was supported PAPIIT grant IN and the first author s research was also partially supported by CONACYT grant F. 1

2 2 MICHAEL HRUŠÁK AND CARLOS MARTÍNEZ RANERO Following [5], to every C-sequence C α : α < ω 1 we associate two functions ρ 0, ρ 1. The function ρ 0 = ρ 0 (C α : α < ω 1 ) : [ω 1 ] 2 ω <ω is defined recursively as follows { C β α) ρ 0 (α, min(c β \ α)) if α < β ρ 0 (α, β) =. if α = β Even though, ρ 0 is an important function on its own, we use it only as an auxiliar tool in some proofs of the theorems in this article. The function ρ 1 = ρ 1 (C α : α < ω 1 ) : [ω 1 ] 2 ω is defined recursively by { max{ C β α, ρ 1 (α, min(c β \ α))} if α < β ρ 1 (α, β) =, 0 if α = β Thus, ρ 1 (α, β) is simply the maximal integer appearing in the sequence ρ 0 (α, β). We will focus on the function ρ 1. Basic properties of the ρ 1 function are mentioned in the next lemma. Lemma 2 (Todorčević [5]). For all α < β < ω 1 and n < ω, (a) {ξ α : ρ 1 (ξ, α) n} is finite, (b) {ξ α : ρ 1 (ξ, α) ρ 1 (ξ, β)} is finite. Let ρ 1α : α ω be defined by ρ 1α (ξ) = ρ 1 (ξ, α) for every ξ < α. Then it follows from the previous lemma that the sequence ρ 1α : α ω (α < ω 1 ) of finite-to-one functions is coherent in the sense that ρ 1α = ρ 1β α whenever α β. (Here = means the fact that the functions agree on all but finitely many arguments). The corresponding tree T (ρ 1 ) = {ρ 1β α : α < β ω 1 } is a coherent Aronszajn tree. The following two theorems show the relevance of the guessing principles 0 and 1. Theorem 3. 0 implies that there is a non special coherent Aronszajn tree. Proof. Let T = T (ρ 1 ) be the coherent Aronszajn tree constructed from a 0 -sequence S α : α < ω 1 i.e. ρ 1 = ρ 1 (S α : α < ω 1 ). To prove the theorem it is enough to check that A = {ρ 1α : α Λ} T is not a countable union of antichains. Given any partion ϕ : A ω of A, we define a new function ˆϕ : Λ ω by ˆϕ(α) = ϕ(ρ 1α ) for every α Λ. It follows, using 0, that there are α, β Λ such that ˆϕ(α) = ˆϕ(β), S β α S α and α S β. Then let us check that ρ 1α ρ 1β. Let {ξ k : k n} be the increasing enumeration of S β α. The proof proceeds by cases: Case 1. If ξ [0, ξ 0 ] then ρ 0 (ξ, β) = 0 ρ 0 (ξ, ξ 0 ). Since S β α S α the same holds for ρ 0 (ξ, α). Then by the definition of ρ 1 we have that ρ 1 (ξ, α) = ρ 1 (ξ, β).

3 SOME REMARKS ON NON-SPECIAL COHERENT ARONSZAJN TREES 3 Case 2. If ξ (ξ k, ξ k+1 ] then ρ 0 (ξ, β) = S β ξ ρ 0 (ξ, min (S β \ ξ)). However, S β α S α implies that ξ k+1 = min (S β \ ξ) = min (S α \ ξ) and S β ξ = S α ξ so ρ 1 (ξ, β) = ρ 1 (ξ, α). Case 3. If ξ (ξ n, α) then ρ 0 (ξ, β) = n ρ 0 (ξ, α), and ρ 0 (ξ, α) = S α ξ ρ 0 (ξ, min (S α \ ξ)). However, since S β α S α, n S α ξ so we have that ρ 1 (ξ, α) = ρ 1 (ξ, β). Then ξ < α (ρ 1α (ξ) = ρ 1β (ξ)). So we are done. Theorem 4. 1 implies that there is a coherent Aronszajn tree T which does not have stationary antichains. Proof. Let T = T (ρ 1 ) be the coherent Aronszajn tree constructed from a 1 -sequence S α : α < ω 1 i.e. ρ 1 = ρ 1 (S α : α < ω 1 ). The result follows using the same argument as in the previous theorem and the following claim. Claim. T has a stationary antichain if and only if {ρ 1α : α ω 1 } has one. Let us prove the claim. Let A = {t α : α S} be a stationary antichain of T, we may assume that T α A = 1 and ht(t α ) = α for every α S. Note that S is a stationary set. For each t α A there is an F α [α] <ω such that t α (ξ) = ρ 1α (ξ) for every ξ (α \ F α ). By the pressing down lemma, we can find a stationary set S S such that F α = F for every α S. Using again the pressing down lemma we can find a stationary set Ŝ S such that t α F = t β F for every α < β Ŝ. Then α < β Ŝ there is a ξ (α\f ) such that t α (ξ) t β (ξ). This implies that t α (ξ) = ρ 1α (ξ) ρ 1β (ξ) = t β (ξ). So {ρ 1α : α Ŝ} is a stationary antichain in {ρ 1α : α < ω 1 }, and this finishes the proof. As we have seen, the principles 0 and 1 are guessing principles which imply the existence of non-special Aronszajn trees. In order to have a better understanding of these principles we will compare them with some well known principles in set theory, summed up in the following diagram. + 1 w 1 0 w 0 MA ω1 There is a NSAT Here NST A is an abbreviation for non-special Aronszajn tree. As the following theorem shows all the principles are relatively consistent with ZF C, even with MA σ centered. Theorem 5. If V [G] is the generic extension obtained by adding a single Cohen real then V [G] = 1.

4 4 MICHAEL HRUŠÁK AND CARLOS MARTÍNEZ RANERO Proof. From now on assume that c : ω [ω] <ω is a Cohen-generic real and e α : α ω (α < ω 1 ) is a coherent sequence of finite-to-one functions. Let C α : α < ω 1 be an arbitrary C-sequence. We change this C-sequence to a C-sequence S α : α < ω 1 in the following way: S α = {ξ < α : C α (n) ξ < C α (n + 1), e α (ξ) c(n)}, where C α (0) = 0 and C α (n) is the nth element of C α for 0 < n < ω. Note that since e αs are finite-to-one ot(s α ) = ω. Let us check that S α : α < ω 1 is a 1 -sequence. Assume that A is a stationary subset of ω 1. Note that if A is stationary in V [G], then there is a stationary set A 0 V such that A 0 A. So without loss of generality we may assume that A is in the ground model. Fix p F n(ω, [ω] <ω ) with dom(p) ω, use the pressing down lemma to find a stationary set S A such that S α agree with S β in all the places decided by p for every α, β S. Pick an accumulation point β of S, now choose an α S in such a way that C β (n 0 ) < α C β (n 0 + 1) where dom(p) < n 0. Let q be defined by p(n) if n dom(p) q(n) = if dom(p) < n < n 0 {e β (α)} if n = n 0 then q Ṡβ α Ṡα & α Ṡβ. Corollary 6. 1 (and hence also 0, w 0 and w 1 ) are relatively consistent with MA σ centered. Proof. Let V be a model of MA and P a forcing which adds a single Cohen real. By the previous theorem if G is a P-generic filter then M[G] = 1 and by the theorem of Roitman (see [1]) the extension M[G] = MA σ centered. The fact that after adding a single Cohen real there is a coherent Aronszajn tree without stationary antichains was first observed by B. König in [3]. The following propositions give us some relationship beetwen and + with our guessing principles. Proposition 7. implies w 0. Proof. Let ϕ α : α ω 1 be a -sequence which guesses elements of ω1 ω (i.e. ϕ α ω α ). Define X α = {n : ϕ 1 α (n) is cofinal in α} for every limit α. For every α Λ choose S α α of order type ω such that S α ϕ 1 α (n) is a cofinal in α for every n X α. This is very easy to do. Let us check that the C-sequence S α : α < ω 1 has the required properties. Now, let ϕ : Λ ω be given. Set X = {n ω : ϕ 1 (n) is cofinal in ω 1 } and C = {α : n X (ϕ 1 (n) is cofinal in α)}. It is easy to see that C is a club in ω 1. Let be ξ 0 = max{ϕ 1 (n) : n / X} + 1 and S = {α : ϕ α = ϕ α}. Pick any β C S [ξ 0, ω 1 ) then ϕ(β) = n 0 X β. It follows from the properties of S β that there is an α S β such that ϕ(α) = n 0.

5 SOME REMARKS ON NON-SPECIAL COHERENT ARONSZAJN TREES 5 Proposition 8. + implies w 1. Proof. Let A α : α ω 1 be a + -sequence. For each α, let S α α be a sequence of order-type ω such that S α A for every A A α (this can be done by an easy induction). Let us verify that S α : α ω 1 is a w 1 -sequence. Given a stationary set S, there is a club C such that α C (S α A α ). Pick any β (C S) then S β (S β), now choose α S β (S β). Then α, β S and α S β. So we are done. We do not know if in the previous propositions we can replace the weak versions for the stronger ones. However, we have some limitations as the following theorem shows. Theorem 9. does not implies w 1. To prove the theorem we need the following lemmas. Lemma 10. For every C-sequence S α : α ω 1 there is an α such that for every β > α, {γ : (S γ \ α) β = } is stationary. Proof. Suppose that this is not the case. Then for every α there is a β(α) and a club C α such that (S γ \ α) β(α), whenever γ C α. Pick α 0 ω 1 and define α n+1 = β(α n ). Let ξ n ω C α n be greater than α = sup{α n : n ω}. Since S ξ intersects each interval [α n, a n+1 ), α is an accumulation point of S ξ, so the order-type of S ξ is greater than ω, which is a contradiction. The following lemma is a well known fact. Lemma 11. (1) Countable support iteration of σ-closed forcings is σ- closed, (2) Every σ-closed forcing preserves. Proof of theorem 9. For every C-sequence C = C α : α ω 1, define the notion of forcing P C where P C = {p 2 <ω 1 : α p 1 (1), C α p 1 (1) = and p α C 0} Here α C is the α in the previous lemma which correspond to the C-sequence C, and the order is by extension. Claim 1. P C is a σ-closed forcing. Let p n be a decreasing sequence of conditions in P C and set p ω = n ω p n. Obviously, p ω 2 <ω 1 and p ω α C 0. Suppose that there are α, β p 1 ω (1) (1), then there are n, m ω such that α dom(p n ) and β dom(p m ) but this implies that α, β p 1 m+n (1) and C β p 1 m+n (1) which is a contradiction. Claim 2. P C forces that C is not a w 1 -sequence. Let f G be the P C -generic function and S = f 1 G (1). To see that C is not a witness for w 1 in M[G] it suffices to prove that S is stationary such that α C β i.e. C β p 1 ω

6 6 MICHAEL HRUŠÁK AND CARLOS MARTÍNEZ RANERO in M[G]. Let Ċ be a name for a club and p P C a condition such that p Ċ is a club. By Lemma 10, we can find a sequence M 0 M 1... M n... of countable elementary submodels of H(θ) for θ large enough, such that p, C α : α ω 1, Ċ M 0 and moreover, (C δn \α C ) dom(p) =, where δ n = M n ω 1 and we may assume that δ n M n+1. Set M ω = n ω M n and δ = M ω ω 1. We will construct a sequence p n of conditions such that p n+1 p n, p n δ n Ċ, p 1 n (1) C δ = and p n M n by recursion as follows: Let ξ 0 = max(c δ δ 0 ), and extend p to a condition q = p {(α, 0) : α [dom(p), ξ 0 ]}. Note that q M 0. Since M 0 [G] = C is a club there is an η 0 ω 1 M 0 and a p 0 P C M 0, p 0 q such that p 0 η 0 Ċ. For the inductive step assume that we have constructed p k for k n with the required properties. Pick ξ n+1 < δ n+1 such that ξ n+1 > max(c δ δ n+1 ).Then q = p n {(α, 0) : α [dom(p n ), ξ n+1 ] M n+1 is a condition. As q Ċ is a club there is a η n+1 < δ n+1 and a condition q p n+1 M n+1 such that p n+1 η n+1 Ċ. Finally, let p ω = n ω p n {(δ, 1)}. Note that p ω is a condition as p 1 ω (1) C δ =. As p ω p n for all n ω, p ω {η n : n ω} Ċ. As δ = sup n ωη n and since Ċ is a name for a club p ω δ Ṡ Ċ. So S is stationary and Claim 2 holds. Let V = L and construct a countable support iteration P = P α, Q α : α < ω 2 so that Pα Q α = PC for some C sequence C. By a standard bookkeeping argument one can make sure that all C-sequences in the intermediate models are listed. Let G be a P-generic. Since every C-sequence C in M[G] has a P α -name for some α < ω 2, and at some stage β < ω 2 we have that Q β = PC then C is not a w 1-sequence. So M[G] = w 1 and by the Lemma 11. M[G] =. Finally we show that none of the principles is consistent with Martin s Axiom. Theorem 12. MA(ω 1 ) implies w 0. Proof. Let C α : α < ω 1 be a C-sequence. Define P = {p : A ω : A [Λ] <ω, ( α < β)(p(α) = p(β) α / C β )} ordered by inverse inclusion. It is easy to see that, if f G is the generic function, then f G is defined on Λ and forces that C α : α < ω 1 is not a w 0 -sequence, to assure both we need to meet only ω 1 many dense sets. To finish the proof it suffices to check that: Claim P is a c.c.c. forcing. Suppose that {p α : α ω 1 } is an antichain. By a standard -system type argument, we can assume that their domains form a -system with root r,

7 SOME REMARKS ON NON-SPECIAL COHERENT ARONSZAJN TREES 7 such that there is a N ω with dom(p α ) = N for each α ω 1 and all the functions agree on r. Moreover, we can assume that dom(p α ) dom(p b ) = for every α, β ω 1, and max(dom(p α )) < min(dom(p β )) if α < β. Now, set dom(p ω N+1 ) = {ξ 1,..., ξ N }. Since p ω N+1 is incompatible with p α for every α < ω N +1, ( N i=1 C ξ i ) dom(p α ) for every α < ω N +1. Then by the pigeon hole principle there is a i such that ot(c ξi ) ω + 1. However, this contradicts the fact that C α : α < ω 1 is a C-sequence, so we are done. We conclude with some open problems. Questions 13. (1) Does w 1 imply 1? (2) Does w 0 imply 0? (3) Does imply 0? An early version of this paper contained also the following questions: (4) Does + imply 1? and (5) Does imply 0?. These questions were answered by Paul Larson. we present the proof with his kind permission. Theorem 14 (Larson). (i) implies 0. (ii) + implies 1. Proof. Fix for every limit ordinal α < ω 1 a strictly increasing sequence {α n : n ω} such that sup n ω α n = α and let θ be a sufficiently large regular cardinal. To prove (i) let ϕ α : α < ω 1 } be a -sequence which guesses elements of ω ω 1 (i.e. ϕ α ω α ). Construct recursively a C-sequence S α : α Λ and a sequence e α n : α Λ, n ω of finite subsets of α with the following properties: (i) S α = n ω eα n, (ii) e α n e α n+1, max(eα n+1 ) > α n, (iii) e α n+1 = eα n {ξ}, where ξ = min{η : η > max(e α n {a n }) e α n S η ϕ α (ξ) = n} if such ξ exists, otherwise ξ = α N, where N = min{k : α k > (e α n {α n })}. Now let us check that S α : α Λ is a 0 -sequence. It follows from (ii) that S α is cofinal in α with order type ω. Let ϕ : ω 1 ω be given. Set S = {α < Λ : ϕ α = ϕ α}, since ϕ α : α < ω 1 is a -sequence S is a stationary set. Since C = {ω 1 M : M H(θ) such that ϕ, S α : α < Λ, ϕ α : α < ω 1 M} is a club there is an M C such that M ω 1 = δ S. Suppose that ϕ(δ) = n and let e δ n 1 = S δ n 1 (here e δ 1 = ) then for every α M H(θ) = β > α(ϕ(β) = n e δ n 1 S β. So, there is an α δ, α > (e δ n {α n }) such that n = ϕ(α) = ϕ δ (α) and e δ n 1 S α. It follows for the construction of e δ n+1 that eδ n+1 = eδ n {ξ} for some ξ with the same properties of α Then we have that ϕ(ξ) = ϕ(δ), ξ S δ and S δ α = e δ n 1 S ξ.

8 8 MICHAEL HRUŠÁK AND CARLOS MARTÍNEZ RANERO To prove (ii) let A α : α < ω 1 } be a + -sequence. Enumerate A α as {A α n : n ω}. Construct recursively a C-sequence S α : α Λ and a sequence e α n : α Λ, n ω of finite subsets of α with the following properties: (i) S α = n ω eα n, (ii) e α n e α n+1, max(eα n+1 ) > α n, (iii) e α n+1 = eα n {ξ}, where ξ = min{η : η > max(e α n {a n }) e α n S η (ξ) A α n+1} if such ξ exists, otherwise ξ = α N, where N = min{k : α k > (e α n {α n })}. Now let us check that S α : α Λ is a 1 -sequence. It follows from (ii) that S α is cofinal in α with order type ω. Let S a stationary. Set D = {α < Λ : S α A α }, since A α : α < ω 1 is a + -sequence D is a club. Since C = {ω 1 M : M H(θ) such thats, S α : α < Λ, A α : α < ω 1 M} is a club there is an M C such that M ω 1 = δ S D. Suppose that S δ = A δ n and let e δ n 1 = S δ n 1 then for every α M H(θ) = β > α(β S e δ n 1 S β. So, there is an α δ, α > (e δ n {α n }) such that α A δ n+1 and eδ n S α. It follows for the construction of e δ n+1 that eδ n+1 = eδ n {ξ} for some ξ with the same properties of α. Then we have that ξ S, ξ S δ and S δ ξ = e δ n 1 S ξ. Acknowledgments. The work contained in this paper is part of the second author s Master s thesis at Universidad Nacional Autónoma de México, written under the supervision of the first author. References [1] T. Bartoszyński and H. Judah, Set Theory on the Structure of the Real Line, A K Peters, [2] T. Jech, Set Theory, Springer Monographs in Mathematics, [3] B. König, A non-special tree, Unpublished Notes. [4] K. Kunen, Set Theory: An Introduction to Independence Proofs, North-Holland, Amsterdam, [5] S. Todorčević, Partitioning pairs of countable ordinals, Acta Math., 159 (3-4): , [6] S. Todorčević, Coherent sequences, In Handbook of Set Theory, North-Holland, (in preparation). Instituto de Matemáticas, UNAM (Morelia), Apartado postal 61-3 (Xangari), Morelia Michoacán, México, address: michael@matmor.unam.mx Instituto de Matemáticas, UNAM (Morelia), Apartado postal 61-3 (Xangari), Morelia Michoacán, México, address: azarel@matmor.unam.mx

A NOTE ON THE EIGHTFOLD WAY

A NOTE ON THE EIGHTFOLD WAY THOMAS GILTON AND JOHN KRUEGER Abstract. Assuming the existence of a Mahlo cardinal, we construct a model in which there exists an ω 2 -Aronszajn tree, the ω 1 -approachability