Annals of Pure and Applied Logic. On guessing generalized clubs at the successors of regulars

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1 Annals of Pure and Applied Logic 162 (2011) Contents lists available at ScienceDirect Annals of Pure and Applied Logic journal homepage: On guessing generalized clubs at the successors of regulars Assaf Rinot School of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel a r t i c l e i n f o a b s t r a c t Article history: Available online 5 February 2011 MSC: primary 03E35 secondary 03E05 03E65 Keywords: Club guessing Generalized clubs Souslin tree Diamond Non-saturation Uniformization König, Larson and Yoshinobu initiated the study of principles for guessing generalized clubs, and introduced a construction of a higher Souslin tree from the strong guessing principle. Complementary to the author s work on the validity of diamond and non-saturation at the successor of singulars, we deal here with a successor of regulars. It is established that even the non-strong guessing principle entails non-saturation, and that, assuming the necessary cardinal arithmetic configuration, entails a diamond-type principle which suffices for the construction of a higher Souslin tree. We also establish the consistency of GCH with the failure of the weakest form of generalized club guessing. This, in particular, settles a question from the original paper Elsevier B.V. All rights reserved. 1. Background Given a regular cardinal, and a stationary subset S, Jensen s diamond principle S asserts the existence of a sequence A δ δ S such that {δ S A δ = A δ } is stationary for every A. Jensen discovered this principle in the course of establishing the existence of a Souslin tree in Gödel s constructible universe. He proved: Theorem 1.1 (Jensen, [7]). If = < is a regular cardinal for which E + tree. 1 holds, then there exists a -complete + -Souslin A famous, long-standing, open problem is whether GCH entails the existence of an ℵ 2 -Souslin tree. As GCH is consistent together with the failure of ℵ E 2 (see [8]), throughout the years, weakenings of the diamond principle has been suggested ℵ 1 and studied. Here is one. Given regular cardinals κ <, and a stationary set S Eκ, Shelah s club guessing principle asserts the existence of a sequence C δ δ S satisfying: 1. for every δ S, C δ is a club subset of δ; 2. {δ S C δ D} is stationary for every club D. Clearly, S entails the existence of a guessing sequence as above, but Shelah [13] proved that if κ + <, then the existence of such a sequence is provable just in ZFC. As for the remaining case of = κ +, he obtained the following approximation. address: assaf@rinot.com. URL: 1 Here, E κ := {δ < cf(δ) = κ}; For the definition of a Souslin tree, etc, see the end of this section /$ see front matter 2011 Elsevier B.V. All rights reserved. doi: /j.apal

2 A. Rinot / Annals of Pure and Applied Logic 162 (2011) Theorem 1.2 (Shelah, [13]). Suppose that is a regular uncountable cardinal, and S E + is stationary. Then there exists a sequence C δ δ S satisfying: 1. for every δ S, C δ is a club in δ of order-type ; 2. {δ S sup(c δ D E + ) = δ} is stationary for every club D. Shelah s club guessing principles were found to be fruitful in establishing many non-trivial ZFC results (see, for instance, [2]). Naturally, several variations of these principles have been studied, as well. To mention two of them: Definition 1.3 (Folklore). Suppose that κ < are regular cardinals, S Eκ is a stationary set, and C sequence such that C δ is a club in δ for all δ S. = C δ δ S is a C is a strong club guessing sequence iff for every club D, there exists a club C such that: S C {δ S β < δ(c δ \ β D)}. C is a tail club guessing sequence iff for every club D, the following set is stationary: {δ S β < δ(c δ \ β D)}. Having the Souslin problem in mind, König et al. considered in [10] a generalization of club guessing, for the notion of generalized club. Recall that, given cardinals κ, a set C is a (generalized) club in [] <κ := {x x < κ}, iff there exists a function f : [] <ω such that C = {x [] <κ f [x] <ω x}. Definition 1.4 (Strong Generalized Club Guessing, [10]). Suppose that κ are given cardinals, and that S is a stationary subset of = cf() > ω. (κ, S) asserts the existence of a sequence C δ δ S such that 1. for every δ S, C δ is a club in [δ] <κ ; 2. for every club D in [] <κ, there exists a club C such that: S C {δ S x C δ (x y C δ y D)}. In [10], Theorem 1.2 above has been utilized to prove: Theorem 1.5 (König-Larson-Yoshinobu, [10]). Suppose that = < is a regular uncountable cardinal, and 2 = +. If (, E + ) holds, then there exists a -closed + -Souslin tree. In this paper, we shall mostly be studying the non-strong, and the tail version of the generalized guessing principle. Definition 1.6. Suppose that κ are given cardinals, and that S is a stationary subset of = cf() > ω. (κ, S) asserts the existence of a sequence C δ δ S such that: 1. for every δ S, C δ is a club in [δ] <κ ; 2. for every club D in [] <κ, the following set is stationary: {δ S C δ D}. (κ, S) asserts the existence of a matrix Cδ i δ S, i < δ such that: 1. for every δ S and i < δ, C i δ is cofinal in [δ]<κ ; 2. for every club D in [] <κ, the following set is stationary: {δ S i < δ (C i δ D)}. Remark. If S +, then (κ, S) entails (κ, S) whenever cf([] <κ, ) =, and in particular, (, S) entails (, S). We shall link between generalized club guessing and the usual club guessing, and shall show that a diamond-type principle, which we denote by T S (see Definition 2.2 below), can be derived from the generalized principles. Here is a special case of these theorems. Theorem 1. Suppose that S is a stationary subset of E + for a given successor cardinal, and that (κ, S) holds for some κ. Then: 1. There exists a sequence C δ δ S such that the following holds: (a) C δ is a club subset of E< δ for all δ S; (b) {δ S C δ D} is stationary for every club D If 2 = +, then T S holds for every stationary T.

3 568 A. Rinot / Annals of Pure and Applied Logic 162 (2011) Remark. Compare with Theorem 1.2 above. Then we shall exemplify how even the weakest form of T S can be utilized to construct an higher Souslin tree: Theorem 2. If < = is an uncountable cardinal and E + Remark. Compare with Theorem 1.1 above. holds, then there exists a -complete + -Souslin tree. We also introduce sufficient conditions for the principles T S and S, as well as (κ, S) and + S, to coincide.2 As consequences, we get: Theorem 3. Suppose that = < is a successor cardinal, and 2 = +. If (κ, E + ) holds for some κ, then at least one of the following holds: There exists a -Kurepa tree; There exists a + -Kurepa tree, and moreover, + + holds. Theorem 4. Suppose that is a regular uncountable cardinal, T is a stationary subset of, and S is a stationary subset of E +. If T S holds, while S fails, then NS T is non-saturated. So, altogether, we see that generalized club guessing principles at + can witness non-saturation at. Let us remind our reader that NS S is said to be non-saturated whenever there exists a collection of more than many pairwise-nonstationary stationary subsets of S. Note that non-saturation is yet another weakening of diamond. To see this, suppose that S is stationary and S holds; let A α α S be a S -sequence. Then {{δ S A δ = A δ } A } is a 2 -sized collection witnessing that NS S is non-saturated. By Shelah [13], if is a regular cardinal and S is a stationary subset of E< +, then NS + S is non-saturated, hence, we should focus our attention on subsets of E +. Probably, the main result of this paper is the introduction of the following GCH-free sufficient condition for non-saturation at stationary subsets of E +. Theorem 5. Suppose that S is a stationary subset of E +, for a given regular uncountable cardinal. If (κ, S) holds for some cardinal κ <, then NS + S is non-saturated. Note that by the work of Woodin and others, it is indeed consistent, modulo large cardinals, that NS S is saturated for inaccessible, successor of singular, successor of regular, and some S (see [5]). So, in general, GCH does not entail (κ, S). In [10], the authors prove that GCH does not imply (κ, E + ) for any ℵ 0 < κ, and write that they do not know whether GCH is moreover consistent with the failure of (κ, S) for every stationary S E +. Here, this is established (without appealing to large cardinals). Theorem 6. Suppose that is successor cardinal in Gödel s constructible universe, L. Then, in some forcing extension of L, we have: GCH is preserved; the cardinal structure and cofinalities are preserved; (κ, E + ) fails for every κ, and hence (κ, S) fails for every κ and every stationary S E +. Let us explain the value of getting the failure of (κ, E + ) in a model that shares the cardinal structure with L. For brevity, we assume GCH throughout the whole paragraph. In [9, Theorem 2], Kojman and Shelah introduced a certain club-guessing principle let us denote it by KS and demonstrated that KS suffices for the construction of a -complete + -Souslin tree. In light of Theorem 1.5, let us point out that KS is strictly weaker than (, E + ). ( ) By the proof of Theorem 1, (κ, E + ) (κ, E + ) KS. In fact, for a successor, we can just invoke the statement of Theorem 1, yielding (κ, E + ) (κ, E + ) E + KS, (κ ). ( ) By [9, Theorem 3], the existence of a non-reflecting stationary subset of E< + entails the validity of KS. As is wellknown, if the universe shares the cardinal s structure with L, then for every uncountable cardinal, there exists a nonreflecting stationary subset of E< +. It now follows from Theorem 6 that (κ, E + ) KS for all κ, and in particular, (, E + ) KS. 2 For + S, see Definition 2.8 below.

4 A. Rinot / Annals of Pure and Applied Logic 162 (2011) Terminology concerning trees We adopt the trees terminology of [1]. A tree T = T, is a partially ordered-set such that for every x T, the set x := {y T y x & y x} is well-ordered by. The order-type of the set x under is called the height of x in T, denoted by ht T (x). If α is an ordinal, the α th level of T is the set T α := {x T ht T (x) = α}. We let T α := {x T ht T (x) < α}, and denote by T α the restriction of the structure T to this set. A subset B of T is a branch if it is linearly ordered by and if x B x B for all x T. A set A T is an antichain if x y and y x for all distinct x, y A. The antichain A is maximal, if A {x} is not an antichain for any x T \ A. A tree T is said to be an (α, )-tree iff it is of height α and of width <, that is, T α =, while 0 < T β < for all β < α. A branch B T is cofinal iff B T β for all β < α. For a cardinal, a + -Kurepa tree is a ( +, + )-tree that has more than + many cofinal branches, and a -Souslin tree is a (, )-tree having no antichains of size. A tree T = T, is µ-complete if for every -increasing elements of the tree x i i < µ, there exists some y T for which i<µ (x i x). 2. Diamonds and trees 2.1. Club guessing We commence by giving a simple demonstration of the flavor of arguments for deriving consequences from generalized club guessing. Theorem 2.1. Suppose that S is a stationary subset of E + for a given regular uncountable cardinal,, If (κ, S) holds for some κ, then there exists a sequence C δ δ S such that the following holds: 1. C δ is a club subset of E< δ for all δ S; 2. {δ S C δ D} is stationary for every club D +. Proof. Let Cδ i i <, δ S witness (κ, S). Let be some well-ordering of [ + ] <κ. For all i < and δ S, define a function h i δ : δ δ by letting: h i δ min (β) := sup {x Ci δ β x}, (β < δ). It is not hard to verify that h i δ is well-defined, and that Cδ i := {α < δ hi δ [α] α & α is a limit ordinal} is a club subset of δ. Claim There exists some i < such that {δ S C i δ D} is stationary for every club D +. Proof. Suppose not. Then, for every i <, let us fix D i and E i, club subsets of + such that C i δ D i for all δ S E i. Put D := i< D i and E := i< E i. Define a function f : + + by letting: f (β) := min(d \ β + 1), (β < + ). Put D := {x [ + ] <κ f [x] x}. Then D is a club in [ + ] <κ, so let us pick some δ S E and i < such that Cδ i D. We now show that Cδ i D i, contradicting the fact that δ E i. Suppose α Cδ i. Then α is a limit ordinal, and so, to see that α D i, it suffices to prove that for every β < α, there exists some γ D with β < γ < α. Fix β < α, and let x := min{x Cδ i β x}. Then β x D, and hence f (β) x. Put γ := min(d \ β + 1). Then: β < γ = f (β) sup(x) = h i δ (β) hi δ [α] α. So, γ D and β < γ < α, as desired. Let i < be given by the previous claim. For all δ S, let C δ be some club subset of Cδ i Eδ <. Then C δ δ S is as requested Reflected diamond Definition 2.2 (Reflected Diamond). Suppose that is a regular uncountable cardinal, T is a stationary subset of, and S is a stationary subset of E +. T S asserts the existence of sequences C δ δ S and A δ i δ S, i < such that: 1. for all δ S, C δ is a club subset of δ of order-type ; 2. if for all δ S, {δ i i < } denotes the increasing enumeration of C δ, then for every club D + and every subset A +, there exist stationarily many δ S for which: {i T δ i+1 D & A δ i+1 = A δ i+1 } is stationary in.

5 570 A. Rinot / Annals of Pure and Applied Logic 162 (2011) To motivate the above definition, we refer the reader to the proofs of Theorems 2.5 and 2.7 below. Theorem 2.3. Suppose that S is a stationary subset of E + for a given regular uncountable cardinal,. If (κ, S) holds for some cardinal κ <, and 2 = +, then T S holds for every stationary T. Proof. We may assume that S =. Let C δ δ S witness (κ, S), and let {X β β < + } be some enumeration of [κ + ]. Fix δ S. For all α < δ, pick x δ α C δ such that {0, α} x δ α. Define a function h δ : δ δ by letting: h δ (α) := sup(x δ α ), (α < δ). It is not hard to verify that h δ is well-defined, and that {α < δ h δ [α] α} is a club in δ. Thus, pick {δ i i < } in such a way that: {δ i i < } is the increasing enumeration of a set, C δ ; C δ is a club subset of δ, consisting of limit ordinals; h δ [α] α for all α C δ. For all i <, let {B δ[j] i j < κ} be some enumeration (possibly, with repetition) of {X β β x δ δ i }, and let A δ[j] := {γ < δ (j, γ ) i Bδ i [j]} for all j < κ. Claim For every stationary T, there exists some j < κ such that C δ δ S and A δ i [j] δ S, i < witnesses T S. Proof. Suppose not, and let T be a counterexample. It follows that for every j < κ, we may pick a club D j +, and a subset A j + such that for every δ D j S: {i T δ i+1 D j & A j δ i+1 = A δ i+1 [j]} is non-stationary. Put D := j<κ D j, and A := j<κ {j} A j. Define a function f : [ + ] <ω +, by letting for all σ [ + ] <ω : 0, σ = f (σ ) := min{β < + A (κ α) = X β }, σ = {α} min(d \ sup(σ ) + 1), σ > 1. Put D := {x [ + ] <κ f [x] <ω x}. As D is a club, let us pick some δ S D for which C δ D. Subclaim {δ i+1 i T} D. Proof. It suffices to show that sup(d α) = α for all α C δ. Fix α C δ and β < α, and let us find some γ (D α) above β. By α C δ, we have h δ [α] α, and that α is a limit ordinal. Clearly, we may assume that β > 0. By {0, β} x δ β, we have γ := f δ ({0, β}) = min(d \ β + 1) x δ β. But γ sup(xδ β ) = h δ(β) h δ [α] α. So γ D is above β and below α. Subclaim There exists a function g : κ such that for all i < : A (κ δ i ) = B δ i [g(i)]. Proof. Fix i <. Put β := min{β < + A (κ δ i ) = X β }. Then f ({δ i }) = β, and as δ i x δ δ i C δ D, we have β x δ δ i. So X β is a member of {B δ i [j] j < κ}. Let g be given by the previous claim. Define h : T κ by letting: h(i) := g(i + 1), (i T). Fix j < κ for which T := h 1 {j} is stationary. Then: T {i T δ i+1 D & A (κ δ i+1 ) = B δ i+1 [j]}. In particular, T {i T δ i+1 D j & A j δ i+1 = A δ i+1 [j]}. The indirect assumption lead to a contradiction, so we are done. The claim completes the proof. Note that a complication of the preceding arguments yields the following. Theorem 2.4. Suppose that S is a stationary subset of E + for a given uncountable cardinal,, and that at least one of the following holds: (κ, S) for some κ < ; (, S), and is a successor cardinal. If 2 = +, then T S holds for every stationary T.

6 A. Rinot / Annals of Pure and Applied Logic 162 (2011) We omit the proof, but refer the reader to the proof of Theorem 3.2, for arguments of a similar flavor When the principles coincide Theorem 2.5. Suppose that is a regular uncountable cardinal, T is a stationary subset of, and S is a stationary subset of E +. Then (1) (2) (3) (4), where: 1. S ; 2. (κ, S) for all κ, and 2 = + ; 3. (κ, S) for some κ <, and 2 = + ; 4. T S. If NS T is saturated, then moreover (4) (1). Proof. (1) (2) This is similar to the proof in [10] that S (see Definition 2.8 below) entails (κ, S). By S, it is possible to find a sequence f δ : [δ] <ω δ δ S such that for every f : [ + ] <ω +, there exist stationarily many δ S for which f [δ] <ω = f δ. Let f δ δ S be as above. Given κ, let C δ := {x [δ] <κ f δ [x] <ω x}, and simply notice that C δ δ S witnesses (κ, S). The implication (2) (3) is trivial, and the implication (3) (4) is the content of Theorem 2.4 above. Finally, suppose that NS T is saturated and let us establish that (4) (1). Let C δ δ S and A δ i δ S, i < witness T S. Fix δ S. For every X δ, let T δ := X {i T X δ i+1 = A δ i+1 }. Then, put: A δ := {X δ T δ X is stationary}. Clearly, if X, Y are distinct elements of A δ, and j < is large enough such that X δ j Y δ j, then T δ X T δ Y j. So, elements of A δ may be identified with stationary subset of T, where distinct elements corresponds to sets whose pairwise intersection is nonstationary. As NS T is saturated, we conclude that A δ. Thus, by a famous argument of Kunen, to prove S, it now suffices to establish that for every A +, the following set is stationary: {δ S A δ A δ }. Fix A + and a club D +. We shall find δ D S for which A δ A δ. Pick δ S such that: {i T δ i+1 D & A δ i+1 = A δ i+1 } is stationary in. In particular, δ = sup(c δ D), and hence δ D S. Let X := A δ. As {i T δ i+1 D & X δ i+1 = A δ i+1 } is stationary, we get that T δ X is stationary, so A δ = X A δ. Theorem 2.6. Suppose that S is a stationary subset of E +, for a given successor cardinal. If there exist no -Kurepa trees, then the following are equivalent: 1. S holds; 2. (κ, S) holds for all κ, and 2 = + ; 3. (κ, S) for some κ, and 2 = +. Proof. (3) (1). Let Cδ i i <, δ S witness (, S). Let {X β β < + } be some enumeration of [ + ], and let be some well-ordering of [ + ] <κ. For every δ S, put: Fδ X i (α) := β β min {x Ci δ α x}, (i <, α < δ), Fδ i i := {X δ α < δ X α Fδ (α)}, A δ := {Fδ i i < }. (i < ), Thus, to establish S, it now suffices to prove the next two claims. Claim A δ for all δ S. Proof. Fix δ S and ι <. We argue that Fδ ι. Let {δ i i < } denote the increasing enumeration of some club subset of δ. For every X Fδ ι, define g X letting: g X (i) := min{β < + X δ i = X β }, (i < ). : + by Clearly, {g X j j <, X Fδ ι }, is a tree of height. Since there exists no -Kurepa tree, it now suffices to prove that {g X j X Fδ ι ι } Fδ (δ j) < for unboundedly many j <.

7 572 A. Rinot / Annals of Pure and Applied Logic 162 (2011) Fix a limit j <. By definition of Fδ ι(δ j), there exists some y C δ and an injection from Fδ ι(δ j) to y. So Fδ ι(δ j) < κ. Finally, it is easy to see that the mapping h {X h(i) i < j} forms an injection from {g X j X Fδ ι} to F δ ι(δ j), so we are done. Claim For every A +, the following set is stationary: {δ S A δ A δ }. Proof. Fix a set A +. Let f : + + be the function satisfying: f (α) := min{β < + A α = X β }, (α < + ). Put D := {x [ + ] < f [x] x}. Since D is a club, let us fix some i < such that the following set is stationary: S := {δ S C i δ D}. It now suffices to prove that A δ A δ for all δ S. Fix δ S. We shall show that A δ Fδ i. For this, we need to show that A α Fδ i(α) for all α < δ. But this is easy; if α < δ, and x = min {x Cδ i α x}, then α x D and hence f (α) f [x] x. So, X α = X f (α) Fδ i(α). This completes the proof Trees The next theorem exemplifies the utility of the principle T S. Theorem 2.7. If < = is an uncountable cardinal and E + holds, then there exists a -complete + -Souslin tree. Remark. This can be considered as a special case of Theorem 2 from [9]. Proof. Let C δ δ E + and Aδ i δ E +, i < witness E + enumeration of C δ. Recall that a (δ, + )-tree T is said to be normal iff all of the following holds:. For all δ E +, let {δ i i < } denote the increasing T 0 = 1; for all α < δ and x T α, there are distinct y, z T α+1 with x y z, provided that α + 1 < δ; for all α < β < δ and x T α, there exists some y T β with x y; for all limit α < δ, the mapping x x is injective over T α. In other words, the elements of T α are unique limits. So, we build a -complete + -Souslin tree, T, by recursion on the levels, in such a way that T δ is a normal (δ, + )-tree. The elements of T will be the ordinals in +, and we shall ensure that α < T β α < β. Set T 0 := { }. If T δ + 1 is defined for an ordinal δ < +, T δ+1 is obtained by using new ordinals from + to provide each element of T δ with two successors in T δ+1. If T δ is defined for an ordinal δ E< +, T δ is obtained by using new ordinals from + to provide a unique extension in T δ to each cofinal branch in T δ. Since T δ cf(δ) < =, the resulting T δ + 1 is indeed a normal (δ + 1, + )-tree. Suppose that T δ is defined for an ordinal δ E +. Fix x T δ. We now define an increasing sequence xδ = x δ (i) i <, by induction on i <, in such a way that for all i <, either j i x δ(j) = x, or ht T δ (x δ (i)) = δ i. Base case: Let x δ (0) := x. Successor step: Suppose that x δ (j) j i is defined for an ordinal i <. If ht T δ (x) δ i, let x δ (i+1) := x. Now, suppose ht(x) T δ < δ i. Let: a x δ (i) := {y T δ i+1 x δ (i) < T δ y}, b x δ (i) := {y ax δ (i) z Aδ i+1 (z < y)}. Clearly, a x δ (i) is non-empty. If bx δ (i), let x δ(i + 1) := min(b x δ (i)), otherwise, let x δ(i + 1) := min(a x δ (i)). Limit case: Suppose that x δ (j) j < i is defined for a limit ordinal i <. If x δ (j) j < i is a constant sequence, let x δ (i) := x. Otherwise, sup{ht(x δ (j)) j < i} = δ i and δ i E< +, so let x δ(i) be the unique element of T δi that extends the branch {x δ (j) j < i}. Once the construction of the sequence x δ is over, let: B x δ := {x δ (j) j < }.

8 A. Rinot / Annals of Pure and Applied Logic 162 (2011) Finally, T δ is obtained by using new ordinals from + to provide a unique extension in T δ to Bδ x for each x T δ. Since T δ, we get that T δ + 1 is indeed a normal (δ + 1, + )-tree. This completes the construction of the normal ( +, + )-tree, T. Thus, we are left with establishing the following. Claim T has no antichains of size +. Proof. Suppose not, and let A + denote a maximal antichain of size +. Put D := {δ < + T δ δ & A δ is a maximal antichain of T δ}. Since D is a club, let us pick some δ E + such that: H := {i < δ i+1 D & A δ i+1 = A δ i+1 } is stationary in. Since A = +, let us pick some a A with ht T (a) > δ. Let b be the unique element of T δ a. Since δ E +, the definition of T δ entails that there exists some x T δ such that b is the extension of Bδ x. Fix a large enough i H such that ht T (x) < δ i. As A δ i+1 = A δ i+1 is a maximal antichain of T δ i+1, we get that b x δ (i) is non-empty. Hence, by definition of x δ (i), we may pick some z A δ i+1 such that z < T x δ (i). So z < T x δ (i) < T b < T a, a contradiction to the fact that z and a are elements of the antichain, A. This completes the proof. Definition 2.8 (Jensen). Suppose S is a stationary subset of a given regular uncountable cardinal, and that A = A δ δ S is a sequence such that A δ for all δ S. A witnesses S if for every A, there exists a club C such that: C S {δ S A δ A δ }. A witnesses + S if for every A, there exists a club C such that: C S {δ S {C δ, A δ} A δ }. By Jensen and Solovay, for every successor cardinal,, + Consequently: Theorem 2.9. Suppose that = < is a successor cardinal, and 2 = +. If (κ, E + ) holds for some κ, then at least one of the following holds: There exists a -Kurepa tree; There exists a + -Kurepa tree, and moreover, + + holds. entails the existence of a -Kurepa tree (see [11, II.ğ5]). Proof. Suppose (κ, E + ) holds for some κ, and that there exist no -Kurepa trees. By <κ = and 2 = +, the proof of Theorem 2.6 yields that holds. By a theorem of Gregory from [6], < < + = 2 entails E +. So, altogether, E < + + holds. Let A δ δ < + witness +. For all δ < +, put: B δ := A δ C C [A δ ] ω. It now suffices to prove the next claim. Claim B δ δ < + witnesses + +. Proof. By ω =, we have B δ for all δ <. Now, suppose A is a given subset of +. Pick a club C 0 + such that A δ A δ for all δ C 0. Suppose n < ω and that C n is defined; pick a club C n+1 + such that C n δ A δ for all δ C n+1. Put C := n<ω C n. Then A δ A δ and {C n δ n < ω} [A δ ] ω for all δ C. In particular, A δ B δ, and C δ = {C n δ n < ω} B δ, for all δ C. So, we are done. Remark. The preceding proof shows that S is equivalent to + S, for all stationary S + with ℵ 0 =. 3. Non-trivial consequences of guessing 3.1. The failure of uniformization Definition 3.1 (Shelah, See [4]). A set S of limit ordinals is said to have the κ-uniformization property iff there exists a sequence A δ δ S such that: 1. A δ is a cofinal subset of δ for all δ S;

9 574 A. Rinot / Annals of Pure and Applied Logic 162 (2011) for any sequence of functions f δ : A δ κ δ S, there exists some function f : δ S A δ κ such that for every δ S: sup{α A δ f δ (α) f (α)} < δ. The uniformization property is an anti-diamond principle, and indeed, the canonical models for the failure of diamond at successors of regulars, are, in fact, models of the uniformization property. In [10], König et al. established the consistency of GCH with the failure of (κ, E + ) for all uncountable κ. They wrote that they do not know whether GCH is consistent with the failure of (κ, S) for every stationary S E +, but conjectured that the forcing techniques of [8], that yields the consistency of GCH with the 2-uniformization property for E ℵ 2 ℵ 1, can be applied to achieve this. Motivated by this question, we were hoping to supply an affirmative answer by moreover establishing the consistency of GCH with the failure of the weakest among these principles, (κ, E + ).3 For a while, we were trying to attack this problem by contrasting Theorem 2.1 with the techniques of [14], but this was not fully successful. So we went back and reconsidered the uniformization property. It is easy to see that if (κ, S) holds, then S fails to have the κ-uniformization property, but deriving the failure of the 2-uniformization property seems harder. Fortunately, the next finding served as the key to that door. Theorem 3.2. Suppose that S is a stationary subset of E + for a given regular uncountable cardinal,, 2 = +, and that at least one of the following holds: (κ, S) for some κ < ; (, S), and is a successor cardinal. If A δ δ S is a sequence such that for all δ S, A δ is a cofinal subset of δ, then there exists a sequence f δ : A δ δ δ S such that for every function f : + +, the following set is stationary: {δ S sup{α A δ f δ (α) = f (α)} = δ}. Proof. If (κ, S) holds for some cardinal κ <, let Cδ i i <, δ S witness that. Otherwise, is a successor cardinal and (, S) holds. In this case, let κ be the cardinal such that = κ +, and let Cδ i i <, δ S witness (, S). In either case, κ is a well-defined cardinal <, and Cδ i [δ] κ for all (i, δ) S. Next, suppose that a sequence A δ δ S as above is given. We shall derive the existence of a sequence f δ : A δ δ δ S with the required guessing property. Let be some well-ordering of [ + ] κ. Fix a bijection ψ : κ + +. For all (i, j) κ, define a function ψ i,j : + + by letting: ψ i,j (α) := ψ 1 (α)(i, j), (α < + ). Fix δ S. For all α < δ, pick a function ϕ α δ : κ δ satisfying: ϕδ α [{i} κ] = min {x Ci δ α x}, (i < ). Finally, for all (i, j) κ, define f i,j δ : A δ δ by letting for all α A δ : f i,j 0, δ (α) := ψ i,j (ϕδ α (i, j)), otherwise. ψi,j (ϕδ α (i, j)) δ Claim There exists (i, j) κ such that f i,j δ δ S has the required properties. Proof. Suppose not. So, for every (i, j) κ, let us pick a function f i,j : + + and a club D i,j + such that: D i,j {δ S sup{α A δ f i,j δ (α) = f i,j(α)} = δ} =. For all α < +, let g α : κ + be the function satisfying g α (i, j) = f i,j (α) for all (i, j) κ. Define f : + + by letting f (α) := ψ(g α ) for all α < +. Put D := (i,j) κ {δ D i,j f i,j [δ] δ} and D := {x [ + ] < f [x] x}. Then D is a club in +, and D is a club in [ + ] <. Thus, let us pick some δ S D and i < such that Cδ i D. Then α ϕδ α[{i} κ] D for all α A δ, and hence f (α) ϕδ α[{i} κ] for all α A δ. Define h : A δ κ by letting: h(α) := min{j < κ f (α) = ϕδ α (i, j)}, (α A δ). 3 Note that under GCH, (κ, S) entails (κ, S) for every κ.

10 A. Rinot / Annals of Pure and Applied Logic 162 (2011) Since δ = cf() > κ, let us pick some j < κ such that h 1 {j} is a cofinal subset of A δ. So: sup{α A δ f (α) = ϕδ α (i, j)} = δ, and in particular sup(b δ ) = δ, where B δ := {α A δ ψ i,j (ϕ α δ (i, j)) = ψ i,j(f (α))}. Since δ D, we get that f i,j [δ] δ, and hence, for all α B δ : f i,j δ (α) = ψ i,j(ϕ α δ (i, j)) = ψ i,j(f (α)) = ψ 1 (f (α))(i, j) = g α (i, j) = f i,j (α), so, sup{α A δ f i,j δ (α) = f i,j(α)} = δ, contradicting the fact that δ D i,j. This completes the proof. Back to the question that was left open in [10], we conclude: Corollary 3.3. Suppose that is a successor cardinal in L. Then, in some forcing extension of L, we have: GCH is preserved; the cardinal structure and cofinalities are preserved; (κ, E + ) fails for every κ, and hence: (κ, S) fails for every κ and every stationary S E +. Proof. Work in L. By the technique of [8], there exists a forcing notion that preserves cofinalities and the GCH, such that in the generic extension, E + has the 2-uniformization property. More specifically, there exists a sequence A δ δ E + such that: for every δ E +, A δ is a cofinal subset of δ, with otp(δ) = ; in the generic extension, for every sequence of functions g δ : A δ 2 δ E +, there exists some function f : + 2 such that sup{α A δ g δ (α) f (α)} < δ for all δ E +. Now, work in this generic extension and assume indirectly that κ is a cardinal for which (κ, E + ) holds. By GCH, we may appeal to Theorem 3.2 to obtain a sequence of functions f δ : A δ δ δ E + such that for every function f : + +, the following set is stationary: {δ S sup{α A δ f δ (α) = f (α)} = δ}. Now, for all δ E +, define g δ : A δ 2 by letting for every α A δ : 0, fδ (α) = 1 g δ (α) := 1, otherwise. By the 2-uniformization property, let us fix a function f : + 2 such that sup{α A δ g δ (α) f (α)} < δ for all δ E +. By the choice of f δ δ E +, we may pick some δ E + such that sup{β A δ f δ (β) = f (β)} = δ. In particular, we have: δ {δ E + This is a contradiction Non-saturation sup{β A δ g δ (β) f (β)} < δ}. We commence with a simple corollary to a result of a previous section, and then turn to the main result of this paper. Theorem 3.4. Suppose that is a regular uncountable cardinal, T is a stationary subset of, and S is a stationary subset of E +. If T S holds, while S fails, then NS T is non-saturated. Proof. By Theorem 2.5 above. Theorem 3.5. Suppose that S is a stationary subset of E +, for a given regular uncountable cardinal. If (κ, S) holds for some cardinal κ <, then NS + S is non-saturated. Proof. We extend the arguments of [3] to apply to stationary subsets of E +. Let Ci δ i <, δ S exemplify (κ, S). For all δ S, fix a function h δ : δ [δ] <κ satisfying j h δ (i, j) Cδ i for all j < δ and i <. Put µ := +. For functions f, g µ µ, denote by f < g the statement that sup {j < µ f (j) g(j)} < µ. Let f α α < µ + be a sequence of functions

11 576 A. Rinot / Annals of Pure and Applied Logic 162 (2011) in µ µ, which is increasing with respect to <. For all δ S and α < µ +, define a function f δ α f δ α (i, j) := min((h δ(i, j) {µ}) \ f α (j)), (i <, j < δ). : δ δ {µ} as follows: Notice that if f α (j) f β (j), then fα δ(i, j) f β δ (i, j). For all α < β < µ + and i <, consider the set: Sα,β i := {δ S sup{j < δ f α δ (i, j) = f β δ (i, j)} < δ}. Claim For every α < β < γ < µ + and i <, there exists some ε < + such that the two holds: (1) S i α,β \ ε Si α,γ ; (2) S i β,γ \ ε Si α,γ. Proof. Fix α < β < γ < µ +. By the choice of the sequence f α α < µ +, let us pick some ε < µ such that f α (j) < f β (j) < f γ (j) whenever ε j < µ. Fix also i <. (1) Suppose δ Sα,β i \ ε. Put A := {j < δ f α δ(i, j) = f β δ (i, j)} ε. Then A is a bounded subset of δ. Fix j δ \ A. Then f α (j) < f β (j) < f γ (j) and fα δ(i, j) < f β δ(i, j). In particular, f α δ(i, j) < f β δ(i, j) f γ δ(i, j). It follows that {j < δ f α δ (i, j) = fγ δ(i, j)} A, and hence δ Si α,γ. (2) Suppose δ Sβ,γ i \ ε. Put A := {j < δ f β δ(i, j) = f γ δ (i, j)} ε. Then A is a bounded subset of δ. Fix j δ \ A. Then f α (j) < f β (j) < f γ (j) and fβ δ(i, j) < f γ δ(i, j). In particular, f α δ(j) f β δ(j) < f γ δ(j). It follows that {j < δ f α δ(i, j) = f γ δ (i, j)} A, and hence δ S i α,γ. Claim There exists a function g : µ + such that S g(α) α,β is stationary for every α < β < µ+. Proof. Fix α < µ +. We would like to define g(α). By Claim 3.5.1(1), if i < is such that S i α,α+1 is stationary, then Si α,β is stationary for all β > α. Thus, let us argue the existence of some i < such that S i α,α+1 is stationary. Put j := sup {j < µ f α (j) f α+1 (j)} + 1. Next, define a function f : [ + ] <ω + as follows: 0, σ = f (σ ) = f α (min(σ )), σ = 1, (σ [ + ] <ω ). f α+1 (min(σ )), otherwise Put D := {x [ + ] <κ f [x] <ω x}. Then D is a club in [ + ] <κ, and hence there exists some i < such that {δ S Cδ i D} is stationary. Fix such i <, and let us show that T S i α,α+1, for the stationary set T := {δ S \ j Cδ i D}. Fix δ T. For j < δ, we have j h δ (i, j) C i j D, and hence f α (j) = f ({j}) h δ (i, j). Likewise, f α+1 (j) h δ (i, j) for all j < δ. So fα δ(i, j) = f α(j) < f α+1 (j) = f δ (i, j) α+1 for all j δ \ j, testifying that δ S i α,α+1. Let g be given by the previous claim. Fix an ordinal i < such that g 1 {i } is cofinal in µ +. By thinning out the sequence f α α < µ +, we may assume without loss of generality that g(α) = i for all α < µ +. For a set S S, let [S ] denote its equivalence class in P (S)/NS + S. By Claim 3.5.1, [Sα,β i ] α < β < µ+ is weakly increasing for all α < µ +, and [Sβ,γ i ] β < γ is weakly decreasing for all γ < µ+. Assume now that NS + S is saturated. Then, for all α < µ +, the sequence [Sα,β i ] α < β < µ+ is eventually constant, thus for all α < µ +, let us pick a large enough α < µ +, for which the sequence [Sα,β i ] α < β < µ + is constant. Now, if α < β < µ +, and γ > max{α, β }, then [Sβ,β i ] = [Si β,γ ] [Si α,γ ] = [Si α,α ], and hence the sequence [Si β,β ] β < µ+ is weakly decreasing and so eventually constant. Pick a large enough γ < µ + at which this sequence is constant. By the choice of i, Sγ i,γ is stationary, so let us fix a cardinal θ < κ such that the following set is stationary: S := {δ S i γ,γ sup{j < δ h δ(i, j) = θ} = δ}. Recursively define an increasing function g : θ + µ + by letting g(0) := γ + 1 and g(ς) := sup{g(ι) ι < ς} + 1 for all ς < θ +. Notice that if ι < ς < θ +, then [Sg(ι),g(ς) i ] = [Si g(ι),g(ι) ] = [Si γ,γ ], and we may find a club C ι,ς such that Sg(ι),g(ς) i C ι,ς = Sγ i,γ C ι,ς. Let C := C ι,ς ι < ς < θ +, then we have: S i g(ι),g(ς) C = Si γ,γ C, (ι < ς < θ + ).

12 A. Rinot / Annals of Pure and Applied Logic 162 (2011) In particular: S i g(ι),g(ς) S C, (ι < ς < θ + ). Put j := sup j < µ ι < ς < θ + = f g(ι) (j) f g(ς) (j) + 1, and pick δ S C above j. Let L := {j δ \ j h δ (i, j) = θ}. Then sup(l) = δ, and for all j L, the sequence f g(ι) (j) ι < is strictly increasing. Hence, for all j L, f δ g(ι) (i, j) ι < θ + is a weakly increasing sequence of elements of h δ (i, j), and as h δ (i, j) = θ < θ +, the sequence is eventually constant. Thus, for every j L, let us pick a large enough ι(j) < θ + at which this sequence is constant. As δ = cf() > κ θ +, let A be a cofinal subset of L and let ι < θ + be such that ι(j) = ι for all j A. Finally, fix ι < ς < θ + with ι > ι. Since ι < ι < ς, we get that f δ g(ι) ({i } A) = f δ δ Sg(ι),g(ς) i δ, that is, sup{j < δ fg(ι) (i, j) = fg(ς) δ (i, j)} < δ. This is a contradiction. 4. Open problems g(ς) ({i } A), while, by δ S C and (1) above, we get that Question 1. Is it consistent with GCH that for some regular uncountable cardinal,, T E + fails for all stationary T? Note that by Theorem 2.7, a negative answer would have yielded that GCH implies the existence of an ω 2 -Souslin tree. But we expect a positive answer, and have already obtained a few results in this direction. For instance, the following two are consistent: GCH, and for some stationary S E ω 2 ω 1 for which (E ω 2 ω 1 \ S) is stationary as well, T S fails for all stationary T ω 1 ; GCH, and for some stationary T ω 1 for which (ω 1 \ T) is stationary as well, T ω E 2 fails. ω 1 On another front, Alex Primavesi considered a generalization for stationary sets of the tail club guessing principle (Definition 1.3), and asked about its consistency. Definition (Primavesi). For a regular cardinal and a stationary T, Stat T asserts the existence of a sequence A δ δ T such that: A δ δ and sup(a δ ) = δ for every limit δ T ; for every stationary S T, {δ S β < δ(a δ \ β S)} is stationary. We have noticed that if > ω 1, then Stat T fails for every stationary T. Thus, the remaining question is as follows. Question 2. Is it consistent that Stat T holds for some stationary T ω 1? (1) Acknowledgements The author would like to thank his Ph.D. advisor, Moti Gitik, for his comments and remarks. I thank the organizers of the Logic Colloquium 2009 for a successful meeting, and for inviting me to present the results of my dissertation [12] in the Set Theory special session of that meeting. References [1] Keith J. Devlin, Constructibility, in: Perspectives in Mathematical Logic, Springer-Verlag, Berlin, [2] Mirna Džamonja, Club guessing and the universal models, Notre Dame J. Formal Logic 46 (3) (2005) (electronic). [3] Mirna Džamonja, Saharon Shelah, Saturated filters at successors of singular, weak reflection and yet another weak club principle, Ann. Pure Appl. Logic 79 (3) (1996) [4] Paul C. Eklof, Alan H. Mekler, Almost free modules, in: North-Holland Mathematical Library, vol. 46, North-Holland Publishing Co., Amsterdam, Set-theoretic methods. [5] Matthew Foreman, Ideals and generic elementary embeddings, in: Matthew Foreman, Akihiro Kanamori (Eds.), in: Handbook of Set Theory, vol. II, Springer-Verlag, 2010, pp [6] John Gregory, Higher Souslin trees and the generalized continuum hypothesis, J. Symbolic Logic 41 (3) (1976) [7] R. Björn Jensen, The fine structure of the constructible hierarchy, Ann. Math. Logic 4 (1972) ; R. Björn Jensen, The fine structure of the constructible hierarchy, Ann. Math. Logic 4 (1972) 443 (erratum). With a section by Jack Silver. [8] James H. King, Charles I. Steinhorn, The uniformization property for ℵ 2, Israel J. Math. 36 (3 4) (1980) [9] Menachem Kojman, Saharon Shelah, µ-complete Souslin trees on µ +, Arch. Math. Logic 32 (3) (1993) [10] Bernhard König, Paul Larson, Yasuo Yoshinobu, Guessing clubs in the generalized club filter, Fund. Math. 195 (2) (2007) [11] Kenneth Kunen, Set theory, in: Studies in Logic and the Foundations of Mathematics, vol. 102, North-Holland Publishing Co., Amsterdam, An introduction to independence proofs. [12] Assaf Rinot, Squares, diamonds and stationary reflection. Ph.D. Thesis, Tel Aviv University, [13] Saharon Shelah, Cardinal arithmetic, in: Oxford Logic Guides, vol. 29, The Clarendon Press Oxford University Press, New York, 1994, Oxford Science Publications. [14] Saharon Shelah, Not collapsing cardinals κ in (<κ)-support iterations, Israel J. Math. 136 (2003)