aa r X i v : . [ m a t h . L O ] F e b A LARGE PAIRWISE FAR FAMILY OF ARONSZAJN TREES
JOHN KRUEGER
Abstract.
We construct a large family of normal κ -complete R κ -embeddable κ + -Aronszajn trees which have no club isomorphic subtrees using an instanceof the proxy principle of Brodsky-Rinot [6]. Two trees are said to be club isomorphic if there exists a club subset of theirheight and an isomorphism between the trees restricted to that club. Abraham-Shelah [1] proved a number of essential results about club isomorphisms of ω -Aronszajn trees. They showed that under PFA , any two normal ω -Aronszajn treesare club isomorphic. In the other direction, they proved that the weak diamondprinciple on ω implies the existence of a family of 2 ω many pairwise non-clubisomorphic ω -Aronszajn trees. A natural problem is to generalize these theoremsto higher Aronszajn trees. Krueger [4] constructed a model in which any twonormal countably closed ω -Aronszajn trees are club isomorphic. Chavez-Krueger[2] proved that for any regular uncountable cardinal κ satisfying κ <κ = κ and ♦ ( κ + ), there exists a pairwise far family of 2 ( κ + ) many special κ + -Aronszajn trees,meaning that any two trees in the family have no club isomorphic subtrees. Theproof of the latter result used a technique of sealing potential club isomorphismsusing diamond.In the present article we build a pairwise far family of 2 ( κ + ) many normal κ -complete R κ -embeddable κ + -Aronszajn trees, where κ is a regular uncountablecardinal satisfying κ <κ = κ . There are two main differences of this constructionfrom that of [2]. First, instead of using diamond, we build the trees using theproxy principle of Brodsky-Rinot [6]. This principle implies the existence of auseful combinatorial object which is a combination of a club guessing sequence anda diamond sequence, and has been used to construct a variety of Aronszajn andSuslin trees ([6], [5], [7]). Secondly, rather than sealing potential club isomorphisms,we create a more explicit distinction between the different trees by ensuring thenon-existence of stationary antichains below certain stationary sets.In Section 1 we provide the necessary information about the linear orders Q κ and R κ which we use to build our trees, review a standard construction of a higherAronszajn tree, and describe the material on the proxy principle which we willneed. In Section 2 we show how the proxy principle can be used to produce twopairwise non-club isomorphic Aronszajn trees by killing potential club antichains.In Section 3 we present the main result of the article, which is the construction of alarge pairwise far family of Aronszajn trees which generalizes the proof of Section2. Date : February 2021.2020
Mathematics Subject Classification : Primary 03E05, 03E65.
Key words and phrases : Aronszajn tree, club isomorphism, proxy principle.This material is based upon work supported by the Simons Foundation under Grant 631279.
I would like to thank Assaf Rinot for his helpful and extensive comments on anearlier draft of this article.1.
Aronszajn trees and the proxy principle
We assume that the reader is familiar with the basic definitions and facts abouttrees, although we will carefully review the notation used in this article. Let (
T, < T )be a tree. A chain is a linearly ordered subset of T , and an antichain is a set ofpairwise incomparable elements of T . A branch is a maximal chain. For each x ∈ T ,let ht T ( x ) denote the height of x in T . For each ordinal δ , let T ( δ ) := { x ∈ T :ht T ( x ) = δ } denote level δ of T , and T ↾ δ := { x ∈ T : ht T ( x ) < δ } . More generally,if A is a subset of the height of a tree T , T ↾ A := { x ∈ T : ht T ( x ) ∈ A } . A branch b of T is cofinal if b ∩ T ( δ ) = ∅ for all δ less than the height of T . For an infinitecardinal κ , T is κ -complete if every chain of T whose order type is less than κ hasan upper bound.Let λ be a regular uncountable cardinal. A λ -tree is a tree of height λ such thatfor all δ < λ , level T ( δ ) has size less than λ . A λ -tree is Aronszajn if it has nocofinal branch and is
Suslin if it has no chains or antichains of size λ . If λ = µ + isa successor cardinal, a tree T of height λ is special if there exists a function from T to µ which is injective on chains. Note that a λ -tree being special implies thatit is Aronszajn and not Suslin. If S ⊆ λ is a stationary set and T ↾ S is special,then there exists an antichain A ⊆ T ↾ S such that the set { ht T ( x ) : x ∈ A } is astationary subset of S .A subtree of a tree T is any subset U of T considered as a tree with the inheritedorder < T ∩ ( U × U ). A set X ⊆ T is downward closed if for all x ∈ X , { y ∈ T : y < T x } ⊆ X . Note that branches are downward closed. If U is a downwardclosed subtree of T , then for all x ∈ U , ht T ( x ) = ht U ( x ). For any set Y ⊆ T , the downward closure of Y is the set { z ∈ T : ∃ y ∈ Y z < T y } .A λ -tree T is normal if:(1) for every x ∈ T and every γ < λ above ht T ( x ), there exists y ∈ T such that x < T y and ht T ( y ) = γ ;(2) if x and y are distinct nodes of T with the same limit height δ , then thesets { z ∈ T : z < T x } and { z ∈ T : z < T y } are not equal;(3) for every node x of T , there are incomparable nodes y and z above x .Let T and U be λ -trees. A function f : T → U is an isomorphism if f is abijection and for all x and y in T , x < T y iff f ( x ) < U f ( y ). We say that T and U are isomorphic if there exists an isomorphism from T to U . A club isomorphism of T and U is an isomorphism f : T ↾ C → U ↾ C , where C ⊆ λ is some club. Ifthere exists a club isomorphism of T and U , then T and U are club isomorphic . Itis easy to verify that if f : T ↾ C → U ↾ C is a club isomorphism of T and U , thenfor all x ∈ T ↾ C , ht T ( x ) = ht U ( f ( x )). Define T and U to be near if there existdownward closed subtrees of T and U which are club isomorphic, and otherwise T and U are far .In order to construct Aronszajn trees, we will employ a generalized version ofthe rationals. Definition 1.1.
Let κ be a regular cardinal. Define Q κ as the set of all functions f : κ → such that the set { α < κ : f ( α ) = 1 } is non-empty and has size less than κ , ordered lexicographically. LARGE PAIRWISE FAR FAMILY OF ARONSZAJN TREES 3
Observe that Q κ is a linear order of size 2 <κ which is dense and without end-points. If κ <κ = κ , then Q κ has size κ .We list the basic properties of Q κ which we will use in our Aronszajn tree con-structions. Lemma 1.2.
Let κ be a regular uncountable cardinal. (1) Between any two elements of Q κ there exists an increasing sequence of ordertype κ ; (2) any increasing sequence of Q κ with order type less than κ has a least upperbound in Q κ ; (3) any decreasing sequence of Q κ whose order type is a limit ordinal less than κ does not have a greatest lower bound. The proof of (1) is routine. (2) and (3) are proven in [3, Lemma 3.4].
Definition 1.3.
Let κ be a regular cardinal. Define R κ as the Dedekind completionof Q κ . Note that R κ is a dense complete linear order without endpoints in which Q κ isdense. Lemma 1.4.
Let κ be a regular uncountable cardinal. (1) Between any two elements of R κ there exists an increasing sequence of ordertype κ ; (2) any increasing sequence of R κ with order type a limit ordinal less than κ has a least upper bound in Q κ .Proof. (1) Consider q < R κ r . By the density of Q κ , we can fix q < R κ q < R κ r < R κ r where q and r are in Q κ . Now apply Lemma 1.2(1) to q and r .(2) Suppose that h r i : i < δ i is an increasing sequence in R κ , where δ < κ is alimit ordinal. By the density of Q κ , for each i < δ we can choose q i ∈ Q κ suchthat r i < R κ q i < R κ r i +1 . Applying Lemma 1.2(2), let q be the least upper bound of { q i : i < δ } in Q κ . Then q is also the least upper bound of { r i : i < δ } in R κ . (cid:3) Definition 1.5.
Let T be a tree and L a linear order. We say that T is L -embeddable if there exists a function f : T → L such that x < T y implies f ( x ) < L f ( y ) for all x, y ∈ T . Suppose that κ is a regular uncountable cardinal such that κ <κ = κ . Then Q κ has size κ . Hence, if T is a κ + -tree which is Q κ -embeddable, then T is special.Also, T being R κ -embeddable implies that T is not Suslin. One way to see this isto note that T ↾ ( κ + ∩ cof( < κ )) is special by Lemma 1.4(2).We now review a standard construction of a normal κ -complete special κ + -Aronszajn tree which uses the linear order Q κ . This construction will be a blueprintfor more complicated constructions given later in the article, so we will providemany of the details.Fix a regular cardinal κ and assume that κ <κ = κ . We will define by recursiona κ + -tree T together with a map π : T → Q κ . We will maintain several propertiesof T and π :(1) For each α < κ + , the elements of T ( α ) will be the ordinals in the interval[ κ · α, κ · ( α + 1));(2) x < T y implies π ( x ) < Q κ π ( y ), for all x, y ∈ T ; JOHN KRUEGER (3) if δ ∈ κ + ∩ cof( < κ ), then every cofinal branch b of T ↾ δ has a unique upperbound y in T ( δ ), and π ( y ) = sup { π ( x ) : x ∈ b } ;(4) for all x ∈ T , β < κ + , and q ∈ Q κ such that ht T ( x ) < β and π ( x ) < Q κ q ,there exists y ∈ T ( β ) above x satisfying that π ( y ) ≤ Q κ q .We will abbreviate the restriction of π to T ↾ α by π ↾ α , for all α < κ + .For the base case, let T (0) consist of the ordinals in λ , and define π on T (0) tobe any bijection between T (0) and Q κ .For the successor case, let α < κ + and assume that T ↾ ( α +1) and π ↾ ( α +1) aredefined. Let the elements of T ( α + 1) consist of the ordinals in [ κ · α, κ · ( α + 1)). Leteach node of T ( α ) have exactly κ many immediate successors in T ( α + 1). Define π ↾ ( α + 1) so that for each x ∈ T ( α ), π is a bijection from the set of immediatesuccessors of x onto the set { q ∈ Q κ : π ( x ) < Q κ q } . The inductive hypotheses areeasy to check.For the limit case, let δ < κ + be a limit ordinal and assume that T ↾ δ and π ↾ δ are defined. First, assume that δ has cofinality less than κ . Since κ <κ = κ andcf( δ ) < κ , there are exactly κ many cofinal branches of T ↾ δ . Let the nodes of T ( δ ) be the ordinals in [ κ · δ, κ · ( δ + 1)). Place exactly one node above each cofinalbranch of T ↾ δ . For each x ∈ T ( δ ), define π ( x ) := sup { π ( y ) : y < T x } .Let us verify the inductive hypotheses. (1), (2), and (3) are immediate. For (4),let x ∈ T ↾ δ and π ( x ) < Q κ q . Applying Lemma 1.2(1), fix an increasing sequence h q i : i < cf( δ ) i of elements of Q κ between π ( x ) and q . Fix an increasing andcontinuous sequence h δ i : i < cf( δ ) i of ordinals cofinal in δ such that ht T ( x ) < δ .Using inductive hypotheses (3) and (4), recursively build a chain h x i : i < cf( δ ) i above x such that x i ∈ T ( δ i ) and π ( x i ) ≤ Q κ q i for all i < cf( δ ). Let y be an upperbound of this chain in T ( δ ). Then x < T y and π ( y ) ≤ Q κ q .Secondly, assume that δ has cofinality κ . In this case, there are 2 κ many cofinalbranches of T ↾ δ . The definition of T at this level depends on selecting whichcofinal branches of T ↾ δ will have upper bounds. It is this part of the constructionwhich will vary in later constructions.Consider x ∈ T ↾ δ and π ( x ) < Q κ q . Using Lemma 1.2(1) and inductive hy-potheses (3) and (4) as in the previous case, recursively construct a chain above x of length κ whose elements have heights cofinal in δ and whose values under π arebelow q . Let b ( x, q ) be the downward closure of this chain. Then b ( x, q ) is a cofinalbranch of T ↾ δ . Using the fact that each node in T ↾ δ has κ many immediatesuccessors, it is easy to arrange the function which maps ( x, q ) to b ( x, q ) to beinjective. Let the nodes of T ( δ ) be the ordinals in the interval [ κ · δ, κ · ( δ + 1)).For each cofinal branch of the form b ( x, q ), place one node above b ( x, q ) and mapit under π to q . An argument similar to that in the previous case shows that theinductive hypotheses are maintained.This completes the construction of T and π . Observe that since Q κ has size κ , T is special, and it is clearly normal and κ -complete.Let us make an additional observation about T which we will need later. Weclaim that there exists an antichain A ⊆ T such that { ht T ( y ) : y ∈ A } = κ + ∩ cof( κ ).Namely, fix any q ∈ Q κ and fix x ∈ T (0) such that π ( x ) < Q κ q . For each β ∈ κ + ∩ cof( κ ), choose y β ∈ T ( β ) above x such that π ( y β ) = q . Let A := { y β : β ∈ κ + ∩ cof( κ ) } . By property (2) of T , A is an antichain as described in the claim.We now turn to the combinatorial principles which we will use to constructnon-club isomorphic Aronszajn trees. LARGE PAIRWISE FAR FAMILY OF ARONSZAJN TREES 5
Let κ be an infinite cardinal and S ⊆ κ + a stationary set. Recall that ♦ ( S ) is thestatement that there exists a sequence h s α : α ∈ S i , where each s α ⊆ α , satisfyingthat for any set X ⊆ κ + the set { α ∈ S : X ∩ α = s α } is stationary. And ♦ ∗ ( S )is the statement that there exists a sequence hS α : α ∈ S i , where each S α ⊆ P ( α )has size κ , satisfying that for any set X ⊆ κ + there is a club D ⊆ κ + such that forall α ∈ D ∩ S , X ∩ α ∈ S α . We have that ♦ ∗ ( S ) implies ♦ ( S ) and ♦ ( S ) implies2 κ = κ + . Theorem 1.6 (Shelah [8]) . Let κ be a regular uncountable cardinal and assumethat κ = κ + . Then ♦ ( κ + ) holds. In our Aronszajn tree constructions we will use the following combinatorial prin-ciple of Brodsky-Rinot [6].
Definition 1.7.
Let κ be a regular uncountable cardinal and S a family of sta-tionary subsets of κ + ∩ cof( κ ) . Then ⊠ ∗ κ ( S ) is the statement that there exists asequence h C δ : δ < κ + i , where each C δ is a club subset of δ of order type less thanor equal to κ , satisfying that for any unbounded set X ⊆ κ + and any S ∈ S , thereare stationarily many δ ∈ S for which the set ( C δ ∩ X ) \ lim( C δ ) is cofinal in δ . Inthe case S = { S } is a singleton, ⊠ ∗ κ ( S ) abbreviates ⊠ ∗ κ ( { S } ) . Note that if S ′ ⊆ S , then ⊠ ∗ κ ( S ) implies ⊠ ∗ κ ( S ′ ).The statement ⊠ ∗ κ ( S ) is a special case of the very general proxy principle whichwas introduced in [6] and used to construct a wide variety of Suslin trees. Inparticular, ⊠ ∗ κ ( S ) is the instance P − ( κ + , , κ ⊑ , , S , , , E κ ) of [6, Definition 1.5].For any stationary set S ⊆ κ + , ♦ ( S ) implies ⊠ ∗ κ ( S ) by [6, Theorem 5.1(2)], and ♦ ∗ ( κ + ∩ cof( κ )) implies ⊠ ∗ κ (NS + ↾ ( κ + ∩ cof( κ ))) by [5, Theorem 6.1], where NS + ↾ T is the collection of all stationary subsets of T for any stationary set T ⊆ κ + .Rather than using the principle ⊠ ∗ κ ( S ) directly in our Aronszajn tree construc-tions, we will use consequences of ⊠ ∗ κ ( S ) which are a kind of combination of dia-mond and club guessing principles. These consequences are described in the nexttwo lemmas, whose proofs follow easily from the methods of [6]. Lemma 1.8.
Let κ be a regular uncountable cardinal satisfying κ = κ + . Let S bea stationary subset of κ + ∩ cof( κ ) and assume that ⊠ ∗ κ ( S ) holds. Then there existsequences h C δ : δ ∈ S i and h s α : α < κ + i , where C δ = { α δ,i : i < κ } is a clubsubset of δ for every δ ∈ S and s α ⊆ α for every α < κ + , satisfying that for anyclub C ⊆ κ + and any set X ⊆ κ + , there are stationarily many δ ∈ S such that sup { i ∈ κ ∩ Succ : α δ,i ∈ C, X ∩ α δ,i = s α δ,i } = κ .Proof. Fix a sequence h C δ : δ < κ + i which witnesses ⊠ ∗ κ ( S ). For each δ ∈ S ,the fact that cf( δ ) = κ and ot( C δ ) ≤ κ implies that ot( C δ ) = κ . So write each C δ = { α δ,i : i < κ } in increasing order.Since 2 κ = κ + , ♦ ( κ + ) holds, so fix a diamond sequence h s α : α < κ + i . We claimthat h C δ : δ ∈ S i and h s α : α < κ + i are as desired. So let C ⊆ κ + be a club and X ⊆ κ + a set.Define U := { α ∈ κ + : X ∩ α = s α } , which is stationary by the diamondproperty. So U ∩ C is stationary. In particular, U ∩ C is an unbounded subset of κ + . By the choice of h C δ : δ < κ + i , there are stationarily many δ ∈ S for whichthe set ( C δ ∩ U ∩ C ) \ lim( C δ ) is cofinal in δ . For any such δ , there are cofinallymany i < κ such that i ∈ Succ and α δ,i ∈ U ∩ C , and hence by the choice of U , X ∩ α δ,i = s α δ,i . (cid:3) JOHN KRUEGER
The proof of the next lemma is similar and we leave it for the interested reader.
Lemma 1.9.
Let κ be a regular uncountable cardinal satisfying κ = κ + . Let S be a stationary subset of κ + ∩ cof( κ ) and assume that ⊠ ∗ κ ( NS + ↾ S ) holds. Thenthere exist sequences h C δ : δ ∈ S i and h s α : α < κ + i , where C δ = { α δ,i : i < κ } isa club subset of δ for every δ ∈ S and s α ⊆ α for every α < κ + , satisfying that forany club C ⊆ κ + and any set X ⊆ κ + , there exists a club D ⊆ κ + such that for all α ∈ D ∩ S , sup { i ∈ κ ∩ Succ : α δ,i ∈ C, X ∩ α δ,i = s α δ,i } = κ . Distinct Aronszajn Trees
In this section we show how to use the proxy principle to construct non-clubisomorphic Aronszajn trees. This construction can be thought of as a warm up forthe main result in Section 3.
Theorem 2.1.
Let κ be a regular uncountable cardinal such that κ <κ = κ and κ = κ + . Assume that ⊠ ∗ κ ( κ + ∩ cof( κ )) holds. Then there exist two normal κ -complete R κ -embeddable κ + -Aronszajn trees which are not club isomorphic. In particular, the conclusion of the theorem follows from κ <κ = κ and ♦ ( κ + ∩ cof( κ )).An alternative proof of Theorem 2.1 based on a result of Brodsky-Rinot [7] issketched at the end of the section. Lemma 2.2.
Let λ be a regular uncountable cardinal. Assume that T and U areclub isomorphic λ -trees. If A ⊆ U is an antichain such that S := { ht U ( x ) : x ∈ A } is a stationary subset of λ , then there exists a club C ⊆ λ and an antichain B ⊆ T such that { ht T ( y ) : y ∈ B } = S ∩ C .Proof. Fix a club isomorphism f : U ↾ C → T ↾ C . Let A ′ := { x ∈ A : ht U ( x ) ∈ C } .Note that { ht U ( x ) : x ∈ A ′ } = S ∩ C . Since f is an isomorphism and A ′ ⊆ dom( f ), B := f [ A ′ ] is an antichain of T . And { ht T ( y ) : y ∈ B } = { ht T ( f ( x )) : x ∈ A ′ } = { ht U ( x ) : x ∈ A ′ } = S ∩ C . (cid:3) We now prove Theorem 2.1. Fix a regular cardinal κ such that κ <κ = κ and2 κ = κ + . Assume that ⊠ ∗ κ ( κ + ∩ cof( κ )) holds. Fix sequences h C δ : δ ∈ κ + ∩ cof( κ ) i and h s α : α < κ + i satisfying the description given in Lemma 1.8. Our goal is toproduce normal κ -complete R κ -embeddable κ + -Aronszajn trees T and U which arenot club isomorphic.By the construction in Section 1, we can fix a normal κ -complete κ + -Aronszajntree U which is Q κ -embeddable and satisfies that for some antichain A ⊆ U , { ht U ( x ) : x ∈ A } = κ + ∩ cof( κ ). Then U is also R κ -embeddable. We will build anormal κ -complete R κ -embeddable κ + -Aronszajn tree T satisfying that there doesnot exist an antichain B ⊆ T and a club C ⊆ κ + such that C ∩ cof( κ ) ⊆ { ht T ( y ) : y ∈ B } . Lemma 2.2 then implies that T and U are not club isomorphic, whichcompletes the proof.We construct T together with a function π : T → R κ by recursion. We willmaintain the following properties:(1) for each α < κ + , T ( α ) consists of the ordinals in the interval [ κ · α, κ · ( α +1));(2) x < T y implies π ( x ) < R κ π ( y ), for all x, y ∈ T ;(3) for each x ∈ T , the restriction of π to the immediate successors of x is abijection from that set onto the set { q ∈ Q κ : π ( x ) < R κ q } ; LARGE PAIRWISE FAR FAMILY OF ARONSZAJN TREES 7 (4) if δ ∈ κ + ∩ cof( < κ ), then every cofinal branch b of T ↾ δ has a unique upperbound y in T ( δ ), and π ( y ) = sup { π ( x ) : x ∈ b } ;(5) for all x ∈ T , β < κ + , and q ∈ Q κ with ht T ( x ) < β and π ( x ) < R κ q , thereexists y ∈ T ( β ) above x such that π ( y ) ≤ R κ q .We will abbreviate the restriction of π to T ↾ δ by π ↾ δ for all δ < κ + .The base case, successor steps, and limit stages of cofinality less than κ arehandled in basically the same way as in the construction of the Aronszajn treein Section 1. The only difference is that we are mapping the nodes of T into R κ instead of Q κ . But nodes of successor height will have their values under π in Q κ by (3), and by Lemma 1.4(2) nodes whose heights are limit ordinals of cofinalityless than κ also have their values under π in Q κ . Since Q κ is dense in R κ , property(5) is easily shown to hold at successor levels and at limit levels of cofinality lessthan κ . Thus, the inductive hypotheses hold at these types of levels.Assume that δ < κ + has cofinality κ and we have defined T ↾ δ and π ↾ δ asrequired. We would like to associate to each pair ( x, q ), where x ∈ T ↾ δ and q ∈ Q κ with π ( x ) < R κ q , a cofinal branch b ( x, q ) of T ↾ δ . Then we will add an upper bound y to b ( x, q ) at level δ and define π ( y ) so that π ( y ) ≤ R κ q and π ( z ) < R κ π ( y ) for all z ∈ b . If we succeed in doing this, the inductive hypotheses are clearly maintained.Since each node has κ many immediate successors and there are only κ many pairs( x, q ) to handle, it is easy to arrange that the function mapping ( x, q ) to b ( x, q ) isinjective, so we will neglect this point.Fix x ∈ T ↾ δ and q ∈ Q κ with π ( x ) < R κ q . Recall that C δ = { α δ,i : i < κ } is aclub subset of δ . Let i ∗ < κ be the least ordinal such that ht T ( x ) < α δ,i ∗ . We willrecursively define sequences h x i : i < κ i and h q i : i < κ i satisfying:(1) for all i ≤ i ∗ , x i = x and q i = q ;(2) for all i ∗ < i < κ , x i is a node above x on level α δ,i of T and q i ∈ Q κ ;(3) q j < Q κ q i for all i < j < κ ;(4) for all i, j < κ , π ( x i ) < R κ q j .Begin by setting x i := x and q i := q for all i ≤ i ∗ . Now let i > i ∗ be given andassume that for all j < i we have defined x j and q j as required.Suppose that i = j + 1 is a successor ordinal, and we will define x i and q i .Consider the following statements:(A) α δ,i = κ · α δ,i .(B) s α δ,i is an antichain of T ↾ α δ,i .(C) there exists some y ∈ s α δ,i above x j in T such that π ( y ) < R κ q j .First, assume that these statements are not all true. In that case, choose x i to beany node above x j on level α δ,i of T ↾ δ such that π ( x i ) < R κ q j using inductiveproperty (5). Then choose q i ∈ Q κ such that π ( x i ) < R κ q i < R κ q j . Secondly,assume that all three statements are true. Fix y as in (C). Then use inductiveproperty (5) to choose x i above y on level α δ,i of T ↾ δ satisfying that π ( x i ) < R κ q j .Then choose q i ∈ Q κ such that π ( x i ) < R κ q i < R κ q j .Suppose that i is a limit ordinal. Since i < κ , there exists a unique node x i onlevel α δ,i of T which is above x j for all j < i , and π ( x i ) = sup { π ( x j ) : j < i } .By Lemma 1.4(2), π ( x i ) is in Q κ . By Lemma 1.2(3), the descending sequence h q j : i ∗ < j < i i does not have a greatest lower bound in Q κ . In particular, π ( x i )is a lower bound but not a greatest lower bound of this sequence. Thus, we can fix q i ∈ Q κ such that π ( x i ) < Q κ q i and for all j < i , q i < Q κ q j . JOHN KRUEGER
Let b ( x, q ) be the downward closure in T ↾ δ of the chain { x i : i < κ } . Then b ( x, q ) is a cofinal branch of T ↾ δ . We place some y above this branch on level δ ,and define π ( y ) := inf { q j : j < κ } . This makes sense because the set { q i : i < κ } is bounded below by π ( x ) and R κ is complete. Now for all i < κ and j < κ , π ( x i ) < R κ π ( x i +1 ) < R κ q j , and hence π ( x i ) < R κ inf { q j : j < κ } = π ( y ). It followsthat for all z < T y , π ( z ) < R κ π ( y ). Also, π ( y ) < R κ q .This completes the construction of T . Let us prove that T is as required. Clearly, T is a normal κ -complete R κ -embeddable κ + -Aronszajn tree. We claim that theredoes not exist an antichain B ⊆ T and a club C ⊆ κ + such that C ∩ cof( κ ) ⊆{ ht T ( y ) : y ∈ B } .Let B ⊆ T be an antichain and C ⊆ κ + a club. For each q ∈ Q κ , let C q be theclub set of α < κ + such that for all x ∈ T ↾ α , if there exists some z ∈ B such that x < T z and π ( z ) < R κ q , then there exists such a z which is in T ↾ α . Let D be theclub of all α < κ + such that α = κ · α . Define E := C ∩ D ∩ T { C q : q ∈ Q κ } . Then E is club in κ + since Q κ has cardinality κ .Let S be the set of δ ∈ κ + ∩ cof( κ ) such that the set { i ∈ κ ∩ Succ : α δ,i ∈ E, B ∩ α δ,i = s α δ,i } is cofinal in κ . By the ♦ -club guessing property, S is stationary.Fix δ in S ∩ E . Then δ ∈ C ∩ cof( κ ).Suppose for a contradiction that C ∩ cof( κ ) ⊆ { ht T ( y ) : y ∈ B } . Then inparticular, δ ∈ { ht T ( y ) : y ∈ B } . Fix y ∗ ∈ B such that ht T ( y ∗ ) = δ . By theconstruction of T , y ∗ is an upper bound of a branch b ( x, q ) for some x ∈ T ↾ δ and q ∈ Q κ with π ( x ) < R κ q . Fix i ∗ < κ and sequences h x i : i < κ i and h q j : j < κ i asdescribed in the definition of b ( x, q ).As δ ∈ S , we can find i = j + 1 ∈ κ ∩ Succ greater than i ∗ such that α δ,i ∈ E and B ∩ α δ,i = s α δ,i . Since α δ,i ∈ D , α δ,i = κ · α δ,i . And s α δ,i = B ∩ α δ,i is an antichainof T ↾ α δ,i . Thus, statements (A) and (B) in the case division of the definition of x i and q i are met.We claim that statement (C) holds as well. The node y ∗ is an element of B which is above x j . By definition, π ( y ∗ ) = inf { q i : i < κ } , and in particular, π ( y ∗ ) < R κ q j . Since α δ,i ∈ C q j , there exists some y ∈ B ∩ ( T ↾ α δ,i ) above x j suchthat π ( y ) < R κ q j . Then y ∈ B ∩ α δ,i = s α δ,i , proving (C).By the definition of x i , there exists some y ∈ s α δ,i = B ∩ α δ,i below x i . But x i < T y ∗ , so y < T y ∗ . This is a contradiction since B is an antichain and both y and y ∗ are in B . This completes the proof of Theorem 2.1.Assaf Rinot has pointed out that Theorem 2.1 can also be proven using [7,Theorem 5.4]. This theorem states that from the assumptions of Theorem 2.1,there exist normal κ + -Aronszajn trees T and T such that T is special and T isnot special (these trees are T ( ρ ) and T ( ρ ) associated with walks on ordinals). Itis not hard to show that the trees defined in [7, Theorem 5.4] are R κ -embeddable.Now using the method of the proof of [4, Proposition 3.7], we can enlarge T and T to normal κ -complete R κ -embeddable κ + -Aronszajn trees U and U such that U is special and U is not special. But a special tree cannot be club isomorphicto a non-special tree, since by a straightforward argument if a tree is special on aclub of levels, then it is special.Note however that this alternative proof does not obviously generalize to con-struct a large pairwise far family of Aronszajn trees, as we will do in the nextsection. LARGE PAIRWISE FAR FAMILY OF ARONSZAJN TREES 9 A large pairwise far family of Aronszajn trees
We now present the main result of the article.
Theorem 3.1.
Let κ be a regular uncountable cardinal such that κ <κ = κ and κ = κ + . Assume that ⊠ ∗ κ ( NS + ↾ ( κ + ∩ cof( κ ))) holds. Then there exists a pairwisefar family of ( κ + ) many normal κ -complete R κ -embeddable κ + -Aronszajn trees. In particular, the conclusion of the theorem follows from κ <κ = κ and ♦ ∗ ( κ + ∩ cof( κ )). Proposition 3.2.
Let κ be a regular uncountable cardinal such that κ <κ = κ and κ = κ + . Let S ⊆ κ + ∩ cof( κ ) be stationary and assume that ⊠ ∗ κ ( NS + ↾ S ) holds. Then there exists a normal κ -complete R κ -embeddable κ + -Aronszajn tree T S satisfying: (1) the subtree T S ↾ ( κ + \ S ) is special; (2) for any antichain A ⊆ T S , the set { ht T S ( x ) : x ∈ A } ∩ S is non-stationary. We claim that Theorem 3.1 follows from Proposition 3.2. To see this, let κ be aregular uncountable cardinal such that κ <κ = κ and 2 κ = κ + . Fix a family { S i : i < ( κ + ) } of stationary subsets of κ + ∩ cof( κ ) such that for all distinct i, j < ( κ + ) , S i \ S j is stationary. The existence of such a family can be proven by a slightvariation of the proof of [9, Proposition 1.1]. Assume that ⊠ ∗ κ (NS + ↾ ( κ + ∩ cof( κ )))holds. Then for all i < ( κ + ) , ⊠ ∗ κ (NS + ↾ S i ) holds. For each i < ( κ + ) , fix a κ + -Aronszajn tree T i satisfying the conclusions of Proposition 3.2 for the set S i .Suppose for a contradiction that T i and T j are near for some distinct i, j < ( κ + ) .Fix downward closed subtrees T ′ i and T ′ j of T i and T j respectively which are clubisomorphic. Fix a club isomorphism f : T ′ i ↾ C → T ′ j ↾ C .Let S ∗ := C ∩ ( S j \ S i ), which is a stationary subset of κ + . As S ∗ ⊆ κ + \ S i , T i ↾ S ∗ is special. Therefore, T ′ i ↾ S ∗ is special. Fix an antichain A ⊆ T ′ i ↾ S ∗ suchthat S ∗∗ := { ht T ′ i ( x ) : x ∈ A } is stationary. Note that S ∗∗ ⊆ S ∗ ⊆ C ∩ S j .Now S ∗ ⊆ C , so A ⊆ dom( f ). Hence, f [ A ] is an antichain of T ′ j ↾ C . So f [ A ] isan antichain of T j . Also { ht T j ( y ) : y ∈ f [ A ] } = { ht T ′ j ( f ( x )) : x ∈ A } = { ht T ′ i ( x ) : x ∈ A } = S ∗∗ , which is a stationary subset of S j . Thus, f [ A ] is an antichain of T j for which { ht T j ( y ) : y ∈ f [ A ] } is a stationary subset of S j , contradicting the choice of T j .It remains to prove Proposition 3.2. Assume that κ <κ = κ and 2 κ = κ + , and let S ⊆ κ + ∩ cof( κ ) be a stationary set such that ⊠ ∗ κ (NS + ↾ S ) holds. Fix sequences h C δ : δ ∈ S i and h s α : α < κ + i satisfying the description given in Lemma 1.9.We construct T S together with a function π S : T S → R κ by recursion. We willmaintain the following properties:(1) For each α < κ + , T ( α ) consists of the ordinals in [ κ · α, κ · ( α + 1));(2) x < T S y implies π S ( x ) < R κ π S ( y ), for all x, y ∈ T S ;(3) for each x ∈ T S , π S restricted to the immediate successors of x is a bijectionfrom that set onto the set { q ∈ Q κ : π S ( x ) < R κ q } ;(4) if δ ∈ κ + ∩ cof( < κ ), then every cofinal branch b of T S ↾ δ has a uniqueupper bound y , and π S ( y ) = sup { π S ( x ) : x ∈ b } ;(5) for all x ∈ T , β < κ + , and q ∈ Q κ with ht T S ( x ) < β and π S ( x ) < R κ q ,there exists y ∈ T S ( β ) above x such that π S ( y ) ≤ R κ q . We will abbreviate the restriction of π S to T S ↾ δ by π S ↾ δ .As in the construction of Section 2, it suffices to define the tree at levels ofcofinality κ , the other cases being easy. For all δ ∈ κ + ∩ cof( κ ) which are not in S ,build a family of distinct cofinal branches b ( x, q ) of T S ↾ δ , for each x ∈ T S ↾ δ and q ∈ Q κ with π S ( x ) < R κ q . Then put an ordinal y above each such branch b ( x, q )on level δ and define π S ( y ) := q .This completes the description of T S on levels not in S . Since π S ( x ) will be in Q κ for all x ∈ T S of height not in S , T S ↾ ( κ + \ S ) will be special.It remains to define T S and π S at levels which are in S . So assume that δ ∈ S and T S ↾ δ and π S ↾ δ have been defined. We will associate to each pair ( x, q ),where x ∈ T S ↾ δ and q ∈ Q κ with π S ( x ) < R κ q , a cofinal branch b ( x, q ) of T S ↾ δ ,and then add an upper bound y to b ( x, q ) on level δ and define π S ( y ). Since eachnode of T S ↾ δ has κ many immediate successors, it is easy to arrange that thefunction which maps ( x, q ) to b ( x, q ) is injective.Fix x ∈ T S ↾ δ and q ∈ Q κ such that π S ( x ) < R κ q , and we will define b ( x, q ).Recall that C δ = { α δ,i : i < κ } is a club subset of δ . Let i ∗ < κ be the least ordinalsuch that ht T S ( x ) < α δ,i ∗ .We will define sequences h x i : i < κ i and h q i : i < κ i satisfying:(1) for all i ≤ i ∗ , x i = x and q i = q ;(2) for all i ∗ < i < κ , x i is a node on level α δ,i of T S ↾ δ above x ;(3) for all i < κ , q i ∈ Q κ ;(4) q j < Q κ q i for all i ∗ ≤ i < j < κ ;(5) for all i, j < κ , π S ( x i ) < R κ q j .Begin by setting x i := x and q i := q for all i ≤ i ∗ . Now let i < κ be greater than i ∗ and assume that for all j < i we have defined x j and q j as required.Case 1: Assume that i = j + 1 is a successor ordinal. Consider the followingstatements: • α δ,i = κ · α δ,i ; • s α δ,i is an antichain of T S ↾ α δ,i ; • there exists z ∈ s α δ,i above x j such that π S ( y ) < R κ q j .First, assume that at least one of these statements is false. Since π S ( x j ) < R κ q j ,we can choose x i above x j on level α δ,i of T S ↾ δ such that π S ( x i ) < R κ q j . Thenchoose q i ∈ Q κ such that π S ( x i ) < R κ q i < R κ q j .Secondly, assume that the three statements are true. Fix z as in (3). Then fix x i above z on level α δ,i of T S ↾ δ such that π S ( x i ) < R κ π S ( z ). Finally, choose q i ∈ Q κ such that π S ( x i ) < R κ q i < R κ π S ( z ). This completes the definition of x i and q i when i < κ is a successor ordinal.Case 2: Assume that i < κ is a limit ordinal and x j and q j have been defined forall j < i . Let x i be the least upper bound of { x j : j < i } , which is on level α δ,i of T S ↾ δ . Since cf( α δ,i ) < κ , π S ( x i ) ∈ Q κ . Let q ∗ i be the greatest lower bound of { q j : j < i } . Then π S ( x i ) ≤ R κ q ∗ i . By Lemma 1.2(3), q ∗ i / ∈ Q κ , so π S ( x i ) < R κ q ∗ i .Choose q i ∈ Q κ with π S ( x i ) < R κ q i < R κ q ∗ i .This completes the construction of the sequences h x i : i < κ i and h q i : i < κ i .Define b ( x, q ) to be the downward closure of the chain { x i : i < κ } . Note that b ( x, q ) is a cofinal branch of T S ↾ δ . Now put some y above the branch b ( x, q ) on LARGE PAIRWISE FAR FAMILY OF ARONSZAJN TREES 11 level δ of T S and define π S ( y ) = sup { π S ( x i ) : i < κ } . Note that π S ( y ) < R κ q i forall i < κ , and hence π S ( y ) < R κ q .This completes the construction of T S and π S . It is clear that T S is a normal κ -complete R κ -embeddable κ + -Aronszajn tree satisfying that T S ↾ ( κ + \ S ) is special.It remains to prove that for any antichain A ⊆ T S , the set { ht T S ( x ) : x ∈ A } ∩ S isnon-stationary.Let A ⊆ T S be an antichain. For each q ∈ Q κ , define C q to be the set of all δ < κ + satisfying that for all x ∈ T S ↾ δ , if there exists some y ∈ A above x suchthat π S ( y ) < R κ q , then there exists such a y in T S ↾ δ . Let C := T { C q : q ∈ Q κ } ∩ { α ∈ κ : κ · α = α } . Since Q κ has size κ , C is a club subset of κ + .By the property described in Lemma 1.9, fix a club D ⊆ κ + such that for all δ ∈ D ∩ S , the set { i ∈ κ ∩ Succ : α δ,i ∈ C, A ∩ α δ,i = s α δ,i } is cofinal in κ .We claim that { ht T S ( x ) : x ∈ A } ∩ S ∩ D = ∅ , which will complete the proof.Suppose for a contradiction that δ is in this intersection. Fix y ∈ A such thatht T S ( y ) = δ . By the definition of level δ of T S , there exists some x ∈ T S ↾ δ and q ∈ Q κ such that y is the unique upper bound of b ( x, q ) on level δ .Let i ∗ , h x i : i < κ i , and h q i : i < κ i be as in the definition of b ( x, q ). Recall that π S ( y ) < R κ q i for all i < κ . Fix i ∈ κ ∩ Succ greater than i ∗ such that α δ,i ∈ C and A ∩ α δ,i = s α δ,i .Now y ∈ A , x j < T S y , and π S ( y ) < R κ q j . Since α δ,i ∈ C ⊆ C q , there exists z ∈ A ∩ ( T S ↾ α δ,i ) = A ∩ α δ,i = s α δ,i such that x j < T S z and π S ( z ) < R κ q j . Bydefinition, x i is above such a z . So z < T S x i < T S y , and hence z < T S y . This is acontradiction because both y and z are in A and A is an antichain. This completesthe proof of Theorem 3.1.Finally, we point out that the assumptions of Theorem 3.1 hold in a variety ofmodels in addition to one satisfying ♦ ∗ ( κ + ∩ cof( κ )). Assuming κ <κ = κ , [5, The-orem 6.1] shows that ⊠ ∗ κ (NS + ↾ S ) holds in any of the following generic extensions:(1) any generic extension by Add( κ, χ > κ is strongly inaccessible, any generic extension by a ( < κ )-distributive χ -c.c. forcing which collapses χ to become κ + ;(3) if 2 κ = κ + , any generic extension by a κ + -c.c. forcing of size ≤ κ + whichpreserves the regularity of κ and is not κ κ -bounding.In particular, in any such generic extension there exists a pairwise far family of2 ( κ + ) many normal κ -complete R κ -embeddable κ + -Aronszajn trees. References [1] U. Abraham and S. Shelah. Isomorphism types of Aronszajn trees.
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John Krueger, Department of Mathematics, University of North Texas, 1155 UnionCircle
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