aa r X i v : . [ m a t h . L O ] O c t CAN YOU TAKE KOMJATH’S INACCESSIBLE AWAY?
HOSSEIN LAMEI RAMANDI AND STEVO TODORCEVIC
Abstract.
In this paper we aim to compare Kurepa trees andAronszajn trees. Moreover, we analyze the affect of large cardinalassumptions on this comparison. Using the the method of walkson ordinals, we will show it is consistent with ZFC that there is aKurepa tree and every Kurepa tree contains a Souslin subtree, ifthere is an inaccessible cardinal. This is stronger than Komjath’stheorem in [3], where he proves the same consistency from two in-accessible cardinals. We will show that our large cardinal assump-tion is optimal, i.e. if every Kurepa tree has an Aronszajn subtreethen ω is inaccessible in the constructible universe L . Moreover,we prove it is consistent with ZFC that there is a Kurepa tree T such that if U ⊂ T is a Kurepa tree with the inherited orderfrom T , then U has an Aronszajn subtree. This theorem uses nolarge cardinal assumption. Our last theorem immediately impliesthe following: assume MA ω holds and ω is not a Mahlo cardinalin L . Then there is a Kurepa tree with the property that everyKurepa subset has an Aronszajn subtree. Our work entails provinga new lemma about Todorcevic’s ρ function which might be usefulin other contexts. Introduction
In this paper we aim to compare Kurepa trees and Aronszajn trees.Moreover, we analyze the affect of large cardinal assumptions on thiscomparison. We are interested in the question that to what extent doKurepa trees contain Aronszajn subtrees. The first result regardingthis question is due to Jensen. He showed that there is a Kurepatree in the constructible universe L , which has no Aronszajn subtrees.Todorcevic showed that there is a countably closed forcing which addsa Kurepa tree with no Aronszajn subtree. Both of these results are inthe negative direction. In the positive direction, Komjath proved thefollowing theorem. Theorem 1.1. [3]
It is consistent relative to the existence of two in-accessible cardinals that there is a Kurepa tree and every Kurepa treehas an Aronszajn subtree.
Key words and phrases.
Aronszajn trees, Kurepa trees .
It is natural to ask whether or not the large cardinal assumptions inTheorem 1.1 is sharp. In other words, assume every Kurepa tree hasan Aronszajn subtree, then is it consistent that there are at least twoinaccessible cardinals?In addition to addressing this question, our work reveals a new factabout Todorcevic’s ρ function. Based on this fact about ρ and a notionof capturing which was introduced in [1], we find Arinszajn suborders insome canonical Kurepa trees without any large cardinal assumptions.It is worth mentioning that although we analyze some ω -trees to provethis fact about ρ , the function ρ is defined in terms of ordinals with noreference to ω -trees.In this paper we will show the following theorem, which is strongerthan Komjath’s Theorem. Theorem 1.2.
Assume there is an inaccessible cardinal. Then it isconsistent that there is a Kurepa tree and every Kurepa tree contains aSouslin subtree.
Moreover, we will show that our large cardinal strength is sharp inthe following theorem.
Theorem 1.3.
Assume that every Kurepa tree contains an Aronszajnsubtree. Then ω is inaccessible in the constructible universe L . Roughly speaking, Theorem 1.3 asserts that it is impossible to reacha model in which every Kurepa tree has an Aronszajn subtree, unless atsome point some large cardinal is collapsed. This theorem does not sayanything about the situation where there is a Kurepa tree all of whoseKurepa subtrees have Aronszajn subtrees. By analyzing Todorcevic’s ρ function and a ccc poset which is a defined based on ρ we prove thefollowing theorem. Theorem 1.4.
It is consistent that there is a Kurepa tree T such thatwhenever U ⊂ T is a Kurepa tree when it is considered with the inher-ited order from T , then U has an Aronszajn subtree. In [4] by using ρ , Tododrcevic introduces a forcing which satisfiesthe Knaster condition and which adds a Kurepa tree which has a nicestructure related to ρ . We use this forcing to prove Lemma 5.3. Thenwe use this lemma to show Theorem 1.4. Since the tree T can be forcedto exist in any model of (cid:3) ω , the following corollary trivially followsfrom Theorem 1.4. Corollary 1.5.
Assume MA ω holds and ω is not a Mahlo cardinalin L . Then there is a Kurepa tree with the property that every Kurepasubset has an Aronszajn subtree. AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 3 Preliminaries
As we mentioned in the introduction, we will show that the asser-tion every Kurepa tree has an Aronszajn subtree has large cardinalstrength. For this purpose we will work in L [ X ], the constructible uni-verse relativised to the set X . Here we review some definitions andfacts regarding L [ X ]. Definition 2.1.
Assume M is a transitive and X is a set. The set ofall definable subsets of M from X , which is denoted by def X ( M ), isthe set of all D ⊂ M such that D is definable over ( M, ∈ , X ∩ M ). Inother words, X ∩ M is considered a unary predicate in M . Definition 2.2.
The constructible universe relativised to X is denotedby L [ X ] and defined by transfinite induction as follows: • L [ X ] = ∅ , • L α +1 [ X ] = def X ( L α [ X ]), and • if α is a limit ordinal then L α [ X ] = S ξ ∈ α L ξ [ X ]The universe L [ X ] is the union of all L α [ X ] where α is an ordinal.It is also useful to recall the following fact regarding relativized con-densation. Fact 2.3.
Assume M ≺ ( L β [ X ] , ∈ , X ∩ L β [ X ]) where β is a limitordinal. Then the transitive collapse of M is L δ [ X ] for some δ ≤ β . The following fact is well known.
Fact 2.4.
Assume ω V is not inaccessible in L . Then there is X ⊂ ω V such that L [ X ] computes ω , ω as V does. We will use the following Theorem in order to show that if everyKurepa tree has an Aronszajn subtree then there is an inaccessiblecardinal in the constructible universe L . Theorem 2.5 (Todorcevic ) . Assume M is an inner model of set the-ory which correctly computes ω . Then M contains a partition of ω into infinitely many sets which are stationary in the universe of all sets V . In [1] a notion of capturing is defined for linear orders. This notioncan be used for ω -trees as well. We will use this notion and Proposition2.8 in order to characterize when an ω -tree contains an Aronszajnsubtree. Let’s fix some notation regarding ω -trees. Assume T is an ω -tree. For any α ∈ ω , T α denotes the set of all elements of T which See Corollary 2.3.5 in [4].
H. LAMEI RAMANDI, S. TODORCEVIC have height α . T <α is the set of all members of T which have heightless than α . B ( T ) is the set of all cofinal branches of T . If b is a cofinalbranch in T and α ∈ ω , b ( α ) refers to the element in b which is ofheight α . If t ∈ T and α ∈ ω then t ↾ α refers to the set of all elements x ≤ T t whose height is less than α . For any x ∈ T , T x is the set of all t ∈ T that are comparable with x . Definition 2.6. [1] Assume T is an ω -tree, κ is a large enough regularcardinal, t ∈ T ∪ B ( T ), and N ≺ H κ is countable such that T ∈ N . Wesay that N captures t if there is a chain c ⊂ T in N which contains allelements of T Assume T = ( ω , < ) is an ω -tree , x ∈ T ∪ B ( T ) and N ≺ H θ is countable with T ∈ N . We say that x is weakly external to N if there is a stationary Σ ⊂ [ H ω ] ω in N such that ∀ M ∈ N ∩ Σ, M does not capture x. Note that there is a major difference between the definition above andDefinition 3.1 in [1]. If we require Σ to be a club we obtain the definitionof external elements in [1]. This is why we call x weakly external in ourdefinition. The purpose of this definition is to find Aronszajn suborders.It turns out that the existence of weakly external elements is enoughfor an ω -tree to have Aronszajn subtrees. This should be comparedwith Theorem 4.1 in [1], where the existence of external elements isrequired for finding Aronszajn suborders. The proof we present hereuses the ideas in the proof of Theorem 4.1 in [1], but we will include itfor more clarity. Proposition 2.8. Let T = ( ω , < ) be an ω -tree, κ = 2 ω + and Σ ⊂ [ H κ ] ω be stationary. Assume for all large enough regular cardinal θ there are x ∈ T and countable N ≺ H θ such that x is weakly externalto N witnesses by Σ . In other words, for all M ∈ Σ ∩ N , M does notcapture x . Then T has an Aronszajn subtree.Proof. Fix θ as in the proposition. For each t ∈ T let W t be the set ofall countable N ′ ≺ H θ such that Σ , T are in N ′ and there is s > t suchthat for all M ∈ Σ ∩ N ′ , M does not capture s . Let A be the set of all t ∈ T such that W t is stationary. We will show that A is Aronszajn.First note that A is downward closed. This is because if t < t ′ then W t ′ ⊂ W t . Moreover, if t ∈ T , δ ∈ ω and ht( t ) < δ then W t = S { W s : s > t and ht( s ) = δ } . In other words, if A = ∅ then A isuncountable. So it suffices to show that A = ∅ and A does not containany uncountable branch of T . AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 5 First we will show that A = ∅ . Fix a regular cardinal λ > θ suchthat θ is definable in H λ . Let P ≺ H λ be countable such that for some x ∈ T Σ witnesses that x is weakly external to P . Let t ∈ T ∩ P and t < x . Then P ∩ H θ ∈ W t . Hence W t is stationary and A = ∅ .In order to see A contains no uncountable branch of T , assume for acontradiction that b ⊂ A is a cofinal branch. Let M ≺ H κ be countablesuch that T, A, b, are in M and M ∈ Σ . Let δ = M ∩ ω and t = b ( δ ).Let N ≺ H θ be countable such that N ∈ W t . This is possible because t ∈ A and W t is a stationary subset of [ H θ ] ω . Let s > t be the elementin T such that for all Z ∈ Σ ∩ N , Z does not capture s . But M ∈ Σ ∩ N and it captures s via b . This is a contradiction. (cid:3) As mentioned in [1], it is easy to see that if T is an ω -tree with anAronszajn subtree A , then for every countable N ≺ H θ with A ∈ N ,and for every x ∈ A \ N , we have that x is external to N . Therefore,we can conclude the following Corollary. Corollary 2.9. Assume T = ( ω , < ) is an ω -tree. Then the followingare equivalent: • T has an Aronszajn subtree. • For all large enough regular cardinal θ there are x ∈ T andcountable N ≺ H θ such that x is external to N . • For all large enough regular cardinal θ there are x ∈ T andcountable N ≺ H θ such that x is weakly external to N . We will use the following facts from [2] which is due to Jensen andSchlechta in the last section. For more clarity we will include the sketchof their proof. Fact 2.10. [2] Assume A ∈ V is a countably closed poset, F ⊂ A is V -generic, B ∈ V is a ccc poset and G ⊂ B is V [ F ] -generic. Let T ∈ V [ G ] be a normal ω -tree. (1) If b ∈ V [ F ][ G ] is a cofinal branch in T , then b ∈ V [ G ] . (2) If S ∈ V [ F ][ G ] is a downward closed Souslin subtree of T then S ∈ V [ G ] .Proof. Assume for a contradiction that b ∈ V [ F ][ G ] \ V [ G ] is a branchin T , and let ˙ b be the name which is forces by to be outside of V [ G ].For k ∈ , let j k : A × B −→ A × B be the injections which take ( p, q )to ( , p, q ) and ( p, , q ). Obviously, these injections naturally induceinjections on ( A × B )-names. We will abuse the notation and use j k forthe injections on names too. Let j k (˙ b ) = τ k for k ∈ 2. Since b / ∈ V [ G ], A × B (cid:13) τ = τ . H. LAMEI RAMANDI, S. TODORCEVIC Note that the set D = { ( a , a ) ∈ A : ∃ α ∈ ω ( a , a , B ) (cid:13) τ ( α ) = τ ( α ) } is dense in A . This uses an argument similar to theproof of fullness lemma and the fact that countably closed posets donot add new countable subsets of the ground model. Similarly, the set D α = { a ∈ A : for some B -name ˙ x , ( a, B ) (cid:13) τ ( α ) = ˙ x } is dense in A .Now construct an increasing sequence α n , n ∈ ω and a s , ˙ x s for s ∈ <ω such that: • ( a s , ) (cid:13) τ ( α | s | ) = ˙ x s where ˙ x s is a B -name in V , • a s⌢ , a s⌢ are both below a s and ( a s⌢ , a s⌢ , ) (cid:13) ˙ x s⌢ = ˙ x s⌢ .For each r ∈ ω ∩ V let a r be the lower bound for h a s : s ⊂ r i . In V [ G ]let y r be the element which is forced by a r to be the element on topof h x s : s ⊂ r i . This means that T has an uncountable level in V [ G ]which is a contradiction.The proof of the statement for Souslin subtrees uses similar ideasand the following facts, which we briefly mention. First note that if X is a countable subset of V which is in V [ F ][ G ] then X ∈ V [ G ]. Also,if S is a Souslin subtree of T in V [ G ] then it is Souslin in V [ F ][ G ]. If S ∈ V [ F ][ G ] \ V [ G ] is a Souslin subtree of T then there is S ′ ⊂ S suchthat every cone S ′ x is outside of V [ G ] for all x ∈ S ′ .Now assume for a contradiction that S is a Souslin subtree of T which is in V [ F ][ G ] \ V [ G ] . Without loss of generality we can assumethat every cone S x is outside of V [ G ], for every x ∈ S . Assume ˙ S isthe name which is forced by to be outside of V [ G ]. Again let τ k bethe corresponding names j k ( ˙ S ) as above.Let S k be the Souslin tree for τ k , for k ∈ , in the extension by( F , F , G ) ⊂ A × B which is V -generic. Note that S ∩ S ⊂ T <α for some α ∈ ω . In order to see this assume S ∩ S is uncount-able. Then for some x ∈ S ∩ S , ( S ) x = ( S ) x . But then ( S ) x =( S ) x ∈ V [ F ][ G ] ∩ V [ F ][ G ] = V [ G ], which is a contradiction. Chooseincreasing sequences h α | s | , a s , ˙ x s : s ∈ <ω i such that: • ( a s , ) (cid:13) ˙ S ∩ T <α | s | = ˙ x s , where ˙ x s is a B -name in V , • a s⌢ , a s⌢ are both below a s , • ( a s⌢ , a s⌢ , ) (cid:13) T α s ∩ ˙ x s⌢ ∩ ˙ x s⌢ = ∅ .For each r ∈ ω ∩ V , let a r be a lower bound for h a s : s ⊂ r i . Alsolet α = sup { α n : n ∈ ω } . Now we work in V [ G ]. For each r let x r = S s ⊂ r ˙ x s [ G ]. Note that if r = r ′ then T α ∩ x r ∩ x r ′ = ∅ . For each r ,let y r ∈ T α such that { t ∈ T : t < y r } ⊂ x r . But this means that T α isuncountable which is a contradiction. (cid:3) We will use (cid:3) ω in order to have the structure of walks on ordinalsup to ω . The following is the standard definition of (cid:3) ω . AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 7 Definition 2.11. A sequence h C α : α is limit and ω < α < ω i is saidto be a (cid:3) ω -sequence if • C α is a closed unbounded subset of α , • otp( C α ) < α and • if α is a limit point of C β then C β ∩ α = C α .The assertion that there is a (cid:3) ω -sequence is called (cid:3) ω . The following proposition is obtained from standard argument using (cid:3) ω -sequences. Proposition 2.12. If (cid:3) ω holds then there is a sequence h C α : α ∈ ω i such that • C α is a closed unbounded subset of α , • for all α , otp ( C α ) ≤ ω and if cf ( α ) = ω then otp ( C α ) < ω , • if α ∈ C β then cf ( α ) ≤ ω , • if α is a limit point of C β then C β ∩ α = C α . We only consider (cid:3) ω -sequences which have the properties mentionedin the proposition above. We will also use the following standard fact. Fact 2.13. Assume λ is a regular cardinal which is not Mahlo in L .Let G ⊂ coll ( ω , < λ ) be L -generic. Then (cid:3) ω holds in L [ G ] . Now we briefly review some definitions and facts about walks onordinals, from [4]. We fix a (cid:3) ω sequence h C α : α ∈ ω i which satisfiesthe properties in Proposition 2.12.We will use the following notation in the rest of the paper. For all X , α X = sup( X ∩ ω ). For each α ∈ ω we let L α be the set of all β ∈ ω such that α ∈ lim( C β ). For each α < β in ω , let Λ( α, β ) bethe maximal limit point of C β ∩ ( α + 1). Definition 2.14. [4] The function ρ : [ ω ] −→ ω is defined recur-sively as follows: for α < β , ρ ( α, β ) = max { otp( C β ∩ α ) , ρ ( α, min( C β \ α )) , ρ ( ξ, α ) : ξ ∈ C β ∩ [ λ ( α, β ) , α ) } .We define ρ ( α, α ) = 0 for all α ∈ ω . Lemma 2.15. [4] Assume ξ ∈ α and α is a limit point of C β . Then ρ ( ξ, α ) = ρ ( ξ, β ) . Lemma 2.16. [4] If α < β , α is a limit ordinal such that there is acofinal sequence of ξ ∈ α , with ρ ( ξ, β ) ≤ ν then ρ ( α, β ) ≤ ν . Lemma 2.17. [4] For all ν ∈ ω and α ∈ ω , the set { ξ ∈ α : ρ ( ξ, α ) ≤ ν } is countable. H. LAMEI RAMANDI, S. TODORCEVIC Lemma 2.18. [4] Assume α ≤ β ≤ γ . Then • ρ ( α, γ ) ≤ max { ρ ( α, β ) , ρ ( β, γ ) } , • ρ ( α, β ) ≤ max { ρ ( α, γ ) , ρ ( β, γ ) } . Lemma 2.19. [4] Assume α < β < γ . We have ρ ( α, γ ) = ρ ( α, β ) , if ρ ( β, γ ) < max { ρ ( α, β ) , ρ ( α, γ ) } . Lemma 2.20. [4] Assume β ∈ lim( ω ) , and γ > β . Then there is β ′ ∈ β such that for all α ∈ ( β ′ , β ) , ρ ( α, γ ) ≥ ρ ( α, β ) . When we work with sets of ordinals like x, y , we say x < y if everyelement of x is below every element of y . Lemma 2.21. Assume A is an uncountable family of finite subsets of ω and ν ∈ ω . Then there is an uncountable B ⊂ A such that B formsa ∆ -system with root r and for all a, b in B : • a \ r < b \ r implies that for all α ∈ a \ r and β ∈ b \ r , ρ ( α, β ) > ν , • r < a \ r < b \ r implies that for all α ∈ a \ r , β ∈ b \ r , and γ ∈ r , ρ ( α, β ) ≥ min { ρ ( γ, α ) , ρ ( γ, β ) } . The following forcing, which we will use with minor modifications invarious parts of this paper, is equivalent to the forcing which is usedin the proof of Theorem 7.5.9 in [4]. Definition 2.22. [4] Q is the poset consisting of all finite functions p such that the following holds.(1) dom( p ) ⊂ ω .(2) For all α ∈ dom( p ), p ( α ) ∈ [ ω ] <ω such that for all ν ∈ ω , p ( α ) ∩ [ ν, ν + ω ) has at most one element.(3) For all α, β in dom( p ), p ( α ) ∩ p ( β ) is an initial segment of both p ( α ) and p ( β ).(4) For all α < β in dom( p ), max( p ( α ) ∩ p ( β )) ≤ ρ ( α, β ).We let q ≤ p if dom( p ) ⊂ dom( q ) and ∀ α ∈ dom( p ), p ( α ) ⊂ q ( α ). Definition 2.23. Assume G is generic for Q . Then b ξ = S { p ( ξ ) : p ∈ G } .Recall that a poset P satisfies the Knaster condition if every un-countable subset A of P contains an uncountable subset B such thatthe elements of B are pairwise compatible. Note that Knaster condi-tion is stronger than ccc. Moreover, if P satisfies the Knaster conditionthen it does not add new cofinal branches to ω -trees and its iterationwith any ccc poset is ccc. Proposition 2.24. [4] The poset Q satisfies the Knaster condition. See the proof of Theorem 7.5.9. AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 9 Large Cardinal Strength In this section, using Theorem 2.5 which is due to Todorcevic, wewill show the following theorem. Theorem 3.1. If every Kurepa tree has an Aronszajn subtree then ω is inaccessible in L .Proof. Assume that ω , of V , is a successor cardinal in L . Then there is X ⊂ ω such that L [ X ] computes ω and ω correctly. Using Theorem2.5 there are two disjoint subsets S, S ′ of lim( ω ) in L [ X ] such that S ∪ S ′ = lim( ω ) and they are stationary in V .Let f be a function on lim( ω ) which is defined as follows. If α ∈ S then f ( α ) is the lease ξ > α such that L ξ [ X ] ≺ L ω [ X ]. If α ∈ S ′ then f ( α ) is the least ξ > α such that for some countable N ≺ L ω [ X ], N is isomorphic to L ξ [ X ]. Let T be an ω -tree in L [ X ] such that: • T = ( ω , < ) and 0 is the smallest element of T , • T α +1 is the first ω ordinals after T α , • every node t ∈ T α has two extensions in T α +1 , • if α ∈ lim( ω ) then a cofinal branch b ⊂ T <α has a top elementin T α if and only if b ∈ L f ( α ) [ X ]. Claim 3.2. The tree T is a Kurepa tree.Proof. It suffices to show that T is Kurepa in L [ X ]. Assume for acontradiction that h b ξ : ξ ∈ ω i is an enumeration of all branches of T . Let N ≺ L ω [ X ] be countable such that T, h b ξ : ξ ∈ ω i , X are in N , α = N ∩ ω ∈ S and ϕ : N −→ L δ [ X ] is the transitive collapseisomorphism. Note that δ < f ( α ). Then ϕ ( h b ξ : ξ ∈ ω i ) = h b ξ ↾ α : ξ ∈ α i ∈ L δ [ X ] . But L δ [ X ] ⊂ L f ( α ) [ X ]. Therefore, using the sequence h b ξ ↾ α : ξ ∈ α i one can find a cofinal branch b of T <α which is in L f ( α ) [ X ] and whichis different from all h b ξ ↾ α : ξ ∈ α i . This contradicts the fact that h b ξ : ξ ∈ ω i an enumeration of all branches of T . (cid:3) Now we will show that T has no Aronszajn subtree in V . Note thatthere are stationary many countable N ≺ H ω such that N ∩ L [ X ] ≺ L ω [ X ] and α = N ∩ ω ∈ S ′ . Assume for a contradiction that A isan Aronszajn subtree of T . Fix N as above such that A ∈ N . Wewill show that N captures all elements of T α . In order to see that thisleads to a contradiction, let t ∈ A be of height α . Since N capturesall elements of height α , let c be a chain of T which is in N and whichcontains all elements of T be below t . Because c ∈ N and the ordertype of c is at least α , c has to be uncountable. Hence, N is a model in which c ⊂ A and c is an uncountable chain. Then by elementarity, A contains the uncountable chain c , which contradicts the fact that itwas and Aronszajn subtree.In order to see that N captures all elements of T α , note that thetransitive collapse of N ∩ L [ X ] is equal to L δ [ X ] for some δ ≥ f ( α ).Let π be the transitive collapse map for N ∩ L [ X ]. Assume t ∈ T α .Then there is a branch b ⊂ T <α which is in L δ [ X ] such that t is thetop element of b . Since α is the first uncountable ordinal in L δ [ X ], ω L δ [ X ] = α . Since π is an isomorphism there is b ∈ N such that b is an uncountable branch of T and π ( b ) = b . But then b ∩ T <α = b .This means that N captures t via b , as desired. (cid:3) Analyzing Quotients of the Forcing Q When we analyze subtrees of the generic tree T , which is added by Q , it will be useful to know if there is a quotient forcing which addsthe tree T but does not add certain branches. In this section we willfind some subsets of Q which are complete suborders of it. Definition 4.1. Q c is the poset consisting of all conditions p in Q with the additional condition that for all α ∈ dom( p ), cf ( α ) ≤ ω . Ingeneral, if A ⊂ ω , Q A denotes the set of all conditions q in Q suchthat dom( q ) ⊂ A . Lemma 4.2. The poset Q c is a complete suborder of Q . Moreover, if X ⊂ ω is a set of ordinals of cofinality ω , then Q ω \ X is a completesuborder of Q .Proof. We only prove the first part of the lemma. The second part canbe verified by a similar argument. Assume q ∈ Q . We will show thatthere is q ′ ∈ Q c such that for all extensions p ≤ q ′ in Q c , the conditions p, q are compatible. Assume { β i : i ∈ n } is the increasing enumerationof all ordinals in dom( q ) which have cofinality ω . Also let C be theset of all ordinals in dom( q ) which have countable cofinality. Define β ′ i ,for each i ∈ n , to be the least ordinal ξ such that:(1) ξ is a limit point of C β i ,(2) ξ is strictly above all elements of dom( q ) ∩ β i ,(3) for all α ∈ dom( q ) \ β , ρ ( ξ, α ) = ρ ( β, α ),(4) otp( C ξ ) > max( q ( β i ))Let q ′ be the condition in Q c such that dom( q ′ ) = C ∪ { β ′ i : i ∈ n } , q ′ ( α ) = q ( α ) for all α ∈ C , and q ′ ( β ′ i ) = q ( β i ). It is easy to see that q ′ ∈ Q c . AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 11 Now let p < q ′ be in Q c . Let r be the condition in Q such thatdom( r ) = dom( p ) ∪ { β i : i ∈ n } , r ( α ) = p ( α ) for each α ∈ dom( p ), and r ( β i ) = p ( β ′ i ) ∩ (max( q ( β i )) + 1). It is easy to see that r is a commonextension of p, q , provided that it is in Q . We only verify a specificcase of Condition 4, of Definition 2.22. Assume α ∈ dom( p ) and β isone of the β i . If α < β ′ , ρ ( α, β ′ ) = ρ ( α, β ) and there is nothing toshow. If β ′ ≤ α < β , there is nothing to show because otp( C β ′ ) > max( q ( β i )). If β < α note that by lemma 2.19 either ρ ( β, α ) ≥ ρ ( β ′ , α )or ρ ( β ′ , β ) = ρ ( β ′ , α ). In the first case there is nothing to show, andfor the second case we have ρ ( β ′ , α ) = ρ ( β ′ , β ) ≥ max( q ( β ))), whichfinishes the proof. (cid:3) It is well known that if there is a ccc poset P which adds a branch b to an ω -tree U , then { u ∈ U : ∃ p ∈ P, p (cid:13) u ∈ ˙ b } is a Souslinsubtree of U . Here, Q is a ccc poset and Q c is a complete suborder of Q . Moreover, if G ⊂ Q is generic then G ∩ Q c knows the generic tree T . Since there is a ccc poset R such that Q is equivalent to Q c ∗ ˙ R , T has lots of Souslin subtrees in any extension by Q c . This leads to thefollowing corollary. In the next section we prove a stronger statementwhich we will use to prove a fact about ρ . For now, this corollary helpsus to have a better picture of the forcing Q . Corollary 4.3. The generic tree for Q c has Souslin subtrees. Lemma 4.4. Assume CH . Let h N ξ : ξ ∈ ω i be a continuous ∈ -chainof countable elementary submodels of H θ where θ is a regular largeenough cardinal, N ω = S ξ ∈ ω N ξ , and µ = sup( N ω ∩ ω ) . Then Q µ isa complete suborder of Q .Proof. We need to show that for all q ∈ Q there is p ∈ Q µ such thatif r ≤ p and r ∈ Q µ then r is compatible with q . Let R = S range( q ), L = dom( q ) ∩ µ , and H = dom( q ) \ µ = { β i : i ∈ k } such that β i isincreasing. Fix ¯ ν ∈ ω which is above all elements of R and all ρ ( α, β )where α, β are in dom( q ). Using Lemma 2.17, fix µ ∈ µ above max( L )such that for all β ∈ H and for all γ ∈ µ \ µ , ρ ( γ, β ) > ¯ ν . For each β ∈ H and ν ∈ ¯ ν let A ν,β = { α ∈ µ : ρ ( α, β ) = ν } . Again by Lemma 2.17, for all ν ∈ ¯ ν and β ∈ H , A ν,β is a count-able subset of µ . Since CH holds, we can fix N = N ξ such that µ , ¯ ν, L, R, h A ν,β : β ∈ H, ν ∈ ¯ ν i are in N . By elementarity, there is H ′ = { β ′ i : i ∈ k } which is in N and(1) β ′ i is increasing,(2) min( H ′ ) > µ (3) for all i ∈ k and for all ν ∈ ¯ ν , A ν,β i = { α ∈ µ : ρ ( α, β ′ i ) = ν } ,and (4) for all i < j in k , ρ ( β i , β j ) = ρ ( β ′ i , β ′ j ).Let p be the condition that dom( p ) = L ∪ H ′ , for all ξ ∈ L , p ( ξ ) = q ( ξ ) and for all i ∈ k , p ( β ′ i ) = q ( β i ). Suppose r ≤ p is in Q µ . Wewill find s ∈ Q which is a common extension of r, q . Pick s such thatdom( s ) = dom( r ) ∪ H , s ↾ dom( r ) = r , and for all i ∈ k s ( β i ) = r ( β ′ i ) ∩ (max( q ( β i )) + 1).We need to show that s is a condition in Q . All of the conditions inDefinition 2.22 obviously hold, except for condition 4. If α < β are in H , by the last requirement for H ′ and the fact that r is a condition,max( s ( α ) ∩ s ( β )) ≤ ρ ( α, β ).Now assume that α ∈ dom( r ) and β = β i ∈ H . If ρ ( α, β ) ≥ ¯ ν ,everything is obvious because max( s ( β )) < ¯ ν . Assume ρ ( α, β ) = ν < ¯ ν . So α ∈ A ν,β . Since r ∈ Q µ , we have max( s ( α ) ∩ s ( β i )) < max( r ( α ) , r ( β ′ i )) ≤ ν . (cid:3) Lemma 4.4 shows under CH that for many ordinals µ with cofinality ω , Q µ is a complete suborder of Q . It is natural to ask the samequestion for ordinals of countable cofinality. The following fact showsthat quit often Q µ is not a complete suborder of Q , when µ varies overordinals of countable cofinality. Fact 4.5. Assume cf ( µ ) = ω , µ ∈ ω , for some β > µ , µ is a limitpoint of C β and the set of all limit points of C µ is cofinal in µ . Then Q µ is not a complete suborder of Q .Proof. Assume β > µ such that µ is a limit point of C β and cf( β ) = ω .Let ν = otp( C β ) and q = { ( β, { ν } ) } . We claim that for all p ∈ Q µ there is an extension ¯ p ≤ p in Q µ such that ¯ p is incompatible with q . Fix p ∈ Q µ . Without loss of generality ν ∈ S range( p ) and p is compatible with q . Let ξ ∈ dom( p ) such that ν ∈ p ( ξ ). Then p ∪ { ( β, p ( ξ ) ∩ ( ν + 1)) } ∈ Q . Let α be a limit point of C µ which isabove all elements of dom( p ). Then ¯ p = p ∪ { ( α, p ( ξ ) ∩ ( ν + 1)) } is acondition in Q µ . But ρ ( α, β ) = otp( C α ) < ν and ν ∈ ¯ p ( α ). Hence ¯ p, q are incompatible. (cid:3) Climbing Souslin Trees to See ρ In this section we analyzing the external elements of the genericKurepa tree that is added by the poset Q c . The aim is to prove Lemma5.3, which is a general fact about the function ρ . We use Lemma 5.3to find more weakly external elements in the tree which is generic for Q . Proposition 5.1. Fix κ a regular cardinal greater than ω + . Assume S is the set of all X ∈ [ ω ] ω such that C α X ⊂ X and lim( C α X ) is cofinal AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 13 in X . Define Σ = { M ≺ H κ : M ∩ ω ∈ S ∧ | L α M | = ℵ } . Then Σ isstationary in [ H κ ] ω .Proof. Let E ⊂ [ H κ ] ω be a club, θ be a regular cardinal above 2 κ + ,and h M ξ : ξ ∈ ω i be a continuous ∈ -chain of countable elementarysubmodels of H θ such that for all ξ ∈ ω , M ξ ∩ ω ∈ S . Let α ξ =sup( M ξ ∩ ω ) and α = sup { α ξ : ξ ∈ ω } . By thinning out if necessary,without loss of generality we can assume that for all ξ ∈ ω , α ξ is alimit point of C α .Let f : { η ∈ ω : | L η | ≤ ℵ } −→ ω by f ( η ) = sup( L η ), and C f bethe set of all ordinals that are f -closed. Obviously f ∈ M and for all ξ , α ξ ∈ C f . But for any ξ ∈ ω , sup L α ξ / ∈ M ξ +1 . So, for all ξ ∈ ω , M ξ ∩ H κ ∈ E ∩ Σ. (cid:3) Lemma 5.2. Assume G ⊂ Q c is generic and T is the Kurepa treethat is added by G . Assume Q/G is the quotient poset such that Q isequivalent to Q c ∗ ( Q/G ) . For each α of cofinality ω , let A α = { x ∈ T : ∃ q ∈ Q/G q (cid:13) “ x ∈ ˙ b α ” } . Then each A α is a Souslin subtree of T .Moreover, there is α ∈ ω of cofinality ω such that for all x ∈ A α , T x contains ℵ many b ξ with cf ( ξ ) = ω .Proof. It is trivial that A α is a Souslin subtree of T . For the rest of thelemma, let θ > ω + be a regular cardinal, and Assume S is the set of all X ∈ [ ω ] ω such that C α X ⊂ X and lim( C α X ) is cofinal in X . Let h M ξ : ξ ∈ ω i be a continuous ∈ -chain of countable elementary submodels of H θ such that for all ξ ∈ ω , M ξ ∩ ω ∈ S . Let α ξ = sup( M ξ ∩ ω ) and α = sup { α ξ : ξ ∈ ω } . Also fix q ∈ Q with α ∈ dom( q ), t ∈ q ( α ) , and ξ ∈ ω . We find β > ξ and p ≤ q such that cf( β ) = ω , β ∈ dom( p ) , and t ∈ p ( β ). Find α ′ ∈ lim( C α ) such that(1) dom( q ) ∩ [ α ′ , α ) = ∅ ,(2) tp( C α ′ ) is above all elements of S range( q ) and all ρ ( { α, β } ) for β ∈ dom( q ),(3) for all β ∈ dom( q ) \ α, ρ ( α, β ) = ρ ( α ′ , β ),(4) | L α ′ | = ℵ Now pick β ∈ L α ′ which is above ξ and all elements of dom( q ). Define p by dom( p ) = dom( q ) ∪ { β } , q ( ζ ) = p ( ζ ), for all ζ ∈ dom( q ) and p ( β ) = q ( α ) ∩ ( t + 1). (cid:3) Now we are ready to prove the main lemma of this section. Lemma 5.3. Let ω + < κ < κ < θ be regular cardinals such that κ + < κ , and κ + < θ . Let S be the set of all X ∈ [ ω ] ω such that C α X ⊂ X and lim( C α X ) is cofinal in X . Assume A is the set of allcountable N ≺ H θ with the property that if N ∩ ω ∈ S then there is a club of countable elementary submodels E ⊂ [ H κ ] ω in N such that forall M ∈ E ∩ N , ρ ( α M , α N ) ≤ M ∩ ω . Then A contains a club.Proof. Assume G is the V -generic filter over Q c and T be the tree thatis introduces by G . Assume ˙ A is a Q c -name for an Aronszajn subtreeof T with the property that for all t ∈ ˙ A , |B ( T t ) | = ω . Fix N ≺ H θ ,in V , with ˙ A ∈ N and N ∩ ω ∈ S . Suppose for a contradiction that( ∗ ):for all clubs E ⊂ [ H κ ] ω in N there is M ∈ E ∩ N such that ρ ( α M , α N ) > M ∩ ω .Let δ M , δ N be M ∩ ω and N ∩ ω respectively. Fix t ∈ [ δ N , δ N + ω ), q ∈ Q c such that q forces that t ∈ ˙ A . Obviously, q forces that t is externalto N [ ˙ T ]. In other words, q forces that there is a club E ⊂ [ H κ [ ˙ G ]] ω in N [ ˙ T ] such that for all Z ∈ E ∩ N [ ˙ T ], Z does not capture t . Let ˙ E bea name for the witness E above. So q forces that for all Z ∈ ˙ E ∩ N [ ˙ T ], Z does not capture t . In order to reach a contradiction, it suffices toshow ( ∗ ) implies that there are M ≺ H κ in N and p ≤ q in Q c suchthat:(1) ˙ E ∈ M and(2) p forces that M [ ˙ T ] captures t.We consider three cases. First, consider the case where t / ∈ S range( q ).Let γ ∈ ( M ∩ ω ) \ dom( q ), with cf ( γ ) = ω . Let M ≺ H κ be in N such that γ, ˙ E are in M . Let p be the condition such that dom( p ) =dom( q ) ∪ { γ } , ∀ ξ ∈ dom( q ) p ( ξ ) = q ( ξ ), and p ( γ ) = { t } . It is obviousthat p is an extension of q and it forces that M [ ˙ T ] captures t via ˙ b γ .Now suppose for some ξ ∈ dom( q ) ∩ N , t ∈ q ( ξ ). In this case assume M ≺ H κ is in N such that ˙ E, ξ are in M . Then q forces that M [ ˙ T ]captures t via ˙ b ξ .For the last case, suppose t ∈ S range( q ) but ∀ ξ ∈ dom( q ) ∩ Nt / ∈ q ( ξ ). Since any element of ˙ A is an element of ℵ many branches of T ,by extending q if necessary, we can assume that there is τ ∈ dom( q ) \ α N such that t ∈ q ( τ ). We consider the partition dom( q ) = H ∪ L ∪ R where R = dom( q ) ∩ N (rudimentary ordinals w.r.t. N ), L = (dom( q ) ∩ α N ) \ R (low ordinals), and H = dom( q ) \ α N (high ordinals). Let B t be theset of all ξ ∈ dom( q ) such that t ∈ q ( ξ ). So B t ∩ R = ∅ and τ ∈ B t .By Lemma 2.20 we have the following about the ordinals in H : ∃ γ ∈ N ∩ ω ∀ γ ∈ N \ γ ∀ ξ ∈ H ρ ( γ, ξ ) ≥ ρ ( γ, α N ).The following holds for ordinals in L : AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 15 ∃ γ ∈ ( N ∩ ω ) \ max( L ∪ R ) ∀ γ > γ ∀ ξ ∈ L ρ ( ξ, γ ) ≥ N ∩ ω .In order to see this, fix ξ ∈ L . Let α = min[( N ∩ ω ) \ ξ ]. Since ξ < α N is outside of N , cf( α ) = ω . Let γ ∈ N be above α . We show that ρ ( ξ, γ ) ≥ N ∩ ω . Note that there is α ′ ∈ α such that for all η ∈ ( α ′ , α )otp( C α ∩ η ) appears in the definition of ρ ( η, γ ). Since γ, α are in N , byelementarity, the ordinal α ′ exists in N . Since ξ ∈ ( α ′ , α ), otp( C α ∩ ξ )appears in the definition of ρ ( ξ, γ ). But otp( C α ∩ ξ ) ≥ N ∩ ω . Thisargument shows that γ = max { min(( N ∩ ω ) \ ξ ) : ξ ∈ L } works.Now using ( ∗ ) choose M ≺ H κ in N such that ρ ( α M , α N ) > δ M andsuch that M has γ , γ , dom( q ) ∩ N, S range( q ) ∩ N, ˙ E as elements. Let γ > max { γ , γ } be in M such that for all γ ∈ M that are above γ , ρ ( γ, α N ) > δ M .Note that max( q ( ξ ) ∩ q ( η )) = max( q ( τ ) ∩ q ( ξ )) for each ξ ∈ R and η ∈ B t . If max( q ( ξ ) ∩ q ( τ )) / ∈ M for some ξ ∈ R , we are done. Soassume max( q ( ξ ) ∩ q ( τ )) ∈ M . By elementarity, fix γ > γ in M suchthat cf ( γ ) = ω and for all ξ ∈ R, ρ ( ξ, γ ) ≥ max( q ( τ ) ∩ q ( ξ )). Nowdefine p ≤ q as follows: • dom( p ) = dom( q ) ∪ { γ } , • ∀ ξ ∈ dom( q ) \ B t p ( ξ ) = q ( ξ ), • ∀ ξ ∈ B t p ( ξ ) = q ( ξ ) ∪ { δ M } , • p ( γ ) = p ( τ ) ∩ ( δ M + 1).Obviously, p forces that M [ ˙ T ] captures t via ˙ b γ , provided that p ∈ Q c .We only check condition 4, in order to show p ∈ Q c . This conditiontrivially holds for pairs of ordinals in dom( q ). If β ∈ H , then ρ ( γ, β ) ≥ ρ ( γ, α N ) > δ M = max( p ( γ )). So condition 4 holds for γ and any β ∈ H . If α ∈ L , then ρ ( α, γ ) ≥ δ N > max( p ( γ )). If α ∈ R then ρ ( α, γ ) ≥ max( q ( τ ) ∩ q ( α )) = max( p ( γ ) ∩ p ( α )). Therefore, p ∈ Q c . (cid:3) ρ Introduces Aronszajn Subtrees Everywhere in T In this section we will use Lemma 5.3 to show that every Kurepasubset of the generic Kurepa tree has an Aronszajn subtree. Recallthat a subset Y of T is said to be a Kurepa subset if it is a Kurepa treewhen it is considered with the order inherited from T . Note that Y isnot necessarily downward closed. The theorems in this section are notusing any large cardinal assumption. Definition 6.1. P is the poset as in Definition 2.22, but instead ofcondition 4, the elements p ∈ P have the property that for all α < β indom( p ), max( p ( α ) ∩ p ( β )) < ρ ( α, β ) . Moreover, b ξ = S { p ( ξ ) : p ∈ G } ,whenever G is a generic filter for P . Remark 6.2. Note that P satisfies the Knaster condition. Althoughwe are not using this, it is worth mentioning that all the statementsregarding complete suborders of Q hold for P as well.The proof of the following Lemma uses ideas from the proof of The-orem 7.5.9 in [4]. This lemma determines the set of all branches of thegeneric tree created by T . We will use it in order to determine whichmodels capture certain branches or elements of T . Lemma 6.3. Assume T is the generic tree for P . Then { b ξ : ξ ∈ ω } is the set of all branches of T .Proof. Assume π is a P -name for a branch that is different from all b ξ , ξ ∈ ω . Inductively construct a sequence h p η : η ∈ ω i as follows. p ∈ P is arbitrary. If h p η : η < α i is given, find P α ∈ P such that: • p α decides min( π \ S { b ξ : ξ ∈ S { dom( p η ) : η ∈ α }} ) to be t p α , • t p α ∈ S range( p α ), • for every β ∈ dom( p α ), ht(max( p α ( β ))) > ht( t p α ) . Let A = { p α : α ∈ ω } . By going to a subset of A if necessary, wemay assume that { dom( p ) : p ∈ A } forms a ∆-system with root d .Also { S range( p ) : p ∈ A } forms a ∆-system with root c . Moreover,we may assume that elements of A are pairwise isomorphic structuresand the isomorphism between them fixes the root. By Lemma 2.21there is an uncountable set B ⊂ A such that for every p, q in B if α ∈ dom( p ) \ dom( q ), β ∈ dom( q ) \ dom( p ), γ ∈ d , and all elements of a = S range( p ) \ c are below all elements of b = S range( q ) \ c , then ρ ( α, β ) > max( c ) and ρ ( α, β ) ≥ min { ρ ( γ, α ) , ρ ( γ, β ) } .We claim that elements of B are pairwise compatible. In order tosee this, fix p, q in B as above and assume c < a = S range( p ) \ c < b = S range( q ) \ c. We define their common extension r on dom( p ) ∪ dom( q ) as follows.For γ ∈ d let r ( γ ) = p ( γ ) ∪ q ( γ ), and for α ∈ a let r ( α ) = p ( α ). For β ∈ b we have two cases. Either for all γ ∈ d , max( q ( γ ) ∩ q ( β )) ∈ c or there is a unique γ ∈ d such that max( q ( γ ) ∩ q ( β )) ∈ b . In the firstcase let r ( β ) = q ( β ) and in the second case let r ( β ) = p ( γ ) ∪ q ( β ).First we will show that r satisfies Condition 3. Note that if γ , γ areboth in d then p ( γ ) ∩ p ( γ ) ⊂ c and q ( γ ) ∩ q ( γ ) ⊂ c . In order to seethis, assume this is not the case. Then by the fact that the conditionsin B are pairwise isomorphic, sup { max( s ( γ ) ∩ s ( γ )) : s ∈ B } = ω which implies that ρ ( γ , γ ) ≥ ω . But this is absurd. Now assume i ∈ ( p ( γ ) ∪ q ( γ )) ∩ ( p ( γ ) ∪ q ( γ )), j < i and j ∈ ( p ( γ ) ∪ q ( γ )). We Note that the levels of the generic tree are in the ground model. AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 17 will show that j ∈ p ( γ ) ∪ q ( γ ). Note that j ∈ c . Then j ∈ p ( γ ) ∩ c = q ( γ ) ∩ c . Since p, q both satisfy Condition 3 and i ∈ p ( γ ) ∪ q ( γ ), wehave j ∈ p ( γ ) ∪ q ( γ ). If α ∈ dom( p ) \ dom( q ) and γ ∈ d note that r ( α ) ∩ r ( γ ) = p ( α ) ∩ ( p ( γ ) ∪ q ( γ )) = p ( α ) ∩ p ( γ ) . But p ( α ) ∩ p ( γ ) is an initial segment of both p ( α ) and r ( γ ) because a < b . If β ∈ dom( q ) \ dom( p ) and for all γ ∈ d , max( q ( γ ) ∩ q ( β )) ∈ c the argument is the same. So assume that for a unique γ β , max( q ( β ) ∩ q ( γ β )) ∈ b . Then it is easy to see that r ( β ) ∩ r ( γ β ) = p ( γ β ) ∪ ( q ( β ) ∩ q ( γ β )) is an initial segment of both r ( β ) , r ( γ β ). If β ∈ dom( q ) \ dom( p )and γ = γ β , in order to see r ( β ) ∩ r ( γ ) is an initial segment of both r ( β ) , r ( γ ), note that r ( β ) ∩ r ( γ ) = ( p ( γ β ) ∪ q ( β )) ∩ ( p ( γ ) ∪ q ( γ )) ⊂ c. Then r ( β ) ∩ r ( γ ) = p ( γ β ) ∩ p ( γ ) which makes Condition 3 trivial. Therest of the cases use the same type of argument.For Condition 4, we only prove the case α ∈ dom( p ) \ dom( q ) and β ∈ dom( q ) \ dom( p ). The rest of the cases use similar arguments. Iffor all γ ∈ d , max( q ( γ ) ∩ q ( β )) ∈ c , there is nothing to prove. Assumefor some unique γ ∈ d , max( q ( γ ) ∩ q ( β )) ∈ b . If ρ ( α, β ) ≥ ρ ( γ, α ),everything is trivial. Assume ρ ( α, β ) ≥ ρ ( γ, β ). Since q ( γ ) ∩ q ( β ) ∩ b = ∅ , we have max( p ( α )) < ρ ( α, β ). But max( p ( α )) ≥ max( r ( α ) ∩ r ( β )).So r satisfies Condition 4.Now we have two cases: either there is an uncountable C ⊂ B such that for all p ∈ C , there is γ ∈ d with t p ∈ p ( γ ), or there areonly countable many p ∈ B such that for some γ ∈ d , t p ∈ p ( γ ) . Ifsuch an uncountable C exists, let s ∈ P such that s forces that thegeneric filter intersects C on an uncountable set. Then for some γ ∈ d , s (cid:13) | π ∩ b γ | = ℵ . But this contradicts the fact that π was a name fora branch that is different from all b ξ ’s.Now assume that there are only countable many p ∈ B such thatfor some γ ∈ d , t p ∈ p ( γ ) . Let p, q be in B both of which are outsideof this countable set. Let t p ∈ p ( α ) , α ∈ dom( p ) \ dom( q ) and t q ∈ q ( β ) , β ∈ dom( q ) \ dom( p ). Also let a, b, c, d be as above. Note that wecan choose p, q such that t p , t q are not in c . Moreover, p forces that t p is not in the branches that are indexed by ordinals in d . We will finda common extension of p, q which forces that t p is not below t q . Thiscontradicts the fact that π is a name for a branch.First consider the case max( q ( β ) ∩ q ( γ )) ∈ c . Let r be the commonextension of p, q described as above. Recall that r ( β ) = q ( β ) in thiscase. Let ξ ∈ ( t p , t p + ω ) \ ( a ∪ b ). Note that ξ > max( c ). Let X = { η ∈ dom( r ) : max( r ( β ) ∩ r ( η )) > ξ } . Obviously, X ∩ dom( p ) = ∅ and β ∈ X . Extend r to r ′ such that dom( r ′ ) = dom( r ), r and r ′ agreeon any element of their domain which is not in X , and r ′ ( η ) = r ( η ) ∪{ ξ } for all η ∈ X . Checking r ′ is a condition is routine. The condition r ′ forces that in the generic tree ht( ξ ) = ht( tp ) and they are distinct.Therefore, it forces that ξ < t q and that t p is not below t q .Now assume for some γ ∈ d , max( q ( β ) ∩ q ( γ )) ∈ b . Again assumethat r is the common extension described above. So r ( β ) = p ( γ ) ∪ q ( β ),and r forces that max( p ( γ )) is bellow t q in the generic tree. Recall thatht(max( p ( γ ))) > ht( t p ) and p forces that t p is not in the branchesindexed by the ordinals in the root d . Hence p forces that t p is notbelow max( p ( γ )). Since r ≤ p , it forces that t p is not bellow t q in thegeneric tree. (cid:3) Now we are ready to prove the main theorem of this section. Theorem 6.4. It is consistent that there is a Kurepa tree T such thatevery Kurepa subset of T has an Aronszajn subtree.Proof. Assume G is a generic filter for the forcing P , and T is the treeintroduced by G . Since P is ccc, it preserves all cardinals and T is aKurepa tree.Assume U is a Kurepa subset of T , and X is the set of all ξ ∈ ω suchthat b ξ ∩ U is uncountable. Let h N ξ : ξ ∈ ω i be a continuous ∈ -chainof countable elementary submodels of H θ such that U ∈ N and let N ω = S ξ ∈ ω N ξ , µ = N ω ∩ ω . Fix η ∈ X be above µ . By Proposition2.8, it suffices to show that for some ξ ∈ ω , the first element of b η ∩ U whose height is more than N ξ ∩ ω is weakly external to N ξ witnessedby some stationary set Σ.Without loss of generality we can assume that for all ξ ∈ ω : • N ξ ∩ ω is a cofinal subset of C α Nξ , • ξ ∈ ω , α N ξ = sup( N ξ ∩ ω ) is a limit point of C µ , and • ξ ∈ ω , N ξ ∩ ω ⊃ C α Nξ .In order to see this, let f from ω to µ be the function such that foreach ξ ∈ ω , f ( ξ ) is the least ζ ∈ ω with N ζ ⊃ C µ ∩ α N ξ . Now observethat if ξ is f -closed then it satisfies all the three properties above.Let ξ ∈ ω be such that the lower trace of the walk from η to µ is in N ξ . Then note that ρ ( µ, η ) ∈ N ξ and ρ ( α N ξ , η ) = ρ ( µ, η ). UseLemma 5.3 to find E ∈ N ξ which is a club of countable elementarysubmodels of H ω such that for all M ∈ E ∩ N ξ , ρ ( µ, η ) ∈ M and ρ ( α M , α N ξ ) ≤ M ∩ ω . Now let Σ be the set of all M ∈ E such that M ∩ ω is a cofinal subset of C α M . Obviously Σ is stationary and in N ξ . Let M ∈ Σ ∩ N ξ . We want to show that M does not capture b η , AN YOU TAKE KOMJATH’S INACCESSIBLE AWAY? 19 as a branch of T . Equivalently, for all b ∈ M which is a cofinal branchof T , ∆( b, b η ) ∈ M . By the lemma above, it suffices to show that forall γ ∈ M , ρ ( γ, η ) ≤ M ∩ ω . Recall that: ρ ( γ, η ) ≤ max { ρ ( γ, α M ) , ρ ( α M , α N ξ ) , ρ ( α N ξ , η ) } ≤ M ∩ ω . Fix β which is a limit point of C α M and which is above γ . Since β ∈ M and ρ ( γ, β ) = ρ ( γ, α M ), we have that ρ ( γ, α M ) ∈ M . Since M ∈ E , ρ ( α M , α N ξ ) ≤ M ∩ ω . Recall that ρ ( µ, η ) ∈ N ξ . Since ρ ( α N ξ , η ) = ρ ( µ, η ) ∈ M , we have ρ ( γ, η ) ≤ M ∩ ω and for all γ ∈ M ∩ ω ∆( b γ , b η ) < M ∩ ω .Now assume M ∈ Σ ∩ N ξ , t is the first element of b η whose heightis more than N ξ ∩ ω . It suffices to show that M does not capture t as an element in U . Assume b ⊂ U is a cofinal branch of U whichis in M and b contains { s ∈ U ∩ M : s < t } . Since t / ∈ M, the set { s ∈ U ∩ M : s < t } has order type M ∩ ω . Let b γ be the downwardclosure of b in T . Then obviously γ ∈ M . But then the order type of b γ ∩ b η is at least M ∩ ω , which is a contradiction. (cid:3) We finish this section by a corollary which relates the theorem aboveto Martin’s Axiom. Corollary 6.5. Assume MA ω holds and ω is not a Mahlo cardinalin L . Then there is a Kurepa tree with the property that every Kurepasubset has an Aronszajn subtree. Taking Komjath’s Inaccessible Away In this section we will show that if there is an inaccessible cardinalin L then there is a model of ZFC in which every Kurepa tree has aSouslin subtree. The strategy can be described as follows. Assume λ ∈ L is the first inaccessible cardinal. First we collapse every κ ∈ λ to ω using the standard Levy collapse. Then we force with Q = Q λ . Weuse quotients of Q to analyze how often Souslin subtrees are producedand how often they are killed, and conclude that every Kurepa treemust have Souslin subtrees which survive.Let’s fix some notation. λ is the first inaccessible cardinal in L . A µ = coll( ω , < µ ) for every uncountable cardinal µ ∈ λ , F is L -generic for A = A λ and F µ = F ∩ A µ . V µ = L [ F µ ], and V = L [ F ]. Let G ⊂ Q be V -generic and for each µ ∈ λ, let G µ = G ∩ Q µ . Obviously,the following claim finishes the proof of Theorem 1.2. Claim 7.1. Every Kurepa tree in V [ G ] , has a Souslin subtree.Proof. Assume K ∈ V [ G ] is an ω -tree with no Souslin subtrees. Wewill show that K is not Kurepa. Let H be the set of all α ∈ λ such that Q α is a complete suborder of Q . Obviously, H ∈ V and it is stationaryin V [ G ] by Lemma 4.4.For each α ∈ H with K ∈ V [ G α ], there is β ∈ H such that whenever U ⊂ K is a downward closed subtree which is Souslin in V [ G α ], then U is not Soulsin in V [ G β ]. In order to see this, fix α ∈ H with K ∈ V [ G α ].Using Fact 2.10, if U ⊂ K is a downward closed subtree which is Souslinin V [ G α ] then U ∈ V α [ G α ]. So there are only ℵ many such U ’s in V [ G ]. Since H is cofinal in λ , there is β ∈ P such that if U ⊂ K isa downward closed subtree which is Souslin in V [ G α ] then U is notSouslin in V [ G β ].Now define f : H −→ H in V [ G ], by letting f ( α ) be the smallest β ∈ H such that every Souslin U ⊂ K which is in V [ G α ] is killed in V [ G β ]. Let µ ∈ H be an f -closed ordinal. Also note that Q = Q µ ∗ R µ for some Q µ -name for a ccc poset R µ . Since K has no Souslin subtreein V [ G µ ], R µ can not add branches to K . Moreover, there is no cccposet of size ℵ in V which creates a Kurepa tree. Therefore, K has atmost ℵ many branches. (cid:3) Acknowledgement. The research on the second author is partiallysupported by grants from NSERC(455916) and CNRS(UMR7586). References [1] T. Ishiu and J. T. Moore. Minimality of non σ -scattered orders. Fund. Math. ,205(1):29–44, 2009.[2] R. B. Jensen and K. Schlechta. Results on the generic Kurepa hypothesis. Arch.Math. Logic , 30(1):13–27, 1990.[3] P. Komj´ath. Morasses and the L´evy-collapse. J. Symbolic Logic , 52(1):111–115,1987.[4] S. Todorcevic. Walks on ordinals and their characteristics , volume 263 of Progress in Mathematics . Birkh¨auser, 2007. Department of Mathematics, University of Toronto, Toronto, Canada Email address : [email protected] Department of Mathematics, University of Toronto, Toronto, Canada Email address : [email protected] Institut de Math´ematiques de Jussieu, Paris, France Email address ::