Renling Jin & Saharon Shelah- Can a Small Forcing Create Kurepa Trees

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Can a Small Forcing Create Kurepa Trees 1

Renling Jin & Saharon Shelah 2

AbstractIn the paper we probe the possibilities of creating a Kurepa tree in a genericextension of a model of CH plus no Kurepa trees by an ω 1-preserving forcingnotion of size at most ω 1. In the rst section we show that in the Levy modelobtained by collapsing all cardinals between ω 1 and a strongly inaccessiblecardinal by forcing with a countable support Levy collapsing order many ω 1-preserving forcing notions of size at most ω 1 including all ω -proper forcingnotions and some proper but not ω -proper forcing notions of size at mostω 1 do not create Kurepa trees. In the second section we construct a modelof CH plus no Kurepa trees, in which there is an ω -distributive Aronszajntree such that forcing with that Aronszajn tree does create a Kurepa treein the generic extension. At the end of the paper we ask three questions.

0. Introduction

By a model we mean a model of ZFC . By a forcing notion we mean a separativepartially ordered set P with a largest element 1 P used for a corresponding forcingextension. Given a model V of CH , one can create a generic Kurepa tree by forcingwith an ω1-closed, ω2-c.c. forcing notion no matter whether or not V contains Kurepatrees [Je1]. One can also create a generic Kurepa tree by forcing with a c.c.c. forcing

notion provided V satises2

ω1 in addition [V]. Both forcing notions mentioned herehave size at least ω2. The size being at least ω2 seems necessary for guaranteeing thegeneric trees have at least ω2 branches. On the other hand, a Kurepa tree has a baseset of size ω1, so it seems possible to create a Kurepa tree by a forcing notion of size

ω1. In this paper we discuss the following question: Given a model of CH plus noKurepa tree, whether can we nd an ω1-preserving forcing notion of size ω1 suchthat the forcing creates Kurepa trees?

This question is partially motivated by a parallel result about Souslin tree. Givena ground model V . A Souslin tree could be created by a c.c.c. forcing notion of sizeω1 [ST]. There is also an ω1-closed forcing notion of size ω1 which creates Souslin treeprovided V satises CH [Je1]. The question whether a Souslin trees could be created

1 Mathematics Subject Classication Primary 03E35.2 The research of the second author was supported by Basic Research Foundation of The Israel

Academy of Sciences and Humanities. Publication No 563.1



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by a countable forcing notion (equivalent to adding a Cohen real) turns out to bemuch harder. It was answered positively by the second author [S1] ten years ago.

We call a forcing notion ω1-preserving if ω1 in the ground model is still a cardinal inthe generic extension. In this paper we consider only ω1-preserving forcing notion by

the following reason. Let V be the Levy model. In V there are no Kurepa trees andCH holds. Notice also that there is an ω2-Kurepa tree in V . If we simply collapse ω1

by forcing with the collapsing order Coll (ω, ω1), the set of all nite partial functionfrom ω to ω1 ordered by reverse inclusion, in V , then the ω2-Kurepa tree becomes aKurepa tree in V Coll (ω,ω 1 ) . Notice also that Coll (ω, ω1) has size ω1 in V . So we requirethe forcing notions under consideration be ω1-preserving to avoid the triviality.

In the rst section we show some evidence that in the Levy model it is extremelyhard to nd a forcing notion, if it ever exists, of size ω1 which could create a Kurepa

tree in the generic extension. Assume our ground model V is the Levy model. Weshow rst an easy result that any forcing notion of size ω1 which adds no reals couldnot create Kurepa trees. Then we prove two main results: (1) For any stationary setS ⊆ ω1, if P is an (S, ω )-proper forcing notion of size ω1, then there are no Kurepatrees in the generic extension V P . Note that all axiom A forcing notions are ( S, ω )-proper. (2) Some proper forcing notions including the forcing notion for adding a clubsubset of ω1 by nite conditions do not create Kurepa trees in the generic extension.

In the second section we show that there is a model of CH plus no Kurepa trees,in which there is an ω-distributive Aronszajn tree T such that forcing with T doescreate a Kurepa tree in the generic extension. We start with a model V containinga strongly inaccessible cardinal κ . In V we dene an ω1-strategically closed, κ-c.c.forcing notion P such that forcing with P creates an ω-distributive Aronszajn tree T

and a T -name K for a Kurepa tree K . Forcing with P collapses also all cardinalsbetween ω1 and κ so that κ is ω2 in V P . Take V = V P as our ground model. Forcingwith T in V creates a Kurepa tree in the generic extension of V . So the model V iswhat we are looking for except that we have to prove that there are no Kurepa treesin V , which is the hardest part of the second section.

We shall write V , V , etc. for (countable) transitive models of ZFC . For a forcingnotion P in V we shall write V P for the generic extension of V by forcing with P .Sometimes, we write also V [G] instead of V P for a generic extension when a particulargeneric lter G is involved. We shall x a large enough regular cardinal λ throughoutthis paper and write H (λ ) for the collection of sets hereditarily of power less than λ



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equipped with the membership relation. In a forcing argument with a forcing notionP we shall write a for a P -name of a and a for a P -name of a which is again a Q -nameof a for some forcing notion Q . If a is already in the ground model we shall writesimply a for a canonical name of a . Let P be a forcing notion and p ∈P . We shall

write q p to mean q ∈ P and q is a condition stronger than p. We shall oftenwrite p “. . . ” for some p ∈P instead of p V

P “. . . ” when the ground model V andthe forcing notion P in the argument is clear. We shall also write “. . . ” instead of 1P “. . . ”. In this paper all of our trees are subtrees of the tree 2<ω 1 , ⊆ . So if C isa linearly ordered subset of a tree T , then C is the only possible candidate of theleast upper bound of C in T . In this paper all trees are growing upward. If a treeis used as a forcing notion we shall put the tree upside down. Let T be a tree andx ∈T . We write ht (x) = α if x ∈T ∩ 2α . We write T α or (T )α , the α -th level of T ,

for the set T ∩ 2α

and write T α or (T ) α for the set β<α T β . We write ht (T ) forthe height of T , which is the smallest ordinal α such that T α is empty. By a normaltree we mean a tree T such that (1) for any α < β < ht (T ), for any x ∈T α there isan y ∈T β such that x < y ; (2) for any α such that α + 1 < ht (T ) and for any x ∈T αthere is β < ht (T ) and there are distinct y1, y2 ∈T β such that x < y 1 and x < y 2.Given two trees T and T . We write T end T for T being an end-extension of T ,i.e. T ht (T ) = T . By a branch of a tree T we mean a totally ordered set of T

which intersects every non-empty level of T . By an ω1-tree we mean a tree of heightω

1with each of its levels at most countable. A Kurepa tree is an ω

1-tree with more

than ω1 branches. To see [J], [K] and [S2] for more information on forcing, iteratedforcing, proper forcing, etc. and to see [T] for more information on trees.

Acknowledgements The rst part of this paper is originated in 1993, when therst author was a Morrey assistant professor in University of California-Berkeley.He thanks H. Woodin for some inspiring discussion. The rst author thanks alsothe Department of Mathematics, Rutgers University for offering free housing duringhis one week visit there at October, 1994, when the second part of the paper was

developed.

1. Creating Kurepa Trees By a Small Forcing Is Hard

First, we would like to state a theorem in [S2, 2.11] without proof as a lemma whichwill be used in this section.



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Lemma 1. In a model V let P be a forcing notion and let N be a countable elementary submodel of H (λ ). Suppose G ⊆ P is a V -generic lter. Then

N [G] = {aG : a is a P -name and a ∈N }

is a countable elementary submodel of (H (λ )) V [G ].

We choose the Levy model V = V Lv (κ,ω 1 ) as our ground model throughout thissection, where κ is a strongly inaccessible cardinal in V and Lv (κ, ω 1), the Levycollapsing order, is the set

{ p⊆ (κ × ω1) × κ : p is a countable function and(∀(α, β ) ∈dom ( p))( p(α, β ) ∈α )}

ordered by reverse inclusion. For any A ⊆ κ we write Lv (A, ω ) for the set of all

p∈Lv (κ, ω 1) such that dom ( p) ⊂A × ω1.We now prove an easy result.

Theorem 2. Let P be a forcing notion of size ω1 in V . If forcing with P does not add new countable sequences of ordinals, then there are no Kurepa trees in V P .

Proof: Since P has size ω1, there is an η < κ such that P ∈V Lv (η,ω 1 ) . HenceV P = V (Lv (η,ω 1 )∗P )× Lv (κ η,ω 1 ) . But Lv (κ η, ω1) in V is again a Levy collapsing orderin V Lv (η,ω 1 )∗P because P adds no new countable sequences of ordinals, so that the

forcing notion Lv (κ η, ω1) is absolute between V and V Lv (η,ω 1 )∗P. Hence there is no

Kurepa trees in V P . 2

Next we prove the results about ( S, ω )-proper forcing notions.

Denition 3. A forcing notion P is said to satises property ( †) if for any x ∈H (λ ),there exists a sequence N i : i ∈ω of elementary submodels of H (λ ) such that

(1) N i ∈N i+1 for every i ∈ω,(2) {P , x } ⊆ N 0,

(3) for every p ∈ P ∩ N 0 there exists a q p and q is (P , N i )-generic for every i ∈ω.

Lemma 4. Let V be any model. Let P and Q be two forcing notions in V such that P has size ω1 and satises property ( †), and Q is ω1-closed (in V ). Suppose T isan ω1-tree in V P . Then T has no branches which are in V P × Q but not in V P .



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Proof: Suppose, towards a contradiction, that there is a branch b of T in V P × Q V P .Without loss of generality, we can assume that

P Q (b is a branch of T in V P × Q V P ).

Claim 4.1 For any p ∈P , q ∈Q , n ∈ω and α ∈ω1, there are p p, q j q for j < n and β ∈ω1 α such that

p ((∃{t j : j < n } ⊆ T β )(( j = j → t j = t j ) ∧ j<n

(q j t j ∈b))) .

Proof of Claim 4.1: Since

p P q Q (b is a branch of T in V P × Q V P ),

then p forces that q can’t determine b. Hence

p ((∃β ∈ω1 α )(∃q j q for j < n )(∃t j ∈T β for j < n )

(( j = j → t j = t j ) ∧ j<n

(q j t j ∈b))) .

Now the claim is true by a fact about forcing (see [K, pp.201]).

Claim 4.2 Let η ∈ω1 and let q ∈Q . There exists a ν ω1, a maximal antichain pα : α < ν of P , two decreasing sequences q j

α : α < ν , j = 0 , 1, in Q and anincreasing sequence ηα : α < ν in ω1 such that q0

0 , q10 < q , η0 > η and for any α < ν

pα ((∃t0 , t 1 ∈

˙T ηα )( t0 = t1 ∧ (q

0α t0 ∈

¨b) ∧ (q

1α t1 ∈

¨b))) .

Proof of Claim 4.2: We dene those sequences inductively on α . First let’s xan enumeration of P in order type ζ ω1, say, P = {xγ : γ < ζ }. For α = 0 we applyClaim 4.1 for p = 1 P and n = 2 to obtain p0 , q0

0 , q10 and η0. Let α be a countable

ordinal. Suppose we have found pβ : β < α , q0β : β < α , q1

β : β < α andηβ : β < α . If pβ : β < α is already a maximal antichain in P , then we stop and

let ν = α . Otherwise choose a smallest γ < ζ such that x γ is incompatible with all pβ ’s for β < α . Pick q j

∈Q which are lower bounds of q jβ : β < α for j = 0 , 1,

respectively, and pick η ∈ω1 which is an upper bound of ηβ : β < α . By applyingClaim 4.1 twice we can nd

p xγ , q00 , q0

1 q0, q10 , q1

1 q1, t00, t0

1, t10, t1

1 and ηα > η

such that p ( t0

0, t01 ∈T ηα ∧ t0

0 = t01 ∧ (q0

0 t00 ∈b) ∧ (q0

1 t01 ∈b))



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and

p ( t10, t1

1 ∈T ηα ∧ t10 = t1

1 ∧ (q10 t1

0 ∈b) ∧ (q11 t1

1 ∈b)) .

If p t00 = t1

0, then let pα = p , q0α = q0

0 and q1α = q1

0 . Otherwise we can nd a pα < p such that pα t0

0= t1

1. Then let q0

α= q0

0and q1

α= q1

1. If for any countable

α , the set { pβ ∈ P : β < α } has never been a maximal antichain, then the set{ pβ ∈P : β < ω 1} must be a maximal antichain of P by the choice of pβ ’s accordingto the xed enumeration of P = {x γ : γ < ζ = ω1}. In this case we choose ν = ω1.

The lemma follows from the construction. Let n ∈ ω, δn = ω1 ∩ N n and letδ = n∈ω δn . For each s ∈2n we construct, in N n , a maximal antichain ps

α : α < ν s

of P , two decreasing sequences qs jα : α < ν s for j = 0 , 1, and an increasing sequence

ηsα : α < ν s in δn such that ν s δn , qs j

0 are lower bounds of qsα : α < ν s n − 1 for

j = 0 , 1, ηs

0 = δn − 1

and ps

α ((∃t0, t 1 ∈T ηsα )( t0 = t1 ∧ (qs 0

α t0 ∈b) ∧ (qs 1α t1 ∈b))) .

Each step of the construction uses Claim 4.2 relative to N n for some n ∈ω. We canchoose qs 0

0 and qs 10 to be lower bounds of qs

α : α < ν s n − 1 because qsα : α < ν s n − 1 is

constructed in N n − 1 and hence, is countable in N n . Here we use the fact N n − 1 ∈N n .Let ¯ p 1P be (P , N n )-generic for every n ∈ω. Since Q is ω1-closed in V , for every

f ∈ 2ω there is a qf which is a lower bound of qf n0 : n ∈ ω . Let G ⊆ P be a

V -generic lter such that p ∈G . We claim that T δ is uncountable in V [G]. Thiscontradicts that T is an ω1-tree in V P . Notice that 2 ω ∩ V is uncountable in V [G].In V [G] for each f ∈2ω ∩ V there is a qf qf and a t f ∈T δ such that qf t f ∈ b.Suppose f, g ∈2ω ∩V are different and n = min {i ∈ω : f (i) = g(i)}. If t f = tg , thenthere is a p ∈G , p ¯ p such that

p ((∃t ∈T δ)(( qf t ∈b) ∧ (qg t ∈b))) .

Suppose f n = s = g n , f (n ) = 0 and g(n ) = 1. Since p is (P , N n )-generic and p∈G , there is a ps

α ∈G for some α ν s . Let p p,psα . Then

p ((∃t0, t 1 ∈T ηsα )( t0 = t1 ∧ (qf t0 ∈b) ∧ (qg t1 ∈b))) .

But this contradicts the following:

p ( t0 ∈T ηsα ∧ t ∈T δ ∧ (qf t0, t ∈b) → t0 t),

p ( t1 ∈T ηsα ∧ t ∈T δ ∧ (qf t1, t ∈b) → t1 t),



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and

p ( t0 , t1 ∈T ηsα ∧ t0 t ∧ t1 t → t0 = t1).

Hence in V [G] different f ’s in 2ω ∩V correspond to different t f ’s in T δ . Therefore T δ

is uncountable.2

A forcing notion P is called ω-proper if for any ω-sequence N n : n ∈ω of countableelementary submodels of H (λ ) such that N n ∈N n +1 for every n ∈ω and P ∈N 0,for any p ∈P ∩ N 0 there is a ¯ p p such that p is (P , N n )-generic for every n ∈ω.Let S be a stationary subset of ω1. A forcing notion P is called S -proper if for anycountable elementary submodel N of H (λ ) such that P ∈N and N ∩ ω1 ∈S , andfor any p ∈P ∩ N there is a ¯ p p such that p is (P , N )-generic. A forcing notionP is called (S, ω )-proper if for any ω-sequence N n : n ∈ω of countable elementary

submodels of H (λ ) such that N n ∈N n +1 for every n ∈ω, N n ∩ ω1 ∈ S for everyn ∈ω, N ∩ ω1 ∈S , where N = n∈ω N n , and P ∈N 0, for any p ∈P ∩ N 0 there is a¯ p p such that p is (P , N n )-generic for every n ∈ω.

Theorem 5. Let S be a stationary subset of ω1 and let P be an (S, ω )-proper forcing notion of size ω1 in V . Then there are no Kurepa trees in V P .

Proof: Choose an η < κ such that S and P are in V Lv (η,ω 1 ) . Then

V P = V (Lv (η,ω 1 )∗P )× Lv (κ η,ω 1 )

and Lv (κ η, ω1) is ω1-closed in V Lv (η,ω 1 ) . By Lemma 4 it suffices to show that P

satises property ( †) in V Lv (η,ω 1 ) . Working in V Lv (η,ω 1 ) . Let x ∈ H (λ ). Since S

is also stationary in V Lv (η,ω 1 ) , we can choose a sequence N n : n ∈ω of countableelementary submodels of H (λ ) such that N n ∈N n +1 , {P , x } ⊆N 0 and N n ∩ω1 ∈S forevery n ∈ω. Since the forcing Lv (κ η, ω1) is countably closed, then we can choose adecreasing sequence qn : n ∈ω in Lv (κ η, ω1) such that qn is a (Lv (κ η, ω1), N n )-master condition ( q is a (Q , N )-master condition iff for every dense open subset D of Q there exists a d ∈D such that q d). Let q be a lower bound of qn : n ∈ω . LetG ⊆ Lv (κ η, ω1) be V Lv (η,ω 1 )-generic such that q ∈G . By Lemma 1 every N n [G] is acountable elementary submodel of ( H (λ )) V . It is also easy to see that {P , x } ⊆N 0[G].Now we have N n [G]∈N n +1 [G] and N n [G] ∩ ω1 ∈S because q (N n = N n [G]).

Pick a p ∈P ∩ N 0. Since P is (S, ω )-proper in V , there exists a p p such that pis (P , N n [G])-generic for every n ∈ω. It is easy to see that p is also (P , N n )-generic



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because a maximal antichain of P in N n is also a maximal antichain in N n [G]. Thisshows that P satises property ( †) in V Lv (η,ω 1 ) . 2

Remarks (1) If P satises Baumgartner’s axiom A, then P is ω-proper or (ω1, ω)-proper. Hence forcing with a forcing notion of size ω1 satisfying axiom A in V doesnot create Kurepa trees. Notice also that all c.c.c. forcing notions, ω1-closed forcingnotions and the forcing notions of tree type such as Sack’s forcing, Laver forcing,Miller forcing, etc. satisfy axiom A.

(2) The idea of the proof of Lemma 4 is originally from [D]. A version of Theorem5 for axiom A forcing was proved in [J].

(3) The ω-properness implies the ( S, ω )-properness and the ( S, ω )-properness im-plies the property ( †).

Now we prove the results about some non-( S, ω )-proper forcing notions.The existence of a Kurepa tree implies that there are no countably complete, ℵ2-

saturated ideals on ω1. Therefore, one can destroy all those ideals by creating ageneric Kurepa tree [V]. But one don’t have to create Kurepa trees for this purpose.Baumgartner and Taylor [BT] proved that adding a club subset of ω1 by nite condi-tions destroys all countably complete, ℵ2-saturated ideals on ω1. The forcing notionfor adding a club subset of ω1 by nite conditions has size ω1 and is proper butnot (S, ω )-proper for any stationary subset S of ω1.. We are going to prove next thatthis forcing notion and some other similar forcing notions do not create Kurepa treesif our ground model is the Levy model V . Notice also that the ideal of nonstationarysubsets of ω1 could be ℵ2-saturated in the Levy model obtained by collapsing a su-percompact cardinal down to ω2 [FMS]. As a corollary we can have a ground model V

which contains countably complete, ℵ2-saturated ideals on ω1 such that forcing withsome small proper forcing notion P in V destroys all countably complete, ω2-saturatedideals on ω1 without creating Kurepa trees.

We rst dene a property of forcing notions which is satised by the forcing notionfor adding a club subset of ω1 by nite conditions.

Denition 6. A forcing notion P is said to satisfy property (#) if for any x ∈H (λ )there exists a countable elementary submodel N of H (λ ) such that {P , x } ⊆ N and for any p0 ∈P ∩N there exists a ¯ p p0, ¯ p is (P , N )-generic, and there exists a countablesubset C of P such that for any ¯ p ¯ p there is a c ∈C and a p ∈P ∩ N , p p0

such that



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(1) for any dense open subset D of P below p in N there is an d ∈D ∩ N such that d is compatible with c, and

(2) for any r ∈P ∩ N and r p , r is compatible with c implies r is compatiblewith ¯ p .

Let’s call the pair ( p , c) a related pair corresponding to ¯ p .

Examples 7. Following three examples are the forcing notions which satisfy property (#).

(1) LetP = { p⊆ω1 × ω1 : p is a nite function which can be extended to

an increasing continuous function from ω1 to ω1.}and let P be ordered by reverse inclusion. P is one of the simplest proper forcing notion

which does not satisfy axiom A [B2]. Forcing with P creates a generic club subset of ω1 and destroys all ℵ2-saturated ideals on ω1 [BT]. It is easy to see that P satisesproperty (#) dened above. For any x ∈H (λ ) we can choose a countable elementarysubmodel N of H (λ ) such that {P , x } ⊆ N and N ∩ ω1 = δ is an indecomposableordinal. For any p0 ∈P ∩ N let ¯ p = p0 ∪ (δ, δ) and let C = {¯ p}. Then for any p ¯ pthere is a p = p δ and a c = p ∈C such that all requirements for the denition of property (#) are satised.

(2) Let S be a stationary subset of ω1. If we deneP S = { p : p is a nite function such that there is an increasing continuous

function f from some countable ordinal to S such that p ⊆ f. }and let P S be ordered by reverse inclusion, then P S is S -proper [B2]. Forcing withP S adds a club set inside S . It is also easy to check that P S satises (#). For anyx ∈H (λ ). Let N be a countable elementary submodel of H (λ ) such that {x, P S } ⊆ N ,N ∩ ω1 = δ is an indescomposable ordinal and δ ∈S . Then for any p0 ∈P S ∩ N theelement p = p0 ∪ {(δ, δ)} is (P S , N )-generic. Now N , ¯ p and C = {¯ p} witness that P S

satises property (#).

(3) Let T and U be two normal Aronszajn trees such that every node of T orU has innitely many immediate successors. Let P be the forcing notion such that p = ( A p, f p) ∈P iff

(a) A p is a nite subset of ω1,(b) f p is a nite partial isomorphism from T A p into U A p,



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(c) dom (f p) is a subtree of T A p in which every branch has cardinality |A p|.P is ordered by p q iff A p ⊇ Aq and f p ⊇ f q. P is proper [T]. P is used in [AS] forgenerating a club isomorphism from T to U . For any x ∈H (λ ), for any countableelementary submodel N of H (λ ) such that {P , x } ⊆ N and for any p0 ∈ P ∩ N ,

let δ = N ∩ ω1, let A ¯ p = A p0 ∪ {δ} and let f ¯ p be any extension of f p0 such thatT δ ∩ dom (f ¯ p) = ∅. Then p = ( A ¯ p, f ¯ p) is a (P , N )-generic condition. Let

C = {d : d is a nite isomorphism from T δ to U δ}.

Then C is countable. For any p ¯ p let c = ( f ¯ p {δ}) ∈C , let α < δ , α > max( A ¯ p ∩δ)and

gα = {(t, u ) ∈T α × U α : (∃(t , u ) ∈(f ¯ p {δ}))( t < t ∧u < u )}

be such that gα and f ¯ p {δ} have same cardinality, let A p = ( A ¯ p ∩ δ) ∪ {α }, letf p = ( f ¯ p (A ¯ p ∩ δ)) ∪ gα , and let p = ( A p , f p ). Then ( p , c) is a related paircorresponding to p [AS] and N, ¯ p, C witness that P satises property (#). For anystationary set S we can also dene an S -proper version of this forcing notion.

Lemma 8. Let V be a model. Let P and Q be two forcing notions in V such that P

has size ω1 and satises property (#), and Q is ω1-closed (in V ). Suppose T is an ω1-tree in V P . Then T has no branches which are in V P× Q but not in V P .

Proof: Suppose, towards a contradiction, that there is a branch b of T in V P × Q V P .Without loss of generality, we assume that

P Q (b is a branch of T in V P× Q V P ).

Following the denition of property (#), we can nd a countable elementary submodelN of H (λ ) such that {P , Q , T , b} ⊆ N , a ¯ p 1P which is (P , N )-generic and acountable set C ⊆ P such that N , ¯ p and C witness that P satises property (#). Let( pi , ci ) : i ∈ω be a listing of all related pairs in ( P ∩N ) × C with innite repetition,

i.e. every related pair ( p,c) in (P ∩ N ) × C occurs innitely ofter in the sequence.We construct now, in V , a set {qs ∈Q ∩ N : s ∈2<ω } and an increasing sequence

δn : n ∈ω such that(1) s ⊆ t implies qt qs ,(2) δn ∈δ = N ∩ ω1,



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(3) for every n ∈ω there is a p ∈P ∩ N, p pn such that p is compatible withcn , and

p ((∃{t s : s ∈2n } ⊆ T δn )(( s = s → t s = t s ) ∧s∈2n

(qs t s ∈b))) .

The lemma follows from the construction. Let G ⊆ P be a V -generic lter and¯ p∈G . We want to show that

V [G] |= T δ is uncountable.

For any f ∈2ω ∩ V let qf ∈Q be a lower bound of the set {qf n : n ∈ω} such thatthere is a t f ∈T δ such that qf t f ∈ b. Suppose T δ is countable. Then there aref, g ∈2ω ∩V such that t f = tg . Let t f , tg be P -names for t f , t g and let p ¯ p be suchthat

¯ p (˙t f =

˙tg ∧ (qf

˙t f ∈

¨b) ∧ (qg

˙tg ∈

¨b)) .

Let m = min {i ∈ω : f (i) = g(i)}. By the denition of property (#) we can nda related pair ( p,c) corresponding to p . Choose an n ∈ ω such that n m and( p,c) = ( pn , cn ). Since (1) of Denition 6 is true, there is a p ∈ P ∩ N such that p p, p is compatible with cn and

p ((∃{t s : s ∈2n } ⊆ T δn )(( s = s → t s = t s ) ∧s∈2n

(qs t s ∈b))) .

Since qf qf n and qg qg n , then

¯ p ((∃t0, t 1 ∈T δn )( t0 = t1 ∧ (qf t0 ∈b) ∧ (qg t1 ∈b))) .

But also¯ p ((∃t ∈T δ)(( qf t ∈b) ∧ (qg t ∈b))) .

By the fact that any two nodes in T δn which are below a node in T δ must be same,and that p is compatible with p , we have a contradiction.

Now let’s inductively construct {δi : i ∈ω} and {qs : s ∈2<ω }. Suppose we havehad {qs : s ∈2 n } and {δi : i n}. let D ⊆ P be such that r ∈D iff

(1) r pn (recall that ( pn , cn ) is in the enumeration of all related pairs in ( P ∩N ) × C ),

(2) there exists η > δ n and there exists {qs qs n : s ∈2n +1 } such that

r ((∃{t s : s ∈2n +1 } ⊆ T η)(( s = s → t s = t s ) ∧s∈2n +1

(qs t s ∈b))) .

It is easy to see that D is open and D ∈N .



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Claim 8.1 D is dense below pn .Proof of Claim 8.1: Suppose r 0 pn . It suffices to show that there is an r r 0

such that r ∈D . Applying Claim 4.1, for any s ∈2n we can nd r s r 0, ηs > δ n

and {qs j qs : j < 2n +1 } such that

r s ((∃{t j : j < 2n +1 } ⊆ T ηs )(( j = j → t j = t j ) ∧ j< 2n +1

(qs j t j ∈b))) .

Let {s i : i < 2n } be an enumeration of 2 n . By applying Claim 4.1 2n times as abovewe obtained r 0 r s 0 r s 1 . . . r s 2 n − 1 such that above arguments are true for anys ∈2n . Pick η = max {ηs : s ∈2n }. Then we extend r s 2 n − 1 to r , and extend qs

j to qs j

for every such s and j such that for each s ∈2n

r ((∃{t j : j < 2n +1 } ⊆ T η)(( j = j → t j = t j ) ∧ j< 2n +1

(qs j t j ∈b))) .

Now applying an argument in Claim 4.2 repeatedly we can choose {qs 0 , qs 1 } ⊆{qs

j : j < 2n +1 } for every s ∈2n and extend r to r such that

r ((∃{t s : s ∈2n +1 } ⊆ T η)(( s = s → t s = t s ) ∧s∈2n +1

(qs t s ∈b))) .

This showed that D is dense below pn .

Notice that since N is elementary, then η exists in N and all those qs ’ for s ∈2n +1

exist in N . Choose r ∈D such that r, c n are compatible and let δn +1 be correspondentη. This ends the construction. 2

Theorem 9. If P in V is a forcing notion dened in (1), (2) or (3) of Examples 7,then forcing with P does not create any Kurepa trees.

Proof: Suppose T is a Kurepa tree in V P . Let η < κ be such that P , T ∈V Lv (η,ω 1 ) .Since the denition of P is absolute between V and V Lv (η,ω 1 ) , then P satises property

(#) in V Lv (η,ω 1 ) . Since T has less than κ branches in V Lv (η,ω )∗P , there exist branchesof T in V P which are not in V Lv (η,ω 1 )∗P . This contradicts Lemma 8. 2

Remark: The forcing notions in Examples 7, (1), (2) and (3) are not ( S, ω )-properfor any stationary S .



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2. Creating Kurepa Trees By a Small Forcing Is Easy

In this section we construct a model of CH plus no Kurepa trees, in which thereis an ω-distributive Aronszajn tree T such that forcing with T does create a Kurepatree in the generic extension.

Let V be a model and κ be a strongly inaccessible cardinal in V . Let T be the setof all countable normal trees. Given a set A and a cardinal λ . Let [A]<λ = {S ⊆A :|S | < λ } and [A] λ = {S ⊆A : |S | λ}. We dene a forcing notion P as following:

Denition 10. p is a condition in P iff

p = α p, t p, k p, U p, B p, F p

where(a) α p ∈ω1,(b) t p ∈ T and ht (t p) = α p + 1 ,(c) k p is a function from t p to T such that for any x ∈ t p, ht (k p(x)) = ht (x) + 1 ,

and for any x, y ∈t p, x < y implies k p(x) end k p(y),(d) U p ∈[κ] ω1 ,(e) B p = {b p

γ : γ ∈ U p} where b pγ is a function from t p (β pγ + 1) to ω<ω 1

1 for some β pγ α p such that for any x ∈ t p (β pγ + 1) , b p

γ (x) ∈ (k p(x))ht (x ) and for any x, y ∈t p (β pγ ), x y implies b p

γ (x) b pγ (y),

(f ) F p = {f pγ : γ ∈U p} where f pγ is a function from δ pγ to γ for some δ p

γ α p,

(g) for any x ∈t p α p, for any nite U 0 ⊆ U p and for any such that ht (x) < α p, there exists an x ∈ (t p) such that x > x and for any γ 1, γ 2 ∈U 0 either one of β pγ 1 , β pγ 2 is less than or b p

γ 1 (x) = b pγ 2 (x) implies b p

γ 1 (x ) = b pγ 2 (x ).

In the condition (g) of the denition we call x a conservative extension of x at levelwith respect to U 0 (or with respect to {b p

γ : γ ∈U 0}).Generally we have the following notation. Suppose t ∈ T and B is a set of functions

such that for each b∈B there is a β b ht (t) such that domain (b) = t β . We say t

is consistent with respect to B if for any x ∈ t ht (t), for any nite B 0 ⊆ B and forany such that ht (x) < ht (t), there exists an x ∈t such that x > x and for anyb1, b2 ∈B 0 either one of β b1 , β b2 is less than or b1(x) = b2(x) implies b1(x ) = b2(x ).So p ∈P implies that t p is consistent with respect to B p.

For any p, q ∈P we dene the order of P by letting p q iff (1) α q α p, tq end t p, kq ⊆ k p and U q ⊆U p,



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(2) for any γ ∈U q, bqγ ⊆ b p

γ and f qγ ⊆ f pγ ,

(3) {γ ∈U q : β pγ > β qγ } is at most countable,

(4) {γ ∈U q : δ pγ > δ q

γ } is at most countable.

Remarks: In the denition of P the part t p is used for creating an ω-distributiveAronszajn tree T . The part k p is used for creating a T -name of an ω1-tree K . Thepart B p is used for adding κ branches to K so that K becomes a Kurepa tree in thegeneric extension by forcing with T . The part F p is used for collapsing all cardinalsbetween ω1 and κ .

For any ∈ω1, γ ∈κ and η ∈γ , let

D 1 = { p∈P : α p },

D2γ = { p∈P : γ ∈U p},

D 3η,γ = { p∈P : γ ∈U p and η ∈range (f pγ )},

D 4,γ = { p∈P : γ ∈U p and β pγ }.

Lemma 11. The sets D 1, D 2γ , D 3

η,γ and D 4,γ are open dense in P .

Proof: It is easy to see that all four sets are open. Let’s show they are dense. Theproofs of the denseness of the rst three sets are easy.

Given p0 ∈P . We need to nd a p p0 such that p ∈D 1. Pick an α p and

α p α p0 . Let t p ∈ T be such that ht (t p) = α p + 1 and t p0 end t p. Let k p : t p → T be any suitable extension of k p0 . Let U p = U p0 . For any γ ∈U p let b p

γ = b p0γ and

f pγ = f p0γ . Then p p0 and p ∈D 1.

Given p0 ∈P . We need to nd a p p0 such that p ∈D 2γ . If γ ∈U p0 , let p = p0.

Otherwise, let

p = α p0 , t p0 , k p0 , U p0 ∪ {γ }, B p0 ∪ {b pγ }, F p0 ∪ {f pγ } ,

where b pγ and f pγ are empty functions. Then p p0 and p ∈D 2

γ .Given p0 ∈P . We need to nd a p p0 such that p ∈D 3

η,γ . First, pick p ∈D 1α 0 +1

such that p p0 and f pγ = f p0γ . Then extend f pγ to f pγ on α 0 + 1 arbitrary except

assigning f pγ (α 0) = η. Let everything else keep unchanged. Then p p and p ∈D 3η,γ .

Proving the denseness of D 4,γ is not trivial due to the condition (g) of Denition

10. Given p0 ∈ P . Without loss of generality we assume that p0 ∈D 1 ∩ D 2γ and

> β p0

δ for all δ ∈ U p0 . We need to nd a p p0 such that p ∈ D 4,γ . Choose



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α p = α p0 , t p = t p0 , k p = k p0 , U p = U p0 , b pδ = b p0

δ for all δ ∈U p0 {γ } and f pδ = f p0

δ forall δ ∈U p0 . Let β pγ = . We need to extend b p0

γ to b pγ on t p ( + 1) such that p ∈P .

For each x ∈t p ( + 1) t p β p0γ and for each µ Let C x,µ be the cone above x

up to level µ, i.e.C x,µ = {y ∈t p : x < y and ht (y) µ}.

We construct t0 ⊆ t1 ⊆ . . . with t0 = t p β p0γ and dene b p

γ on tn inductively. Supposewe have had t n and b p

γ tn . For any maximal node x of t n we dene a subset tnx above

x . It will be self-clear from the construction that for any n ∈ω and for any x ∈ t n

there is a maximal node x of tn such that x x . Our tn +1 will be the union of tn

and those tnx ’s. Let x be a maximal node of tn . Let

U x = {β pδ : δ ∈U p {γ }, β pδ > ht (x) and b pδ (x) = b p

γ (x)}.

Case 1: U x = ∅. Let tnx = ∅. This means any choice of b pn above x will not violatethe condition (g).

Case 2: U x has a largest element, say β pδ . Let tnx = C x,β p

δand let b p

γ tnx = b p

δ t nx .

Case 3: U x is a limit ordinal. Fix a strictly increasing sequence ν x,m : m ∈ω

of ordinals such that m∈ω ν x,m = U x . Let x0 x1 . . . xn = x be such that x i

is a maximal node of t i for i = 0 , 1, . . . , n . Notice that if i < n , then U x i U x ,and if U x i is a limit ordinal, then ν x i ,m : m ∈ω has already been dened. Let

l = min {i :m∈ω

ν x i ,m =m∈ω

ν x n ,m }

and letν = max {ν x i ,n : l i n}.

Choose δ ∈U x such that β pδ ν and let b pγ C x,β p

δ= b p

δ C x,β pδ. Let tn

x = C x,β pδ. Now

we take

tn +1 = tn ∪ ( {tnx : x is a maximal node of tn .}

and dene b pγ tn +1 accordingly. Let t = n∈ω tn . Notice that t may not be equal to

t p ( + 1). But it is no problem because any extension of b pγ t to t p ( + 1) following

the condition (e) will not violate the condition (g). Let b pγ be such an extension of b p

γ t .

Claim 11.1 p ∈P .Proof of Claim 11.1: We need only to check that the condition (g) of Denition

10 is satised. Pick x ∈ t p and pick a nite subset U 0 of U p. Pick also an such



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that ht (x) < . First, we assume that x ∈t n tn − 1 for some n ∈ω (let t− 1 = ∅).Without loss of generality we assume that x is a maximal node of t n .

Case 1: Every β ∈U x is less than . Then the condition (g) is trivially satisedbecause any conservative extension of x at level with respect to U 0 {γ } is a

conservative extension of x with respect to U 0.Case 2: There is a largest ordinal β pδ in U x such that

b pγ C x,β p

δ= b p

δ C x,β pδ

.

Then a conservative extension of x at level with respect to ( U 0 {γ }) ∪ {δ } is aconservative extension of x with respect to U 0.

Case 3: U x is a limit ordinal greater than . First, choose β pδ > in U x .Suppose ν x,m β pδ < ν x,m +1 . Then choose a maximal node x1 of tn +1 such that x1 is

a conservative extension of x with respect to U 0 ∪ {γ, δ }. Now we haveU x 1 β pδ > .

Notice that ht (x1) ν x,n . We are done if U x 1 has a largest ordinal. Otherwise werepeat the same procedure to get x2. Eventually, we can nd an xk such that xk is aconservative extension of x with respect to U 0 ∪ {δ } and ht (xk ) ν x,m +1 > . Letx x and ht (x ) = . It is easy to see that x is a conservative extension of x atlevel with respect to U 0.

Suppose x ∈t . Then U x = ∅. So every x x , x ∈t is a conservative extensionof x with respect to U 0.

This ends the proof of the claim. It is easy to see that p ∈D 4,γ . 2

Next we want to prove that P is ω1-strategically closed. Let Q be a forcing notion.Two players, I and II , play a game G(Q ) by I choosing pn ∈ Q and II choosingqn ∈Q alternatively such that

p0 q0 p1 q1 . . . .

II wins the game G(Q ) if and only if the sequence p0, q0, p1, q1, . . . has a lower boundin Q . A forcing notion Q is called ω1-strategically closed if II wins the game G(Q ).Note that any ω1-strategically closed forcing notion does not add new countablesequences of ordinals to the generic extension.

Lemma 12. P is ω1-strategically closed.



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Proof: We choose qn inductively for Player II after Player I choose any pn qn − 1.Suppose pi , qi have been chosen for i < n . Let pn qn − 1 be any element chosen byPlayer I . Player II want to choose qn pn . Let

U n = {γ ∈U pn : (∃i < n )(β pi

γ = β qi

γ ) or (∃i n )(β pi

γ = β qi − 1

γ )}.Choose qn pn such that α qn > α pn and for any γ ∈U n , β qn

γ = α qn . This can bedone by repeating the steps countably many times used in the proof of the densenessof D 4

,γ in Lemma 11. This nishes the inductive step of the construction. Let

α q =n∈ω

α qn , t =n∈ω

, k =n∈ω

kqn , U q =n∈ω

U qn

and for each γ ∈U q

bγ

= {bqn

γ : n ∈ω, γ ∈U qn }

and

f qγ = {f qn

γ : n ∈ω, γ ∈U qn }.

We need now to add one more level on the top of t and extend k and bγ ’s accordingly.The main difficulty here is to make the condition (g) of Denition 10 true. Remember

U ω =n∈ω

U n ⊆U p

is the set of allγ

’s such thatβ qn

γ grows for somen

. The setU

ω is at most countabledue to the denition of the order of P . Note that α qn is strictly increasing. Note alsothat for each γ ∈U q U ω the sequence

{bqnγ : n ∈ω, γ ∈U qn }

is a constant sequence. So the top level we are going to add does not affect those b pγ ’s

for γ ∈U q U ω .Let { xm , Γm : m ∈ω} be an enumeration of t × [U ω]<ω . For each xm , Γm we

choose an increasing sequence ym,i : i ∈ω such that

xm = ym, 0 < y m, 1 < . . . ,

ym,i +1 is a conservative extension of ym,i with respect to Γ m and

i∈ω

ht (ym,i ) = α q .



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Now let ym = i∈ω ym,i and let tq = t ∪ {ym : m ∈ω}. It is easy to see that t q ∈ T .For each γ ∈U ω we dene bq

γ to be an extension of bγ on tq such that

bqγ (ym ) =

i∈ω

bγ (ym,i )

for all m ∈ω. We dene also kq to be an extension of k on tq such that for eachm ∈ ω, the tree kq(ym ) is in T , ht (kq(ym )) = α q + 1, kq(ym ) is an end-extensionof i∈ω k (ym,i ) and bq

γ (ym ) ∈kq(ym ) for all γ ∈U ω . It is easy to see now that theelement q is in P and is a lower bound of pn ’s and qn ’s. 2

Lemma 13. The forcing notion P satises κ-c.c..

Proof: Let { pη : η ∈ κ} ⊆ P . By a cardinality argument and ∆-system lemmathere is an S ⊆ κ , |S | = κ and there is a triple α 0, t 0, k0 such that for every η ∈S

α pη , t pη , k pη = α 0, t 0, k0 ,

and {U pη : η ∈S } forms a ∆-system with the root U 0. Furthermore, we can assumethat for each γ ∈U 0,

b pηγ = b

pηγ and f pη

γ = f pηγ

for any η, η ∈S . Since there are at most ( |ω α 01 | | t 0 |)ω1 = 2 ω1 sequences of length ω1

of the functions from t 0 to ω α 01 , there are η, η ∈S such that

{b pηγ : γ ∈U pη U 0} and {b

pηγ : γ ∈U pη

U 0}

are same set of functions. It is easy to see now that the element

p = α 0, t 0, k0, U pη ∪U pη, B pη ∪B pη

, F pη ∪F pη

is a common lower bound of pη and pη . 2

Lemma 14. All cardinals between ω1 and κ in V are collapsed in V P .

Proof: For any γ ∈κ let

f γ = {f pγ : p ∈G and γ ∈U p}

where G ⊆ P is a V -generic lter. It is easy to check that range (f γ ) = γ . Alsodom (f γ ) ⊆ ω1. So in V P we have |γ | ω1. 2

Remark: By Lemma 12, Lemma 13 and Lemma 14 we have

V P |= (2 ω = ωV 1 = ω1 and 2ω1 = κ = ω2).



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Lemma 15. Let G ⊆ P be a V -generic lter and let T G = {t p : p ∈G}. Then T Gis an ω-distributive Aronszajn tree in V [G].

Proof: It is easy to see that T G is an ω1-tree. Suppose there is a p0 ∈P such that

p0 B is a branch of T G .We construct p0 q0 p1 q1 . . . similar to the construction in Lemma 12 suchthat

pn +1 zn ∈B ∩ (tqn )α qn

for some zn ∈ωα qn1 . For constructing qn +1 we use almost same method as in Lemma

12 except that we require qn +1 satisfy the following condition (g’):

For any x ∈ t pn +1 and Γ ∈[U n +1 ]<ω (see Lemma 12 for the denitionof U n +1 ) there are innitely many x ∈ (tqn +1 )α qn +1

such that x is aconservative extension of x with respect to Γ.

This can be done just by stretching t qn +1 a little bit higher and manipulating thosebqn +1

γ (t qn +1 t pn +1 ) for γ ∈U n +1 more carefully. Let q be a lower bound of qn : n ∈ω

constructed same as in Lemma 12 except that for any x m , Γm the sequence ym,i :i ∈ω is chosen such that i∈ω ym,i is different from n∈ω zn . This is guaranteed bythe condition (g’). Now

n∈ω

zn ∈(tq)α q .

Hence q B ⊆ tq.

This contradicts that B is a branch of T G in V [G].Next we prove that T G is ω-distributive. Let Q = T G , be the forcing notion by

reversing tree order ( = T G ). Given any τ ∈2ω in V P∗Q . It suffices to show thatτ ∈V . We construct a decreasing sequence

p0, x0 q0, x0 p1, x1 q1, x1 . . .

in P ∗Q such that

p0, x0 τ is a function from ω to 2, pn xn ∈ωα p n

1 ,

qn τ (n ) = ln

for some ln ∈ {0, 1} andqn xn = xn



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for some xn ∈ (t pn )α p n . In addition we can extend qn so that the requirements forPlayer II to win the game are also satised. Now we can construct a lower bound q

of qn same as we did in Lemma 12 except that we put also x = n∈ω xn into the toplevel of tq . It is easy to see that q, x ∈P ∗Q and there is a σ = l0, l1, . . . ∈2ω inV such that

q, x τ = σ. 2

Lemma 16. Let G ⊆ P be a V -generic lter and let kG = {k p : p ∈G}. Let T Gand Q be same as in Lemma 15. Suppose H ⊆ Q is a V [G]-generic lter. Then K H = {kG (x) : x ∈H } is a Kurepa tree in V [G][H ].

Proof: It is easy to see that K H is an ω1-tree. For any γ ∈κ let

bγ = {b p

γ : p ∈G and γ ∈U p}.Then bγ is a function with domain T G . Let

W γ = {bγ (x) : x ∈H }.

Then it is easy to see that W γ is a branch of K H . We need now only to show thatW γ and W γ are different branches for different γ, γ ∈κ . Given distinct γ and γ inκ . Let

D 5γ,γ = { p∈P : β pγ = β pγ = α p and

(∀x ∈t p α p)(∃y ∈t p)(y x and b pγ (y) = b p

γ (y))}.

Claim 16.1 The set D 5γ,γ is dense in P .

Proof of Claim 16.1: Given p0 ∈ P . Without loss of generality we assume that p0 ∈D 2

γ ∩ D 2γ and β p0

γ = β p0

γ = α p0 . First, we extend t p0 to t p ∈ T such that

α p = ht (t p) = α p0 + ω + 1 .

Then, we choose one extension k p of k p0 on t p. Now we can easily extend b p0γ and

b p0

γ to b pγ and b p

γ on t p while keeping other things unchanged such that the resultingelement p is in P and for each x ∈t p α p there is an y ∈(t p)α p and y > x such that

b pγ (y) = b pγ . It is easy to see the element p is less than p0 and is in D 5γ,γ . This ends

the proof of the claim.

We need to prove W γ and W γ are different branches of K H in V [G][H ]. Supposex ∈H and

x W γ = W γ



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in V [G]. Let p0 ∈G be such that x ∈ t p0 . By the claim we can nd a p p0 and p∈G ∩ D 5

γ,γ such that α p > ht (x). Then we can choose y ∈t p and y > x such thatb p

γ (y) = b pγ (y). Therefore

y W γ = W γ ,

which contradicts thatx W γ = W γ . 2

The next lemma is probably the hardest part of this section.

Lemma 17. There are no Kurepa trees in V P .

Proof: Suppose

P T is a Kurepa tree with κ branches C = {cγ : γ ∈κ}.

For each γ ∈κ such that cof (γ ) = (2 ω1 )+ we choose an elementary submodel A γ of H (λ ) such that

(a) |A γ | 2ω1 ,(b) {T , C, P , γ } ⊆ A γ ,(c) [A γ ] ω1 ⊆ A γ .

By the Pressing Down Lemma we can nd a set

S ⊆ {γ ∈κ : cof (γ ) = (2 ω1 )+ }

with |S | = κ such that(d) {A γ : γ ∈S } forms a ∆-system with the common root B ,(e) there is a η0 ∈κ such that η0 = {η ∈κ : η ∈A γ ∩ γ } for every γ ∈S ,(f) for any γ, γ ∈S there is an isomorphism h γ,γ from A γ to A γ such that h γ,γ B

is an identity map.Notice that ω1 ⊆ B and ω<ω 1

1 ⊆ B . So for any x ∈ω<ω 11 we have h γ,γ (x) = x .

Let γ 0 be the minimal ordinal in S . For any p, p ∈P we write p A γ = p to meanα p, t p, k p = α p , t p , k p , U p ∩ A γ = U p , b p

γ = b pγ and f pγ = f pγ for each γ ∈U p .

We write also p B = p to mean the same thing as above except replacing A γ byB . Notice that for p, p ∈A γ the sentence p B = p is rst-order with parametersin A γ , i.e. the term B could be eliminated. Next we are going to do a complicatedinductive construction of several sequences.

We construct inductively the sequences pn ∈P : n ∈ω ,



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ps ∈P : s ∈2<ω ,ηn ∈ω1 : n ∈ω andx s ∈ω<ω 1

1 : s ∈2<ω

in A γ 0 such that

(1) pn +1 < p n and α pn < α pn +1 for every n ∈ω,(2) ps ps for any s, s ∈2<ω and s ⊆ s ,(3) ps B = pn for any n ∈ω and s ∈2n ,(4) ηn < η n +1 for every n ∈ω,(5) x s x s for any s, s ∈2<ω and s ⊆ s ,(6) ht (x s ) = ηn for any s ∈2n ,(7) x s = x s for any s, s ∈2n and s = s ,(8) ps x s ∈cγ 0 for every s ∈2<ω ,

(9) t pn is consistent with respect to {b p

sγ : γ ∈ s∈2n U ps } for each n ∈ω,(10) β ps

γ = α ps for all γ ∈U ps such that β psγ = β ps

γ for some s ⊆ s ⊆ s ,(11) {b ps

γ : γ ∈U ps U pn } and {b psγ : γ ∈U ps

U pn } are the same set of functionsfor all s, s ∈2n .

We need to add more requirements for those sequences along the inductive con-struction.

For any s ∈2<ω let

U s = {γ ∈U ps : ∃s , s (s ⊆ s ⊆ s and β psγ = β ps

γ )}.

Let’s x an onto function j : ω → ω × ω such that j (n ) = a, b implies a n . Letπ1, π 2 be projections from ω × ω to ω such that π1( a, b ) = a and π2( a, b ) = b. Let

ξn : ω → t pn × ([s∈2n

U s ]<ω )

andζ n : ω →

s∈2n

U s

be two onto functions for each n ∈ω. Let e be a function with domain (e) = ω such

thate(n ) = ξπ 1 ( j (n )) (π2( j (n ))) .

The functions ξn ’s, ζ n ’s and e are going to be used for bookkeeping purpose. Fors ∈2m and m < n let

C s,n = {s ∈2n : s ⊆ s }.



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For any m, n ∈ω, m n let

Z nm = {b psγ : s ∈2π 1 ( j (m )) , γ ∈π2(e(m )) ∩ U s and s ∈C s,n }∪

{b psγ : s ∈2π 1 ( j (m )) , γ ∈U s and γ = ζ π 1 ( j (m )) (i) for some i n}.

Note that Z nm is nite and for each b pγ ∈Z nm we have β pγ = α pn . For each m, n ∈ω we

need also construct another set

Y nm = {ym,i : m i n}.

Then Z nm ’s and Y nm ’s and other four sequences should satisfy two more conditions.(12) ym,m = π1(e(m )) and ym,i ∈(t pi )α p i

for m < i n ,(13) ym,i +1 is a conservative extension of ym,i with respact to Z i+1

m .Next we do the inductive construction. Suppose we have had sequences

pn ∈P : n < l ,

ps ∈P : s ∈2<l

,ηn ∈ω1 : n < l ,x s ∈ω<ω 1

1 : s ∈2<l ,{Z nm : n < l, m n} and{Y nm : n < l, m n}.

We rst choose distinct {γ s : s ∈2l} ⊆S . For any s ∈2l let ps = h γ 0 ,γ s ( ps l). Notethat

ps = α ps l , t ps l , k ps l , U ps , B ps , F ps

whereU ps = {h γ 0 ,γ s (γ ) : γ ∈U ps l },

B ps = {b ps

h γ 0 ,γ s (γ ) : b ps

h γ 0 ,γ s (γ ) = h γ 0 ,γ s (b ps lγ ) and γ ∈U ps l }

andF ps = {h γ 0 ,γ s (f ps l

γ ) : γ ∈U ps l }.

Notice that b ps

h γ 0 ,γ s (γ ) and b ps lγ are same functions with different indices. Notice also

that

α ps l = α pl − 1 , t ps l = t pl − 1 , k ps l = k pl − 1

andU ps = {h γ 0 ,γ s (γ ) : γ ∈U ps l U pl − 1 }∪U pl − 1 .

Let

¯ pl− 1 = α pl − 1 , t pl − 1 , k pl − 1 , U pl − 1 , B ¯ pl − 1 , F ¯ pl − 1



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whereU pl − 1 =

s∈2l

U ps ,

B ¯ pl − 1 = {b ps

γ : s ∈2l , γ ∈U ps }

andF pl − 1 = {f p

s

γ : s ∈2l , γ ∈U ps }.

Since t p− 1 is consistent with s∈2l − 1 B ps by (9), then we have pl− 1 ∈P . Since

¯ pl− 1 {cγ s : s ∈2l} is a set of distinct branches of T ,

then there exist pl ¯ pl− 1, ηl ∈ω1 such that ηl > η l− 1, and there exist distinct

{x s : s ∈2l} ⊆ωηl

such that¯ pl x s ∈cγ s

for all s ∈ 2l . We can also require that α ¯ pl > α ¯ pl − 1 and β plγ = α ¯ pl for all γ ∈U pl

such that β plγ > β ¯ pl − 1

γ , or for all γ ∈ s∈2l − 1 hγ 0 ,γ s [U s l]. For each s ∈ 2l let U s bea set of ω1 ordinals such that U s ⊆ A γ s B and U s ∩ U pl = ∅. Since B ¯ pl has only

ω1 functions, we can use the ordinals in U s to re-index all functions in B ¯ pl , say B ¯ pl

and {b¯ plγ : γ ∈U s } are same set of functions. Let f ¯ pl

γ be an empty function for eachγ ∈U s . We now construct a p such that

¯ p = α ¯ pl , t ¯ pl , k ¯ pl , U p, B ¯ p, F ¯ p ,

whereU p = U pl ∪ (

s∈2l

U s ),

B ¯ p = B ¯ pl ∪ (s∈2l

{b¯ plγ : γ ∈U s })

andF p = F pl ∪ (

s∈2l

{f ¯ plγ : γ ∈U s }).

It is easy to see that p∈P and ¯ p ¯ pl .

Claim 16.2 For each s ∈2l let ¯ ps = p A γ s . Then ps x s ∈cγ s .Proof of Claim 16.2: It is true that ps ∈A γ s because (A γ s ) ω1 ⊆ A γ s . Suppose

¯ ps x s ∈cγ s .



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Then there is a ps ¯ ps such that

ps x s ∈cγ s .

Since A γ s H (λ ), we can choose ps ∈A γ s . It is now easy to see that ps and ¯ p are

compatible (here we use the fact that every function in B ¯ p is also in B ¯ ps with possiblydifferent index). This derives a contradiction.

Let pl = p B and ps = h γ s ,γ 0 (¯ ps ). Then pn : n l , ps : s ∈2 l ,ηn : n l andx s : s ∈2 l

satisfy conditions (1)—(11). For example, we have

ps x s ∈cγ 0

because ps = h γ s ,γ 0 (¯ ps ), γ 0 = h γ s ,γ 0 (γ s ) and x s = h γ s ,γ 0 (x s ). We have also that t pl isconsistent with {b ps

γ : s ∈2l and γ ∈U ps } because ¯ p∈P .We need to deal with the conditions (12) and (13).For each m l the set Z lm has been dened before. For m < l since Z lm is nite,

there exists a ym,l ∈ (t pl )α p lsuch that ym,l is a consistent extension of ym,l − 1 with

respect to Z lm . Let yl,l = π1(e(l)). It is not hard to see that those sequences up tostage l satisfy conditions (12) and (13). This ends the construction.

We want to draw the conclusion now.For each m ∈ω let ym = i∈ω ym,i and let

t pω = (n∈ω

t pn ) ∪ {ym : m ∈ω}.

It is easy to see that t pω ∈ T . Let α pω = n∈ω α pn . Then ht (t pω ) = α pω + 1. Let

U =s∈2<ω

U s = {γ : ∃τ ∈2ω such that {β pτ nγ : n ∈ω and γ ∈U pτ n } = α pω }.

Then U is a countable set. Notice that for any s ∈2<ω and γ ∈U ps

U , for anys ⊇ s we have β sγ = β sγ . Let k = n∈ω k pn . For each τ ∈2ω and γ ∈ s∈2<ω U ps let

bτ γ = {b pτ n

γ : γ ∈U pτ n , n ∈ω}

and letf τ

γ = {f pτ nγ : γ ∈U pτ n , n ∈ω}.



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For each m ∈ω and γ ∈U we dene

bτ γ (ym ) = {bτ

γ (ym,i ) : m i < ω }.

Since for each γ ∈U and m ∈ω there exists an n such that for any s, s ∈2l for l n

and s n = s n we have b psγ (ym ) = b

ps

γ (ym ). This is guaranteed by the constructionof Z nm ’s and Y nm ’s. So for any τ, τ ∈2ω ,

τ n = τ n implies bτ γ (ym ) = bτ

γ (ym ).

Hence for each m ∈ω the set

{bτ γ (ym ) : τ ∈2ω , γ ∈U }

is countable. (This is why the condition (g) of Denition 10 is needed.) Let k (ym )be in T such that

i∈ωk (ym,i ) end k (ym )

and{bτ

γ (ym ) : τ ∈2ω , γ ∈U } ⊆ (k (ym ))α p ω .

Then let k pω = k . For each τ ∈2ω let x τ = n∈ω x τ n . Then x τ ∈ωα p ω1 . For each

τ ∈2ω let pτ be the lower bound of { pτ n : n ∈ω} constructed same as in Lemma 12.Then we have pτ ∈P and

pτ x τ ∈cγ 0 .

Choose distinct ordinals {γ τ

: τ ∈ O} ⊆ S for some O ⊆ 2ω and |O | = ω1. Let

pτ = h γ 0 ,γ τ ( pτ ). Then pτ x s ∈cγ τ

for any τ ∈O . Letq = α pω , t pω , k pω , U q , B q, F q ,

whereU q =

τ ∈O

{h γ 0 ,γ τ (γ ) : γ ∈U pτ },

B q = τ ∈O{h γ 0 ,γ τ (bτ

γ ) : γ ∈U pτ }and

F q =τ ∈O

{h γ 0 ,γ τ (f τ γ ) : γ ∈U pτ }.

Claim 16.3 The element q is in P and q pτ for all τ ∈O .



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Proof of Claim 16.3: It is easy to see that |U q| ω1 (the condition |U p| ω1

for p ∈P in Denition 10 is needed here since if we require only |U p| < ω 1, then q

wouldn’t be in P ). It is also easy to see that for each τ ∈O we have q A γ τ = pτ .Hence it suffices to show that t pω is consistent with B q. But this is guaranteed by

condition (9) and the construction of ym ’s.

Claim 16.4 q (T )α p ω is uncountable.Proof of Claim 16.4: This is because of the facts x τ = x τ for different τ, τ ∈O ,

|O | = ω1,q x τ ∈cγ τ

andq cγ τ ⊆ T .

By above claim we have derived a contradiction that

(T is a Kurepa tree)

butq (T is not a Kurepa tree) . 2

3. Questions

We would like to ask some questions.

Question 1. Suppose our ground model is the Levy model dened in the rst section.Can we nd a proper forcing notion such that the forcing extension will contain Kurepa trees? If the answer is ‘no’, then we would like to know if there are any forcing notionsof size ω1 which preserve ω1 such that the generic extension contains Kurepa trees?

Question 2. Suppose the answer of one of the questions above is Yes. Is it true that given any model of CH there always exists an ω1-preserving forcing notion of size

ω1 such that forcing with that notion creates Kurepa trees in the generic extension?

Question 3. Does there exist a model of CH plus no Kurepa trees, in which there isa c.c.c.-forcing notion of size ω1 such that forcing with that notion creates Kurepa trees in the generic extension? If the answer is Yes, then we would like to ask thesame question with c.c.c. replaced by one of some nicer chain conditions such asℵ1-caliber, Property K , etc.



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References

[AS] U. Abraham and S. Shelah, “Isomorphism types of Aronszajn trees”, Israel Journal of Mathe-matics, 50 (1985), pp. 75—113.

[B1] J. Baumgartner, “Iterated forcing”, pp. 1—59 in Surveys in Set Theory , ed. by A. R. D.Mathias, Cambridge University Press, 1983.

[B2] , “Applications of the proper forcing axiom”, pp. 913—959 in Handbook of SetTheoretic Topology , ed. by K. Kunen and J. E. Vaughan, North–Holland, Amsterdam, 1984.

[BT] J. Baumgartner and A. Taylor, “Saturation properties of ideals in generic extension. II”, Trans-action of the American Mathematical Society, 271 , No. 2 (1982), pp. 587—609.

[D] K. J. Devlin, “ℵ 1 –trees”, Annals of Mathematical Logic, 13 (1978), pp. 267—330.[FMS] M. Foreman, M. Magidor and S. Shelah, “Martin’s Maximum, saturated ideals, and non-

regular ultralters. Part I”, Annals of Mathematics, 127 (1988), pp. 1—47.[Je1] T. Jech, “Trees”, The Journal of Symbolic Logic, 36 (1971), pp. 1—14.[Je2] , “Set Theory ”, Academic Press, New York, 1978.[Ji] R. Jin “The differences between Kurepa trees and Jech–Kunen trees”, Archive For Mathematical

Logic, 32 (1993), pp. 369—379.[JS] R. Jin and S. Shelah, “Essential Kurepa trees versus essential Jech–Kunen trees”, Annals of

Pure and Applied Logic, to appear.[K] K. Kunen, “ Set Theory , an introduction to independence proofs”, North–Holland, Amster-

dam, 1980.[S1] S. Shelah, “Can you take Solovay’s inaccessible away?”, Israel Journal of Mathematics, 48 , No.

1 (1984), pp. 1—47.[S2] , “Proper Forcing ”, Lecture Notes in Mathematics, 940 (Springer, Berlin, 1982).[ST] R. Solovay and S. Tennenbaum, “Iterated Cohen extensions and Souslin’s Problem”, Annals

of Mathematics, 94 (1971), pp. 201—245.[T] S. Todorcevic, “Trees and linearly ordered sets”, pp. 235—293 in Handbook of Set Theoretic

Topology , ed. by K. Kunen and J. E. Vaughan, North–Holland, Amsterdam, 1984.[V] B. Velickovic, “Forcing axioms and stationary sets”, Advances in Mathematics, 94 (1992), pp.

256—284.

Department of Mathematics, University of IllinoisUrbana, IL 61801, USAe-mail: [email protected]

Institute of Mathematics, The Hebrew UniversityJerusalem, Israel

Department of Mathematics, Rutgers UniversityNew Brunswick, NJ, 08903, USA

Sorting: The rst address is the rst author’s and the last two are the secondauthor’s.

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Renling Jin & Saharon Shelah- Can a Small Forcing Create Kurepa Trees