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Planting Kurepa Trees And Killing Jech–Kunen Trees

arXiv:math/9211214v1 [math.LO] 15 Nov 1992Planting Kurepa Trees And Killing Jech–Kunen TreesIn a Model By Using One Inaccessible Cardinal 1Saharon Shelah2 and Renling JinAbstractBy an ω1–tree we mean a tree of power ω1 and height ω1 .Under CH and2ω1 > ω2 we call an ω1–tree a Jech–Kunen tree if it has κ many branches forsome κ strictly between ω1 and 2ω1 . In this paper we prove that, assuming theexistence of one inaccessible cardinal, (1) it is consistent with CH plus 2ω1 > ω2that there exist Kurepa trees and there are no Jech–Kunen trees, which answersa question of [Ji2], (2) it is consistent with CH plus 2ω1 = ω4 that only Kurepatrees with ω3 many branches exist, which answers another question of [Ji2].An partially ordered set, or poset for short, ⟨T,

The order type of that setis called the height of t in T, denoted by ht(t). We will not distinguish a tree fromits base set.

For every ordinal α, let Tα, the α–th level of T, = {t ∈T : ht(t) = α}and T ↾α = Sβ<α Tβ. Let ht(T), the height of T, is the smallest ordinal α such thatTα = ∅.

By a branch of T we mean a linearly ordered subset of T which intersectsevery nonempty level of T. Let B(T) be the set of all branches of T. T ′ is called asubtree of T if T ′ ⊆T, ω1 and for every α ∈ω1, |Tα| < ω1.

An ω1–tree is called a Jech–Kunentree if ω1 < |B(T)| < 2ω1.T. Jech in [Je1] constructed by forcing a model of CH plus 2ω1 > ω2, in whichthere is a Jech–Kunen tree.

In fact, it is a Kurepa tree with fewer than 2ω1–manybranches. Later, K. Kunen [K1] found a model of CH plus 2ω1 > ω2, in which thereare neither Kurepa trees nor Jech–Kunen trees.

In his paper he gave an equivalentform of Jech–Kunen trees in terms of compact Hausdorffspaces. The detailed proofcan be found in [Ju, Theorem 4.8].The second author in [Ji1] started discussing the differences between the existence ofKurepa trees and the existence of Jech–Kunen trees.

He showed that it is independentof CH plus 2ω1 > ω2 that there exists a Kurepa tree which has no Jech–Kunen11980 Mathematics Subject Classification (1985 Revision). Primary 03E35.2The research of the first author was partially supported by the United States–Israel BinationalScience Foundation, publ.

469.1

2subtrees.He also showed that it is independent of CH plus 2ω1 > ω2 that thereexists a Jech–Kunen tree which has no Kurepa subtrees. In his proofs some stronglyinaccessible cardinals were assumed and later, Kunen eliminated the large cardinalassumption for one of the proofs.In [SJ], the both authors answered a question of [Ji2] by proving that, assumingthe existence of one inaccessible cardinal, it is consistent with CH plus 2ω1 > ω2 thatthere exist Jech–Kunen trees and there are no Kurepa trees.In [Ji2], the second author proved that, assuming the existence of two inaccessiblecardinals, it is consistent with CH plus 2ω1 > ω2 that there exist Kurepa trees andthere are no Jech–Kunen trees.Since the consistency of the nonexistence of Jech–Kunen trees implies the consis-tency of the existence of an inaccessible cardinal [Ju, Theorem 4.10], we have to useat least one inaccessible cardinal to build a model of CH plus 2ω1 > ω2, in whichthere are Kurepa trees but there are no Jech–Kunen trees.

The question whetherone inaccessible cardinal is enough was asked in [Ji2]. In this paper, we will give apositive answer to the question.

We also discover that the same techniques can beused to answer another question in [Ji2] by constructing a model of CH plus 2ω1 = ω4,in which only the Kurepa trees with ω3–many branches exist.First let’s look at the second author’s original idea in [Ji2] to construct a model ofCH plus 2ω1 > ω2, in which there are Kurepa trees and there are no Jech–Kunen trees,by using two inaccessible cardinals. Let κ1 < κ2 be two strongly inaccessible cardinalsin a model M. First, Jin collapses κ2 down to κ+1 by forcing with a < κ1–support L´evycollapsing order.

Next, he collapses κ1 down to ω1 by forcing with a finite supportL´evy collapsing order. This step will create a very homogeneous Kurepa tree.

Thenhe force with that Kurepa tree λ–many times for some regular cardinal λ which isgreater than κ2. In the resulting model, that Kurepa tree has λ–many branches andλ = 2ω1.

In that model there are no Jech–Kunen trees.If we want to obtain the same result by using only one inaccessible cardinal, wemay have to find a way to create a homogeneous ω1–tree with every level countable,without the assistance of large cardinals. Then the questions arise.

First, how can wecreate the desired tree? Second, can we force with that tree for multiple times (withcountable supports) without collapsing ω1.

(Note that a normal ω1–tree with everylevel countable is never ω1–closed.)

3In this paper, we construct a homogeneous generic ω1–tree with every level count-able by forcing with an ω1–closed poset, whose elements are countable homogeneousnormal subtrees of ⟨2<ω1, ⊆⟩. The generic tree is, in fact, a Suslin tree.

Then we forcewith that generic tree λ–many times with countable supports. We will prove thatthis two–step forcing adds no new countable sequences of ordinals, hence it will notcollapse ω1.

We will also prove that if the ground model is Silver’s model (see [K2,pp. 259]), then in the final model there are no Jech–Kunen trees.Before proving our results we need more notations and definitions.A tree T is called normal if,(1) every t ∈T, which is not in the top level of T, has at least two immediatesuccessors,(2) for every limit ordinal α < ht(T) and every B ∈B(T ↾α), there exist at mostone least upper bound of B in T,(3) for every t ∈T and α such that ht(t) < α < ht(T), there exists t′ ∈Tα suchthat t

Let T be a tree and B ⊆T be a totally ordered subset of T.Then S B is the only candidate for the least upper bound of B in T.Let α ∈ω1 and s, t ∈2α. We define a map Fs,t from 2<ω1 to 2<ω1.

Let u ∈2β forsome β < ω1. The domain of Fs,t(u) is β and for every γ < β, if γ < α, then letFs,t(u)(γ) = u(γ) + t(γ) −s(γ) (mod 2),otherwise let Fs,t(u)(γ) = u(γ).Lemma 1.

Fs,t(s) = t, Fs,t(t) = s and Fs,t is an isomorphism from ⟨2<β, ⊆⟩to⟨2<β, ⊆⟩for any β ≤ω1.Proof:Trivial.✷A normal tree T is called homogeneous if for any α < ht(T), for any s, t ∈Tα,Fs,t↾T is an isomorphism from T to T.LetPhom = {T : T is a countable homogeneous normal subtree of ⟨2<ω1, ⊆⟩}be a poset ordered by letting T < T ′ iffht(T ′) < ht(T) and T ′ = T ↾ht(T ′).

4Lemma 2. Let T be a totally ordered subset of Phom.

Then S T is a homogeneoustree. Moreover, if T is countable, then S T ∈Phom.Proof:Trivial.✷Remark:Above lemma says that Phom is ω1–closed, which means that every count-able decreasing sequence in Phom has a lower bound in Phom.Lemma 3.

Let T ∈Phom and ht(T) = α for some limit ordinal α < ω1. Let C be acountable subset of B(T).

Then there exists T ∈Phom such that T < T and for everyC ∈C, S C ∈T α.Proof:Without loss of generality, we can assume that for every t ∈T, there existsC ∈C such that t ∈C. (This will guarantee the normality of the resulting tree.) Wenow construct inductively a sequence of countable trees ⟨Tn : n ∈ω⟩such that:(1) T0 = T S{S C : C ∈C},(2) for every n ∈ω, ht(Tn) = α + 1 and(3) for every n ∈ω,Tn+1 = Tn[{Fs,t(u) : s, t ∈Tn, ht(s) = ht(t) and u ∈(Tn)α}.Note that if I is an isomorphism from T to T, then for every B ∈B(T), I[B] ∈B(T).Let T = Sn∈ω Tn.

It is obvious that T is countable and for any s, t ∈T such thatht(s) = ht(t), Fs,t is an isomorphism from T to T. Hence T ∈Phom, T < T and forevery C ∈C, S C ∈T0 ⊆T.✷Next we discuss forcing method. For the terminology and basic facts of forcing,see [K2] and [Je2].

We always assume the consistency of ZFC and let M be alwaysa countable transitive model of ZFC. In the forcing arguments, we always let ˙a be aname of a.

For every element a in the ground model, we may use a itself as its name.Let P be a poset in a model M, ˙a be a P–name for a and G be a P–generic filter overM. Then ˙aG is the value of ˙a in M[G] (see [K2, pp.

189] for the definition of ˙aG).Let I, J be two sets. LetFn(I, J, ω1) = {p : p ⊆I × J is a function and |p| < ω1}be a poset ordered by reverse inclusion.

Let I be a subset of a cardinal κ. LetLv(I, ω1) =

5{p : p ⊆(I × ω1) × κ is a function, |p| < ω1 and ∀⟨α, β⟩∈dom(p)(p(α, β) ∈α)}be a poset ordered by reverse inclusion. The poset Lv(κ, µ) for some regular cardinalsκ > µ is usually called a < µ–support L´evy collapsing order.

Let T be a tree and Ibe an index set. For a function p from I to T, let supt(p), the support of p, be theset {i ∈I : p(i) ̸= ∅}.

LetP(T, I, ω1) = {p : p ∈T I, |supt(F)| < ω1}.For any p, p′ ∈P(T, I, ω1), define p ≤p′ ifffor every i ∈I, p′(i) ≤T p(i). Let R be aposet and ˙T be an R–name for a tree T. LetP( ˙T, I, ω1) = { ˙q : ˙q ∈( ˙T)I, |supt( ˙q)| < ω1}.Then P( ˙T, I, ω1) is an R–name for the poset P(T, I, ω1).

Let Q = P(T, I, ω1) (orP( ˙T, I, ω1)) and J ⊆I. We denote Q ↾J for the set {p ↾J : p ∈Q}.

If H is aQ–generic filter, then let HJ = {p↾J : p ∈H}.Lemma 4. Let T be an ω1–tree and P be an ω1–closed poset in a model M. Let Gbe a P–generic filter over M. Assume that there exists a branch of T in M[G] ∖M.Then T is neither a Kurepa tree nor a Jech–Kunen tree in M. Moreover, there existsan isomorphic embedding from ⟨2<ω1, ⊆⟩into T.Proof:See [K2, pp.

260] and [Ju, Theorem 4.9].✷Lemma 5. Let M be a model, P = (Phom)M and G be a P–generic filter over M.Let TG = S G. Then the generic tree TG is a homogeneous normal ω1–tree with everylevel countable.

In fact, TG is a Suslin tree.Proof:See [Je2, Theorem 48] for the proof. The homogeneity of TG follows fromLemma 2.✷Lemma 6.

Let M be a model, I be an index set in M, P = (Phom)M, T ˙G be P–namefor the P–generic tree TG, and ˙Q = P(T ˙G, I, ω1), which is a P–name for P(TG, I, ω1).Let G ∗H be a P ∗˙Q–generic filter over M. Then Mω T M[G ∗H] ⊆M.Proof:Suppose that there is an f ∈Mω T M[G ∗H] such that f ̸∈M.Let⟨p, ˙q⟩∈P ∗˙Q such that⟨p, ˙q⟩⊩˙f ∈Aω ∖Mfor some A ∈M.

6We now want to construct a sequence ⟨⟨pn, ˙qn⟩∈P ∗˙Q : n ∈ω⟩in M such that forevery n ∈ω,(1) ⟨pn+1, ˙qn+1⟩≤⟨pn, ˙qn⟩≤⟨p, ˙q⟩,(2) ∃an ∈A (⟨pn, ˙qn⟩⊩˙f(n) = an),(3) ∀i ∈supt( ˙qn) ∃tn(i) ∈pn ∖pn−1 (pn ⊩˙qn(i) = tn(i)).The contradiction follows from the construction. Let pω = Sn∈ω pn.

For everyi ∈Sn∈ω supt( ˙qn), letCi = {t ∈pω : ∃n ∈ω, t < tn(i)}.By (3), Ci ∈B(pω). By Lemma 3, there is pω ∈P, pω ≤pω such that S Ci ∈pω.Define ˙qω from I to T ˙G such that ˙qω(i) = S Ci if i ∈Sn∈ω supt( ˙qn) and ˙qω(i) = ∅otherwise.

(In fact, q is in M.) Then ⟨pω, ˙qω⟩≤⟨pn, ˙qn⟩for every n ∈ω. Hence, forevery n ∈ω,⟨pω, ˙qω⟩⊩˙f(n) = an.This contradicts f ̸∈M.Assume that we have already had ⟨pn, ˙qn⟩for every n ≤m.First, let ⟨r, ˙s⟩≤⟨pm, ˙qm⟩and am+1 ∈A such that⟨r, ˙s⟩⊩˙f(m + 1) = am+1.For every i ∈supt( ˙s),r ⊩∃α ∈ω1 ( ˙s(i) ∈2α).Then there exist α ∈ω1 and r′ ≤r such thatr′ ⊩˙s(i) ∈2α.Since P is ω1–closed andr′ ⊩The domain of ˙s(i) is α, a countable ordinal.then there exist t(i) ∈2α and r′′ ≤r′ such thatr′′ ⊩˙s(i) = t(i).Let r′′′ ≤r′′ such that ht(r′′′) > max{α, ht(pm)}.

Thenr′′′ ⊩˙s(i) = t(i) ∈r′′′because ⊩˙s(i) ∈T ˙G.

7Since supt( ˙s) is countable and P is ω1–closed, then we can find pm+1 ≤r′′′ suchthat∀i ∈supt( ˙s) ∃α < ht(pm+1) ∃t(i) ∈(pm+1)α (pm+1 ⊩˙s(i) = t(i)).Let tm+1(i) ∈pm+1 ∖pm such that t(i) ≤tm+1(i) and define ˙qm+1(i) = tm+1(i) ifi ∈supt( ˙s) and ˙qm+1(i) = ∅otherwise. This ends the construction and the sequencewe have constructed does obviously satisfy (1), (2) and (3).✷Remark:The poset P ∗˙Q in Lemma 6 is, in fact, strategically complete.

Let Rbe any poset. Two players, I and II, choose from R successively the members of adecreasing sequencea0 ≥b0 ≥a1 ≥b1 ≥· · · ≥an ≥bn ≥· · · .I chooses the an’s and II chooses the bn’s.

II wins the game if and only if the sequencehas a lower bound in R. We call R strategically complete if II has a winning strategy.It has been shown that R is strategically complete if and only if there exists a posetS such that R × S has a dense subset which is ω1–closed (see [Je3, pp. 90]).Theorem 7.

Assuming the existence of an inaccessible cardinal, it is consistent withCH plus 2ω1 > ω2 that there exist Kurepa tree but there are no Jech–Kunen trees.Proof:Let M be a model of GCH, κ be an inaccessible cardinal and λ > κ be aregular cardinal in M. In M, let P1 = Lv(κ, ω1), P2 = Phom, T ˙G2 be a P2–name for theP2–generic tree TG2 = S G2, where G2 is a P2–generic filter, and ˙Q = P(T ˙G2, λ, ω1).Let G1 × (G2 ∗H) be a P1 × (P2 ∗˙Q)–generic filter over M.We will show thatM[G1 × (G2 ∗H)] = M[G1][G2 ∗H] is the model we are looking for.Claim 7.1Mω T M[G1][G2 ∗H] ⊆M.Proof of Claim 7.1 :By Lemma 6, Mω T M[G2∗H] ⊆M. This implies that P1is still ω1–closed in M[G2 ∗H].

Hence (M[G2 ∗H])ω T M[G2 ∗H][G1] ⊆M[G2 ∗H].So for every f ∈Mω T M[G2 ∗H][G1], f is in M[G2 ∗H] and hence, f is in M. Theclaim is true because M[G1][G2 ∗H] = M[G2 ∗H][G1].Claim 7.2P1 × (P2 ∗˙Q) has the κ–c.c..Proof of Claim 7.2 :A poset R is called λ–centered for some regular cardinalλ if for any subset S ⊆R and |S| ≥λ, there exists S′ ⊆S, |S′| ≥λ, such that anytwo elements in S′ are compatible. By a simple ∆–system lemma argument, we can

8show that P1 is κ–centered. Since |P2| = ω1, then |T ˙G2| ≤(|P2|ω1)ω1 = ω2.

Again bya simple ∆–system lemma argument, we can show that P2 ∗˙Q is κ–centered. In fact,it is also ω3–centered.

Hence P1 × (P2 ∗˙Q) is κ–centered, which implies the κ–c.c..Remark:By Claim 1 and Claim 2 and the fact that M[G1] |= [CH +2ω1 = ω2 = κ],we know that ω1 and all the cardinals greater than or equal to κ in M is preservedin M[G1][G2 ∗H]. We also know that M[G1][G2 ∗H] |= [CH + 2ω1 = λ > κ].Claim 7.3TG2 is a Kurepa tree with λ–many branches in M[G1][G2 ∗H].Proof of Claim 7.3 :It is obvious that TG2 is an ω1–tree with every levelcountable (in fact, it is a Suslin tree in M[G2]).

In M[G1][G2], ˙QG2 = P(TG2, λ, ω1)is a countable support (note that no new countable sequences of ordinals are added)product of λ–many copies of TG2. Then forcing with Q will add at least λ–many newbranches to TG2.

Hence λ ≤|B(TG2)| ≤2ω1 = λ.Claim 7.4There are no Jech–Kunen trees in M[G1][G2 ∗H].Proof of Claim 7.4 :Suppose that T is a Jech–Kunen tree in M[G1][G2 ∗H].Since |T| = ω1, then there exists a cardinal θ < κ and a subset I of λ with |I| ≤ω2such that T ∈M[G′1][G2 ∗HI], where G′1 = G1T Lv(θ, ω1) and HI = H T Q↾I. Thisis true because P1 has the κ–c.c.

and P2 ∗˙Q has the ω3–c.c.. In M[G′1][G2 ∗HI],2ω1 < κ, so that there exists a branch b of T in M[G1][G2 ∗H] ∖M[G′1][G2 ∗HI].Since Lv(κ ∖θ, ω1) in M is still ω1–closed in M[G′1][G2 ∗HI] and T is a Jech–Kunentree in M[G1][G2 ∗H], then by Lemma 4, b ̸∈M[G1][G2 ∗HI].Let M′ = M[G1][G2 ∗HI].

We now work in M′. In M′, Q↾(λ ∖I) has the ω1–c.c..Then there exists J ⊆λ ∖I with |J| = ω1 in M′ such that b ∈M′[HJ].

Let r ∈HJbe such thatr ⊩Q↾J ∃b ∈B(T) ∖M′.Since TG2 is homogeneous (here we use the homogeneity of the tree), then we canassume that⊩Q↾J ∃b ∈B(T) ∖M′.By the maximal principle we can find a Q↾J–name ˙b for b in M′ such that⊩Q↾J ˙b ∈B(T) ∖M′.Since b ̸∈M′, then in M′, the sentence Φ(Q↾J, T, ˙b) is true, where Φ(X, Y, Z) is∀s ∈X ∃s0, s1 ≤s ∃α ∈ω1 ∃t0, t1 ∈Yα, t0 ̸= t1, (si ⊩ti ∈Z) for i = 0, 1.

9In M[G1][G2 ∗H] T has less than λ–many branches, so there exists µ < λ such thatI S J ⊆µ and every branch of T in M[G1][G2 ∗H] is already in M′[Hµ∖I]. LetJ′ ⊆λ ∖µ be such that |J′| = |J| and let π be the natural isomorphism from Q↾Jto Q↾J′.

Then in M′⊩Q↾J′ π∗(˙b) ∈B(T)is true andM′ |= Φ(Q↾J′, T, π∗(˙b)),where π∗is the map from Q ↾J–names to Q ↾J′–names induced by π (see [K2, pp.222] for the definition of π∗).Subclaim 7.4.1M′[Hµ∖I] |= [Φ(Q↾J′, T, π∗(˙b)) and ⊩Q↾J′ π∗(˙b) ∈B(T)].Proof of Subclaim 7.4.1 :Let HJ′ be a Q ↾J′–generic filter over M′[Hµ∖I].Then HJ′ is also a Q↾J′–generic filter over M′. Hence in M′[HJ′], (π∗(˙b))HJ′ ∈B(T).If si ∈HJ′, then ti ∈(π∗(˙b))HJ′ is also true in M′[HJ′].In M′[HJ′], forcing with Q↾(µ∖I) will not change the truth of the above sentences.Hence in M′[HJ′][Hµ∖I] = M′[Hµ∖I][HJ′], (π∗(˙b))HJ′ ∈B(T) and ti ∈(π∗(˙b))HJ′ arealso true.

This implies thatM′[Hµ∖I] |= [Φ(Q↾J′, T, π∗(˙b)) and ⊩Q↾J′ π∗(˙b) ∈B(T)].This ends the proof of Subclaim 7.4.1.Since forcing with Q↾J′ will not add any new branches of T, then B = (π∗(˙b))HJ′is already in M′[Hµ∖I]. In M′[Hµ∖I], letD = {r ∈Q↾J′ : ∃t ̸∈B (r ⊩Q↾J′ t ∈π∗(˙b))}.Then D is dense in Q ↾J′ because Φ(Q ↾J′, T, π∗(˙b)) is true in M′[Hµ∖I].

If r0 ∈D T HJ′, then r0 ⊩π∗(˙b) ̸= B. This contradicts (π∗(˙b))HJ′ = B.✷Theorem 8.

Assuming the existence of an inaccessible cardinal, it is consistent withCH plus 2ω1 = ω4 that only the Kurepa trees with ω3–many branches exist.Proof:Let’s follow the notation of the proof of Theorem 7. Let λ = κ+ in M. LetP3 = Fn(κ++, 2, ω1) = Fn(ω4, 2, ω1)in M[G1][G2 ∗H] (note that P3 is absolute with respect to M and M[G1][G2 ∗H]).Let G3 be a P3–generic filter over M[G1][G2 ∗H].

In M[G1][G2 ∗H][G3], the numberof the branches of TG2 is λ = κ+ = ω3 by Lemma 4.

10Let T be any ω1–tree in M[G1][G2 ∗H][G3]. Then there exists K ⊆κ++ with|K| = ω1 such that T ∈M[G1][G2 ∗H][G′3], where G′3 = G3T Fn(K, 2, ω1).If |B(T)| = ω4 in M[G1][G2 ∗H][G3], then forcing with Fn(κ++ ∖K, 2, ω1) will addnew branches to T. This implies T is not a Kurepa tree by Lemma 4.If |B(T)| = ω2 in M[G1][G2 ∗H][G3], then by Lemma 4, T is already a Jech–Kunentree with ω2–many branches in M[G1][G2 ∗H][G′3].

Without loss of generality we canassume that K = ω1. SoM[G1][G2 ∗H][G′3] |= “There exists a Jech–Kunen tree with ω2–many branches”.ButM[G1][G2 ∗H][G′3] = M[G′3][G1][G2 ∗H] = M[G1][G2 ∗H],where M = M[G′3].

By the same proof of Theorem 7, we can also show that thereare no Jeck–Kunen trees in M[G1][G2 ∗H], a contradiction.✷References[Je1] T. Jech, “Trees”, The Journal of Symbolic Logic, 36 (1971), pp. 1—14.

[Je2], “Set Theory, Academic Press, New York, 1978.[Je3]. “Multiple Forcing, Cambridge University Press, 1986.

[Ji1] R. Jin, “Some independence results related to the Kurepa tree”, Notre Dame Journal of FormalLogic, 32, No 3 (1991), pp. 448—457.

[Ji2], “A model in which every Kurepa tree is thick”, Notre Dame Journal of Formal Logic,33, No 1 (1992), pp. 120—125.

[Ju] I. Juh´asz, “Cardinal functions II”, pp. 63—110 in Handbook of Set Theoretic Topology,ed.

by K. Kunen and J. E. Vaughan, North–Holland, Amsterdam, 1984. [K1] K. Kunen, “On the cardinality of compact spaces”, Notices of The American MathematicalSociety, 22 (1975), 212.

[K2], “Set Theory, an introduction to independence proofs”, North–Holland, Amsterdam,1980. [S1] S. Shelah, “Proper Forcing”, Springer–Verlag, 1982.

[S2], New version of “Proper Forcing”, to appear. [SJ] S. Shelah and R. Jin, “A model in which there are Jech–Kunen trees but there are no Kurepatrees”, preprint.

[T] S. Todorˇcevi´c, “Trees and linearly ordered sets”, pp. 235—293 in Handbook of Set TheoreticTopology, ed.

by K. Kunen and J. E. Vaughan, North–Holland, Amsterdam, 1984.Institute of Mathematics,The Hebrew University,Jerusalem, Israel.Department of Mathematics,Rutgers University,

11New Brunswick, NJ, 08903, USA.Department of Mathematics,University of Wisconsin,Madison, WI 53706, USA.Sorting: The first two addresses are the first author’s; the last one is the secondauthor’s.

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