Additivity of the two-dimensional Miller ideal

Otmar Spinas Sonja Thiele

May 5, 2010

Abstract

Let J (M2) denote the σ-ideal associated with two-dimensional Miller forcing.

We show that it is relatively consistent with ZFC that the additivity of J (M2)

is bigger than the covering number of the ideal of the meager subsets of ωω. We

also show that Martin’s Axiom implies that the additivity of J (M2) is 2ω.

Finally we prove that there are no analytic infinite maximal antichains in any

finite product of P(ω)/fin.

Introduction

For several of the classical tree forcings, such as Sacks forcing S, Laverforcing L and Miller forcing M, there are corresponding σ-ideals. Al-ready in 1935 Marczewski [M] studied the σ-ideal J (S) := {X ⊆ ω2 | ∀p ∈S∃q ∈ S (q ≤ p ∧ [q] ∩ X = ∅)}. Later, for example, Velickovic, Ju-dah, Miller and Shelah [Ve, JuMiSh] continued this investigation of J (S).Brendle, Goldstern, Johnson, Judah, Miller, Repicky, Shelah and Spinas[Br, JuMiSh, GoReShSp, GoJSp, Sp1] studied the σ-ideals J (Q) := {X ⊆ωω | ∀p ∈ Q ∃q ∈ Q (q ≤ p ∧ [q]∩X = ∅)} for Q ∈ {M,L}. For every idealI one can define the additivity add(I) and the covering number cov(I) ofI. The additivity is the minimal cardinality of a subset of the ideal whoseunion is not in the ideal, and the covering number is the least cardinalityof a subset of the ideal whose union is the whole set on which the ideal isdefined. We always assume that all singletons belong to I. Clearly, thefollowing inequalities hold for every σ-ideal I on ω2 or ωω:

ω1 ≤ add(I) ≤ cov(I) ≤ 2ω

and a typical question is whether any of the above inequalities could consis-tently be strict. Judah, Miller and Shelah built a model for add(J (S)) <cov(J (S)) and Goldstern, Repicky, Shelah and Spinas built models foradd(J (Q)) < cov(J (Q)) for Q ∈ {M,L}. In [Sp1, Sp2, Sp3] Spinasstarted to develop a combinatorial theory for the two-dimensional Millerforcing M2, where Mn for n ∈ ω \ {0} consists of all n-tuples of superper-fect trees and carries the coordinatewise ordering. It turned out that M2

is much closer to M than it is to M3. On the one hand, for example, both,M and M2 do not add a Cohen-real (see [Mi, Sp2]), while on the otherhand, by an unpublished result of Shelah, M3 does add a Cohen-real.

1

2

Jossen and Spinas [JoSp] started the investigation of the n-dimensionalideals

J (Qn) := {X ⊆ (ωω)n | ∀(p0, . . . , pn−1) ∈ Qn ∃(q0, . . . , qn−1) ∈ Qn

((q0, . . . , qn−1) ≤ (p0, . . . , pn−1) ∧ ([q0]× . . .× [qn−1]) ∩X = ∅)}

for Q ∈ {L,M}. They showed that only the two-dimensional Millerideal J (M2) is a σ-ideal, but neither the higher dimensional Miller idealsJ (Mn) for n ≥ 3 nor the Laver ideals J (Lm) for m ≥ 2 are σ-ideals,and thus the additivity of the higher dimensional ideals is ω. There-fore, it is interesting to pay more attention to J (M2). Analogously tothe one-dimensional case, Jossen and Spinas [JoSp] have built a model foradd(J (M2)) < cov(J (M2)). It is obtained – starting with a model of ZFCand the continuum hypothesis – by a countable support iteration of M2

of length ω2. Since no Cohen-reals are added, cov(M) < cov(J (M2))holds in their model, hence, since the continuum equals ω2, we havecov(M) = add(J (M2)) = ω1 there, where M is the ideal of all mea-ger subsets of ωω.Here we prove the consistency of cov(M) < add(J (M2)) (Theorem 1.2).The natural forcings to increase the additivity numbers are called amoebaforcings. It is well known, that in order to increase cov(M) Cohen realsmust be added. In general, given any definable tree forcing Q, an amoebaforcing A(Qn) for Qn is a forcing adding some (p0, . . . , pn−1) ∈ Qn suchthat every n-tuple of branches (x0, . . . , xn−1) ∈ [p0] × . . . × [pn−1] is Qn-generic. We construct an amoeba forcing A(M2) for M2 that does notadd Cohen-reals. Our construction is inspired by the work of Spinas [Sp1],where, in the one-dimensional situation, amoeba forcings for L andM havebeen constructed, which have the Laver property. The Laver property is acombinatorial property ruling out that Cohen-reals are added. By a resultof Shelah, the Laver property is preserved under countable support forcingiterations. Our A(M2) will also have the Laver property, and hence we canincrease add(J (M2)) without adding Cohen-reals. Let us remark that thenatural amoeba forcings for L, M and M2 that come to one’s mind are notsuitable for our purpose, as they add lots of Cohen-reals (see [Sp1]).Similar ideas as in [Sp1] have been used independently in [LoShVe], whereimplicitly an amoeba forcing for Sacks forcing has been constructed whichdoes not add Cohen-reals.Judah, Miller, Shelah and Velickovic [JuMiSh, Ve] have independentlyshown that Martin’s Axiom does not imply that add(J (S)) = 2ω. In con-trast to this, it is possible to blow up add(J (L)) and add(J (M)) by aforcing fulfilling the countable chain condition. In fact, in [JuMiSh] it wasshown that Martin’s Axiom (t = 2ω is enough, where t is the tower num-ber) implies add(J (L)) = 2ω. The analogous result for M was claimedas well, but the proof was faulty. It was later corrected in [GoJSp]. Herewe show that Martin’s Axiom for σ-centered forcings (MA(σ-centered))implies that add(J (M2)) = 2ω (Theorem 2.1) by combining the ideas ofthese two papers with the combinatorial properties of M2. As a corollaryit turns out that under the assumption that MA(σ-centered) is true forc-ing with M2 does not collapse cardinals. By using the same methods wecan prove that MA(σ-centered) implies Martin’s Axiom for the forcing M2

(Theorem 2.17).In chapter 3 we will treat a completely different problem with similarmethods, in the sense that carefully chosen finite products of Mathias

3

forcing are used that have the Laver property. In [Ma], Mathias intro-duced his famous Mathias forcing M(U) restricted to a Ramsey ultrafil-ter U , i.e. M(U) := {(s, S) ∈ [ω]<ω × [ω]ω | (max(s) < min(S) or s =∅) and S ∈ U} and the ordering is defined by (s, S) ≤ (t, T ) if and onlyif s ⊇ t and S ∪ (s \ t) ⊆ T . The analysis of this forcing led him, amongmany other things, to the result that there are no analytic infinite maxi-mal almost disjoint families in P(ω). Note that maximal almost disjointfamilies in P(ω) correspond to maximal antichains in the Boolean alge-bra P(ω)/fin. Here we investigate infinite maximal antichains in finiteproducts of P(ω)/fin. Using similar but simpler ideas as in [LoShVe] and[Sp1], Shelah and Spinas [ShSp2] introduced an n-dimensional version ofMathias forcing with the Laver property. Combining their analysis of thisforcing with some combinatorial facts about infinite maximal antichainsin (P(ω)/fin)n we prove that for every n ∈ ω \ {0, 1} there are no analyticinfinite maximal antichains in (P(ω)/fin)n (Theorem 3.1). Curiously, indimension ω there exists a perfect (thus closed) partition, as the follow-ing example shows: choose a ∈ [ω]ω with ω \ a ∈ [ω]ω, too, and considerA := {(ai)i∈ω | ∀i ∈ ω (ai ∈ {a, ω \ a})}.

0 Preliminaries

We fix our notation: For a set A let [A]<ω denote the collection of all finitesubsets of A and let [A]ω denote the collection of all countably infinitesubsets of A. <ωA denotes the set of all functions s : n → A for somen ∈ ω, ωA is the set of all functions f : ω → A. For s ∈ <ωA for any set Awe write |s| = n (the length of s) if s : n → A. For a ∈ A let s_〈a〉 be thefunction s ∪ {(|s|, a)}.Trees: A set p ⊆ <ωω is called a tree if for every σ ∈ p and τ ⊆ σ we haveτ ∈ p. Given a tree p ⊆ <ωω, the set of all infinite branches through pis denoted by [p]. For σ ∈ p let (p)σ be the subtree of p consisting of allν ∈ p which are comparable with σ. For σ ∈ p let succp(σ) designate theset of all extensions of the form σ_〈n〉, for n ∈ ω, with σ_〈n〉 ∈ p. Callσ ∈ p a splitnode if |succp(σ)| > 1, an infinite splitnode if |succp(σ)| isinfinite. The set of all infinite splitnodes of a tree p is denoted by split(p).However, if p is a finite tree we also use split(p) to denote the set of allsplitnodes of p. For σ ∈ split(p) define Succp(σ) := {τ ∈ split(p) |σ $τ ∧ ∀ρ(σ $ ρ $ τ ⇒ ρ /∈ split(p))}, the set of all direct successors of σ inthe tree sense in split(p). By st(p) we denote the stem of p. This is theshortest splitnode in p. A tree p ⊆ <ωω is called a Laver tree if it has astem st(p) and for every σ ∈ p with σ ⊇ st(p) we have σ ∈ split(p). Laverforcing is the set of all Laver trees L ordered by inclusion. A tree p ⊆ <ωωis called a Miller or superperfect tree, if it has a stem and for every σ ∈ pthere exists an extension τ % σ in p which is an infinite splitnode. LetM denote the set of all superperfect trees p with the additional propertythat every splitnode of p is an infinite one. Miller forcing is M orderedby inclusion. Notice that M is dense in the forcing sense in the set of allsuperperfect trees. During the whole paper, if we write p is “superperfect”or “Miller” we mean p ∈M. LetM2 and more generallyMn denote the setof all n-tuples of superperfect trees carrying the coordinatewise ordering.We write p ≤0 q if p ≤ q and additionally we have st(p) = st(q) and(p′, q′) ≤0 (p, q) if (p′, q′) ≤ (p, q) and we have that st(p′) = st(p) andst(q′) = st(q) hold.

4

A set S ⊆ ωω is called superperfect, if there exists a superperfect treep ∈M with S = [p]. For (p, q) ∈M2 by [p]×+ [q] we denote the upper halfof the superperfect rectangle [p] × [q], i.e. the set of all (x, y) ∈ [p] × [q]with x(|st(p)|) < y(|st(q)|). Similarly, [p] ×− [q] denotes the lower half of[p]× [q].We use the following well-ordering ≺ of <ωω:

σ ≺ τ :⇔ max{|σ|, max ran(σ)} < max{|τ |,max ran(τ)}∨ (max{|σ|,max ran(σ)} = max{|τ |, max ran(τ)} ∧ |σ| < |τ |)∨ (max{|σ|,max ran(σ)} = max{|τ |, max ran(τ)} ∧ |σ| = |τ |

∧σ precedes τ lexicographically) .

Let E : ω → <ωω be the order preserving enumeration of (<ωω,≺) andwrite #ρ = n if E(n) = ρ, so we have #ρ ≥ max{|ρ|, max ran(ρ)}.We repeat the following definition and fact from [Sp2]:

Definition 0.1 Let (σ, τ) ∈ (<ωω)2 and (x, y) ∈ (ωω)2 with σ ⊆ x andτ ⊆ y. We say that (x, y) oscillates infinitely often above (σ, τ) if thereexists a strictly increasing sequence (ki)i∈ω in ω such that the followinghold for all n ∈ ω:

k0 = |τ |, k1 > |σ|;k2n = min{i ∈ ω | y(i) > #x¹ k2n+1};k2n+1 = min{i ∈ ω |x(i) > #y ¹ k2n+2};k2n+1 < y(k2n) < k2n+2 < x(k2n+1) < k2n+3.

The sequence(σ, τ, x¹ k1, y ¹ k2, x¹ k3, y ¹ k4, . . .)

is called the typeσ,τ -sequence of the pair (x, y).

Fact 0.2 [Sp2] For every (p, q) ∈ M2 there exists (p′, q′) ≤0 (p, q) suchthat for every (x, y) ∈ [p′] ×+ [q′], (x, y) oscillates infinitely often above(st(p), st(q)). Moreover, if (µ0, ν0, µ1, ν1, . . .) is the typest(p),st(q)-sequenceof (x, y), we have {µn |n ∈ ω} ⊆ split(p′) and {νn |n ∈ ω} ⊆ split(q′). 2

Hence for every (x, y) ∈ [p′]× [q′], (p′, q′) as in fact 0.2, there is a uniqueassociated sequence (ki)i∈ω in ω which is determined solely by (x, y) and(st(p), st(q)) =: (σ, τ). Define the 0-type pair of the pair of branches (x, y)by tpσ,τ -0-pair(x, y) := (σ, τ), and define tpσ,τ -(2n + 1)-pair(x, y) = (x ¹k2n+1, y ¹ k2n) and tpσ,τ -(2n + 2)-pair(x, y) = (x ¹ k2n+1, y ¹ k2n+2) forevery n ∈ ω. Using this we can define a partial function

tpp′,q′σ,τ : (<ωω)2 → ω

by letting tpp′,q′σ,τ (µ, ν) = n if and only if there exists (x, y) ∈ [p′] ×+ [q′]

such that (µ, ν) = tpσ,τ -n-pair(x, y).For (µ, ν) ∈ p′ × q′ with tpp′,q′

σ,τ (µ, ν) = 2n there exists a unique sequence(µ0, ν0, . . . , µn, νn), which is the initial sequence of length 2n of the typeσ,τ -sequence of (x, y) for some infinitely oscillating (x, y) ∈ [p′] ×+ [q′] over(σ, τ). We call this sequence the typeσ,τ -sequence of (µ, ν) and definetpσ,τ -i-pair(µ, ν) = tpσ,τ -i-pair(x, y) for all i ≤ 2n. And similarly for(µ, ν) ∈ p′ × q′ with tpp′,q′

σ,τ (µ, ν) odd.If (σ, τ) or (p′, q′) are clear from context, we omit them in the above

5

notation.For (p, q) ∈M2 and (σ, τ) ∈ p× q let

TPσ,τ (p, q) := dom(tpp,qσ,τ ) ∩ (split((p)σ)×+ split((q)τ )),

the set of all type pairs of (p, q), and for n ∈ ω let TPnσ,τ (p, q) denote the

set of all type pairs of (p, q) of type n. For (µ, ν) ∈ TPnσ,τ with n even let

Sopp,qσ,τ (µ, ν) := {µ′ ∈ split(pµ) | tpp,q

σ,τ (µ′, ν) = n + 1 ∧tpσ,τ -n-pair(µ′, ν) = (µ, ν)},

the set of all possible successor oscillation points, and similarly, we defineSopp,q

σ,τ (µ, ν), if n is odd.Note that for (p′, q′) as in fact 0.2 and (u, v) ≤ (p′, q′) and (σ, τ) ∈ u × vwe in general have TPσ,τ (p′, q′) ∩ (u × v) 6= TPσ,τ (u, v), but there ex-ists (u′, v′) ≤0 (u, v) such that for almost all n ∈ ω and every (x, y) ∈[u′]×+ [v′] we have tpσ,τ -n-pair(x, y) ∈ TPσ,τ (u′, v′) (and therefore tpσ,τ -n-pair(x, y) ∈ split(u′)× split(v′)).

For the rest of the paper we always assume that we work with elements(p, q) ∈ M2, which have the property of (p′, q′) of fact 0.2. By the fact,the set of all such (p, q) is dense in M2, so with this partial ordering weget forcing extensions isomorphic to those of M2.We will need the following result of Miller, which is not very difficult toprove:

Fact 0.3 [Mi] For every colouring of the splitnodes of a Miller tree p infinitely many colours, there exists a q ∈ M with q ≤ p and such that allsplitnodes of q have the same colour. 2

Forcing: For a forcing P , an M -generic filter G for P and a P -name a wewrite valG(a) or aG for the value of a. For elements x of the ground modelwe do not distinguish between the canonical name x and x itself.Recall that a forcing P has the Laver property if there exists F ∈ ωω ∩ Vsuch that for every P -name f for an element of ωω, every p ∈ P andevery g ∈ ωω ∩ V with p °P ∀n ∈ ω (f(n) ≤ g(n)), there exist q ≤ p and(Hn)n∈ω in V such that |Hn| ≤ F (n) and q °P ∀n (f(n) ∈ Hn).

1 The consistency of cov(M) < add(J (M2))

We repeat the definition of the introduction:

Definition 1.1 Let

J (Mn) := {X ⊆ (ωω)n | ∀(p0, . . . , pn−1) ∈Mn ∃(q0, . . . , qn−1) ∈Mn

((q0, . . . , qn−1) ≤ (p0, . . . , pn−1) ∧ ([q0]× . . .× [qn−1]) ∩X = ∅)}.It is easy to see that J (M) is a σ-ideal. Jossen and Spinas [JoSp] haveshown that J (M2) is a σ-ideal but J (Mn) is not for n > 2. And theypointed out parallelisms between the behaviour of J (M) and J (M2). Here,we underpin this by showing that, analogously to the one-dimensional case,we have the following:

Theorem 1.2 It is relatively consistent with ZFC that add(J (M2)) = ω2

and cov(M) = ω1.

6

For this define an amoeba forcing A(M2) for M2 as follows:

Definition 1.3 Let A(M2) be the set of all pairs ((s, p), (t, q)) such that

(i) (p, q) ∈ M2, s : n → split(p) and t : m → split(q) for some n,m ∈ω \ {0} with n = m or n = m + 1, and s(0) = st(p) and t(0) = st(q);

(ii) the downward closure S of ran(s) is a finite subtree of p and split(S) ⊆ran(s); the downward closure T of ran(t) is a finite subtree of q withsplit(T ) ⊆ ran(t);

(iii) the mappings Φ : (ran(s),⊆) → (ran(E ¹ n),⊆) and Ψ : (ran(t),⊆) →(ran(E ¹ m),⊆) defined by s(k) 7−→ E(k) and t(k) 7−→ E(k) are bothisomorphisms, where E is the order preserving enumeration of ourwell-ordering ≺ of <ωω fixed in the preliminaries;

(iv) if s(j) is the immediate predecessor in the tree sense of s(k + 1) insplit(p), then s(k+1)(|s(j)|) > max({#s(i) | i ≤ k}∪{#t(i) | i ≤ k})and |s(k + 1)| > s(k + 1)(|s(j)|);if t(j) is the immediate predecessor in the tree sense of t(k + 1) insplit(q), then t(k + 1)(|t(j)|) > max({#s(i) | i ≤ k + 1} ∪ {#t(i) | i ≤k}) and |t(k + 1)| > t(k + 1)(|t(k)|).

Define partial orderings ≤ and 60 on A(M2) by

((s, p), (t, q)) ≤ ((s′, p′), (t′, q′)) :⇔ s ⊇ s′, t ⊇ t′, p ⊆ p′ and q ⊆ q′.

((s, p), (t, q) 60 ((s′, p′), (t′, p′)) :⇔ ((s, p), (t, q) 6 ((s′, p′), (t′, p′))

and s = s′ and t = t′.

Often we will write s and t as tuples. To rule out that A(M2) adds Cohen-reals, we need the following decision property of A(M2):

Lemma 1.4 A(M2) has the pure decision property.

Here, the pure decision property means that for every A(M2)-name Θ andevery ((s, p), (t, q)) ∈ A(M2) such that

((s, p), (t, q)) °A(M2) Θ ∈ {0, 1}

there exists ((s′, p′), (t′, q′)) ∈ A(M2) with ((s′, p′), (t′, q′)) ≤0 ((s, p), (t, q))and ((s′, p′), (t′, q′)) deciding Θ (this means there exists j < 2 such that((s′, p′), (t′, q′)) °A(M2) Θ = j).

Proof of lemma 1.4: Let Θ be an A(M2)-name and ((s, p), (t, q)) ∈A(M2) such that ((s, p), (t, q)) °A(M2) Θ ∈ {0, 1}. Our first goal is toconstruct by a fusion a pair (p′, q′) of superperfect trees with (p′, q′) ≤0

(p, q), ((s, p′), (t, q′)) ∈ A(M2) and with the following property:

(∗) For all (s′, t′) ∈ <ω(split(p′))× <ω(split(q′)), if there is some (u, v) ≤(p′, q′) with ((s′, u), (t′, v)) ∈ A(M2), ((s′, u), (t′, v)) ≤ ((s, p′), (t, q′))and ((s′, u), (t′, v)) deciding Θ, then already ((s′, p′), (t′, q′)) decidesΘ.

Suppose |s| = |t|, so we have to extend s first. The case |s| = |t| + 1 isvery similar and is not written down. We want to construct sequences(σn)n∈ω, (τn)n∈ω in <ωω and ((u′n, v′n))n∈ω\{0}, ((un, vn))n∈ω in M2 suchthat, letting sn := s_〈σ0, . . . , σn−1〉 and tn := t_〈τ0, . . . , τn−1〉, for everyn ∈ ω the following hold:

7

(a) (p, q) ≥ (u0, v0) ≥ (u′n+1, v′n+1) ≥ (un+1, vn+1) ≥ (u′n+2, v

′n+2);

(b) ((sn, un), (tn, vn)) ∈ A(M2) and ((sn+1, u′n+1), (tn, v′n+1)) ∈ A(M2);

(c) for every (s′, t′) ∈ <ω(ran(sn+1)) × <ω(ran(tn)) with s′(|s′| − 1) =σn, if there exists (u, v) ∈ M2 such that ((s′, u), (t′, v)) ∈ A(M2),((s′, u), (t′, v)) ≤ ((s, u′n+1), (t, v

′n+1)) and ((s′, u), (t′, v)) decides Θ,

then ((s′, u′n+1), (t′, v′n+1)) decides Θ as well;

and for every (s′, t′) ∈ <ω(ran(sn+1)) × <ω(ran(tn+1)) with t′(|t′| −1) = τn, if there exists (u, v) ∈ M2 with ((s′, u), (t′, v)) ∈ A(M2),((s′, u), (t′, v)) ≤ ((s, un+1), (t, vn+1)) and ((s′, u), (t′, v)) decides Θ,then ((s′, un+1), (t′, vn+1)) decides Θ as well.

For beginning the construction define s0 := s, t0 := t, u0 := p and v0 := q.Suppose sn = s_ 〈σ0, . . . , σn−1〉, tn = t_ 〈τ0, . . . , τn−1〉, un and vn arealready defined. Fix µ ∈ ran(sn) maximal such that whenever the element((sn

_〈σ〉, un), (tn, vn)) ∈ A(M2) extends ((sn, un), (tn, vn)) we have µ $ σ.Later, we shall often refer to this µ as the place where we have to extendsn. Choose σn ∈ Succun

(µ) such that ((sn_〈σn〉, un), (tn, vn)) ∈ A(M2)

and define sn+1 := sn_〈σn〉.

Claim 1: There exists (u′n+1, v′n+1) ∈ M2 such that (u′n+1, v

′n+1) ≤

(un, vn), ((sn+1, u′n+1), (tn, v′n+1)) is an element of A(M2) and the first

part of (c) holds.

Proof of claim 1: Let ((si, ti))i<N for some N ∈ ω enumerate allpairs (s′, t′) ∈ <ω(ran(sn+1)) × <ω(ran(tn)) with s′(|s′| − 1) = σn. Wewant to consider successively all pairs (si, ti) and thin out (un, vn) to get(u′n+1, v

′n+1). Let p0 := un and q0 := vn and suppose we have already

constructed (pi, qi) ∈M2 for some i < N .Consider (si, ti) and suppose there exists a pair (u, v) ≤ (pi, qi) such that((si, u), (ti, v)) ∈ A(M2), ((si, u), (ti, v)) ≤ ((s, pi), (t, qi)) and the pair((si, u), (ti, v)) decides Θ (if there does not exist such a pair (u, v) we letpi+1 := pi and qi+1 := qi). Let Si be the downward closure of ran(si) inthe tree sense and define

rp :=⋃{(pi)σ |σ ∈ ran(sn+1) \ Si},

r′p :=⋃{(pi)σ_〈k〉 |σ ∈ (ran(sn+1) \ ran(si)) ∩ Si ∧ σ_〈k〉 /∈ Si}

and letpi+1 := u ∪ rp ∪ r′p.

Then pi+1 ∈ M and note that every σ ∈ ran(sn+1) \ ran(si) remains aninfinite splitnode of pi+1. Analogously let Ti be the downward closure ofran(ti) and define

rq :=⋃{(qi)τ | τ ∈ ran(tn+1) \ Ti},

r′q :=⋃{(qi)τ_〈k〉 | τ ∈ (ran(tn+1) \ ran(ti)) ∩ Ti ∧ τ_〈k〉 /∈ Ti)}

and letqi+1 := v ∪ rq ∪ r′q.

Again, every τ ∈ ran(tn) \ ran(ti) remains an infinite splitnode of qi+1.Easily, ((sn+1, pi+1), (tn, qi+1)) ∈ A(M2) and ((sn+1, pi+1), (tn, qi+1)) ≤((sn+1, pi), (tn, qi)). And ((si, pi+1), (ti, qi+1)) decides Θ as well, because

8

for ((s, u), (t, v)) ≤ ((si, pi+1), (ti, qi+1)) our construction and (iv) of thedefinition of A(M2) guarantees that we have s \ si ⊆ u and t \ ti ⊆ v (re-member that si(|si − 1|) = σn). So for ¯u := u\(rp∪r′p) and ¯v := v\(rq∪r′q)we have ((s, ¯u), (t, ¯v)) ∈ A(M2) and ((s, ¯u), (t, ¯v)) ≤ ((si, u), (ti, v)), andhence ((s, u), (t, v)) and ((si, u), (ti, v)) are compatible as required.Finally let u′n+1 := pN and v′n+1 := qN . Then (u′n+1, v

′n+1) is as desired.

2(claim1)Fix now ν ∈ ran(tn), the point where we have to extend the finite treegenerated by ran(tn). Choose an element τn ∈ Succv′n+1

(ν) such that((sn+1, u

′n+1), (tn

_ 〈τ〉)) ∈ A(M2) and define tn+1 := tn_ 〈τn〉. In the

same way as we got claim 1, we get

Claim 2: There exists (un+1, vn+1) ∈ M2 such that (un+1, vn+1) ≤(u′n+1, v

′n+1), ((sn+1, un+1), (tn+1, vn+1)) ∈ A(M2) and for every (s′, t′) ∈

<ω(ran(sn+1))×<ω(ran(tn+1)) with t′(|t′|−1) = τn, if there exists (u, v) ∈M2 with ((s′, u), (t′, v)) ∈ A(M2), ((s′, u), (t′, v)) ≤ ((s, un+1), (t, vn+1))and ((s′, u), (t′, v)) decides Θ, then ((s′, un+1), (t′, vn+1)) decides Θ aswell. 2(claim 2)This finishes our construction.Let p′ be the unique superperfect tree determined by

split(p′) =⋃n∈ω

ran(sn)

and q′ the unique superperfect tree determined by

split(q′) =⋃n∈ω

ran(tn).

Claim 3: (p′, q′) satisfies property (∗) (see at the beginning of this proof).

Proof of claim 3: Suppose (s′, t′) ∈ <ω(split(p′))×<ω(split(q′)) and thereexists some (u, v) ≤ (p′, q′) with ((s′, u), (t′, v)) ∈ A(M2), ((s′, u), (t′, v)) ≤((s, p′), (t, q′)) and ((s′, u), (t′, v)) decides Θ. If |s′| = |t′|, there exists n ∈ ωwith (s′, t′) ∈ <ω(ran(sn+1)) × <ω(ran(tn+1)) and t′(|t′| − 1) = τn. Andif |s′| = |t′| + 1 there exists n ∈ ω such that (s′, t′) ∈ <ω(ran(sn+1)) ×<ω(ran(tn)) and s′(|s′| − 1) = σn. Since

((s′, u), (t′, v)) ≤ ((s′, p′), (t′, q′)) ≤((s′, un+1), (t′, vn+1)) ≤ ((s′, u′n+1), (t

′, v′n+1)),

property (c) gives us that ((s′, u′n+1), (t′, v′n+1)) or ((s′, un+1), (t′, vn+1))

decides Θ, so ((s′, p′), (t′, q′)) decides Θ as well. 2(claim 3)

Claim 4: ((s, p′), (t, q′)) decides Θ.

Proof of claim 4: Suppose not. And suppose again that we have |s| = |t|;the case |s| = |t| + 1 is once more similar. We will construct sequences((un, vn))n∈ω, ((u′n, v′n))n∈ω in M2 and (sn)n∈ω and (tn)n∈ω in <ω(<ωω)with sn = s_〈σ0, . . . σn−1〉 and tn = t_〈τ0, . . . , τn−1〉 such that for everyn ∈ ω the following hold:

(a’) s ⊆ sn ⊆ sn+1 and t ⊆ tn ⊆ tn+1;

(b’) (p′, q′) ≥ (u′n, v′n) ≥ (un, vn) ≥ (u′n+1, v′n+1);

9

(c’) ((sn+1, p′), (tn+1, q

′)) ∈ A(M2) and ((sn+1, p′), (tn, q′)) ∈ A(M2);

(d’) if (s′, t′) ∈ <ω(ran(sn))×<ω(ran(tn)) such that |s′| = |t′|, t′(|t′|−1) =τn−1 for n > 0, ((s′, u′n), (t′, v′n)) ∈ A(M2) and ((s′, u′n), (t′, v′n)) ≤((s, u′n), (t, v′n)), then there is no σ ∈ split(u′n) such that we have((s′_〈σ〉, u′n), (t′, v′n)) ∈ A(M2) and ((s′_〈σ〉, u′n), (t′, v′n)) decides Θ;

(e’) if (s′, t′) ∈ <ω(ran(sn+1)) × <ω(ran(tn)) such that |s′| = |t′| + 1,s′(|s′| − 1) = σn, ((s′, un), (t′, vn)) ∈ A(M2) and ((s′, un), (t′, vn)) ≤((s, un), (t, vn)), then there is no τ ∈ split(vn) such that we have((s′, un), (t′_〈τ〉, vn)) ∈ A(M2) and ((s′, un), (t′_〈τ〉, vn)) decides Θ.

Define s := s0 and t := t0. For beginning the induction fix µ ∈ ran(s),the unique point where we can extend the finite subtree of p generated byran(s), and prune p′ by defining

p :=⋃{(p′)µ_〈k〉 | ∀σ ∈ ran(s) \ {µ} (µ_〈k〉 * σ)}.

Then p is a superperfect tree with stem µ and for ρ ∈ Succp(µ) we have(p)ρ = (p′)ρ. For every ρ ∈ Succp(µ) we want to colour the set

Sρ := {σ ∈ split(p′)ρ | ((s_〈σ〉, p′), (t, q′)) ∈ A(M2)}in three colours by defining

ϕρ : Sρ −→ {0, 1, 2}

σ 7−→

0, if ((s_〈σ〉, p′), (t, q′)) °A(M2) Θ = 0,1, if ((s_〈σ〉, p′), (t, q′)) °A(M2) Θ = 1,2, if ((s_〈σ〉, p′), (t, q′)) does not decide Θ.

Fact 0.3 gives us for every ρ ∈ Succp(µ) a superperfect tree pρ ⊆ (p′)ρ

whose splitnodes all have colour ερ.If there are an infinite set A ⊆ Succp(µ) and an ε ∈ {0, 1} such that forevery ρ ∈ A we have ερ = ε, then we can define an element of A(M2)which forces Θ to be this ε as follows: Define

p′ :=⋃{pρ | ρ ∈ A} ∪ (p′ \ p);

then clearly ((s, p′), (t, q′)) °A(M2) Θ = ε.By the property (∗) of our constructed pair (p′, q′) this implies that((s, p′), (t, q′)) decides Θ, a contradiction to our assumption.So there must exist a finite set F ⊆ Succp(µ) such that for every ρ ∈Succp(µ) \ F we have ερ = 2. Define

u′0 :=⋃{pρ | ρ ∈ Succp(µ) \ F} ∪ (p′ \ p)

and v′0 := q′. Then for no σ ∈ split(u′0) with ((s_〈σ〉, u′0), (t, v′0)) ∈ A(M2)the element ((s_〈σ〉, u′0), (t, v′0)) decides Θ. Choose σ0 ∈ Succu′0(µ) with((s_〈σ0〉, u′0), (t, v′0)) ∈ A(M2) and define s1 := s_〈σ0〉.Now fix ν ∈ ran(t), the unique point where we can extend the finite subtreegenerated by ran(t). We want to do the same as before at the right side:Prune v′0 by defining

v :=⋃{(v′0)ν_〈k〉 | ∀τ ∈ ran(t) \ {ν} (ν_〈k〉 * τ)}

and colour for every ρ ∈ Succv(ν) the set

Tρ := {τ ∈ split(v′0)ρ | ((s1, u′0), (t

_〈τ〉, v′0)) ∈ A(M2)}

10

by defining

ψρ : Tρ −→ {0, 1, 2}

τ 7−→

0, if ((s1, u′0), (t

_〈τ〉, v′0)) °A(M2) Θ = 0,1, if ((s1, u

′0), (t

_〈τ〉, v′0)) °A(M2) Θ = 1,2, if ((s1, u

′0), (t

_〈τ〉, v′0)) does not decide Θ.

Again we get for every ρ ∈ Succv(ν) a superperfect tree qρ ⊆ (v′0)ρ, whosesplitnodes all have colour ερ. If there exist an infinite set A ⊆ Succv(ν) andε ∈ {0, 1} such that for every ρ ∈ A we have ερ = ε, we can define v′ suchthat ((s1, u

′0), (t, v

′)) °A(M2) Θ = ε, and therefore – again by property (∗)– ((s1, p

′), (t, q′)) decides Θ. But ((s1, u′0), (t, v

′0)) ≤ ((s1, p

′), (t, q′)), hence((s1, u

′0), (t, v

′0)) decides Θ, and this is a contradiction to the choice of

(u′0, v′0) (property (d’) for n = 0). Hence we get a finite set F ⊆ Succv(ν)

with ερ = 2 for every ρ ∈ Succv(ν) \ F . Define

v0 :=⋃{qρ | ρ ∈ Succv(ν) \ F} ∪ (v′0 \ v)

and u0 := u′0. Then for no τ ∈ split(v0) with ((s1, u0), (t_ 〈τ〉, v0)) ∈A(M2) the element ((s1, u0), (t_〈τ〉, v0)) decides Θ. Choose τ0 ∈ Succv0(ν)with ((s1, u0), (t_〈τ0〉, v0)) ∈ A(M2) and define t1 := t_〈τ0〉.

Suppose we have constructed ((ui, vi))i≤n, ((u′i, v′i))i≤n, (si)i≤n+1 and

(ti)i≤n+1 with si = s_ 〈σ0, . . . σi−1〉 and ti = t_ 〈τ0, . . . , τi−1〉 for somen ∈ ω. First we will describe how to get (u′n+1, v

′n+1): Let ((si, ti)i<N

be an enumeration of all (s′, t′) ∈ <ω(ran(sn+1)) × <ω(ran(tn+1)) with|s′| = |t′| and t′(|t′|−1) = τn. By considering successively all pairs (si, ti),we will thin out (un, vn) step by step to make (d’) true. Let p0 := un andq0 := vn. Suppose (pi, qi) is already constructed for some i < N . Consider(si, ti) and fix µ ∈ ran(si), the unique point where the finite subtree of un

generated by si can be extended. Prune pi by defining

pi :=⋃{(pi)µ_〈k〉 | ∀σ ∈ ran(sn) \ {µ} (µ_〈k〉 * σ)}.

For every ρ ∈ Succp(µ) we colour the set

Siρ := {σ ∈ split(pi)ρ | ((si

_〈σ〉, pi), (ti, qi)) ∈ A(M2)}with the three colours 0, 1 and 2 as in the first step of the induction. Weget superperfect trees pρ ⊆ (pi)ρ whose splitnodes all have colour ερ.If there is an ε ∈ {0, 1} such that for infinitely many ρ ∈ Succpi(µ) we haveερ = ε, we can easily define (u, v) such that ((si, u), (ti, v)) °A(M2) Θ = ε.By property (∗) we have that ((si, p

′), (ti, q′)) decides Θ. Fix k ≤ n suchthat si(|si| − 1) = σk. Then ((si, uk), (ti, vk)) ≤ ((si, p

′), (ti, q′)), andtherefore ((si, uk), (ti, vk)) decides Θ as well. But we have (si, ti ¹ (|ti| − 1))∈ <ω(ran(sk+1)) × <ω(ran(tk)) and |si| = |ti ¹ (|ti| − 1))| + 1, and hencethe element ((si, uk), (ti, vk)) does not decide Θ by property (e′) for k, acontradiction.So for all but finitely many ρ ∈ Succpi(µ) we have ερ = 2. Define

pi+1 :=⋃{pρ | ρ ∈ Succpi(µ) ∧ ερ = 2} ∪ (pi \ pi)

and qi+1 := qi.In the end, we define u′n+1 := pN and v′n+1 := qN . Then (u′n+1, v

′n+1)

11

is as desired. Choose an element σn+1 ∈ split(u′n+1) such that we have((sn+1

_〈σn+1〉, u′n+1), (tn+1, v′n+1)) ∈ A(M2) and let sn+2 := sn+1

_〈σn+1〉.Analogously we get (un+1, vn+1) ≤ (u′n+1, v

′n+1) in M2 such that prop-

erty (e′) holds. Choose an element τn+1 ∈ split(vn+1) such that we have((sn+2, un+1), (tn+1

_〈τn+1〉, vn+1)) ∈ A(M2) and let tn+2 := tn+1_〈τn+1〉.

This finishes the construction. Now we can define u and v to be the uniquesuperperfect trees with

split(u) =⋃{ran(sn) |n ∈ ω}

andsplit(v) =

⋃{ran(tn) |n ∈ ω}.

Then ((s, u), (t, v)) ∈ A(M2) and the construction and property (∗) guar-antee that no extension of ((s, u), (t, v)) decides Θ, which is a contradiction.So claim 4 is proven and we are done. 2(claim 4)

2

Corollary 1.5 A(M2) has the following decision property:If Θ is an A(M2)-name and ((s, p), (t, q)) ∈ A(M2) such that for somen ∈ ω

((s, p), (t, q)) °A(M2) Θ < n,

then there exists an element ((s, p′), (t, q′)) ∈ A(M2) with ((s, p′), (t, q′)) ≤((s, p), (t, q)) and ((s, p′), (t, q′)) decides Θ.

Proof: Suppose ((s, t), (t, q)) °A(M2) Θ < n. Choose an A(M2)-name a0

such that for every A(M2)-generic filter G we have (a0)G = 0 if and only ifΘG = 0 and (a0)G = 1 if and only if ΘG 6= 0. As °A(M2) a0 ∈ {0, 1}, there-fore ((s, p), (t, q)) °A(M2) a0 ∈ {0, 1}. The pure decision property of A(M2)gives us a pair (p0, q0) ∈ M2 such that (p0, q0) ≤ (p, q), ((s, p0), (t, q0)) ∈A(M2) and ((s, p0), (t, q0)) decides a0. If ((s, p0), (t, q0)) °A(M2) a0 = 0define p′ := p0 and q′ := q0; if ((s, p0), (t, q0)) °A(M2) a0 = 1 choosean A(M2)-name a1 such that for every A(M2)-generic filter G we have(a1)G = 0 if and only if ΘG = 1 and (a1)G = 1 if and only if ΘG 6= 1.Again the pure decision property of A(M2) gives us (p1, q1) ≤ (p0, q0) suchthat ((s, p1), (t, q1)) decides a1. If ((s, p1), (t, q1)) °A(M2) a1 = 0 let p′ := p1

and q′ := q1; if ((s, p1), (t, q1)) °A(M2) a0 = 1 choose an A(M2)-name a2

with (a2)G = 0 if and only if ΘG = 2 and (a2)G = 1 if and only if ΘG 6= 2and so on. Since ((s, p), (t, q)) °A(M2) Θ < n, this procedure stops afterfinitely many steps and we get (p′, q′) as desired. 2

Lemma 1.6 A(M2) has the Laver property.

As an abbreviation we will use the following:

Definition 1.7 For ((s, p), (t, q)) ∈ A(M2) and s′ ∈ <ω(ran(s)) with((s′, p), (t′, q)) ∈ A(M2) for some t′ define

p(s, s′) :=⋃{(p)σ_〈k〉 |σ ∈ ran(s′) ∧ ∀σ′ ∈ ran(s) \ ran(s′) (σ_〈k〉 * σ′)}

and analogously define a superperfect tree q(t, t′) for ((s, p), (t, q)) ∈ A(M2)and t′ ∈ <ω(ran(t)) with ((s′, p), (t′, q)) ∈ A(M2) for some s′.

12

Proof of lemma 1.6: Suppose ((s, p), (t, q)) ∈ A(M2), g ∈ ωω ∩ V and fis an A(M2)-name for an element of ωω such that

((s, p), (t, q)) °A(M2) ∀n (f(n) < g(n)).

We want to construct by a fusion (p′, q′) ∈ A(M2) and a sequence (Hn)n∈ω

in V such that ((s, p′), (t, q′)) ∈ A(M2), ((s, p′), (t, q′)) ≤ ((s, p), (t, q)),|Hn| ≤ 2mn for a strictly increasing sequence (mn)n∈ω of natural numbersnot depending on ((s, p), (t, q)), ((s, p′), (t, q′)), g and f and ((s, p′), (t, q′))°A(M2) ∀n (f(n) ∈ Hn). Often mn equals n, here we have to choose it alittle bit larger, but as the specific value of mn is insignificant, we avoidthe finitary combinatorics.Suppose |s| = |t|; the case |s| = |t|+ 1 is similar. We construct sequences(sn)n∈ω, (tn)n∈ω in <ω(<ωω), ((un, vn))n∈ω in M2, (mn)n∈ω in ω and asequence (Hn)n∈ω of subsets of ω in V by induction such that for everyn ∈ ω the following hold:

(i) (p, q) ≥ (un, vn) ≥ (un+1, vn+1);(ii) ((sn, un), (tn, vn)) ∈ A(M2);(iii) s ⊆ sn ⊆ sn+1 and t ⊆ tn ⊆ tn+1; and if n is even we have sn+1 = sn

and if n is odd we have tn+1 = tn;(iv) ((s, un), (t, vn)) °A(M2) f(n) ∈ Hn;

(v) the sequence (mn)n∈ω is strictly increasing and only depends on nand is independent of ((s, p′), (t, q′)), g and f and |Hn| ≤ 2mn .

For beginning the induction we find a K ∈ ω and (u, v) ∈ M2 with((s, u), (t, v)) ≤ ((s, p), (t, q)) and ((s, u), (t, v)) °A(M2) f(0) = K by corol-lary 1.5. Define (u0, v0) := (u, v), s0 := s, t0 := t, m0 := 0 and H0 := {K}.Suppose we have already constructed ((si, ui), (ti, vi)) ∈ A(M2), mi and Hi

for i ≤ n for some n ∈ ω and suppose first that n is odd. Fix µ ∈ ran(sn),the node where we have to extend the finite tree generated by ran(sn).Choose σ ∈ Succun(µ) such that ((sn

_ 〈σ〉, un), (tn, vn)) ∈ A(M2) anddefine sn+1 := sn

_〈σ〉 and tn+1 := tn.

Claim 1: There exists a pair (u, v) ∈ M2 such that (u, v) ≤ (un, vn)and ((sn+1, u), (tn+1, v)) ∈ A(M2) and for every (s′, t′) ∈ <ω(ran(sn+1))×<ω(ran(tn+1)) with ((s′, u), (t′, v)) ∈ A(M2) and ((s′, u), (t′, v)) ≤((s, un), (t, vn)) there exists some K such that

((s′, u(sn+1, s′)), (t′, v(tn+1, t

′))) °A(M2) f(n + 1) = K.

Proof of claim 1: Let ((si, ti))i<N for some N ∈ ω enumerate all pairs(s′, t′) as in claim 1. We want to consider successively all pairs (si, ti) andthin out (un, vn) step by step to get (u, v): Define p0 := un and q0 := vn

and suppose we have constructed (pi, qi) for some i < N . Consider (si, ti).Suppose there exists (u, v) ∈ M2 with ((si, u), (ti, v)) ≤ ((s, pi), (t, qi))(if not, we let pi+1 := pi and qi+1 := qi). We have ((s, pi), (t, qi)) ≤((s, p), (t, q)), hence ((si, u), (ti, v)) °A(M2) f(n + 1) < g(n + 1), and thepure decision property of A(M2), corollary 1.5 to be precise, gives us a pair(p, q) ∈ M2 with ((si, p), (ti, q)) ∈ A(M2), ((si, p), (ti, q)) ≤ ((si, u), (ti, v))and ((si, p), (ti, q)) °A(M2) f(n + 1) = K for some K. Clearly, we can thinout p and q to ¯p and ¯q such that (¯p, ¯q) ≤ (pi(sn+1, si), qi(tn+1, ti)) and((si, ¯p), ti, ¯q)) ∈ A(M2). Define

pi+1 := ¯p ∪ (pi \ pi(sn+1, si))

13

andqi+1 := ¯q ∪ (qi \ qi(tn+1, ti)).

Then clearly we have

((si, pi+1(sn+1, si)), (ti, qi+1(tn+1, ti))) °A(M2) f(n + 1) = K,

as well.In the end, define u := pN and v := qN . 2(Claim 1)

Fix (u, v) ∈ M2 as in claim 1 and define H ′ to be the collection of all Kwith ((s′, u(sn+1, s

′)), (t′, v(tn+1, t′))) °A(M2) f(n+1) = K for some (s′, t′)

as in claim 1, then H ′ is finite and the cardinality of H ′ only depends onthe number of σi’s and τi’, which we have already chosen, hence on n.Now we thin out u and v so that for every pair (s′, t′) as in claim 1 with|s′| = |t′| and every ρ ∈ u(sn+1, s

′) such that ((s′_〈ρ〉, u), (t′, v)) ∈ A(M2)and for every (s′, t′) as in claim 1 with |s′| = |t′|+1 and every ρ ∈ v(tn+1, t

′)such that ((s′, u), (t′_〈ρ〉, v)) ∈ A(M2), respectively, we have

min{|ρ|, ρ(|µ|)} > max({#σ |σ ∈ ran(sn+1)} ∪ {#τ | τ ∈ ran(tn+1)}),

where µ is the immediate predecessor of ρ in the tree sense in ran(s′), ran(t′),respectively. For this, we have to consider all pairs (s′, t′) and cut offfinitely often finitely many superperfect trees, so there is no problem forthe remaining tree to be a superperfect one. This implies the followingproperty for (s′, t′) as in claim 1:

(∗) Suppose ρ ∈ u(sn+1, s′) such that ((s′ _ 〈ρ〉, u), (t′, v)) ∈ A(M2).

If s ⊇ s′ _ 〈ρ〉 and t ⊇ t′ with ((s, u), (t, v)) ∈ A(M2) and with((s, u), (t, v)) ≤ ((s′_ 〈ρ〉, u), (t′, v)), then we have s ⊆ u(sn+1, s

′)and t ⊆ v(tn+1, t

′).And similarly suppose ρ ∈ v(tn+1, t

′) such that ((s′, u), (t′_〈ρ〉, v)) ∈A(M2). If t ⊇ t′_〈ρ〉 and s ⊇ s′ with ((s, u), (t, v)) ∈ A(M2) and((s, u), (t, v)) ≤ ((s′, u), (t′_ 〈ρ〉, v)), then we have s ⊆ u(sn+1, s

′)and t ⊆ v(tn+1, t

′).

By a similar fusion as in the construction of the pair (p′, q′) in the proofof lemma 1.4, using the pure decision property of A(M2), we get (u′, v′) ∈M2 such that ((sn+1, u

′), (tn+1, v′)) ∈ A(M2), ((sn+1, u

′), (tn+1, v′)) ≤

((sn+1, u), (tn+1, v)) and for every (s′, t′) ∈ <ω(split(u′)) × <ω(split(v′))with ran(s′) * ran(sn), ((s′, u′), (t′, v′)) ∈ A(M2) and ((s′, u′), (t′, v′)) ≤((s, u′), (t, v′)), the element ((s′, u′), (t′, v′)) decides f(n + 1).

Claim 2: There exists (u′′, v′′) ∈ M2 with (u′′, v′′) ≤ (u′, v′) such that((sn+1, u

′′), (tn+1, v′′)) ∈ A(M2) and for every (s′, t′) ∈ <ω(ran(sn+1)) ×

<ω(ran(tn+1)) with ((s′, u′′), (t′, v′′)) ∈ A(M2) and ((s′, u′′), (t′, v′′)) ≤((s, u′), (t, v′)) we have the following:

If τ ∈ ran(sn+1) with ((s′_〈τ〉, u′′), (t′, v′′)) ∈ A(M2) then there ex-ists K such that for every ρ ∈ split(u′′(sn+1, s

′_〈τ〉)) which properlyextends τ we have ((s′_〈ρ〉, u′′), (t′, v′′)) °A(M2) f(n + 1) = K;

if τ ∈ ran(tn+1) with ((s′, u′′), (t′_〈τ〉, v′′)) ∈ A(M2) then there ex-ists K such that for every ρ ∈ split(v′′(tn+1, t

′_〈τ〉)) which properlyextends τ we have ((s′, u′′), (t′_〈ρ〉, v′′)) °A(M2) f(n + 1) = K.

14

Proof of claim 2: Let ((si, ti, τi))i<N for some N ∈ ω enumerate alltriples (s′, t′, τ) as in the claim. Define p0 := u′ and q0 := v′ and supposewe have constructed (pi, qi) for some i < N . Consider (si, ti, τi) and sup-pose τi ∈ ran(sn+1), so |si| = |ti|; the case τi ∈ ran(tn+1) is similar.For each ρ ∈ split(pi(sn+1, si

_〈τi〉)) which properly extends τi and with((si

_〈ρ〉, pi), (ti, qi)) ∈ A(M2) we have ran(si_〈ρ〉) * ran(sn) (by prop-

erty (∗)), ((si_ 〈ρ〉, u′), (ti, v′)) ∈ A(M2) and ((si

_ 〈ρ〉, u′), (ti, v′)) ≤((s, u′), (t, v′)). The choice of (u′, v′) gives us that ((si

_ 〈ρ〉, pi), (ti, qi))decides f(n + 1), say ((si

_〈ρ〉, pi), (ti, qi)) °A(M2) f(n + 1) = Kρ.By colouring splitnodes as in the proof of claim 4 in lemma 1.4, we canfind for every above considered ρ ∈ Succpi(sn+1,si

_〈τi〉)(τi) a Kρ and atree pρ ∈ M such that pρ ⊆ (pi)ρ and for every ν ∈ split(pρ) we have((si

_〈ν〉, pi), (ti, qi)) °A(M2) f(n + 1) = Kρ. There are infinitely many ρbut finitely many Kρ, so there is an infinite set A of ρ’s and a Ki withKρ = Ki for every ρ ∈ A. Fix µ ∈ ran(si), the immediate predecessor inthe tree sense of τi and define

pi+1 :=⋃{pρ | ρ ∈ A} ∪ {σ ∈ pi |µ * σ ∨

∃ν ∈ ran(sn+1) (µ $ ν ⊆ σ ∨ µ ⊆ σ $ ν)}and qi+1 := qi.In the end, define u′′ := pN and v′′ := qN . 2(Claim 2)

Define un+1 := u′′ and vn+1 := v′′ and let H ′′ be the collection of all Kas in claim 2. Clearly, H ′′ is finite and the cardinality only depends on n.Define Hn+1 := H ′ ∪H ′′ then we can choose mn+1 ∈ ω with mn+1 > mn

and |H ′′| ≤ 2mn+1 .

Claim 3: ((s, un+1), (t, vn+1)) °A(M2) f(n + 1) ∈ Hn+1.

Proof of claim 3: Let ((s′, u′), (t′, v′)) ∈ A(M2) with ((s′, u′), (t′, v′)) ≤((s, un+1), (t, vn+1)) and ((s′, u′), (t′, v′)) °A(M2) f(n+1) = K for some K.Let s be the maximal initial segment of s′ with ran(s) ⊆ ran(sn+1) and tthe maximal initial segment of t′ with ran(t) ⊆ ran(tn+1). Without lossof generality, the length of s′ is longer than the length of s and the lengthof t′ is longer than the length of t. Define µ := s′(|s|) and ν := t′(|t|).Distinguish the following two cases:First case: µ ∈ un+1(sn+1, s) and ν ∈ vn+1(tn+1, t), so by property (∗)we have ran(s′) ⊆ un+1(sn+1, s) and ran(t′) ⊆ vn+1(tn+1, t), and hence((s′, u′), (t′, v′)) and ((s, un+1(sn+1, s)), (t, vn+1(tn+1, t))) are compatible.Since

((s, un+1(sn+1, s)), (t, vn+1(tn+1, t))) ≤ ((s, u(sn+1, s)), (t, v(tn+1, t)))

and ((s, u(sn+1, s)), (t, v(tn+1, t))) °A(M2) f(n + 1) ∈ H ′ we conclude that((s′, u′), (t′, v′)) °A(M2) f(n + 1) ∈ H ′.Second case: µ /∈ un+1(sn+1, s) or ν /∈ vn+1(tn+1, t). Suppose µ /∈un+1(sn+1, s), and therefore µ extends an element of ran(sn+1) \ ran(s);the case ν /∈ vn+1(tn+1, t) is similar. Choose σ ∈ ran(sn+1) with maximallength such that µ % σ. It is easy to see that we have ((s_ 〈σ〉, un+1),(t, vn+1)) ∈ A(M2), and hence (s, t, σ) is among the enumerated triplesof claim 2 above, say (s, t, σ) = (si, ti, τi). Then µ is an element of

15

split(pi(sn+1, si_ 〈τi〉)) with µ % τi and ((si

_ 〈µ〉, pi), (ti, qi)) °A(M2)

f(n + 1) = Ki ∈ H ′′. Since ((s′, u′), (t′, v′)) ≤ ((si_ 〈µ〉, pi), (ti, qi)),

also ((s′, u′), (t′, v′)) °A(M2) f(n + 1) ∈ H ′′. 2(claim 3)

If n is even, we have to do the analogous step at the right side by choosingτ ∈ Succvn

(ν), where ν ∈ ran(tn) is the node where we have to extendran(tn). Then we let tn+1 := tn

_ 〈τ〉 and sn+1 := sn. This step isvery similar to the even case, we do not carry it out. This finishes theconstruction.In the end, define p′ and q′ to be the superperfect trees determined by

split(p′) =⋃{ran(sn) |n ∈ ω}

andsplit(q′) =

⋃{ran(tn) |n ∈ ω}.

Then (p′, q′) and (Hn)n∈ω are as desired. 2

It is well known that the Laver property of a forcing P implies that forcingwith P does not add Cohen-reals.

Definition 1.8 Let G be an A(M2)-generic filter over V . Define (pG, qG)to be the unique pair (p′, q′) ∈M2 such that

split(p′) =⋃{ran(s) | ∃p, q ∈M ∃t ∈ <ω(<ωω) (((s, p), (t, q)) ∈ G)}

and

split(q′) =⋃{ran(t) | ∃p, q ∈M ∃s ∈ <ω(<ωω) (((s, p), (t, q)) ∈ G)}.

Clearly, pG and qG are well defined.

Up to this point we could have worked with any finite power of M andproven the analogous results. It is now that the restriction to dimension 2becomes necessary. For justifying the name amoeba for our forcing A(M2)we want to prove the following lemma:

Lemma 1.9 Suppose that D ⊆M2 is open and dense and in V and G isA(M2)-generic over V .Then there exists a countable F ⊆ D in V such that in V [G] we have[pG]× [qG] ⊆ ⋃{[p]× [q] | (p, q) ∈ F}.Moreover, every pair of branches of (pG, qG) is M2-generic over V .

For the proof we use the following:

Fact 1.10 [Sp2] Suppose D ⊆M2 is open and dense and (p, q) ∈M2.Then there exists (p′, q′) ∈M2 with (p′, q′) ≤0 (p, q) and there exists somecountable F ⊆ D such that for every (x, y) ∈ [p′]× [q′] there is (u, v) ∈ Fwith (x, y) ∈ [u]× [v]. 2

This fact is the result of a difficult induction on types invented by Spinas.

Proof of lemma 1.9: So suppose G is A(M2)-generic over V and D ⊆M2

is open and dense. Define

D := {(p′, q′) ∈M2 | there exists a countable F ⊆ D such that

[p′]× [q′] ⊆⋃{[u]× [v] | (u, v) ∈ F}}.

16

Then D is ≤0-dense in M2 by fact 1.10; recall that this means that forevery (p, q) ∈M2 there exists (p′, q′) ≤ (p, q) with (p′, q′) ∈ D and st(p′) =st(p) and st(q′) = st(q). So, given ((s, p), (t, q)) ∈ A(M2) it is easy tofind (p′, q′) ≤0 (p, q) such that ((s, p′), (t, q′)) ∈ A(M2) and there exists acountable F ⊆ D such that [p′] × [q′] ⊆ ⋃{[u] × [v] | (u, v) ∈ F}. Hencethe set

D′ := {((s, p′), (t, q′)) ∈ A(M2) | ∃F ⊆ D countable

([p′]× [q′] ⊆⋃{[u]× [v] | (u, v) ∈ F})}

is ≤0-dense in A(M2), i.e. for every ((s, p), (t, q)) ∈ A(M2) exists (p′, q′) ≤0

(p, q) such that ((s, p′), (t, q′)) ∈ D′. Choose ((s, p), (t, q)) ∈ D′∩G and letF ∈ V be the countable witness for this. Notice that the property of F is aΠ1

1 one and is therefore absolute for transitive models of ZFC, and hence Fhas the same property in V [G]. For every (x, y) ∈ [pG]×[qG] and every n ∈ω there is an element ((sn, pn), (tn, qn)) ∈ G such that for every n ∈ ω wehave ((sn+1, pn+1), (tn+1, qn+1)) ≤ ((sn, pn), (tn, qn)) ≤ ((s, p), (t, q)) andx ¹ n ∈ Sn and y ¹ n ∈ Tn, where Sn and Tn are the downward closuresof ran(sn), ran(tn), respectively. Hence (x, y) ∈ [p] × [q], and therefore[pG]× [qG] ⊆ ⋃{[u]× [v] | (u, v) ∈ F}.It remains to prove that every pair of branches in (pG, qG) is M2-generic.For this we have to prove that every pair (x, y) ∈ ωω × ωω which has thefollowing property:

(∗) for every dense D ⊆ M2 in V there exists (p, q) ∈ D with (x, y) ∈[p]× [q]

is a pair of Miller-reals, i.e. H := {(p, q) ∈ (M2)V | (x, y) ∈ [p] × [q]} isM2-generic over V .Clearly, H is upwards closed. Suppose there are incompatible (p, q), (p′, q′)in H. By Shoenfield’s absoluteness theorem they are incompatible in V ,i.e. [p]∩ [p′] or [q]∩ [q′] (or both) does not contain the branches of a Millertree, without loss of generality, [p]∩ [p′] does not. Since [p]∩ [p′] is a closedsubset of ωω, we can apply a fact of Kechris:

Fact 1.11 [Ke] Every closed set A ⊆ ωω is bounded with respect to ≤∗ orcontains the branches of a Miller tree.

Choose by this fact an upper bound f ∈ V ∩ ωω of [p] ∩ [p′]. It is easy tosee that the set E := {(p, q) ∈ M2 | ∀h ∈ [p] (h 6≤∗ f)} is dense in M2, soby property (∗) of (x, y) there exists (p, q) ∈ E with (x, y) ∈ [p]× [q], andhence we have x 6≤∗ f . But by choice x ∈ [p] ∩ [p′], therefore x ≤∗ f , andthis is a contradiction. Hence H is a filter.The genericity of H is clear by property (∗). 2

Now we are ready to prove the main theorem of this chapter:

Proof of theorem 1.2: Suppose that ((Pα)α≤ω2 , (Qα)α<ω2) is a count-able support iteration of A(M2), i.e. for every α < ω2, Qα is a Pα-namefor A(M2) defined in the model V Pα . Let G be Pω2-generic over V .By lemma 1.6, every iterand of our iteration has the Laver property andby a result of Shelah ([Sh], see also [Go]) the Laver property is preservedunder countable support iterations. So Pω2 has the Laver property, andhence V [G] |= cov(M) = ω1. It remains to prove that, in V [G], we haveadd(J (M2)) = ω2. So suppose, in V [G], (Xα)α<ω1 is a family of elements

17

of J (M2) and let X :=⋃{Xα |α < ω1}. Suppose (p, q) ∈ M2. We have

to find (p′, q′) ≤ (p, q) in M2 such that ([p′]× [q′]) ∩X = ∅.For α < ω1 define

Dα := {(u, v) ∈M2 | ([u]× [v]) ∩Xα = ∅}.

Clearly, Dα is open and dense inM2. Our iteration has countable support,and hence for every α < ω2 with cf(α) = ω1 we have for every (u, v) ∈(M2)V [Gα] that (u, v) already belongs to V [Gβ ] for some β < α (see forexample [Ku], ch. VIII, lemma 5.14). With this fact, by a Lowenheim-Skolem-argument, the set

Cα := {β < ω2 |Dα ∩ V [Gβ ] ∈ V [Gβ ] and

Dα ∩ V [Gβ ] is open and dense in (M2)V [Gβ ]}

is ω1-club in ω2, i.e. it is unbounded and closed under increasing sequencesof length ω1. Hence C :=

⋂α<ω1

Cα is ω1-club in ω2, too. But G is generic,and hence by the above mentioned lemma of [Ku] we find γ ∈ C and some(s, t) ∈ <ωsplit(p)× <ωsplit(q) such that ((s, p), (t, q)) ∈ G(γ), where G(γ)is the Qγ [Gγ ]-generic filter induced by G. By lemma 1.9 in V [Gγ+1] wehave for every α < ω1

[pG(γ)]× [qG(γ)] ⊆⋃{[u]× [v] | (u, v) ∈ D′

α ∩ V [Gγ ]}

for some countable D′α ⊆ Dα. By absoluteness this is also true in V [G],

and hence V [G] |= ([pG(γ)]× [qG(γ)])∩X = ∅ and as (pG(γ), qG(γ)) ≤ (p, q)we are done. 2

2 Martin’s Axiom and the additivity of J (M2)

In this section, we want to prove the following theorem:

Theorem 2.1 MA(σ-centered) implies that add(J (M2)) equals 2ω.

Here σ-centered denotes the class of all forcings which are σ-centered,where a forcing P is σ-centered if it is a countable union of centered setsPn, i.e. for p, q ∈ Pn there exists r ∈ Pn with r ≤ p, q.We will use the following:

Fact 2.2 [JoSp] For (p, q) ∈ M2 and X ∈ J (M2) there exists (p′, q′) ≤0

(p, q) such that ([p′]× [q′]) ∩X = ∅. 2

In other words, the set {(p′, q′) ∈ M2 | ([p′] × [q′]) ∩ X = ∅} is ≤0-densein M2 for fixed X ∈ J (M2). This fact was the main step in proving thatJ (M2) is a σ-ideal.For the proof of theorem 2.1 we need the concept of good sequences andthe corresponding sequences of superperfect trees defined in the next def-initions:

Definition 2.3 Suppose (p, q) ∈M2 and (p, q) has the property of (p′, q′)formulated in fact 0.2 (recall that we always assume this).Let I(p,q) := TPst(p),st(q)(p, q), then we have I(p,q) ⊆ split(p) ×+ split(q).A sequence A = (A(σ,τ))(σ,τ)∈I(p,q)

is called good for (p, q) or a good (p, q)-sequence if the following hold for every (σ, τ) ∈ I(p,q):

(i) A(σ,τ) ⊆ <ωω is infinite;

18

(ii) if tp(σ, τ) is even, then ∀ρ ∈ A(σ,τ)(ρ % σ ∧ |ρ| > ρ(|σ|)) and ∀ρ, ρ′ ∈A(σ,τ)(ρ 6= ρ′ ⇒ ρ(|σ|) 6= ρ′(|σ|));if tp(σ, τ) is odd, then ∀ρ ∈ A(σ,τ)(ρ % τ ∧ |ρ| > ρ(|τ |)) and ∀ρ, ρ′ ∈A(σ,τ)(ρ 6= ρ′ ⇒ ρ(|τ |) 6= ρ′(|τ |));

(iii) A(σ,τ) ⊆ Sopp,qst(p),st(q)(σ, τ);

(iv) if tp(σ, τ) is even and σ′, σ′′ ∈ A(σ,τ) are such that σ′(|σ|) < σ′′(|σ|),then we have A(σ′,τ) ⊇ A(σ′′,τ),if tp(σ, τ) is odd and τ ′, τ ′′ ∈ A(σ,τ) are such that τ ′(|τ |) < τ ′′(|τ |),then we have A(σ,τ ′) ⊇ A(σ,τ ′′);

(v) if tp(σ, τ) is even, then ∀σ′ ∈ A(σ,τ) ∃σ′′ ∈ A(σ,τ) (σ′′(|σ|) > σ′(|σ|) ∧A(σ′,τ) % A(σ′′,τ)),if tp(σ, τ) is odd, then ∀τ ′ ∈ A(σ,τ) ∃τ ′′ ∈ A(σ,τ) (τ ′′(|τ |) > τ ′(|τ |) ∧A(σ,τ ′) % A(σ,τ ′′)).

Thus, if tp (σ, τ) = 2n, then A(σ,τ) is a thinned out choice of extensionsof σ that are left legs of 2n + 1-type-pairs whose 2n-type-pair is (σ, τ);similarly if tp(σ, τ) is odd. Hence, e.g. in (iv), tp (σ′, τ) = tp (σ′′, τ) is oddand A(σ′,τ) and A(σ′′,τ) are both (coherent) choices of extensions of τ .

If (p, q) is clear from context, we sometimes omit the mention of (p, q).For every good sequence A = (A(σ,τ))(σ,τ)∈I(p,q)

there is a correspondingsequence ((pA(σ,τ), q

A(σ,τ)))(σ,τ)∈I(p,q)

in M2:We just describe how to get (pA(σ,τ), q

A(σ,τ)) in the case where tp(σ, τ) is

even, the odd case is similar. So suppose tp(σ, τ) is even. Define pA(σ,τ)

and qA(σ,τ) by defining

splitn(pA(σ,τ)) := {σ ∈ split(pA(σ,τ)) | σ has exactly n different

predecessors in the tree sense in split(pA(σ,τ))}and

splitn(qA(σ,τ)) :={τ ∈ split(qA(σ,τ)) | τ has exactly n different

predecessors in the tree sense in split(qA(σ,τ))}by induction on n.For n = 0 let split0(pA(σ,τ)) = {st(pA(σ,τ))} := {σ} and split0(qA(σ,τ)) ={st(qA(σ,τ))} := {τ}. Suppose for k ≤ n we have already defined splitk(pA(σ,τ))and splitk(qA(σ,τ)). For defining splitn+1(pA(σ,τ)) we have to define what isin SuccpA(σ,τ)

(σ) for every σ ∈ splitn(pA(σ,τ)): Let

SuccpA(σ,τ)(σ) :=

⋃{A(σ,t) | t ∈

⋃

k≤n

splitk(qA(σ,τ)) ∧ tp(σ,τ)(σ, t) = 2n}

for every σ ∈ splitn(pA(σ,τ)). And then define

SuccqA(σ,τ)(τ) :=

⋃{A(s,τ) | s ∈

⋃

k≤n+1

splitk(pA(σ,τ)) ∧

tp(σ,τ)(s, τ) = 2n + 1}

for every τ ∈ splitn(qA(σ,τ)).

Notice that (pA(σ,τ), qA(σ,τ)) ≤ (p, q), even (pA(σ,τ), q

A(σ,τ)) ≤0 ((p)σ, (q)τ ), for

every (σ, τ) ∈ I(p,q).

19

pA(σ,τ)

σ

®­

©ª

A(σ,τ)qσ0

qσ1

qσ2. . .

®­

©ªq

σ′0q q

σ′2σ′1A(σ0,τ0)\A(σ0,τ1)

®­

©ª. . .

A(σ0,τ1)\A(σ0,τ2) ®­

©ª

A(σ1,τ2)\A(σ1,τ3)®­

©ª. . .

®­

©ª

A(σ′0,τ ′0)\A(σ′0,τ ′1)®­

©ª

A(σ′0,τ ′1)\A(σ′0,τ ′2)

. . .

AA

AA

¢¢¢¢

bb

bb

bbb

@@

@@

EEEE

¢¢¢¢

JJ

JJ

EEEE

qA(σ,τ)

τ

®­

©ªq q

τ0τ1A(σ0,τ)\A(σ1,τ)

®­

©ª

A(σ1,τ)\A(σ2,τ)®­

©ª. . .

A(σ2,τ)\A(σ3,τ)

®­

©ªq

τ ′0

A(σ′0,τ0)\A(σ′1,τ0)®­

©ª

A(σ′1,τ0)\A(σ′2,τ0)

. . .®­

©ª

A(σ′2,τ1)\A(σ′3,τ1)®­

©ª. . .

@@

@@

BB

BB

££££

@@

@@

AA

AA

¦¦¦¦

­­

­­

Remark 2.4 Fix (p, q) ∈ M2. Let (p′, q′) ∈ M2 with (p′, q′) ≤ (p, q) andsuch that for almost all n ∈ ω and every (x, y) ∈ [p′] ×+ [q′] we havetp-n-pair(x, y) ∈ split(p′) × split(q′) (recall that for every (u, v) ∈ M2,(u, v) ≤ (p, q), there exists (p′, q′) ≤0 (u, v) with this property (com-pare the preliminaries)). Then we can define a partial good sequenceA = (A(σ,τ))(σ,τ)∈I′ for (p, q), where I ′ := I(p,q) ∩ (split(p′) × split(q′)):For (σ, τ) ∈ I ′ with tp(σ, τ) even let A(σ,τ) be the subset of Succp′(σ) ∩{ρ | ρ(|σ|) > #σ} thinned out so that for every ρ ∈ A(σ,τ) we have |ρ| >ρ(|σ|), min{ρ(|σ|), |ρ|} > #τ and ρ(|σ|) > #ρ′ for every ρ′ ∈ A(σ,τ) withρ′(|σ|) < ρ(|σ|). Analogously, if tp(σ, τ) is odd let A(σ,τ) be the subset ofSuccq′(τ) ∩ {ρ | ρ(|τ |) > #τ} thinned out such that for every ρ ∈ A(σ,τ)

we have |ρ| > ρ(|τ |), min{ρ(|τ |), |ρ|} > #σ and ρ(|τ |) > #ρ′ for everyρ′ ∈ A(σ,τ) with ρ′(|τ |) < ρ(|τ |). It is easy to see that (i)-(v) of defini-tion 2.3 hold for A = (A(σ,τ))(σ,τ)∈I′ .

Definition 2.5 Fix (p, q) ∈M2. We define the following relations for anygood (p, q)-sequences A = (A(σ,τ))(σ,τ)∈I(p,q)

and B = (B(σ,τ))(σ,τ)∈I(p,q):

A ≤ B :⇔ (pA(σ,τ), qA(σ,τ)) ≤ (pB(σ,τ), q

B(σ,τ)) for every (σ, τ) ∈ I(p,q).

A ≈ B :⇔ A(σ,τ) =∗ B(σ,τ) for every (σ, τ) ∈ I(p,q) andA(σ,τ) = B(σ,τ) for all but finitely many (σ, τ) ∈ I(p,q).

A ≤∗B :⇔ there exists a good sequence C such that A ≈ C ≤ B.

The good sequence C in the definition of A ≤∗ B we often call a witnessfor A ≤∗ B. Notice that A ≤∗ B implies A(σ,τ) ⊆∗ B(σ,τ) for every(σ, τ) ∈ I(p,q). Clearly, ≈ is an equivalence relation on the set of goodsequences for (p, q). We have the following long but elementary lemma:

Lemma 2.6 Suppose (p, q) ∈M2, C is good for (p, q) and (σ, τ) ∈ I(p,q).

(a) Suppose µ ∈ split(pC(σ,τ)) with µ % σ. Then we have the following:If tp(σ, τ) is even, there exist sequences

µ = s0 % . . . % sk = σ and t1 % . . . % tk = τ

such that si ∈ C(si+1,ti+1) for i ∈ {0, . . . , k − 1} and ti ∈ C(si,ti+1)

for i ∈ {1, . . . , k − 1}. If we choose t1 ≺-maximal with µ ∈ C(s1,t1)

20

(notice there exist only finitely many t with µ ∈ C(s1,t)), where s1 isthe immediate predecessor in the tree sense of µ in split(pC(σ,τ)), thenthe sequences are unique.If tp(σ, τ) is odd, there exist sequences

µ = s0 % . . . % sk = σ and t1 % . . . % tk+1 = τ

such that si ∈ C(si+1,ti+1) for i ∈ {0, . . . , k − 1} and ti ∈ C(si,ti+1)

for i ∈ {1, . . . , k}. Again, if we choose t1 ≺-maximal with µ ∈C(s1,t1), where s1 is the immediate predecessor in the tree sense of µin split(pC(σ,τ)), then the sequences are unique.

(b) Suppose ν ∈ split(qC(σ,τ)) with ν % τ . Then we have the following:If tp(σ, τ) is even, there exist sequences

s0 % . . . % sk = σ and ν = t0 % . . . % tk = τ

such that ti ∈ C(si,ti+1) and si ∈ C(si+1,ti+1) for i ∈ {0, . . . , k − 1}.If we choose s0 ≺-maximal with ν ∈ C(s0,t1), where t1 is the directpredecessor in the tree sense of ν in split(qC(σ,τ)), then the sequencesare unique.If tp(σ, τ) is odd, there exist sequences

s0 % . . . % sk = σ and ν = t0 % . . . % tk+1 = τ

such that ti ∈ C(si,ti+1) for i ∈ {0, . . . , k} and si ∈ C(si+1,ti+1) for i ∈{0, . . . , k − 1}. Again, if we choose s0 ≺-maximal with ν ∈ C(s0,t1),where t1 is the direct predecessor in the tree sense of ν in split(qC(σ,τ)),then the sequences are unique.

Proof: We only prove the first case of (a), the rest is similar. So supposethat tp(σ, τ) is even. Let s0 := µ and choose n ∈ ω such that s0 ¹ n isthe immediate predecessor in the tree sense of s0 in split(pC(σ,τ)). Thenthere exists t ∈ split(qC(σ,τ)) with tp(s0¹n, t) is even and s0 ∈ C(s0¹n,t).Choose the ≺-maximal t with that property (property (v) of the definitionof good sequences ensures that there are only finitely many such t) anddefine s1 := s0 ¹ n and t1 := t.Suppose now, sj and tj are defined for some j ∈ ω. If sj = σ we havetj = τ and k = j and we are done. Otherwise, there exists n ∈ ω andm ∈ ω such that tj ∈ C(sj ,tj¹m) and sj ∈ C(sj¹n,tj¹m). Define sj+1 := sj ¹ nand tj+1 := tj ¹ m. After finitely many steps the construction stops. 2

Lemma 2.7 Suppose (p, q) ∈M2 and A and B are good (p, q)-sequences.If there exists a good (p, q)-sequence C with A ≤ C ≈ B, then there existsa good (p, q)-sequence D with A ≈ D ≤ B, and hence we have A ≤∗ B.Moreover, D can be chosen such that D ≤ A.

Proof: Suppose A, B and C are good (p, q)-sequences with A ≤ C ≈ B.Define

D(σ,τ) :=

{A(σ,τ) ∩ split(pB(σ,τ)), if tp(σ, τ) is even,

A(σ,τ) ∩ split(qB(σ,τ)), if tp(σ, τ) is odd.

Claim: D := (D(σ,τ))(σ,τ)∈I(p,q)is a good (p, q)-sequence such that

A ≈ D ≤ B.

21

Proof of the claim: For proving that D is good it is sufficient to showD(σ,τ) =∗ A(σ,τ) for every (σ, τ) ∈ I(p,q).If tp(σ, τ) is even, we prove |A(σ,τ)\split(pB(σ,τ))| < ω: For fixed µ ∈ A(σ,τ)\split(pB(σ,τ)), and therefore µ ∈ split(pC(σ,τ)), we have the unique sequencesµ = s0 % . . . % sk = σ and t1 % . . . % tk = τ as in lemma 2.6. By choice,there must exist i ∈ {0, . . . , k} such that si ∈ C(si+1,ti+1) \B(si+1,ti+1). De-fine Φ(σ,τ)(µ) to be the shortest such si. Then Φ(σ,τ) ¹ (A(σ,τ)\split(pB(σ,τ)))is an injective function, because for µ, µ′ ∈ A(σ,τ) we have µ(|σ|) 6= µ′(|σ|).So we have |A(σ,τ) \ split(pB(σ,τ))| ≤ Σ(s,t)∈I(p,q)

|C(s,t) \B(s,t)| < ω; the lastinequality holds because we have B ≈ C, and therefore B(σ,τ) = C(σ,τ) foralmost every (σ, τ) ∈ I(p,q).If tp(σ, τ) is odd, we get |A(σ,τ) \ split(qB(σ,τ))| < ω by a similar proof.It remains to prove that we have A(σ,τ) = D(σ,τ) for all but finitely many(σ, τ) ∈ I(p,q). Define

E := {(s, t) ∈ (<ωω)2 | ∃(σ, τ) ∈ I(p,q) ((σ, τ) ≥ (s, t) ∧ B(σ,τ) 6= C(σ,τ))}.

E is finite and for all (s, t) in I(p,q)\E we have (pB(s,t), qB(s,t)) = (pC(s,t), q

C(s,t)),

and hence A(s,t) = D(s,t). 2

Corollary 2.8 Fix (p, q) ∈M2. Then our relation ≤∗ is transitive on theset of good (p, q)-sequences.

Proof: Suppose A ≤∗ B ≤∗ C, so there exist good sequences A′ and B′witnessing these. Hence we have the following relationships:

A ≈ A′ ≈ D

≤ ≤

B ≈ B′

≤

C

D exists by lemma 2.7 and is a witness for A ≤∗ C. 2

Lemma 2.9 Suppose (p, q) ∈ M2, A is a good (p, q)-sequence and X ∈J (M2). Then there exists a good (p, q)-sequence B ≤ A such that ([pB(σ,τ)]×[qB(σ,τ)]) ∩X = ∅ for every (σ, τ) ∈ I(p,q).

Proof: Let ((σn, τn))n∈ω enumerate all pairs in I(p,q) starting with (σ0, τ0) =(st(p), st(q)). Let n0 := 0 and choose by fact 2.2 (p(σn0 ,τn0 ), q(σn0 ,τn0 )) inM2 such that (p(σn0 ,τn0 ), q(σn0 ,τn0 )) ≤0 (pA(σn0 ,τn0 ), q

A(σn0 ,τn0 )) and

([p(σn0 ,τn0 )]× [q(σn0 ,τn0 )]) ∩X = ∅.

By a remark in the preliminaries, there exists (p′(σn0 ,τn0 ), q′(σn0 ,τn0 )) ∈

M2 such that (p′(σn0 ,τn0 ), q′(σn0 ,τn0 )) ≤0 (p(σn0 ,τn0 ), q(σn0 ,τn0 )) and for al-

most all n ∈ ω and every (x, y) ∈ [p′(σn0 ,τn0 )] ×+ [q′(σn0 ,τn0 )] we havetp(σn0 ,τn0 )-n-pair(x, y) ∈ split(p′(σn0 ,τn0 ))×split(q′(σn0 ,τn0 )). By remark 2.4we get a partial good sequence B0 = (B0

(σ,τ))(σ,τ)∈I0 , where I0 := I(p,q) ∩(split(p′(σn0 ,τn0 ))× split(q′(σn0 ,τn0 ))). Suppose we have already constructeda strictly ascending sequence (ni)i≤k of natural numbers and a sequence

22

(Bi)i≤k of partial good sequences Bi = (Bi(σ,τ))(σ,τ)∈Ii

such that I0 ⊆ . . . ⊆Ik−1 ⊆ Ik and Bi+1 extends the sequence Bi for every i < k, this meansBi+1

(σ,τ) = Bi(σ,τ) for every (σ, τ) ∈ Ii. Choose nk+1 ∈ ω minimal such that

(σnk+1 , τnk+1) /∈ Ik. Choose, by remark 2.2, (p(σnk+1 ,τnk+1 ), q(σnk+1 ,τnk+1 ))inM2 with (p(σnk+1 ,τnk+1 ), q(σnk+1 ,τnk+1 )) ≤0 (pA(σnk+1 ,τnk+1 ), q

A(σnk+1 ,τnk+1 ))

and([p(σnk+1 ,τnk+1 )]× [q(σnk+1 ,τnk+1 )]) ∩X = ∅.

Thin out p(σnk+1 ,τnk+1 ) and q(σnk+1 ,τnk+1 ) so that whenever σnk+1 ∈p(σni

,τni)\split(p(σni

,τni)) for some i ≤ k we have [p(σnk+1 ,τnk+1 )]∩[p(σni

,τni)]

= ∅ and whenever τnk+1 ∈ q(σni,τni

) \ split(q(σni,τni

)) for some i ≤ kwe have [q(σnk+1 ,τnk+1 )] ∩ [q(σni

,τni)] = ∅ (Notice that we only have to

cut out one single Miller tree for every relevant i). Again by the re-mark in the preliminaries there exists (p′(σnk+1 ,τnk+1 ), q

′(σnk+1 ,τnk+1 )) ≤0

(p(σnk+1 ,τnk+1 ), q(σnk+1 ,τnk+1 )) such that for almost all n ∈ ω and every(x, y) ∈ [(p′(σnk+1 ,τnk+1 )] ×+ [q′(σnk+1 ,τnk+1 )] we have tp(σnk+1 ,τnk+1 )-n-pair

(x, y) ∈ split(p′(σnk+1 ,τnk+1 ))×split(q′(σnk+1 ,τnk+1 )). By remark 2.4 there ex-

ists a partial good sequence C = (C(σ,τ))(σ,τ)∈J for the pair (p(σnk+1 ,τnk+1 ),

q(σnk+1 ,τnk+1 )), where J := I(p,q)∩(split(p′(σnk+1 ,τnk+1 ))×split(q′(σnk+1 ,τnk+1 )).Define Ik+1 := Ik ∪ J and Bk+1 by

Bk+1(σ,τ) :=

{Bk

(σ,τ), if (σ, τ) ∈ Ik,

C(σ,τ), if (σ, τ) ∈ Ik+1 \ Ik.

It is easy to see that Bk+1 is a partial good sequence which extends Bk.In the end, we get a good sequence B = (B(σ,τ))(σ,τ)∈⋃

n∈ω In. We also have⋃

n∈ω In = I(p,q), because for every (σ, τ) ∈ I(p,q) there exists nk ∈ ω with(σ, τ) ∈ split(p(σnk

,τnk)) × split(q(σnk

,τnk)), and hence B(σ,τ) is defined.

Clearly, we have ([pB(σ,τ)]× [qB(σ,τ)]) ∩X = ∅ for every (σ, τ) ∈ I(p,q). 2

Lemma 2.10 Suppose MAκ(σ-centered) is true and fix (p, q) ∈M2.If (Aα)α<κ is a ≤∗-descending sequence of good (p, q)-sequences, then thereexists a good (p, q)-sequence B such that B ≤∗ Aα for all α < κ.

Proof: Define a forcing P as follows:Conditions in P are of the form (F, {T(σ,τ) | (σ, τ) ∈ S}) such that

(a) F is a finite subset of κ;

(b) S is a finite subset of I(p,q);

(c) T(σ,τ) is a finite subset of <ωω;

(d) if tp(σ, τ) is even, then ∀s ∈ T(σ,τ)(s % σ ∧ |s| > s(|σ|)) and ∀s, s′ ∈T(σ,τ)(s 6= s′ ⇒ s(|σ|) 6= s′(|σ|)),if tp(σ, τ) is odd, then ∀t ∈ T(σ,τ)(t % τ ∧ |t| > t(|τ |)) and ∀t, t′ ∈T(σ,τ)(t 6= t′ ⇒ t(|τ |) 6= t′(|τ |));

(e) T(σ,τ) ⊆ Sop(p,q)st(p),st(q)(σ, τ);

(f) if tp(σ, τ) is even and (σ, τ), (ρ, τ), (ρ′, τ) are in S such that ρ, ρ′ ∈T(σ,τ) and ρ(|σ|) < ρ′(|σ|), then we have T(ρ,τ) ⊇ T(ρ′,τ);if tp(σ, τ) is odd and (σ, τ), (σ, ρ), (σ, ρ′) are in S such that ρ, ρ′ ∈T(σ,τ) and ρ(|τ |) < ρ′(|τ |), then we have T(σ,ρ) ⊇ T(σ,ρ′).

23

Define (F, {T(σ,τ) | (σ, τ) ∈ S}) ≤ (F ′, {T ′(σ,τ) | (σ, τ) ∈ S′}) :⇔

F ⊇ F ′,S ⊇ S′,∀ (σ, τ) ∈ S′(T(σ,τ) ⊇ T ′(σ,τ)) and∀(σ, τ) ∈ S ∀α ∈ F ′ (T(σ,τ) \ T ′(σ,τ) ⊆ split(rA

α

(σ,τ))),

where r ={

p, if tp(σ, τ) is even,q, if tp(σ, τ) is odd; and T ′(σ,τ) = ∅ if (σ, τ) /∈ S′.

Then P is σ-centered because

P =⋃{P{T(σ,τ) | (σ,τ)∈S} | {T(σ,τ) | (σ, τ) ∈ S} is as in the definition of P},

where P{T(σ,τ) | (σ,τ)∈S} := {(F, {T(σ,τ) | (σ, τ) ∈ S}) ∈ P |F ⊆ κ finite}.There are only countably many {T(σ,τ) | (σ, τ) ∈ S} and obviously the setsP{T(σ,τ) | (σ,τ)∈S} are centered.In the following we define suitable dense subsets of P : For α < κ let

Dα := {(F, {T(σ,τ) | (σ, τ) ∈ S}) ∈ P |α ∈ F}.

Clearly, Dα is dense for every α < κ.We define

Es,t,n := {(F, {T(σ,τ) | (σ, τ) ∈ S}) | (s, t) ∈ S ∧ ∃µ ∈ T(s,t) (µ(|s|) > n)}

for n ∈ ω and s, t ∈ <ωω such that tp(s, t) is even and

Fs,t,m := {(F, {T(σ,τ) | (σ, τ) ∈ S}) | (s, t) ∈ S ∧ ∃ν ∈ T(s,t) (ν(|t|) > m)}

for m ∈ ω and s, t ∈ <ωω such that tp(s, t) is odd.

Claim 1: The set Es,t,n is dense for every n ∈ ω and every s, t ∈ <ωωsuch that tp(s, t) is even and Fs,t,m is dense for every m ∈ ω and everys, t ∈ <ωω such that tp(s, t) is odd.

Proof of claim 1: We only prove the even case, as the odd case issimilar. Suppose n ∈ ω and s, t are in <ωω such that tp(s, t) is even and(F, {T(σ,τ) | (σ, τ) ∈ S}) ∈ P \ Es,t,n. Define S′ := S ∪ {(s, t)} (maybeS = S′). Let (α0, . . . , αk−1) be an ascending enumeration of F . Thenwe have Aα0 ≥∗ . . . ≥∗ Aαk−1 and this implies that the set

⋂i<k Aαi

(s,t)

is infinite. We have to take care of property (f) in the definition of P :Suppose there exists τ with (s, τ) ∈ S and t ∈ T(s,τ) (τ is unique); elsewe have nothing to do. Let (tj)j<N be an enumeration of all t′ witht′ ∈ T(s,τ), (s, t′) ∈ S and t′(|τ |) < t(|τ |) such that tj(|τ |) < tj+1(|τ |) forevery j < N−1. Then we have Aαi

(s,t0)⊇∗ Aαi

(s,t1)⊇∗ . . . ⊇∗ Aαi

(s,t) for everyi < k, and therefore

⋂i<k Aαi

(s,t) ⊆∗⋂

i<k Aαi

(s,tN−1)⊆∗ . . . ⊆∗ ⋂

i<k Aαi

(s,t0).

So the set ⋂

i<k

Aαi

(s,t) ∩⋂

i<k

Aαi

(s,tN−1)∩ . . . ∩

⋂

i<k

Aαi

(s,t0)

is infinite. Choose µ ∈ ⋂i<k Aαi

(s,t)∩⋂

i<k Aαi

(s,tN−1)∩ . . .∩⋂

i<k Aαi

(s,t0)with

µ(|s|) > n and µ /∈ π1(S) = {s | ∃t ((s, t) ∈ S)}. Define T ′(s,t) := T(s,t)∪{µ}(T(s,t) = ∅ if (s, t) /∈ S), T ′(s,tj)

:= T(s,tj) ∪ {µ} for every j < N andT ′(σ,τ) := T(σ,τ) for every (σ, τ) ∈ S \ ({(s, t)} ∪ {(s, tj) | j < N}). We have

24

T ′(s,t′) \ T(s,t′) = {µ} ⊆ split(pAαi

(s,t′)) for every t′ ∈ {tj | j < N − 1} ∪ {t},and hence (F, {T ′(σ,τ) | (σ, τ) ∈ S}∪{T ′(s,t)}) ≤ (F, {T(σ,τ) | (σ, τ) ∈ S}) and(F, {T ′(σ,τ) | (σ, τ) ∈ S} ∪ {T ′(s,t)}) is in Dα. 2(claim1)

By MA(σ-centered) let G be a generic filter for the following dense sets:Es,t,n for s, t ∈ <ωω such that tp(s, t) is even and n ∈ ω, Fs,t,m fors, t ∈ <ωω such that tp(s, t) is odd and m ∈ ω and Dα for α < κ. For(s, t) ∈ I(p,q) define

B(s,t) := {ρ ∈ <ωω | ∃(F, {T(σ,τ) | (σ, τ) ∈ S}) ∈ G ((s, t) ∈ S ∧ ρ ∈ T(s,t))}.Claim 2: B := (B(s,t))(s,t)∈I(p,q)

is a good (p, q)-sequence.

Proof of claim 2: First suppose tp(s, t) is even and ρ, ρ′ ∈ B(s,t) withρ 6= ρ′. So there are (F, {T(σ,τ) | (σ, τ) ∈ S}) and (F ′, {T ′(σ,τ) | (σ, τ) ∈ S′})in G with ρ ∈ T(s,t) and ρ′ ∈ T ′(s,t), and there exists (F ′′, {T ′′(σ,τ) | (σ, τ) ∈S′′}) ≤ (F, {T(σ,τ) | (σ, τ) ∈ S}), (F ′, {T ′(σ,τ) | (σ, τ) ∈ S′}) in G. So wehave T ′′(s,t) ⊇ T(s,t), T

′(s,t), and hence ρ(|s|) 6= ρ′(|s|). The case that tp(s, t)

is odd is analogous, so (ii) of the definition of a good sequence (defini-tion 2.3) holds.For (i) we have to prove that |B(s,t)| is infinite. If tp(s, t) is even, for everyn ∈ ω there exists (F, {T(σ,τ) | (σ, τ) ∈ S}) ∈ G ∩ Es,t,n, hence (s, t) ∈ Sand there is a µ ∈ T(s,t) with µ(|s|) > n. If tp(s, t) is odd we can find forevery n ∈ ω a ν ∈ T(s,t) with ν(|t|) > n.(iii) and (iv) are easy to verify. For (v) suppose tp(s, t) is even andρ ∈ B(s,t). Fix ν ∈ B(ρ,τ). Since G ∩ Es,t,n 6= ∅ for every n ∈ ω, weget ρ′ ∈ B(s,t) such that ρ′(|s|) > #ν, hence ν /∈ B(ρ′,t), and hence wehave B(ρ,t) % B(ρ′,t). 2(claim 2)

Claim 3: B ≤∗ Aα for every α < κ.

Proof of claim 3: Suppose α < κ and fix (F, {T(σ,τ) | (σ, τ) ∈ S}) ∈G ∩Dα.

Subclaim 1: For every (s, t) ∈ I(p,q) \ S we have B(s,t) ⊆ split(rAα

(s,t))where r = p if tp(s, t) is even and r = q if tp(s, t) is odd.

Proof of subclaim 1: Suppose (s, t) ∈ I(p,q) \ S and tp(s, t) is even. Letµ ∈ B(s,t); then there exists (F ′, {T ′(σ,τ) | (σ, τ) ∈ S′}) ∈ G with (s, t) ∈ S′

and µ ∈ T ′(s,t). Choose (F ′′, {T ′′(σ,τ) | (σ, τ) ∈ S′′}) ≤ (F, {T(σ,τ) | (σ, τ) ∈S}), (F ′, {T ′(σ,τ) | (σ, τ) ∈ S′}) in G. Then we have (s, t) ∈ S′′, µ ∈ T ′′(s,t),(s, t) /∈ S and α ∈ F and these imply µ ∈ T ′′(s,t) \ T(s,t) = T ′′(s,t) ⊆split(pA

α

(s,t)), as desired. The odd case is once more analogous.2(subclaim 1)

Subclaim 2: For the finitely many (s, t) ∈ S we have B(s,t) ⊆∗ split(rAα

(s,t))for r = p if tp(s, t) is even and r = q if tp(s, t) is odd.

Proof of subclaim 2: Suppose (s, t) ∈ S and tp(s, t) is even. We provethat we have B(s,t)\T(s,t) ⊆ split(pA

α

(s,t)). Suppose µ ∈ B(s,t)\T(s,t), so thereexists (F ′, {T ′(σ,τ) | (σ, τ) ∈ S′}) ∈ G with (s, t) ∈ S′ and µ ∈ T ′(s,t). Choose(F ′′, {T ′′(σ,τ) | (σ, τ) ∈ S′′}) ≤ (F, {T(σ,τ) | (σ, τ) ∈ S}), (F ′, {T ′(σ,τ) | (σ, τ) ∈

25

S′}) in G. Then we have µ ∈ T ′′(s,t), µ /∈ T(s,t) and α ∈ F and these implyµ ∈ T ′′(s,t) \ T(s,t) ⊆ split(pA

α

(s,t)). The odd case is analogous. 2(subclaim 2)Subclaim 1 and 2 together easily imply B ≤∗ Aα. 2(claim 3)

2

Now we are able to prove the main theorem of this chapter, but we firstwrite down a generalized version:

Theorem 2.11 MAκ(σ-centered) implies add(J (M2)) > κ.

Proof of theorem 2.11 and therefore of theorem 2.1: SupposeMAκ(σ-centered) is true and Xα ∈ J(M2) for α < κ and (p, q) ∈ M2. Wesearch for a pair (p′, q′) ≤ (p, q) inM2 such that ([p′]×[q′])∩⋃

α<κ Xα = ∅.Define by induction a ≤∗-descending sequence (Aα)α≤κ of good (p, q)-sequences: For α = 0 define A0 to be the good sequence for (p, q) definedin remark 2.4. Suppose Aβ for β < α is already defined and α < κ isa limit ordinal. Choose by lemma 2.10 (we use MAκ(σ-centered) here) agood (p, q)-sequence Aα with Aα ≤∗ Aβ for every β < α.For the successor step, suppose Aα is defined. Choose a good (p, q)-sequence Aα+1 ≤∗ Aα with ([pA

α+1

(σ,τ) ]× [qAα+1

(σ,τ) ])∩Xα = ∅ for every (σ, τ) ∈I(p,q) by lemma 2.9.

Notice that the pair (pAκ

(st(p),st(q)), qAκ

(st(p),st(q))) of Miller trees does not nec-essarily have the desired property, because for α < κ we only have Aκ ≤∗Aα+1, hence there could exist a pair (f, g) ∈ [pA

κ

(st(p),st(q))] × [qAκ

(st(p),st(q))]

with (f, g) /∈ [pAα+1

(st(p),st(q))] × [qAα+1

(st(p),st(q))], and therefore we do not knowwhether (f, g) is in X or not. We will define a forcing Q to get (p′, q′) ≤(pA

κ

(st(p),st(q)), qAκ

(st(p),st(q))) such that for every (f, g) ∈ [p′] × [q′] and every

α < κ we have (f, g) ∈ [pAα+1

(f¹n,g¹m)] × [qAα+1

(f¹n,g¹m)] for suitable n, m ∈ ω. Inthis situation the choice of Aα+1 guarantees (f, g) /∈ X.

Define Q in the following way: Q is the set of all (u, v, X) with

(a’) X ⊆ κ is finite;

(b’) u : m → split(pAκ

(st(p),st(q))) and v : m′ → split(qAκ

(st(p),st(q))), m,m′ ∈ ω

with m = m′ or m = m′ + 1;

(c’) the downward closures U and V of ran(u) and ran(v) are finite sub-trees of pA

κ

(st(p),st(q)), qAκ

(st(p),st(q)), respectively, with split(U) ⊆ ran(u)and split(V ) ⊆ ran(v);

(d’) Φ : (ran(u),≤) → (T ¹ m,≤) defined by u(k) 7−→ T (k) and Ψ :(ran(v),≤) → (T ¹ m′,≤) defined by v(k) 7−→ T (k) are both isomor-phisms;

(e’) if u(j) is the immediate predecessor in the tree sense of u(k + 1)in ran(u), then we have u(k + 1)(|u(j)|) > max({#u(i) | i ≤ k} ∪{#v(i) | i ≤ k}) and |u(k + 1)| > u(k + 1)(|u(j)|),if v(j) is the immediate predecessor in the tree sense of v(k + 1) inran(v), then we have v(k + 1)(|v(j)|) > max({#u(i) | i ≤ k + 1} ∪{#v(i) | i ≤ k}) and |v(k + 1)| > v(k + 1)(|v(j)|).

Define (u, v,X) ≤ (u′, v′, X ′) :⇔u ⊇ u′, v ⊇ v′, X ⊇ X ′ and

26

for every α ∈ X ′ and every (σ, τ) ∈ (ran(u)× ran(v)) \(ran(u′)× ran(v′)) we have (σ, τ) ∈ pA

α

(σ¹n,τ¹m) × qAα

(σ¹n,τ¹m)

for every n,m ∈ ω such that (σ ¹ n, t¹ m) ∈ I(p,q).

Q is σ-centered, because Q =⋃{Q(u,v) | (u, v) is as in the definition of Q},

where Q(u,v) := {(u, v, X) ∈ Q |X ⊆ κ is finite} is obviously centered andthere are only countably many (u, v).In the following, we will define suitable dense subsets of Q: For α < κdefine

Dα := {(u, v,X) ∈ Q |α ∈ X}.Clearly, Dα is dense for every α < κ.We define

Es,t,n := {(u, v, X) ∈ Q | (s, t) ∈ ran(u)× ran(v) ∧∃µ ∈ ran(u) (s $ µ ∧ µ(|s|) > n)}

for (s, t) ∈ I(p,q) with tp(s, t) even and n ∈ ω and

Fs,t,m := {(u, v,X) ∈ Q | (s, t) ∈ ran(u)× ran(v) ∧∃ν ∈ ran(v) (t $ ν ∧ ν(|t|) > m)}

for (s, t) ∈ I(p,q) with tp(s, t) odd and m ∈ ω. For n ∈ ω and (s, t) ∈ I(p,q)

with tp(s, t) even the set Es,t,n is dense below every (u, v, X) with (s, t) ∈ran(u) × ran(v). And for m ∈ ω and (s, t) ∈ I(p,q) with tp(s, t) odd theset Fs,t,m is dense below every (u, v, X) with (s, t) ∈ ran(u)× ran(v). Theproof of this equals the proof of claim 1 in lemma 2.10 and is skipped.Letting

Es,t,n := {(u, v, X) ∈ Q | ∃(u′, v′, X ′) ∈ Es,t,n ((u, v, X) ≤ (u′, v′, X ′))}∪{(u, v,X) ∈ Q | (u, v, X) is incompatible with every element of Es,t,n}

and

Fs,t,m := {(u, v, X) ∈ Q | ∃(u′, v′, X ′) ∈ Fs,t,m ((u, v,X) ≤ (u′, v′, X ′))}∪{(u, v, X) ∈ Q | (u, v,X) is incompatible with every element of Fs,t,m}

we get dense subsets of Q. By MAκ(σ-centered) let G be a generic filterfor the dense sets Dα for α < κ, Es,t,n for (s, t) ∈ I(p,q) with tp(s, t) evenand n ∈ ω and Fs,t,m with (s, t) ∈ I(p,q) with tp(s, t) odd and m ∈ ω.Define p′ and q′ to be the superperfect trees determined by

split(p′) =⋃{ran(u) | ∃v∃X ((u, v,X) ∈ G)}

and bysplit(q′) =

⋃{ran(v) | ∃u∃X ((u, v, X) ∈ G)}.

Claim 1: (p′, q′) ∈M2.

Proof of claim 1: Suppose (s, t) ∈ p′×q′. Then there exists (u, v,X) ∈ Gand s′ ∈ ran(u) and t′ ∈ ran(v) with s′ ⊇ s, t′ ⊇ t, (s′, t′) ∈ ran(u)×ran(v)and tp(s′, t′) exists. If tp(s′, t′) is even, we have G ∩ Es′,t′,n 6= ∅ for everyn ∈ ω, therefore s′ is an infinite splitnode above s, and hence p′ ∈ M. Iftp(s′, t′) is odd we have G∩ Fs′,t′,m 6= ∅ for some m ∈ ω and it is possibleto choose extensions s′′ of s and t′′ of t′ with tp(s′′, t′′) is even. Again,

27

we have G ∩ Es′′,t′′,n 6= ∅ for every n ∈ ω, and therefore s′′ is an infinitesplitnode above s. Analogously, we get q′ ∈M. 2(claim 1)

Obviously, we have (p′, q′) ≤ (p, q).

Claim 2: ([p′]× [q′]) ∩⋃α<κ Xα = ∅.

Proof of claim 2: It suffices to prove ([p′] × [q′]) ∩ Xα = ∅ for everyα < κ. Fix α < κ and (u, v,X) ∈ G ∩Dα+1, so α + 1 ∈ X and let U andV be the downward closures of ran(u), ran(v), respectively.

Subclaim: For every (s, t) ∈ (p′ \U)× (q′ \V ) we have (s, t) ∈ pAα+1

(s¹n,t¹m)×qA

α+1

(s¹n,t¹m) for every n,m ∈ ω such that (s¹ n, t¹ m) ∈ I(p,q).

Proof of the subclaim: Suppose (s, t) ∈ (p′ \ U) × (q′ \ V ) and choose(s′, t′) ≥ (s, t) with (s′, t′) ∈ ((p′ \ U) × (q′ \ V )) ∩ (split(pA

κ

(st(p),st(q))) ×split(qA

κ

(st(p),st(q)))). There are (u′, v′, X ′), (u′′, v′′, X ′′) ∈ G such that s′ ∈ran(u′) and t′ ∈ ran(v′′). Choose (u, v, X) ≤ (u′, v′, X ′), (u′′, v′′, X ′′),(u, v, X) in G. Then we have s′ ∈ ran(u) \ ran(u), t′ ∈ ran(v) \ ran(v)and α + 1 ∈ X and these imply (s′, t′) ∈ (pA

α+1

(s′¹n,t′¹m) × qAα+1

(s′¹n,t′¹m)) for ev-ery n,m ∈ ω such that (s′ ¹ n, t′ ¹ m) ∈ I(p,q). Therefore, (s, t) is inpA

α+1

(s′¹n,t′¹m) × qAα+1

(s′¹n,t′¹m), as well. 2(subclaim)

Now we are able to prove claim 2: For (f, g) ∈ [p′] × [q′] we get (f, g) ∈[pA

α+1

(f¹n,g¹m)]× [qAα+1

(f¹n,g¹m)] for suitable n,m ∈ ω and we have chosen Aα+1 in

such a way that ([pAα+1

(f¹n,g¹m)]× [qAα+1

(f¹n,g¹m)]) ∩Xα = ∅ hold. 2(claim 2)2(theorems 2.11 and 2.1)

Since MAκ(σ-centered) is true for every κ < p, where p is the least cardi-nality of a filter base on ([ω]ω,⊆∗) without any lower bound, theorem 2.11implies the following:

Corollary 2.12 add(J (M2)) ≥ p. 2

Notice also that we have cov(J (M2)) ≤ d as in the one-dimensional case,where d is the dominating number, i.e. the least cardinality of a dominatingfamily in (ωω,≤∗). For this, let (fα)α<d be a dominating family in ωω anddefine Xα := {(f, g) ∈ ωω × ωω | f, g ≤∗ fα}. Then it is easy to see that⋃

α<d Xα = ωω × ωω and clearly, each Xα is in J (M2). This argumentactually proves cov(J (Mn)) ≤ d for every n ∈ ω. Summarizing, we have

ω1 ≤ p ≤ add(J (M2)) ≤ cov(J (M2)) ≤ d ≤ 2ω.

Let us remark that, in the model built for proving the main theorem 1.2in the previous chapter, we have add(J (M2)) > cov(M) ≥ t ≥ p, wheret is the tower number, that is the least cardinality of a decreasing chainin ([ω]ω,⊆∗) without any lower bound. The last inequality is trivial, thesecond one is a result of Rothberger [R].

The following two definitions are due to Velickovic [Ve].

Definition 2.13 For a forcing P and a dense set D ⊆ P we say that asubordering Q ⊆ P captures the density of D if D ∩Q is dense in Q.

28

If D is a family of dense sets of P , we say that Q captures the density ofD if Q captures the density of each D ∈ D.

Definition 2.14 For a forcing P and a class Z of forcings let Z(P ) bethe following statement:

Whenever D is a family of at most 2ω dense subsets of P and p ∈ P ,then there exists a Q ∈ Z such that p ∈ Q ⊆ P and Q captures thedensity of D.

Definition 2.15 For a forcing P define MA(P ) to be the statement:

For any collection of less than 2ω dense subsets of P there exists afilter on P intersecting all of them.

For a class Z of forcings MA(Z) means that MA(P ) is true for every P inZ.

Thus, the usual MA is MA(ccc), where ccc denotes the class of all forcingssatisfying the countable chain condition.It is easy to prove the following:

Fact 2.16 For a class Z of forcings and a single forcing P , MA(Z) andZ(P ) together imply MA(P ). 2

We want to prove the following result:

Theorem 2.17 MA(σ-centered) implies MA(M2).

Since MA implies MA(σ-centered) we get that MA implies MA(M2).Because of fact 2.16, the following theorem implies theorem 2.17.

Theorem 2.18 MA(σ-centered) implies σ-centered(M2).

For the proof we need the following definition and lemmata:

Definition 2.19 Suppose (p, q) ∈ M2. For a good (p, q)-sequence A de-fine

M2A := {(u, v) ∈M2 | (st(u), st(v)) ∈ I(p,q) ∧

∃A′ ≈ A∀(σ, τ) ∈ I(p,q) ∩ (split(u)× split(v))

((pA′

(σ,τ), qA′(σ,τ)) ≤ (u, v))}.

Lemma 2.20 Suppose (p, q) ∈ M2 and suppose A and B are good (p, q)-sequences. Then the following hold:

(a) M2A is σ-centered.

(b) A ≈ B implies M2A =M2

B.

(c) A ≤∗ B implies M2A ⊇M2

B.

Proof: For (a): For (s, t) ∈ I(p,q) define

MA(s,t) := {(u, v) ∈M2

A | st(u) = s ∧ st(v) = t}.

Then we have M2A =

⋃{MA(s,t) | (s, t) ∈ I(p,q)} and since I(p,q) is countable,

it suffices to prove that each MA(s,t) is centered. Fix (s, t) and assume

(u, v), (u′, v′) ∈ MA(s,t). Then there are good sequences B and B′ witnessing

29

this. Suppose tp(s, t) is even, the odd case is similar. It is easy to find aset C(s,t) such that C(s,t) =∗ A(s,t), C(s,t) =∗ B(s,t) and

∀σ ∈ C(s,t) ∀σ′ ⊇ σ ∀τ ⊇ t (tp(σ′, τ) exists ⇒ B(σ′,τ) = B′(σ′,τ)).

Define C(σ,τ) := B(σ,τ) for (σ, τ) 6= (s, t). Then C = (C(σ,τ))(σ,τ)∈I(p,q)is

a good sequence with B ≈ C ≈ B′ and C ≈ A. And (pC(s,t), qC(s,t)) is an

element of MA(s,t) smaller than (u, v) and (u′, v′).

(b) is clear.For (c): Suppose we have A ≤∗ B, so there exists a good (p, q)-sequenceA′ with A ≈ A′ ≤ B, and suppose (u, v) ∈ M2

B, witnessed by B′. Bylemma 2.7 there exists a good (p, q)-sequence A′′ such that A′ ≈ A′′ ≤ B′.Then for every ∀(σ, τ) ∈ I(p,q)∩(split(u)×split(v)) we have (pA

′′(σ,τ), q

A′′(σ,τ)) ≤

(pB′

(σ,τ), qB′(σ,τ)) ≤ (u, v), so A′′ witnesses that (u, v) ∈M2

A. 2

Lemma 2.21 Suppose (p, q) ∈ M2 and A is good for (p, q), (u, v) ∈ M2,D ⊆ M2 is dense and there exists a good (p, q)-sequence B ≤∗ A with(u, v) ∈M2

B.Then there exist C ≤∗ A and a pair (u′, v′) ≤ (u, v) such that (u, v) ∈ M2

Cand (u′, v′) ∈ D ∩M2

C. If D is ≤0-dense, we can choose (u′, v′) ≤0 (u, v).

Proof: Suppose B ≤∗ A and (u, v) ∈ M2B, witnessed by B′. Since D is

dense we find (u′, v′) ≤ (pB′

(st(u),st(v)), qB′(st(u),st(v))) ≤ (u, v) with (u′, v′) ∈ D

(If D is ≤0-dense, we choose (u′, v′) such that st(u′) = st(u) and st(v′) =st(v).). We define C in the following way: If tp(σ, τ) is even and (σ, τ) ∈I(p,q)∩(split(u)×split(v)) let C(σ,τ) be a subset of Succu′(σ)∩{ρ | ρ(|σ|) >#σ} such that for every ρ ∈ C(σ,τ) we have |ρ| > ρ(|σ|), min{ρ(|σ|), |ρ|} >#τ and ρ(|σ|) > #ρ′ for every ρ′ ∈ C(σ,τ) with ρ′(|σ|) < ρ(|σ|). Pro-ceed analogously for (σ, τ) ∈ I(p,q) ∩ (split(u) × split(v)) when tp(σ, τ)is odd. For (σ, τ) ∈ I(p,q) \ (split(u) × split(v)) let C(σ,τ) := B′

(σ,τ). If

tp(σ, τ) is even we have C(σ,τ) ⊆ split(pB′

(σ,τ)); if tp(σ, τ) is odd we have

C(σ,τ) ⊆ split(qB′

(σ,τ)). So C ≤ B′ ≤∗ A, and therefore C ≤∗ A, and wehave (u, v) ∈M2

B′ = M2B ⊆M2

C by lemma 2.20 (c). It is easy to prove that(u′, v′) is in M2

C . 2

Now we are able to prove theorem 2.18:

Proof of theorem 2.18: Suppose MA(σ-centered) is true. Suppose alsoD = {Dα |α < 2ω} is a family of dense subsets of M2 and (p, q) ∈M2. Let

{((pα, qα), Dα) |α < 2ω}

be an enumeration of M2 ×D with (p0, q0) = (p, q). We have to find a σ-centered forcing Q such that (p, q) ∈ Q ⊆ M2 and Q captures the densityof D.Define a ≤∗-descending sequence (Aα)α<2ω of good (p, q)-sequences as fol-lows: For α = 0 define A0 to be the good sequence for (p0, q0) = (p, q)defined in remark 2.4. Then clearly (p, q) ∈M2

A0 .If (Aβ)β<α is already defined and α is a limit ordinal, we choose by MA(σ-centered) and lemma 2.10 a good (p, q)-sequence Aα such that Aα ≤∗ Aβ

for every β < α.For the successor step suppose Aα is already defined. If there is no good(p, q)-sequence A ≤∗ Aα with (pα, qα) ∈M2

A, let Aα+1 := Aα.

30

If there exists A ≤∗ Aα with (pα, qα) ∈ M2A, lemma 2.21 gives us a good

(p, q)-sequence B ≤∗ Aα such that (pα, qα) ∈ M2B and there is a pair

(p′α, q′α) ≤ (pα, qα) with (p′α, q′α) ∈ Dα ∩M2B. Let Aα+1 := B.

In the end, defineQ :=

⋃α<2ω

M2Aα .

Clearly, we have Q ⊆M2.

Claim 1: Q is σ-centered.

Proof of claim 1: For (σ, τ) ∈ I(p,q) define Q(σ,τ) := {(u, v) ∈ Q | st(u) =σ ∧ st(v) = τ}; then we have Q =

⋃{Q(σ,τ) | (σ, τ) ∈ I(p,q)}. I(p,q) iscountable and as in the proof of lemma 2.20 (a) using that (Aα)α<2ω is a≤∗-descending chain, we see that every Q(σ,τ) is centered. 2(claim 1)

Claim 2: Q captures the density of D.

Proof of claim 2: Suppose D ∈ D and (p, q) ∈ Q, so there exists α < 2ω

such that ((p, q), D) = ((pα, qα), Dα). We have to find (p′, q′) ≤ (p, q) suchthat (p′, q′) ∈ D ∩Q.Since (p, q) ∈ Q we have (p, q) ∈M2

Aβ for some β, by lemma 2.20 we have(p, q) ∈ M2

Aγ for every γ ≥ β. Hence, at step α in the construction of thesequence (Aα)α<2ω the second case arises. There we have got (p′α, q′α) ≤(p, q) such that (p′α, q′α) ∈ D∩M2

Aα+1 ⊆ D∩Q. Let (p′, q′) := (p′α, q′α) andwe are done. 2(claim 2)

2

Corollary 2.22 MA(σ-centered) implies that M2 does not collapse car-dinals.

Proof: Since M2 clearly satisfies the (2ω)+-chain-condition, it could onlycollapse cardinals ≤ 2ω. Suppose there exist cardinals κ and λ such thatκ < λ ≤ 2ω and there is a M2-name τ with

°M2 τ : κ → λ is surjective.

Define for each α < λ

Dα := {(p, q) ∈M2 | ∃ξ < κ ((p, q) °M2 τ(ξ) = α)}.

Then Dα is open and dense for every α < λ. As MA(σ-centered) is true,we have σ-centered(M2) and since λ ≤ 2ω there exists a forcing P ⊆ M2

such that each Dα ∩ P is dense in P and P is σ-centered. But then Pcollapses κ to λ, and this is a contradiction to the fact that P fulfills thecountable chain condition. 2

3 Infinite maximal antichains in (P(ω)/fin)n∗

In the sequel, we always identify the elements of P(ω)/fin with theirrepresentatives in P(ω). By (P(ω)/fin)n∗ , for n∗ > 0 in ω, we meanthe full product carrying the coordinatewise ordering. Hence elements of(P(ω)/fin)n∗ are n∗-tuples of subsets of ω, elements of (P(ω)/fin \ {0})n∗

are n∗-tuples of infinite subsets of ω, or we can view the elements as

31

functions from n∗ to P(ω)/fin, P(ω)/fin \ {0}, respectively. For a ∈(P(ω)/fin \ {0})n∗ and F ⊆ n∗ we write a ¹ F for the restricted func-tion a : F → P(ω)/fin \ {0}. If F consists only of one element i, we writeai instead of a¹ {i}, so a = (ai)i<n∗ .Two elements a, b ∈ (P(ω)/fin \ {0})n∗ are called compatible if there ex-ists an element c ∈ (P(ω)/fin \ {0})n∗ with c ≤ a and c ≤ b, i.e. fora = (ai)i<n∗ and b = (bi)i<n∗ that ai ∩ bi is infinite for every i < n∗.As usual, if a and b are not compatible, they are called incompatible. Asubset A ⊆ (P(ω)/fin \ {0})n∗ of pairwise incompatible elements is calledan antichain of (P(ω)/fin)n∗ .Here, we are interested in maximal antichains of (P(ω)/fin)n∗ . Since(P(ω)/fin)n∗ can be densely embedded in the complete Boolean algebraof the regular open subsets of (P(ω)/fin)n∗ (see [Ko], for example), peopleoften use the terminology of Boolean algebras and call maximal antichainsin (P(ω)/fin)n∗ partitions.Of course there are finite maximal antichains of (P(ω)/fin)n∗ for everyn∗ ∈ ω \ {0} and all these are analytic.Our goal is to prove, analogously to Mathias’ one-dimensional result, thefollowing theorem:

Theorem 3.1 Fix n∗ ∈ ω \ {0, 1}.There are no analytic infinite maximal antichains in (P(ω)/fin)n∗ .

Recall that a filter on ω is called Ramsey if for every descending sequence(Xn)n∈ω with Xn ∈ U for every n ∈ ω there exists {xn |n ∈ ω} ∈ U suchthat xn+1 ∈ Xxn for every n ∈ ω. In this case, we say that {xn |n ∈ ω}diagonalizes the sequence (Xn)n∈ω. If f : ω → ω is an enumeratingfunction for the set {xn |n ∈ ω}, hence f(n + 1) ∈ Xf(n) for every n ∈ ω,we say that f diagonalizes (Xn)n∈ω.The Rudin-Keisler ordering ≤RK is defined by the following: For twoultrafilters U and U ′ we have U ≤RK U ′ if there exists a function h :ω → ω such that U = {x ⊆ ω |h−1[x] ∈ U ′}. Two ultrafilters U,U ′ arecalled Rudin-Keisler-equivalent if and only if U ≤RK U ′ and U ′ ≤RK U .Equivalently (see for example [Je]), U and U ′ are Rudin-Keisler-equivalentif there exists a bijection h : ω → ω such that U = {h[x] |x ∈ U ′}.It is well known that Martin’s Axiom implies the existence of a Ramseyultrafilter and the continuum hypothesis (CH) implies the existence as well(this is for example a consequence of lemma 3.12). And also a P(ω)/fin-generic filter over V is a Ramsey ultrafilter. Kunen got, starting with amodel of CH and adding ℵ2 random reals, a model in which there are noRamsey ultrafilters (see for example [Je]).For n∗ ∈ ω \ {0, 1} and an infinite maximal antichain A ⊆ (P(ω)/fin \{0})n∗ define

J (A) := {x ∈ (P(ω)/fin \ {0})n∗ | ∃k ∈ ω ∃a0, . . . , ak ∈ A∀i < n∗

xi \ (a0i ∪ . . . ∪ ak

i ) = 0}and

J (A)+ := (P(ω)/fin \ {0})n∗ \ J (A)

= {x ∈ (P(ω)/fin \ {0})n∗ | ∃∞a ∈ A (a and x are compatible)}.Here, ∃∞ means “there are infinitely many”. For x ∈ (P(ω)/fin \ {0})n∗

defineA¹ x := {a ∈ A | a and x are compatible},

32

the set of all elements of A compatible with x.We will use the following lemma of [Sp4]:

Lemma 3.2 [lemma 1.1 of [Sp4]]Suppose n∗ ∈ ω \ {0, 1} and A ⊆ (P(ω)/fin)n∗ is an infinite maximalantichain. If F ⊆ n∗ is (with respect to ⊆) maximal such that there existsF ∈ [A]ω with the property that {a ¹ F | a ∈ F} has the finite intersectionproperty in (P(ω)/fin \ {0})n∗ , then |F| = n∗ − 1.

Here, the finite intersection property in (P(ω)/fin\{0})n∗ means, that forevery finite subset there is a lower bound in (P(ω)/fin \ {0})n∗ . In otherwords, every finite coordinatewise intersection of elements of F is in everycoordinate infinite. Of course, this lemma also holds for the restrictionA¹ x for every x = (x0, . . . , xn∗−1) ∈ (P(ω)/fin \ {0})n∗ .

Corollary 3.3 Suppose n∗ and A are as in lemma 3.2 and x is an elementof (P(ω)/fin \ {0})n∗ . Then there exist F ∈ [A¹ x]ω and i∗ ∈ n∗ such thatfor every i ∈ n∗ \ {i∗} there exists bi ∈ [ω]ω with bi ⊆∗ ai for every a ∈ F .Clearly, the set {ai∗ | a ∈ F} must be an antichain.

Lemma 3.4 Suppose n∗ ∈ ω \ {0, 1} and A ⊆ (P(ω)/fin)n∗ is an infinitemaximal antichain. Suppose also b = (b0, . . . , bn∗−1) ∈ J (A)+ and h :ω → ω is a bijection.Then for every pair k, l ∈ n∗ with k < l there exist b′k ⊆ bk and b′l ⊆ bl suchthat we still have (b0, . . . , b

′k, . . . , b′l, . . . , bn∗−1) ∈ J (A)+ and additionally

h[b′k] ∩ b′l = ∅ or h[b′l] ∩ b′k = ∅ holds.

Proof: So suppose b ∈ J (A)+, h : ω → ω is bijective and k < l < n∗ aregiven. By corollary 3.3 there exist i∗ < n∗ and F ∈ [A ¹ b]ω such that forevery i ∈ n∗ \{i∗} there exists b∗i ∈ [ω]ω with b∗i ⊆∗ ci for every c ∈ F . Let(cn)n∈ω be an enumeration of F . Notice that the set {cn

i∗ |n ∈ ω} consistsof pairwise almost disjoint subsets of ω, since F is countable without lossof generality we have cn

i∗ ∩ cmi∗ = ∅ for n 6= m.

First case: k, l ∈ n∗ \ {i∗}.Consider the set (b∗l ∩ bl) \ h[b∗k ∩ bk]. If this set is infinite, let b′l :=(b∗l ∩ bl) \ h[b∗k ∩ bk] and b′k := b∗k ∩ bk. Then clearly we have h[b′k] ∩ b′l = ∅and because of b′k ⊆ b∗k, b′l ⊆ b∗l and the choice of b∗k and b∗l we have(b0, . . . , b

′k, . . . , b′l, . . . , b

′n∗−1) is compatible with every c ∈ F , and hence it

is an element of J (A)+.If (b∗l ∩ bl) \ h[b∗k ∩ bk] is finite, divide the infinite set b∗l ∩ bl ∩ h[b∗k ∩ bk]in two infinite pieces b′∪b′′ and let b′l := b′ and b′k := h−1[b′′]. Again, wehave (b0, . . . , b

′k, . . . , b′l, . . . , b

′n∗−1) ∈ J (A)+.

Second case: l = i∗.If there is an infinite subsequence (cnm

i∗ )m∈ω of (cni∗)n∈ω such that cnm

i∗ ∩h[bk ∩ b∗k] ∩ bl is infinite for every m ∈ ω, then let

b′k := h−1[⋃

m∈ω cn2mi∗ ] ∩ (bk ∩ b∗k) and

b′l :=⋃

m∈ω cn2m+1i∗ ∩ bl.

If there are only finitely many cn0i∗ , . . . , cnu

i∗ such that cnmi∗ ∩ h[bk ∩ b∗k] ∩ bl

is infinite for every m ≤ u, then let

b′k := bk ∩ b∗k andb′l := (

⋃{cni∗ |n ∈ ω \ {n0, . . . , nu}} ∩ bl) \ h[bk ∩ b∗k].

Easily, in both subcases b′k and b′l are as claimed.Third case: k = i∗.

33

This case is analogous to the second one, we only have to change the rolesof k and l. 2

Now we introduce a sophisticated n∗-dimensional version of Mathias forc-ing:

Definition 3.5 [ShSp2] Fix n∗ ∈ ω \ {0, 1}. Let

Q := {(s, S) ∈ [ω]<ω × [ω]ω | max(s) < min(S) ∨ s = ∅}.For defining the ordering on Q we need the following: let (lj)j<|s| be theincreasing enumeration of s ∈ [ω]<ω and define si := {lj | j ≡ i(mod n∗)}for i < n∗ and let (kj)j<ω be the increasing enumeration of S ∈ [ω]ω anddefine Si := {kj | j ≡ i(mod n∗)} for i < n∗. Then we define

(s, S) ≤ (t, T ) :⇔ s ∩ (max t + 1) = t or t = ∅,si \ ti ⊆ Ti for every i < n∗ andSi ⊆ Ti for every i < n∗.

We write (s, S) ≤0 (t, T ) if (s, S) ≤ (t, T ) and additionally s = t hold.Notice that (s, S), (t, T ) ∈ Q are compatible if and only if

(∀i < n∗ (si ⊆ ti ∧ ti \ si ⊆ Si)) ∨ (∀i < n∗ (ti ⊆ si ∧ si \ ti ⊆ Ti))

and for every i < n∗ we have |Si ∩ Ti| = ω.

Note that this forcing is like the product of n∗ copies of Mathias forcingbut with the extra requirement that the n∗ components of the generic realare cyclically interleaved.

As for the classical Mathias forcing we have the following relationshipbetween Q-reals and generic filters: If G is Q-generic over some modelM define x :=

⋃{s | ∃S ((s, S) ∈ G)} and xi := {kj | j ≡ i(mod n∗)},where (ki)i<ω is the increasing enumeration of x (notice that we havexi =

⋃{si | ∃S′ ((s′, S′) ∈ G ∧ s′i = si)}). Then G = {(s, S) ∈ QM | ∀i <n∗ (si ⊆ xi ⊆ Si ∪ si)}. The proof of this fact equals the classical proof.Hence a set x is a Q-real if the set Gx := {(s, S) ∈ QM | ∀i < n∗(si ⊆ xi ⊆Si ∪ si)} is a Q-generic filter over M , where si is defined as above.Suppose U0, . . . , Un∗−1 are ultrafilters on ω. Then we define the suborder-ing Q(U0, . . . , Un∗−1) of Q as follows: in Q(U0, . . . , Un∗−1) are only thosepairs (s, S) ∈ Q with Si ∈ Ui for every i < n∗.

In the following we list some of the results of the investigation of Q byShelah and Spinas [ShSp2] which we will need later. All these resultsgeneralize well known facts about the one-dimensional Mathias forcing(see [Ma], for example). Fix n∗ ∈ ω \ {0, 1}.Fact 3.6 [ShSp2] The forcing Q is equivalent to the forcing (P(ω)/fin)n∗ ∗Q(G0, . . . , Gn∗−1), where (G0, . . . , Gn∗−1) is the canonical name for thegeneric object added by the forcing (P(ω)/fin)n∗ consisting of pairwise notRudin-Keisler-equivalent Ramsey ultrafilters. 2

Shelah and Spinas used a helpful game:

Definition 3.7 Let U0, . . . , Un∗−1 be ultrafilters on ω. Then the gameΓ(U0, . . . , Un∗−1) for two players I and II is defined as follows: player Ichooses in the mth-move (Am

0 , . . . , Amn∗−1) ∈ U0 × . . .× Un∗−1 and player

II responds by playing km ∈ Amm mod n∗ . Player II wins the game if and

only if for every i < n∗ the set {kn |n ≡ i(mod n∗)} is an element of Ui.

34

Fact 3.8 [ShSp2] Suppose U0, . . . , Un∗−1 are pairwise not Rudin-Keisler-equivalent Ramsey ultrafilters. Then player I does not have a winningstrategy in the game Γ(U0, . . . , Un∗−1). 2

Here, it is easy to see that it is necessary that the ultrafilters are pairwisenot Rudin-Keisler-equivalent.

Fact 3.9 [ShSp2] Let U0, . . . , Un∗−1 are pairwise not Rudin-Keisler-equi-valent Ramsey ultrafilters on ω. Suppose S ∈ [ω]ω such that for every i <n∗ and for every X ∈ Ui we have Si ⊆∗ X. Then S is Q(U0, . . . , Un∗−1)-generic over V . 2

As a corollary we get:

Fact 3.10 [ShSp2] Let U0, . . . , Un∗−1 are pairwise not Rudin-Keisler-equi-valent Ramsey ultrafilters. If S ∈ [ω]ω is Q(U0, . . . , Un∗−1)-generic over Vand T ∈ [ω]ω with Ti ⊆ Si for every i < n∗, then T is Q(U0, . . . , Un∗−1)-generic over V as well. 2

Fact 3.11 [ShSp2] If U0, . . . , Un∗−1 are pairwise not Rudin-Keisler-equi-valent Ramsey ultrafilters, then the forcing Q(U0, . . . , Un∗−1) has the puredecision property. This means for a given Q(U0, . . . , Un∗−1)-name Θ with°Q(U0,...,Un∗−1) Θ ∈ {0, 1} and given (s, S) ∈ Q(U0, . . . , Un∗−1), thereexist (t, T ) ∈ Q(U0, . . . , Un∗−1) and ε ∈ {0, 1} such that (t, T ) ≤0 (s, S)and (t, T ) °Q(U0,...,Un∗−1) Θ = ε. 2

Suppose A is an infinite maximal antichain in P(ω)/fin. Mathias [Ma]proved that CH implies that there exists a Ramsey ultrafilter U such thatU ⊆ J (A)+. For the proof of Theorem 3.1 we need the following n∗-dimensional generalization:

Lemma 3.12 (CH) Suppose n∗ ∈ ω\{0, 1} and A is an infinite maximalantichain in (P(ω)/fin)n∗ . Then there exist pairwise not Rudin-Keisler-equivalent Ramsey ultrafilters U0, . . . , Un∗−1 such that U0 × . . .×Un∗−1 ⊆J (A)+.

Proof: Assume CH and fix the following enumerations: Let (xα)α<ω1

enumerate [ω]ω; (hα)α<ω1 enumerate all bijections from ω to ω and let(αX)α∈Lim(ω1) enumerate all n∗-tuples of decreasing sequences (of lengthω) in [ω]ω, such that every αX occurs cofinally often.We want to construct sequences (aα

0 )α<ω1 , . . . , (aαn∗−1)α<ω1 in [ω]ω by in-

duction such that we get Ui as the filter generated by {aαi |α < ω1} for

i < n∗.For beginning the induction choose (a0

0, . . . , a0n∗−1) ∈ J (A)+ arbitrarily.

Suppose (aν0)ν<α, . . . , (aν

n∗−1)ν<α are chosen such that {aβi |β < ν} gen-

erates a filter called F νi for every i < n∗ and every ν < α and we have

F ν0 × . . .× F ν

n∗−1 ⊆ J (A)+ for every ν < α.First, suppose that α is a successor ordinal, say α = β + 1. Consider theinfinite subset xβ of ω. Let

0bi := aβi ∩ xβ and 1bi := aβ

i \ xβ

for i < n∗. Because of (aβ0 , . . . , aβ

n∗−1) ∈ J (A)+ there exists at leastone sequence (ε0, . . . , εn∗−1) ∈ {0, 1}n∗ such that (ε0b0, . . . ,

εn∗−1 bn∗−1) ∈J (A)+. Now consider the function hβ . By applying lemma 3.4 finitelymany times we get the following: For every i < n∗ there is an infinite

35

ai ⊆ εibi such that (a0, . . . , an∗−1) ∈ J (A)+ and for every pair k, l < n∗

with k 6= l we have hβ [ak] ∩ al = ∅ or hβ [al] ∩ ak = ∅. Let aαi := ai for

every i < n∗.Second, suppose that α is a limit ordinal. Consider the tuple αX =((αXn

0 )n∈ω, . . . , (αXnn∗−1)n∈ω) of decreasing sequences in [ω]ω.

First case: Πi<n∗{αXni |n ∈ ω} ⊆ Πi<n∗

⋃ν<α F ν

i . We want to constructfor every i < n∗ strictly increasing functions fi : ω → ω such that fi(n +1) ∈ αX

fi(n)i for every i < n∗ and every n ∈ ω, (f0[ω], . . . , fn∗−1[ω]) ∈

J (A)+ and Πi<n∗(⋃

α<ν F νi ∪{fi[ω]}) has the finite intersection property

in J (A)+.Fix for every i < n∗ an enumeration (a′ni )n∈ω of {aν

i | ν < α} .Choose g0

i (0) ∈ αX0i ∩ a′0i for every i < n∗. If g0

0(k), . . . , g0n∗−1(k) are

already chosen for some k ∈ ω, take for every i < n∗

g0i (k + 1) ∈ αX0

i ∩ . . . ∩ αXg0

i (k)i ∩ a′0i ∩ . . . ∩ a′k+1

i

with g0i (k + 1) > g0

i (k) (notice that in this first case the set αX0i ∩ . . . ∩

αXk+1i ∩ a′0i ∩ . . .∩ a′k+1

i is infinite for every i < n∗). Hence we get strictlyincreasing functions g0

i : ω → ω for i < n∗ with (g00 [ω], . . . , g0

n∗−1[ω]) ∈([ω]ω)n∗ . As A is a maximal antichain there is a c0 = (c0

0, . . . , c0n∗−1) ∈

A such that (g00 [ω], . . . , g0

n∗−1[ω]) and c0 are compatible in (P(ω)/fin)n∗ .Define

εY ni :=

{αXn

i \ c0i , if ε = 0;

αXni , if ε = 1.

Since we have (αXn0 , . . . , αXn

n∗−1) ∈ J (A)+ for every n ∈ ω and of course(c0

0, . . . , c0n∗−1) is not in J (A)+, for every n ∈ ω there exists a sequence

(εn0 , . . . , εn

n∗−1) ∈ {0, 1}n∗ such that εni = 0 for at least one i < n∗ and we

have (εn0Y n

0 , . . . , εnn∗−1Y n

n∗−1) ∈ J (A)+. Because the sequences (αXni )n∈ω

are decreasing we have the following: If (ε0Y n0 , . . . , εn∗−1Y n

n∗−1) /∈ J (A)+

for some sequence (ε0, . . . , εn∗−1) ∈ {0, 1}n∗ , then for every m > n wehave (ε0Y m

0 , . . . , εn∗−1Y mn∗−1) /∈ J (A)+. So we can choose an n ∈ ω large

enough such that for every m > n we have εmi = 1 whenever εn

i = 1.Let εi := εn

i for every i < n∗; then εi = 0 for at least one i < n∗ and(ε0Y n

0 , . . . , εn∗−1Y nn∗−1) ∈ J (A)+ for every n ∈ ω.

Let 1Zni := εiY n

i for every i < n∗ and every n ∈ ω. So we have an n∗-tuple1Z = ((1Zn

0 )n∈ω, . . . , (1Znn∗−1)n∈ω) of descending sequences in [ω]ω such

that (1Zn0 , . . . , 1Zn

n∗−1) ∈ J (A)+ for every n ∈ ω. Notice that we haveΠi<n∗{1Zn

i |n ∈ ω} ⊆ Πi<n∗⋃

ν<α F νi .

Now we construct functions g10 , . . . , g1

n∗−1 : ω → ω in the same way asbefore but for 1Z instead of αX. Choose g1

i (0) ∈ 1X0i ∩ a′0i for every i < n∗

and if g10(k), . . . , g1

n∗−1(k) are already chosen take for every i < n∗

g1i (k + 1) ∈ 1Z0

i ∩ . . . ∩ 1Zg1

i (k)i ∩ a′0i ∩ . . . ∩ a′k+1

i

with g1i (k + 1) > g1

i (k). Then there exists an element c1 ∈ A such that(g1

0 [ω], . . . , g1n∗−1[ω]) and c1 are compatible in (P(ω)/fin)n∗ . As there is at

least one index i < n∗ with 1Zni = αXn

i \ c0i for every n ∈ ω the elements

c0 and c1 are different. As before we let 0Y ni := 1Zn

i \ c1i and 1Y n

i := 1Zni

and we find a sequence (ε0, . . . , εn∗−1) ∈ {0, 1}n∗ with εi = 0 for at leastone i < n∗ such that (ε0Y n

0 , . . . , εn∗−1Y nn∗−1) ∈ J (A)+ for every n ∈ ω. For

every i < n∗ and every n ∈ ω let 2Zni := εiY n

i .Hence we can find a sequence (lZ)l∈ω = ((lZn

0 )n∈ω, . . . , (lZnn∗−1)n∈ω) of

36

n∗-tuples of descending chains in [ω]ω with (lZn0 , . . . , lZn

n∗−1) ∈ J (A)+

for every n ∈ ω, a sequence ((gl0, . . . , g

ln∗−1))l∈ω of n∗-tuples of increasing

functions from ω to ω with gli(k + 1) ∈ lZ

gli(k)

i and (gl0[ω], . . . , gl

n∗−1[ω]) ∈([ω]ω)n∗ and a third sequence (cl)l∈ω of pairwise different elements of Asuch that |cl

i ∩ gli[ω]| = ω for every i < n∗ and every l ∈ ω.

Now we are ready to define diagonalizing functions f0, . . . , fn∗−1 : ω → ωfor (αXn

0 )n∈ω, . . . , (αXnn∗−1)n∈ω such that (f0[ω], . . . , fn∗−1[ω]) ∈ J (A)+:

Choose fi(0) ∈ αX0i for i < n∗. For k > 0 there exist j, l ∈ ω such that

k = 2j(2l + 1). Choose for every i < n∗

fi(k) ∈ cji ∩ gj

i [ω] ∩ αX0i ∩ . . . ∩ αX

fi(k−1)i ∩ a′0i ∩ . . . ∩ a′k−1

i

with fi(k) > fi(k − 1).Now we define aα

i := fi[ω] for every i < n∗. Hence for every i < n∗ wehave that aα

i diagonalizes (αXni )n∈ω; and as (aα

0 , . . . , aαn∗−1) cuts infinitely

many cj we have (aα0 , . . . , aα

n∗−1) ∈ J (A)+ and Πi<n∗{aνi | ν ≤ α} has the

finite intersection property in J (A)+.

Second case: Πi<n∗{αXni |n ∈ ω} is not a subset of Πi<n∗

⋃ν<α F ν

i .Then choose aα

i ∈ [ω]ω arbitrarily such that aαi ⊆∗ aν

i for every ν < α andevery i < n∗.

This finishes our construction. Our construction guarantees that for everyi < n∗ the set {aα

i |α < ω1} has the finite intersection property, and thuswe can define Ui to be the filter generated by {aα

i |α < ω1}. We have tocheck that Ui is an ultrafilter, that this ultrafilter is Ramsey and that Ui

and Uj for i 6= j are not Rudin-Keisler-equivalent.First we check that each Ui is an ultrafilter: Suppose a ∈ [ω]ω, so thereexists α < ω1 with a = xα. If xα /∈ Ui we have aα+1

i * xα, hence inthe successor step we have chosen aα+1

i ⊆ aαi \ xα ⊆ ω \ xα, and hence

ω \ xα ∈ Ui.Check that each Ui is Ramsey: Suppose (Xn)n∈ω is a descending chain inUi. Then there exists α ∈ Lim(ω1) such that (Xn)n∈ω = (αXn

i )n∈ω andΠi<n∗{αXn

i |n ∈ ω} ⊆ Πi<n∗⋃

ν<α F νi , and hence we are in the first case

of the limit step of the construction and aαi ∈ Ui diagonalizes (αXn

i )n∈ω.It remains to check that for i, j < n∗ with i 6= j that the ultrafilters Ui

and Uj are not Rudin-Keisler-equivalent: Suppose there exists a bijectionh : ω → ω such that Ui = {h[x] |x ∈ Uj}. But there is an α < ω1 withh = hα and in the successor step of our construction we have chosen aα+1

i

and aα+1j such that hα[aα+1

i ]∩aα+1j = ∅ or hα[aα+1

j ]∩aα+1i = ∅, and hence

we get a contradiction. 2

Proof of theorem 3.1: Suppose n∗ ∈ ω \{0, 1} and A ⊆ (P(ω)/fin)n∗ isan infinite maximal antichain. Since we can collapse 2ω to ω1 by a σ-closedforcing, hence without changing the set P(ω), we can assume withoutloss of generality that CH is true. By lemma 3.12 there are pairwisenot Rudin-Keisler-equivalent Ramsey ultrafilters U0, . . . , Un∗−1 such thatU0 × . . .× Un∗−1 ⊆ J (A)+.Suppose A is analytic. Notice that in this case J (A) is analytic, too.Choose a countable elementary submodel N of HΘ for a sufficiently largeΘ such that J (A), U0, . . . , Un∗−1, code(J (A)) ∈ N , where code(J (A)) isthe Σ1

1-code of the analytic set J (A). Consider the image of the Mostowskicollapse ϕ(N ) =: N of N , hence N is transitive, J (A)∩N is analytic in N

37

by the same code as in V , Ui := ϕ(Ui) = Ui∩N , hence Ui ∈ N for every i <n∗ and Q(U0, . . . , Un∗−1) = Q(U0, . . . , Un∗−1)∩N = Q(U0, . . . , Un∗−1)∩N ,and hence Q(U0, . . . , Un∗−1) ∈ N . Let (g0, . . . , gn∗−1) be a name for aQ(U0, . . . , Un∗−1)-generic real over N . By the pure decision property ofQ(U0, . . . , Un∗−1) (fact 3.11) there exists (∅, S) ∈ Q(U0, . . . , Un∗−1) suchthat

(i) (∅, S) °Q(U0,...,Un∗−1)(g0, . . . , gn∗−1) ∈ J (A)

or(ii) (∅, S) °Q(U0,...,Un∗−1)

(g0, . . . , gn∗−1) /∈ J (A).

Claim: There exists a ∈ [ω]ω, which is Q(U0, . . . , Un∗−1)-generic over N ,such that ai ⊆ Si and ai ∈ Ui for every i < n∗ .

Proof of the claim: By fact 3.10 every a ∈ [ω]ω such that ai diagonalizesUi (i.e. ∀ b ∈ Ui ai ⊆∗ b) for every i < n∗ is Q(U0, . . . , Un∗−1)-generic overN . As each Ui is countable we can easily find such a. 2(claim)

Now, if (i) is true we have

(i)′ N [a] |= (a0, . . . , an∗−1) ∈ J (A)

and if (ii) is true we have

(ii)′ N [a] |= (a0, . . . , an∗−1) /∈ J (A).

But by fact 3.10 this implies

(i)′′ N [c] |= (c0, . . . , cn∗−1) ∈ J (A)

for every c with ci ∈ [ai]ω for every i < n∗ or

(ii)′′ N [c] |= (c0, . . . , cn∗−1) /∈ J (A)

for every c with ci ∈ [ai]ω for every i < n∗. But Σ11-definitions are absolute

for transitive models, hence

(i)′′′ V |= (c0, . . . , cn∗−1) ∈ J (A)

for every c with ci ∈ [ai]ω for every i < n∗ or

(ii)′′′ V |= (c0, . . . , cn∗−1) /∈ J (A)

for every c with ci ∈ [ai]ω for every i < n∗. But both are impossible. 2

References

[BaJu] T. Bartoszynski, H. Judah, Set Theory. On the Structure ofthe Real line; A. K. Peters, Wellesley, Massachusetts (1995).

[Br] J. Brendle, Combinatorial aspects of the meager and null ideals andof other ideals on the reals; Thesis for Habilitation, Eberhard-Karls-Universitat Tubingen, 1994/95.

[Go] M. Goldstern, Tools for your forcing construction; Set theoryof the reals (Ramat Gan, 1991), Israel Mathematical ConferenceProoceedings, vol. 6, Bar Ilan University, Ramat Gan, 1992, pp. 307-362.

38

[GoJSp] M. Goldstern, M. J. Johnson, O. Spinas, Towers on trees;Proceedings of the American Mathematical Society 122, vol. 2 (1994),pp. 557-564.

[GoReShSp] M. Goldstern, M. Repicky, S. Shelah, O. Spinas, Ontree ideals; Proceedings of the American Mathematical Society 123,vol. 5 (1995), pp. 1573-1581.

[Je] T. Jech, Set Theory; New York, Academic Press (1978).

[JoSp] S. Jossen, O. Spinas, A two-dimensional tree ideal; Berichtsreihedes Mathematischen Seminars Kiel, Vol. 00-32 (2000).

[JuMiSh] H. Judah, A. Miller, S. Shelah, Sacks forcing, Laver forc-ing, and Martin’s axiom; Arch. Math. Logic 31, vol. 3 (1992), pp. 145-161.

[JuSh] H. Judah, S. Shelah, The Kunen-Miller chart; The Journal ofSymbolic Logic, vol. 55 (1990), pp. 909-927.

[Ke] A. Kechris, On a notion of smallness for subsets of the Baire space;Transactions of the American Mathematical Society, vol. 229 (1977),pp. 191-207.

[Ko] S. Koppelberg, Handbook of Boolean algebras Vol. 1 (edited byJ .D. Monk); North Holland, Amsterdam, New York, Oxford, 1989.

[Ku] K. Kunen, Set Theory. An Introduction to Independence Proofs,North Holland, Amsterdam, New York, Oxford, 1980.

[LoShVe] A. Louveau, S. Shelah, B. Velickovic, Borel partitions ofinfinite subtrees of a perfect tree; Annals of Pure and Applied Logic63, vol. 3 (1993), pp. 271-281.

[M] E. Marczewski, Sur une classe de fonctions de W. Sierpinski etla classe correspondante d’ensembles; Fundamenta Mathematicae,vol. 24 (1935), pp. 17-37.

[Ma] A. R. D. Mathias, Happy families, Ann. Math. Logic 12 (1977),pp. 59-111.

[Mi] A. Miller, Rational perfect set forcing; Contemporary Mathemat-ics, vol. 31 (1984), pp. 143-159.

[R] F. Rothberger, On the property C and a problem of Hausdorff,Canadian J. Math. 4 (1952), pp. 111-116.

[Sh] S. Shelah, Proper forcing; Lecture Notes in Mathematics, vol. 940,Springer Verlag, Berlin, 1982.

[ShSp1] S. Shelah, O. Spinas, The distributivity numbers of P(ω)/finand its square; Transactions of the American Mathematical Society352, no. 5 (2000), pp. 2023-2047.

[ShSp2] S. Shelah, O. Spinas, The distributivity numbers of finite prod-ucts of P(ω)/fin; Fundamenta Mathematicae 158 (1998), pp. 81-93.

[Sp1] O. Spinas, Generic trees; The Journal of Symbolic Logic, vol. 60(1995), pp. 705-726.

[Sp2] O. Spinas, Ramsey and freeness properties of Polish planes; Pro-ceedings of the London Mathematical Society, vol. 82 (2001), pp. 31-63.

[Sp3] O, Spinas, Canonical behaviour of Borel functions on superperfectrectangles; J. Math. Logic 1, vol. 2 (2001), pp. 173-220.

39

[Sp4] O. Spinas, Partitioning products of P(ω)/fin; Pacific Journal ofMathematics, vol. 176, no. 1 (1996), pp. 249-262.

[Ve] B. Velickovic, ccc posets of perfect trees; Compositio Math. 79(1991), pp. 279-294.