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166 (2000)

Cellularity of free products of Boolean algebras (or topologies)

by

Saharon S h e l a h (Jerusalem and New Brunswick, NJ)

Abstract. The aim this paper is to present an answer to Problem 1 of Monk [10], [11]. We do this by proving in particular that if µ is a strong limit singular cardinal, θ = (2 cf(µ) ) + and 2 µ = µ + then there are Boolean algebras B 1 , B 2 such that

c( B 1 ) = µ, c( B 2 ) < θ but c( B 1 B 2 ) = µ + .

Further we improve this result, deal with the method and the necessity of the assumptions.

In particular we prove that if B is a ccc Boolean algebra and µ i

ω

≤ λ = cf(λ) ≤ 2 µ then B satisfies the λ-Knaster condition (using the “revised GCH theorem”).

0. Introduction

Notation 0.1. (1) In the present paper all cardinals are infinite so we will not repeat this additional demand. Cardinals will be denoted by λ, µ, θ (with possible indices) while ordinal numbers will be called α, β, ζ, ξ, ε, i, j. Usually δ will stand for a limit ordinal (we may forget to repeat this assumption).

(2) Sequences of ordinals will be called η, ν, % (with possible indices). For sequences η 1 , η 2 their longest common initial segment is denoted by η 1 ∧ η 2 . The length of the sequence η is lg(η).

(3) Ideals are supposed to be proper and contain all singletons. For a limit ordinal δ the ideal of bounded subsets of δ is denoted by J δ bd . If I is an ideal on a set X then I + is the family of I-large sets, i.e.

a ∈ I + if and only if a ⊆ X & a 6∈ I, and I c is the dual filter of sets with the complements in I.

2000 Mathematics Subject Classification: 54A25, 06E99, 03E04.

Key words and phrases: set theory, pcf, Boolean algebras, cellularity, product, colour- ings.

The research was partially supported by the Israel Science Foundation, founded by the Israel Academy of Sciences and Humanities. Publication 575. We thank Andrzej Ros lanowski for writing Sections 1–5 from lectures, and 6–7 from notes.

[153]

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Notation 0.2. (1) In a Boolean algebra we denote the Boolean opera- tions by ∩ (and T), ∪ (and S), −. The distinguished elements are 0 and 1.

In the cases which may be confusing we will add indices to underline in which Boolean algebra the operation (or element) is considered, but generally we will not do it.

(2) For a Boolean algebra B and an element x ∈ B we write x 0 = x and x 1 = −x.

(3) The free product of Boolean algebras B 1 , B 2 is denoted by B 1 ∗ B 2 . We will use F to denote the free product of a family of Boolean algebras.

Definition 0.3. (1) A Boolean algebra B satisfies the λ-cc if there is no family F ⊆ B + := B \ {0} such that |F| = λ and any two members of F are disjoint (i.e., their meet in B is 0).

(2) The cellularity of the algebra B is

c(B) = sup{|F| : F ⊆ B + & (∀x, y ∈ F )(x 6= y ⇒ x ∩ y = 0)}, c + (B) = sup{|F| + : F ⊆ B + & (∀x, y ∈ F )(x 6= y ⇒ x ∩ y = 0)}.

(3) For a topological space (X, τ ),

c(X, τ ) = sup{|U | : U is a family of pairwise disjoint

non-empty open sets}.

The problem can be posed in each of the three ways (λ-cc is the way of forcing, the cellularity of Boolean algebras is the approach of Boolean algebraists, and the cellularity of a topological space is the way of general topologists). It is well known that the three are equivalent, though (1) makes the attainment problem more explicit. We use the second approach.

A stronger property than λ-cc is the λ-Knaster property. This property behaves nicely in free products—it is productive. We will use it in our construction.

Definition 0.4. A Boolean algebra B has the λ-Knaster property if for every sequence hz ε : ε < λi ⊆ B + there is A ∈ [λ] λ such that

ε 1 , ε 2 ∈ A ⇒ z ε

1

∩ z ε

2

6= 0.

We are interested in the behaviour of the cellularity of Boolean algebras when their free product is considered.

Thema 0.5. When, for Boolean algebras B 1 , B 2 ,

c + (B 1 ) ≤ λ 1 & c + (B 2 ) ≤ λ 2 ⇒ c + (B 1 ∗ B 2 ) ≤ λ 1 + λ 2 ?

There are a lot of results about it, particularly if λ 1 = λ 2 (see [22] or [10], more [24]). It is well known that if

+ 1 + λ + 2 ) → (λ + 1 , λ + 2 ) 2

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then the answer is “yes”. These are exactly the cases for which the “yes”

answer is known. Under GCH the only problem which remained open was the one presented below:

The Problem We Adddress 0.6 (posed by D. Monk as Problem 1 in [10], [11] under GCH). Are there Boolean algebras B 1 , B 2 and cardinals µ, θ such that:

(1) λ 1 = µ is singular, µ > λ 2 = θ > cf(µ) and (2) c(B 1 ) = µ, c(B 2 ) ≤ θ but c(B 1 ∗ B 2 ) > µ?

We will answer this question proving in particular the following result (see 4.4):

• If µ is a strong limit singular cardinal, θ = (2 cf(µ) ) + and 2 µ = µ + then there are Boolean algebras B 1 , B 2 such that

c(B 1 ) = µ, c(B 2 ) < θ but c(B 1 ∗ B 2 ) = µ + . Later we deal with better results by refining the method.

Remark 0.7. On products of many Boolean algebras and square bracket arrows see [17, 1.2A, 1.3B].

If λ → [µ] 2 θ , is the cardinal θ is possibly finite, B i (for i < θ) are Boolean algebras such that for each j < θ the free product F i∈θ\{j} B i satisfies the µ-cc then the algebra B = F i<θ B i satisfies the λ-cc.

[Why? Assume ha ζ i : i < θi ∈ Q

i<θ B + i (for ζ < λ) such that for every ζ < ξ < λ, for some i = i(ζ, ξ), B i |=“a ζ i ∩ a ξ i = 0”. We can find A ∈ [λ] µ and i < θ such that i(ζ, ξ) 6= i for ζ < ξ from A. Then ha ζ i : i < θ, i 6= i i for ζ ∈ A exemplifies that F i∈θ\{i

} B i fails the µ-cc. We can also deal with ultraproducts and other products similarly.]

1. Preliminaries: products of ideals

Notation 1.1. For an ideal J on δ the quantifier (∀ J i < δ) means “for all i < δ except a set from the ideal J ”, i.e.,

(∀ J i < δ)ϕ(i) ≡ {i < δ : ¬ϕ(i)} ∈ J.

The dual quantifier (∃ J i < δ) means “for a J -positive set of i < δ”.

Proposition 1.2. Assume that λ 0 > λ 1 > . . . > λ n−1 are cardinals, I l are ideals on λ l (for l < n) and B ⊆ Q

l<n λ l . Further suppose that : (α) (∃ I

0

γ 0 ) . . . (∃ I

n−1

γ n−1 )(hγ l : l < ni ∈ B),

(β) the ideal I l is (2 λ

l+1

) + -complete (for l + 1 < n).

Then there are sets X l ⊆ λ l , X l 6∈ I l such that Q

l<n X l ⊆ B.

(4)

[Note that this translates the situation to arity 1; it is a kind of polarized (1, . . . , 1)-partition with ideals.]

P r o o f. We show it by induction on n. Define

E 0 := {(γ 0 , γ 00 ) : γ 0 , γ 00 < λ 0 and for all γ 1 < λ 1 , . . . , γ n−1 < λ n−1 , (hγ 0 , γ 1 , . . . , γ n−1 i ∈ B ⇔ hγ 00 , γ 1 , . . . , γ n−1 i ∈ B)}.

Clearly E 0 is an equivalence relation on λ 0 with ≤ 2 Q

0<m<n

λ

m

= 2 λ

1

equiv- alence classes. Hence the set

A 0 := [

{A : A is an E 0 -equivalence class, A ∈ I 0 } is in the ideal I 0 . Let

A 0 := {γ 0 < λ 0 : (∃ I

1

γ 1 ) . . . (∃ I

n−1

γ n−1 )(hγ 0 , γ 1 , . . . , γ n−1 i ∈ B)}.

The assumption (α) implies that A 0 6∈ I 0 and hence we may choose γ 0 ∈ A 0 \ A 0 . Let

B 1 := {¯ γ ∈ Q n−1

k=1 λ k : hγ 0 i _ ¯ γ ∈ B}.

Since γ 0 ∈ A 0 we are sure that

(∃ I

1

γ 1 ) . . . (∃ I

n−1

γ n−1 )(hγ 1 , . . . , γ n−1 i ∈ B 1 ).

Hence we may apply the inductive hypothesis for n − 1 and B 1 to find sets X 1 ∈ (I 1 ) + , . . . , X n−1 ∈ (I n−1 ) + such that Q n−1

l=1 X l ⊆ B 1 , so then (∀γ 1 ∈ X 1 ) . . . (∀γ n−1 ∈ X n−1 )(hγ 0 , γ 1 , . . . , γ n−1 i ∈ B).

Take X 0 to be the E 0 -equivalence class of γ 0 (so X 0 ∈ (I 0 ) + as γ 0 6∈ A 0 ).

By the definition of the relation E 0 and the choice of the sets X l we see that for each γ 0 ∈ X 0 ,

(∀γ 1 ∈ X 1 ) . . . (∀γ n−1 ∈ X n−1 )(hγ 0 , γ 1 , . . . , γ n−1 i ∈ B), which means that Q

l<n X l ⊆ B.

Proposition 1.3. Assume that λ 0 > λ 1 > . . . > λ n−1 ≥ σ are cardinals, I l are ideals on λ l (for l < n) and B ⊆ Q

l<n λ l . Further suppose that : (α) (∃ I

0

γ 0 ) . . . (∃ I

n−1

γ n−1 )(hγ l : l < ni ∈ B),

(β) I l is ((λ l+1 ) σ ) + -complete for each l < n − 1, and [λ n−1 ] ⊆ I n−1 . Then there are sets X l ∈ [λ l ] σ such that Q

l<n X l ⊆ B.

P r o o f. The proof is by induction on n. If n = 1 then there is nothing to do as I n−1 contains all subsets of λ n−1 of size < σ and λ n

1

≥ σ so every A ∈ I n +

1

has cardinality ≥ σ.

Let n > 1 and let

a 0 := {γ ∈ λ 0 : (∃ I

1

γ 1 ) . . . (∃ I

n−1

γ n−1 )(hγ, γ 1 , . . . , γ n−1 i ∈ B)}.

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By our assumptions we know that a 0 ∈ (I 0 ) + . For each γ ∈ a 0 we may apply the inductive hypothesis to the set

B γ := {hγ 1 , . . . , γ n−1 i ∈ Q

0<l<n λ l : hγ, γ 1 , . . . , γ n−1 i ∈ B}

and get sets X 1 γ ∈ [λ 1 ] σ , . . . , X n−1 γ ∈ [λ n−1 ] σ such that Y

0<l<n

X l γ ⊆ B γ .

There are at most (λ 1 ) σ possible sequences hX 1 γ , . . . , X n−1 γ i, and the ideal I 0 is ((λ 1 ) σ ) + -complete, so for some sequence hX 1 , . . . , X n−1 i and a set a ⊆ a 0 , a ∈ (I 0 ) + we have

(∀γ ∈ a )(X 1 γ = X 1 & . . . & X n−1 γ = X n−1 ).

Choose X 0 ∈ [a ] σ (remember that I 0 contains singletons and it is complete enough to make sure that σ ≤ |a |). Clearly Q

l<n X l ⊆ B.

Remark 1.4. We can use σ 0 ≥ σ 1 ≥ . . . ≥ σ n−1 , I l is (λ σ l+1

l+1

) + - complete, [λ l ]

l

⊆ I l .

Proposition 1.5. Assume that n < ω and λ m l , χ m l , P l m , I l m , I m and B are such that for l, m ≤ n:

(α) I l m is a χ m l -complete ideal on λ m l (for l, m ≤ n),

(β) P l m ⊆ P(λ m l ) is a family dense in (I l m ) + in the sense that (∀X ∈ (I l m ) + )(∃a ∈ P l m )(a ⊆ X),

(γ) I m = {X ⊆ Q

l≤n λ m l : ¬(∃ I

m0

γ 0 ) . . . (∃ I

nm

γ n )(hγ 0 , . . . , γ n i ∈ X)} [thus I m is the ideal on Q

l≤n λ m l such that the dual filter (I m ) c is the Fubini product of the filters (I 0 m ) c , . . . , (I n m ) c ],

(δ) χ m n−m > P n

l=m+1 (|P n−l l | + P n−l k=0 λ l k ), (ε) B ⊆ Q

m≤n

Q

l≤n λ m l is a set satisfying

(∃ I

0

η 0 )(∃ I

1

η 1 ) . . . (∃ I

n

η n )(hη 0 , η 1 , . . . , η n i ∈ B).

Then there are sets X 0 , . . . , X n such that for m ≤ n:

(a) X m ⊆ Q

l≤n−m λ m l , (b) if η, ν ∈ X m , η 6= ν then

(i) η (n − m) = ν(n − m), (ii) η(n − m) 6= ν(n − m), (c) {η(n − m) : η ∈ X m } ∈ P n−m m , (d) for each hη 0 , . . . , η n i ∈ Q

m≤n X m there is hη 0 , . . . , η n i ∈ B such that

(∀m ≤ n)(η m E η m ).

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Remark 1.5.A. (1) Note that the sets X m in the assertion of 1.5 may be thought of as sets of the form X m = {ν m _ hαi : α ∈ a m } for some ν m ∈ Q

l<n−m λ m l and a m ∈ P n−m m .

(2) We will apply this proposition with λ m l = λ l , I l m = I l and λ l > χ l >

P

k<l λ k .

(3) In the assumption (δ) of 1.5 we may assume that the last sum on the right hand side ranges from k = 0 to n − l − 1. We did not formulate that assumption in this way as with n − l it is easier to handle the induction step and this change is not important for our applications.

(4) In the assertion (d) of 1.5 we can have η l depending on hη 0 , . . . , η l i only.

P r o o f (of Proposition 1.5). The proof is by induction on n. For n = 0 there is nothing to do. Let us describe the induction step.

Suppose 0 < n < ω and λ m l , χ m l , P l m , I l m , I m (for l, m ≤ n) and B satisfy the assumptions (α)–(ε). Let

B := {hη 0 , η 1 n, . . . , η n ni : η m ∈ Q

l≤n λ m l (for m ≤ n) and

hη 0 , η 1 , . . . , η n i ∈ B}, and for η 0 ∈ Q

l≤n λ 0 l let

B η

0

:= {hν 1 , . . . , ν n i ∈ Q n m=1

Q n−1

l=0 λ m l : hη 0 , ν 1 , . . . , ν n i ∈ B }.

Let J m (for 1 ≤ m ≤ n) be the ideal on Q n−1

l=0 λ m l coming from the ideals I l m , i.e., a set X ⊆ Q

l<n λ m l is in J m if and only if

¬(∃ I

0m

γ 0 ) . . . (∃ I

n−1m

γ n−1 )(hγ 0 , . . . , γ n−1 i ∈ X).

Let us call the set B η

0

big if

(∃ J

1

ν 1 ) . . . (∃ J

n

ν n )(hν 1 , . . . , ν n i ∈ B η

0

).

We may write more explicitly what the bigness means: the above condition is equivalent to

(∃ I

01

γ 1 0 ) . . . (∃ I

n−11

γ n−1 1 ) . . .

. . . (∃ I

0n

γ 0 n ) . . . (∃ I

n−1n

γ n−1 n )(hhγ 1 0 , . . . , γ n−1 1 i, . . . hγ n 0 , . . . , γ n−1 n ii ∈ B η

0

), which means

(∃ I

01

γ 0 1 ) . . . (∃ I

n−1n

γ n−1 n )

(∃γ n 1 ) . . . (∃γ n n )(hη 0 , hγ 1 0 , . . . , γ n 1 i, . . . , hγ 0 n , . . . , γ n n ii ∈ B).

By the assumptions (γ) and (ε) we know that (∃ I

00

γ 0 0 ) . . . (∃ I

n0

γ n 0 )(∃ I

01

γ 0 1 ) . . . (∃ I

n1

γ n 1 ) . . .

. . . (∃ I

0n

γ 0 n ) . . . (∃ I

nn

γ n n )(hhγ 0 0 , . . . , γ n 0 i, hγ 0 1 , . . . , γ n 1 i, . . . , hγ 0 n , . . . , γ n n ii ∈ B).

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Obviously any quantifier (∃ I

lm

γ l m ) above may be replaced by (∃γ l m ) and then “moved” right as for as we want. Consequently, we get

(∃γ 0 0 ) . . . (∃γ n−1 0 )(∃ I

n0

γ n 0 )(∃ I

01

γ 0 1 ) . . . (∃ I

n−11

γ n−1 1 ) . . . (∃ I

0n

γ 0 n ) . . . (∃ I

n−1n

γ n−1 n ) (∃γ 1 n ) . . . (∃γ n n )(hhγ 0 0 , . . . , γ n 0 i, hγ 0 1 , . . . , γ 1 n i, . . . , hγ 0 n , . . . , γ n n ii ∈ B),

which means that

(∃γ 0 0 ) . . . (∃γ 0 n−1 )(∃ I

n0

γ n 0 )(B

0

0

,...,γ

nn

i is big).

Hence we find γ 0 0 , . . . , γ n−1 0 and a set a ∈ (I n 0 ) + such that (∀γ ∈ a)(B

0

0

,...,γ

nn

i is big).

Note that the assumptions of the proposition are such that if we know that B η

0

is big then we may apply the inductive hypothesis to λ m l , χ m l , P l m , I l m , J m (for 1 ≤ m ≤ n, l ≤ n − 1) and B η

0

. Consequently, for each γ ∈ a we find sets X 1 γ , . . . , X n γ such that for 1 ≤ m ≤ n:

(a) X m γ ⊆ Q

l≤n−m λ m l , (b) if η, ν ∈ X m γ , η 6= ν then

(i) η (n − m) = ν(n − m), and (ii) η(n − m) 6= ν(n − m), (c) {η(n − m) : η ∈ X m γ } ∈ P n−m m , (d) for all hη 0 , . . . , η n i ∈ Q

m≤n X m γ we have (∃hη 0 , . . . , η n i ∈ B

0

0

,...,γ

0n−1

,γi )(∀1 ≤ m ≤ n)(ν m E ν m ).

Now we may ask how many possibilities for X m γ we have: not too many.

If we fix the common initial segment (see (b) ) the only freedom we have is in choosing an element of P n−m m (see (c) ). Consequently, there are at most

|P n−m m | + P

l≤n−m λ m l possible values for X m γ and hence there are at most

n

X

m=1



|P n−m m | + X

l≤n−m

λ m l



< χ 0 n

possible values for the sequence hX 1 γ , . . . , X n γ i. Since the ideal I n 0 is χ 0 n - complete we find hX 1 , . . . , X n i and a set b ⊆ a, b ∈ (I n 0 ) + , such that

(∀γ ∈ b)(hX 1 γ , . . . , X n γ i = hX 1 , . . . , X n i).

Next choose b 0 n ∈ P n 0 such that b 0 n ⊆ b and put

X 0 = {hγ 0 0 , . . . , γ n−1 0 , γi : γ ∈ b 0 n }.

Now it is a routine to check that the sets X 0 , X 1 , . . . , X n are as required

(i.e., they satisfy clauses (a)–(d)).

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2. Cofinal sequences in trees

Notation 2.1. For a tree T ⊆ δ> µ the set of δ-branches through T is lim δ (T ) := {η ∈ δ µ : (∀α < δ)(η α ∈ T )}.

The ith level (for i < δ) of the tree T is T i := T ∩ i µ and T <i := S

j<i T j .

If η ∈ T then the set of immediate successors of η in T is succ T := {ν ∈ T : η C ν & lg(ν) = lg(η) + 1}.

We shall not distinguish strictly between succ T (η) and {α : η _ hαi ∈ T }.

Definition 2.2. (1) K µ,δ is the family of all pairs (T, ¯ λ) such that T ⊆ δ> µ is a tree with δ levels and ¯ λ = hλ η : η ∈ T i is a sequence of cardinals such that for each η ∈ T we have succ T (η) = λ η (compare the previous remark about not distinguishing succ T (η) and {α : η _ hαi ∈ T }).

(2) For a limit ordinal δ and a cardinal µ we let K µ,δ id := {(T, ¯ λ, ¯ I) : (T, ¯ λ) ∈ K µ,δ , ¯ I = hI η : η ∈ T i,

each I η is an ideal on λ η = succ T (η)}.

Let (T, ¯ λ, ¯ I) ∈ K µ,δ id and let J be an ideal on δ (including J δ bd if we do not say otherwise). Further let ¯ η = hη α : α < λi ⊆ lim δ (T ) be a sequence of δ-branches through T .

(3) We say that ¯ η is J -cofinal in (T, ¯ λ, ¯ I) if (a) η α 6= η β for distinct α, β < λ,

(b) for every sequence ¯ A=hA η : η ∈ T i ∈ Q

η∈T I η there is α < λ such that

α ≤ α < λ ⇒ (∀ J i < δ)(η α (i + 1) 6∈ A η

α

i ).

(4) If I is an ideal on λ then we say that (¯ η, I) is a J -cofinal pair for (T, ¯ λ, ¯ I) if

(a) η α 6= η β for distinct α, β < λ,

(b) for every sequence ¯ A = hA η : η ∈ T i ∈ Q

η∈T I η there is A ∈ I such that

α ∈ λ \ A ⇒ (∀ J i < δ)(η α (i + 1) 6∈ A η

α

i ).

(5) The sequence ¯ η is strongly J -cofinal in (T, ¯ λ, ¯ I) if (a) η α 6= η β for distinct α, β < λ,

(b) for every n < ω and functions F 0 , . . . , F n there is α < λ such

that if m ≤ n, α 0 < . . . < α n < λ, α ≤ α m then the set of i < δ

such that:

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(i) (∀l < m)(λ η

αl

i < λ η

αm

i ) and

(ii) F m (η α

0

(i+1), . . . , η α

m−1

(i+1), η α

m

i, . . . , η α

n

i) ∈ I η

αm

i

(and well defined) but

η α

m

(i+1) ∈ F m (η α

0

(i+1), . . . , η α

m−1

(i+1), η α

m

i, . . . , η α

n

i), is in the ideal J .

[Note: in (b) above we may have α < α 0 , this causes no real change.]

(6) The sequence ¯ η is stronger J -cofinal in (T, ¯ λ, ¯ I) if (a) η α 6= η β for distinct α, β < λ,

(b) for every n < ω and functions F 0 , . . . , F n there is α < λ such that if m ≤ n, α 0 < . . . < α n < λ, α ≤ α m then the set of i < δ such that:

(ii) F m (η α

0

(i+1), . . . , η α

m−1

(i+1), η α

m

i, . . . , η α

n

i) ∈ I η

αm

i

(and well defined) but

η α

m

(i+1) ∈ F m (η α

0

(i+1), . . . , η α

m−1

(i+1), η α

m

i, . . . , η α

n

i), is in the ideal J .

(7) The sequence ¯ η is strongest J -cofinal in (T, ¯ λ, ¯ I) if (a) η α 6= η β for distinct α, β < λ,

(b) for every n < ω and functions F 0 , . . . , F n there is α < λ such that if m ≤ n, α 0 < . . . < α n < λ, α ≤ α m then the set of i < δ such that:

(i 0 ) (∃l < m)(λ η

αl

i ≥ λ η

αm

i ) or

(ii 0 ) F m (η α

0

(i+1), . . . , η α

m−1

(i+1), η α

m

i, . . . , η α

n

i) ∈ I η

αm

i

(and well defined) but

η α

m

(i+1) ∈ F m (η α

0

(i+1), . . . , η α

m−1

(i+1), η α

m

i, . . . , η α

n

i), is in the ideal J .

(8) The sequence ¯ η is big J -cofinal in (T, ¯ λ, ¯ I) if (a) η α 6= η β for distinct α, β < λ,

(b) for every n < ω and functions F 0 , . . . , F n there is α such that if α 0 < . . . < α n and α ≤ α m , m ≤ n then the set

{i < δ : η α

m

(i) ∈ F m (ν l ) l≤n ∈ I η

αm

i } is in the ideal J , where

ν l =

η α

l

(i+1) if λ η

αl

i < λ η

αm

i or

λ η

αl

i = λ η

αm

i and η α

l

(i) < η α

m

(i),

η α

l

i if not.

(10)

(9) In almost the same way we define “strongly J -cofinal”, “stronger J -cofinal” and “strongest big J -cofinal”, replacing the requirement that α ≤ α m in 5(b), 6(b), 7(b) above (respectively) by α ≤ α 0 .

Remark 2.3. (a) Note that “strongest J -cofinal” implies “stronger J - cofinal” and this implies “strongly J -cofinal”. “Stronger J -cofinal” implies

“J -cofinal”. Also “bigger” ⇒ “big” ⇒ “cofinal”, “big” ⇒ “strongly”.

(b) The different notions of “strong J -cofinality” (the conditions (i) and (i 0 )) are to allow us to carry some diagonalization arguments.

(c) The difference between “strongly J -cofinal” and “strongly J -cofinal”

etc. is, in our context, immaterial. We may in all places in this paper replace the relevant notion with its version with “∗” and no harm will be done.

Remark 2.4. (1) Recall pcf: An important case is when hλ i : i < δi is an increasing sequence of regular cardinals, λ i > Q

j<i λ j , λ η = λ lg(η) , I η = J λ bd

η

and λ = tcf( Q

i<δ λ i /J ).

(2) Moreover we are interested in more complicated I η ’s (as in [23, §5]), connected to our problem, so “the existence of the true cofinality” is less clear. But the assumption 2 µ = µ + will rescue us.

(3) There are natural stronger demands of cofinality since here we are not interested just in x α ’s but also in Boolean combinations. Thus naturally we are interested in behaviours of large sets of n-tuples (see 5.1).

Proposition 2.5. Suppose that (T, ¯ λ, ¯ I) ∈ K id µ,δ , ¯ η = hη α : α < λi ⊆ lim δ (T ) and J is an ideal on δ, J ⊇ J δ bd .

(1) Assume that

( }) if α < β < λ then (∀ J i < δ)(λ η

α

i < λ η

β

i ).

Then the following are equivalent :

• “¯ η is strongly J -cofinal for (T, ¯ λ, ¯ I)”,

• “¯ η is stronger J -cofinal for (T, ¯ λ, ¯ I)”,

• “¯ η is strongest J -cofinal for (T, ¯ λ, ¯ I)”,

• “¯ η is big J -cofinal for (T, ¯ λ, ¯ I)”.

(2) If I ν ⊇ J λ bd

ν

and λ ν = λ lg(ν) for each ν ∈ T and the sequence ¯ η is stronger J -cofinal for (T, ¯ λ, ¯ I) then for some α < λ the sequence hη α : α ≤ α < λi is < J -increasing.

(3) If η ∈ T i ⇒ λ η = λ i and ¯ η is < J -increasing in Q

i<δ λ i then “big”

is equivalent to “stronger”.

Proposition 2.6. Suppose that :

(1) hλ i : i < δi is an increasing sequence of regular cardinals, where δ < λ 0 is a limit ordinal ,

(2) T = S

i<δ

Q

j<i λ j , I η = I lg(η) = J λ bd

lg(η)

and λ η = λ lg(η) ,

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(3) J is an ideal on δ, λ = tcf( Q

i<δ λ i /J ) and it is exemplified by a sequence ¯ η = hη α : α < λi ⊆ Q

i<δ λ i ,

(4) |{η α i : α < λ}| < λ i for each i < δ (so, e.g., λ i > Q

j<i λ j suffices).

Then the sequence ¯ η is J -cofinal in (T, ¯ λ, ¯ I).

P r o o f. First note that our assumptions imply that each ideal I η = I lg(η)

is |{η α  lg(η) : α < λ}| + -complete. Hence for each sequence ¯ A = hA η : η ∈ T i ∈ Q

η∈T I η and i < δ the set A i := [

{A η

α

i : α < λ}

is in the ideal I i , i.e., it is bounded in λ i (for i < δ). (We should remind here our convention that we do not distinguish λ i and succ T (η) if lg(η) = i, see 2.1.) Take η ∈ Q

i<δ λ i such that for each i < δ we have A i ⊆ η (i). As the sequence ¯ η realizes the true cofinality of Q

i<δ λ i /J we find α < λ such that α ≤ α < λ ⇒ {i < δ : η α (i) < η (i)} ∈ J,

which allows us to finish the proof.

It follows from the above proposition that the notion of J -cofinal se- quence is not empty. Of course, it is better to have “strongly (or even:

stronger) J -cofinal” sequences ¯ η. So it is nice to find that sometimes the weaker notion implies the stronger one.

Proposition 2.7. Assume that δ is a limit ordinal , µ is a cardinal , and (T, ¯ λ, ¯ I) ∈ K id µ,δ . Let J be an ideal on δ such that J ⊇ J δ bd (which is our standard hypothesis). Further suppose that

( ~) if η ∈ T i then the ideal I η is (|T i | + P{λ ν : ν∈T i & λ ν <λ η }) + - complete.

Then each J -cofinal sequence ¯ η for (T, ¯ λ, ¯ I) is strongly J -cofinal for (T, ¯ λ, ¯ I).

If , in addition, η 6= ν ∈ T i ⇒ λ η 6= λ ν then ¯ η is big J -cofinal for (T, ¯ λ, ¯ I). Also, if in addition

η ∈ T i ⇒ (∃ !1 ν ∈ T i )(λ ν = λ η ) ∨ [(∃ ≤λ

η

ν ∈ T i )(λ ν = λ η ) & I η normal ] then ¯ η is big J -cofinal.

P r o o f. Let n < ω and F 0 , . . . , F n be (n + 1)-place functions. First we define a sequence ¯ A = hA η : η ∈ T i. For m ≤ n and a sequence hη m , . . . , η n i ⊆ T i we put

A m

m

,...,η

n

i = [

{F m (ν 0 , . . . , ν m−1 , η m , . . . , η n ) : ν 0 , . . . , ν m−1 ∈ T i+1 , (ν 0 , . . . , ν m−1 , η m , . . . , η n ) ∈ dom(F ),

λ ν

0

i < λ η , . . . , λ ν

m−1

i < λ η

m

and F (ν 0 , . . . , ν m−1 , η m , . . . , η n ) ∈ I η

m

},

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and next for η ∈ T i let A η = [

{A m hη,η

m+1

,...,η

n

i : m ≤ n & η m+1 , . . . , η n ∈ T i }.

Note that the assumption ( ~) was set up so that A m

m

,...,η

n

i ∈ I η

m

and the sets A η are in I η (for η ∈ T ).

By the J -cofinality of ¯ η, for some α < λ we have

α ≤ α < λ ⇒ (∀ J i < δ)(η α (i + 1) 6∈ A η

α

i ).

We are going to prove that this α is as required in the definition of strongly J -cofinal sequences. So suppose that m ≤ n, α 0 < . . . < α n < λ and α ≤ α m . By the choice of α the set A := {i < δ : η α

m

(i + 1) ∈ A η

αm

i } is in the ideal J . But if i < δ is such that

• (∀l < m)(λ η

αl

i < λ η

αm

i ),

• F (η α

0

(i + 1), . . . , η α

m−1

(i + 1), η α

m

i, . . . , η α

n

i) ∈ I η

αm

i , but

• η α

m

(i + 1) ∈ F (η α

0

(i + 1), . . . , η α

m−1

(i + 1), η α

m

i, . . . , η α

n

i) then clearly η α

m

(i + 1) ∈ A m

αm

i,...,η

αn

ii and so i ∈ A.

The “big” version should be clear too.

Proposition 2.8. Assume that µ is a strong limit uncountable cardinal and hµ i : i < δi is an increasing sequence of cardinals with limit µ. Further suppose that (T, ¯ λ, ¯ I) ∈ K id µ,δ , |T i | ≤ µ i (for i < δ), λ η < µ and each I η

is µ + lg(η) -complete and contains all singletons (for η ∈ T ). Finally assume 2 µ = µ + and let J be an ideal on δ, J ⊇ J δ bd . Then there exists a stronger J -cofinal sequence ¯ η for (T, ¯ λ, ¯ I) of length µ + (even for J = J δ bd ). We can get “big” if

% 6= η ∈ T i & λ % = λ η ⇒ (∃ ≤λ

η

ν ∈ T i )(λ ν = λ η ) & I η normal.

P r o o f. This is a straight diagonal argument. Put

Y := {hF 0 , . . . , F n i : n < ω and each F l is a function with dom(F ) ⊆ T n+1 , rng(F ) ⊆ S

η∈T I η }.

Since |Y | = µ µ = µ + (remember that µ is strong limit and λ η < µ for η ∈ T ) we may choose an enumeration Y = {hF 0 ξ , . . . , F n ξ

ξ

i : ξ < µ + }. For each ζ < µ + choose an increasing sequence hA ζ i : i < δi such that |A ζ i | ≤ µ i

and ζ = S

i<δ A ζ i . Now we choose by induction on ζ < µ + branches η ζ such that for each ζ the restriction η ζ i is defined by induction on i as follows.

If i = 0 or i is limit then there is nothing to do.

Suppose now that we have defined η ζ i and η ξ for ξ < ζ. We find η ζ (i)

such that:

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(α) η ζ (i) ∈ λ η

ζ

i ,

(β) if ε ∈ A ζ i , m ≤ n ε , α 0 , . . . , α m−1 ∈ A ζ i (hence α l < ζ so η α

l

are already defined), ν m+1 , . . . , ν n ∈ T i and

F m ε (η α

0

(i + 1), . . . , η α

m−1

(i + 1), η ζ i, ν m+1 , . . . , ν n ) ∈ I η

ζ

i

and well defined, then

η ζ (i + 1) 6∈ F m ε (η α

0

(i + 1), . . . , η α

m−1

(i + 1), η ζ i, ν m+1 , . . . , ν n ), (γ) η ζ (i + 1) 6∈ {η ε (i + 1) : ε ∈ A ζ i }.

Why is it possible? Note that there are ≤ ℵ 0 + |A ζ i | + |A ζ i | <ℵ

0

+ |T i | ≤ µ i negative demands and each of them says that η ζ (i + 1) is in no set from I η

ζ

i (remember that we have assumed that the ideals I η

ζ

i contain singletons). Consequently, using the completeness of the ideal we may satisfy the requirements (α)–(γ) above.

Now of course η ζ ∈ lim δ (T ). Moreover if ε < ζ < µ + then (∃i < δ)(ε ∈ A ζ i ), which implies (∃i < δ)(η ε (i + 1) 6= η ζ (i + 1)). Consequently,

ε < ζ < µ + ⇒ η ε 6= η ζ .

Checking the demand (b) of “stronger J -cofinal” is straightforward: for functions F 0 , . . . , F n (and n ∈ ω) take ε such that

hF 0 , . . . , F n i = hF 0 ε , . . . , F n ε

ε

i

and put α = ε + 1. Suppose now that m ≤ n, α 0 < . . . < α n < λ, α ≤ α m . Let i < δ be such that for i > i we have

ε, α 0 , . . . , α m−1 ∈ A α i

m

.

Then by the choice of η α

m

(i + 1) we see that for each i > i , if

F m ε (η α

0

(i + 1), . . . , η α

m−1

(i + 1), η ζ i, η α

m+1

i, . . . , η α

n

i) ∈ I η

αm

i , then

η α

m

i 6∈ F m ε (η α

0

(i + 1), . . . , η α

m−1

(i + 1), η ζ i, η α

m+1

i, . . . , η α

n

i).

Remark 2.9. The proof above can be carried out for functions F which depend on (η α

0

, . . . , η α

m−1

, η α

m

i, . . . , η α

n

i). This will be natural later.

Let us note that if the ideals I η are sufficiently complete then J -cofinal sequences cannot be too short.

Proposition 2.10. Suppose that (T, ¯ λ, ¯ I) ∈ K id µ,δ is such that for each η ∈ T i , i < δ, the ideal I η is (κ i ) + -complete ([λ η ] κ

i

⊆ I η is enough). Let J ⊇ J δ bd be an ideal on δ and let ¯ η = hη α : α < δ i be a J -cofinal sequence for (T, ¯ λ, ¯ I). Then

δ > lim sup

J

κ i and consequently cf(δ ) > lim sup

J

κ i .

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P r o o f. Fix an enumeration δ = {α ε : ε < |δ |} and for α < δ let ζ(α) be the unique ζ such that α = α ζ . For η ∈ T i , i < δ, put

A η := {ν ∈ succ T (η) : (∃ε ≤ κ i )(ν C η ε )}.

Clearly |A η | ≤ κ i and hence A η ∈ I η . Apply the J -cofinality of ¯ η to the sequence ¯ A = hA η : η ∈ T i. Thus there is α < δ such that for each α ∈ [α , δ ) we have

(∀ J i < δ)(η α (i + 1) 6∈ A η

α

i ) and hence (∀ J i < δ)(ζ(α) > κ i ) and consequently

ζ(α) ≥ lim sup

J

κ i . Hence we conclude that |δ | > lim sup J κ i .

For the “consequently” part of the proposition note that if hη α : α < δ i is J -cofinal (in (T, ¯ λ, ¯ I)) and A ⊆ δ is cofinal in δ then hη α : α ∈ Ai is J -cofinal too.

Remark 2.11. (1) So if we have a J -cofinal sequence of length δ then we also have one of length cf(δ ). Thus assuming regularity of the length is natural.

(2) Moreover the assumption that the length of the sequence is above |δ|+

|T | is very natural and in most cases it will follow from the J -cofinality (and completeness assumptions). However we will try to state this condition in the assumptions whenever it is used in the proof (even if it can be concluded from the other assumptions).

3. Getting (κ, notλ)-Knaster algebras

Proposition 3.1. Let λ, σ be cardinals such that (∀α < σ)(2 |α| < λ) and σ is regular. Then there are a Boolean algebra B, a sequence hy α : α < λi ⊆ B + and an ideal I on λ such that :

(a) if X ⊆ λ, X 6∈ I then (∃α, β ∈ X)(B |= y α ∩ y β = 0), (b) the ideal I is σ-complete,

(c) the algebra B satisfies the µ-Knaster condition for any regular un- countable µ (actually, B is free).

P r o o f. Let B be the Boolean algebra freely generated by {z α : α < λ}

(so the demand (c) is satisfied). Let A = {(α, β) : α < β < λ} and y (α,β) = z α − z β (6= 0) (for (α, β) ∈ A). The ideal I of subsets of A is defined by:

• a set X ⊆ A is in I if there are ζ < σ, X ε ⊆ A (for ε < ζ) such that X ⊆ S

ε<ζ X ε and for every ε < ζ no two y

1

1

) , y

2

2

) ∈ X ε are

disjoint in B.

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First note that Claim 3.1.1. A 6∈ I.

P r o o f. If not then we have witnesses ζ < σ and X ε (for ε < ζ) for it. So A = S

ε<ζ X ε and hence for (α, β) ∈ A we have ε(α, β) such that y (α,β) ∈ X ε(α,β) . So ε(·, ·) is actually a function from [λ] 2 to ζ < σ. By the Erd˝ os–Rado theorem we find α < β < γ < λ such that ε(α, β) = ε(β, γ).

But

y (α,β) ∩ y (β,γ) = (z α − z β ) ∩ (z β − z γ ) = 0, so (α, β), (β, γ) cannot be in the same X ε —a contradiction.

To finish the proof note that I is σ-complete (as σ is regular), and if X 6∈ I then, by the definition of I, there are two disjoint elements in {y (α,β) : (α, β) ∈ X}. Finally |A| = λ.

Definition 3.2. (a) A pair (B, ¯ y) is called a λ-marked Boolean algebra if B is a Boolean algebra and ¯ y = hy α : α < λi is a sequence of non-zero elements of B.

(b) A triple (B, ¯ y, I) is called a (λ, χ)-well marked Boolean algebra if (B, ¯ y) is a λ-marked Boolean algebra, χ is a regular cardinal and I is a (proper) χ-complete ideal on λ such that

{A ⊆ λ : (∀α, β ∈ A)(B |= y α ∩ y β 6= 0)} ⊆ I.

By a λ-well marked Boolean algebra we will mean a (λ, ℵ 0 )-well marked one.

As in the above situation λ can be read off from ¯ y (as λ = lg(¯ y)) we may omit it and then we may speak just about well marked Boolean algebras.

Remark 3.3. Thus Proposition 3.1 says that if λ, σ are regular cardinals and

(∀α < σ)(2 |α| < λ)

then there exists a (λ, σ)-well marked Boolean algebra (B, ¯ y, I) such that B has the κ-Knaster property for every κ.

Definition 3.4. (1) For cardinals µ and λ and a limit ordinal δ, a (δ, µ, λ)-constructor is a system

C = (T, ¯ λ, ¯ η, h(B η , ¯ y η ) : η ∈ T i) such that:

(a) (T, ¯ λ) ∈ K µ,δ ,

(b) ¯ η = hη i : i ∈ λi where η i ∈ lim δ (T ) (for i < λ) are distinct δ-branches through T ,

(c) for each η ∈ T , (B η , ¯ y η ) is a λ η -marked Boolean algebra, i.e.,

¯

y η = hy η

_

hαi : α < λ η i ⊆ B + η (usually this will be an enumera-

tion of B + η ).

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(2) Let C be a constructor (as above). We define Boolean algebras B 2 = B red = B red (C) and B 1 = B green = B green (C) as follows.

B red is the Boolean algebra freely generated by {x i : i < λ} except that if i 0 , . . . , i n−1 < λ, ν = η i

0

ζ = η i

1

ζ = . . . = η i

n−1

ζ, B ν |= \

l<n

y η

il

(ζ+1) = 0 then T

l<n x i

l

= 0. [Note: we may demand that the sequence hη i

l

(ζ) : l < ni is strictly increasing, this will cause no difference.]

B green is the Boolean algebra freely generated by {x i : i < λ} except that if

ν = η i ζ = η j ζ, η i (ζ) 6= η j (ζ), B ν |= y η

i

(ζ+1) ∩ y η

j

(ζ+1) 6= 0 then x i ∩ x j = 0.

Remark 3.5. (1) The equations for the green case can look strange but they have to be dual to the ones of the red case.

(2) “Freely generated except . . .” means that a Boolean combination is non-zero except when some (finitely many) conditions imply it. For this it is enough to look at elements of the form

x t i

00

∩ . . . ∩ x t i

n−1

n−1

where t l ∈ {0, 1}.

(3) Working in the free product B red ∗B green we will use the same notation for elements (e.g., generators) of B red as for elements of B green . Thus x i may stand either for the corresponding generator in B red or B green . We hope that this will not be confusing, as one can easily decide in which algebra the element is considered from the place of it (if x ∈ B red , y ∈ B green then (x, y) will stand for the element x ∩ B

red

∗B

green

y ∈ B red ∗ B green ). In particular we may write (x i , x i ) for an element which could be denoted by x red i ∩ x green i .

Remark 3.6. If the pair (B red , B green ) is a counterexample with the free product B red ∗ B green failing the λ-cc but each of the algebras satisfying that condition then each of the algebras fails the λ-Knaster condition. But B red is supposed to have the κ-cc (κ smaller than λ). This is known to restrict λ.

Proposition 3.7. Assume that C = (T, ¯ λ, ¯ η, h(B η , ¯ y η ) : η ∈ T i) is a (δ, µ, λ)-constructor and J ⊇ J δ bd is an ideal on δ such that:

(a) ¯ η = hη i : i ∈ T i is J -cofinal for (T, ¯ λ, ¯ I), (b) if X ∈ I η + then

(∃α, β ∈ X)(B η |= y η

_

hαi ∩ y η

_

hβi = 0).

Then the sequence hx red α : α < λi exemplifies that B red (C) fails the λ-Knaster

condition.

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Explanation. The above proposition is not just something in the di- rection of Problem 0.6. The tuple (B red , ¯ x, J λ bd ) is like (B η , ¯ y η , I η ), but J λ bd is nicer than the ideals given by previous results. Using such objects makes building examples for Problem 0.6 much easier.

P r o o f (of Proposition 3.7). It is enough to show that for each Y ∈ [λ] λ one can find ε, ζ ∈ Y such that

B η

ε

i |= y η

ε

(i+1) ∩ y η

ζ

(i+1) = 0 where i = lg(η ε ∧ η ζ ). For this, for each ν ∈ T we put

A ν := {α < λ ν : (∃ε ∈ Y )(ν _ hαi C η ε )}.

Claim 3.7.1. There is ν ∈ T such that A ν 6∈ I ν .

P r o o f. First note that by the definition of A ν , for each ε ∈ Y we have (∀i < δ)(η ε _ hii ∈ A η

ε

i ).

Now, if we had A ν ∈ I ν for all ν ∈ T then we could apply the assumption that ¯ η is J -cofinal for (T, ¯ λ, ¯ I) to the sequence hA ν : ν ∈ T i. Thus we would find α < λ such that

α ≤ α < λ ⇒ {i < δ : η α (i) 6∈ A η

α

i } ∈ J, which contradicts our previous remark (remember |Y | = λ).

Due to the claim we find ν ∈ T such that A ν 6∈ I ν . By part (b) of our assumptions we find α, β ∈ A ν such that

B ν |= y ν

_

hαi ∩ y ν

_

hβi = 0.

Choose ε, ζ ∈ Y such that ν _ hαi C η ε , ν _ hβi C η ζ (see the definition of A ν ). Then ν = η ε ∧ η ζ and

B ν |= y η

ε

(i+1) ∩ y η

ζ

(i+1) = 0 (where i = lg(ν)), finishing the proof of the proposition.

Lemma 3.8. Let C = (T, ¯ λ, ¯ η, h(B η , ¯ y η ) : η ∈ T i) be a (δ, µ, λ)-constructor such that

( F) for η ∈ T , the Boolean algebras B η satisfy the (2 |δ| ) + -Knaster con- dition.

Then the Boolean algebra B red (C) satisfies the (2 |δ| ) + -Knaster condition. In fact we may replace (2 |δ| ) + above by any regular cardinal θ such that

(∀α < θ)(|α| |δ| < θ).

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To deduce that B red (C) satisfies the (2 |δ| ) + -cc it is enough to assume, instead of ( F),

( FF) for η ∈ T , every free product of finitely many of the Boolean alge- bras B η satisfies the (2 |δ| ) + -cc.

Remark. (1) Usually we will have δ = cf(µ).

(2) Later we will get more (e.g., |δ| + -Knaster if (T, ¯ η) is hereditarily free, see 5.12, 5.13).

P r o o f (of Lemma 3.8). Let θ = (2 |δ| ) + and assume ( F) (the other cases have the same proofs). Suppose that z ε ∈ B red \ {0} (for ε < θ). We start with a series of reductions which we describe fully here but later, in similar situations, we will state the result of the procedure only.

Standard cleaning. Each z ε is a Boolean combination of some genera- tors x i

0

, . . . , x i

n−1

. But, as we want to find a subsequence with non-zero intersections, we may replace z ε by any non-zero z ≤ z ε . Consequently, we may assume that each z ε is an intersection of some generators or their com- plements. Further, as cf(θ) = θ > ℵ 0 we may assume that the number of generators needed for this representation does not depend on ε and is equal to, say, n . Thus we have two functions

i : θ × n → λ and t : θ × n → 2 such that for each ε < θ,

z ε = \

l<n

(x i(ε,l) ) t(ε,l)

and there is no repetition in hi(ε, l) : l < n i. Moreover we may assume that t(ε, l) does not depend on ε, i.e., t(ε, l) = t(l). By the ∆-system lemma for finite sets we may assume that hhi(ε, l) : l < n i : ε < θi is a ∆-system of sequences, i.e.:

(∗) 1 i(ε, l 1 ) = i(ε, l 2 ) ⇒ l 1 = l 2 , (∗) 2 for some w ⊆ n we have

(∃ε 1 < ε 2 < θ)(i(ε 1 , l) = i(ε 2 , l)) iff (∀ε 1 , ε 2 < θ)(i(ε 1 , l) = i(ε 2 , l)) iff l ∈ w.

Now note that, by the definition of the algebra B red , (∗) 3 z ε

1

∩ z ε

2

= 0 if and only if

\ {x t(l) i(ε

1

,l) : l < n , t(l) = 0} ∩ \ {x t(l) i(ε

2

,l) : l < n , t(l) = 0} = 0.

Consequently, we may assume that

(∀l < n )(∀ε < θ)(t(l) = 0).

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Explanation of what we are going to do now. We want to replace the sequence hz ε : ε < θi by a large subsequence such that the places of splitting between two branches used in two different z ε ’s will be uniform. Then we will be able to translate our θ-cc problem to one on the algebras B η .

Let

A ε := {ν ∈ δ> µ : (∃j < ε)(∃l < n )(ν C η i(j,l) )}

and let B ε be the closure of A ε :

B ε := {% ∈ δ≥ µ : % ∈ A ε or lg(%) is a limit ordinal and (∀ζ < lg(%))(% ζ ∈ A ε )}

Note that |A ε | ≤ |ε| · |δ| and hence |B ε | ≤ |A ε | ≤|δ| < θ. Next we define (for ε < θ, l < n )

ζ(ε, l) := sup{ζ < δ : η i(ε,l) ζ ∈ B ε }.

Thus ζ(ε, l) ≤ lg(η i(ε,l) ) = δ. Let S = {ε < θ : cf(ε) > |δ|}. For each ε ∈ S we necessarily have

η i(ε,l) ζ(ε, l) ∈ B ε and B ε = [

ξ<ε

B ξ

(remember that cf(ε) > |δ| and for limit ε we have A ε = S

ξ<ε A ξ ) and hence

η i(ε,l) ζ(ε, l) ∈ B ξ(ε,l) for some ξ(ε, l) < ε.

Let ξ(ε) = max{ξ(ε, l) : l < n }. By the Fodor lemma we find ξ < θ such that the set

S 1 := {ε ∈ S : ξ(ε) = ξ }

is stationary. Thus η i(ε,l) ζ(ε, l) ∈ B ξ

for each ε ∈ S 1 , l < n . Since

|B ξ

|, |δ| < θ we find ν 0 , . . . , ν n

−1 ∈ B ξ

and α(l 1 , l 2 ) ≤ δ (for l 1 ≤ l 2 < n ) such that the set

S 2 := {ε ∈ S 1 : (∀l < n )(η i(ε,l) ζ(ε, l) = ν l )

& (∀l 1 ≤ l 2 < n )(lg(η i(ε,l

1

) ∧ η i(ε,l

2

) ) = α(l 1 , l 2 ))}

is stationary. Further, applying the ∆-system lemma we find a set S 3 ∈ [S 2 ] θ such that

{hη i(ε,l) (lg(ν l )) : l < n i : ε ∈ S 3 } forms a ∆-system of sequences.

For ε ∈ S 3 and ν ∈ T define b ε ν := \

{y η

i(ε,l)

(lg(ν)+1) : l < n , ν C η i(ε,l) } ∈ B ν .

Claim 3.8.1. For each ε ∈ S 3 , ν ∈ T the element b ε ν (of the algebra B ν )

is non-zero.

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P r o o f. This follows from the definition of B red and the fact that z ε 6= 0, as

b ε ν = 0 ⇒ \

{x η

i(ε,l)

: l < n , ν C η i(ε,l) } = 0 ⇒ z ε = 0.

Since for each l < n the algebra B ν

l

has the θ-Knaster property we find a set S 4 ∈ [S 3 ] θ such that for each l < n and ε 1 , ε 2 ∈ S 4 we have

ε 1 6= ε 2 ⇒ b ε ν

1l

∩ b ε ν

2

l

6= 0 in B ν

l

. Now we may finish by proving the following claim.

Claim 3.8.2. For each ε 1 , ε 2 ∈ S 4 ,

B red |= z ε

1

∩ z ε

2

6= 0.

P r o o f. Since z ε

1

∩ z ε

2

is just the intersection of generators it is enough to show that (remember the definition of B red ):

(⊗) for each ε 1 , ε 2 ∈ S 4 and for every ν ∈ T , B ν |= \

{y η

i

(lg(ν)+1) : i ∈ {i(ε 1 , l), i(ε 2 , l) : l < n } and ν C η i } 6= 0.

If ν = ν l , l < n then the intersection is b ε ν

1l

∩ b ε ν

2

l

, which is not zero by the choice of S 4 . So suppose that ν 6∈ {ν l : l < n }. Put

u ν := {i : ν C η i and for some l < n either i = i(ε 1 , l) or i = i(ε 2 , l)}.

If

i (lg(ν)) : i ∈ u ν } ⊆ {η i(ε

2

,l) (lg(ν)) : l < n & ν C η i(ε

2

,l) }

then we are done as b ε ν

2

6= 0. So there is l 1 < n such that ν C η i(ε

1

,l

1

) and η i(ε

1

,l

1

) (lg(ν) + 1) 6∈ {η i(ε

2

,l) (lg(ν) + 1) : l < n & ν C η i(ε

2

,l) }.

Similarly we may assume that there is l 2 < n such that ν C η i(ε

2

,l

2

) and η i(ε

2

,l

2

) (lg(ν) + 1) 6∈ {η i(ε

1

,l) (lg(ν) + 1) : l < n & ν C η i(ε

1

,l) }.

By symmetry we may assume that ε 1 < ε 2 . Then ν = η i(ε

2

,l

2

)  lg(ν) ∈ A ε

1

+1 ⊆ B ε

2

and hence ζ(ε 2 , l 2 ) ≥ lg(ν). By the choice of S 2 (remember ε 1 , ε 2 ∈ S 4 ⊆ S 2 ), we get ν E ν l

2

. But we have assumed that ν 6= ν l

2

, so ν C ν l

2

. Hence (once again due to ε 1 , ε 2 ∈ S 2 )

η i(ε

2

,l

2

) (lg(ν) + 1) = η i(ε

1

,l

2

) (lg(ν) + 1) = ν l

2

(lg(ν) + 1), which contradicts the choice of l 2 .

This completes the proof of Lemma 3.8.

Remark 3.9. We can strengthen “θ-Knaster” in the assumption and

conclusion of 3.8 in various ways. For example we may have “intersection

of any n members of the final set is non-zero”.

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Definition 3.10. Let (B, ¯ y) be a λ-marked Boolean algebra, κ ≤ λ. We say that:

(1) (B, ¯ y) the κ-Knaster property if B satisfies the condition in the def- inition of the κ-Knaster property (see 0.4) with restriction to subsequences of ¯ y.

(2) (B, ¯ y) is (κ, notλ)-Knaster if

(a) the algebra B has the κ-Knaster property, but

(b) the sequence ¯ y witnesses that the λ-Knaster property fails for B.

Conclusion 3.11. Assume that µ is a strong limit singular cardinal , λ = 2 µ = µ + and θ = (2 cf(µ) ) + . Then there exists a λ-marked Boolean algebra (B, ¯ y) which is (θ, notλ)-Knaster.

P r o o f. Choose cardinals µ 0 i , µ i < µ (for i < cf(µ)) such that:

(α) cf(µ) < µ 0 0 , (β) Q

j<i µ j < µ 0 i , µ i = (2 µ

0i

) + ,

(γ) hµ i : i < cf(µ)i and hµ 0 i : i < cf(µ)i are increasing cofinal in µ.

(Possible as µ is strong limit singular.) By Proposition 3.1 we find µ i -marked Boolean algebras (B i , ¯ y i ) and (µ 0 i ) + -complete ideals I i on µ i (for i < δ) such that:

(a) if X ⊆ µ i , X 6∈ I i then (∃α, β ∈ X)(B i |= y α i ∩ y i β = 0), (b) the algebra B i has the (2 cf(µ) ) + -Knaster property.

Let T = S

i<cf(µ)

Q

j<i µ j and for ν ∈ T i (i < cf(µ)) let I ν = I i , B ν = B i ,

¯

y ν = ¯ y i and λ ν = µ i . Now we may apply Proposition 2.8 to µ, hµ 0 i : i <

cf(µ)i and (T, ¯ λ, ¯ I) to find a stronger J cf(µ) bd -cofinal sequence ¯ η for (T, ¯ λ, ¯ I) of length λ. Consider the (cf(µ), µ, λ)-constructor C = (T, ¯ λ, ¯ η, h(B ν , ¯ y ν ) : ν ∈ T i). By (b) above we may apply Lemma 3.8 to deduce that the algebra B red (C) satisfies the (2 cf(µ) ) + -Knaster condition. Finally we use Propo- sition 3.7 (and (a) above) to conclude that (B red (C), hx red α : α < λi) is (θ, notλ)-Knaster.

Proposition 3.12. Assume that κ is a regular cardinal such that (∀α <

κ)(|α| |δ| < κ), ¯ λ = hλ i : i < δi is an increasing sequence of regular cardinals such that κ ≤ λ 0 , Q

j<i λ j < λ i (or just max pcf{λ j : j < i} < λ i ) for i < δ and λ ∈ pcf{λ i : i < δ}. Further suppose that for each i < δ there exists a λ i -marked Boolean algebra which is (κ, notλ i )-Knaster. Then there exists a λ-marked Boolean algebra which is (κ, notλ)-Knaster.

P r o o f. If λ = λ i for some i < δ then there is nothing to do. If λ < λ i

for some i < δ then let α < δ be the maximal limit ordinal such that

(∀i < α)(λ i < λ) (it necessarily exists). Now we may replace hλ i : i < δi by

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i : i < αi. Thus we may assume that (∀i < δ)(λ i < λ). Further we may assume that

λ = max pcf{λ i : i < δ}

(by [22, I, 1.8]). Now, due to [22, II, 3.5, p. 65], we find a sequence ¯ η ⊆ Q

i<δ λ i and an ideal J on δ such that:

(1) J ⊇ J δ bd and λ = tcf( Q

i<δ λ i /J )

(naturally: J = {a ⊆ δ : max pcf{λ i : i ∈ a} < λ}), (2) ¯ η = hη ε : ε < λi is < J -increasing cofinal in Q

i<δ λ i /J , (3) for each i < δ, |{η ε i : ε < λ}| < λ i .

Let T = S

i<δ

Q

j<i λ j and for ν ∈ T i (i < δ) let λ ν = λ i , I ν = J λ bd

i

.

It follows from the choice of ¯ η, J above and our assumptions that we may apply Proposition 2.6 and hence ¯ η is J -cofinal for (T, ¯ λ, ¯ I). For ν ∈ T let (B ν , ¯ y ν ) be a λ ν -marked (κ, notλ ν )-Knaster Boolean algebra (exists by our assumptions). Now we may finish using 3.8 and 3.7 for C = (T, ¯ λ, ¯ η, h(B η , ¯ y η ) : η ∈ T i), ¯ I and J (note the assumption (b) of 3.7 is satisfied as I η = J λ bd

η

; remember the choice of (B η , ¯ y η )).

Remark 3.13. Note that from the cardinal arithmetic hypothesis cf(µ)

= χ, χ < χ < µ, µ + = λ < 2 χ alone we cannot hope to build a coun- terexample. This is because of [15, §4], particularly Lemma 4.13 there. It was shown in that paper that if χ < χ 1 = χ 1

1

then there is a χ + -cc χ-complete forcing notion P of size χ 1 such that

P “if |B| < χ 1 , B |= χ-cc then B + is the union of χ ultrafilters”.

More on this in Section 8. So the centrality of λ ∈ Reg ∩(µ, 2 µ ], µ strong limit singular, is very natural.

4. The main result

Proposition 4.1. Suppose that C is a (δ, µ, λ)-constructor. Then the free product B red (C)∗B green (C) fails the λ-cc (so c(B red (C)∗B green (C)) ≥ λ).

P r o o f. Look at the elements (x i , x i ) ∈ B red ∗ B green for i < λ. It follows directly from the definition of the algebras that for each i < j < λ,

either B red |= x red i ∩ x red j = 0 or B green |= x green i ∩ x green j = 0.

Consequently, the sequence h(x i , x i ) : i < λi witnesses the assertion.

Proposition 4.2. Suppose that n < ω and for l ≤ n:

(1) χ l , λ l are regular cardinals, χ l < λ l < χ l+1 ,

(2) (B l , ¯ y l , I l ) is a (λ l , χ l )-well marked Boolean algebra (see Defini-

tion 3.2), ¯ y l = hy l i : i < λ l i,

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(3) B is the Boolean algebra freely generated by {y η : η ∈ Q

l≤n λ l } except that if

η i

0

, . . . , η i

k−1

∈ Y

l≤n

λ l , η i

0

l = η i

1

l = . . . = η i

n−1

l, B l |= \

m<k

y η l

im

(l) = 0 then T

m<k y η

im

= 0. [Compare the definition of the algebras B red (C).]

(4) I = {B ⊆ Q

l≤n λ l : ¬(∃ I

0

γ 0 ) . . . (∃ I

n

γ n )(hγ 0 , . . . , γ n i ∈ B)}.

Then:

(a) if all the algebras B l (for l ≤ n) have the θ-Knaster property and θ is a regular uncountable cardinal then B has the θ-Knaster property,

(b) I is a χ 0 -complete ideal on Q

l≤n λ i , (c) if Y ⊆ ( Q

l≤n λ l ) n is such that

(∃ I η 0 ) . . . (∃ I η n )(hη 0 , . . . , η n i ∈ Y )

then there are hη 0 0 , . . . , η 0 n i, hη 00 0 , . . . , η n 00 i ∈ Y such that for all l ≤ n, B |= y η

0l

∩ y η

00

l

= 0.

P r o o f. (a) The proof that the algebra B satisfies the θ-Knaster condi- tion is exactly the same as that of 3.8 (actually it is a special case).

(b) Should be clear.

(c) For l, m ≤ n put χ m l = χ l , λ m l = λ l , I l m = I l , P l m = {{α, β} ⊆ λ l : B l |= y α l ∩ y l β = 0}, B = Y . It is easy to check that the assumptions of Proposition 1.5 are satisfied. Applying it we find sets X 0 , . . . , X n satisfying the appropriate versions of clauses (a)–(d) there. Note that our choice of the sets P l m and clauses (b), (c) of 1.5 imply that

X m = {ν m 0 , ν m 00 } ⊆ Y

l≤n−m

λ l , ν m 0 (n − m) = ν m 00 (n − m), B n−m |= y ν n−m

0

m

(n−m) ∩ y n−m ν

00

m

(n−m) = 0.

Look at the sequences hν 0 0 , . . . , ν n 0 i, hν 0 00 , . . . , ν n 00 i. By clause (d) of 1.5 we find hη 0 0 , . . . , η 0 m i, hν 0 00 , . . . , ν n 00 i ∈ Y such that for each m ≤ n,

ν m 0 E η 0 m , ν m 00 E η 00 m .

Now, the properties of ν m 0 , ν m 00 and the definition of the algebra B imply that for each m ≤ n,

B |= y η

m0

∩ y η

00m

= 0.

Lemma 4.3. Assume that λ is a regular cardinal , |δ| < λ, J is an ideal on δ extending J δ bd , C = (T, ¯ λ, ¯ η, h(B η , ¯ y η ) : η ∈ T i) is a (δ, µ, λ)-constructor and ¯ I is such that (T, ¯ λ, ¯ I) ∈ K id δ,µ . Suppose that ¯ η = hη α : α < λi is a sequence stronger (or big) J -cofinal in (T, ¯ λ, ¯ I) such that

(∀i < δ)(|{η α i : α < λ}| < λ).

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