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146 (1995)

Linear orders and MA + ¬wKH

by

Zoran S p a s o j e v i ´ c (Madison, Wisc.)

Abstract. I prove that the statement that “every linear order of size 2

ω

can be embedded in (ω

ω

, )” is consistent with MA + ¬wKH.

Let ϕ

κ

denote the statement that every linear order of size κ can be embedded in (ω

ω

, ) for regular κ ≤ 2

ω

= c where ω

ω

denotes the set of all functions from ω to ω and  is a partial order on ω

ω

defined as follows: for f, g ∈ ω

ω

let f  g if and only if ∃n < ω ∀i ≥ n (f (i) ≤ g(i)) and f (i) < g(i) on an infinite set. Under CH, ∀κ ≤ c (ϕ

κ

), which basically follows from the fact that there are no (ω, ω)-gaps in (ω

ω

, ). If CH fails then (ω

ω

, ) may not even contain a well order of type ω

2

regardless of what c is. On the other hand, MA + ¬CH → ∀κ < c (ϕ

κ

). Kunen constructed a model for MA + ¬CH + ¬ϕ

c

and Laver [L] constructed a model for ¬CH + ϕ

c

. For a while, the question was whether MA + ¬CH is strong enough to decide ϕ

c

. Woodin [W] constructed a model for MA + c = ω

2

+ ϕ

c

, therefore, together with Kunen’s result, showing that ϕ

c

is independent of MA + ¬CH.

On the other hand, PFA → MA + ¬wKH → MA + ¬CH and neither of the implications is reversible. Therefore MA + ¬wKH is in strength some- where between PFA and MA + ¬CH. But also PFA → c = ω

2

+ ¬ϕ

c

. Therefore, it is reasonable to ask whether MA + ¬wKH is strong enough to decide ϕ

c

. This question is the main consideration of this paper. The main result is Theorem 3.2 which states that if M is a countable transitive model (c.t.m.) for ZFC + V=L and κ is the first inaccessible cardinal in M then there is an extension N[J] of M which is a model for ZFC + MA + ¬wKH + c = ω

2

+ ϕ

c

. The existence of an inaccessible cardinal is necessary to show the consistency of ¬wKH, as shown by Mitchell [M]. Todorˇcevi´c [T]

constructed a model for MA + ¬wKH + c = ω

2

, and I will use this result together with the result of Laver to construct the model N[J]. Therefore, when combined with PFA → MA + ¬wKH + c = ω

2

+ ¬ϕ

c

, it shows that

1991 Mathematics Subject Classification: Primary 03E35.

[215]

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MA + ¬wKH is still not strong enough to decide ϕ

c

. Woodin’s construc- tion cannot easily be modified to fit the additional arguments required in showing ¬wKH because his construction is completed in ω

2

· ω

2

stages. In order to show that ¬wKH holds in the final model the construction here has to be finished in ω

2

steps. However, the treatment of stages of cofinality ω

1

resembles those in Woodin’s construction. Consequently, the construction here can be regarded as an amalgamation of the constructions mentioned above.

To construct a model N[J], I start with a c.t.m. M for ZFC + V=L in which κ is the first inaccessible cardinal. Then, as in [M], extend M with a partial order to obtain a model N such that N |= “ ¬wKH + c = κ = ω

2

”.

In N, I perform an iterated ccc forcing construction with finite supports of length ω

2

. In the process I construct a c-saturated linearly ordered subset (L, ) of (ω

ω

, ). At the successor stages I alternate between ccc partial orders to make MA true and splitting partial orders for pregaps in L. A difficulty occurs in splitting (ω

1

, ω

1

)-gaps. However, the construction is ar- ranged in such a way that these gaps appear in L only at the limit stages of cofinality ω

1

; at these stages I split all such gaps, all at once. The elements of ω

ω

obtained at these stages will not be used directly, but they are needed to ensure that the splitting orders for all the pregaps in L continue to have the ccc until they are filled, one by one, at the later successor stages. The partial orders at these limit stages have cardinality ω

2

, which causes some difficulty in the proof of ¬wKH. This difficulty is overcome by reducing the argument to suborders of size ω

1

of these partial orders.

Since trees and gaps play a central role in the construction, I begin with some notions and results on trees and gaps that are needed here. Many results included here are already known, however I present a different view point. Notation and terminology are adapted from [K], especially the part on iterated forcing.

1. Trees. A tree is a partial order in the strict sense, hT, ≤i, such that for each x ∈ T, b x = {y ∈ T : y < x} is well ordered by <. If x ∈ T, the height of x in T, ht(x, T), is the ordinal α which is the order type of b x and T

x

= {y ∈ T : y ≤ x ∨ x < y}. For each ordinal α, the αth level of T, Lev

α

(T), is the set {x ∈ T : ht(x, T) = α}. The height of T, ht(T), is the least α such that Lev

α

(T) = ∅. A chain in T is a set C ⊆ T which is totally ordered by <. If C intersects every level of T then C is called a path through T. A ⊆ T is an antichain iff ∀x, y ∈ A (x 6= y → (x 6≤ y ∧ y 6≤ x)). I will only consider well pruned trees. A well pruned tree is a tree T such that

(i) |Lev

0

(T)| = 1,

(ii) ∀α < β < ht(T) ∀x ∈ Lev

α

(T) ∃y

1

, y

2

∈ Lev

β

(T) (y

1

6= y

2

∧ x ≤

y

1

, y

2

),

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(iii) ∀α < ht(T) ∀x, y ∈ Lev

α

(T) (lim(α) → (x = y ↔ b x = b y)).

From now on any mention of a tree T will automatically mean that T is a well pruned tree. An ω

1

-tree is a tree T such that ht(T) = |T| = ω

1

. An ω

1

-tree is a weak Kurepa tree if it has at least ω

2

paths. The assertion that there is a weak Kurepa tree is denoted by wKH and ¬wKH denotes its negation. An Aronszajn tree is an ω

1

-tree T without any paths such that

∀α < ω

1

(|Lev

α

(T)| ≤ ω). A Suslin tree is an Aronszajn tree with no un- countable antichains. If T is an ω

1

-tree and ∃(f : T → ω) (∀x, y ∈ T (x < y

→ f (x) 6= f (y))) then T is called a special ω

1

-tree and f a specializing function for T. It follows that if T is a special Aronszajn tree with a special- izing function f then for some n ∈ ω, f

−1

(n) is uncountable and as such an uncountable antichain in T. Therefore neither T nor any subtree of T can be Suslin. Next, I define a partial order S

T

, due to Baumgartner, which is intended to add a specializing function for T.

Definition 1.1. Let T be an ω

1

-tree. Then

S

T

= {p : ∃x ∈ [T]

(p : x → ω) ∧ ∀s, t ∈ x (s < t → p(s) 6= p(t))}

with p

1

≤ p

2

iff p

1

⊇ p

2

.

The symbol “⊥” denotes incompatibility in any partial order P and “Y”

will be used to denote incompatibility in a tree T, i.e.

∀x, y ∈ T (x Y y ↔ (x 6≤ y ∧ y 6≤ x)).

Then Y extends to incompatibility in [T]

as follows:

∀a, b ∈ [T]

(a Y b ↔ (a ∩ b = ∅ ∧ ∀x ∈ a ∀y ∈ b (x Y y))).

Also note that if p, q ∈ S

T

and dom(p) Y dom(q) then p and q are compatible in S

T

.

Lemma 1.2. If T is an Aronszajn tree then (S

T

, ≤) has the ccc.

P r o o f. By way of contradiction assume that A = {p

α

: α < ω

1

} ⊆ S

T

is an uncountable antichain. Without loss of generality I may assume

(1) ∀α < ω

1

(|dom(p

α

)| = n) for some n < ω,

(2) ∀α, β < ω

1

(α 6= β → (dom(p

α

) ∩ dom(p

β

) = ∅)).

To see that I may assume (2), first assume, by the ∆-system lemma, that {dom(p

α

) : α < ω

1

} forms a ∆-system with root r. Then, since ω

r

is count- able, I may assume that ∀α, β < ω

1

(p

α

¹r = p

β

¹r). Then (2) is implied at once by the claim below.

Claim. If e

α

= dom(p

α

) \ r then (p

α

¹e

α

⊥ p

β

¹e

β

) ↔ (p

α

⊥ p

β

).

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P r o o f o f C l a i m. Let p

α

⊥ p

β

. Then

∃x ∈ dom(p

α

) ∃y ∈ dom(p

β

)

((x < y ∧ p

α

(x) = p

β

(y)) ∨ (y < x ∧ p

β

(y) = p

α

(x))).

It cannot happen that x, y ∈ dom(p

α

) since p

α

∈ S

T

, and it cannot happen that x, y ∈ dom(p

β

) for the same reason. Therefore x, y 6∈ r so that x ∈ e

α

and y ∈ e

β

. This basically proves the claim since the implication in the other direction is trivial.

Now let dom(p

α

) = {s

α0

, s

α1

, . . . , s

αn−1

}. Finally I may assume that if α < β < ω

1

, p

α

(s

αi

) = p

β

(s

βj

), and s

αi

and s

βj

are comparable (which must happen for some i and j since p

α

⊥ p

β

) then s

αi

< s

βj

. Therefore for each α there must be i(α), j(α) < n such that {β : s

αi(α)

< s

βj(α)

} is uncountable.

Furthermore, there must be i and j such that B = {α : i(α) = i ∧ j(α) = j}

is also uncountable. But now if α

1

, α

2

∈ B there is β > α

1

, α

2

such that s

αi1

, s

αi2

< s

βj

. And since T is a tree, s

αi1

and s

αi2

are comparable. Therefore {s

αi

: α ∈ B} may be extended to a path through T, contradicting the fact that T has no paths. Therefore A cannot be an uncountable antichain.

From the proof above immediately follow the two corollaries below.

Corollary 1.3. Let M be a c.t.m. for ZFC and, in M, suppose that T is an Aronszajn tree and P a ccc partial order with G P-generic over M.

Then S

T

fails to have the ccc in M[G] iff a new path has been added through T in M[G].

Corollary 1.4. Let M be a c.t.m. for ZFC, T an Aronszajn tree in M, and G S

T

-generic over M. Then

M[G] |= “ T is a special Aronszajn tree ”.

Definition 1.5. Let P be a partial order. Then P has the property K iff

∀A ∈ [P]

ω1

∃B ∈ [A]

ω1

∀x, y ∈ B (x 6⊥ y).

Lemma 1.6. If T is a special Aronszajn tree then S

T

has the property K.

P r o o f. Let {p

α

: α < ω

1

} ⊆ S

T

. Then, as in the proof of Lemma 1.2, I may assume

(1) ∀α < ω

1

(|dom(p

α

)| = n) for some n < ω, (2) ∀α, β < ω

1

(α 6= β → dom(p

α

) ∩ dom(p

β

) = ∅).

Let dom(p

α

) = e

α

. To get p

α

and p

β

compatible it suffices to get e

α

Y e

β

. Therefore the proof follows immediately from the following

Claim. ∃A ∈ [ω

1

]

ω1

∀α, β ∈ A (α 6= β → e

α

Y e

β

).

P r o o f o f C l a i m. The proof is by induction on |e

α

| = n. Fix n and

assume the result is true for all m < n.

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C a s e 1: Suppose ∀γ < ω

1

∃x ∈ Lev

γ

(T) ∃α < ω

1

(e

α

⊆ T

x

). Then for µ < ω

1

choose x

µ

∈ Lev

γµ

(T), α

µ

and increasing γ

µ

such that e

αµ

⊆ T

xµ

with γ

µ

> sup{ht(z) : z ∈ S

ν<µ

e

αν

}. Then, by the remarks before Defini- tion 1.1, there is an A ∈ [ω

1

]

ω1

such that {x

µ

: µ ∈ A} is an uncountable antichain in {x

µ

: µ < ω

1

}. But then ∀α, β ∈ A (α 6= β → e

α

Y e

β

).

C a s e 2: This is just ¬Case 1. Fix γ such that ∀x ∈ Lev

γ

(T) ∀α <

ω

1

(e

α

6⊆ T

x

). Then, since each level of T is countable and e

α

are all pairwise disjoint, it follows that n ≥ 2 and only countably many e

α

meet Lev

γ

(T) or below. Therefore without loss of generality I may throw those away and assume that ∀α < ω

1

∀z ∈ e

α

(ht(z) > γ). I may also assume that ∃x ∈ Lev

γ

(T) ∀α < ω

1

(e

α

∩ T

x

6= ∅) since e

α

S

{T

x

: x ∈ Lev

γ

(T)} and

|Lev

γ

(T)| ≤ ω. So fix any such x. Then without loss of generality I may assume that

∀α < ω

1

((|e

α

∩ T

x

| = i > 0) ∧ (|e

α

\ T

x

| = j > 0))

since e

α

6⊆ T

x

. Then 0 < i, j < n and i + j = n. And by the induction hypothesis I may assume that

(∗) ∀α, β < ω

1

(α 6= β → (((e

α

∩ T

x

) Y (e

β

∩ T

x

)) ∧ ((e

α

\T

x

) Y (e

β

\T

x

)))).

But then e

α

are also pairwise incompatible in [T]

. Here I claim that it is not possible to have s ∈ e

α

and t ∈ e

β

with s < t and α 6= β. There are 4 cases to consider. If s, t ∈ T

x

or s, t 6∈ T

x

then I am done by (∗). The cases s ∈ T

x

∧ t 6∈ T

x

or s 6∈ T

x

∧ t ∈ T

x

cannot happen since T is a tree. This proves the claim and hence the lemma.

Lemma 1.7. Let M be a c.t.m. for ZFC and suppose that U and T are Aronszajn trees in M. If G is S

T

-generic over M then M[G] |= “ U is Aronszajn ”.

P r o o f. It suffices to prove that no new paths through U are added in M[G]. So by way of contradiction let p ∈ S

T

and ˙b ∈ M

ST

with p ° “ ˙b is a new path through ˇ U ”. Since U is Aronszajn in M, it follows that ˙b

G

= b 6∈ M.

Let

X = {u ∈ U : ∃p

u

≤ p (p

u

° “ ˇ u ∈ ˙b ”)}.

Let u

α

∈ Lev

α

(U) and p

α

∈ S

T

with p

α

≤ p such that p

α

° “ ˇ u

α

∈ ˙b ”.

Now

M[G] |= “ S

T

has the property K ”

so in M[G] let B ∈ [ω

1

]

ω1

such that {p

α

: α ∈ B} are pairwise compatible.

Then there is a path, d, through U determined by B with d ∈ M[G] and d ⊆ X.

On the other hand, b is a new path through U so for each u ∈ X there

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are s, t ∈ X such that u ≤

U

s, t and t and s are incomparable. Let Y = {u ∈ X : u is ≤

U

-minimal with u 6∈ d}.

Then Y ∈ M[G] and for each u ∈ Y fix a p

u

∈ P such that p

u

≤ p ∧ p

u

°

“ ˇ u ∈ ˙b ” and let A = {p

u

∈ S

T

: u ∈ Y }. Then A ∈ M[G] and A is an uncountable subset of S

T

and any two elements of A are incompatible. Hence A is an uncountable antichain in S

T

, which contradicts the fact that S

T

has the property K in M[G].

Corollary 1.8. M[G] |= “ S

U

has the ccc ”.

Lemma 1.9. Let M be a c.t.m. for ZFC and, in M, suppose that P is a ccc partial order and T an ω

1

-tree. If G is P-generic over M with

M[G] |= “ b is a new path through T ” then there is a Suslin tree U ⊆ T with U ∈ M such that

M[G] |= “ b is a new path through U ”.

P r o o f. Let p ∈ P with p ° “ ˙b is a new path through ˇ T ”. Let U = {u ∈ T : ∃q ≤ p (q ° “ ˇ u ∈ ˙b ”)}.

Clearly U ∈ M, U ⊆ T, and |U| = ω

1

. The fact that b is a new path through T also implies that ht(U) = ω

1

. If U is not Suslin in M then there is an A ⊆ U with A ∈ M and |A| = ω

1

such that any two elements of A are incomparable. For each u ∈ U fix a p

u

∈ P such that p

u

≤ p ∧ p

u

° “ ˇ u ∈ ˙b ” and let

A

P

= {p

u

: u ∈ A ∧ p

u

≤ p ∧ (p

u

° “ ˇ u ∈ ˙b ”)}.

Clearly A

P

∈ M. Then A

P

is an antichain in P. This follows since if p

u

, p

t

A

P

for u 6= t ∈ A and q ∈ P with q ≤ p

u

, p

t

then q ° “ ˇ u ∈ ˙b ∧ ˇt ∈ ˙b ” so that u and t are comparable, which is impossible by the choice of A.

Furthermore, A

P

is uncountable since A is. Hence A

P

is an uncountable antichain in P contradicting the fact that P has the ccc in M. Therefore U is Suslin with M[G] |= “ b ⊆ U ” so that

M[G] |= “ b is a new path through U ”.

And this is precisely what I set out to show.

Let P be a partial order and ˙ Q a P-name for a partial order. Then P ∗ ˙ Q denotes a two-step iteration. The following result is taken from [K] and is needed in the proof of Lemma 1.11.

Lemma 1.10. Assume that in M, P is a ccc partial order and ˙ Q a P-name

for a partial order such that 1 °

P

“ ˙ Q has the ccc ”. Then P ∗ ˙ Q has the ccc

in M.

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Lemma 1.11. Suppose M is a c.t.m. for ZFC and P and Q two ccc partial orders in M. Then P × Q has the ccc iff 1 °

P

“ ˇ Q has the ccc ”.

P r o o f. If 1 °

P

“ ˇ Q has the ccc ” then by Lemma 1.10, P ∗ ˇ Q has the ccc.

Then since P ∗ ˇ Q and P × Q are isomorphic it follows that P × Q has the ccc.

Now suppose that P × Q has the ccc and by way of contradiction assume that

1 °

P

“ ˙ A is an uncountable antichain in ˇ Q ”.

Let τ be a P-name and p

0

∈ P with

p

0

°

P

“ τ : ˇ ω

1

→ ˙ A and τ is one-to-one and onto ”.

Also let p

ξ

≤ p

0

and q

ξ

∈ Q with p

ξ

°

P

“ τ (ξ) = ˇ q

ξ

”. Then B = {hp

ξ

, ˇ q

ξ

i : ξ < ω

1

} is an uncountable antichain in P∗ ˇ Q. To see this suppose that hp

α

, ˇ q

α

i and hp

β

, ˇ q

β

i are compatible for some α 6= β. Let hp, ˇ qi ≤ hp

α

, ˇ q

α

i, hp

β

, ˇ q

β

i.

Then p ≤ p

α

, p

β

and p °

P

“ ˇ q ≤ ˇ q

α

, ˇ q

β

”. But this leads to a contradiction since also p ≤ p

0

so that

p °

P

“ ˙ A is an antichain in ˇ Q and ˇ q

α

, ˇ q

β

∈ ˙ A ”.

Therefore 1 °

P

“ ˇ Q has the ccc ”.

Lemma 1.12. Let M be a c.t.m. for ZFC and, in M, P a ccc partial order and hP

ξ

: ξ ≤ αi an iterated ccc forcing construction with finite supports where α is a limit ordinal. If ∀ξ < α (1 °

Pξ

“ ˇ P has the ccc ”) then 1 °

Pα

“ ˇ P has the ccc ”.

P r o o f. If cf(α) = ω and

1 °

Pα

“ ˙ A is an uncountable antichain in ˇ P ”

then since hP

ξ

: ξ ≤ αi has finite supports it follows that some uncountable subset of A is constructed at some earlier stage. But any subset of A is also an antichain in P. Therefore

∃β < α (1 °

Pβ

“ ˇ P fails to have the ccc ”), contradicting the hypothesis.

If cf(α) > ω

1

then any subset of P of size ω

1

is constructed by some stage β < α. Therefore if

1 °

Pα

“ ˇ P fails to have the ccc ” then

∃β < α (1 °

Pβ

“ ˇ P fails to have the ccc ”),

again contradicting the hypothesis.

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Finally, let cf(α) = ω

1

and suppose that the conclusion of the lemma fails. Therefore

∀β < α (1 °

Pβ

“ ˇ P has the ccc ”) but

1 °

Pα

“ ˇ P fails to have the ccc ”.

Then according to Lemma 1.11, P × P

β

has the ccc for each β < α, but P × P

α

does not have the ccc. Then again by Lemma 1.11,

∀β < α (1 °

P

“ ˇ P

β

has the ccc ”) but

1 °

P

“ ˇ P

α

fails to have the ccc ”.

Let G be P-generic over M and, working in M[G], let A = {p

ξ

: ξ < ω

1

} be an uncountable antichain in P

α

. Then by the ∆-system lemma I may assume that {supp(p

ξ

) : ξ < ω

1

} forms a ∆-system with root r. Let β < α with r ⊆ β. Then since P

β

has the ccc let ξ, η < ω

1

and p ∈ P

β

be such that p ≤ p

ξ

¹β, p

η

¹β. Now define p

as follows:

p

(θ) =

 

 

p(θ) if θ < β,

p

ξ

(θ) if θ ∈ supp(p

ξ

) ∩ (α \ β), p

η

(θ) if θ ∈ supp(p

η

) ∩ (α \ β), 1(θ) otherwise.

Then p

∈ P

α

and p

≤ p

ξ

, p

η

, which contradicts the assumption that A is an uncountable antichain in P

α

. Therefore 1 °

P

“ ˇ P

α

has the ccc ” and hence by Lemma 1.11, 1 °

Pα

“ ˇ P has the ccc ”.

According to the lemma just proved if T is Aronszajn in the ground model and S

T

fails to have the ccc then this cannot happen at a limit stage.

Equivalently, if any new paths are added through T then it can only happen at a successor stage.

This concludes the work on trees required for the final model.

2. Gaps. In the construction of a c-saturated linear order in (ω

ω

, ) gaps occur naturally. This section deals with gaps and their properties that are necessary for the construction in Section 3.

For convenience I choose to work with (Z

ω

, ) rather than (ω

ω

, ) and

construct a c-saturated linear order in (Z

ω

, ) instead of (ω

ω

, ). This

will imply the result for (ω

ω

, ) since (Z

ω

, ) can easily be embedded in

ω

, ). Recall that Z

ω

is the set of all functions that map ω into Z, the

set of integers. This set has a natural partial order , “”, which is defined

as follows: If f, g ∈ Z

ω

then f  g iff ∃n < ω ∀i ≥ n (f (i) ≤ g(i)) and

f (i) < g(i) on an infinite set.

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Definition 2.1. Let I, J be two linearly ordered sets and hf, gi = hf

ξ

, g

η

: ξ ∈ I, η ∈ Ji ⊆ Z

ω

such that ∀ξ, η ∈ I (ξ ≤ η → f

ξ

 f

η

) and

∀ζ, θ ∈ J (ζ ≤ θ → g

θ

 g

ζ

) and ∀ξ ∈ I ∀η ∈ J (f

ξ

 g

η

). Then hf, gi is called an (I, J)-pregap in Z

ω

. If ∃h ∈ Z

ω

∀ξ ∈ I ∀η ∈ J (f

ξ

 h  g

η

) then h splits hf, gi. If no such h exists then hf, gi is called an (I, J)-gap.

Definition 2.2. Let I, J, I

0

, J

0

be linearly ordered sets and hf, gi an (I, J)-pregap and hf

0

, g

0

i an (I

0

, J

0

)-pregap. Then hf, gi and hf

0

, g

0

i are equivalent iff ∀ξ ∈ I ∃ζ ∈ I

0

∀η ∈ J ∃θ ∈ J

0

(f

ξ

 f

ζ0

∧ g

0θ

 g

η

) and

∀ξ ∈ I

0

∃ζ ∈ I ∀η ∈ J

0

∃θ ∈ J (f

ξ0

 f

ζ

∧ g

θ

 g

η0

).

Let hf, gi and hf

0

, g

0

i be two equivalent gaps. Then h ∈ Z

ω

splits hf, gi if and only if h splits hf

0

, g

0

i. From this fact it easily follows that there is a ccc partial order that splits hf, gi if and only if there is a ccc partial order that splits hf

0

, g

0

i. Therefore considering splitting orders for an (I, J)-pregap is equivalent to considering splitting orders for an (I

0

, J

0

)-pregap where I

0

is a cofinal well ordered subset of I and J

0

is a cofinal well ordered subset of J. Thus in considering splitting orders for pregaps I can use ordinals for indexing sets and an (I, J)-pregap will also be called a (λ, κ)-pregap if cf(I) = λ and cf(J) = κ. One such splitting order is given by the following

Definition 2.3. Let hf, gi = hf

ξ

, g

η

: ξ < λ, η < κi be a (λ, κ)-pregap where λ, κ are ordinals. Set

S

hf,gi

= {hx, y, n, si : x ∈ [λ]

∧ y ∈ [κ]

∧ n < ω

∧(s : n → Z) ∧ ∀ξ ∈ x ∀η ∈ y ∀i ≥ n (f

ξ

(i) ≤ g

η

(i))}

with hx

2

, y

2

, n

2

, s

2

i ≤ hx

1

, y

1

, n

1

, s

1

i iff

(1) x

1

⊆ x

2

, y

1

⊆ y

2

, n

1

≤ n

2

, s

1

= s

2

¹n

1

,

(2) ∀ξ ∈ x

1

∀η ∈ y

1

∀i < ω (n

1

≤ i < n

2

→ (f

ξ

(i) ≤ s

2

(i) ≤ g

η

(i))).

The splitting function h for hf, gi is given by h = [

{s : ∃x, y, n (hx, y, n, si ∈ G)}

where G is S

hf,gi

-generic. Note that if λ = κ = 0 then S

hf,gi

is isomorphic to the partial order that adds a generic element to Z

ω

.

Definition 2.4. Let hf, gi = hf

ξ

, g

η

: ξ < λ, η < κi be a (λ, κ)-pregap where λ, κ are ordinals. Then the function h is S

hf,gi

-generic if the filter

G = {hx, y, n, si ∈ S

hf,gi

: (s = h¹n) ∧ ∀ξ ∈ x ∀η ∈ y ∀i ≥ n

(f

ξ

(i) ≤ h(i) ≤ g

η

(i))}

is S

hf,gi

-generic.

Note that h is S

hf,gi

-generic if and only if −h is S

h−g,−f i

-generic where

h−g, −f i = h−g

η

, −f

ξ

: η < κ, ξ < λi. This fact will be used later and this is

precisely the reason why I chose to work with (Z

ω

, ) rather than (ω

ω

, ).

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The partial order in Definition 2.3 is due to Kunen as is the following Lemma 2.5. Let hf, gi be a (λ, κ)-pregap.

(1) If the pregap is split then S

hf,gi

has the property K.

(2) If cf(λ) 6= ω

1

or cf(κ) 6= ω

1

then S

hf,gi

has the property K.

(3) If λ = κ = ω

1

and S

hf,gi

fails to have the ccc then there is an m < ω and there are X, Y ∈ [ω

1

]

ω1

with X = {ξ

α

: α < ω

1

} and Y = {η

α

: α < ω

1

} such that

(i) ∀α < ω

1

∀i ≥ m (f

ξα

(i) ≤ g

ηα

(i)) and

(ii) ∀α, β < ω

1

(α 6= β → ∃i ≥ m (f

ξα

(i) 6≤ g

ηβ

(i)∨f

ξβ

(i) 6≤ g

ηα

(i))).

P r o o f. (1) Let {p

α

: α < ω

1

} ⊆ S

hf,gi

where p

α

= hx

α

, y

α

, n

α

, s

α

i.

Suppose h splits hf, gi. For each α < ω

1

fix k

α

< ω such that

∀ξ ∈ x

α

∀η ∈ y

α

∀i ≥ k

α

(f

ξ

(i) ≤ h(i) ≤ g

η

(i)).

By extending each p

α

if necessary I may assume that ∀α < ω

1

(k

α

≤ n

α

).

Then it is easily seen that

∃A ∈ [ω

1

]

ω1

∃n < ω ∃ (s : n → Z) ∀α ∈ A (n

α

= n ∧ s

α

= s).

Now it clearly follows that ∀α, β ∈ A (p

α

6⊥ p

β

) so that S

hf,gi

has the property K.

(2) Let {p

α

: α < ω

1

} ⊆ S

hf,gi

where p

α

= hx

α

, y

α

, n

α

, s

α

i. First assume that cf(λ) > ω

1

. Then there exists µ < λ such that ∀α < ω

1

(x

α

⊆ µ).

Therefore {p

α

: α < ω

1

} ⊆ S

hfξ,gη:ξ<µ,η<κi

. Then the result follows from (1) since f

µ

splits hf

ξ

, g

η

: ξ < µ, η < κi. If cf(λ) < ω

1

then ∃µ < λ such that x

α

⊆ µ for uncountably many α and this is sufficient to obtain the result as above. The case cf(κ) 6= ω

1

is handled in the same way.

(3) Let A = {p

α

= hx

α

, y

α

, n

α

, s

α

i : α < ω

1

} be an uncountable anti- chain in S

hf,gi

. For each α < ω

1

fix k

α

such that

∀ξ, ζ ∈ x

α

∀i ≥ k

α

(ξ ≤ ζ → f

ξ

(i) ≤ f

ζ

(i)) and

∀η, θ ∈ y

α

∀i ≥ k

α

(η ≤ θ → g

θ

(i) ≤ g

η

(i)).

Then without loss of generality I may make the following assumptions:

(a) ∀α < ω

1

(k

α

= k ∧ n

α

= n ∧ s

α

= s), (b) n ≥ k (by extending each p

α

if necessary), (c) ∀α, β < ω

1

(α < β → (max(x

α

) < max(x

β

))), (d) ∀α, β < ω

1

(α < β → (max(y

α

) < max(y

β

))).

Let m = n, ξ

α

= max(x

α

) and η

α

= max(y

α

). Now it easily follows from

the fact that A is an uncountable antichain that if X = {ξ

α

: α < ω

1

} and

Y = {η

α

: α < ω

1

} then both (i) and (ii) hold.

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In the discussion that follows I will usually work with equivalent gaps.

Therefore when referring to the lemma above I may without loss of generality assume that X = Y = ω

1

and m = 0.

Lemma 2.6. Let M be a c.t.m. for ZFC and assume that, in M, hf, gi is a (λ, κ)-pregap, for regular λ, κ, such that S

hf,gi

has the ccc, and T is an Aronszajn tree. If G is S

hf,gi

-generic over M then M[G] |= “ S

T

has the ccc ”.

P r o o f. According to Corollary 1.3 it is sufficient to show that no new paths are added through T in M[G]. So by way of contradiction assume that

˙b is an S

hf,gi

-name for a new path through T and p ∈ S

hf,gi

such that p ° “ ˙b is a new path through ˇ T ”.

Let

X = {t ∈ T : ∃p

t

≤ p (p

t

° “ ˇt ∈ ˙b ”)}.

Since b is a new path through T, for each s ∈ X there are t, u ∈ X such that s ≤

T

t, u and t and u are incomparable in T. Working in M[G], let

Y = {t ∈ X : t is ≤

T

-minimal with t 6∈ b}.

Then Y ∈ M[G] and for each t ∈ Y fix a p

t

≤ p with p

t

° “ ˇt ∈ ˙b ” and let A = {p

t

∈ S

hf,gi

: t ∈ Y }. Then A is an uncountable subset of S

hf,gi

in M[G]

and any two elements of A are incompatible. Hence A is an uncountable antichain in S

hf,gi

which contradicts the fact that S

hf,gi

has the property K in M[G]. Therefore M[G] |= “ S

T

has the ccc ”.

Let L ⊆ Z

ω

such that (L, ) is a linear order. Then I ⊆ L is an interval in L iff

∀x, y ∈ I ∀z ∈ L (x  z  y → z ∈ I).

If hf

ξ

, g

η

: ξ < λ, η < κi ⊆ L is a (λ, κ)-pregap and I is an interval in L then hf

ξ

, g

η

: ξ < λ, η < κi ⊆ I will mean that

∃α < λ ∃β < κ (hf

ξ

, g

η

: α ≤ ξ < λ, β ≤ η < κi ⊆ I).

Lemma 2.7. Let M be a c.t.m. for ZFC and suppose that, in M, P is a ccc partial order and (L, ) a linear order in (Z

ω

, ). If G is P-generic over M with

M[G] |= “ hf, gi is a new (ω

1

, ω

1

)-gap in L ” then, in M, there is a Suslin tree T and a P-name ˙b such that

M[G] |= “ b is a new path through T ”.

P r o o f. Let p

0

∈ G with

() p

0

° “ h ˙ f , ˙gi is a new (ˇ ω

1

, ˇ ω

1

)-gap in ˇ L ”.

(12)

By recursion on α < ω

1

I construct sequences hS

α

: α < ω

1

i and hA

αI

: α < ω

1

, I ∈ S

α

i where for each α < ω

1

each element of S

α

is a non-empty interval in L such that

(1) ∀I, J ∈ S

α

(I 6= J → I ∩ J = ∅), (2) S

{A

αI

: I ∈ S

α

} is a maximal antichain in P below p

0

, (3) ∀p ∈ A

αI

(p ≤ p

0

∧ p ° “ h ˙ f , ˙gi ⊆ ˇ I ”),

(4) ∀I ∈ S

α

∀β ≥ α ∃I

1

, I

2

∈ S

β+1

(I

1

∩ I

2

= ∅ ∧ I

1

∪ I

2

⊆ I), (5) ∀β > α ∀I ∈ S

β

∃J ∈ S

α

(I ⊆ J).

Let S

0

= {L} and A

0L

= {p

0

}. Fix α < ω

1

and assume that ∀ξ < α, S

ξ

is constructed together with A

ξI

, for each I ∈ S

ξ

, such that (1)–(5) are satisfied.

First assume α = β + 1. Note that (1)–(3) and the fact that P has the ccc imply that |S

β

| ≤ ω. Choose I ∈ S

β

and q ∈ A

βI

. Then since

q ° “ h ˙ f , ˙gi is a new (ˇ ω

1

, ˇ ω

1

)-gap in ˇ L ”

there are r

1

, r

2

≤ q and disjoint intervals I

0

, I

1

⊆ I with r

i

° “ h ˙ f , ˙gi ⊆ ˇ I

i

”, for i < 2, and I

0

∪ I

1

= I. Let B

Ii

be a maximal antichain below p

0

such that

r

i

∈ B

Ii

∧ ∀r ∈ B

Ii

∃q ∈ A

βI

(r ≤ q ∧ r ° “ h ˙ f , ˙gi ⊆ ˇ I

i

”).

Now repeat this construction for each I ∈ S

β

. Then S

α

= {I

i

: I ∈ S

β

i < 2} and for each i < 2 and I ∈ S

β

let A

αIi

= B

Ii

. Note that hS

ξ

: ξ ≤ αi and hA

ξI

: ξ ≤ α, I ∈ S

ξ

i satisfy (1)–(5). This finishes the construction for successor stages.

Now suppose cf(α) = ω. Let S be the set of all intervals in L such that for each I ∈ S there is a p ≤ p

0

and an increasing sequence hα

n

: n < ωi with sup{α

n

: n < ω} = α and for each n < ω an I

n

∈ S

αn

such that (m < n → I

n

⊆ I

m

) and I = T

n<ω

I

n

with p ° “ h ˙ f , ˙gi ⊆ ˇ I ”. Note that

∀I, J ∈ S (I 6= J → I ∩ J = ∅) and (() → S 6= ∅). Furthermore, S is countable since P has the ccc. Let S

α

= S and for each I ∈ S let A

αI

be a maximal antichain below p

0

such that ∀p ∈ A

αI

(p ≤ p

0

∧ p ° “ h ˙ f , ˙gi ⊆ ˇ I ”).

Then by the definition of S, each A

αI

is non-empty and by maximality of S, S

{A

αI

: I ∈ S} is a maximal antichain in P below p

0

. This finishes the construction.

It is easy to see now that hS

α

: α < ω

1

i and hA

αI

: α < ω

1

, I ∈ S

α

i satisfy (1)–(5). Furthermore, () implies that T = h S

α<ω1

S

α

, ⊇i is a Suslin tree in M. However, in M[G], hf, gi is a new (ω

1

, ω

1

)-gap in L so that hf, gi determines a path, b, through T.

The results so far are all that is necessary for treatment of successor

stages in the construction of the final model. Now I present several results

that will enable me to go beyond the limit stages. Lemmas 1.9 and 1.12

(13)

are used to show, as indicated earlier, that no new paths can be added through existing ω

1

-trees at limit stages. For gaps the situation is slightly different. In the construction of a c-saturated linear order L in (Z

ω

, ), new gaps can appear at limit stages in the portion of L constructed by that stage. According to Lemma 2.5 there is no problem with non-(ω

1

, ω

1

)-gaps.

But (ω

1

, ω

1

)-gaps can be somewhat problematic. However, with the aid of Lemma 2.7 the construction will be arranged in such a way that such gaps can only occur at stages of cofinality ω

1

and the splitting orders for such gaps will have the ccc. The next sequence of results is a formalization of the facts just stated. But first some terminology.

In the discussion that follows nice names play an important role. Let M be a c.t.m. for ZFC and P ∈ M a partial order. If σ ∈ M

P

, a nice P-name for a subset of σ is τ ∈ M

P

of the form S

{{π} × A

π

: π ∈ dom(σ)}, where each A

π

is an antichain in P. It is shown in [K] that if σ, µ ∈ M

P

then there is a nice P-name τ ∈ M

P

for a subset of σ such that 1 °

P

“ µ ⊆ σ → µ = τ ”.

Since isomorphic partial orders lead to the same generic extensions, it is then justified to use cardinals κ for base sets of partial orders and subsets of κ × κ for ordering relations. Therefore, the phrase “ let ˙ Q be a nice P-name . . . ” will mean that ˙ Q is of the form (ˇ κ, σ), where κ is some cardinal and σ is a nice P-name for a subset of (κ × κ)ˇ. Now, in M, let α be a limit ordinal and hP

ξ

: ξ ≤ αi an iterated forcing construction with finite supports where the limit stages are handled in the usual way and the successor stages are obtained as follows: Let Λ = {ξ : ξ < α ∧ ξ is even ∧ cf(ξ) 6= ω

1

} and let P

0

be the trivial partial order. Let γ + 1 = β < α and assume that hP

ξ

: ξ < βi has been constructed together with the sequence hf

ξ

: ξ ∈ Λ∩βi of functions in Z

ω

linearly ordered by . For the simplicity of notation denote “ °

Pξ

” by “ °

ξ

”.

If γ is an odd ordinal, let ˙ Q

γ

be a nice P

γ

-name for a partial order such that 1 °

γ

“ ˙ Q

γ

has the ccc ” and let P

β

= P

γ

∗ ˙ Q

γ

. At this point it is not important how ˙ Q

γ

are selected, but in the final construction ˙ Q

γ

will be chosen in a way that will ensure Martin’s Axiom holds in the final model.

If γ is an even ordinal and not of cofinality ω

1

(i.e. γ ∈ Λ), then choose

a pregap C

γ

in hf

ξ

: ξ ∈ Λ ∩ βi and let P

β

= P

γ

∗ ˙S

γ

where ˙S

γ

is a nice

P

γ

-name for the partial order that splits C

γ

and let f

γ

be an element of

Z

ω

obtained in such a way. The function f

γ

will be a part of L and only at

these stages new elements are added to L. At this point also assume that

1 °

γ

“ ˙S

γ

has the ccc ”. Once again, at this point it is not important how

C

γ

are selected, but in the final construction, C

γ

will be chosen in a way

that will ensure L = hf

ξ

: ξ ∈ Λi is a c-saturated linear order. However,

the description of stages γ, where γ is a limit ordinal of cofinality ω

1

(which

follows next), will imply at once that 1 °

γ

“ ˙S

γ

has the ccc ”.

(14)

Finally, let γ be a limit ordinal of cofinality ω

1

. Let ˙R

γ

be a nice P

ξ

-name for the partial order obtained by taking the product of all the splitting orders for (ω

1

, ω

1

)-pregaps in hf

ξ

: ξ ∈ Λ ∩ βi which are also gaps in (Z

ω

, ), and let P

β

= P

γ

∗ ˙R

γ

. The rest of this section is devoted to precisely defining this product and showing that 1 °

γ

“ ˙R

γ

has the ccc ” so that at the end P

α

will have the countable chain condition. No element of Z

ω

obtained at this stage will be a part of L. Their existence only ensures that each (ω

1

, ω

1

)- pregap in the portion of L constructed by this stage can be split by a ccc partial order at some later stage. Now let G be P

α

-generic over M, with G

ξ

= G ∩ P

ξ

and M

ξ

= M[G

ξ

]. Let θ < β < α with β ∈ Λ and A ⊆ θ ∩ Λ with A ∈ M

θ

. Then C

β

also defines a pregap in hf

ξ

: ξ ∈ Ai. For p ∈ S

β

let p¹A = hx

p

∩ A, y

p

∩ A, n

p

, s

p

i and S

β,A

= {q : ∃p ∈ S

β

(q = p¹A)} and assume that S

β,A

∈ M

θ

.

Lemma 2.8. Let M, α, hP

ξ

: ξ ≤ αi, and G be as above with cf(α) 6= ω

1

and L = hf

ξ

: ξ ∈ Λi the linear order in M[G] obtained by the construction.

If hf, gi is an (ω

1

, ω

1

)-pregap in L, in M[G], then there is a β < α and an equivalent (ω

1

, ω

1

)-pregap hf

0

, g

0

i such that hf

0

, g

0

i is added to L at stage β.

P r o o f. If cf(α) = ω then the result follows from the fact that if A is a set of size ω

1

constructed at stage α then there is a B ∈ [A]

ω1

and β < α such that B is constructed at stage β.

If cf(α) > ω

1

then the result follows from the fact that all sets of size ω

1

constructed at stage α are in fact constructed at some earlier stage.

Proposition 2.9. In M, let l < ω and let hP

ξ

: ξ ≤ αi and G be as before with cf(α) = ω

1

. In M[G], let A

i

, B

i

⊆ α, with A

i

∪ B

i

cofinal in α, and let hf

ai

, f

bi

: a

i

∈ A

i

, b

i

∈ B

i

i be (ω

1

, ω

1

)-gaps with the corresponding splitting orders S

i

, for i < l. Then S

0

× . . . × S

l−1

has the countable chain condition in M[G].

The next two lemmas are needed in the proof of this proposition.

Lemma 2.10. f

β

is S

β,A

-generic over M

θ

.

P r o o f. It suffices to show that the filter obtained from f

β

, in S

β,A

, intersects each dense subset of S

β,A

in M

θ

. So let D be a dense subset of S

β,A

, in M

θ

. By recursion I define a sequence of sets hD

ξ

: ξ ≤ βi in M

β

as follows: Let S

β,ξ

be the partial order that fills the pregap in hf

ζ

: ζ ∈ ξ ∩ Λi determined by C

β

. Then

D

0

= {q ∈ S

β,0

: ∃p ∈ D (q ≤ p¹0)}.

Fix ξ < β and assume D

ζ

has been defined for each ζ < ξ. If ξ = ζ + 1 then D

ξ

= {q ∈ S

β,ξ

: ∃q

1

∈ D

ζ

∃p ∈ D (q ≤ q

1

, p¹ξ)}.

And if ξ is a limit then D

ξ

= S

ζ<ξ

D

ζ

.

(15)

Then by induction I show that each D

ξ

is dense in S

β,ξ

. Since D is dense it follows that D

0

is also dense. If ξ is a limit ordinal then the result also follows easily from the definition of D

ξ

and the induction hypothesis. Now assume that D

ξ

is dense and show that D

ξ+1

is also dense. If ξ 6∈ Λ then D

ξ+1

= D

ξ

and the result follows from the induction hypothesis. So assume ξ ∈ Λ.

C a s e 1: C

ξ

and C

β

define the same pregap in hf

ζ

: ζ ∈ ξ ∩ Λi. In this case S

β,ξ

= S

ξ

so that f

ξ

is S

β,ξ

-generic over M

ξ

. Let p ∈ S

β,ξ+1

and by extending p if necessary I may assume that ξ ∈ x

p

∪ y

p

, say ξ ∈ y

p

. Let q = hx

p

, y

p

\ {ξ}, n

p

, s

p

i and note that q ∈ S

ξ

. Now D

ξ

is dense in S

ξ

, so let q

1

∈ D

ξ

and p

1

∈ D from which q

1

is defined (q

1

≤ p

1

¹ξ) such that q

1

≤ q.

Note that D

ξ

may not be in M

ξ

, but q

1

is. Now f

ξ

is S

ξ

-generic over M

ξ

so that f

ξ0

is also S

ξ

-generic over M

ξ

where f

ξ0

is just f

ξ

modified by s

q1

. So let q

2

be an element in the S

ξ

-generic filter over M

ξ

determined by f

ξ0

with q

2

≤ q

1

. Then it is easily seen that q

3

= hx

q2

, y

q2

∪ {ξ}, n

q2

, s

q2

i ∈ S

β,ξ+1

with q

3

≤ q

1

, p. But also q

3

≤ p

1

¹(ξ + 1) so that q

3

∈ D

ξ+1

, showing that D

ξ+1

is dense in S

β,ξ+1

.

C a s e 2: C

ξ

and C

β

do not define the same pregap in hf

ζ

: ζ ∈ ξ ∩ Λi.

Then there is a ζ

0

< ξ such that f

ζ0

is between the pregaps C

ξ

and C

β

in hf

ζ

: ζ ∈ ξ ∩Λi. I may assume C

ξ

is to the right of C

β

. Let p ∈ S

β,ξ+1

and by extending p if necessary I may assume that ξ ∈ y

p

. In addition I may assume that ζ

0

∈ y

p

and that n

0

< ω is such that ∀i ≥ n

0

(f

ζ0

(i) ≤ f

ξ

(i)) with n

0

n

p

. Let q = hx

p

, y

p

\ {ξ}, n

p

, s

p

i and choose q

1

∈ D

ξ

and p

1

∈ D from which q

1

is defined (q

1

≤ p

1

¹ξ) such that q

1

≤ q. Let q

2

= hx

q1

, y

q1

∪ {ξ}, n

q1

, s

q1

i.

Then it is clear that q

2

∈ S

β,ξ+1

with q

2

≤ q

1

, p, p

1

¹(ξ +1) so that q

2

∈ D

ξ+1

, showing that D

ξ+1

is dense in S

β,ξ+1

.

And now I conclude that D

β

is dense in S

β

. Therefore let q be an element in the intersection of D

β

and the S

β

-generic filter determined by f

β

. By definition of D

β

, let p ∈ D with q ≤ p. Then p is also in the filter obtained from f

β

in S

β,A

.

Lemma 2.11. Let M, hP

ξ

: ξ ≤ αi, and G be as before with cf(α) = ω

1

in M. In addition, assume that P

α

has the ccc in M. Let A, B ⊆ α ∩ Λ with each A and B of order type ω

1

, A∪B cofinal in α and hf

a

, f

b

: a ∈ A, b ∈ Bi an (ω

1

, ω

1

)-gap in hf

ξ

: ξ ∈ α ∩ Λi, in M[G], with its splitting order S

A,B

. Then S

A,B

has the ccc in M[G].

P r o o f. Working in M[G], let A = ha

ξ

: ξ < ω

1

i and B = hb

ξ

: ξ < ω

1

i be

increasing enumerations of A and B. By way of contradiction assume that

the conclusion of the lemma is false. Then by restricting the discussion to

an equivalent gap or to h−f

b

, −f

a

: b ∈ B, a ∈ Ai I may assume, according

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