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]

*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

**1. Trees. A tree is a partial order in the strict sense, hT, ≤i, such that**

*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}

## ),

*(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]*

^{<ω}*(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]

^{<ω}## as follows:

*∀a, b ∈ [T]*

^{<ω}*(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}*).*

*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*

^{α}_{0}

*, s*

^{α}_{1}

*, . . . , s*

^{α}_{n−1}*}. Finally I may assume that if* *α < β < ω*

_{1}

*, p*

_{α}*(s*

^{α}_{i}*) = p*

_{β}*(s*

^{β}_{j}*), and s*

^{α}_{i}*and s*

^{β}_{j}## are comparable (which must *happen for some i and j since p*

_{α}*⊥ p*

_{β}*) then s*

^{α}_{i}*< s*

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

^{α}_{i(α)}*< s*

^{β}_{j(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*

^{α}_{i}^{1}

*, s*

^{α}_{i}^{2}

*< s*

^{β}_{j}*. And since T is a tree, s*

^{α}_{i}^{1}

*and s*

^{α}_{i}^{2}

## are comparable. Therefore *{s*

^{α}_{i}*: 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]*

^{ω}^{1}

*∃B ∈ [A]*

^{ω}^{1}

*∀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}

## ]

^{ω}^{1}

*∀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.*

*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*

_{γ}

_{µ}*(T), α*

_{µ}*and increasing γ*

_{µ}*such that e*

_{α}

_{µ}*⊆ T*

_{x}

_{µ}*with γ*

_{µ}*> sup{ht(z) : z ∈* S

*ν<µ*

*e*

_{α}

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

1## ]

^{ω}^{1}

*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]

^{<ω}## . 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*

^{S}

^{T}

*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}

## ]

^{ω}^{1}

*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*

*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**for a partial order such that 1**

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

*in M.*

*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** °

**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** °

**where α is a limit ordinal. If ∀ξ < α (1**

_{P}

_{ξ}## “ ˇ **P has the ccc ”) then 1** °

**P has the ccc ”) then 1**

_{P}

_{α}## “ ˇ *P has the ccc ”.*

*P r o o f. If cf(α) = ω and*

**1** °

_{P}

_{α}## “ ˙ *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** °

**∃β < α (1**

_{P}

_{β}## “ 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** °

_{P}

_{α}## “ ˇ P fails to have the ccc ” then

**∃β < α (1** °

**∃β < α (1**

_{P}

_{β}## “ ˇ *P fails to have the ccc ”),*

## again contradicting the hypothesis.

*Finally, let cf(α) = ω*

1 ## and suppose that the conclusion of the lemma fails. Therefore

**∀β < α (1** °

P**∀β < α (1**

_{β}## “ ˇ P has the ccc ”) but

**1** °

_{P}

_{α}## “ 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**∀β < α (1**

## “ 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.

**1(θ)**

*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** °

_{P}

_{α}## “ 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**

**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*

^{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.*

*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*

_{ζ}^{0}*∧ g*

^{0}_{θ}* g*

*η*

## ) and

*∀ξ ∈ I*

^{0}*∃ζ ∈ I ∀η ∈ J*

^{0}*∃θ ∈ J (f*

_{ξ}^{0}* f*

_{ζ}*∧ g*

_{θ}* g*

_{η}^{0}## ).

*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]*

^{<ω}*∧ y ∈ [κ]*

^{<ω}*∧ 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}*, ).*

## 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}

## ]

^{ω}^{1}

*with X = {ξ*

_{α}*: α < ω*

_{1}

*} and* *Y = {η*

_{α}*: α < ω*

_{1}

*} such that*

*(i) ∀α < ω*

_{1}

*∀i ≥ m (f*

_{ξ}

_{α}*(i) ≤ g*

_{η}

_{α}*(i)) and*

*(ii) ∀α, β < ω*

_{1}

*(α 6= β → ∃i ≥ m (f*

_{ξ}

_{α}*(i) 6≤ g*

_{η}

_{β}*(i)∨f*

_{ξ}

_{β}*(i) 6≤ g*

_{η}

_{α}*(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## ]

^{ω}^{1}

*∃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*

_{hf}

_{ξ}

_{,g}

_{η}*: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.*

## 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 ”.*

*By recursion on α < ω*

1 *I construct sequences hS*

*α*

*: α < ω*

1*i and hA*

^{α}_{I}## : *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*

^{α}_{I}*: I ∈ S*

*α*

*} is a maximal antichain in P below p*

0## , *(3) ∀p ∈ A*

^{α}_{I}*(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*

^{0}

_{L}

*= {p*

0*}. Fix α < ω*

1 *and assume that ∀ξ < α,* *S*

*ξ*

*is constructed together with A*

^{ξ}_{I}*, 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*

^{β}_{I}## . 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*

*I*

_{i}*be a maximal antichain below p*

0 ## such that

*r*

_{i}*∈ B*

_{I}

_{i}*∧ ∀r ∈ B*

_{I}

_{i}*∃q ∈ A*

^{β}_{I}*(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*

^{α}_{I}

_{i}*= B*

*I*

_{i}*. Note that hS*

*ξ*

*: ξ ≤ αi* *and hA*

^{ξ}_{I}*: 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*

_{α}

_{n}## 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*

^{α}_{I}## be a *maximal antichain below p*

0*such that ∀p ∈ A*

^{α}_{I}*(p ≤ p*

0*∧ p* *° “ h ˙* *f , ˙gi ⊆ ˇ* *I ”).*

*Then by the definition of S, each A*

^{α}_{I}## is non-empty and by maximality of *S,* S

*{A*

^{α}_{I}*: 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*

^{α}_{I}*: α < ω*

1*, I ∈ S*

*α*

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

*α<ω*_{1}

*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

## 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** °

**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 “* °

_{P}

_{ξ}## ” 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 ”.

*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]*

^{ω}^{1}

*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*

_{a}*i*

*, f*

_{b}*i*

*: 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*

## .

*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*

_{ξ}^{0}## is also S

_{ξ}## -generic over M

_{ξ}*where f*

_{ξ}^{0}*is just f*

_{ξ}*modified by s*

_{q}_{1}

## . So *let q*

_{2}

## be an element in the S

_{ξ}## -generic filter over M

_{ξ}*determined by f*

_{ξ}^{0}## with *q*

2 *≤ q*

1*. Then it is easily seen that q*

3 *= hx*

*q*

_{2}

*, y*

*q*

_{2}

*∪ {ξ}, n*

*q*

_{2}

*, s*

*q*

_{2}

*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*

_{ζ}_{0}

*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*

_{ζ}_{0}

*(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*

_{q}_{1}

*, y*

_{q}_{1}

*∪ {ξ}, n*

_{q}_{1}

*, s*

_{q}_{1}

*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}