the special aronszajn tree property at 2 and gchbfe12ncu/special-aleph-2-a-trees-gch.pdf ·...

THE SPECIAL ARONSZAJN TREE PROPERTY AT @2

AND GCH

DAVID ASPERO AND MOHAMMAD GOLSHANI

Abstract. Starting from the existence of a weakly compact car-dinal, we build a generic extension of the universe in which GCHholds and all @2-Aronszajn trees are special and hence there are no@2-Souslin trees. This result answers a longstanding open questionfrom the 1970’s.

1. introduction

Let be a regular uncountable cardinal. Let us recall that a -treeis a tree T of height all of whose levels are smaller than , and thata -tree is called a -Aronszajn tree if it has no -branches. Also, T iscalled a -Souslin tree if it has no -branches and no antichains of size. When = �+ is a successor cardinal, a -Aronszajn tree is said tobe special if and only if it is a union of � antichains.1 Let us make thefollowing definition:

Definition 1.1. (1) Souslin’s Hypothesis at , SH

, is the state-ment “there are no -Souslin trees”.

(2) The special Aronszajn tree property at = �+, SATP

, is thestatement “there exist -Aronszajn trees and all such trees arespecial” (see [5]).

Aronszajn trees were introduced by Aronszajn (see [9]), who provedthe existence, in ZFC, of a special @1-Aronszajn tree. Later, Specker([17]) showed that 2<� = � implies the existence of special �+-Aronszajntrees for � regular, and Jensen ([7]) produced special �+-Aronszajntrees for singular � in L.

In [16], Solovay and Tennenbaum proved the consistency of Martin’sAxiom + 2@0 > @1 and showed that this implies SH@1 . This was laterextended by Baumgartner, Malitz and Reinhardt [3], who showed that

2010 Mathematics Subject Classification. 03E05, 03E35, 03E50, 03E55.Key words and phrases. Special @2-Aronszajn trees, GCH, weakly compact car-

dinals, iterated forcing with side conditions.1A ✓ T is an antichain of T i↵ A consists of pairwise incomparable nodes.

1

2 D. ASPERO AND M. GOLSHANI

Martin’s Axiom + 2@0 > @1 implies SATP@1 . Later, Jensen (see [4] and[13]) produced a model of GCH in which SATP@1 holds.

The situation at @2 turned out to be more complicated. In [7], Jensenproved that the existence of an @2-Souslin tree follows from each of thehypotheses CH + }(S2

1) and ⇤!1 + }(S2

0) (where, given m < n < !,Sn

m

= {↵ < @n

| cf(↵) = @m

}). The second result was improved byGregory in [6], where he proved that GCH together with the existenceof a non–reflecting stationary subset of S2

0 yields the existence of an@2-Souslin tree. In [10], Laver and Shelah produced, relative to theexistence of a weakly compact cardinal, a model of ZFC+CH in whichthe special Aronszajn tree property at @2 holds. But in their model2@1 > @2, and the task of finding a model of ZFC+GCH+SATP@2 ,or even of ZFC+GCH+SH@2 , remained a major open problem. Theearliest published mention of this problem seems to appear in [8] (seealso [10], [18], [15], [14], or [11]).

In this paper we solve the above problem by proving the followingtheorem.

Theorem 1.2. Suppose is a weakly compact cardinal. Then thereexists a set–generic extension of the universe in which GCH holds, =@2, and the special Aronszajn tree property at @2 (and hence Souslin’sHypothesis at @2) holds.

Remark 1.3. (1) Our argument can be easily extended to deal withthe successor of any regular cardinal.

(2) By results of Shelah and Stanley ([15]) and of Rinot ([12]), ourlarge cardinal assumption is optimal. Specifically:(a) It is proved in [15] that if !2 is not weakly compact in L,

then either ⇤!1 holds or there is a non–special @2-Aronszajn

tree; in particular, GCH+SATP@2 implies that !2 is weaklycompact in L by one of Jensen’s results mentioned above.

(b) Rinot proved in [12] that if GCH holds, � � !1 is a cardi-nal, and ⇤(�+) holds, then there is a �-closed �+-Souslintree; on the other hand, Todorcevic ([19]) proved that if � !2 is a regular cardinal and ⇤() fails, then is weaklycompact in L.

The rest of the paper is devoted to the proof of Theorem 1.2. Wewill next give an (inevitably) vague and incomplete description of theforcing witnessing the conclusion of this theorem.

The construction of the forcing witnessing Theorem 1.2 combines anatural iteration for specializing @2-Aronszajn trees, due to Laver andShelah ([10]), with ideas from [2]. More specifically, we build a certaincountable support forcing iteration hQ

�

| � +i with side conditions.

THE SPECIAL ARONSZAJN TREE PROPERTY AT @2 AND GCH 3

The first step of the construction is essentially the Levy–collapse ofthe weakly compact cardinal to become !2. At subsequent stages,we consider forcings for specializing @2-Aronszajn trees by countableapproximations. Conditions in a given Q

�

, for � > 0, will consist of aworking part f

q

, together with a certain side condition. The workingpart f

q

will be a countable function with domain contained in � suchthat for all ↵ 2 dom(f

q

),

• fq

(↵) is a condition in the Levy–collapse if ↵ = 0, and• if ↵ > 0, f

q

(↵) is a countable subset of ⇥ !1.

Letting ↵ = ↵0 + ⌫, where ↵0 is a multiple of !1 and ⌫ < !1, any twodistinct members of f

q

(↵), when ↵ > 0, will be forced to be incompa-rable nodes in a certain -Aronszajn tree T⇠↵0 on ⇥ !1 chosen via agiven bookkeeping function � : X �! H(+), where X denotes theset of multiples of !1 below +.

The side condition will be a countable directed graph ⌧q

whose ver-tices are ordered pairs of the form (N, �), where N is an elementarysubmodel of H(+) such that |N | = |N\| and <|N |N ✓ N , and where� is an ordinal in the closure of N \ {⇠+1 | ⇠ < �} in the order topo-logy. Given any such (N, �), � is to be seen as a marker for N in q,telling us up to which stage isN ‘active’ as a model. We will tend to callsuch pairs (N, �) models with markers. Whenever h(N0, �0), (N1, �1)i isan edge in ⌧

q

, for a condition q, (N0,2) and (N1,2) are 2–isomorphicvia a (unique) isomorphism

N0,N1 such that N0,N1(⇠) ⇠ for every

ordinal ⇠ 2 N0 and such that N0,N1 is in fact an isomorphism be-

tween the structures (N0,2,�,�↵

) and (N1,2,�,� N0,N1 (↵)) for every

↵ 2 N0 \ �0 such that N0,N1(↵) is less than �1, for a certain sequence

(�↵

)↵<

+ of increasingly expressive predicates contained inH(+) fixedat the outset.

In the above situation, N0 and N1 are to be seen as ‘twin models’,relative to q, with respect to all stages ↵ and

N0,N1(↵) such that↵ 2 N0 \ �0 and

N0,N1(↵) < �1. What this means is that the naturalrestriction of f

q

(↵) to N0—which of course is nothing but fq

(↵)\N0—is to be copied over, via

N0,N1 , into the restriction of fq

( N0,N1(↵)) to

N1, i.e., we require that

N0,N1(fq(↵) \N0) = f

q

(↵) \N0 ✓ fq

( N0,N1(↵)),

and similarly for the restriction of ⌧q

� ↵+1 to N0 (with the restriction⌧q

� ↵ + 1 being defined naturally).We can describe our copying procedure by saying that we are copying

into the past information coming from the future via the edges in ⌧q

.Given an edge h(N0, �0), (N1, �1)i as above and some ↵ 2 N0 \ �0


such that ↵ = N0,N1(↵) is less than �1, the intersection of f

q

(↵) with�N0 ⇥ !1 may certainly contain more information than the intersectionof f

q

(↵) with �N0 ⇥ !1. Thanks to the way we are setting up the

copying procedure—namely, only copying from the future into the pastvia edges as we have described—, it is straightforward to see that ourconstruction is in fact a forcing iteration, in the sense that Q

↵

is acomplete suborder of Q

�

for all ↵ < �. This does not need be true ingeneral, in forcing constructions of this sort, if we allow also to copy‘from the past into the future’.2

Given an edge h(N0, �0), (N1, �1)i in ⌧q and a stage ↵ 2 N0 \ �0, wewould like to require that Q

↵+1 \N0 be a complete suborder of Q↵+1;

indeed, having this would be useful in the proof that our constructionhas the -chain condition. This cannot be accomplished, while definingQ

↵+1, on pain of circularity. However, a certain approximation to theabove situation can be stipulated, which we do, and this su�ces forour purposes.

Our construction is �-closed for all � +. In particular, forcingwith Q

+ preserves !1 and CH. The preservation of all higher cardinalsproceeds by showing that the construction has the -chain condition.For this, we use the weak compactness of in an essential way. Thefact that the length of our iteration is not longer than + seems to beneeded in this proof. Finally, the edges occurring in ⌧

q

, for a conditionq, are crucially used in the proof that our forcing preserves 2@1 = @2.

Side conditions are often employed in forcing constructions with thepurpose of guaranteeing that certain cardinals are preserved. In thepresent construction, on the other hand, they are used to ensure thatthe relevant level of GCH3 be preserved. This use of side conditionsis taken from [2], where they are crucially used in the proof of CH-preservation. It is worth observing that, while in the construction from[2] a certain account of structure is needed for the models occurring inthe side condition,4 no structure whatsoever (for the underlying set ofmodels) is needed in the present construction. We should point out thateven if it preserves 2@1 = @2, our construction does add new subsets of!1 after collapsing to become !2, although only @2-many of them (cf.

2It turns out that, in our specific construction, we could as well have required toalso copy information ‘from the past into future’, i.e. that, in the above situation,full symmetry obtains, below �N0 ⇥!1, between stages ↵ and N0,N1(↵) (s. Lemma2.17). However, the current presentation, only deriving full symmetry for a denseset of conditions, seems to be cleaner.

32@1 = @2.4Using the terminology of [1], they need to come from a symmetric system.


the construction in [2], where CH is preserved but @1-many new realsare added).

The paper is organized as follows. In Section 2 we define our mainforcing construction and prove some if its basic properties. In Section3 we show that the forcing has the -chain condition. This is the mostelaborate proof in the paper.5 Finally, in Section 4 we complete theproof of Theorem 1.2. The main argument in this section is to showthat our forcing preserves 2@1 = @2.

2. Definition of the forcing and its basic properties

In this section we define our forcing and prove some of its basicproperties.

Let us fix, for the remainder of this paper, a weakly compact cardinal, and let us assume, without loss of generality, that 2µ = µ+ for everycardinal µ � .6

Throughout the paper, if N is a set such that N \ is an ordinal,we denote this ordinal by �

N

and call it the height of N . If X is a set,we set

cl(X) = X [ {↵ 2 Ord | ↵ = sup(X \ ↵)}

If, in addition, � is an ordinal, we let �X

be the highest ordinal ⇠ 2cl(X) such that ⇠ �.

Given a function f and a set x, if x /2 dom(f), we let f(x) denotethe empty set.

Given an ordered pair q = (fq

, ⌧q

), where fq

is a function, and givena model N , we denote by q � N the ordered pair (f, ⌧

q

\ N), where fis the function with domain dom(f

q

) \N such that f(x) = fq

(x) \Nfor every x 2 dom(f).

Let X = {!1 · ↵ | ↵ < +}, where !1 · ↵ denotes ordinal multi-plication. Given an ordinal ↵ < +, there is a unique representation↵ = ↵0 + ⌫, where ↵0 2 X and ⌫ < !1. We will denote the aboveordinal ↵0 by µ(↵).

Let

� : X �! H(+)

5Cf. the proof in [10], where the hardest part is to prove that the forcing is -c.c.,or the proof in [2], where the hardest part is to prove that the forcing is proper.

6In fact, if is weakly compact, then GCH at every cardinal µ � can be easilyarranged by collapsing cardinals with conditions of size � , which will preservethe weak compactness of .


be such that for each x 2 H(+), ��1(x) is an unbounded subset ofX . This function � exists by 2 = +. Also, let F : + �! H(+) bea bijection canonically defined from �.7

Let (�↵

)↵<

+ be the following sequence of subsets of H(+).

• �0 is the satisfaction predicate of the structure

hH(+),2,�i

• If ↵ > 0, then �↵

is the satisfaction predicate of the structure

hH(+),2, ~�↵

i,

where ~�↵

= {(↵0, x) | ↵0 < ↵, x 2 �↵

0}.We will call ordered pairs of the form (N, �), where

• N ✓ H(+), N \ 2 , |N | = |N \ |, and <|N |N ✓ N ,• � 2 cl(N) \ +, and• (N,2,�0,�↵

) is an elementary submodel of (H(+),2,�0,�↵

)for every ↵ 2 N \ �,8

models with markers. We will often use, without mention, the fact that(N, �) 2 N 0 whenever (N, �) and (N 0, �0) are models with markers andN 2 N 0.

Given models N0 and N1 such that (N0,2) ⇠= (N1,2), we will denotethe unique 2-isomorphism

: (N0,2) ! (N1,2)

by N0,N1 .

Given any nonzero ordinal ⌘ < +, we let e⌘

be the F -least surjectionfrom onto ⌘. Let ~e = he

⌘

| 0 < ⌘ < +i. We will say that a modelN ✓ H(+) is closed under ~e if e

⌘

(⇠) 2 N for every nonzero ⌘ 2 N \+and every ⇠ 2 \N .

In the fact below (and elsewhere in the paper), we use the convention,given functions f0, . . . , fn, for n < !, to let

fn

� . . . � f0

denote the function f with domain the set of x 2 dom(f0) such thatfor every i < n, (f

i

� . . . � f0)(x) 2 dom(fi+1), and such that for every

x 2 dom(f), f(x) = (fn

� . . . � f0)(x).

7For example, F can be defined so that for each x 2 H(+), F�1(x) = min{↵ |�(!1 · ↵) = x}.

8Given a model N and predicates P0, . . . , Pn ✓ H(+), we will tend to write(N,2, P0, . . . , Pn) as short-hand for (N,2, P0 \N, . . . , Pn \N).


Fact 2.1. Suppose N0 and N1 are models closed under ~e of the sameheight. Then N0 \ N1 \ + is an initial segment of both N0 \ + andN1 \ +. In particular, if (N i

0, �i

0) and (N i

1, �i

1) (for i n) are modelswith markers such that (N i

0,2,�) ⇠= (N i

1,2,�) for all i n, then

( N

n0 ,N

n1� . . . �

N

00 ,N

01)(x) = x

for every x 2 dom( N

n0 ,N

n1� . . . �

N

00 ,N

01) \Nn

1 .

Proof. Let us first prove the first assertion. Given any nonzero ⌘ 2N0 \N1 \ + and any ↵ 2 N0 \ ⌘ there is some ⇠ 2 N0 \ such thate⌘

(⇠) = ↵. But since ⌘ and ⇠ are both members of N1, we also havethat ↵ = e

⌘

(⇠) 2 N1.As to the second assertion, let us first consider the case in which

�N

i0= �

N

i00for all i, i0. Note that if x 2 dom(

N

n0 ,N

n1� . . .�

N

00 ,N

01)\Nn

1

and ↵ = F�1(x), then ↵ 2 N00 \Nn

1 and

↵0 := ( N

n0 ,N

n1� . . . �

N

00 ,N

01)(↵) = ↵

since ot(↵ \ Nn

1 ) = ot(↵ \ N00 ) = ot(↵0 \ Nn

1 ), where the first equal-ity follows from the first assertion and the second equality from thedefinition of ↵0 as (

N

n0 ,N

n1� . . . �

N

00 ,N

01)(↵). It follows that

( N

n0 ,N

n1� . . . �

N

00 ,N

01)(x) = (

N

n0 ,N

n1� . . . �

N

00 ,N

01)(F (↵))

= F ( N

n0 ,N

n1� . . . �

N

00 ,N

01(↵)) = F (↵) = x

Suppose now that �N

i0< �

N

i00for some i 6= i0 and let i⇤ be such that

�N

i⇤0

= min{�N

i0| i n}. Since each N i

✏

is an elementary submodel

of (H(+),2,�0) closed under sequences of length less than |N i

✏

| and|N i

⇤⇤✏

| = |�N

i⇤0| for every ✏ 2 {0, 1} and every i⇤⇤ such that �

N

i⇤⇤0

= �N

i⇤0,

it is easy to see that there is a sequence (hN i

0, Ni

1i)in

of pairs of modelswith the following properties.

• For all i n, (N i

0,2,�) and (N i

1,2,�) are isomorphic elemen-tary submodels of (H(+),2,�) closed under ~e.

• �N

i0= �

N

i00= �

N

i⇤0

for all i, i0 n.

• For every i n and every ✏ 2 {0, 1}, N i

✏

= N i

✏

or N i

✏

2 N i

✏

.• x 2 dom(

N

n0 ,N

n1� . . . �

N

00 ,N

11) \ Nn

1

• ( N

n0 ,N

n1� . . . �

N

00 ,N

01)(x) = (

N

n0 ,N

n1� . . . �

N

00 ,N

11)(x)

But now we are done by the previous case. ⇤It will be convenient to use the following pieces of terminology: given

models with markers (N0, �0), (N1, �1), we will say that (N0, �0) and(N1, �1) are twin models (with markers) if and only if (N0,2,�) ⇠=(N1,2,�). If

N0,N1(↵) ↵ for every ordinal ↵ 2 N0, then we saythat N1 is a projection of N0.


We will call ordered pairs of the form

h(N0, �0), (N1, �1)isuch that

(1) (N0, �0) and (N1, �1) are twin models with markers,(2) N1 is a projection of N0, and such that(3) for every ↵ 2 N0 \ �0 such that ↵ :=

N0,N1(↵) < �1, N0,N1

is an isomorphism between the structures (N0,2,�,�↵

) and(N1,2,�,�↵

)

edges. If, in the above situation,

�0 2 cl(N0 \ {⇠ + 1 | ⇠ < �})and

�1 2 cl(N1 \ {⇠ + 1 | ⇠ < �}),then we call h(N0, �0), (N1, �1)i an edge below �. Given an edge e =h(N0, �0), (N1, �1)i and an ordinal ↵, let e � ↵ denote the edge

h(N0,min{↵, �0}N0), (N1,min{↵, �1}N1)i.9

Given a collection ⌧ of edges and given an ordinal ↵, we denote by⌧ � ↵ the set {e � ↵ | e 2 ⌧}. Note that ⌧ � ↵ is a collection of edgesbelow ↵.

Given models with markers (N, �), (N0, �0) and (N1, �1), if N 2 N0

and (N0,2) ⇠= (N1,2), then we let ⇡�,N

N0,�0,N1,�1denote the supremum of

the set of ordinals N0,N1(⇠) + 1 such that

• ⇠ 2 N \ �,• ⇠ < �0, and•

N0,N1(⇠) < �1We will say that a set ⌧ of edges is closed under copying in case for

all edges e = h(N0, �0), (N1, �1)i and e0 = h(N 00, �

00), (N

01, �

01)i in ⌧q such

that e0 2 N0 there are ordinals �⇤0 � ⇡�

00,N

00

N0,�0,N1,�1and �⇤1 � ⇡

�

01,N

01

N0,�0,N1,�1

such thath(

N0,N1(N00), �

⇤0), ( N0,N1(N

01), �

⇤1)i 2 ⌧

q

We shall define a sequence hQ�

| � +i. For each ↵ 2 X , assumingQ

↵

has been defined and there is a Q↵

-name T⇠ 2 H(+) for a -Arosnzajn tree, we let T⇠↵

be such a Q↵

-name. Further, if �(↵) is aQ

↵

-name for a -Aronszajn tree, then we let T⇠↵

= �(↵). For simplicityof exposition we will assume that the universe of T⇠↵

is forced to be⇥!1 and that for each ⇢ < , its ⇢-th level is {⇢}⇥!1. We will oftenrefer to members of ⇥ !1 as nodes.

9Recall that min{↵, �✏}N✏ , for ✏ 2 {0, 1}, is the highest ordinal ⇠ 2 cl(N✏) suchthat ⇠ min{↵, �✏}.


As we will see, each forcing notion Q�

in our construction will consistof ordered pairs of the form q = (f

q

, ⌧q

), where fq

is a function and ⌧q

isa set of edges below �. Given a nonzero ordinal ↵ < + and an ordinal� < , we will write Q�

↵+1 to denote the suborder of Q↵+1 consisting of

those conditions q such that �N0 � for every h(N0,↵+ 1), (N1, �1)i 2

⌧q

.Now suppose that � + and that Q

↵

has been defined for all↵ < �. Given an ordered pair q = (f

q

, ⌧q

), where fq

is a function and⌧q

is a set of edges, and given an ordinal ↵, we denote by q � ↵ theordered pair (f

q

� ↵, ⌧q

� ↵).We are now ready to define Q

�

.A condition in Q

�

is an ordered pair of the form q = (fq

, ⌧q

) withthe following properties.

(1) fq

is a countable function such that

dom(fq

) ✓ �

and such that the following holds for every ↵ 2 dom(fq

).(a) If ↵ = 0, then f

q

(↵) is a condition in Col(!1, <), the Levycollapse turning into @2, i.e., fq(0) is a countable functionwith domain included in ⇥!1 such that (f

q

(0))(⇢, ⇠) < ⇢for all (⇢, ⇠) 2 dom(f

q

(0)).(b) If ↵ > 0, then f

q

(↵) 2 [⇥ !1]@0 .(2) ⌧

q

is a countable set of edges below �.(3) q � ↵ 2 Q

↵

for all ↵ < �.(4) For every nonzero ↵ < � such that T⇠µ(↵) is defined,10 if x0 6=

x1 are nodes in fq

(↵), then q � µ(↵) forces x0 and x1 to beincomparable in T⇠µ(↵).

(5) ⌧q

is closed under copying.(6) For every edge h(N0, �0), (N1, �1)i 2 ⌧

q

and every ↵ 2 N0 \ �0such that ↵ :=

N0,N1(↵) < �1, if ↵ 6= 0, ↵ 2 dom(fq

), andx 2 f

q

(↵) \N0, then(a) ↵ 2 dom(f

q

), and(b) x 2 f

q

(↵)(7) Suppose h(N0, �0), (N1, �1)i 2 ⌧

q

and ↵ 2 dom(fq

) \ N0 \ �0 isa nonzero ordinal such that T⇠µ(↵) is defined. Suppose

• ⌧ ◆ ⌧q

� ↵ + 1 is a countable set of edges below ↵ + 1,

• r0 2 Q�N0↵+1 extends q � ↵ and (q � ↵ + 1) � N0, and is such

that fq

(↵) ✓ fr0(↵),

• ⌧r0 [ ⌧ is closed under copying, and

10We will see that in fact each T⇠µ(↵) is defined.


• if h(M0,↵ + 1), (M1, ⌘1)i 2 ⌧ , x 2 fr0(↵) \ M0, and ↵ :=

M0,M1(↵) < ⌘1, then ↵ 2 dom(f

r0) and x 2 fr0(↵).

Then there is an extension q0 2 Q↵+1\N0 of r0 � N0 with the

property that for every extension q1 2 Q↵+1 of q0 in N0 there is

a condition r1 2 Q�N0↵+1 such that

(a) r1 extends both r0 and q1,(b) ⌧

r1 [ ⌧ is closed under copying, and(c) if h(M0,↵ + 1), (M1, ⌘1)i 2 ⌧ , x 2 f

r1(↵) \ M0, and ↵ :=

M0,M1(↵) < ⌘1, then ↵ 2 dom(fr1) and x 2 f

r1(↵).

The extension relation on Q�

is defined in the following way:Given q1, q0 2 Q

�

, q1 Q�q0 (q1 is an extension of q0) if and only if

the following holds.

(1) dom(fq0) ✓ dom(f

q1)(2) For every ↵ 2 dom(f

q0), fq0(↵) ✓ fq1(↵).

(3) For every h(N0, �0), (N1, �1)i 2 ⌧q0 there are �

00 � �0 and �01 � �1

such that h(N0, �00), (N1, �

01)i 2 ⌧

q1 .

Remark 2.1. Given an ordinal ↵ < +, the definition of Q↵+1 can be

seen, because of clause (7), as being by recursion on the supremum ofthe set of heights of models N0 occurring in edges h(N0, �0), (N1, �1)i in⌧q

.

2.1. Basic properties of hQ�

| � +i. Our first lemma followsfrom the choice of the predicates �

↵

.

Lemma 2.2. For every nonzero ↵ < +, Q↵

is definable over thestructure

(H(+),2,�↵

)

without parameters. Moreover, this definition can be taken to be uni-form in ↵.

Our next lemma follows from the fact that Q1 is essentially the Levycollapse turning into !2.

Lemma 2.3. Q1 forces = @2.

The following lemma is also an easy consequence of the definition ofcondition.

Lemma 2.4. For every � +, Q↵

✓ Q�

for all ↵ < �.

Lemma 2.5 follows easily from the definition of hQ↵

| ↵ +i.Lemma 2.5. For all ↵ < � +, q 2 Q

�

, and r 2 Q↵

, if r Q↵ q � ↵,then

(fr

[ fq

� [↵, �), ⌧q

[ ⌧r

)

is a common extension of q and r in Q�

.


Proof. Let p = (fr

[fq

� [↵, �), ⌧q

[⌧r

). We show that p satisfies clauses(1)–(7) of the definition of a Q

�

-condition. It su�ces to consider (5)and (6), as all other clauses can be proved easily.

We first show that p satisfies clause (5). Thus let

e = h(N0, �0), (N1, �1)i 2 ⌧q

[ ⌧r

and

e0 = h(N 00, �

00), (N

01, �

01)i 2 (⌧

q

[ ⌧r

) \N0.

We have to show that there are ordinals �⇤0 � ⇡�

00,N

00

N0,�0,N1,�1and �⇤1 �

⇡�

01,N

01

N0,�0,N1,�1such that

h( N0,N1(N

00), �

⇤0), ( N0,N1(N

01), �

⇤1)i 2 ⌧

q

[ ⌧r

.

We divide the proof into three cases:

(1) Both e and e0 belong to ⌧q

(resp. ⌧r

). Then the conclusion isimmediate as q (resp. r) is a condition in Q

�

.(2) e 2 ⌧

q

and e0 2 ⌧r

. Then e � ↵ 2 ⌧r

, so we can find �⇤0 �⇡�

00,N

00

N0,min{�0,↵}N0 ,N1,min{�1,↵}N1and �⇤1 � ⇡

�

01,N

01

N0,min{�0,↵}N0 ,N1,min{�1,↵}N1

such that

h( N0,N1(N

00), �

⇤0), ( N0,N1(N

01), �

⇤1)i 2 ⌧

r

.

As r 2 Q↵

, e0 2 ⌧r

, and �00, �01 ↵, one can easily show that

⇡�

00,N

00

N0,min{�0,↵}N0 ,N1,min{�1,↵}N1= ⇡

�

00,N

00

N0,�0,N1,�1

and

⇡�

01,N

01

N0,min{�0,↵}N0 ,N1,min{�1,↵}N1= ⇡

�

01,N

01

N0,�0,N1,�1.

The result follows.(3) e 2 ⌧

r

and e0 2 ⌧q

. Then e0 � ↵ 2 ⌧r

, and hence we can find

�⇤0 � ⇡min{�0

0,↵}N00,N

00

N0,�0,N1,g1and �⇤1 � ⇡

min{�01,↵}N0

1,N

01

N0,�0,N1,g1such that

h( N0,N1(N

00), �

⇤0), ( N0,N1(N

01), �

⇤1)i 2 ⌧

r

.

As r 2 Q↵

, e 2 ⌧r

, and �0, �1 ↵, again one can easily showthat

⇡min{�0

0,↵}N00,N

00

N0,�0,N1,g1= ⇡

�

00,N

00

N0,�0,N1,g1

and

⇡min{�0

1,↵}N01,N

01

N0,�0,N1,g1= ⇡

�

01,N

01

N0,�0,N1,g1,

from which the result follows.


To show that p satisfies clause (6), let e = h(N0, �0), (N1, �1)i 2 ⌧q

[ ⌧r

,⌘ 2 (dom(f

r

)[dom(fq

))\N0\�0, ⌘ 6= 0, and x 2 (fr

(⌘)[ fq

(⌘))\N0.We have to show that ⌘ 2 dom(f

r

) [ dom(fq

) and x 2 fr

(⌘) [ fq

(⌘),where ⌘ =

N0,N1(⌘). If ⌘ < ↵, then ⌘ 2 dom(fr

) and x 2 fr

(⌘).But then ⌘ 2 dom(f

r

) and x 2 fr

(⌘) as r ↵

q � ↵, and thereforee � ↵ 2 ⌧

r

. Next suppose that ⌘ � ↵. In this case we must have e 2 ⌧q

and ⌘ 2 dom(fq

) \ dom(fr

). But then ⌘ 2 dom(fq

) and x 2 fq

(⌘). ⇤The following corollary is a trivial consequence of Lemma 2.5.

Corollary 2.6. hQ↵

| ↵ +i is a forcing iteration, in the sense thatQ

↵

is a complete suborder of Q�

for all ↵ < � +.

We say that a partial order P is �-closed if every descending sequence(p

n

)n<!

of P-conditions has a lower bound in P .

Lemma 2.7. Q�

is �-closed for every � +. In fact, every decreas-ing !-sequence of Q

�

-conditions has a greatest lower bound in Q�

. Inparticular, forcing with Q

�

does not add new !-sequences of ordinals,and therefore this forcing preserves both !1 and CH.

Proof. Given a decreasing sequence (qn

)n<!

of Q�

-conditions, it is im-mediate to check that q = (f,

Sn<!

⌧qn) is the greatest lower bound of

the set {qn

| n < !}, where dom(f) =S

n<!

dom(fqn) and, for each

n < ! and ↵ 2 dom(fqn), f(↵) =

S{f

qm(↵), m � n}. For this oneproves, by induction on ↵, that q � ↵ 2 Q

↵

for every ↵ �. ⇤Lemma 2.7, or rather its proof, will be used, often without mention,

in several places in which we run some construction, in ! steps, alongwhich we build some decreasing sequence (q

n

)n<!

of conditions. Atthe end of such a construction we will have that the ordered pair q =(f,

Sn<!

⌧qn), where f is given as in the above proof, is the greatest

lower bound of (qn

)n<!

.Given an ordered pair e = h(N0, �0), (N1, �1)i of models with mar-

kers, let us denote by e�1 the ordered pair h(N1, �1), (N0, �0)i. We saythat an ordered pair e is a generalized edge if either

• e is an edge, or• e�1 is an edge.

If ⌧ is a set of edges, we say that a generalized edge e comes from ⌧in case e 2 ⌧ (if e is an edge) or e�1 2 ⌧ (if, on the other hand, e�1 isan edge).

Given a sequence ~E = (h(N i

0, �i

0), (Ni

1, �i

1)i | i < n) of generalizededges, we will tend to denote the expression

N

n�10 ,N

n�11

� . . . � N

00 ,N

01


by ~E . If

~E is the empty sequence, we let ~E be the identity function.

Given a set ⌧ of edges, a sequence ~E = (h(N i

0, �i

0), (Ni

1, �i

1)i | i < n)of generalized edges coming from ⌧ , and x 2 N0

0 , we will call h~E , xia ⌧ -thread in case x 2 dom(

~E). We will sometimes just say threadwhen ⌧ is not relevant. In the above situation, if x = (y,↵), wherey 2 H(+) and ↵ < +, we call h~E , xi a correct ⌧ -thread if and only iffor every i < n,

(1) ~E�i(↵) 2 �i0, and

(2) ~E�i+1(↵) 2 �i1

We will be using the following straightforward remark repeatedlywithout mention.

Remark 2.8. If x = (y,↵) 2 H(+) ⇥ + and h~E , xi is a correct ⌧ -thread, then

~E is a (partially defined) elementary embedding from thestructure

(N00 ,2,�,�↵

)

into the structure(Nn�1

1 ,2,�,� ~E(↵))

Given ↵ 2 X , a node x = (⇢, ⇣) in ⇥ !1 and an ordinal ⇢ ⇢, wedenote by A↵

x,⇢

the F -first maximal antichain of Q↵

in H(+) consistingof conditions deciding, for some ordinal ⇣ < !1, that (⇢, ⇣) is T⇠↵

-belowx. If x0 = (⇢0, ⇣0) and x1 = (⇢1, ⇣1) are nodes, ⇢ ⇢0, ⇢1, r0 2 A↵

x0,⇢,

r1 2 A↵

x1,⇢, and there are ordinals ⇣0 6= ⇣1 in !1 such that r0 forces that

(⇢, ⇣0) is T⇠↵

-below x0 and r1 forces that (⇢, ⇣1) is T⇠↵

-below x1, then wesay that r0 and r1 force x0 and x1 to be incomparable in T⇠↵

.11

The following lemma will be used often.

Lemma 2.9. Suppose q is a Q

+-condition, ↵ 2 dom(fq

), ↵ 6= 0,T⇠µ(↵) is defined, h~E , (⇢,↵)i is a correct ⌧

q

-thread, where ⇢ < , andx0 = (⇢0, ⇣0) and x1 = (⇢1, ⇣1) are two nodes such that

• ⇢0, ⇢1 ⇢, and• there is some ⇢ ⇢0, ⇢1 such that q � µ(↵) extends conditions

r0 2 Aµ(↵)x0,⇢

and r1 2 Aµ(↵)x1,⇢

forcing x0 and x1 to be incomparablein T⇠µ(↵).

Let ↵ = ~E(↵). Then

(1) r0 and r1 are in dom( ~E), and

(2) ~E(r0) and ~E(r1) force x0 and x1 to be incomparable in T⇠µ(↵).

11Indeed, since for each ⇢ < , {⇢}⇥ !1 is forced to be the ⇢-th level of T⇠↵, wehave that every condition in Q↵ extending both of r0 and r1 must force that x0

and x1 are incomparable nodes in T⇠↵.


Proof. Let ~E = (h(N i

0, �i

0), (Ni

1, �i

1)i | i n). Since !1 [ ⇢ + 1 ✓dom(

~E) and ~E is a partially defined elementary embedding from

(N00 ,2,�↵

) into (Nn

1 ,2,�↵

) we have, by definability of Aµ(↵)x0,⇢

and Aµ(↵)x1,⇢

over the structure (H(+),2,�↵

) by formulas '0 and '1, respectively,

with x0, x1 and ⇢ as parameters, and definability of Aµ(↵)x0,⇢

and Aµ(↵)x1,⇢

over (H(+),2,�↵

), also by formulas '0 and '1, respectively, that

Aµ(↵)x0,⇢

, Aµ(↵)x1,⇢

2 dom( ~E), A

µ(↵)x0,⇢

= ~E(A

µ(↵)x0,⇢

), and Aµ(↵)x1,⇢

= ~E(A

µ(↵)x1,⇢

).

Again by a definability argument, since |Aµ(↵)x0,⇢

|, |Aµ(↵)x1,⇢

| < , we also

have that Aµ(↵)x0,⇢

and Aµ(↵)x1,⇢

are both subsets of dom( ~E). Finally, we

have ⇣0 6= ⇣1 in !1 such that r0 forces in Qµ(↵) that (⇢, ⇣0) is below x0 in

T⇠µ(↵) and r1 forces in Qµ(↵) that (⇢, ⇣1) is below x1 in T⇠µ(↵). But since

~E � Aµ(↵)

⇠0,⇢[A

µ(↵)x1,⇢

is a partial elementary embedding from (N00 ,2,�↵

)into (Nn

1 ,2,�↵

), by Lemma 2.2 we have that ~E(r0) forces in Q

µ(↵)

that (⇢, ⇣0) is below x0 in T⇠µ(↵) and that ~E(r1) forces in Q

µ(↵) that(⇢, ⇣1) is below x1 in T⇠µ(↵). ⇤

Given generalized edges

e = h(N0, �0), (N1, �1)iand

e0 = h(N 00, �

00), (N

01, �

01)i

such that e0 2 N0, we denote

h( N0,N1(N

00), ⇡

�

00,N

00

N0,�0,N1,�1), (

N0,N1(N01), ⇡

�

01,N

01

N0,�0,N1,�1)i

by e

(e0). Note that e

(e0) is a generalized edge.Given sets ⌧ 0 and ⌧ 1 of edges, we denote by

⌧ 0 � ⌧ 1

the natural amalgamation of ⌧ 0 and ⌧ 1 obtained by taking copies ofedges as dictated by suitable functions

~E , so that ⌧ 0 � ⌧ 1 is closedunder copying (where the edges generated by this copying procedurehave minimal marker so that ⌧ 0 � ⌧ 1 is closed under copying). To bemore specific, ⌧ 0�⌧ 1 =

Sn<!

⌧n

, where (⌧n

)n

is the following sequence.

(1) ⌧0 = ⌧ 0 [ ⌧ 1(2) For each n < !, ⌧

n+1 = ⌧n

[ ⌧ 0n

, where ⌧ 0n

is the set of edges ofthe form

e

(e0), for edges e = h(N0, �0), (N1, �1)i and e0 in ⌧n

such that e0 2 N0.

Given ⌧ 0 and ⌧ 1, two sets of edges, the construction of ⌧ 0 � ⌧ 1 asSn<!

⌧n

gives rise to a natural notion of rank on the set of generalizededges coming from ⌧ 0 � ⌧ 1. Specifically, given n < !, a generalizededge e = h(N0, �0), (N1, �1)i coming from ⌧ 0� ⌧ 1 has (⌧ 0, ⌧ 1)-rank n i↵


n is least such that e comes from ⌧n

. Alternatively, we may define the(⌧ 0, ⌧ 1)-rank of e as follows.

• e has (⌧ 0, ⌧ 1)-rank 0 if e comes from ⌧ 0 [ ⌧ 1.• For every n < !, a generalized edge e coming from ⌧ 0 � ⌧ 1

has (⌧ 0, ⌧ 1)-rank n + 1 i↵ e does not have (⌧ 0, ⌧ 1)-rank m forany m n and there are edges e0 = h(N0, �0), (N1, �1)i and e1coming from ⌧ 0 � ⌧ 1 and such that

– the maximum of the (⌧ 0, ⌧ 1)-ranks of e0 and e1 is n,– e1 2 N0, and– e =

e0(e1).

The following lemma can be easily prove by induction on the (⌧ 0, ⌧ 1)-rank of the members of ⌧ 0 � ⌧ 1.

Lemma 2.10. Let ⌧ 0 and ⌧ 1 be sets of edges, and let � < be anordinal such that all edges in ⌧ 0 involve models of height less than�. Suppose ⌧ 1 is closed under copying. Then all members of ⌧ 0 � ⌧ 1

involving models of height at least � are edges in ⌧ 1.

As we will see, the following lemma will enable us to ease our paththrough the proof of Claim 3.6, in Section 3, in a significant way.

Lemma 2.11. For all sets ⌧ 0 and ⌧ 1 of edges, every set x, and every⌧ 0 � ⌧ 1-thread h~E , xi there is a ⌧ 0 [ ⌧ 1-thread h~E⇤, xi such that

~E(x) = ~E⇤(x)

Furthermore,

(1) if x = (y,↵⇤) 2 H(+)⇥ + and h~E , xi is correct, then ~E⇤ maybe chosen to be correct as well, and

(2) if � < , ~E consists of edges, all members of ⌧ 0 involve modelsof height less than �, and ⌧ 1 is closed under copying, then allmembers of ~E⇤ involving models of height at least � are edges in⌧ 1.

Proof. Let ~E = (ei

| i n), where ei

= h(N i

0, �i

0), (Ni

1, �i

1)i for each i.We aim to prove that there is a ⌧ 0�⌧ 1-thread h~E⇤, xi with the followingproperties.

(1) ~E(x) = ~E⇤(x)

(2) If every ei

has (⌧ 0, ⌧ 1)-rank 0, then ~E⇤ = ~E .(3) If some e

i

has positive (⌧ 0, ⌧ 1)-rank, then the maximum (⌧ 0, ⌧ 1)-rank of the members of ~E⇤ is strictly less than the maximum(⌧ 0, ⌧ 1)-rank of the members of ~E .

(4) The following holds, where ~E⇤ = (e⇤i

| i n⇤).


(a) e⇤0 = e0 if e0 comes from ⌧ 0 [ ⌧ 1.(b) If there are generalized edges e = h(N0, �0), (N1, �1)i and

e0 coming from ⌧ 0�⌧ 1, both of rank less than e0, such thate0 2 N0 and e0 = e

(e0), then

e⇤0 = h(N1, �1), (N0, �0)i,

where h(N0, �0), (N1, �1)i is some generalized edge as above.(c) e⇤

n

⇤ = en

if en

comes from ⌧ 0 [ ⌧ 1.(d) If there are generalized edges e = h(N0, �0), (N1, �1)i and e0

coming from ⌧ 0 � ⌧ 1, both of rank less than en

, such thate0 2 N0 and e

n

= e

(e0), then

e⇤n

⇤ = h(N0, �0), (N1, �1)i,

where h(N0, �0), (N1, �1)i is some generalized edge as above.(5) If � < , ~E consists of edges, all members of ⌧ 0 involve models

of height less than �, and ⌧ 1 is closed under copying, then allmembers of ~E⇤ involving models of height at least � are edgesin ⌧ 1.

It easily follows from the fact that (1) and (4) holds for all threads,

together with the choice of the ordinals ⇡N

00,�

00

N0,�0,N1,�1and ⇡

N

01,�

01

N0,�0,N1,�1, for

relevant generalized edges h(N0, �0), (N1, �1)i and h(N 00, �

00), (N

01, �

01)i,

that if x = (y,↵⇤) 2 H(+)⇥ + and h~E , xi is correct, then so is ~E⇤.The proof of (1)–(5) will be by induction on n. We may obviously

assume that there is some i < n such that ei

does not come from⌧ 0 [ ⌧ 1. Then there are generalized edges e = h(N0, �0), (N1, �1)i ande0 2 N0 coming from ⌧ 0 � ⌧ 1, both of rank less than e

i

, and such thatei

= e

(e0).By induction hypothesis there is a ⌧ 0 � ⌧ 1-thread h~E0, xi, together

with a ⌧ 0 � ⌧ 1-thread of the form h~E2, ~E�i+1(x)i, such that

~E0(x) = ~E�i(x)

and

~E2( ~E�i+1(x)) = ~E�[i+1,n)( ~E�i+1(x)) = ~E(x),

and such that the relevant instances of (1)–(5) hold for h~E0, xi andh~E2, ~E�i+1(x)i. Also, by the choice of e

i

, the thread h~E1, ~E�i(x)i sa-tisfies the instances of (1) and (3)–(5) corresponding to h(e

i

), ~E�i(x)i,

where (ei

) is the sequence whose only member is ei

, and where ~E1 =(e�1, e0, e). But now we may take ~E⇤ to be the concatenation of ~E0, ~E1,and ~E2. ⇤


Given functions f and g, let us momentarily denote by f + g thefunction with dom(f + g) = dom(f) [ dom(g) defined by letting

(f + g)(x) = f(x) [ g(x)

for all x 2 dom(f) [ dom(g).Given Q

+-conditions q0 and q1, let q0 � q1 denote the natural amal-gamation of q0 and q1; to be more specific, q0 � q1 is the orderedpair (f, ⌧

q0 � ⌧q1), where f is the closure of f

q0 + fq1 with respect

to relevant (restrictions of) functions of the form N0,N1 , for edges

h(N0, �0), (N1, �1)i in ⌧q0 � ⌧

q1 , so that clause (6) in the definition ofcondition holds for q0 � q1. Even more precisely, we define

q0 � q1 = ((fq0 + f

q1) + f, ⌧q0 � ⌧

q1),

where f is the function with domain X—for X being the collection ofall ordinals of the form

~E(↵), for a correct ⌧q0 � ⌧

q1-thread h~E , (⇢,↵)isuch that ~E consists of edges, and such that (⇢, ⇣) 2 f

q0(↵)[ fq1(↵) for

some ⇣ < !1—, and such that for every ↵ 2 X, f(↵) is the collectionof all nodes (⇢, ⇣), for correct ⌧

q0 � ⌧q1-threads h~E , (⇢,↵)i such that ~E

consists of edges, (⇢, ⇣) 2 fq0(↵) [ f

q1(↵), and ~E(↵) = ↵.Lemma 2.12 holds by the construction of q0 � q1.

Lemma 2.12. Let q0 and q1 be Q�

-conditions and let q = q0�q1. Thenthe following holds.

(1) ⌧q

is closed under copying.(2) For every ↵ 2 N0 \ �0 such that ↵ :=

N0,N1(↵) < �1, if ↵ 6= 0,↵ 2 dom(f

q

), and x 2 fq

(↵) \N0, then(a) ↵ 2 dom(f

q

), and(b) x 2 f

q

(↵)

The following lemma is a trivial consequence of Lemmas 2.10 and2.11.

Lemma 2.13. For every two Q

+-conditions q0 and q1, if q0 � q1 =(⌧, f), then for every ↵ 2 dom(f) and every x = (⇢, ⇣) 2 f(↵) suchthat x /2 f

q0(↵) [ fq1(↵) there is some ↵⇤ 2 dom(f

q0) [ dom(fq1) such

that x 2 fq0(↵

⇤) [ fq1(↵

⇤) and some correct ⌧q0 [ ⌧q1-thread h~E , (⇢,↵⇤)i

such that ~E(↵

⇤) = ↵. Furthermore, if � < is such that all edges in

⌧q0 involve models of height less than �, then all members of ~E involvingmodels of height at least � are edges in ⌧

q1.

It will be necessary, in the proof of Lemma 4.1, to adjoin a certainedge to some given condition. This will be accomplished by means ofthe following lemma.


Lemma 2.14. Let ↵ < + be a nonzero ordinal such that T⇠µ(↵) isdefined and let N � H(+) be a model such that ↵ 2 N and Q

↵+1 \Nis a complete suborder of Q

↵+1. Suppose

• ⌧ is a countable set of edges below ↵ + 1,• r0 2 Q�N

↵+1,• ⌧

r0 [ ⌧ is closed under copying, and• if h(M0,↵ + 1), (M1, ⌘1)i 2 ⌧ , x 2 f

r0(↵) \ M0, and ↵ :=


r0(↵).

Then there is an extension q0 2 Q↵+1\N of r0 � N with the property

that for every extension q1 2 Q↵+1 of q0 in N there is a condition

r1 2 Q�N↵+1 such that

(1) r1 extends both r0 and q1,(2) ⌧

r1 [ ⌧ is closed under copying, and(3) if h(M0,↵ + 1), (M1, ⌘1)i 2 ⌧ , x 2 f

r1(↵) \ M0, and ↵ :=


r1(↵).

Proof. It is straightforward to verify that the ordered pair

q⇤ = (fr0 , ⌧r0 [ ⌧)

is a Q↵+1-condition. Since Q

↵+1 \N lQ↵+1, q⇤ � N may be extended

to a condition q0 in Q↵+1\N forcing that every condition in GQ↵+1\N is

compatible with q⇤. But then q0 realizes the conclusion of the lemma.To see this, first note that q0 Q↵+1 r0 � N . Now let q1 2 Q

↵+1 \Nbe such that q1 Q↵+1 q0. By the choice of q0 there exists q⇤1 2 Q

↵+1

such that q⇤1 Q↵+1 q1, q⇤. Let r1 = (f

q

⇤1, ⌧

q

⇤1\ ⌧ ⇤), where

⌧ ⇤ = {h(N0, �0), (N1, �1)i 2 ⌧q

⇤1| �

N0 > �N

and max{�0, �1} = ↵ + 1}

Then r1 2 Q�N↵+1 and (1)–(3) hold for this condition. ⇤

Lemma 2.15 will be used in the proof of Lemma 2.17.

Lemma 2.15. Given ↵ < +, q 2 Q↵+1 and h(N0,↵+1), (N1, �1)i 2 ⌧

q

,there is an extension q0 2 Q

↵+1 of q � N0, q0 2 N0, such that everycondition in Q

↵+1 \N0 extending q0 is compatible with q.

Proof. We apply clause (7) in the definition of condition to ↵, the edgeh(N0,↵ + 1), (N1, �1)i,⌧ = {h(M0, �0), (M1, �1)i 2 ⌧

p

| �M0 > � and max{�0, �1} = ↵ + 1},

and r0 = (fq

, ⌧q

\ ⌧). Let q0 2 Q↵+1 \ N0 be the resulting condition.

Let q1 2 Q↵+1 \N0 be any condition extending q0. We show that q1 is

compatible with q. Let r1 2 Q�N0↵+1 be such that

(1) r1 extends both r0 and q1,


(2) ⌧r1 [ ⌧ is closed under copying, and such that

(3) if h(M0,↵ + 1), (M1, ⌘1)i 2 ⌧ , x 2 fr1(↵) \ M0, and ↵ :=

M0,M1(↵) < ⌘1, then ↵ 2 dom(f

r1) and x 2 fr1(↵).

Then q⇤ = (fr1 , ⌧r1 [ ⌧) is a Q

↵+1–condition which extends both ofq1 and q. Clauses (5) and (6) in the definition of Q

↵+1-condition holdfor q⇤ by (2) and (3) above. We show that q⇤ satisfies clause (7) in thedefinition of Q

↵+1-condition as well, as all other clauses are obvious.Let e = h(N 0

0,↵ + 1), (N 01, �

01)i 2 ⌧

r1 [ ⌧ , and suppose

• ⌧ 0 ◆ ⌧r1 [ ⌧ is a countable set of edges below ↵ + 1,

• r00 2 Q�N0

0↵+1 extends both r1 � ↵ and q⇤ � N 0

0 and is such thatfr1(↵) ✓ f

r

00(↵),

• ⌧r

00[ ⌧ 0 is closed under copying, and

• if h(M0,↵ + 1), (M1, ⌘1)i 2 ⌧ 0, x 2 fr

00(↵) \ M0, and ↵ :=

M0,M1(↵) < ⌘1, then ↵ 2 dom(f

r

00) and x 2 f

r

00(↵).

We need to find some extension q00 2 Q↵+1 \ N 0

0 of r00 � N0 withthe property that for every extension q01 2 Q

↵+1 of q00 in N 00 there is a

condition r01 2 Q�N0

0↵+1 such that

(1) r01 extends both r00 and q01,(2) ⌧

r

01[ ⌧ 0 is closed under copying, and

(3) if h(M0,↵ + 1), (M1, ⌘1)i 2 ⌧ 0, x 2 fr

01(↵) \ M0, and ↵ :=

M0,M1(↵) < ⌘1, then ↵ 2 dom(f

r

01) and x 2 f

r

01(↵).

Let us suppose first that e 2 ⌧r1 . Since of course r00 extends r1 � N 0

0,q00 can be found by an application of clause (7) in the definition ofcondition to r1 with respect to h(N 0

0,↵ + 1), (N 01, �

01)i, ⌧ 0, and r00.

Suppose now that e 2 ⌧ . Since ⌧ 0 ◆ ⌧q

and r00 extends both q � ↵and q � N 0

0 and is such that fq

(↵) ✓ fr

00(↵), q00 can be found by an

application of clause (7) in the definition of condition to q with respectto the same objects as in the previous case. ⇤

The last three lemmas in this section will be used in the proof ofLemma 3.1.

Lemma 2.16. Let � +, and suppose q0, q1 2 Q�

are such that forevery ↵ < �, if

(q0 � ↵)� (q1 � ↵) 2 Q↵

,

then(q0 � ↵ + 1)� (q1 � ↵ + 1) 2 Q

↵+1

Then q0 � q1 2 Q�

.

Proof. The proof is by induction on �. We only need to argue for theconclusion in the case that � is a nonzero limit ordinal. But in that


case the conclusion follows easily from the induction hypothesis andthe fact that for every ↵ < �, (q0 � q1) � ↵ is the greatest lower boundof the set of Q

↵

-conditions given by

X = {((q0 � ↵0)� (q1 � ↵0)) � ↵ | ↵ ↵0 < �}

It is easy to see that there is a decreasing sequence (rn

)n<!

of Q↵

-conditions in X with the property that every r 2 X is such that r

n

Q↵

r for some n, and thus this greatest lower bound exists. ⇤

We will call a generalized edge e an anti-edge if e�1 is an edge.Given � < + and given q 2 Q

�

, we will say that q is adequate incase (1) and (2) below hold.

(1) For every ↵ 2 dom(fq

) such that T⇠µ(↵) is defined and for alldistinct x0 = (⇢0, ⇣0), x1 = (⇢1, ⇣1) 2 f

q

(↵) there is some ⇢ ⇢0,

⇢1, together with conditions r0 2 Aµ(↵)x0,⇢

and r1 2 Aµ(↵)x1,⇢

weakerthan q � µ(↵) and such that r0 and r1 force x0 and x1 to beincomparable in T⇠µ(↵).

(2) The following holds for all ordinals ↵ ↵ < �, and for everycorrect ⌧

q

-thread h(~E , (y, ↵)i such that• ~E consists of anti–edges, and•

~E(↵) = ↵.(a) If y is an edge e = h(N0, �0), (N1, �1)i 2 ⌧

q

and �0, �1 > ↵,then

h( ~E(N0), �

⇤0), ( ~E(N1), �

⇤1)i 2 ⌧

q

for some ordinals �⇤0 and �⇤1 such that ↵ < �⇤0 , �⇤1 .

(b) If ↵ 6= 0, ↵ 2 dom(fq

), y is an ordinal ⇢ < , ⇣ < !1, andx = (⇢, ⇣) 2 f

q

(↵), then(i) ↵ 2 dom(f

q

), and(ii) x 2 f

q

(↵)

We call a condition weakly adequate if it satisfies clause (1) in theabove definition. The set of weakly adequate conditions is clearly densein Q

�

for each � +. To see this, let q 2 Q�

. By a book-keepingargument, we can define a decreasing sequence (q

i

)i<!

of Q�

-condition,with q0 = q, such that for all i < !, ↵ 2 dom(f

qi), ↵ 6= 0, and alldistinct x0 = (⇢0, ⇣0), x1 = (⇢1, ⇣1) 2 f

qi(↵) there are m � i, ⇢ ⇢0,

⇢1, together with conditions r0 2 Aµ(↵)x0,⇢

and r1 2 Aµ(↵)x0,⇢

weaker thanqm

� µ(↵) and such that r0 and r1 force x0 and x1 to be incomparablein T⇠µ(↵). Then q⇤ is weakly adequate, where q⇤ is the greatest lowerbound of the sequence (q

i

)i<!

.We in fact have the following.


Lemma 2.17. For every � +, the set of adequate Q�

-conditions isdense in Q

�

.

Proof. Let q 2 Q�

. We will find an adequate Q�

-condition q⇤ strongerthan q. We prove this by induction on �.

First, suppose that � is a limit ordinal of countable cofinality and let(�

i

)i<!

be an increasing sequence of ordinals cofinal in �. We define,by induction on i < !, two sequences (q

i

)i<!

and (ri

)i<!

of conditionssuch that q0 = q and such that for all i < !,

(1) qi

2 Q�

,(2) r

i

is an adequate Q�i-condition,

(3) ri

Q�iqi

� �i

, and(4) q

i+1 = (fri [ f

qi � [�i, �), ⌧ri [ ⌧qi).12

Let q⇤ be the greatest lower bound of the sequence (qi

)i<!

, which existsby Lemma 2.7. Then q⇤ is an adequate Q

�

-condition extending q.If � is a limit ordinal of uncountable cofinality, then we proceed

exactly as above, but at each step we choose the ordinal �i

to be suchthat dom(f

qi) ✓ �i

.Finally suppose that � = ↵ + 1 is a successor ordinal. By induction

hypothesis together with Lemma 2.5, we may assume that q � ↵ isadequate. We may also assume that there is some edge in ⌧

q

of theform h(N0, �), (N1, �1)i, as otherwise we are done. Let (�i)i<�

, for some� < !1, be the strictly increasing enumeration of the heights of themodels occurring in the edges from ⌧

q

of the above form. We buildq⇤ as the greatest lower bound of a suitably constructed descendingsequence (q

i

)i<!

of conditions extending q0 = q and such that qi

� ↵ isadequate for each i.

Let us start by describing how to construct q1. For this, we first letp0 = q0, pick some edge e = h(N0, �0), (N1, �)i 2 ⌧

p0 with N0 of mini-mum height and, using Lemma 2.15, extend p0 to a condition p+0 suchthat every condition in Q

�

\N0 extending p+0 � N0 is compatible withp0. By further extending p+0 if necessary using the induction hypothe-sis, we may assume that p+0 � ↵ is adequate. Let ↵ =

N0,N1(↵). Wethen find p00 2 Q

�

\N0 extending p+0 � N0, and satisfying the conclusion

of clause (2) in the definition of adequate condition for every relevant⌧p

+0-thread h~E , (y,↵)i such that the last member of ~E is e�1. For this it

su�ces to simply

(1) add to fp

+0(↵) all nodes of level less than �

N0 in fp

+0(↵), and

12Note that by Lemma 2.5 (and Lemma 2.4), each qi+1 is indeed a Q�-condition.


(2) add h(N 00,min{

e

�1(�0), �}N

00), (N 0

1,min{ e

�1(�1), �}N

01)i to ⌧

p

+0

for every h(N0, �0), (N1, �1)i 2 ⌧p

+0\ N1, where N 0

0 = e

�1(N0)

and N 01 = e

�1(N1).

In order to see that the result of building p00 in this way will indeedbe a condition in Q

�

, it is enough, thanks to the adequacy of p+0 � ↵,to show that if x0 = (⇢0, ⇣0) and x1 = (⇢1, ⇣1) are distinct nodes infp

+0(↵) of level less than �

N0 , then p+0 � µ(↵) forces x0 and x1 to be

incomparable in T⇠µ(↵). By weak adequacy of p+0 � ↵ we know that

there is some ⇢ ⇢0, ⇢1, together with conditions r0 2 Aµ(↵)x0,⇢

\N1 and

r1 2 Aµ(↵)x1,⇢

\N1 such that r0 and r1 force x0 and x1 to be incomparablein T⇠µ(↵). By clause (2) in the definition of adequate condition, we havethat p+0 � µ(↵) extends

N1,N0(r0) and N1,N0(r1). But now we are donesince

N1,N0(r0) and N1,N0(r1) force x0 and x1 to be incomparable inT⇠µ(↵) by Lemma 2.9.Next we let p1 be a common extension of p+0 and p00. We repeat the

above construction, dealing in turn with all remaining edges in ⌧p

+0of

the form h(N 00, �), (N

01, �

01)i with �

N

00= �0, starting in the next step

with p1 instead of p0, and taking greatest lower bounds at limit stages.We thus end up with a condition p

l1 2 Q�

satisfying the conclusion ofclause (2) in the definition of adequate condition for all correct threadsin which the last anti–edge involves a model of height �0 and marker�.

If �1 exists, we repeat the above construction, starting now with pl1 .

This time we deal with all edges in ⌧p

+0in which the first edge has height

�1 and marker �. We iterate this construction, handling all heights �i

in increasing order, taking greatest lower bounds at limit stages. Wefinally let q1 be the greatest lower bound of the sequence of conditionspi

thus constructed. In general, for every n < ! for which qn

has beendefined, we construct q

n+1 from qn

in exactly the same way that wehave constructed q1 from q0. In the end, we let q⇤ be the greatest lowerbound of {q

n

| n < !}. By our construction it is easy to verify thatq⇤ is adequate. ⇤

We conclude this section with the following trivial lemma.

Lemma 2.18. If q is an adequate Q

+-condition, ↵ 2 dom(fq

), x =(⇢, ⇣) 2 f

q

(↵), and h~E , (⇢,↵)i is a correct ⌧q

-thread, then ~E(↵) 2

dom(fq

) and x 2 fq

( ~E(↵)).

3. The chain condition

This section is devoted to proving Lemma 3.1.


Lemma 3.1. For each � +, Q�

has the -chain condition.

As we will see, the weak compactness of is used crucially in orderto prove Lemma 3.1. Let F be the weak compactness filter on , i.e.,the filter on generated by the sets

{� < | (V�

,2, B \ V�

) |= },

where B ✓ V

and where is a ⇧11 sentence for the structure (V

,2, B)such that (V

,2, B) |= . F is a proper normal filter on . Let also Sbe the collection of F -positive subsets of , i.e.,

S = {X ✓ | X \ C 6= ; for all C 2 F}

We will call a model Q suitable if Q is an elementary submodel ofcardinality of some high enough H(✓), closed under <-sequences,and such that hQ

↵

| ↵ < +i 2 Q. Given a suitable model Q, abijection ' : ! Q, and an ordinal � < , we will denote '“� by M'

�

.It is easily seen that

{� < | M'

�

� Q, M'

�

\ = � and <�M'

�

✓ M'

�

} 2 F .

Given � +, we will say that Q�

has the strong -chain conditionif for every X 2 S, every suitable model Q such that �, X 2 Q, everybijection ' : ! Q, and every two sequences (q0

�

| � 2 X) 2 Q and(q1

�

| � 2 X) 2 Q of adequate Q�

-conditions, if M'

�

\ = � andq0�

� M'

�

= q1�

� M'

�

for every � 2 X, then there is some Y 2 S,Y ✓ X, together with sequences (r0

�

| � 2 Y ) and (r1�

| � 2 Y ) ofadequate Q

�

-conditions with the following properties.

(1) r0�

Q�q0�

and r1�

Q�q1�

for every � 2 Y .(2) For all �0 < �1 in Y , r0

�0�r1

�1is a common extension of r0

�0and

r1�1.

The following lemma is an immediate consequence of Lemma 2.17.

Lemma 3.2. For every � +, if Q�

has the strong -chain condition,then Q

�

has the -chain condition.

Following [5], given � +, a suitable model Q such that � 2 Q, abijection ' : ! Q, � < , and a Q

�

-condition q 2 Q, let us say thatq is �-compatible with respect to ' and � if, letting Q⇤

�

= Q�

\ Q, wehave that

• Q⇤�

\M'

�

lQ⇤�

,• q � M'

�

2 Q⇤�

, and


• q � M'

�

forces in Q⇤�

\ M'

�

that q is in the quotient forcing

Q⇤�

/GQ⇤�\M

'�; equivalently, for every r Q⇤

�\M'�

q � M'

�

, r is

compatible with q.13

Adopting the approach from [10], rather than proving Lemma 3.1we will prove the following more informative lemma.

Lemma 3.3. The following holds for every � < +.

(1)�

Q�

has the strong -chain condition.(2)

�

Suppose D 2 F , Q is a suitable model, �, D 2 Q, ' : ! Qis a bijection, and (q0

�

| � 2 D) 2 Q and (q1�

| � 2 D) 2 Qare sequences of Q

�

-conditions. Then there is some D0 2 F ,D0 ✓ D, such that for every � 2 D0 and for all q0

0�

Q�q0�

andq1

0�

Q�q1�

, if q00

�

� M'

�

2 Q�

and q00

�

� M'

�

= q10

�

� M'

�

, thenthere are conditions r0

�

Q�q0

0�

and r1�

Q�q1

0�

such that(a) r0

�

� M'

�

= r1�

� M'

�

and(b) r0

�

and r1�

are both �-compatible with respect to ' and �.

Corollary 3.4. Q

+ has the -c.c.

Proof. Suppose qi

, for i < , are conditions in Q

+ . By Lemma 2.17,we may assume that each q

i

, for i < , is adequate. We may thenfix � < + such that q

i

2 Q�

for all i < . But by Lemma 3.3 (1)�

together with Lemma 3.2 there are i 6= i0 in such that qi

and qi

0 arecompatible in Q

�

and hence in Q

+ . ⇤The rest of the section is devoted to proving the above lemma.

Proof. (of Lemma 3.3) The proof is by induction on �. Let � < +

and suppose (1)↵

and (2)↵

holds for all ↵ < �. We will show that (1)�

and (2)�

also hold.There is nothing to prove for � = 0, and the case � = 1 is trivial,

using the inaccessibility of .Let us proceed to the case when � > 1. We start with the proof of

(1)�

.Let X 2 S be given, together with a suitable model Q such that

�, X 2 Q, a bijection ' : ! Q, and sequences ~�0 = (q0�

| � 2 X) 2 Qand ~�1 = (q1

�

| � 2 X) 2 Q of adequate Q�

-conditions such thatM'

�

\ = � and q0�

� M'

�

= q1�

� M'

�

for every � 2 X. We needto prove that there is some Y 2 S, Y ✓ X, together with sequences(r0

�

| � 2 Y ) and (r1�

| � 2 Y ) of Q�

-conditions such that the followingholds.

(1) r0�

Q�q0�

and r1�

Q�q1�

for every � 2 Y .

13In [10], this situation is denoted by ⇤��(q0, q0 � M'� ).


(2) For all �0 < �1 in Y , r0�0�r1

�1is a common extension of r0

�0and

r1�1.

In what follows, we will write M�

instead of M'

�

.Let Q⇤

↵

= Q↵

\ Q for every ↵ 2 Q \ (� + 1). By the inductionhypothesis, Q

↵

has the -c.c. for every ↵ 2 Q\ �. Hence, since <Q ✓Q, we have that Q⇤

↵

lQ↵

for every such ↵; in particular, we have thatfor every ↵ 2 Q \ X \ �, Q⇤

↵

forces over V that T⇠↵

does not have-branches.

Given

• conditions q0, q1 in Q�

,• nonzero stages ↵ 2 dom(f

q

0) and ↵0 2 dom(fq

1),14

• nodes x = (⇢0, ⇣0) and y = (⇢1, ⇣1) such that x 2 fq

0(↵) andy 2 f

q

1(↵0),15 and• � < ,

we will say that x and y are separated below � at stages µ(↵) and µ(↵0)by q0 � µ(↵) and q1 � µ(↵0) (via x, y) if there are ⇢ < � and ⇣ 6= ⇣ 0 in!1 such that x = (⇢, ⇣) and y = (⇢, ⇣ 0), and such that

(1) q0 � µ(↵) extends a condition in Aµ(↵)x,⇢

forcing x to be below xin T⇠µ(↵) and

(2) q1 � µ(↵0) extends a condition in Aµ(↵)y,⇢

forcing y to be below yin T⇠µ(↵0).

Definition 3.5. Given Y 2 S such that Y ✓ X and such that M�

� Q,M

�

\ = �, and <�M�

✓ M�

for all � 2 Y , and given two sequences~�⇤0 = (r0

�

| � 2 Y ), ~�⇤1 = (r1

�

| � 2 Y ) of adequate Q⇤�

-conditions, wesay that ~�⇤

0, ~�⇤1 is a separating pair for ~�0 and ~�1 if the following holds.

(1) For every � 2 Y , r0�

Q�q0�

, r1�

Q�q1�

, and dom(fr

0�) =

dom(fr

1�).

(2) For every � 2 Y , every ↵ 2 dom(fr

0�) \ M

�

, every nonzero↵0 2 dom(f

r

1�) such that ↵0 ↵, and for all

x 2 fr

0�(↵) \ (�⇥ !1)

andy 2 f

r

1�(↵0) \ (�⇥ !1),

x and y are separated below � at stages µ(↵) and µ(↵0) by r0�

and r1�

via some pair �0(x, y,↵,↵0,�), �1(x, y,↵,↵0,�) of nodes.(3) The following holds for all �0 < �1 in Y .

14Note that, by induction hypothesis, both Qµ(↵) and Qµ(↵0) have the -c.c. andhence T⇠µ(↵) and T⇠µ(↵0) are both defined.

15↵ and ↵0 may or may not be equal and the same applied to x and y.


(a) r0�0

� M�0 = r1

�1� M

�1 and dom(fr

0�0)\M

�0 = dom(fr

1�1)\

M�1.

(b) r0�0

2 M�1

(c) Let N✏

and ⌅✏

, for ✏ 2 {0, 1}, be defined as follows.• N

✏

is the union of the sets of the form N0[N1, whereh(N0, �0), (N1, �1)i 2 ⌧

r

0�✏[ ⌧

r

1�✏

and �N0 < �

✏

.

• ⌅✏

is the collection of ordinals of the form ~E(↵),

where h~E , (⇢, ↵)i is a correct ⌧r

1�✏-thread such that

(⇢,↵) 2 N \ ( ⇥ +) for some model N of heightless than �

✏

coming from some edge in ⌧r

1�✏.

Then(i) N0 \M

�0 = N1 \M�1 and

(ii) ⌅0 \M�0 = ⌅1 \M

�1.(4) For all �1 > �0 in Y , all ordinals ↵ 2 dom(f

r

0�0) \ M

�0 and

↵0 2 dom(fr

1�1) such that 0 < ↵0 ↵, and all nodes

x 2 fr

0�0(↵) \ (�0 ⇥ !1)

and

y0 2 fr

1�1(↵0) \ (�1 ⇥ !1)

there are• a node x0 2 f

r

0�1(↵) \ (�1 ⇥ !1),

• a stage ↵⇤ 2 dom(fr

1�0) such that ↵⇤ ↵, and

• a node y 2 fr

1�0(↵⇤) \ (�0 ⇥ !1)

such that

�0(x, y,↵,↵⇤,�0) = �0(x

0, y0,↵,↵0,�1)

and

�1(x, y,↵,↵⇤,�0) = �1(x

0, y0,↵,↵0,�1)

Let us now prove the following.

Claim 3.6. Let Y 2 S be such that M�

\ = � for all � 2 Y , andsuppose ~�⇤

0 = (r0�

| � 2 Y ), ~�⇤1 = (r1

�

| � 2 Y ) is a separating pairfor ~�0 and ~�1. Then for all �0 < � < �1 in Y , r0

�0� r1

�1is a common

extension of r0�0

and r1�1

in Q�

.

Proof. Suppose, towards a contradiction, that there are �0 < � < �1in Y such that r0

�0� r1

�1is not a common extension of r0

�0and r1

�1. It

then follows that r0�0

� r1�1

is not a condition. Hence, by Lemma 2.16,there is an ordinal ↵ < � such that (r0

�0� ↵)� (r1

�1� ↵) is a condition


yet (r0�0

� ↵ + 1) � (r1�1

� ↵ + 1) is not. Assuming that we are in thissituation, we will derive a contradiction. Set q = (r0

�0� ↵)� (r1

�1� ↵).

To start with, note that ↵ > 0. We will need the following subclaim.

Subclaim 3.7. Suppose

(1) �1 �0 are ordinals in M�

,(2) h(h(N i

0, �i

0), (Ni

1, �i

1)i | i n), �0i is a ⌧r

1�1-thread,

(3) �1 = ~E(�0), where~E = (h(N i

0, �i

0), (Ni

1, �i

1)i | i n), and(4) �

N

i0� �1 for all i n.

Then �1 = �0.

Proof. Suppose first that �N

i0= �

N

i00for all i, i0. By correctness of the

structure (N00 ,2,�0) within (H(+),2,�0), we may pick some model

M 2 N00 closed under ~e such that �0 2 M , �

M

= �, and |M | = �(since M

�

is such a a model). Given that �1 �0 are both in M�

,�M�

= �, and �0 2 M , by the first part of Fact 2.1 we then have that�1 2 M ✓ N0

0 . But that means, by the second part of Fact 2.1, that(

N

n0 ,N

n1� . . . �

N

00 ,N

01)(�0) = �1 is in fact �0 since �1 2 N0

0 \ Nn

1 .Suppose now that �

N

i0< �

N

i00for some i 6= i0 and let i⇤ be such that

�N

i⇤0

= min{�N

i0| i n}. The rest of the proof is like in the proof

of the second part of Fact 2.1 in the general case. Since each N i

✏

(for✏ 2 {0, 1}) is an elementary submodel of (H(+),2,�0) closed undersequences of length less than |N i

✏

| and |N i

⇤⇤0 | = |N i

⇤⇤1 | = |�

N

i⇤0| for every

i⇤⇤ such that �N

i⇤⇤0

= �N

i⇤0, there is a sequence (hN i

0, Ni

1i)in

of pairs ofmodels such that the following holds.

• For all i n, (N i

0,2,�) and (N i

1,2,�) are isomorphic elemen-tary submodels of (H(+),2,�) closed under ~e.

• �N

i0= �

N

i00= �

N

i⇤0

for all i, i0 n.

• For every i n and every ✏ 2 {0, 1}, N i

✏

= N i

✏

or N i

✏

2 N i

✏

.• �0 2 dom(

N

n0 ,N

n1� . . . �

N

00 ,N

11) \ Nn

1

• �1 = ( N

n0 ,N

n1� . . . �

N

00 ,N

01)(�0) = (

N

n0 ,N

n1� . . . �

N

00 ,N

11)(�0)

But now we are done by the previous case. ⇤We will also be using the following subclaim.

Subclaim 3.8. Suppose ↵⇤ 2 dom(fr

0�0) [ dom(f

r

1�1), x = (⇢, ⇣) 2

fr

0�0(↵⇤) [ f

r

1�1(↵⇤), h~E⇤, (⇢,↵⇤)i is a correct ⌧

r

0�0

[ ⌧r

1�1-thread, and all

members h(N0, �0), (N1, �1)i of ~E⇤ such that �N0 � �1 are edges from

⌧ 1�1. Then at least one of the following holds, where ↵ = ~E⇤(↵⇤).

(1) ↵ 2 dom(fr

0�0) and x 2 f

r

0�0(↵).


(2) ↵ 2 dom(fr

1�1) and x 2 f

r

1�1(↵).

(3) There is some ↵⇤⇤ 2 dom(fr

0�0) such that x 2 f

r

0�0(↵⇤⇤) and

some correct ⌧r

1�1-thread h~E , (⇢,↵⇤⇤)i such that

• ~E(↵

⇤⇤) = ↵ and

• all members of ~E are edges h(N0, �0), (N1, �1)i such that�N0 � �1.

Proof. We prove this by induction on |~E⇤|, which we may obviouslyassume is nonzero. Let ~E⇤ = (h(N i

0, �i

0), (Ni

1, �i

1) | i m) and e =h(Nm

0 , �m0 ), (Nm

1 , �m1 )i. By induction hypothesis, one of (1)–(3) holdsfor h~E⇤ � m, (⇢,↵⇤)i. Let ↵† =

~E⇤�m(↵⇤).

Suppose (1) holds for h~E⇤ � m, (⇢,↵⇤)i. We have two cases. If e 2⌧r

0�0, then (1) holds trivially for h~E⇤, (⇢,↵⇤)i by Lemma 2.18 applied to

r0�0. The other case is that e 2 ⌧

r

1�1. If �

N

m0

� �1, then obviously (3)

holds for h~E⇤, (⇢,↵⇤)i as witnessed by ↵† and the thread h(e), (⇢,↵†)i.We may thus assume that �

N

m0

< �1. By, first, clause (3)(c)(i) inDefinition 3.5 (for the pair r0

�1, r1

�1), and then by clause (3)(a), ↵† 2

dom(fr

1�1) and x 2 f

r

1�1(↵†). But then ↵ 2 dom(f

r

1�1) and x 2 f

r

1�1(↵)

by Lemma 2.18 applied to r1�1

and so (2) holds for h~E⇤, (⇢,↵⇤)i.Next suppose (2) holds for h~E⇤ � m, (⇢,↵⇤)i. Suppose e 2 ⌧

r

0�0. Since

e 2 M�1 by clause (3)(b) in Definition 3.5, it follows from (3)(a) in Def-

inition 3.5 that ↵† 2 dom(fr

0�0) and x 2 f

r

0�0(↵†). But then, by Lemma

2.18 applied to r0�0, ↵ 2 dom(r0

�0) and x 2 f

r

0�0(↵). Hence (1) holds for

h~E⇤, (⇢,↵⇤)i. If e 2 ⌧r

1�1, then (2) holds trivially for h~E⇤, (⇢,↵⇤)i, again

by Lemma 2.18 applied to r1�1.

Finally, suppose (3) holds for h~E⇤ � m, (⇢,↵⇤)i, as witnessed by ↵⇤⇤ 2dom(f

r

0�0) together with a correct ⌧

r

1�1-thread h~E , (⇢,↵⇤⇤)i. Suppose

e 2 ⌧r

0�0. Then, since all models occurring in the edges in ~E are of

height at least �1 and since r0�0

2 M�

by clause (3)(b) in Definition3.5, by Subclaim 3.7 we have that ↵† = ↵⇤⇤. But then ↵ 2 dom(f

r

0�0)

and x 2 fr

0�0(↵) by an application of Lemma 2.18 to r0

�0, and so (1)

holds for h~E⇤, (⇢,↵⇤)i. Finally, suppose e 2 r1�1. If �

N

m0

� �1, then (3)

holds trivially for h~E⇤, (⇢,↵⇤)i as witnessed by h~E_hei, (⇢,↵⇤⇤)i. In theother case, by clauses (3)(c)(ii) and (3)(a) in Definition 3.5 for the pairr0�1, r1

�1, we have that ↵⇤⇤ 2 dom(f

r

1�1) and x 2 f

r

1�1(↵⇤⇤). But then


↵ 2 dom(fr

1�1) and x 2 f

r

1�1(↵) by Lemma 2.18 applied to r1

�1. Thus

we have that (2) holds for h~E⇤, (⇢,↵⇤)i, which finishes the proof of thesubclaim. ⇤

Since (r0�0

� ↵ + 1) � (r1�1

� ↵ + 1) is not a condition, there are ✏,✏0 2 {0, 1}, together with ↵0 2 dom(f

r

✏�✏), x0 = (⇢0, ⇣0) 2 f

r

✏�✏(↵0),

↵1 2 dom(fr

✏0�✏0), x1 = (⇢1, ⇣1) 2 f

r

✏0�✏0(↵1), ↵1, ↵0 ↵, and a nonzero

ordinal ↵ such that there are correct ⌧r

0�0

�r

1�1-threads h~E⇤

0 , (⇢0,↵0)i andh~E⇤

1 , (⇢1,↵1)i, respectively, such that

• ↵ = ~E⇤0(↵0) = ~E⇤

1(↵1),

• both ~E⇤0 and ~E⇤

1 consist of edges in ⌧r

0�0

�r

1�1, and such that

• q � µ(↵) does not force x0 and x1 to be incomparable in T⇠µ(↵).

By Lemma 2.13, we may replace h~E⇤0 , (⇢0,↵0)i and h~E⇤

1 , (⇢1,↵1)i bycorrect ⌧

r

0�0

[r1�1-threads h~E0, (⇢0,↵0)i and h~E1, (⇢1,↵1)i all of whose mem-

bers involving models of height at least �1 are edges in ⌧r

1�1.16 By Sub-

claim 3.8, and after changing some of the above objects if necessary,we may assume that one of the following holds.

(1) ↵ 2 dom(fr

0�0) and x0, x1 2 f

r

0�0(↵).

(2) ↵ 2 dom(fr

1�1) and x0, x1 2 f

r

1�1(↵).

(3) ↵ 2 dom(fr

0�0) \ dom(f

r

1�1), x0 2 f

r

0�0(↵), and x1 2 f

r

1�1(↵).

(4) ↵0 2 dom(fr

0�0), x0 2 f

r

0�0(↵0), and there is a correct ⌧

r

1�1-thread

h~E0, (⇢0,↵0)i such that•

~E0(↵0) = ↵ and

• all members of ~E0 are edges involving models of height atleast �1,

and such that one of the following holds.(a) ↵ 2 dom(f

r

0�0) and x1 2 f

r

0�0(↵).

(b) ↵ 2 dom(fr

1�1) and x1 2 f

r

1�1(↵).

(c) ↵1 2 dom(rf

0�0), x1 2 f

r

0�0(↵1) , and there is a correct ⌧

r

1�1-

thread h~E1, (⇢1,↵1)i such that•

~E1(↵1) = ↵ and

• all members of ~E1 are edges involving models of heightat least �1.

We may clearly rule out (1) and (2) since both r0�0

� µ(↵) + 1 andr1�1

� µ(↵) + 1 are conditions and q � µ(↵) extends their restriction to

16~E0 and ~E1 may of course involve anti–edges.


µ(↵). Let us assume that (4) holds.17 We will first consider the subcasewhen (a) holds. By Subclaim 3.7 applied to the fact that the height ofall models occurring in ~E0 is at least �1 and the fact that both ↵ and↵0 are in M

�

, we get that ↵ = ↵0. But then we get a contradiction asin case (1).

Let us now consider the subcase when (b) holds. By Lemma 2.18applied to r1

�1it follows that ↵0 2 dom(f

r

1�1). If ⇢1 < �1, then by

another application of Lemma 2.18 to r1�1

we get that x1 2 fr

0�0(↵0).

But then r0�0

� µ(↵0) extends conditions r0 2 Aµ(↵0)x0,⇢

and r1 2 Aµ(↵0)x1,⇢

, forsome ⇢ min{⇢0, ⇢1}, forcing x0 and x1 to be incomparable in T⇠µ(↵0)

and, by Lemma 2.9, ~E0(r0) and

~E0(r1) are conditions weaker thanq � µ(↵) and forcing x0 and x1 to be incomparable in T⇠µ(↵). We maythus assume that ⇢1 � �1. Now suppose ⇢0 < �0. Then x0 2 f

r

1�1(↵0)

by clause (3)(a) in Definition 3.5, and hence x0 2 fr

1�1(↵) by Lemma

2.18 applied to r1�1. We again reach a contradiction as in case (2).

Hence we may assume �0 ⇢0. The rest of the argument, in this case,is now essentially as in the corresponding proof in [10]. Since ↵ ↵0

due to the fact that all members of ~E0 are edges, by a suitable instanceof clause (4) in Definition 3.5 we may pick

• a node x00 = (⇢0, ⇣ 00) 2 f

r

0�1(↵0) \ (�1 ⇥ !1),

• a stage ↵⇤ 2 dom(fr

1�0) such that ↵⇤ ↵0, and

• a node x⇤1 = (⇢⇤1, ⇣

⇤1 ) 2 f

r

1�0(↵⇤) \ (�0 ⇥ !1)

such that

�0(x0, x⇤1,↵0,↵

⇤,�0) = �0(x00, x1,↵0, ↵,�1)

and

�1(x0, x⇤1,↵0,↵

⇤,�0) = �1(x00, x1,↵0, ↵,�1)

(where �0 and �1 are the projections in Definition 3.5). Let ⇢ be suchthat �0(x0, x

⇤1,↵0,↵

⇤,�0) = (⇢, ⇣0) and �1(x00, x1,↵0, ↵,�1) = (⇢, ⇣1) for

some ⇣0 6= ⇣1 in !1. We have that q � µ(↵0) extends a condition r0 2A

µ(↵0)x0,⇢

forcing �0(x0, x⇤1,↵0,↵

⇤,�0) to be below x0 in T⇠µ(↵0) (becausethis is true about r0

�0� µ(↵0)). Also, r1

�1� µ(↵) extends a condition

r1 2 Aµ(↵)x1,⇢

forcing that �1(x00, x1,↵0, ↵,�1) is below x1 in T⇠µ(↵), and

therefore so does q � µ(↵). We also have, by an argument as in theproof of Lemma 2.9, that r1

�1� µ(↵)—and therefore also q � µ(↵)—

extends ~E0(r0), and this condition forces �0(x0, x

⇤1,↵0,↵

⇤,�0) to be

17We are considering this case before the case that (3) holds since the proof inthe latter case will be a simpler variant of an argument we are about to see.


below x0 in T⇠µ(↵). But now we get a contradiction since ⇣0 6= ⇣1 andhence q � µ(↵) forces x0 and x1 to be incomparable in T⇠µ(↵).

It remains to consider the subcase that (c) holds. Since all modelsoccurring in members of ~E0 or of ~E1 are of height at least �1 and both↵0 and ↵1 are in M

�

, by Subclaim 3.7 we get that ↵0 = ↵1. But now weget a contradiction by the same argument we have already encounteredusing Lemma 2.9.

We finally handle the case when (3) holds. In this case we mayassume that both ⇢0 � �0 and ⇢1 � �1 hold, as otherwise we get, byan application of clause (3)(a) in Definition 3.5, that at least one ofx0, x1 is in f

r

0�0(↵)\f

r

1�1(↵), which immediately yields a contradiction.

But now, since ⇢0 � �0 and ⇢1 � �1, we obtain a contradiction by aseparation argument using clause (4) in Definition 3.5—with both ↵and ↵0, in that definition, being ↵—like the one we have already seen.This final contradiction concludes the proof of the claim. ⇤

The following technical fact appears essentially in [10].

Claim 3.9. Suppose Z 2 S, (p0�

| � 2 Z) 2 Q and (p1�

| � 2 Z) 2 Qare sequences of conditions in Q⇤

�

such that p0�

� M�

and p1�

� M�

arecompatible conditions in Q⇤

�

\ M�

for every � 2 Z, and suppose thatfor every � 2 Z,

• p0�

and p1�

are �-compatible with respect to ' and ↵ for all ↵ 2� \Q,

• ↵�

2 dom(fp

0�) \M

�

,• ↵0

�

2 dom(fp

1�) is a nonzero ordinal such that ↵0

�

↵�

, and• x

�

= (⇢0�

, ⇣0�

) and y�

= (⇢1�

, ⇣1�

) are nodes of level at least � suchthat x

�

2 fp

0�(↵

�

) and y�

2 fp

1�(↵0

�

).

Then there is D 2 F , together with two sequences (p2�

| � 2 Z \ D),(p3

�

| � 2 Z \D) of conditions in Q⇤�

such that

(1) for each � 2 Z \D, p2�

q0�

and p3�

p1�

,(2) for each � 2 Z \ D, p2

�

� M�

and p3�

� M�

are compatible inQ⇤

�

\M�

, and(3) for each � 2 Z \D, x

�

and y�

are separated below � at stagesµ(↵

�

) and µ(↵0�

) by p2�

� µ(↵�

) and p3�

� µ(↵0�

).

Proof. Let B ✓ V

code ', (Q⇤↵

)↵2(�+1)\Q, the collection of maximal

antichains of Q⇤↵

, for ↵ 2 � \ Q, and (T⇠↵

)↵2X\�\Q. By a reflection

argument with a suitable ⇧11 sentence over the structure (V

,2, B),together with the fact that Q⇤

↵

has the -c.c. for every ↵ 2 �\Q, thereis a set D 2 F consisting of inaccessible cardinals � < for which M

�


is a model such that M�

\ = �, M�

is closed under <�-sequences,and such that for every ↵ 2 M

�

\ �,(1) Q⇤

µ(↵) \M�

forces, over V , that T⇠µ(↵) \M�

has no �-branches,(2) Q⇤

↵

\M�

has the �-c.c., and(3) Q⇤

↵

\M�

lQ⇤↵

Fix � 2 Z \D. Thanks to Lemma 2.5, it su�ces to show that thereare extensions p2

�

and p3�

of p0�

� ↵�

and p1�

� ↵�

, respectively, such thatp2�

� M�

and p3�

� M�

are compatible in Q⇤↵�

\ M�

, and such that x�

and y�

are separated below � at stages µ(↵�

) and µ(↵0�

) by p2�

� µ(↵�

)and p3

�

� µ(↵0�

). By (3) we may view Q⇤↵�

as a two-step forcing iteration(Q⇤

↵�\M

�

) ⇤ S⇠. By �-compatibility we may then identify p0�

� ↵�

andp1�

� ↵�

with, respectively, hr0, s⇠0i and hr1, s⇠

1i, both in (Q⇤↵�\M

�

)⇤ S⇠.Note that r0 and r1 are compatible in Q⇤

↵�\ M

�

. Working in an(Q⇤

↵�\M

�

)-generic extension V [G] of V containing r0 and r1, we notethat there have to be

• extensions hr00, s⇠00i and hr01, s⇠

01i of hr0, s⇠0i and

• an extension hr3, s⇠3i of hr1, s⇠

1isuch that r00, r01 and r3 are all in G, together with some ⇢ < � forwhich there is a pair ⇣00 6= ⇣01 of ordinals in !1 and there is ⇣3 2 !1

such that, identifying T⇠µ(↵�) and T⇠µ(↵0�)

with (Q⇤↵�

\M�

) ⇤ S⇠-names,18

we have the following.

• hr00, s⇠00i extends a condition in A

µ(↵�)x�,⇢

forcing that (⇢, ⇣00) isbelow x

�

in T⇠µ(↵�).


µ(↵�)x�,⇢

forcing that (⇢, ⇣01) isbelow x

�

in T⇠µ(↵�).


µ(↵0�)

y�,⇢forcing that (⇢, ⇣3) is be-

low y�

in T⇠µ(↵0�).

Indeed, any condition hr, s⇠i in (Q⇤↵�

\M�

) ⇤ S⇠ such that r 2 G canbe extended, for any ⇢ < �, to a condition hr+, s⇠

+i such that

• hr+, s⇠+i is stronger than some condition in A

µ(↵�)x�,⇢

decidingsome node (⇢, ⇣) to be below x

�

in T⇠µ(↵�), and• r+ 2 G,

and similarly with y�

and T⇠µ(↵0�)

in place of x�

and T⇠µ(↵�). Hence,if the above were to fail, then there would be some ⇢⇤ < � with thefollowing property.

18T⇠µ(↵0�)

is of course a Q⇤↵�

-name since Q⇤↵0

�✓ Q⇤

↵�, so this identification makes

sense.


• For every ⇢ < � above ⇢⇤ there is exactly one ⇣ < !1 for whichthere is some condition hr, s⇠i in (Q⇤

↵�\ M

�

) ⇤ S⇠ with r 2 G

and such that hr, s⇠i extends a condition in Aµ(↵�)x�,⇢

forcing that(⇢, ⇣) is below x

�

in T⇠µ(↵�).

It would then follow that T⇠µ(↵�) has a �-branch in V [G], which con-tradicts (1).

Let ⇣3 < !1 be such that some condition hr3, s⇠3i extending hr1, s⇠

1iis such that

• hr3, s⇠3i � µ(↵0

�

) extends a condition in Aµ(↵0

�)y�,⇢

forcing (⇢, ⇣3) tobe below y

�

in T⇠µ(↵0�), and

• r3 2 G

But now, given conditions hr0i, s⇠0ii as above (for i 2 {0, 1}) there

must be i 2 {0, 1} such that ⇣0i 6= ⇣3. We may then set p2�

= hr0i, s⇠0ii

and p3�

= hr3, s⇠3i. ⇤

By Claim 3.6, in order to conclude the proof of the current instanceof (1)

�

, it su�ces to prove the following.

Claim 3.10. There is a separating pair for ~�0 and ~�1.

Proof. This follows from first applying Claim 3.9 and (2)↵

, for ↵ < �,countably many times, using the normality of F , and then running apressing–down argument again using the normality of F .

To be more specific, we start by building sequences

~�0,n = (q0�,n

| � 2 X \Dn

)

and

~�1,n = (q1�,n

| � 2 X \Dn

),

for a ✓-decreasing sequence (Dn

)n<!

of sets in F , such that ~�0,0 = ~�0and ~�1,0 = ~�1, and such that for every n < !, ~�0,n+1 and ~�1,n+1 areobtained from ~�0,n and ~�1,n in the following way.

We first let

~�0,n,+ = (q0�,n,+ | � 2 X \D

n

)

and

~�1,n,+ = (q1�,n,+ | � 2 X \D

n

)

be sequences of Q⇤�

-conditions, �-compatible with respect to ' and ↵,for ↵ 2 � \ Q, such that q0

�,n,+ Q�q0�,n

and q0�,n,+ Q�

q1�,n

for all� 2 X \ D

n

. Given �, q0�,n,+ and q0

�,n,+ can be found by a simpleconstruction in countably many steps, along which we apply (2)

↵

, forsome ↵ 2 � \Q.


Now we find Dn+1 and ~�0,n+1, ~�1,n+1 by an application of Claim

3.9 to ~�0,n,+ and ~�1,n,+ with a suitable sequence ↵�

, ↵0�

, x�

, y�

(for� 2 X \D

n

).By a standard book-keeping argument we can ensure that all relevant

objects have been chosen in such a way that in the end, letting r0�

andr1�

be the greatest lower bound of, respectively, (q0�,n

)n<!

and (q1�,n

)n<!

,for � 2 X \

Tn

Dn

, (r0�

| � 2 X \T

n

Dn

) and (r1�

| � 2 X \T

n

Dn

)satisfy clause (2) in Definition 3.5.

Finally, by a standard pressing–down argument using the normalityof F , we may find Y 2 S, Y ✓ X\

Tn

Dn

, such that ~�⇤0 = (r0

�

| � 2 Y )and ~�⇤

1 = (r1�

| � 2 Y ) satisfy clauses (3) and (4) in Definition 3.5. ⇤

We are left with proving (2)�

. This is established with a similarargument as in the corresponding proof in [10]. Suppose D 2 F , Q isa suitable model such that �, D 2 Q, ' : ! Q is a bijection, and(q0

�

| � 2 D) 2 Q and (q1�

| � 2 D) 2 Q are sequences of Q�

-conditions.By shrinking D if necessary we may assume that M

�

\ = � for each� 2 D. It su�ces to show that there is some D0 2 F , D0 ✓ D, withthe property that for every � 2 D0, if q0

0�

�

q0�

and q10

�

�

q1�

are suchthat q0

0�

� M�

2 Q�

and q00

�

� M�

= q10

�

� M�

, then there is a conditionr�

Q�q0

0�

� M�

such that every condition in Q�

\M�

extending r�

iscompatible with both q0

0�

and q10

�

.The case when � is a limit ordinal follows from the induction hypo-

thesis, using the normality of F (cf. the proof in [10]). Specifically,we fix an increasing sequence (�

i

)i<cf(�) of ordinals in Q converging to

�. If cf(�) = , we take each �i

to be sup(M�

\ �) for some � 2 D.For each i < cf(�) we fix some D

i

2 F , Di

✓ D, witnessing (2)�i for

(q0�

� �i

| � 2 D) and (q1�

� �i

| � 2 D). We make sure that (Di

)i<cf(�)

is ✓-decreasing. If cf(�) < , then D0 =T

i<cf(�) Di

will witness (2)�

for (q0�

| � 2 D) and (q1�

| � 2 D), and if cf(�) = , D0 = �i<

Di

willwitness (2)

�

for these objects. This can be easily shown, using the factthat each M

�

is closed under !-sequences in the case when cf(�) = !.To see this, suppose � 2 D0, q0

0�

Q�q0�

, q10

�

Q�q1�

, q00

�

� M�

2 Q�

,and q0

0�

� M�

= q10

�

� M�

. Suppose first that cf(�) > !. In this case, wepick any i 2 cf(�)\M

�

such that �i

is above (dom(fq

00�)[dom(f

q

10�))\

sup(M�

\ �) and find a condition r 2 M�

\Q�i with the property that

every condition in Q�i\M�

is compatible with both q00

�

� �i

and q10

�

� �i

.Let r

�

be any condition in Q�

\M�

extending r and q00

�

� M�

. It thenfollows that every condition in Q

�

\M�

extending r�

is compatible withboth q0

0�

and q10

�

.


Now suppose � has countable cofinality. Since � 2 M�

, we may buildsequences (q0,i

�

| i < !), (q1,i�

| i < !) and (ri�

| i < !) such that foreach i,

(1) q0,i�

and q1,i�

are conditions in Q�i extending q0

0�

� �i

and q10

�

� �i

,respectively,

(2) ri�

2 Q�i \M

�

,(3) every condition in Q

�i \ M�

extending ri�

is compatible withboth q0,i

�

and q1,i�

,(4) q0,i+1

�

� �i

extends q0,i�

and ri�

,(5) q1,i+1

�

� �i

extends q1,i�

and ri�

, and(6) ri+1

�

� �i

extends ri�

.

Let r�

2 Q�

be the greatest lower bound of {ri�

| i < !}, and note thatr�

2 M�

since M�

is closed under sequences of length !. But now it isstraightforward to verify that every condition in Q

�

\M�

extending r�

is compatible with q00

�

and q10

�

.It remains to consider the case that � is a successor ordinal, � =

�0 + 1. Assuming the desired conclusion fails, there is some X 2 S,X ✓ D, together with sequences (q0

0�

| � 2 X) and (q00

�

| � 2 X) ofconditions in Q

�

such that for every � 2 X,

• q00

�

extends q0�

and q10

�

extends q1�

,• q0

0�

� M�

2 Q�

,• q0

0�

� M�

= q10

�

� M�

, and• for every condition r in Q

�

\ M�

extending q00

�

� M�

there isa condition in Q

�

\M�

extending r and incompatible with atleast one of q0

0�

, q10

�

.

Thanks to the induction hypothesis applied to �0 and to the factthat (1)

�

holds we may assume, after shrinking X to some Y 2 Sand extending the corresponding conditions if necessary, that for each� 2 Y ,

• q00

�

� �0 and q10

�

� �0 are both �-compatible with respect to 'and �0, and

• q00

�

� q10

�

⇤ is a condition for each �⇤ 2 Y , �⇤ > �.

By our assumption above we may then assume, after shrinking Yif necessary, that for each � 2 Y there is a maximal antichain A

�

ofQ

�

\ M�

below q0�

� M�

consisting of conditions r such that at leastone of the following statements holds.

✓r,0,�: r is incompatible with q0

0�

.✓r,1,�: r is incompatible with q1

0�

.


By the definition of F coupled with a suitable ⇧11-reflection argu-

ment, we may further assume that each A�

is in fact a maximal an-tichain of Q

�

below q00

�

� M�

and that it has cardinality less than �(cf. the proof of Claim 3.9). Hence, after shrinking Y yet another timeusing the normality of F , we may assume, for all � < �⇤ in Y , that

• A�

= A�

⇤ and that• for every r 2 A

�

, ✓r,0,� holds if and only if ✓

r,0,�⇤ does, and ✓r,1,�

holds if and only if ✓r,1,�⇤ does.

Let us now fix any � < �⇤ in Y . Since A�

is a maximal antichain ofQ

�

below q00

�

� M�

, we may find some r 2 A�

compatible with q00

�

�q10

�

⇤ .We have that ✓

r,0,� cannot hold since q00

�

� q10

�

⇤ extends q0�

. Therefore✓r,1,� holds, and hence also ✓

r,1,�⇤ does. But that is also a contradictionsince q0

0�

� q10

�

⇤ extends q10

�

⇤ .This contradiction concludes the proof of (2)

�

, and hence the proofof the lemma. ⇤

4. Completing the proof of Theorem 1.2

In this final section we conclude the proof of Theorem 1.2. By Lemma2.7, Q

+ does not add new !-sequences of ordinals and hence it pre-serves CH. We will start this section by proving that Q

+ also preserves2@1 = @2.

Lemma 4.1. �Q+2@1 =

Proof. Suppose, towards a contradiction, that there is a condition q 2Q

+ and a sequence ( r⇠i

)i<

+ of Q

+-names for subsets of !1 such that

q �Q+r⇠i

6= r⇠i

0 for all i < i0 < +

By Lemma 3.1 we may assume, for each i, that r⇠i

2 H(+) and r⇠i

isa Q

�i-name for some �i

< +.Let ✓ be a large enough regular cardinal. For each i < + let N⇤

i

�H(✓) be such that

(1) |N⇤i

| = |N⇤i

\ |,(2) N⇤

i

is closed under sequences of length less than |N⇤i

|,(3) q, r⇠i

, �i

, (�↵

)↵<

+ , (Q↵

)↵<

+ 2 N⇤i

, and(4) Q

↵

\N⇤i

lQ↵

for every ↵ 2 + \N⇤i

.

N⇤i

can be found by a ⇧11-reflection argument, using the weak compact-

ness of and the -chain condition of each Q↵

, as in the proof of Claim3.9. Let N

i

= N⇤i

\H(+) for each i.Let now P be the satisfaction predicate for the structure

hH(+),2, ~�i,


where ~� ✓ H(+) codes (�↵

)↵<

+ in some canonical way, and let M bean elementary submodel of H(✓) containing q, r⇠i

, (�i

)i<

+ , (Q↵

)↵

+ ,(N⇤

i

)i<

+ and P , and such that |M | = and <M ✓ M .Let i0 2 + \ M . By a standard reflection argument we may find

i1 2 + \M for which there exists an isomorphism

: (Ni0 ,2, P, r⇠i0 , �i0 , q) ⇠= (N

i1 ,2, P, r⇠i1 , �i1 , q),

such that (⇠) ⇠ for every ordinal in Ni0 . Indeed, the existence of

such an i1 follows from the correctness of M in H(✓) about a suitablestatement with parameters (N

i

)i<

+ , q, P , (�i

)i<

+ , ( r⇠i

)i<

+ , Ni0 \M ,

and the isomorphism type of the structure (Ni0 ,2, P, r⇠i0 , �i0 , q), all of

which are in M . Let q = (fq

, ⌧q

), where

⌧q

= ⌧q

[ {h(Ni0 , �i0), (Ni1 , �i1)i}

Thanks to the choice of N⇤i0and N⇤

i1, together with Lemma 2.14, it

follows that q 2 Q

+ . We show that q �Q+r⇠i0 = r⇠i1 .

Suppose not, and we will derive a contradiction. Thus we can find⌫ < !1 and q0

+ q such that

q0 �Q+“⌫ 2 r⇠i0 () ⌫ /2 r⇠i1”.

Let us assume, for concreteness, that q0 �Q+“⌫ 2 r⇠i0 and ⌫ /2 r⇠i1”

(the proof in the case that q0 �Q+“⌫ 2 r⇠i1 and ⌫ /2 r⇠i0” is exactly

the same). By correctness of N⇤i0

we have that this model containsa maximal antichain A of conditions in Q

�i0deciding the statement

“⌫ 2 r⇠i0”. By Lemma 3.1 we know that |A| < and hence, sinceN⇤

i0\ 2 , A ✓ N⇤

i0\ H(+) = N

i0 (cf. the proof of Lemma 2.9).Hence, we may find a common extension q00 of q0 and some r 2 N

i0 \Asuch that r �Q+

“⌫ 2 r⇠i0”.Also, note that, since is an isomorphism between the structures

(Ni0 ,2, P, r⇠i0 , �i0 , q) and (N

i1 ,2, P, r⇠i1 , �i1 , q), and by the choice of P ,we have that

(r) �Q�i1“⌫ 2 ( r⇠i0) = r⇠i1 ”

But then, by clauses (5) and (6) in the definition of condition, togetherwith the fact that (⇠) ⇠ for every ordinal ⇠ 2 N

i0 , we have thatq00 (r). We thus obtain that q00 �Q+

“⌫ 2 r⇠i1”, which is impossibleas q0 �Q+

“⌫ /2 r⇠i1” and q00 q0.We get a contradiction and the lemma follows.19 ⇤

Corollary 4.2. Q

+ forces GCH.

19Note the resemblance of this proof with the proof of Lemma 2.9.


Lemma 4.3, which completes the proof of Theorem 1.2, follows im-mediately from earlier lemmas, together with a standard density argu-ment.

Lemma 4.3. Q

+ forces SATP@2.

Proof. Let G be Q

+-generic over V . Since CH holds in V [G], thereare @2-Aronszajn trees there. Hence, it su�ces to prove that, in V [G],every @2-Aronszajn tree is special.

Let T 2 V [G] be an @2-Aronszajn tree. Note that @2 = in V [G] byLemmas 2.3 and 3.1. We need to prove that T is special in V [G]. Let usgo down to V and let us note there that, by the -chain condition ofQ

+

together with the choice of �, we may find some nonzero ↵ 2 X suchthat �(↵) is a Q

↵

-name for an @2-Aronszajn tree such that �(↵)G

= T .We then have that T⇠↵

= �(↵).For every ⌫ < !1, let A⌫

=S{f

q

(↵+ ⌫) | q 2 G}. By the definitionof the forcing, we have that A

⌫

is an antichain of T . Also, given anycondition q 2 Q

+ and any node x 2 ⇥!1 such that x /2 fq

(↵+⌫) forany ⌫ < !1, it is easy to see that we may extend q to a condition q⇤ suchthat x 2 f

q

⇤(↵+ ⌫) for some ⌫ < !1; indeed, it su�ces for this to pickany ⌫ < !1 such that ↵0+ ⌫ /2 dom(f

q

) for any ↵0 2 X , which of courseis possible since dom(f

q

) is countable, extend fq

to a function f suchthat ↵+ ⌫ 2 dom(f) and f(↵+ ⌫) = {x}, and close under the relevant(restrictions of) functions

N0,N1 for edges h(N0, �0), (N1, �1)i 2 ⌧q

.20

The above density argument shows that every node in T is in some A⌫

.It follows that T is special in V [G], which concludes the proof. ⇤Acknowledgements. The first author acknowledges support of EP-SRC Grant EP/N032160/1. The second author’s research has beensupported by a grant from IPM (No. 97030417). Much of the work onthis paper was done while the first author was visiting the Institute forResearch in Fundamental Sciences (IPM), Tehran, in December 2017.He thanks the institute for their warm hospitality. We also thank As-saf Rinot for a comprehensive list of historical remarks on GCH vs.SATP@2 ; we have only included the remarks pertaining to the earlyhistory of the problem, leading to the Laver–Shelah result from [10].

References

[1] Aspero, David; Mota, Miguel Angel, Forcing consequences of PFA togetherwith the continuum large, Trans. Amer. Math. Soc., vol. 367 (2015), no. 9,6103–6129.

20In other words, we may take q⇤ = q�q0, for q0 = (f, ;), where dom(f) = {↵+⌫}and fq(↵+ ⌫) = {x}.


[2] Aspero, David; Mota, Miguel Angel, Few new reals. Submitted.[3] Baumgartner, J.; Malitz, J.; Reinhardt, W., Embedding trees in the rationals,

Proc. Nat. Acad. Sci. U.S.A., vol. 67 (1970), 1748–1753.[4] Devlin, Keith J.; Johnsbraten, Havard, The Souslin problem, Lecture Notes in

Mathematics, vol. 405, Springer–Verlag, Berlin, New York, 1974. viii+132 pp.[5] Golshani, Mohammad; Hayut, Yair, The Special Aronszajn tree property. To

appear in J. Math. Logic.[6] Gregory, John, Higher Souslin trees and the generalized continuum hypothesis,

J. Symbolic Logic, vol. 41 (1976), no. 3, 663–671.[7] Jensen, Ronald, The fine structure of the constructible hierarchy. With a sec-

tion by Jack Silver, Ann. Math. Logic, vol. 4 (1972), 229–308; erratum, ibid.4 (1972), 443.

[8] A. Kanamori; M. Magidor, The evolution of large cardinal axioms in set theory,in Higher set theory (Proc. Conf. Math. Forschungsinst., Oberwolfach, 1977),Lecture Notes in Mathematics, 669, Springer, 9–275.

[9] Kurepa, Djuro, Ensembles ordonnees et ramifies. These, Paris. Published as:Publications mathematiques de l’Universite de Belgrade, vol. 4 (1935). 1–138.

[10] Laver, Richard; Shelah, Saharon, The @2-Souslin hypothesis, Trans. Amer.Math. Soc., vol. 264 (1981), no. 2, 411–417.

[11] Rinot, Assaf, Jensen’s diamond principle and its relatives, in Set theory and itsapplications, 125–156, Contemp. Math., 533, Amer. Math. Soc., Providence,RI, 2011.

[12] Rinot, Assaf, Higher Souslin trees and the GCH, revisited, Adv. Math., vol.311 (2017), 510–531.

[13] Shelah, Saharon, Proper and improper forcing. Second edition, Perspectives inMathematical Logic, Springer–Verlag, Berlin, 1998. xlviii+1020 pp.

[14] Shelah, Saharon, On what I do not understand (and have something to say). I,in Saharon Shelah’s anniversary issue, Fund. Math. 166 (2000), no. 1–2, 1–82.

[15] Shelah, Saharon; Stanley, Lee, Weakly compact cardinals and nonspecial Aron-szajn trees, Proc. of the Amer. Math. Soc., vol. 104 (1988), no. 3, 887–897.

[16] R. Solovay; S. Tennenbaum, Iterated Cohen extensions and Souslins problem,Ann. of Math. (2) 94 (1971), 201-245.

[17] E. Specker, Sur un probleme de Sikorski, Colloquium Math. 2 (1949), 9–12.[18] S. Todorcevic, Trees and linearly ordered sets, in Handbook of set-theoretic

topology, pp. 235–293, North-Holland, Amsterdam, 1984.[19] S. Todorcevic, Partitioning pairs of countable ordinals, Acta Math., vol. 159(3-

4):261–294, 1987.

David Aspero, School of Mathematics, University of East Anglia,Norwich NR4 7TJ, UK

E-mail address: [email protected]

Mohammad Golshani, School of Mathematics, Institute for Researchin Fundamental Sciences (IPM), P.O. Box: 19395–5746, Tehran, Iran.

E-mail address: [email protected]

the special aronszajn tree property at 2 and gchbfe12ncu/special-aleph-2-a-trees-gch.pdf ·...

Documents