Saharon Shelah- Strong Partition Realations Below the Power Set: Consistency Was Sierpinski Right? II

8/3/2019 Saharon Shelah- Strong Partition Realations Below the Power Set: Consistency Was Sierpinski Right? II

http://slidepdf.com/reader/full/saharon-shelah-strong-partition-realations-below-the-power-set-consistency 1/32

288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

C O L L OQ U I A M AT H E M AT I C A S O C I ETAT I S J A N OS B O LYA I

60. SETS, GRAPHS AND NUMBERS, BUDAPEST (HUNGARY), 1991

Strong Partition Realations Below thePower Set: Consistency

Was Sierpinski Right? II.

S. SHELAH∗

We continue here [Sh276] (see the introduction there) but we do notrelay on it. The motivation was a conjecture of Galvin stating that 2 ω ≥ ω2+ ω2 → [ω1 ]nh (n ) is consistent for a suitable h : ω → ω. In section 5we disprove this and give similar negative results. In section 3 we provethe consistency of the conjecture replacing ω2 by 2ω , which is quite large,starting with an Erd˝ os cardinal. In section 1 we present iteration lemmaswhich needs when we replace ω by a larger λ and in section 4 we generalizea theorem of Halpern and Lauchli replacing ω by a larger λ.

0. Preliminaries

Let < ∗χ be a well ordering of H(χ ), where H(χ ) = {x : the transitive closureof x has cardinality < χ }, agreeing with the usual well-ordering of the

∗

Publication no 288, summer 86. The author would like to thank the United States– Israel Binational Science Foundation for partially supporting this research.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

2ordinals. P (and Q, R) will denote forcing notions, i.e. partial orders witha minimal element ∅= ∅P .

A forcing notion P is λ-closed if every increasing sequence of membersof P , of length less than λ, has an upper bound.

If P ∈H(χ ), then for a sequence p = pi : i < γ of members of P letα˜

= α˜ p

def = sup{ j˜

: {β j : j < j˜} has an upper bound in P } and dene the

canonical upper bound of p, &¯ p as follows:

(a) the least upper bound of { pi : i < α˜

} in P if there exists such anelement,

(b) the < ∗χ -rst upper bound of p if (a) can’t be applied but there is such,

(c) p0 if (a) and (b) fail, γ > 0,

(d) ∅P if γ = 0.

Let p0& p1 be the canonical upper bound of p : < 2 .

Take [ a]κ = {b⊆ a : |b| = κ} and [a]<κ = θ<κ [a]θ .For sets of ordinals, A and B , dene H OP

A,B as the maximal orderpreserving bijection between initial segments of A and B , i.e, it is thefunction with domain {α ∈A : otp( α ∩ A) < otp( B )}, and H OP

A,B (α ) = β if and only if α ∈A, β ∈B and otp( α ∩ A) = otp( β ∩ B ).

Denition 0.1 λ →+ (α)<ωµ holds provided whenever F is a function from

[λ]<ω to µ, C ⊆ λ is a club then there is A ⊆ C of order type α such that[w1 , w2 ∈[A]<ω , |w1 | = |w2 | ⇒ F (w1) = F (w2 )].

Denition 0.2 λ → [α ]nκ,θ if for every function F from [λ]

n

to κ there isA ⊆ λ of order type α such that {F (w) : w ∈[A]n } has power ≤ θ.

Denition 0.3 A forcing notion P satises the Knaster condition (has property K) if for any { pi : i < ω 1} ⊂ P there is an uncountable A ⊂ ω1

such that the conditions pi and pj are compatible whenever i, j ∈A.

1. Introduction

Concerning 1.1–1.3 see Shelah [Sh80], Shelah and Stanley [ShSt154, 154a].



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

3Denition 1.1. A forcing notion Q satises ∗ε

µ where ε is a limit ordinal < µ , if player I has a winning strategy in the following game:

Playing: the play nishes after ε moves.in the α th the move:

Player I – if α = 0 he chooses qαζ : ζ < µ + such that qα

ζ ∈Q and(∀β < α )(∀ζ < µ + ) pβ

ζ ≤ qαζ and he chooses a regressive

function f α : µ+ → µ+ (i.e. f α (i) < 1 + i); if α = 0 letqα

ζ = ∅Q , f α = ∅.

Player II – he chooses pαζ : ζ < µ + such that qα

ζ ≤ pαζ ∈Q.

The outcome: Player I wins provided whenever µ < ζ < ξ < µ + , cf(ζ ) =cf(ξ) = µ and ∧β<ε f β (ζ ) = f β (ξ) the set { pα

ζ : α < ε } ∪ { pαξ : α < ε } has

an upper bound in Q.

Denition 1.2. We call P i , Q j : i ≤ i(∗), j < i (∗) a ∗εµ -iteration

provided that:

(a) it is a (< µ )-support iteration ( µ is a regular cardinal)

(b) if i1 < i 2 ≤ i(∗), cf i1 = µ then P i 2 /P i 1 satises ∗εµ .

The Iteration Lemma 1.3. If Q = P i , Qj : i ≤ i(∗), j < i (∗) is a (< µ )-support iteration, (a) or (b) or (c) below hold, then it is a ∗

εµ -iteration.

(a) i(∗) is limit and Q j (∗) is a ∗εµ -iteration for every j (∗) < i (∗).

(b) i(∗) = j (∗) + 1 , Q j (∗) is a ∗εµ -iteration and Q j (∗) satises ∗ε

µ in V P j (∗) .

(c) i(∗) = j (∗) + 1 , cf j (∗) = µ+ , Q j (∗) is a ∗εµ -iteration and for every

successor i < j (∗), P i (∗) /P i satises ∗εµ .

Proof. Left to the reader (after reading [Sh80] or [ShSt154a]).

Theorem 1.4. Suppose µ = µ<µ < χ < λ , and λ is a strongly inaccessible k2

2 -Mahlo cardinal, where k22 is a suitable natural number (see 3.6(2) of

[Sh289]), and assume V = L for the simplicity. Then for some forcing notion P :

(a) P is µ-complete, satises the µ+ -c.c., has cardinality λ , and V P |=”2µ = λ”.

(b) P λ → [µ+ ]23

and even λ → [µ+ ]2κ,2

for κ < µ .

(c) if µ = ℵ0 then “MA χ ”.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

4(d) if µ > ℵ0 then: P “for every forcing notion Q of cardinality ≤ χ , µ-complete satisfying ∗ε

µ , and for any dense sets D i ⊆Q for i < i 0 < λ , there is a directed G ⊆Q, ∧i G ∩ D i = ∅”.

As the proof is very similar to [Sh276], (particularly after reading section

3) we do not give details. We shall dene below just the systems needed tocomplete the proof. More general ones are implicit in [Sh289].

Convention 1.5. We x a one to one function Cd = Cd λ,µ from µ> λ ontoλ.

Remark. Below we could have otp( B x ) = µ+ + 1 with little change.

Denition 1.6. Let µ < χ < κ ≤ λ, λ = λ <µ , χ = χ <µ , µ = µ<µ .1) We call x a (λ,κ,χ,µ )-precandidate if x = a x

u : u ∈I x where for some set Bx (unique, in fact):(i) I x = {s : s ⊆B x , |x | ≤ 2},

(ii) Bx is a subset of κ of order type µ+ ,(iii) ax

u is a subset of λ of cardinality ≤ χ closed under Cd,(iv) ax

u ∩ Bx = u,(v) ax

u ∩ axv ⊆ ax

u ∩v ,(vi) if u, v ∈I x , |u | = |v| then ax

u and axv have the same order type (and

so H OP a x

u ,a xv

maps axu onto ax

v ),(vii) if u , v ∈I x for = 1 , 2, |u1 | = |v1 |, |u2 | = |v2 |, |u1∪u2 | = |v1∪v2 |,

H OP a x

u 1∪a x

u 2,a x

v 1∪a x

v 2maps u onto v for = 1 , 2 then H OP

a xu 1

,a xv 1

andH OP

a xu 2

,a xv 2

are compatible.

2) We say x is a (λ,κ,χ,µ )-candidate if it has the form M xu : u ∈ I xwhere

(α ) (i) |M xu | : u ∈I x is a (λ,κ,χ,µ )-precandidate (with B xdef = ∪I x )

(ii) Lx is a vocabulary with ≤ χ -many < µ -ary placespredicates andfunction symbols,

(iii) each M xu is an Lx -model,(iv) for u, v ∈ I x , |u | = |v|, M xu (|M xu | ∩ |M xv |) is a model, and in fact

an elementary submodel of M xv , M xu and M xu ∩v .(β ) (∗) for u, v ∈I x , |u | = |v|, the function H OP

| M xu | , | M x

v | is an isomorphismfrom M x

uonto M x

v.

3) The set A is a (λ,κ,χ,µ )-system if



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

5(A) each x ∈A is a (λ,κ,χ,µ )-candidate,(B) guessing: if L is as in (2) (α)(ii), M ∗ is an L-model with universe λ

then for some x ∈A , s ∈B x ⇒ M xs M ∗.

Denition 1.7. 1) We call the system A disjoint when:

(∗) if x = y are from A and otp( |M x∅|) ≤ otp( |M y∅|) then for some B1 ⊆Bx ,B2 ⊆By we have a) |B1 | + |B2 | < µ +

b) the sets {|M xs | : s ∈[B x \ B1]≤ 2}

and{|M ys | : s ∈[B y \ B2]≤ 2}

have intersection ⊆M y∅

.

2) We call the system A almost disjoint when:(∗∗) if x, y ∈ A , otp( |M x

∅|) ≤ otp( |M y

∅|) then for some B1 ⊆ Bx ,

B2 ⊆By we have:(a) |B1 | + |B2 | < µ + ,(b) if s ∈[B x \ B1]≤ 2 , t ∈[By \ B2]≤ 2 then |M xs | ∩ |M xt | ⊆ |M y

∅|.

2. Introducing the partition on trees

Denition 2.1. Let1) Per( µ> 2) = {T : where

(a) T ⊆ µ> 2, ∈T,(b) (∀η ∈T ) (∀α < lg(η)) η α ∈T,(c) if η ∈T ∩ α 2, α < β < µ then for some

ν ∈T ∩ β 2, η ν,(d) if η ∈T then for some ν, η ν,

ν 0 ∈T, ν 1 ∈T,(e) if η ∈δ 2, δ < µ is a limit ordinal and

{η α : α < δ } ⊆T then η ∈T.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

6

2) Per f (µ> 2) = T ∈Per( µ> 2) : if α < µ and ν 1 , ν 2 ∈α 2 ∩ T, then

[1

=0

ν 1ˆ ∈T ⇐⇒1

=0

ν 2ˆ ∈T ] .

3) Per u (µ> 2) = {T ∈Per( µ> 2) : if α < µ, ν 1 = ν 2 from α 2 ∩ T,

then1

=0

2

m =1

ν m ˆ /∈T }.

4) For T ∈Per( µ> 2) let lim T = {η ∈µ 2 : (∀α < µ ) η α ∈T }.5) For T ∈ Per f (µ> 2) let clpT : T → µ> 2 be the unique one-to-one

function from sp(T )def = {η ∈ T : ηˆ 0 ∈T, ηˆ 1 ∈T } onto µ> 2,

which preserves and lexicographic order.6) Let SP (T ) = {lg(η) : η ∈sp(T )}, sp(η, ν ) = min {i : η(i) = ν (i) or i =

lg(η) or i = lg( ν )}.

Denition 2.2. 1) For cardinals µ, σ and n < ω and T ∈Per(µ>

2) letColnσ (T ) = {d : d is a function from ∪α<µ [α 2]n ∩T to σ}. We will write

d(ν 0 , . . . , ν n − 1) for d({ν 0 , . . . , ν n − 1}).2) Let < ∗α denote a well ordering of α 2 (in this section it is arbitrary). We

call d ∈Colnσ (T ) end-homogeneous for < ∗α : α < µ provided that: if

α < β are from SP( T ), {ν 0 , . . . , ν n − 1} ⊆ β 2 ∩ T , ν α : < n are pairwise distinct and

,m[ν < ∗β ν m ⇐⇒ ν α < ∗

α ν m α] then

d(ν 0 , . . . , ν n − 1) = d(ν 0 α , . . . , ν n − 1 α).

3) Let EhCol nσ (T ) = {d ∈Coln

σ (T ) : d is end-homogeneous } (for some < ∗α : α < µ ).

4) For ν 0 , . . . , ν n − 1 , η0 , . . . , η n − 1 from µ> 2, we say ν = ν 0 , . . . , ν n − 1 andη = ηo , . . . , ηn − 1 are strongly similar for < ∗α : α < µ if:(i) lg(ν ) = lg( η )

(ii) sp(ν , ν m ) = sp( η , ηm )(iii) if 1 , 2 , 3 , 4 < n and α = sp( ν 1 , ν 2 ) then

ν 3 α < ∗α ν 4 α ⇐⇒ η 3 α < ∗

α η 4 α and ν 3 (α ) = η 3 (α )

5) For ν a0 , . . . , ν an − 1 , ν b0 , . . . , ν bn − 1 from µ> 2 we say ν a = ν a0 , . . . , ν an − 1 and

ν b

= ν b0 , . . . , ν

bn − 1 are similar if the truth values of (i)–(iii) below doe not depend on t ∈ {a, b} for any (1), (2), (3), (4) < n :



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

7(i) lg(ν t (1) ) < lg(ν t (2) )

(ii) sp(ν t (1) , ν t (2) ) < sp(ν t (3) , ν t (4) )

(iii) for α = sp( ν t (1) , ν t (2) ),

ν t(3)

α < ∗

αν t

(4)α

andν t (3) (α ) = 0 .

6) We say d ∈Colnσ (T ) is almost homogeneous [homogeneous] on T 1 ⊆ T

(for < ∗α : α < µ ) if for every α ∈SP( T 1), ν, η ∈[α 2]n ∩ T 1 which are strongly similar [similar] we have d(ν ) = d(η).

7) We say < ∗α : α < µ is nice to T ∈ Per( µ> 2), provided that: if α < β are from SP( T ), (α, β ) ∩ SP( T ) = ∅, η1 = η2 ∈

β 2 ∩ T ,[η1 α < ∗

α η2 α or η1 α = η2 α, η1 (α ) < η 2(α )] then η1 < ∗β η2 .

Denition 2.3. 1) Pr eht (µ,n,σ ) means: for every d ∈Colnσ (µ> 2) for some

T ∈Per( µ> 2), d is end homogeneous on T .2) Pr aht (µ,n,σ ) means for every d ∈Coln

σ (µ> 2) for some T ∈Per( µ> 2), dis almost homogeneous on T .

3) Pr ht (µ,n,σ ) means for every d ∈Colnσ (µ> 2) for some T ∈Per( µ> 2), d

is homogeneous on T .4) For x ∈ {eht, aht, ht }, Pr f

x (µ,n,σ ) is dened like Pr x (µ,n,σ ) but we demand T ∈Per f (µ> 2).

5) If above we replace eht, aht, ht by ehtn, ahtn, htn , respectively, this

means <∗

α : α < µ is xed apriori.6) Replacing n by “ < κ ”, σ by σ = σ : < κ for κ ≤ ℵ0 , means thatdn : n < κ are given, dn ∈Coln

σ (µ> 2) and the conclusion holds for all dn (n < κ ) simultaneously. Replacing “ σ” by “ < σ ” means that the assertion holds for every σ1 < σ .

Denition 2.4. 1) Pr aht (µ,n,σ (1), σ(2)) means: for every d ∈ Colnσ (1)

(µ> 2) for some T ∈Per( µ> 2) and < ∗α : α < µ for every η ∈ {[α 2]n ∩T :α ∈SP( T )},

d(ν ) : ν ∈ {[α 2]n ∩ T 1 : α ∈SP(( T 1 )},

η and ν are strongly similar for < ∗α : α < µ



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

8has cardinality < σ (2) .2) Pr ht (µ,n,σ (1), σ(2)) is dened similarly with “similar” instead of

“strongly similar”.

3) Pr x µ,< κ, σ1 : < κ σ2 : < κ , Pr f x (µ,n,σ (1), σ(2)) , Pr f

x (µ, <

ℵ0 , σ1, σ

2) are dened in the same way.

There are many obvious implications.

Fact 2.5. 1) For every T ∈Per( µ > 2) there is a T 1 ⊆ T , T 1 ∈Per u (µ> 2).2) In dening Pr f

x (µ,n,σ ) we can demand T ⊆ T 0 for any T 0 ∈Per f (µ> 2),similarly for Pr f

x (µ, < κ, σ ).3) The obvious monotonicity holds.

Claim 2.6. 1) Suppose µ is regular, σ ≥ ℵ0 and Pr f eht (µ, n, < σ ). Then

Pr f aht (µ, n, < σ ) holds.

2) If µ is weakly compact and Prf aht (µ, n, < σ ), σ < µ , then Pr

f ht (µ, n, < σ )holds.

3) If µ is Ramsey and Pr f aht (µ, < ℵ0 , < σ ), σ < µ , then Pr f

ht (µ, < ℵ0 , < σ ).4) If µ = ω, in the “nice” version, the orders < ∗α : α < µ disappear.

Proof. : Check it.The following theorem is a quite strong positive result for µ = ω.

Halpern Lauchli proved 2.7(1), Laver proved 2.7(2) (and hence (3)), Pincuspointed out that Halpern Lauchli’s proof can be modied to get 2.7(2), andthen Pr f

eht (ω,n,< σ ) and (by it) Pr f ht (ω,n,< σ ) are easy.

Theorem 2.7. 1) If d ∈Colnσ (ω> 2), σ < ℵ0 , then there are T 0 , . . . , T n − 1 ∈Per f (ω> 2) and k0 < k 1 < . . . < k < . . . and s < σ such that for every

< ω : if µ0 ∈ T 0 , µ1 ∈ T 1 , . . . , ν n − 1 ∈ T n − 1 ,m<n

lg(ν m ) = k , then

d(ν 0 , . . . , ν n − 1) = s.2) We can demand in (1) that

SP( T ) = {k0 , k1 , . . . }

3) Pr f htn (ω,n,σ ) for σ < ℵ0 .

4) Prf

htn ω, < ℵ0 , σ1n : n < ω , σ

2n : n < ω if σ

1n < ℵ0 and σ

2n : n < ωdiverge to innity.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

9Denition 2.8. Let d be a function with domain ⊇ [A]n , A be a set of ordinals, F be a one-to-one function from A to α (∗) 2, < ∗α be a well ordering of α 2 for α ≤ α(∗) such that F (α) < ∗α F (β ) ⇐⇒ α < β , and σ be a cardinal.1) We say d is (F, σ )-canonical on A if for any α 1 < · · · < α n ∈A,

d(β 1 , . . . , β n ) : F (β 1), . . . , F (β n ) similar to

F (α 1), . . . , F (α n ) ≤ σ.

2) We dene “almost (F, σ )-canonical” similarly using strongly similar instead of “similar”.

3. Consistency of a strong partition below the continuum

This section is dedicated to the proof of

Theorem 3.1. Suppose λ is the rst Erd˝ os cardinal, i.e. the rst such thatλ → (ω1)<ω

2 . Then, if A is a Cohen subset of λ , in V [A] for some ℵ1–c.c.forcing notion P of cardinality λ, P “MAℵ1 (Knaster ) + 2 ℵ0 = λ” and:1.) P “ λ → [ℵ1]n

h (n ) ” for suitable h : ω → ω (explicitly dened below).

2.) In V P for any colorings dn of λ, where dn is n-place, and for any diver-gent σn : n < ω (see below), there is a W ⊆ λ, |W | = ℵ1 and a functionF : W → ω 2 such that: dn is (F, σ n ) − canonical on W for each n.(See denition 2.8 above.)

Remark 3.2. h(n) is n! times the number of u ∈ [ω 2]n satisfying (if η1 , η2 , η3 , η4 ∈ u are distinct then sp(η1 , η2), sp(η3 , η4) are distinct) up tostrong similarity for any nice < ∗α : α < ω .

2) A sequence σn : n < ω is divergent if ∀m ∃k ∀n ≥ k σn ≥ m.

Notation 3.3. For a sequence a = α i , e∗i : i < α , we call b⊆ α closed if (i) i ∈b⇒ a i ⊆ b

(ii) if i < α, e ∗i = 1 and sup( b ∩ i) = i then i ∈b.

Denition 3.4. LetK

be the family of ¯Q = P i , Q j , a j , e

∗

j : j < α, i ≤ αsuch that



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

10(a) a i ⊆ i, |a i | ≤ ℵ1 ,(b) a i is closed for a j , e∗j : j < i , e∗i ∈ {0, 1}, and [e∗i = 1 ⇒ cf i = ℵ1](c) P i is a forcing notion, Q

˜ jis a P j -name of a forcing notion of power ℵ1

with minimal element ∅or ∅j and for simplicity the underlying set of Q˜ j

is ⊆ [ω1]< ℵ0 (we do not lose by this).

(d) P β = { p : p is a function whose domain is a nite subset of β and for i ∈ dom( p), P i “ f (i) ∈Q

˜ i”} with the order p ≤ q if and only if for

i ∈dom( p), q i P i “ p(i) ≤ q(i)”.(e) for j < i , Q

˜ jis a P j -name involving only antichains contained in

{ p∈P j : dom( p) ⊆ a j }.

For p ∈P i , j < i , j ∈dom p we let p( j ) = ∅. Note for p ∈P i , j ≤ i, p j ∈P j

Denition 3.5. For Q ∈K as above (so α = lg( Q)):

1) for any b ⊆ β ≤ α closed for a i , e∗

i : i < β we dene P cnb [by

simultaneous induction on β ]:

P cnb = { p∈P β : dom p⊆ b, and for i ∈dom p, p(i) is a canonical name }

i.e., for any x, { p∈P cna i : p P i “ p(i) = x” or p P i “ p(i) = x” } is a predense

subset of P i .2) For Q as above, α = lg( Q), take Q β = P i , Q

˜ j, a j : i ≤ β, j < β for

β ≤ α and the order is the order in P α (if β ≥ α, Q β = Q).

3) “b closed for Q means “ b closed for a i , e∗i

: i < lg Q ”.

Fact 3.6. 1) if Q ∈K then Q β ∈K .2) Suppose b⊆ c⊆ β ≤ lg(θ), b and c are closed for Q ∈K .(i) If p ∈P cn

c then p b∈P cnb .

(ii) If p, q ∈P cnc and p ≤ q then p b ≤ q c.

(iii) P cnc ◦P β . 3) lg Q is closed for Q.

4) if Q ∈K , α = lg Q then P cnα is a dense subset of P α .

5) If b is closed for Q, p, q ∈P cnlg Q , p ≤ q in P lg Q and i ∈dom p then q a i P i

“ p(i) ≤ q(i)” hence P cna i

“ p(i) ≤ Q i q(i)”.

Denition 3.7. Suppose W = ( W, ≤ ) is a nite partial order and¯Q ∈

K .1) IN W (Q) is the set of b-s satisfying (α)– (γ ) below:



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

11(α ) b = bw : w ∈W is an indexed set of Q-closed subsets of lg(Q),(β ) W |= “ w1 ≤ w2” ⇒ bw 1 ⊆ bw 2 ,(γ ) ζ ∈bw 1 ∩bw 2 , w1 ≤ w, w2 ≤ w then (∃u ∈W )ζ ∈bu ∧u ≤ w1∧u ≤ w2 .

We assume b codes (W, ≤ ).2) For b∈IN W (Q), let

Q[b]def = { pw : w ∈W : pw ∈P cn

bw , [W |= w1 ≤ w2 ⇒ pw 2 bw 1 = pw 1 ]}

with ordering Q[b] |= p1 ≤ p2 iff w∈W p1w ≤ p2

w .3) Let K 1 be the family of Q ∈K such that for every β ≤ lg(Q) and (Q β )-closed b, P β and P β /P cn

b satisfy the Knaster condition.

Fact 3.8. Suppose Q ∈K 1 , (W, ≤ ) is a nite partial order, b ∈ IN W (Q)and p∈Q[b].1) If w ∈W , pw ≤ q∈P cn

bwthen there is r ∈Q[b], q ≤ r w , p ≤ r , in fact

r u (γ ) =

pu (γ ) if γ ∈Dom pu \ Dom q pu (γ ) & q(γ ) if γ ∈bu ∩ Dom q and for some v ∈W ,

v ≤ u, v ≤ w and γ ∈bv

pu (γ ) if γ ∈bu ∩ dom q but the previous case fails

2) Suppose (W 1 , ≤ ) is a submodel of (W 2 , ≤ ), both nite partial orders,bl∈IN W l (Q), b1

w = b2w for w ∈W 1 .

(α ) If q∈Q[b2 ] then qw : w ∈W 1 ∈Q[b1].(β ) If p∈Q[b1] then there is q ∈Q[b2], q W 1 = p, in fact qw (γ ) is pu (γ ) if u ∈W 1 , γ ∈bu , u ≤ w, provided that(∗∗) if w1 , w2 ∈W 1 , w ∈W 2 , w1 ≤ w, w2 ≤ w and ζ ∈bw 1 ∩ bw 2 then for some v ∈W 1 , ζ ∈bv , v ≤ w1 , v ≤ w2 .(this guarantees that if there are several u’s as above we shall get the same value).3) If Q ∈K 1 then Q[b] satises the Knaster condition. If ∅is the minimal element of W (i.e. u ∈W ⇒ W |= ∅ ≤u) then Q[b]/P cn

b∅ also satises the Knaster condition and so ◦Q[b], when we identify p ∈P cn

b with p : w ∈W .

Proof. 1) It is easy to check that each r u (γ ) is in P cnbu

. So, in order to prover ∈Q[b], we assume W |= u1 ≤ u2 and has to prove that r u 2 bu 1 = r u 1 . Letζ ∈bu

1.

First case: ζ ∈Dom( pu 1 ) ∪Dom q.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

12So ζ ∈ Dom( r u 1 ) (by the denition of r u 1 ) and ζ ∈ Dom pu 2 (as

p ∈ Q[b]) hence ζ ∈ (Dom pu 2 ) ∪ (Dom q) hence ζ ∈ Dom( r u 2 ) by thechoice of r u 2 , so we have nished.

Second case: ζ ∈Dom pu 1 \ Dom q.As ¯ p∈Q[b] we have pu 1 (ζ ) = pu 2 (ζ ), and by their denition, r u 1 (ζ ) =

pu 1 (ζ ), r u 2 (ζ ) = pu 2 (ζ ).Third case: ζ ∈Dom q and (∃v ∈W ) (ζ ∈bv ∧v ≤ u1 ∧v ≤ w). By the

denition of r u 1 (ζ ), we have r u 1 (ζ ) = pu 1 (ζ )&q(ζ ), also the same v witnessesr u 2 (ζ ) = pu 2 (ζ )&q(ζ ), (as ζ ∈bv∧v ≤ u1∧v ≤ w ⇒ ζ ∈bv∧v ≤ u2∧v ≤ w)and of course pu 1 (ζ ) = pu 2 (ζ ) (as p∈Q[b]).

Fourth case: ζ ∈Dom q and ¬(∃v ∈W ) (ζ ∈bv ∧v ≤ u1 ∧v ≤ w).By the denition of r u 1 (ζ ) we have r u 1 (ζ ) = pu 1 (ζ ). It is enough to prove

that r u 2 (ζ ) = pu 2 (ζ ) as we know that pu 1 (ζ ) = pu 2 (ζ ) (because p ∈ Q[b],u1 ≤ u2). If not, then for some v0 ∈W , ζ ∈bv0 ∧ v0 ≤ u2 ∧ v0 ≤ w. Butb∈INW (Q), hence (see Def. 3.7(1) condition ( γ ) applied with ζ, w1 , w2 , wthere standing for ζ, v0 , u1 , u2 here) we know that for some v ∈W , ζ ∈v ∧ v ≤ v0 ∧ v ≤ u1 . As (W, ≤ ) is a partial order, v ≤ v0 and v0 ≤ w, wecan conclude v ≤ w. So v contradicts our being in the fourth case. So wehave nished the fourth case.

Hence we have nished proving r ∈Q[b]. We also have to prove q ≤ r w ,but for ζ ∈ Dom q we have ζ ∈ bw (as q ∈ P cn

w is on assumption) andr w (ζ ) = q(ζ ) because r w (ζ ) is dened by the second case of the denitionas (∃v ∈W ) (ζ ∈bw ∧v ≤ w ∧v ≤ w), i.e. v = w.

Lastly we have to prove that p ≤ r (in Q[b]). So let u ∈W , ζ ∈Dom pu

and we have to prove r u ζ P ζ “ pu (ζ ) ≤ P ζ r u (ζ )”. As r u (ζ ) is pu (ζ ) or

pu (ζ )&q(ζ ) this is obvoius.

2) Immediate.3) We prove this by induction on |W |.For |W | = 0 this is totally trivial.For |W | = 1 , 2 this is assumed.For |W | > 2 x ¯ pi

∈Q[b] for i < ω 1 . Choose a maximal element v ∈W andlet c = {bw : W |= w < v }. Clearly c is closed for Q.

We know that P cnc , P cn

bv/P cn

c are Knaster by the induction hypothesis.We also know that pi

v c∈P cnc for i < ω 1 , hence for some r ∈P cn

c ,

r ” A˜

def = i < ω 1 : piv c∈G

˜ P cnc

is uncountable”



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

13hence

”there is an uncountable A1⊆A

˜such that

i, j ∈A1⇒ pi

v , pjv are compatible in P cn

bv/ G

˜ P cnc

.

Fix a P cnc -name A

˜1 for such an A1 .

Let A2 = i < ω 1 : ∃q∈P cnc , q i ∈A

˜1 . Necessarily

|A2 | = ℵ1 , and for i ∈ A2 there is qi∈ P cn

c , qi i ∈ A1 , andw.l.o.g. pi

v c ≤ qi . Note that piv &qi

∈P cnc .

For i ∈A2 let, r i be dened using 3.8(1) (with pi , piv &qi ). Let W 1 =

W \ { v}, b = bw : w ∈W 1 .By the induction hypothesis applied to W 1 , b , r i W 1 , for i ∈ A2

there is an uncountable A3⊆ A2 and for i < j in A3 , there is r i,j

∈Q[b ],r i W 1 ≤ r i,j , and r j W 1 ≤ r i,j . Now dene r i,j

c ∈P cnc as follows: its domain

is dom ri,jw : W |= w < v , r

i,jc (dom r

i,jw ) = r

i,jw whenever W |= w < v .Why is this a denition? As if W |= w1 ≤ v ∧w2 ≤ v, ζ ∈bw 1 ∧ ζ ∈bw 2

then for some u ∈W , u ≤ w1 ∧u ≤ w2 and ζ ∈u. It is easy to check thatr i,j

c ∈P cnc . Now r i,j

c P cnc “ pi

bv, pj

bvare compatible in P cn

bv/P cn

c ”.

So there is r ∈P cnbv

such that r i,jc ≤ r , pi

bv≤ r , pj

bv≤ r . As in part (1) of

3.8 we can combine r and r i,j to a common upper bound of pi , pj in Q[b].

Claim 3.9. If e = 0 , 1 and δ is a limit ordinal, and P i , Q˜ i

, α i , e∗i (i < δ ) are such that for each α < δ , Qα = P i , Q

˜ j, α j , e∗j : i ≤ α, j < α belongs to

K , then for a unique P δ , Q = P i , Q˜ j

, α j , e∗j

: i ≤ δ, j < δ belongs to K .

Proof. We dene P δ by (d) of Denition 3.4. The least easy problem is toverify the Knaster conditions (for Q ∈K 1). The proof is like the preservationof the c.c.c. under iteration for limit stages.

Convention 3.9A. By 3.9 we shall not distinguish strictly between P i , Q˜ j

,α j , e∗j : i ≤ δ, j < δ and P i , Q

˜ i, α i , e∗i : i < δ .

Claim 3.10. If Q ∈K , α = lg( Q), a ⊂ α is closed for Q, |a | ≤ ℵ1 , Q˜ 1

is a P cn

a -name of a forcing notion satisfying (in V P α ) the Knaster condition,

its underlying set is a subset of [ω1]< ℵ0

then there is a unique ¯Q

1

∈

K ,lg(Q1) = α + 1 , Q1

α = Q˜

, Q α = Q.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

14Proof. Left to the reader.

Proof of Theorem 3.1.

A Stage: We force by K 1<λ = Q ∈K 1 : lg(Q) < λ, Q ∈H (λ) ordered by

being an initial segment (which is equivalent to forcing a Cohen subset of λ). The generic object is essentially Q∗∈K 1λ , lg(Q∗) = λ, and then we force

by P λ = lim Q∗. Clearly K <λ is a λ-complete forcing notion of cardinality

λ, and P λ satises the c.c.c. Clearly it suffices to prove part (2) of 3.1.

Suppose d˜ n is a name of a function from [ λ]n to k

˜ n for n < ω , σ˜ n < ω ,

σn : n < ω diverges (i.e. ∀m ∃k ∀n ≥ k σn ≥ m) and for some Q0∈K 1

<λ .

Q0K 1

<λ“there is p ∈P

˜ λ [ p P λ d˜ n : n < ω is a

counterexample to (2) of 3.1”].

In V we can dene Qζ : ζ < λ , Qζ∈K 1<λ , ζ < ξ ⇒ Qζ = Qξ lg(Qζ ),

in¯Q

ζ +1

, e∗

lg( Q ζ ) = 1,¯Q

ζ +1

forces (inK 1

<λ ) a value to p and the P ˜ λ -namesd˜ n ζ , σ

˜ n , k˜ n for n < ω , i.e. the values here are still P λ -names. Let Q∗

be the limit of the Qξ -s. So Q∗∈K 1 , lg(Q∗) = λ, Q∗ = P ∗i , Q˜∗

j, α∗j , e∗j :

i ≤ λ, j < λ , and the P ∗λ -names d˜ n , σ

˜ n , k˜ n are dened such that in V P ∗λ ,

d˜ n , σ

˜ n , k˜ n contradict (2) (as any P ∗λ -name of a bounded subset of λ is a

P ∗lg( Q ξ ) -name for some ξ < λ ).

B Stage: Let χ = κ+ and < ∗χ be a well-ordering of H(χ ). Now we can applyλ → (ω1 )<ω

2 to get δ,B,N s (for s ∈[B ]< ℵ0 ) and h s,t (for s, t ∈[B ]< ℵ0 , |s | =|t |) such that:

(a) B ⊆ λ, otp( B ) = ω1 , sup B = δ,(b) N s (H (χ ),∈, < ∗χ ), Q∗∈N s , d

˜ , σ˜ n , k

˜ n : n < ω ∈N s ,

(c) N s ∩ N t = N s ∩t ,

(d) N s ∩ B = s,

(e) if s = t ∩ α, t ∈[B ]< ℵ0 then N s ∩ λ is an initial segment of N t ,

(f) h s,t is an isomorphism from N t onto N s (when dened)

(g) h t,s = h − 1s,t

(h) p0 ∈N s , p0 P λ “ d˜ n , σ

˜ n , k˜ n : n < is a counterexample”,

(i) ω1 ⊆N s , |N s | = ℵ1 and if γ ∈N s , cf γ > ℵ1 then cf(sup( γ ∩ N s )) =ω1 .



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

15Let Q = Q∗ δ, P = P ∗δ and P a = P cn

a (for Q), where a is closed for Q.Note: P ∗λ ∩N s = P ∗δ ∩N s = P sup λ ∩N s ∩N s = P s ∩N s . Note also γ ∈λ ∩N s⇒ a∗γ ⊆ λ ∩ N s .

C Stage: It suffices to show that we can dene Q˜ δ

in V P δ which forces

a subset W of B of cardinality ℵ1 and F ˜ : W →ω

2 which exemplify thedesired conclusion in (2), and prove that Q˜ δ

satises the ℵ1-c.c.c. (in V P δ

(and has cardinality ℵ1)) and moreover (see Denitions 3.4 and 3.7(3)) wealso dene a δ = s∈[B ]< ℵ 0 N s , eδ = 1, Q = Qˆ P ∗δ , Q

˜ δ, a δ , eδ and prove

Q ∈K 1 .We let d

˜(u) = d

˜ | u | (u).

Let F : ω1 → ω 2 be one-to-one such that ∀η ∈ω> 2 ∃ℵ1 α < ω 1 [η F (α)].(This will not be the needed F

˜, just notation).

For s, t ∈ [B ]< ℵ0 , we say s ≡ nF t if |s | = |t | and ∀ξ ∈ s, ∀ζ ∈ t[ξ =

h s,t (ζ ) ⇒ F (ξ) n = F (ζ ) n]. Let I n = I n (F ) = {s ∈[B ]< ℵ0 : (∀ζ = ξ ∈s),

[F (ζ ) n = F (ξ) n]}.We dene R n as follows: a sequence ps : s ∈I n ∈Rn if and only if (i) for s ∈I n , ps ∈P ∗λ ∩ N s ,

(ii) for some cs we have ps “d˜

(s) = cs ”,(iii) for s, t ∈I n , s ≡ n

F t ⇒ h s,t ( pt ) = ps ,(iv) for s, t ∈I n , ps N s ∩t = pt N s ∩t .

R−n is dened similarly omitting (ii).

For x = ps : s ∈ I n let n(x) = n, pxs = ps , and (if dened)

cxs = cs . Note that we could replace x ∈ R n by a nite subsequence.

Let R =n<ω

Rn , R¯ =n<ω

R−n . We dene an order on R¯ : x ≤ y if

and only if n(x) ≤ n(y), and [s ∈I n (x ) ∧ t ∈I n (y ) ∧ s ⊆ t ⇒ pxs ≤ py

t ].

D Stage: Note the following facts::D (α) Subfact: If x ∈R −

n , t ∈ I n and pxt ≤ p1

∈P ∗δ ∩ N t , then there is ysuch that x ≤ y ∈R −

n , pyt = p1 .

Proof. We let for s ∈I n

pys

def = & h s 1 ,t 1 ( p1 N t 1 ) : s1 ⊆ s, t 1 ⊆ t, s 1 ≡ nF t1 & px

s .

(This notation means that pys is a function whose domain is the union

of the domains of the conditions mentioned, and for each coordinate wetake the canonical upper bound, see preliminaries.) Why is pys well dened?



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

16Suppose β ∈N s ∩ λ (for β ∈ λ \ N s , clearly py

s (β ) = ∅β ), s ⊆ s, t ⊆ t ,

s ≡ nF t for = 1 , 2 and β ∈Dom h s ,t ( p1 N t ) , and it suffices to show

that pxs (β ), h s 1 ,t 1 ( p1 N t 1 )(β ), h s 2 ,t 2 ( p1 N t 2 )(β ) are pairwise comparable.

Let u = {v ∈ [B ]< ℵ0 : β ∈ N v }, necessarily u ⊆ s1 ∩ s2 , and let

u = h− 1s ,t (u). As s , t , t ∈ I n , s ≡

nF t and u ⊆ t ⊆ t , necessarily

u1 = u2 . Thus γ def = h − 1u,v (β ) = h − 1

s ,t (β ) and so the last two conditions areequal.

Now pxs (β ) = px

u (β ) = h u,v ( pxs (γ )) ≤ h s ,t (( px

t N t )(γ )) = h s ,t ( pxt

N t ) (β ).

We leave to the reader checking the other requirements.

D (β ) Subfact: If x ∈R−n , t ∈ I then { px

s : s ∈I n , s ⊆ t} (as union of

functions) exists and belongs to P ∗

λ ∩ N t .Proof. See (iv) in the denition of R −

n .

D (γ ) Subfact: If x ≤ y, x ∈R n , y ∈R−n , then y ∈R n .

Proof. Check it.

D (δ) Subfact: If x ∈R −n , n < m , then there is y ∈R m , x ≤ y.

Proof. By subfact D( β ) we can nd x1 = p1t : t ∈ I m ∈inR −

m withx ≤ x1 . Using repeatedly subfact D( α) we can increase x 1 (nitely manytimes) to get y ∈R m .

D (ε) Subfact: If x ∈R −n , s, t ∈I n , s ≡ n

F t , pxs ≤ r 1 ∈P ∗λ ∩ N s , px

t ≤ r 2 ∈

P ∗λ ∩ N t , (∀ζ ∈t) [F (ζ )(n) = ( F (h s,t (ζ )))( n)] ( or just pxs 1 s1 = h s,t ( px

t 1 t1)

where t1def = {ξ ∈ t : F (ξ)(n) = ( F (h s,t (ξ)))( n)}, s1

def = {h s,t (ξ) : ξ ∈ t1}),then there is y ∈R n +1 , x ≤ y such that r 1 = py

s and r 2 = pyt .

Proof. Left to the reader.

E Stage † :

† We will have T ⊂ ω> 2 gotten by 2.7(2) and then want to get a subtree with as

few as possible colors, we can nd one isomorphic to ω> 2, and there restrict ourselves to∪n T ∗n .



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

17

We dene: T ∗k ⊆ 2k ≥ 2 by induction on k as follows:

T ∗0 = { , 1 }T ∗k +1 = {ν : ν ∈T ∗k or 2k < lg(ν ) ≤ 2k +1 , ν 2k

∈T ∗k and

[2k ≤ i < 2k +1∧ν (i) = 1] ⇒ i = 2 k + (

m< 2k

ν (i)2m )]}.

We dene

Tr Emb( k, n ) = h : h a is function from T ∗k into n ≥ 2 such that

for ν, ρ ∈T ∗k :[η = ν ⇔ h(η) = h(ν )][η ν ⇔ h(η) h(ν )][lg(η) = lg( ν ) ⇒ lg(h(η) = lg( h(ν )]

[ν = ηˆ i ⇒ (h(ν ))[lg(h(η))] = i][lg(η) = k 2 ⇒ lg(h(η)) = n] .

T (k, n ) = {Rang h : h ∈Tr Emb( k, n )},

T (∗, n ) =k

T (k, n ),

T (k,∗) =k

T (k, n ).

For T ∈T (k,∗) let n(T ) be the unique n such that T ∈T (k, n ) and let

BT = {α ∈B : F (α) n(T ) is a maximal member of T },

f s T = t ⊆BT : η ∈t ∧ν ∈t ∧η = ν ⇒ η n(T ) = ν n(T )},

ΘT = ps : s ∈fs T : ps ∈P ∩ N s , [s ⊆ t ∧ {s, t } ⊆ fs T ⇒ ps = pt N s ] .

Let furtherΘk = {ΘT : T ∈T (k,∗)}

Θ =k

Θk .

For p∈Θ, n ¯ p = n ( p), T p are dened naturally.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

18For p, q∈Θ, p ≤ q iff n p ≤ n q and for every s ∈fs T p we have ps ≤ qs .

F Stage: Let g˜

: ω → ω, g˜∈N s , g

˜grows fast enough relative σn : n < ω .

We dene a game Gm. A play of the game lasts after ω moves, in the n th

move player I chooses pn∈Θn and a function hn satisfying the restrictions

below and then player II chooses ¯qn ∈Θn , such that pn ≤ qn (so T pn = T qn ).Player I loses the play if sometimes he has no legal move; if he never loses,he wins. The restrictions player I has to satisfy are:

(a) for m < n , qm ≤ pn , pns forces a value to g

˜(n + 1),

(b) hn is a function from [ B T p n]≤ g (n ) to ω,

(c) if m < n ⇒ hn , hm are compatible,(d) If m < n , < g(m), s ∈[B T p n ] , then pn

s d˜

(s) = hn (s),(e) Let s1 , s 2 ∈ Dom hn . Then hn (s1) = hn (s2) whenever s1 , s 2 are

similar over n which means:

(i) F H OP s 2 ,s 1 (ζ ) n [¯ pn ] = F (ζ ) n [¯ pn ] for ζ ∈s1 ,

(ii) H OP s 2 ,s 1 preserves the relations sp F (ζ 1), F (ζ 2 ) < sp F (ζ 3 ),

F (ζ 4) and F (ζ 3 ) sp (F (ζ 1), F (ζ 2 )) = i (in the interesting

case ζ 3 = ζ 1 , ζ 2 implies i = 0).

G Stage/Claim: Player I has a winning strategy in this game.Proof. As the game is closed, it is determined, so we assume player II hasa winning strategy , and eventually we shall get a contradiction. We deneby induction on n, r n and Φ n such that(a) r n

∈Rn , r n ≤ r n +1 ,(b) Φ n is a nite set of initial segments of plays of the game,(c) in each member of Φ n player II uses his winning strategy,(d) if y belongs to Φ n then it has the form py, , h y, , qy, : ≤ m(y) ; let

hy = hy,n y and T y = T qy ,m (y ) ; also T y ⊆n ≥ 2, qy,s ≤ r n

s for s ∈fs T y .(e) Φn ⊆ Φn +1 , Φn is closed under taking the initial segments and the

empty sequence (which too is an initial segment of a play) belongs toΦ0 .

(f) For any y ∈ Φn and T, h either for some z ∈ Φn +1 , n z = n y + 1,y = z (n y + 1), T z = T and hz = h or player I has no legal ( n y + 1) th

move pn

, hn

(after y was played) such that T pn = T , hn

= h, and pns = r n

s for s ∈fs T (or always ≤ or always ≥ ).



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

19There is no problem to carry the denition. Now r n

s : n < ω dene afunction d∗: if η1 , . . . , ηk ∈

m 2 are distinct then d∗( η1 , . . . , ηk ) = c iff forevery (equivalently some) ζ 1 < · · · < ζ k from B , η F (ζ ) and r k

{ζ 1 ,...,ζ k }“d˜ k ({ζ 1 , . . . , ζ k }) = c”.

Now apply 2.7(2) to this coloring, get T ∗⊆ω> 2 as there. Now playerI could have chosen initial segments of this T ∗ (in the n th move in Φn ) andwe get easily a contradiction.

H Stage: We x a winning strategy for player I (whose existence is guar-anteed by stage G).

We dene a forcing notion Q∗. We have ( r,y,f ) ∈Q∗ iff

(i) r ∈P cna δ

(ii) y = p , h , q : ≤ m(y) is an initial segment of a play of Gm in whichplayer I uses his winning strategy

(iii) f is a nite function from B to {0, 1} such that f − 1({1}) ∈fs T y (whereT y = T qm ( y ) ).

(iv) r = qy,m (y )f − 1 ({1}) .

The Order is the natural one.

I Stage: If J ⊆ P cna δ

is dense open then {(r,y,f ) ∈Q∗ : r ∈J } is dense inQ∗.

Proof. By 3.8(1) (by the appropriate renaming).

J Stage: We dene Q δ in V P δ as {(r,y,f ) ∈Q∗ : r ∈G˜

P δ }, the order isas in Q∗.

The main point left is to prove the Knaster condition for the partialordered set Q∗= Qˆ P δ , Q

˜ δ, a δ , eδ demanded in the denition of K 1 . This

will follow by 3.8(3) (after you choose meaning and renamings) as done instages K,L below.

K Stage: So let i < δ , cf(i) = ℵ1 , and we shall prove that P +δ+1 /P i satisesthe Knaster condition. Let pα ∈ P ∗δ+1 for α < ω 1 , and we should nd p ∈P i , p P i “there is an unbounded A ⊆ {α : pα i ∈G

˜ P i } such that for any α, β ∈A, pα , pβ are compatible in P ∗δ+1 / G

˜ P i ”.

Without loss of generality:

(a) pα ∈P cnδ+1 .



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

20(b) for some iα : α < ω 1 increasing continuous with limit δ we have:

i0 > i , cf iα = ℵ1 , pα δ ∈P i α +1 , pα iα ∈P i 0 .Let p0

α = pα i0 , p1α = pα δ = pα iα +1 , pα (δ) = ( r α , yα , f α ), so without

loss of generality(c) r α ∈P i

α +1, r α iα ∈P i

0, m(yα ) = m∗,

(d) Dom f α ⊆ i0 ∪ [iα , iα +1 ),(e) f α i0 is constant (remember otp( B ) = ω1 ,(f) if Dom f α = { j α

0 , . . . j αk α − 1} then kα = k, [ j α < i α ⇔ < k∗],

<k ∗ j α = j , f ( j α ) = f ( j β ), F ( j α )) m(yα ) = F ( j β ) m(yβ ).The main problem is the compatibility of the qy α ,m (y α ) . Now by the

denition Θ α (in stage E) and 3.8(3) this holds.

L Stage: If c ⊂ δ + 1 is closed for Q∗, then P ∗δ+1 /P cnc satises the Knaster

condition.

If c is bounded in δ, choose a successor i ∈(sup c, δ) for Q i ∈K 1 . Weknow that P i /P cnc satises the Knaster condition and by stage K, P ∗δ+1 /P i

also satises the Knaster condition; as it is preserved by composition wehave nished the stage.

So assume c is unbounded in δ and it is easy too. So as seen in stage J,we have nished the proof of 3.1.

Theorem 3.11. If λ ≥ ω , P is the forcing notion of adding λ Cohen reals then

(∗)1 in V P , if n < ω d : [λ]≤ n → σ, σ < ℵ0 , then for some c.c.c. forcing notion Q we have Q “ there are an uncountable A ⊆ λ and an one-to-one F : A →ω 2 such that d is F -canonical on A” (see notation in§2).

(∗)2 if in V , λ ≥ µ →wsp (κ)ℵ0 (see [Sh289]) and in V P , d : [µ]≤ n → σ,σ < ℵ0 then in V P for some c.c.c. forcing notion Q we have Q “ thereare A ∈[µ]κ and one-to-one F : A →ω 2 such that d is F -canonicalon A” (see §2, ).

(∗)3 if in V , λ ≥ µ →wsp (ℵ1)nℵ2

and in V P d : [µ]≤ n → σ, σ < ℵ0 then inV P for every α < ω 1 and F : α →ω 2 for some A ⊆ µ of order type α and F : A →ω 2, F (β )

def = F (otp( A ∩ β )) , d is F -canonical on A.

(∗)4 in V P

, 2ℵ0

→ (α, n )3

for every α < ω 1 , n < ω . Really, assuming V |=GCH, we have ℵn 13

→ (α, n ) see [Sh289].



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

21Proof. Similar to the proof of 3.1. Supercially we need more indiscerni-bility then we get, but getting M u : u ∈[B ]≤ n we ignore d({α, β }) whenthere is no u with {α, β } ∈M u .

Theorem 3.12. If λ is strongly inaccessible ω-Mahlo, µ < λ , then for some

c.c.c. forcing notion P of cardinality λ, V P

satises (a) MA µ

(b) 2ℵ0 = λ = 2 κ for κ < λ(c) λ → [ℵ1 ]nσ,h (n ) for n < ω , σ < ℵ0, h(n) is as in 3.1.

Proof. Again, like 3.1.

4. Partition theorem for trees on large cardinals

Lemma 4.1 Suppose µ > σ + ℵ0 and(∗)µ for every µ-complete forcing notion P , in V P , µ is measurable.Then(1) for n < ω , P r f

eht (µ,n,σ ).

(2) P r f eht (µ, < ℵ0 , σ), if there is λ > µ , λ → µ+

<ω

2.

(3) In both cases we can have the P r f ehtn version, and even choose the

< ∗α : α < µ in any of the following ways.(a) We are given < 0

α : α < µ , and we let for η, ν ∈α 2 ∩T , α ∈SP (T )( T is the subtree we consider):η <∗α ν if and only if clpT (η) < 0

β clpT (ν ) where β = otp( α ∩SP (T ))and clpT (η) = η( j ) : j ∈lg(η), j ∈SP( T ) .

(b) We are given < 0α : α < µ , we let that for ν, η ∈α 2∩T , α ∈SP (T ):

η <∗α ν if and only if n (β + 1) < 0β +1 ν (β + 1) where β = sup( α ∩ SP (T )) .

Remark. 1) (∗)µ holds for a supercompact after Laver treatment. Onhypermeasurable see Gitik Shelah [GiSh344].2) We can in (∗)µ restrict ourselves to the forcing notion P actually used.For it by Gitik [Gi] much smaller large cardinals suffice.

3) The proof of 4.1 is a generalization of a proof of Harrington to HalpernLauchli theorem from 1978.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

22

Conclusion 4.2. In 4.1 we can get P r f ht (µ,n,σ ) (even with (3)).

Proof of 4.2. We do the parallel to 4.1(1). By ( ∗)µ , µ is weakly compacthence by 2.6(2) it is enough to prove P r f

aht (µ,n,σ ). This follows from 4.1(1)by 2.6(1).

Proof of Lemma 4.1. 1), 2). Let κ ≤ ω, σ(n) < µ , dn ∈Colnσ (n ) (µ> 2) for

n < κ .

Choose λ such that λ → (µ+ )< 2κ2µ (there is such a λ by assumption

for (2) and by κ < ω for (1)). Let Q be the forcing notion ( µ> 2, ), andP = P λ be {f : dom(f ) is a subset of λ of cardinality < µ , f (i) ∈ Q}ordered naturally. For i ∈dom( f ), take f (i) = <> ; Let η

˜ ibe the P-name

for {f (i) : f ∈G˜ P }. Let D

˜be a P-name of a normal ultralter over µ (in

V P ). For each n < ω , d ∈Colnσ (n ) (µ> 2), j < σ (n) and u = {α 0 , . . . , α n − 1},

where α 0 < · · · < α n − 1 < λ , let A˜

jd (u) be the P λ -name of the set

Ajd (u) = i < µ : η

˜ αi : < n are pairwise distinct and

j = d(ηα 0 i , . . . , η α n − 1 i) .

So A˜

jd (u) is a P λ -name of a subset of µ, and for j (1) < j (2) < σ (n) we have

P λ “A˜

j (1)d (u) ∩ A

˜j (2)d (u) = ∅, and j<σ (n ) A

˜jd (u) is a co-bounded subset of

µ”. As P “D is µ-complete uniform ultralter on µ”, in V P there is exactlyone j < σ (n) with Aj

d (u) ∈D . Let j˜ d

(u) be the P -name of this j .

Let I d (u) ⊆ P be a maximal antichain of P , each member of I d (u)forces a value to j

˜ d(u). Let W d (u) = {dom( p) : p ∈I d (u)} and W (u) =

{W d n (u) : n < κ }. So W d (u) is a subset of λ of cardinaltiy ≤ µ as well asW (u) (as P satises the µ+ -c.c. and p ∈P ⇒ | dom( p)| < µ ).

As λ → (µ++ )< 2κ2µ , dn ∈ Coln

σ n (µ> 2) there is a subset Z of λ of cardinality µ++ and set W + (u) for each u ∈[Z ]<κ such that:

(i) W + (u1) ∩ W + (u2) = W + (u1 ∩ u2),

(ii) W (u) ⊆W + (u) if u ∈[Z ]<κ ,

(iii) if |u1 | = |u2 | < κ and u1 , u2 ⊆ Z then W + (u1) and W + (u2) have thesame order type and note that H [u1 , u2 ]

def = H OP W + (u 1 ) ,W + (u 2 ) , induces

naturally a map from P u1

def = { p ∈ P : dom( p) ⊆ W + (u1)} to

P u2def = { p∈P : dom( p) ⊆W + (u2)}.





288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

24

β<α F α (η β ) and for u ∈[α 2]<κ , p0u

def = { p0{νβ :ν∈u } : β < α, |u | = |{ν

β : ν ∈u}|}. Clearly p0u ∈R | u | .

Then let h :α 2 → Z be one-to-one, such that η <∗α ν ⇔ h(η) < h (ν ) andlet p

def = { p0u,u (1) : u(1) ∈[Z ]<κ , u ∈[α 2]<κ , |u(1) | = |u |, h (u) = u(1)}.

For any generic G ⊆ P λ to which p belongs, β < α and ordinalsi0 < · · · < i n − 1 from Z such that h− 1(i ) β : < n are pairwise distinctwe have that

B{i : <n } ,β = ξ < µ : dn (ηi 0 ξ , . . . , η i n − 1 ξ) = c(u, h∗) ,

belongs to D [G], where u = {h− 1 (i ) β : < n} and h∗ : u → s(|u |) isdened by h∗(h− 1 (i ) β ) = H OP

{i : <n } ,s (n ) (i ). Really every large enoughβ < µ can serve so we omit it. As D [G] is µ-complete uniform ultralter onµ, we can nd ξ ∈(ζ, κ) such that ξ ∈B u for every u ∈[α 2]n , n < κ . Welet for ν ∈α 2, F α (ν ) = η

˜ h ( i )[G] ξ, and we let pu = p0

u except when u = {ν },

then: pu (i) = p0

u (i) i = γ (0)F α +1 (ν ) i = γ (0)

.

For α + 1, α is a successor: First for η ∈α − 1 2 dene F (ηˆ ) = F α (η)ˆ .Next we let {(u i , h i ) : i < i ∗}, list all pairs ( u, h ), u ∈[α 2]≤ n , h : u → s(|u |),one-to-one onto. Now, we dene by induction on i ≤ i∗, pi

u (u ∈ [α 2]<κ )such that :(a) pi

u ∈R | u | ,(b) pi

u increases with i,(c) for i + 1, (vii) holds for ( u i , h i ),(d) if ν m ∈α 2 for m < n , n < κ , ν m (α − 1) : m < n are pairwise distinct,

then p{ν m (α − 1) : m<n } ≤ p0{ν m :m<n } ,

(e) if ν ∈α 2, ν (α − 1) = then p0{ν } (0) = F α (ν (α − 1))ˆ .

There is no problem to carry the induction.Now F α +1

α 2 is to be dened as in the second case, starting withη → pi∗

{η} (η).

For α = 0, 1: Left to the reader.So we have nished the induction hence the proof of 4 .1(1), (2).

3) Left to the reader ( the only inuence is the choice of h in stage of theinduction).



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

255. Somewhat complimentary negative partition relation inZFC

The negative results here suffice to show that the value we have for 2 ℵ0 in

§3 is reasonable. In particular the Galvin conjecture is wrong and that forevery n < ω for some m < ω, ℵn → [ℵ1]mℵ0

.

See Erdos Hajnal Mate Rado [EHMR] for

Fact 5.1. If 2<µ < λ ≤ 2µ , µ → [µ]nσ then λ → [(2<µ )+ ]n +1

σ .

This shows that if e.g. in 1.4 we want to increase the exponents, to 3(and still µ = µ<µ ) e.g. µ cannot be successor (when σ ≤ ℵ0) (by [Sh276],3.5(2)).

Denition 5.2. P r np (λ,µ, σ), where σ = σn : n < ω , means that

there are functions F n : [λ]n

→ σn such that for every W ∈ [λ]µ

for some n, F n ([W ]n ) = σ(n). The negation of this property is denoted by NP r np (λ,µ, σ).

If σn = σ we write σ instead of σn : n < ω .

Remark 5.2A. 1) Note that λ → [µ]<ωσ means: if F : [λ]<ω → σ then for

some A ∈ [λ]µ , F ([A]<ω ) = σ. So for λ ≥ µ ≥ σ = ℵ0 , λ → [µ]<ωσ , (use

F : F (α ) = |α |) and P r np (λ,µ,σ ) is stronger than λ → [µ]<ωσ .

2) We do not write down the monotonicity properties of P r np — they are obvious.

Claim 5.3 1) We can (in 5.2) w.l.o.g. use F n,m : [λ]n → σn for n, m < ωand obvious monotonicity properties holds, and λ ≥ µ ≥ n.2) Suppose NP r np (λ,µ,κ ) and κ → [κ]nσ or even κ → [κ]<ω

σ . Then the following case of Chang conjecture holds:

(*) for every model M with universe λ and countable vocabulary, there is an elementary submodel N of M of cardinality µ,

|N ∩ κ | < κ

3) If NP r np (λ, ℵ1 ,ℵ0) then (λ, ℵ1) → (ℵ1 ,ℵ0).

Proof. Easy.



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

26Theorem 5.4. Suppose P r np (λ0 , µ, ℵ0 ), µ regular > ℵ0 and λ1 ≥ λ0 , andno µ ∈(λ0 , λ 1) is µ -Mahlo. Then P r np (λ1 , µ, ℵ0).

Proof. Let χ = 8(λ1 )+ , let {F 0n,m : m < ω } list the denable n-place functions in the model ( H (χ ),∈, < ∗

χ ), with λ0 ,µ ,λ 1 as parameters,let F 1

n,m(α

0, . . . , α

n − 1) (for α

0, . . . , α

n − 1< λ

1) be F 0

n,m(α

0, . . . , α

n − 1) if

it is an ordinal < λ 1 and zero otherwise. Let F n,m (α 0 , . . . , α n − 1 ) (forα 0 , . . . , α n − 1 < λ 1 ) be F 0n,m (α 0 , . . . , α n − 1 ) if it is an ordinal < ω and zerootherwise. We shall show that F n,m (n,m < ω ) exemplify P r np (λ1 , µ, ℵ0)(see 5.3(1)).

So suppose W ∈ [λ1 ]µ is a counterexample to P r (λ 1 , µ, ℵ0) i.e. for non,m,F n,m ([W ]n ) = ω. Let W ∗ be the closure of W under F 1n,m (n,m < ω ).Let N be the Skolem Hull of W in (H (χ ),∈, < ∗

χ ), so clearly N ∩ λ 1 = W ∗.Note W ∗⊆ λ1 , |W ∗| = µ. Also as cf(µ) > ℵ0 if A ⊆W ∗, |A| = µ then forsome n,m < ω and u i ∈[W ]n (for i < µ ), F 1n,m (u i ) ∈A and [i < j < µ ⇒

F 1n,m (u i ) = F 1n,m (u i )]. It is easy to check that also W 1 = {F 1n,m (u i ) : i < µ }is a counterexample to P r (λ 1 ,µ ,σ ). In particular, for n, m < ω , W n,m ={F 1n,m (u) : u ∈[W ]n } is a counterexample if it has power µ. W.l.o.g. W is a

counterexample with minimal δ def = sup( W ) = ∪{α +1 : α ∈W }. The abovediscussion shows that |W ∗∩ α | < µ for α < δ . Obviously cf δ = µ+ . Letα i : i < µ be a strictly increasing sequence of members of W ∗, converging

to δ, such that for limit i we have α i = min( W ∗− j<i (α j + 1). LetN = i<µ N i , N i N , |N i | < µ , N i increasing continuous and w.l.o.g.N i ∩ δ = N ∩ α i .

α Fact: δ is > λ 0 .Proof. Otherwise we then get an easy contradiction to P r (λ 0 ,µ ,σ )) aschoosing the F 0n,m we allowed λ0 as a parameter.β Fact: If F is a unary function denable in N , F (α) is a club of α for everylimit ordinal α(< λ 1) then for some club C of µ we have

(∀ j ∈C \ { min C })(∃i1 < j )(∀i ∈(i1 , j ))[i ∈C ⇒ α i ∈F (α j )].

Proof. For some club C 0 of µ we have j ∈C 0 ⇒ (N j , {α i : i < j }, W )(N, {α i : i < µ }, W ).

We let C = C 0 = acc( C ) (= set of accumulation points of C 0).We check C is as required; suppose j is a counterexample. So j =

sup( j ∩ C ) (otherwise choose i1 = max( j ∩ C )). So we can dene, byinduction on n, i n , such that:



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

27(a) in < i n +1 < j(b) α i n ∈F (α j )(c) (α i n , α i n +1 ) ∩ F (α j ) = ∅.

Why ( C 0)? |= “ F (α j ) is unbounded below α j ” hence N |= “ F (α j ) is

unbounded below α j ”, but in N , {α i : i ∈C 0 , i < j } is unbounded belowα j .Clearly for some n,m,α j ∈W n,m (see above). Now we can repeat the

proof of [Sh276,3.3(2)] (see mainly the end) using only members of W n,m .Note: here we use the number of colors being ℵ0 .

β + Fact: Wolog the C in Fact β is µ.Proof: Renaming.

γ Fact: δ is a limit cardinal.Proof: Suppose not. Now δ cannot be a successor cardinal (as cf δ = µ ≤λ0 < δ ) hence for every large enough i, |α i | = |δ|, so |δ| ∈W ∗⊆ N and|δ|+ ∈W ∗.

So W ∗∩ |δ| has cardinality < µ hence order-type some γ ∗< µ . Choosei∗< µ limit such that [ j < i ∗⇒ j + γ ∗< i ∗]. There is a denable functionF of (H (χ ),∈, < ∗χ ) such that for every limit ordinal α, F (α) is a club of α ,0 ∈F (α), if |α | < α , F (α) ∩ |α | = ∅, otp( F (α)) = cf α .

So in N there is a closed unbounded subset C α j = F (α j ) of α j of ordertype ≤ cf α j ≤ | δ|, hence C α j ∩ N has order type ≤ γ ∗, hence for i∗ chosenabove unboundedly many i < i ∗, α i ∈C α i ∗ . We can nish by fact β + .

δ Fact: For each i < µ , α i is a cardinal.Proof: If |α i | < i then |α i | ∈N i , but then |α i |+ ∈N i contradicting to Factγ , by which |α i |+ < δ , as we have assumed N i ∩ δ = N ∩ α i .

ε Fact: For a club of i < µ , α i is a regular cardinal.(Proof: if S = {i : α i singular } is stationary, then the function α i → cf(α i )is regressive on S . By Fodor lemma, for some α∗< δ , {i < µ : cf α i < α ∗} isstationary. As |N ∩ α∗| < µ for some β ∗, {i < µ : cf α i = β ∗} is stationary.Let F 1,m (α) be a club of α of order type cf( α), and by fact β we get acontradiction as in fact γ .

ζ Fact: For a club of i < µ , α i is Mahlo.Proof: Use F 1,m (α ) = a club of α which, if α is a successor cardinal or

inaccessible not Mahlo, then it contains no inaccessible, and continue as infact γ .



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

28ξ Fact: For a club of i < µ , α i is α i -Mahlo.Proof: Let F 1,m (0) (α ) = sup {ζ : α is ζ -Mahlo}. If the set {i < µ : α i is notα i -Mahlo} is stationary then as before for some γ ∈N , {i : F 1,m (0) (α i ) = γ }is stationary and let F 1,m (1) (α ) — a club of α such that if α is not ( γ + 1)-Mahlo then the club has no γ -Mahlo member. Finish as in the proof of fact

δ.

Remark 5.4.A. We can continue and say more.

Lemma 5.5 1) Suppose λ > µ > θ are regular cardinals, n ≥ 2 and(i) for every regular cardinal κ, if λ > κ ≥ θ then κ → [θ]<ω

σ (1) .(ii) for some α(∗) < µ for every regular κ ∈(α(∗), λ ), κ → [α (∗)]n

σ (2) .Then(a) λ → [µ]n +1

σ where σ = min {σ(1) , σ(2)},(b) there are functions d2 : [λ]n +1 → σ(2), d1 : [λ]3 → σ(1) such that for

every W ∈[λ]µ , d1 ([W ]3) = σ(1) or d2 ([W ]n +1 ) = σ(2) .2) Suppose λ > µ > θ are regular cardinals, and(i) for every regular κ ∈[θ, λ ), κ → [θ]<ω

σ (1) ,(ii) sup{κ < λ : κ regular } → [µ]n

σ (2) .Then(a) λ → [µ]2n

σ where σ = min {σ(1) , σ(2)}(b) there are functions d1 : [λ]3 → σ(1), d2 : [λ]2n → σ(2) such that for every W ∈[λ]µ , d1 ([W ]3) = σ(1) or d2 ([W ]2n = σ(2).

Remark. The proof is similar to that of [Sh276] 3.3,3.2.Proof. 1) We choose for each i, 0 < i < λ i , C i such that: if i is a successorordinal, C i = {i − 1, 0}; if i is a limit ordinal, C i is a club of i of order typecf i, 0∈C i , [cf i < i ⇒ cf i < min( C i − { 0})] and C i \ acc(C i ) contains onlysuccessor ordinals.

Now for α < β , α > 0 we dene by induction on , γ + (β, α ), γ − (β, α ),and then κ(β, α ), ε(β, α ).(A) γ +0 (β, α ) = β , γ −0 (β, α ) = 0.(B) if γ + (β, α ) is dened and > α and α is not an accumulation point of

C γ +

(β,α

)then we let γ −

+1(β, α ) be the maximal member of C γ +

(β,α

)which is < α and γ ++1 (β, α ) is the minimal member of C γ + (β,α ) which



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

29is ≥ α (by the choice of C γ + (β,α ) and the demands on γ + (β, α ) they arewell dened).

So(B1) (a) γ − (β, α ) < α ≤ γ + (β, α ), and if the equality holds then γ ++1 (β, α )

is not dened.(b) γ ++1 (β, α ) < γ + (β, α ) when both are dened.

(C) Let k = k(β, α ) be the maximal number k such that γ +k (β, α ) is dened(it is well dened as γ + (β, α ) : < ω is strictly decreasing). So

(C1) γ +k (β,α ) (β, α ) = α or γ +k (β,α ) > α , γ +k (β,α ) is a limit ordinal and α is anaccumulation point of C γ +

k ( β,α )(β, α ).

(D) For m ≤ k(β, α ) let us dene

εm (β, α ) = max {γ − (β, α ) + 1 : ≤ m}.

Note(D1) (a) εm (β, α ) ≤ α (if dened),

(b) if α is limit then εm (β, α ) < α (if dened),(c) if εm (β, α ) ≤ ξ ≤ α then for every ≤ m we have

γ + (β, α ) = γ + (β, ξ ), γ − (β, α ) = γ − (β, ξ ), ε (β, α ) = ε (β, ξ ).

(explanation for (c): if εm (β, α ) < α this is easy (check the denition)and if εm (β, α ) = α , necessarily ξ = α and it is trivial).(d) if ≤ m then ε (β, α ) ≤ εm (β, α )For a regular κ ∈(α (∗), λ ) let g1

κ : [κ]<ω → σ(2) exemplify κ → [θ]<ωσ (1)

and for every regular cardinal κ ∈ [θ, λ ) let g2κ : [κ]n → σ(2) exemplify

κ → [α(∗)]nσ (2) . Let us dene the colourings:

Let α 0 > α 1 > .. . > α n . Remember n ≥ 2.Let n = n(α 0 , α 1 , α 2) be the maximal natural number such that:

(i) εn (α 0 , α 1) < α 0 is well dened,(ii) for ≤ n, γ − (α 0 , α 1) = γ − (α 0 , α 2).

We dene d2(α 0 , α 1 , . . . , α n ) as g2κ (β 1 , . . . , β n ) where

κ = cf ( γ +n (α 0 ,α 1 ,α 2 ) (α 0 , α 1)) ,

β =otp α ∩ C γ +n ( α 0 ,α 1 ,α 2 ) (α 0 ,α 1 ) .



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

30Next we dene d1(α 0 , α 1 , α 2) .

Let i(∗) = sup C γ +n (α 0 ,α 2 ) ∩ C γ +

n (α 1 ,α 2 ) where n = n(α 0 , α 1 , α 2), E bethe equivalence relation on C γ +

n (α 0 ,α 1 ) \ i(∗) dened by

γ 1Eγ 2 ⇔ ∀γ ∈C γ +n (α 0 ,α 2 ) [γ 1 < γ ↔ γ 2 < γ ].

If the set w = γ ∈ C γ +n (α 0 ,α 1 ) : γ > i (∗), γ = min γ/E is nite,

we let d1(α 0 , α 1 , α 2) be g1κ ({β γ : γ ∈w}) where κ = C γ +

n (α 0 ,α 1 ) , β γ =

otp γ ∩ C γ +n (α 0 ,α 1 ) .

We have dened d1 , d2 required in condition ( b) ( though have not yetproved that they work) We still have to dene d (exemplifying λ → [µ]n +1 ).Let n ≥ 3, for α 0 > α 1 > .. . > α n , we let d(α 0 , . . . , α n ) be d1(α 0 , α 1 , α 2) if wdened during the denition has odd number of members and d2(α 0 , . . . , α n )otherwise.

Now suppose Y is a subset of λ of order type µ, and let δ = sup Y . LetM be a model with universe λ and with relations Y and {(i, j ) : i ∈C j }. LetN i : i < µ be an increasing continuous sequence of elementary submodels

of M of cardinality < µ such that α(i) = α i = min( Y \ N i ) belongs to N i +1 ,sup( N ∩ α i ) = sup( N ∩ δ). Let N =

i<µN i . Let δ(i) = δi

def = sup( N i ∩ α i ),

so 0 < δ i ≤ α i , and let n = n i be the rst natural number such that δi anaccumulation point of C i

def = C γ +n (α i ,δ ( i )) , let εi = εn ( i ) (α i , δi ). Note that

γ +n (α i , δi ) = γ +n (α i , ε i ) hence it belongs to N .

Case I: For some (limit) i < µ , cf(i) ≥ θ and (∀γ < i )[γ + α(∗) < i ] suchthat for arbitrarily large j < i , C i ∩ N j is bounded in N j ∩ δ = N j ∩ δj .This is just like the last part in the proof of [Sh276],3.3 using g1

κ and d1 forκ = cf( γ +n i (α i , δi ).

Case II: Not case I .Let S 0 = {i < µ : (∀α < i )[γ + α(∗) < i ], cf(i) = θ}. So for every i ∈S 0

for some j (i) < i , (∀ j ) j ∈( j (i), i) ⇒ C i ∩ N j is unbounded in δj . But as

C i ∩ δi is a club of δi , clearly (∀ j ) j ∈( j (i), i) ⇒ δj ∈C i .

We can also demand j (i) > ε n (α ( i ) ,δ ( i )) (α (i), δ(i)).As S 0 is stationary, (by not case I) for some stationary S 1 ⊆ S 0 and

n(∗), j (∗) we have (∀i ∈S 1) j (i) = j (∗) ∧n(α(i), δi ) = n(∗) .



288

r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

31Choose i(∗) ∈ S 1 , i(∗) = sup( i(∗) ∩ S 1), such that the order type of

S 1 ∩ i(∗) is i(∗) > α (∗). Now if i2 < i 1 ∈S 1 ∩ i(∗) then n(α i (∗) , α i 1 , α i 2 ) =

n(∗). Now L i (∗)def = otp( α i ∩ C i (∗) ) : i ∈S 1 ∩ i(∗) are pairwise distinct

and are ordinals < κdef = |C i (∗) |, and the set has order type α(∗). Now apply

the denitions of d2 and g2κ on L i (∗) .

2) The proof is like the proof of part (1) but for α 0 > α 1 > · · · we letd2(α 0 , . . . , α 2n − 1 ) = g2

κ (β 0 , . . . , β n ) where

β def = otp( C γ +

n (β 2 ,β 2 +1 ) (β 2 , β 2 +1 ) ∩ β 2 +1 )

and in case II note that the analysis gives µ possible β ’s so that we canapply the denition of g2

κ .

Denition 5.7. Let λ →stg [µ]nθ mean: if d : [λ]n → θ, and α i : i < µ is

strictly increasingly continuous and for i < j < µ , γ i,j ∈[α i , α i +1 ) then

θ = d(w) : for some j < µ, w ∈[{γ i,j : i < j }]n .

Lemma 5.8. 1) ℵt → [ℵ1]n +1ℵ0

for n ≥ 1.2) ℵn →stg [ℵ1]n +1

ℵ0for n ≥ 1.

Proof. 1) For n = 2 this is a theorem of Torodcevic, and if it holds forn ≥ 2 by 5.5(1) we get that it holds for n+1 (with n, λ, µ, θ, α (∗), σ(1),σ(2) there corresponding to n + 1, ℵn +1 , ℵ1 , ℵ0 ,ℵ0 , ℵ0 ,ℵ0 here).2) Similar.

References

[EHMR] P. Erd os, A. Hajnal, A. Mate, R. Rado, Combinatorial set theory:partition relations for cardinals, Disq. Math. Hung., no 13, (1984)

[Gi] M. Gitik, Measurability preserved by κ-complete forcing notion[GiSh344] M. Gitik, S. Shelah, On certain indiscernibility of strong cardinals and

a question of Hajnal, Archive f¨ur Math. Logic, 28(1989) 35–42[HL] Halpern, Lauchli,



r e v

i s

i o n :

1993 -

08 -

27

m o

d i f i

e d

: 1 9 9 3

- 0 8

- 2 7

32[Sh80] S. Shelah, A Weak Generalization of MA to Higher Cardinals, Israel J.

of Math, [BSF], 30(1978), 297-306.[Sh289] S. Shelah, Consisting of Partition Theorem for Graphs and Models,

Proc. of the Torontp 8/87 Conference in General Topology.[Sh276] S. Shelah, Was Sierpinsky right I ? Israel J. Math. 62 (1988) 355–380

[ShSt154] S. Shelah and L. Stanley, Generalized Martin Axiom and the SouslinHypothesis for Higher Cardinals, Israel J. of Math, [BSF], 43(1982),225-236.

[ShSt154a] S. Shelah and L. Stanley, Corrigendum to ”Generalized Martin’s Ax-iom and Souslin’s Hypothesis for Higher Cardinal, Israel J. of Math,53(1986), 309-314.

[To] S. Todorcevic, Partitions of pairs of uncountable ordinals, Acta MAth,159 (l987), 261–294

Saharon ShelahInstitute of MathematicsThe Hebrew University Jerusalem, Israel

Documents

Saharon Shelah- Strong Partition Realations Below the Power Set: Consistency Was Sierpinski Right? II