λ we have: η, ν are almost equal if and only if their images are.

166 (2000)

On a problem of Steve Kalikow

by

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

Abstract. The Kalikow problem for a pair (λ, κ) of cardinal numbers, λ > κ (in particular κ = 2) is whether we can map the family of ω-sequences from λ to the family of ω-sequences from κ in a very continuous manner. Namely, we demand that for η, ν ∈

^ω

λ we have: η, ν are almost equal if and only if their images are.

We show consistency of the negative answer, e.g., for ℵ

ω

but we prove it for smaller cardinals. We indicate a close connection with the free subset property and its variants.

0. Introduction. In the present paper we are interested in the following property of pairs of cardinal numbers:

Definition 0.1. Let λ, κ be cardinals. We say that the pair (λ, κ) has the Kalikow property (and then we write KL(λ, κ)) if there is a sequence hF

_n

: n < ωi of functions such that

F

n

:

ⁿ

λ → κ (for n < ω) and if F :

^ω

λ →

^ω

κ is given by

(∀η ∈

^ω

λ)(∀n ∈ ω)(F (η)(n) = F

n

(η n)) then for every η, ν ∈

^ω

λ,

(∀

^∞

n)(η(n) = ν(n)) iff (∀

^∞

n)(F (η)(n) = F (ν)(n)).

In particular we answer the following question of Kalikow:

Kalikow Problem 0.2. Is KL(2

^ℵ0

, 2) provable in ZFC?

The Kalikow property of pairs of cardinals was studied in [Ka90]. Several results are known already. Let us mention some of them. First, one can easily notice that

KL(λ, κ) & λ

⁰

≤ λ & κ

⁰

≥ κ ⇒ KL(λ

⁰

, κ

⁰

).

2000 Mathematics Subject Classification: 03E35, 03E05.

Key words and phrases: set theory, forcing, continuity, Kalikow, free subset.

The research was partially supported by the Israel Science Foundation. Publication 590.

[137]

Also (“transitivity”)

KL(λ

₂

, λ

1

) & KL(λ

1

, λ

0

) ⇒ KL(λ

2

, λ

0

) and

KL(λ, κ) ⇒ λ ≤ κ

^ℵ0

.

Kalikow proved that CH implies KL(2

^ℵ0

, 2) (in fact that KL(ℵ

1

, 2) holds true) and he conjectured that CH is equivalent to KL(2

^ℵ0

, 2).

The question 0.2 is formulated in [Mi91, Problem 15.15, p. 653].

We shall prove that KL(λ, 2) is closely tied with some variants of the free subset property (both positively and negatively). First we present an answer to problem 0.2 proving the consistency of ¬KL(2

^ℵ0

, 2) in 1.1 (see 2.8 too). Later we discuss variants of the proof (concerning the cardinal and the forcing). Then we deal with a positive answer, in particular KL(ℵ

n

, 2), and we show that the negation of a relative of the free subset property for λ implies KL(λ, 2).

We thank the participants of the Jerusalem Logic Seminar 1994/95 and particularly Andrzej Ros lanowski for writing it up so nicely.

Notation. We will use the Greek letters κ, λ, χ to denote (infinite) cardinals and the letters α, β, γ, ζ, ξ to denote ordinals. Sequences of ordinals will be called ¯ α, ¯ β, ¯ ζ with the usual convention that ¯ α = hα

n

: n < lg( ¯ α)i etc. Sets of ordinals will be denoted by u, v, w (with possible indexes).

The quantifiers (∀

^∞

n) and (∃

^∞

n) are abbreviations for “for all but finitely many n ∈ ω” and “for infinitely many n ∈ ω”, respectively.

1. The negative result. For a cardinal χ, the forcing notion C

χ

for adding χ many Cohen reals consists of finite functions p such that for some w ∈ [χ]

^<ω

, n < ω,

dom(p) = {(ζ, k) : ζ ∈ w & k < n} and rang(p) ⊆ 2 ordered by inclusion.

Theorem 1.1. Assume λ → (ω

1

· ω)

^<ω₂κ

, 2

^κ

< λ ≤ χ. Then

Cχ

¬KL(λ, κ) and hence

Cχ

¬KL(2

^ℵ0

, 2).

P r o o f. Suppose that C

^χ

-names F

˜

ⁿ

(for n ∈ ω) and a condition p ∈ C

^χ

are such that

p

Cχ

“hF

˜

ⁿ

: n < ωi exemplifies KL(λ, κ)”.

For ¯ α ∈

ⁿ

λ choose a maximal antichain hp

ⁿ_α,l¯

: l < ωi of C

χ

deciding the values of F

˜

ⁿ

( ¯ α). Thus we have a sequence hγ

_α,lⁿ_¯

: l < ωi ⊆ κ such that p

ⁿ_α,l¯

Cχ

F

˜

ⁿ

( ¯ α) = γ

_α,lⁿ_¯

.

Let χ

^∗

be a sufficiently large regular cardinal. Take an elementary submodel M of (H(χ

^∗

), ∈, <

^∗_χ∗

) such that

• kM k = χ, χ + 1 ⊆ M ,

• hp

ⁿ_α,l¯

: l < ω, n ∈ ω, ¯ α ∈

ⁿ

λi, hγ

_α,lⁿ_¯

: l < ω, n ∈ ω, ¯ α ∈

ⁿ

λi ∈ M .

By λ → (ω

1

· ω)

^<ω₂κ

(see [Sh 481, Claim 1.3]), we find a set B ⊆ λ of indiscernibles in M over

κ ∪ {hp

ⁿ_α,l¯

: l < ω : n ∈ ω, ¯ α ∈

ⁿ

λi, hγ

ⁿ_α,l¯

: l < ω : n ∈ ω, ¯ α ∈

ⁿ

λi, χ, p}

and a system hN

u

: u ∈ [B]

^<ω

i of elementary submodels of M such that (a) B is of order type ω

1

· ω and for u, v ∈ [B]

^<ω

:

(b) κ + 1 ⊆ N

u

,

(c) χ, p, hp

ⁿ_α,l¯

: l < ω, n < ω, ¯ α ∈

ⁿ

λi, hγ

_α,lⁿ_¯

: l < ω, n < ω, ¯ α ∈

ⁿ

λi ∈ N

u

, (d) |N

u

| = κ, N

u

∩ B = u,

(e) N

u

∩ N

_v

= N

u∩v

,

(f) |u| = |v| ⇒ N

u

∼ = N

v

, and let π

u,v

: N

v

→ N

_u

be this (unique) isomorphism,

(g) π

v,v

= id

Nv

, π

u,v

(v) = u, π

u0,u1

◦ π

_u1,u2

= π

u0,u2

, (h) if v

⁰

⊆ v, |v| = |u| and u

⁰

= π

u,v

(v

⁰

) then π

u⁰,v⁰

⊆ π

u,v

. Note that if u ⊆ B is of order type ω then we may define

N

u

= [

{N

_v

: v is a finite initial segment of u}.

Then the models N

u

(for u ⊆ B of order type ≤ ω) have the properties (b)–(h) too.

Let hβ

ζ

: ζ < ω

1

· ωi be the increasing enumeration of B. For a set u ⊆ B of order type ≤ ω let ¯ β

^u

be the increasing enumeration of u (so lg( ¯ β

^u

) = |u|).

Let u

^∗

= {β

ω1·n

: n < ω}. For k ≤ ω and a sequence ¯ ξ = hξ

m

: m < ki ⊆ ω

1

we define

u[ ¯ ξ] = {β

ω1·m+ξm

: m < k} ∪ {β

ω1·n

: n ∈ ω \ k}.

Now, working in V

^Cχ

, we say that a sequence ¯ ξ

˜ is k-strange if (1) ¯ ξ

˜

is a sequence of countable ordinals greater than 0, lg( ¯ ξ

˜

) = k ≤ ω, (2) (∀m < ω)(F

˜

^m

( ¯ β

^{u[ ¯}_˜^ξ]

m) = F

˜

^m

( ¯ β

^u∗

m)).

Claim 1.1.1. In V

^Cχ

, if ¯ ξ

˜

k

are k-strange sequences (for k < ω) such that (∀k < ω)( ¯ ξ

˜

k

C ¯ ξ

˜

k+1

) then the sequence ¯ ξ

˜ := S

k<ω

ξ ¯

˜

k

is ω-strange.

P r o o f. Should be clear (note that in this situation we have ¯ β

^{u[ ¯}_˜^ξ]

m = β ¯

^{u[ ¯}_˜^ξm]

m).

Claim 1.1.2. p

Cχ

“there are no ω-strange sequences”.

P r o o f. Assume not. Then we find a name ¯ ξ

˜

= hξ

˜

^m

: m < ωi for an ω-sequence and a condition q ≥ p such that

q

_Cχ

“(∀m < ω)(0 < ξ

˜

^m

< ω

1

& F

˜

^m

( ¯ β

^{u[ ¯}_˜^ξ]

m) = F

˜

^m

( ¯ β

^u∗

m))”.

By the choice of p and F

˜

^m

we conclude that

q

C^χ

“(∀

^∞

m)( ¯ β

^{u[ ¯}_˜^ξ]

(m) = ¯ β

^u∗

(m))”,

which contradicts the definition of ¯ β

^{u[ ¯}_˜^ξ]

, ¯ β

^u∗

, Definition 0.1 and the fact that

q

Cχ

“(∀m < ω)(0 < ξ

˜

^m

< ω

1

)”.

By 1.1.1, 1.1.2, any inductive attempt to construct (in V

^Cχ

) an ω-strange sequence ¯ ξ

˜

has to fail. Consequently, we find a condition p

^∗

≥ p, an integer k < ω and a sequence ¯ ξ = hξ

l

: l < ki such that

p

^∗

Cχ

“ ¯ ξ is k-strange but ¬(∃ξ < ω

1

)( ¯ ξ

^_

hξi is (k + 1)-strange)”.

Then in particular

( ) p

^∗

Cχ

“(∀m < ω)(F

˜

^m

( ¯ β

^{u[ ¯ξ]}

m) = F

˜

^m

( ¯ β

^u∗

m))”.

[It may happen that k = 0, i.e., ¯ ξ = hi.]

For ξ < ω

1

let u

ξ

= u[ ¯ ξ

^_

hξi] and w

ξ

= u

ξ

∪ (u

^∗

\ {ω

1

· k}). Thus w

0

= u[ ¯ ξ] ∪ u

^∗

and all w

ξ

have order type ω and π

w_ξ1,w_ξ2

is the identity on N

wξ\{ω1·k+ξ2}

. Let q := p

^∗

N

w0

and q

ξ

= π

wξ,w0

(q) ∈ N

wξ

(so q

0

= q). As the isomorphism π

wξ,w0

is the identity on N

w0

∩ N

wξ

= N

w0∩wξ

(and by the definition of Cohen forcing), we see that the conditions q, q

ξ

are compatible.

Moreover, as p

^∗

≥ p and p ∈ N

_∅

, we find that both q and q

ξ

are stronger than p.

Now fix ξ

0

∈ (0, ω

₁

) (e.g. ξ

0

= 1) and look at the sequences ¯ β

^uξ0

and β ¯

^u∗

. They are eventually equal and hence

p

_Cχ

“(∀

^∞

m)(F

˜

^m

( ¯ β

^uξ0

m) = F

˜

^m

( ¯ β

^u∗

m))”.

So we find m

^∗

< ω and a condition q

_ξ⁰₀

≥ q

_ξ0

, q such that (⊗

^ξ_q⁰0^,m∗

ξ0

) q

_ξ⁰₀

Cχ

“(∀m ≥ m

^∗

)(F

˜

^m

( ¯ β

^uξ0

m) = F

˜

^m

( ¯ β

^u∗

m))”

and (as we can increase q

⁰_ξ0

) (⊕

^ξ_q⁰0^,m∗

ξ0

) q

_ξ⁰₀

decides the values of F

˜

^m

( ¯ β

^uξ0

m) and F

˜

^m

( ¯ β

^u∗

m) for all m ≤ m

^∗

.

Note that the condition (⊗

^ξ_q⁰0^,m∗

ξ0

) means that there are NO m ≥ m

^∗

, l

0

, l

1

< ω with γ

^m_¯

β^uξ0m,l⁰

6= γ

^m_¯

β^u∗m,l1

and the three conditions q

⁰_ξ0

, p

^m_¯

β^uξ0m,l⁰

and p

^m_β¯u∗m,l1

have a common upper bound in C

χ

(remember the choice of the p

ⁿ_α,l¯

’s and γ

_α,lⁿ_¯

’s). Similarly, the condition (⊕

^ξ_q⁰0^,m∗

ξ0

) means there are

NO m ≤ m

^∗

, l

0

, l

1

< ω with either γ

_β^m_¯uξ0m,l0

6= γ

^m_¯

β^uξ0m,l1

and both q

_ξ⁰₀

and p

^m_β¯uξ0m,l0

, and q

_ξ⁰₀

and p

^m_β¯uξ0m,l1

are compatible in C

χ

, or γ

_β^m_¯u∗m,l0

6=

γ

_β^m_¯u∗m,l1

and both q

⁰_ξ0

and p

^m_β¯u∗m,l0

, and q

⁰_ξ0

and p

^m_β¯u∗m,l1

are compatible in C

^χ

.

Consequently, the condition q

_ξ^∗₀

:= q

⁰_ξ0

N

w0∪w_ξ0

has both properties (⊗

^ξ_q⁰∗^,m∗

ξ0

) and (⊕

^ξ_q⁰∗^,m∗

ξ0

) (and it is stronger than both q and q

ξ0

).

Now, for 0 < ξ < ω

1

let

q

_ξ^∗

:= π

w0∪wξ,w0∪w_ξ0

(q

_ξ^∗₀

) ∈ N

w0∪wξ

. Then (for ξ ∈ (0, ω

1

)) the condition q

_ξ^∗

is stronger than

both q = π

w0∪wξ,w0∪w_ξ0

(q) and q

ξ

= π

w0∪wξ,w0∪w_ξ0

(q

ξ0

) and it has the properties (⊗

^ξ,m_q∗ ^∗

ξ

) and (⊕

^ξ,m_q∗ ^∗

ξ

). Moreover for all ξ

1

, ξ

2

the conditions q

_ξ^∗₁

, q

^∗_ξ2

are compatible. [Why? By the definition of Cohen forcing, and π

w0∪w_ξ2,w0∪w_ξ1

(q

_ξ^∗₁

) = q

_ξ^∗₂

(chasing arrows) and π

w0∪w_ξ2,w0∪w_ξ1

is the identity on N

w0∪w_ξ2

∩ N

_w0∪w

ξ1

= N

(w0∪w_ξ2)∩(w0∪w_ξ1)

(see clauses (e), (f), (h) above).]

Claim 1.1.3. For each ξ

1

, ξ

2

∈ (0, ω

₁

) the condition q

^∗_ξ1

∪ q

^∗_ξ

2

forces in C

χ

that

(∀m < ω)(F

˜

^m

( ¯ β

^uξ1

m) = F

˜

^m

( ¯ β

^uξ2

m)).

P r o o f. If m ≥ m

^∗

then, by (⊗

^ξ_q¹∗^,m∗

ξ1

) and (⊗

^ξ_q²∗^,m∗

ξ2

) (passing through F ˜ ( ¯ β

^u∗

m)), we get

q

^∗_ξ1

∪ q

_ξ^∗₂

Cχ

“F

˜

^m

( ¯ β

^uξ1

m) = F

˜

^m

( ¯ β

^uξ2

m)”.

If m < m

^∗

then we use (⊕

^ξ_q¹∗^,m∗

ξ1

) and (⊕

^ξ_q¹∗^,m∗

ξ2

) and the isomorphism: the values assigned by q

_ξ^∗₁

, q

^∗_ξ2

to F

˜

^m

( ¯ β

^uξ1

m) and F

˜

^m

( ¯ β

^uξ2

m) have to be equal (remember κ ⊆ N

_∅

, so the isomorphism is the identity on κ).

Look at the conditions

q

ξ1,ξ2

:= q

_ξ^∗₁

N

^w_ξ1

∪ q

_ξ^∗₂

N

^w_ξ2

∈ N

w_ξ1∪w_ξ2

. It should be clear that for each ξ

1

, ξ

2

∈ (0, ω

₁

),

q

ξ1,ξ2

C^χ

“(∀m < ω)(F

˜

^m

( ¯ β

^uξ1

m) = F

˜

^m

( ¯ β

^uξ2

m))”.

Now choose ξ ∈ (0, ω

1

) so large that

dom(p

^∗

) ∩ (N

wξ

\ N

_w0

) = ∅

(possible as dom(p

^∗

) is finite, use (e)). Take any 0 < ξ

1

< ξ

2

< ω

1

and put

q

^∗

:= π

w0∪wξ,w_ξ1∪w_ξ2

(q

ξ1,ξ2

).

(Note: π

w0,w_ξ1

⊆ π

_w0∪wξ,w

ξ1∪w_ξ2

and π

wξ,w_ξ2

⊆ π

_w0∪wξ,w

ξ1∪w_ξ2

.) By the isomorphism we get

q

^∗

_Cχ

“(∀m < ω)(F

˜

^m

( ¯ β

^uξ

m) = F

˜

^m

( ¯ β

^{u[ ¯ξ]}

m))”.

Now look back:

q

^∗_ξ1

≥ q

_ξ1

= π

w0∪w_ξ1,w0∪w_ξ0

(q

ξ0

) = π

w_ξ1,w_ξ0

(q

ξ0

)

= π

w_ξ1,w_ξ0

(π

w_ξ0,w0

(q)) = π

w_ξ1,w0

(q) and hence

q

_ξ^∗₁

N

w_ξ1

≥ π

w_ξ1,w0

(q) and thus

q

^∗

N

w0

≥ π

_w0,w

ξ1

(q

_ξ^∗₁

N

w_ξ1

) ≥ q = p

^∗

N

w0

.

Consequently, by the choice of ξ, the conditions q

^∗

and p

^∗

are compatible (remember the definition of q

ξ1,ξ2

and q

^∗

). Now use ( ) to conclude that

q

^∗

∪ p

^∗

Cχ

“(∀m < ω)(F

˜

^m

( ¯ β

^u∗

m) = F

˜

^m

( ¯ β

^{u[ ¯ξ]}

m) = F

˜

^m

( ¯ β

^uξ

m))”, which implies that q

^∗

∪ p

^∗

C^χ

“ ¯ ξ

^_

hξi is (k + 1)-strange”, a contradiction.

Remark 1.2. About the proof of 1.1:

(1) No harm is done by forgetting 0 and replacing it by ξ

1

, ξ

2

.

(2) A small modification of the proof shows that in V

^Cχ

: If F

n

:

ⁿ

λ → κ (n ∈ ω) are such that

(∀η, ν ∈

^ω

λ)[(∀

^∞

n)(η(n) = ν(n)) ⇒ (∀

^∞

n)(F

n

(η n) = F

n

(ν n))]

then there are infinite sets X

n

⊆ λ (for n < ω) such that (∀n < ω) 

∀ν, η ∈ Y

l<n

X

l



(F

n

(ν) = F

n

(η)).

Say we shall have X

n

= {γ

n,i

: i < ω}. Starting we have γ

₀^∗

, . . . , γ

_n^∗

, . . . In the proof at stage n we have determined γ

l,i

(l, i < n) and p ∈ G, p ∈ N

{γl,i:l,i<ω}∪{γ_n^∗,γ_n+1^∗ ,...}

. For n = 0, 1, 2 as before. For n + 1 > 2 first γ

0,n

, . . . , γ

n−1,n

are easy by transitivity of equalities. Then find γ

n,0

, γ

n,1

as before, and then again duplicate.

(3) In the proof it is enough to use {β

ω·n+l

: n < ω, l < ω}. Hence, by 1.2 of [Sh 481] it is enough to assume λ → (ω

³

)

^<ω₂κ

. This condition is compatible with V = L.

(4) We can use only λ → (ω

²

)

^<ω₂κ

.

Definition 1.3. (1) For a sequence ¯ λ = hλ

n

: n < ωi of cardinals we define the property ( ~)

λ^¯

:

( ~)

^¯λ

for every model M of a countable language with universe sup

_n∈ω

λ

n

and Skolem functions (for simplicity) there is a sequence hX

n

:

n < ωi such that

(a) X

n

∈ [λ

_n

]

^λn

(actually X

n

∈ [λ

_n

]

^ω1

suffices)

(b) for every n < ω and ¯ α = hα

l

: l ∈ [n + 1, ω)i ∈ Q

l≥n+1

X

l

, letting (for ξ ∈ X

n

)

M

_α^ξ_¯

= Sk  [

l<n

X

l

∪ {ξ} ∪ {α

_l

: l ∈ [n + 1, ω)}  we have:

(L) the sequence hM

_α^ξ_¯

: ξ ∈ X

n

i forms a ∆-system with heart N

α¯

and its elements are pairwise isomorphic over the heart N

α¯

.

(2) For a cardinal λ the condition ( ~)

^λ

is:

( ~)

^λ

there exists a sequence ¯ λ = hλ

n

: n < ωi such that P

n<ω

λ

n

= λ and the condition ( ~)

λ^¯

holds true.

In [Sh 76] a condition (∗)

λ

, weaker than ( ~)

^λ

, was considered. Now, [Sh 124] continues [Sh 76] to get stronger indiscernibility. But by the same proof (using ω-measurable) one can show the consistency of ( ~)

^ℵω

+ GCH.

Note that to carry out the proof of 1.1 we need even less than ( ~)

^λ

: the S

l<n

X

l

(in (b) of 1.3) is much more than needed; it suffices to have ¯ β

⁰

∪ ¯ β

¹

where ¯ β

⁰

, ¯ β

¹

∈ Q

l<n

X

l

.

Conclusion 1.4. It is consistent that 2

^ℵ0

= ℵ

ω+1

and ^

n<ω

¬KL(ℵ

_ω

, ℵ

n

) so ¬KL(2

^ℵ0

, 2).

Remark 1.5. Koepke [Ko84] continues [Sh 76] to get equiconsistency.

His refinement of [Sh 76] (for the upper bound) works below too.

2. The positive result. For an algebra M on λ and a set X ⊆ λ the closure of X under functions of M is denoted by cl

M

(X). Before proving our result (2.6) we remind the reader of some definitions and propositions.

Proposition 2.1. For an algebra M on λ the following conditions are equivalent :

( F)

⁰M

for each sequence hα

n

: n ∈ ωi ⊆ λ we have

(∀

^∞

n)(α

n

∈ cl

_M

({α

k

: n < k < ω})), ( F)

¹M

there is no sequence hA

n

: n ∈ ωi ⊆ [λ]

^ℵ0

such that

(∀n ∈ ω)(cl

M

(A

n+1

) cl

M

(A

n

)),

( F)

²M

(∀A ∈ [λ]

^ℵ0

)(∃B ∈ [A]

^ℵ0

)(∀C ∈ [B]

^ℵ0

)(cl

M

(B) = cl

M

(C)).

Definition 2.2. We say that a cardinal λ has the (F)-property for κ

(and then we write Pr

^F

(λ, κ)) if there is an algebra M on λ with vocabulary

of cardinality ≤ κ satisfying one (equivalently: all) of the conditions ( F)

ⁱM

(i < 3) of 2.1. If κ = ℵ

0

we may omit it.

Remember

Proposition 2.3. If V

0

⊆ V

₁

are universes of set theory and V

1

|=

¬Pr

^F

(λ) then V

0

|= ¬Pr

^F

(λ).

P r o o f. By absoluteness of the existence of an ω-branch of a tree.

Remark 2.4. The property ¬Pr

^F

(λ) is a kind of large cardinal property.

It was clarified in L (remember that it is inherited from V to L) by Silver [Si70] to be equiconsistent with “there is a beautiful cardinal” (terminol- ogy of 2.3 of [Sh 110]), another partition property inherited by L. More in [Sh 513].

Proposition 2.5. For each n ∈ ω, Pr

^F

(ℵ

n

).

P r o o f. This was done in [Sh:b, Chapter XIII], see [Sh:g, Chapter VII]

too, and probably earlier by Silver. However, for the sake of completeness we will give the proof.

First note that clearly Pr

^F

(ℵ

0

) and thus we have to deal with the case when n > 0. Let f, g : ℵ

n

→ ℵ

_n

be two functions such that if m < n, α ∈ [ℵ

m

, ℵ

m+1

) then f (α, ·) α : α

^1-1

→ ℵ

_m

, g(α, ·) ℵ

m

: ℵ

m

1-1

→ α are functions inverse to each other.

Let M be the following algebra on ℵ

n

: M = (ℵ

n

, f, g, m)

m∈ω

.

We want to check the condition ( F)

¹M

: assume that a sequence hA

k

: k < ωi

⊆ [ℵ

_n

]

^ℵ0

is such that for each k < ω,

cl

M

(A

k+1

) cl

^M

(A

k

).

For each m < n, the sequence hsup(cl

M

(A

k

) ∩ ℵ

m+1

) : k < ωi is non- increasing and therefore it is eventually constant. Consequently, we find k

^∗

such that

(∀m < n)(sup(cl

M

(A

k^∗+1

) ∩ ℵ

m+1

) = sup(cl

M

(A

k^∗

) ∩ ℵ

m+1

)).

By the choice of hA

k

: k < ωi we have cl

M

(A

k^∗+1

) cl

M

(A

k^∗

). Let α

0

:= min(cl

M

(A

k^∗

) \ cl

M

(A

k^∗+1

)).

As the model M contains individual constants m (for m ∈ ω) we know that ℵ

₀

⊆ cl

_M

(∅) and hence ℵ

0

≤ α

₀

. Let m < n be such that ℵ

m

≤ α

₀

< ℵ

m+1

. By the choice of k

^∗

we find β ∈ cl

M

(A

k^∗+1

) ∩ ℵ

m+1

such that α

0

≤ β. Then necessarily α

0

< β. Look at f (β, α

0

): we know that α

0

, β ∈ cl

M

(A

k^∗

) and therefore f (β, α

0

) ∈ cl

M

(A

k^∗

) ∩ ℵ

m

and f (β, α

0

) < α

0

. The minimality of α

0

implies that f (β, α

0

) ∈ cl

M

(A

k^∗+1

) and hence

α

0

= g(β, f (β, α

0

)) ∈ cl

M

(A

k^∗+1

),

a contradiction.

Explanation. Better think of the proof below from the end. Let ¯ α = hα

_n

: n < ωi ∈

^ω

λ. So for some n(∗), n(∗) ≤ n < ω ⇒ α

n

∈ cl

_M

(α

l

: l > n).

So for some m

n

> n, {α

n(∗)

, . . . , α

n−1

} ⊆ cl

M

(α

n

, . . . , α

mm−1

) and (∀l < n(∗))(α

l

∈ cl

_M

(α

k

: k > n(∗)) ⇒ α

l

∈ cl

_M

(α

k

: k ∈ [n, m

n

))).

Let w = {l < n(∗) : α

l

∈ cl

_M

(α

n

: n ≥ n(∗)). It is natural to aim at:

(∗) for n large enough (say n > m

_n(∗)

), F

n

(hα

l

: l < ni) depends just on {α

l

: l ∈ [n(∗), n) or l ∈ w} and hF

m

( ¯ α m) : m ≥ ni codes

¯

α (w ∪ [n(∗), ω)).

Of course, we are given an n and we do not know how to compute the real n(∗), but we can approximate. Then we look at a late enough end segment where we compute down.

Theorem 2.6. Assume that λ ≤ 2

^ℵ0

is such that Pr

^F

(λ) holds. Then KL(λ, ω) (and hence KL(λ, 2)).

P r o o f. We have to construct functions F

n

:

ⁿ

λ → ω witnessing KL(λ, ω).

For this we will introduce functions k and l such that for ¯ α ∈

ⁿ

λ the value of k( ¯ α) will say which initial segment of ¯ α will be irrelevant for F

n

( ¯ α) and l( ¯ α) will be such that (under certain circumstances) elements α

i

(for k( ¯ α) ≤ i < l( ¯ α)) will be encoded by hα

j

: j ∈ [l( ¯ α), n)i.

Fix a sequence hη

α

: α < λi ⊆

^ω

2 with no repetitions.

Let M be an algebra on λ such that ( F)

⁰M

holds true. We may assume that there are no individual constants in M (so cl

M

(∅) = ∅).

Let hτ

_lⁿ

(x

0

, . . . , x

n−1

) : l < ωi list all n-place terms of the language of the algebra M (and τ

₀¹

(x) is x) when 0 < n < ω. For ¯ α ∈

^ω≥

λ (with α

j

the jth element in ¯ α) let

u( ¯ α) = {l < lg( ¯ α) : α

l

6∈ cl

M

( ¯ α (l, lg(¯ α)))} ∪ {0}

and for l 6∈ u( ¯ α), l < lg( ¯ α) let

f

l

( ¯ α) = min{j : α

l

∈ cl

_M

( ¯ α (l, j))},

g

l

( ¯ α) = min{i : α

l

= τ

_i^{fl( ¯α)−l−1}

( ¯ α (l, f

^l

( ¯ α)))}.

For ¯ α ∈

ⁿ

λ (1 < n < ω) put

k

1

( ¯ α) = min((u( ¯ α (n − 1)) \ u(¯ α)) ∪ {n − 1}), k

0

( ¯ α) = max(u( ¯ α) ∩ k

1

( ¯ α)).

Note that if (n > 1 and) ¯ α ∈

ⁿ

λ then n − 1 ∈ u( ¯ α) (as cl

M

(∅) = ∅) and k

1

( ¯ α) > 0 (as always 0 ∈ u( ¯ β)) and k

0

( ¯ α) is well defined (as 0 ∈ u( ¯ α)∩k

1

( ¯ α)) and k

0

( ¯ α) < k

1

( ¯ α) < n. Moreover, for all l ∈ (k

0

( ¯ α), k

1

( ¯ α)) we have α

l

6∈ u( ¯ α) by the choice of k

0

( ¯ α), hence α

l

6∈ u( ¯ α (n − 1)) by the choice of k

1

( ¯ α) and thus α

l

∈ cl

_M

( ¯ α (l, n − 1)). Now, for ¯ α ∈

^ω>

λ, lg( ¯ α) > 1 we define

l( ¯ α) = max{j ≤ k

1

( ¯ α) : j > k

0

( ¯ α) ⇒ (∀i ∈ (k

0

( ¯ α), j))(g

i

( ¯ α) ≤ lg( ¯ α))},

m( ¯ α) = max{j ≤ l( ¯ α) : j > max{1, k

0

( ¯ α)} ⇒ k

0

( ¯ α j) = k

0

( ¯ α)}, k( ¯ α) = l( ¯ α m(¯ α)) (if m( ¯ α) ≤ 1 then put k( ¯ α) = −1).

Clearly k( ¯ α) < m( ¯ α) ≤ l( ¯ α) ≤ k

1

( ¯ α) < lg( ¯ α).

Claim 2.6.1. For each ¯ α ∈

^ω

λ, the set u( ¯ α) is finite and : (1) The sequence hk

1

( ¯ α n) : n < ωi diverges to ∞.

(2) The sequence hk

0

( ¯ α n) : n < ω & k

0

( ¯ α n) 6= max u(¯ α)i, if infinite, diverges to ∞. There are infinitely many n < ω with k

0

( ¯ α n) = max u(¯ α).

(3) The sequence hl( ¯ α n) : n < ωi diverges to ∞.

(4) The sequences hm( ¯ α n) : n < ωi and hk(¯ α n) : n < ωi diverge to ∞.

P r o o f. Let ¯ α = hα

n

: n < ωi ∈

^ω

λ. By the property ( F)

⁰M

we find n

^∗

< ω such that u( ¯ α) ⊆ n

^∗

. Fix n

0

> n

^∗

and define

n

1

= max{f

n

( ¯ α) + g

n

( ¯ α) + 2 : n ∈ (n

0

+ 1) \ u( ¯ α)}

(so as cl

M

(∅) = ∅ we have n

1

≥ f

_n0

( ¯ α)+2 > n

0

+3 and for l ∈ (n

0

+1)\u( ¯ α), α

l

∈ cl

_M

(α

l+1

, . . . , α

n1−1

) is witnessed by τ

_g^{fl( ¯α)−l−1}

l( ¯α)

(α

l+1

, . . . , α

fl( ¯α)−1

) with f

l

( ¯ α), g

l

( ¯ α) < n

1

− 1).

(1) Note that u( ¯ α n) ∩ (n

⁰

+ 1) = u( ¯ α) for all n ≥ n

1

− 1 and hence for n ≥ n

1

,

u( ¯ α n) ∩ (n

0

+ 1) = u( ¯ α (n − 1)) ∩ (n

0

+ 1).

Consequently, for all n ≥ n

1

we have k

1

( ¯ α n) > n

0

. As we could have chosen n

0

arbitrarily large we may conclude that lim

n→∞

k

1

( ¯ α n) = ∞.

(2) Note that for all n ≥ n

1

,

either k

0

( ¯ α n) = max(u(¯ α)) or k

0

( ¯ α n) > n

0

. Hence, by the arbitrariness of n

0

, we get the first part of (2).

Let l

^∗

= min(u( ¯ α n

1

)\u( ¯ α)) (note that n

1

−1 ∈ u( ¯ α n

1

)\u( ¯ α)). Clearly l

^∗

> n

0

and α

l^∗

6∈ u( ¯ α). Consider n = f

l^∗

( ¯ α) (so l

^∗

≤ n − 2, n

₁

≤ n − 1).

Then l

^∗

∈ u( ¯ α (n − 1)) \ u(¯ α n). As

l

^∗

∩ u( ¯ α n

1

) = l

^∗

∩ u( ¯ α n − 1) = u(¯ α) (remember the choice of l

^∗

) we conclude that

l

^∗

= k

1

( ¯ α n) and k

0

( ¯ α n) = max u(¯ α).

Now, since n

0

was arbitrarily large, we find that for infinitely many n, k

0

( ¯ α n) = max u(¯ α).

(3) Suppose that n ≥ n

1

. Then we know that k

1

( ¯ α n) > n

0

and either k

0

( ¯ α n) = max u(¯ α) or k

0

( ¯ α n) > n

0

(see above). If the first possibility takes place then, as n ≥ n

1

, we may use j = n

0

+ 1 to witness that l( ¯ α n) >

n

0

(remember the choice of n

1

). If k

0

( ¯ α n) > n

0

then clearly l( ¯ α n) > n

0

.

As n

0

could be arbitrarily large we are done.

(4) Suppose we are given m

0

< ω. Take m

1

> m

0

such that for all n ≥ m

1

,

either k

0

( ¯ α n) = max u(¯ α) or k

0

( ¯ α n) > m

0

(possible by (2)) and then choose m

2

> m

1

such that k

0

( ¯ α m

2

) = max u( ¯ α) (by (2)). Due to (3) we find m

3

> m

2

such that for all n ≥ m

3

, l( ¯ α n) > m

2

. Now suppose that n ≥ m

3

. If k

0

( ¯ α n) = max u(¯ α) then, as l( ¯ α n) > m

2

, we get m( ¯ α n) ≥ m

2

> m

0

. Otherwise k

0

( ¯ α n) > m

0

(as n > m

1

) and hence m( ¯ α n) > m

0

. This shows that lim

n→∞

m( ¯ α n) = ∞. Now, immediately by the definition of k and (3) above we conclude that lim

n→∞

k( ¯ α n) = ∞.

Claim 2.6.2. If ¯ α

¹

, ¯ α

²

∈

^ω

λ are such that (∀

^∞

n)(α

¹_n

= α

²_n

) then (∀

^∞

n)(l( ¯ α

¹

n) = l(¯ α

²

n) & m(¯ α

¹

n) = m(¯ α

²

n) & k(¯ α

¹

n) = k(¯ α

²

n)).

P r o o f. Let n

0

be greater than max(u( ¯ α

¹

) ∪ u( ¯ α

²

)) and such that

¯

α

¹

[n

0

, ω) = ¯ α

²

[n

0

, ω).

For k = 1, 2, 3 define n

k

by

n

k+1

= max{f

n

( ¯ α

ⁱ

) + g

n

( ¯ α

ⁱ

) + 2 : n ∈ (n

k

+ 1) \ u( ¯ α

ⁱ

), i < 2}.

As in the proof of 2.6.1, for i = 1, 2 and j < 3 we have:

(⊗

¹

) (∀n ≥ n

j+1

)(k

0

( ¯ α

ⁱ

n) = max u(¯ α

ⁱ

) or k

0

( ¯ α

ⁱ

n) > n

j

),

(⊗

²

) (∀n ≥ n

j+1

)(k

1

( ¯ α

ⁱ

n) > n

^j

& l( ¯ α

ⁱ

n) > n

^j

& m( ¯ α

ⁱ

n) > n

^j

&

h( ¯ α

ⁱ

n) > n

j

),

(⊗

³

) (∃n

⁰

∈ (n

₁

, n

2

))(k

0

( ¯ α

¹

n

⁰

) = max u( ¯ α

¹

) & k

0

( ¯ α

²

n

⁰

) = max u( ¯ α

²

)) (for (⊗

³

) repeat arguments from 2.6.1(2) and use the fact that ¯ α

¹

[n

⁰

, ω) =

¯

α

²

[n

0

, ω)). Clearly

(⊗

⁴

) (∀n > n

0

)(u( ¯ α

¹

n) \ n

⁰

= u( ¯ α

²

n) \ n

⁰

).

Hence, applying (⊗

⁴

) + (⊗

²

) + the definition of k

1

(−), we conclude that (⊗

⁵

) (∀n ≥ n

1

)(k

1

( ¯ α

¹

n) = k

1

( ¯ α

²

n)),

and then applying (⊗

⁴

) + (⊗

²

) + (⊗

⁵

) + the definition of k

0

(−), we get (⊗

⁶

) for all n ≥ n

1

: either k

0

( ¯ α

¹

n) = max u(¯ α

¹

) and k

0

( ¯ α

²

n) =

max u( ¯ α

²

), or k

0

( ¯ α

¹

n) = k

0

( ¯ α

²

n).

Since

(∀n ≥ n

0

)(f

n

( ¯ α

¹

) = f

n

( ¯ α

²

) & g

n

( ¯ α

¹

) = g

n

( ¯ α

²

))

and by (⊗

²

)+(⊗

⁵

)+the choice of n

0

+the definition of l(−), we get (compare the proof of 2.6.1)

(⊗

⁷

) (∀n ≥ n

1

)(l( ¯ α

¹

n) = l(¯ α

²

n)) and by (⊗

²

) + (⊗

⁷

) + (⊗

⁶

) + the definition of m(−),

(∀n ≥ n

3

)(m( ¯ α

¹

n) = m(¯ α

²

n) ≥ n

2

).

Moreover, now we easily get

(∀n ≥ n

3

)(k( ¯ α

¹

n) = k(¯ α

²

n)).

For integers n

0

≤ n

₁

≤ n

₂

we define functions F

_n⁰_0,n1,n2

:

ⁿ²

λ → H(ℵ

0

) by letting F

_n⁰_0,n1,n2

(α

0

, . . . , α

n2−1

) (for hα

0

, . . . , α

n2−1

i ∈

ⁿ²

λ) be the se- quence consisting of:

(a) hn

0

, n

1

, n

2

i,

(b) the set T

n1,n2

of all terms τ

_lⁿ

such that n ≤ n

2

− n

₁

and either l ≤ n

2

(we will call it the simple case) or τ

_lⁿ

is a composition of depth at most n

2

of such terms,

(c) hη

α

n

2

, n, l, hi

0

, . . . , i

n−1

ii for n ≤ n

₂

− n

₁

, i

0

, . . . , i

n−1

∈ [n

₁

, n

2

) and l such that τ

_lⁿ

∈ T

_n1,n2

and α = τ

_lⁿ

(α

i0

, . . . , α

in−1

),

(d) hn, l, hi

0

, . . . , i

n−1

i, ii for n ≤ n

₂

− n

₁

, i

0

, . . . , i

n−1

∈ [n

₁

, n

2

), i ∈ [n

0

, n

1

) and l such that τ

_lⁿ

∈ T

_n1,n2

and α

i

= τ

_lⁿ

(α

i0

, . . . , α

in−1

),

(e) equalities among appropriate terms, i.e. all tuples hn

⁰

, l

⁰

, n

⁰⁰

, l

⁰⁰

, hi

⁰₀

, . . . , i

⁰_n0−1

i, hi

⁰⁰₀

, . . . , i

⁰⁰_n00−1

ii

such that n

1

≤ i

⁰₀

< . . . < i

⁰_n0−1

< n

2

, n

1

≤ i

⁰⁰₀

< . . . < i

⁰⁰_n00−1

< n

2

, n

⁰

, n

⁰⁰

≤ n

₂

− n

₁

, l

⁰

, l

⁰⁰

are such that τ

_lⁿ0⁰

, τ

_lⁿ00⁰⁰

∈ T

_n1,n2

and

τ

_lⁿ0⁰

(α

_i⁰₀

, . . . , α

_i⁰

n0 −1

) = τ

_lⁿ00⁰⁰

(α

_i⁰⁰₀

, . . . , α

_i⁰⁰

n00 −1

).

(Note that the value of F

_n⁰_0,n1,n2

( ¯ α) does not depend on ¯ α n

0

.) Finally we define functions F

n

:

ⁿ

λ → H(ℵ

0

) (for 1 < n < ω) by:

if ¯ α ∈

ⁿ

λ then F

n

( ¯ α) = F

_{k( ¯}⁰_{α),l( ¯α),n}

( ¯ α).

As H(ℵ

0

) is countable we may think that these functions are into ω. We are going to show that they witness KL(λ, ω).

Claim 2.6.3. If ¯ α

¹

, ¯ α

²

∈

^ω

λ are such that (∀

^∞

n)(α

¹_n

= α

²_n

) then (∀

^∞

n)(F

n

( ¯ α

¹

n) = F

n

( ¯ α

²

n)).

P r o o f. Take m

0

< ω such that for all n ∈ [m

0

, ω) we have α

_n¹

= α

²_n

, l( ¯ α

¹

n) = l(¯ α

²

n), k( ¯ α

¹

n) = k(¯ α

²

n) (possible by 2.6.2). Let m

1

> m

0

be such that for all n ≥ m

1

,

k( ¯ α

¹

n) = k(¯ α

²

n) > m

0

(use 2.6.1). Then, for n ≥ m

1

, i = 1, 2 we have

F

n

( ¯ α

ⁱ

n) = F

k( ¯⁰αⁱn),l( ¯αⁱn),n

( ¯ α

ⁱ

n) = F

k( ¯⁰α¹n),l( ¯α¹n),n

( ¯ α

ⁱ

n).

Since the value of F

_n⁰_0,n1,n2

( ¯ β) does not depend on ¯ β n

⁰

and the sequences

¯

α

¹

n, ¯ α

²

n agree on [m

0

, ω), we get

F

_{k( ¯}⁰_α1n),l( ¯α¹n),n

( ¯ α

¹

n) = F

k( ¯⁰α¹n),l( ¯α¹n),n

( ¯ α

²

n) = F

k( ¯⁰α²n),l( ¯α²n),n

( ¯ α

²

n),

and hence

(∀n ≥ m

1

)(F

n

( ¯ α

¹

n) = F

n

( ¯ α

²

n)).

Claim 2.6.4. If ¯ α

¹

, ¯ α

²

∈

^ω

λ and (∀

^∞

n)(F

n

( ¯ α

¹

n) = F

n

( ¯ α

²

n)) then (∀

^∞

n)(α

¹_n

= α

²_n

).

P r o o f. Take n

0

< ω such that

u( ¯ α

¹

) ∪ u( ¯ α

²

) ⊆ n

0

and (∀n ≥ n

0

)(F

n

( ¯ α

¹

n) = F

n

( ¯ α

²

n)).

Then for all n ≥ n

0

we have (by clause (a) of the definition of F

_n⁰_0,n1,n2

) l( ¯ α

¹

n) = l(¯ α

²

n) and k(¯ α

¹

n) = k(¯ α

²

n).

Further, let n

1

> n

0

be such that for all n ≥ n

1

, k( ¯ α

¹

n) > n

0

and k

0

( ¯ α

¹

n

1

) = max u( ¯ α

¹

) (exists by 2.6.1) and choose n

2

> n

1

such that n ≥ n

2

implies m( ¯ α

¹

n) > n

2

.

We are going to show that α

¹_n

= α

²_n

for all n > n

1

. Assume not. Then we have n > n

1

with α

¹_n

6= α

²_n

and thus η

α¹_n

6= η

_α²_n

. Take n

⁰

> n such that η

α¹_n

n

⁰

6= η

_α²

n

n

⁰

. Applying 2.6.1(2) and (4) choose n

⁰⁰

> n

⁰

such that m( ¯ α

¹

n

⁰⁰

) > n

⁰

and k

0

( ¯ α

¹

n

⁰⁰

) = max u( ¯ α

¹

).

Now define inductively: m

0

= n

⁰⁰

, m

k+1

= m( ¯ α

¹

m

k

). Thus n

⁰⁰

= m

0

> l( ¯ α

¹

m

⁰

) ≥ m

1

> l( ¯ α

¹

m

¹

) ≥ m

2

> . . . and (by induction on k)

m

k

> max u( ¯ α

¹

) ⇒ k

0

( ¯ α

¹

m

k

) = max u( ¯ α

¹

)

(see the definition of m). Let k

^∗

be the first such that n ≥ m

k^∗

(so k

^∗

≥ 2, exists by the choice of n

1

). Note that by the choice of n

1

above we necessarily have

m

k^∗

> l( ¯ α

¹

m

k^∗

) = k( ¯ α

¹

m

k^∗−1

) > n

0

. Hence for all k < k

^∗

:

F

mk

( ¯ α

¹

m

k

) = F

mk

( ¯ α

²

m

k

),

l( ¯ α

¹

m

k+1

) = l( ¯ α

²

m

k+1

) = k( ¯ α

¹

m

k

) = k( ¯ α

²

m

k

).

By the definition of the functions l, m, k and the choice of m

0

(remember k

0

( ¯ α

¹

m

0

) = max u( ¯ α

¹

)) we know that for each i ∈ [k( ¯ α

¹

m

k

), l( ¯ α

¹

m

k

)) and k < k

^∗

, for some τ

_l^m

∈ T

_{l( ¯α}¹_mk),mk

and i

0

, . . . , i

m−1

∈ [l( ¯ α

¹

m

^k

), m

k

) we have α

¹_i

= τ

_l^m

(α

¹_i0

, . . . , α

¹_im−1

). Moreover we may demand that τ

_l^m

is a composition of depth at most l( ¯ α

¹

m

k

) − i of simple case terms. Since

F

_{k( ¯}⁰_α1mk),l( ¯α¹mk),mk

( ¯ α

¹

m

k

) = F

_{k( ¯}⁰_α2mk),l( ¯α²mk),mk

( ¯ α

²

m

k

) we conclude that (by clause (d) of the definition of the functions F

_n⁰_0,n1,n2

)

α

²_i

= τ

_l^m

(α

²_i0

, . . . , α

²_im−1

).

Now look at our n. If l( ¯ α

¹

m

k^∗−1

) > n then k( ¯ α

¹

m

k^∗−1

) ≤ n <

l( ¯ α

¹

m

k^∗−1

) and thus we find i

0

, . . . , i

m−1

∈ [l( ¯ α

¹

m

k^∗−1

), m

k^∗−1

) and τ

_l^m

∈ T

l( ¯α¹mk∗ −1),mk∗ −1

such that

α

¹_n

= τ

_l^m

(α

¹_i0

, . . . , α

¹_m−1

) and α

²_n

= τ

_l^m

(α

²_i0

, . . . , α

²_m−1

).

If l( ¯ α

¹

m

k^∗−1

) ≤ n then n ∈ [k( ¯ α

¹

m

k^∗−2

), l( ¯ α

¹

m

k^∗−2

)) (as l( ¯ α

¹

m

k^∗−1

)

= k( ¯ α

¹

m

k^∗−2

) and n < m

k^∗−1

≤ l( ¯ α

¹

m

k^∗−2

)). Hence, for some i

0

, . . . , i

m−1

∈ [l( ¯ α

¹

m

k^∗−2

), m

k^∗−2

) and τ

_l^m

∈ T

_{l( ¯α}¹_{mk∗ −2),mk∗ −2}

, we have α

¹_n

= τ

_l^m

(α

¹_i0

, . . . , α

¹_m−1

) and α

²_n

= τ

_l^m

(α

²_i0

, . . . , α

²_m−1

).

In both cases we may additionally demand that the term τ

_l^m

is a compo- sition of depth l( ¯ α

¹

m

k^∗−1

) − n (or l( ¯ α

¹

m

k^∗−2

) − n, respectively) of terms of the simple case. Now we proceed inductively (taking care of the depth of the terms involved) and we find a term τ ∈ T

_{l( ¯α}¹_m0),m0

(which is a com- position of depth at most l( ¯ α

¹

m

0

) − n of terms of the simple case) and i

0

, . . . , i

m−1

∈ [l( ¯ α

¹

m

0

), m

0

) such that

α

¹_n

= τ (α

¹_i0

, . . . , α

¹_m−1

) and α

²_n

= τ (α

²_i0

, . . . , α

²_m−1

).

But now applying clause (c) of the definition of the functions F

_n⁰_0,n1,n2

we conclude that η

α¹_n

m

0

= η

α²_n

m

0

, contradicting the choice of n

⁰

and the fact that m

0

> n

⁰

.

The last two claims finish the proof of the theorem.

Remark 2.7. If the model M has κ < λ functions (so hτ

iⁿ

(x

0

, . . . , x

n−1

) : i < κi lists the n-place terms) we can prove KL(λ, κ) and the proof is similar.

Final Remarks 2.8. (1) Now we phrase exactly what is needed to carry out the proof of Theorem 1.1 for λ > κ. It is:

( ) for every model M with universe λ and Skolem functions and with countable vocabulary, we can find pairwise distinct α

n,l

< λ (for n < ω, l < ω) such that

(⊗) if m

0

< m

1

< ω and l

⁰_i

< l

⁰⁰_i

for i < m

0

and l

i

< ω for i ∈ [m

0

, m

1

) and k

0

< k

1

< k

2

< ω then the models

(Sk({α

_i,l⁰_i

, α

_i,l⁰⁰_i

: i < m

0

}∪{α

_m0,k0

, α

m0,k1

}∪{α

_i,li

: i ∈ (m

0

, m

1

)}), α

0,l⁰₀

, α

0,l⁰⁰₀

, α

1,l⁰₁

, α

1,l⁰⁰₁

, . . . , α

m0−1,l_m0−1⁰

, α

m0−1,l_m0−1⁰⁰

, α

m0,k0

, α

m0,k1

, α

m0+1,l_m0+1

, . . . , α

m1−1,l_m1−1

)

and

(Sk({α

_i,l⁰_i

, α

_i,l⁰⁰_i

: i < m

0

}∪{α

_m0,k0

, α

m0,k2

}∪{α

_i,li

: i ∈ (m

0

, m

1

)}),

α

0,l⁰₀

, α

0,l⁰⁰₀

, α

1,l⁰₁

, α

1,l⁰⁰₁

, . . . , α

m0−1,l_m0−1⁰

, α

m0−1,l_m0−1⁰⁰

, α

m0,k0

,

α

m0,k2

, α

m0+1,l_m0+1

, . . . , α

m1−1,l_m1−1

)

are isomorphic and the isomorphism is the identity on their intersec- tion and they have the same intersection with κ.

For more details and more related results we refer the reader to [Sh:F254].

(2) Together with 1.5, 2.7 this gives a good bound on the consistency strength of ¬KL(λ, κ).

(3) What if we ask F

n

:

ⁿ

λ →

^ω>

κ such that F

n

(η) E F

n+1

(η) and η ∈

^ω

λ ⇒ F (η) = S F

n

(η n) ∈

^ω

κ? No real change.

References

[Ka90] S. K a l i k o w, Sequences of reals to sequences of zeros and ones, Proc. Amer.

Math. Soc. 108 (1990), 833–837.

[Ko84] P. K o e p k e, The consistency strength of the free-subset property for ω

ω

, J.

Symbolic Logic 49 (1984), 1198–1204.

[Mi91] A. W. M i l l e r, Arnie Miller’s problem list , in: H. Judah (ed.), Set Theory of the Reals (Ramat Gan, 1991), Israel Math. Conf. Proc. 6, Bar-Ilan Univ., Ramat Gan, 1993, 645–654.

[Sh 76] S. S h e l a h, Independence of strong partition relation for small cardinals, and the free-subset problem, J. Symbolic Logic 45 (1980), 505–509.

[Sh 124] —, ℵ

ω

may have a strong partition relation, Israel J. Math. 38 (1981), 283–

288.

[Sh 110] —, Better quasi-orders for uncountable cardinals, ibid. 42 (1982), 177–226.

[Sh:b] —, Proper Forcing , Lecture Notes in Math. 940, Springer, Berlin, 1982.

[Sh:g] —, Cardinal Arithmetic, Oxford Logic Guides 29, Oxford Univ. Press, 1994.

[Sh 481] —, Was Sierpi´ nski right? III Can continuum-c.c. times c.c.c. be continuum- c.c.? Ann. Pure Appl. Logic 78 (1996), 259–269.

[Sh:F254] —, More on Kalikow Property of pairs of cardinals.

[Sh 513] —, PCF and infinite free subsets, Arch. Math. Logic, to appear.

[Si70] J. S i l v e r, A large cardinal in the constructible universe, Fund. Math. 69 (1970), 93–100.

Institute of Mathematics

The Hebrew University of Jerusalem 91904 Jerusalem, Israel

E-mail: shelah@math.huji.ac.il

Department of Mathematics Rutgers University New Brunswick, NJ 08854, U.S.A.

URL: http://www.math.rutgers.edu/∼shelah

Received 2 September 1996;

in revised form 9 August 1999