SOME POSITIVE RESULTS IN THE CONTEXT OF UNIVERSAL MODELS 7

is a condition. The proof of this is by induction on α, the main case being the case

α = β + 1. We have by (i) that p β ∩ N0

β

determines the τ

˜

β

structure of fp(β),

and we also have that N0

β

∈ N so p β ∩ N, which is a condition by the induction

hypothesis is stronger than p β ∩N0

β

and hence it too determines the τ

˜

β

structure

of fp(β). We also need to note that the range of fp(β) is contained in N ∩ ω1, which

follows as fp(β) is ﬁnite.

Lemma 3.2. For any α ≤ ω2, if p ∈ Pα is a complete condition, N ≺ (H(χ), ∈

, N , . . .) countable with α ∈ N and r ∈ Pα ∩ N extends p ∩ N, then p ∪ r can be

extended to a condition in Pα which extends both p and

r3.

Proof. We shall present the main part of the proof, from which it can be seen

where the amalgamation condition is being used. The ﬁnal part of the proof will

only be indicated. Let us assume α ≥ 1, as otherwise the conclusion is trivial.

We shall need to refer to a following observation about elementary submodels.

Claim 3.3. For any γ ≤ ω2, γ ⊆

iω1

Niγ

.

Proof of the Claim. Let β γ. Since γ ∈ N0

γ

there is f ∈ N0

γ

which maps ω1

onto γ. There is i ω1 such that f

−1(β)

∈

Niγ

and then β ∈

Niγ

.

3.3

The construction of the desired condition is by induction on k ω, where the

induction will stop after some ﬁnite number of stages. At each stage we deﬁne

Nk ≺ (H(χ), ∈, . . .) countable, rk ∈ Pα, ak ⊆ dom(p) \ {0}, (sβ)β∈ak

k

, γk

where each sβ

k

is a ﬁnite graph on a subset of ω1. For each k 0, γk is a special

element of ak, called the leading ordinal, with γ0 = ∞. The elements of ak are

called the active ordinals. We denote δk = Nk ∩ω1. The following are our inductive

hypotheses:

• If β ∈ ak then sβ

k

is a subgraph of rk(0) ∪ ran(fp(β)) (which in itself is a

graph), and the universe of sβ

k

∩ rk(0) is exactly sβ

k

∩ δk,

• if k 0 then there is i ω1 such that Nk =

Niγk

,

• rk ∈ Nk ∩ Pγk and it extends p γk ∩ Nk and rm γk for all m k,

• if β ≤ γk is active, then rk β determines the τ

˜fp(β)

β

structure of dom(fp(β))∪

m≤k

dom(frm(β)) ∩ δk and rk β forces that ∪

m≤k

frm(β) δk is

an isomorphism of this structure with sβ,k

• if β = β are both in ak and satisfy Aβ ∩Aβ δk, then sβ

k

∩sβ

k

⊆ p(0).

We say that for β ∈ ak, the elements of dom(fp(β)) that have been used at the stage

k are those in dom(fp(β)) ∩ δm, where m = max{l ≤ k : γl ≥ β} (since γ0 = ∞,

this maximum is well deﬁned). We denote the set of such ordinals by

uβ.k

The induction is not diﬃcult to do: we let N0 = N, r0 = r and a0 = dom(p) \

{0} ∩ N. For β ∈ a0 let sβ 0 = ran(r(β)) with the structure induced by r(0)- this is

well deﬁned as p ∩ N ≤ r. If we have deﬁned all the relevant objects at the stage k,

do the following if possible: choose a minimal γ = γk+1 ∈ ak such that there is an

unused element of dom(fp(γ)). Let i be the minimal such that there is an element

of dom(fp(γk+1 ) \ uγk+1 k in

Niγ

and satisfying Nk ∩ γk+1 ⊆

Niγk+1

(note that such i

exists by Claim 3.3). Let Nk+1 =

Niγk+1

and ak+1 = ak ∪[dom(p)∩Nk+1 ∩γk+1]. If

3By

p ∪ r we mean the function q whose domain is dom(p) ∪ dom(r), with q(0) = p(0) ∪ r(0)

and q(β) = (Xp(β) ∪ Xr(β), fp(β) ∪ fr(β)) for β 0 in dom(q).

7 7

is a condition. The proof of this is by induction on α, the main case being the case

α = β + 1. We have by (i) that p β ∩ N0

β

determines the τ

˜

β

structure of fp(β),

and we also have that N0

β

∈ N so p β ∩ N, which is a condition by the induction

hypothesis is stronger than p β ∩N0

β

and hence it too determines the τ

˜

β

structure

of fp(β). We also need to note that the range of fp(β) is contained in N ∩ ω1, which

follows as fp(β) is ﬁnite.

Lemma 3.2. For any α ≤ ω2, if p ∈ Pα is a complete condition, N ≺ (H(χ), ∈

, N , . . .) countable with α ∈ N and r ∈ Pα ∩ N extends p ∩ N, then p ∪ r can be

extended to a condition in Pα which extends both p and

r3.

Proof. We shall present the main part of the proof, from which it can be seen

where the amalgamation condition is being used. The ﬁnal part of the proof will

only be indicated. Let us assume α ≥ 1, as otherwise the conclusion is trivial.

We shall need to refer to a following observation about elementary submodels.

Claim 3.3. For any γ ≤ ω2, γ ⊆

iω1

Niγ

.

Proof of the Claim. Let β γ. Since γ ∈ N0

γ

there is f ∈ N0

γ

which maps ω1

onto γ. There is i ω1 such that f

−1(β)

∈

Niγ

and then β ∈

Niγ

.

3.3

The construction of the desired condition is by induction on k ω, where the

induction will stop after some ﬁnite number of stages. At each stage we deﬁne

Nk ≺ (H(χ), ∈, . . .) countable, rk ∈ Pα, ak ⊆ dom(p) \ {0}, (sβ)β∈ak

k

, γk

where each sβ

k

is a ﬁnite graph on a subset of ω1. For each k 0, γk is a special

element of ak, called the leading ordinal, with γ0 = ∞. The elements of ak are

called the active ordinals. We denote δk = Nk ∩ω1. The following are our inductive

hypotheses:

• If β ∈ ak then sβ

k

is a subgraph of rk(0) ∪ ran(fp(β)) (which in itself is a

graph), and the universe of sβ

k

∩ rk(0) is exactly sβ

k

∩ δk,

• if k 0 then there is i ω1 such that Nk =

Niγk

,

• rk ∈ Nk ∩ Pγk and it extends p γk ∩ Nk and rm γk for all m k,

• if β ≤ γk is active, then rk β determines the τ

˜fp(β)

β

structure of dom(fp(β))∪

m≤k

dom(frm(β)) ∩ δk and rk β forces that ∪

m≤k

frm(β) δk is

an isomorphism of this structure with sβ,k

• if β = β are both in ak and satisfy Aβ ∩Aβ δk, then sβ

k

∩sβ

k

⊆ p(0).

We say that for β ∈ ak, the elements of dom(fp(β)) that have been used at the stage

k are those in dom(fp(β)) ∩ δm, where m = max{l ≤ k : γl ≥ β} (since γ0 = ∞,

this maximum is well deﬁned). We denote the set of such ordinals by

uβ.k

The induction is not diﬃcult to do: we let N0 = N, r0 = r and a0 = dom(p) \

{0} ∩ N. For β ∈ a0 let sβ 0 = ran(r(β)) with the structure induced by r(0)- this is

well deﬁned as p ∩ N ≤ r. If we have deﬁned all the relevant objects at the stage k,

do the following if possible: choose a minimal γ = γk+1 ∈ ak such that there is an

unused element of dom(fp(γ)). Let i be the minimal such that there is an element

of dom(fp(γk+1 ) \ uγk+1 k in

Niγ

and satisfying Nk ∩ γk+1 ⊆

Niγk+1

(note that such i

exists by Claim 3.3). Let Nk+1 =

Niγk+1

and ak+1 = ak ∪[dom(p)∩Nk+1 ∩γk+1]. If

3By

p ∪ r we mean the function q whose domain is dom(p) ∪ dom(r), with q(0) = p(0) ∪ r(0)

and q(β) = (Xp(β) ∪ Xr(β), fp(β) ∪ fr(β)) for β 0 in dom(q).

7 7