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 ∩ N β 0 determines the τ ˜ β structure of fp(β), and we also have that N β 0 ∈ 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 ⊆ iω 1 N γ i . Proof of the Claim. Let β γ. Since γ ∈ N γ 0 there is f ∈ N γ 0 which maps ω1 onto γ. There is i ω1 such that f −1 (β) ∈ N γ i and then β ∈ N γ i . 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}, (sk)β∈a β k , γk where each sk β 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 sk β 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 ∩ Pγ k and it extends p γk ∩ Nk and rm γk for all m k, • if β ≤ γk is active, then rk β determines the τ ˜ β structure of dom(fp(β))∪ m≤k dom(fr m (β) ) ∩ δk and rk β forces that fp(β) ∪ m≤k fr m (β) δk is an isomorphism of this structure with sk, β • if β = β are both in ak and satisfy Aβ ∩Aβ δk, then sk β ∩sk β ⊆ 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 s0 β = 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(γ k+1 ) \ uk γk+1 in N γ i and satisfying Nk ∩ γk+1 ⊆ N γk+1 i (note that such i exists by Claim 3.3). Let Nk+1 = N γk+1 i and ak+1 = ak ∪[dom(p)∩Nk+1 ∩γk+1]. If 3 By p ∪ r we mean the function q whose domain is dom(p) ∪ dom(r), with q(0) = p(0) ∪ r(0) and q(β) = (X p(β) ∪ X r(β) , f p(β) ∪ f r(β) ) for β 0 in dom(q).

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