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 is a complete condition, N (H(χ),
, N , . . .) countable with α N and r N extends p N, then p r can be
extended to a condition in 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
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
k
is a subgraph of rk(0) ran(fp(β)) (which in itself is a
graph), and the universe of
k
rk(0) is exactly
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 δk, then
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 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
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