8 MIRNA
DZAMONJAˇ
no such choice is possible, stop the induction. Note that m k =⇒ γm γk and
that the induction must be stopped at some stage k∗ as all ordinals that appear as
unused at any stage come from the finite set
γ∈dom(p)\{0}
dom(fp(γ)) (and once
an ordinal is used it remains so). Let
γ∗
= γk∗ .
We shall now proceed to define a condition q in Nγ∗
γ∗.
We shall need the
following claim.
Claim 3.4. For all k
k∗,
m≤k
Nm γk Nk.
Proof of the Claim. The proof is by induction on k. If k = 0, the situation is
clear. For the case of k + 1, we have by the induction hypothesis that for m k,
Nm γk Nk. Since γk+1 γk, certainly Nm γk+1 Nk and hence it suffices to
show that Nk γk+1 Nk+1. This conclusion follows by the choice of Nk+1.
3.4
Suppose that k
k∗
and m k. Then rm Nm so all the finitely many
ordinals involved in the construction of rm γk are in Nm γk, and hence in Nk
by Claim 3.4. Therefore rm γk Nk.
At this point of the proof we shall get to use the Amalgamation Condition
(∗), which we define in terms of a partially ordered class (K, ≤) of structures:
(K, ≤) satisfies a.c. (∗) if for any elements A, B, {Ci : i I} of K satisfying
any two of these structures agree on their intersection
whenever i = j then either Ci Cj A or Ci Cj B,
there is a structure in K which extends all A, B, {Ci : i I}.
Obviously, the class of graphs with the induced subgraph relation satisfies the
a.c. (∗) since we can simply take a union of all the graphs
involved.4
Suppose now that β = β ak∗ . By the induction hypothesis either
δk∗ , in which case
sβ∗
k
sβ∗
k p(0) or δk∗ and so
sβ∗
k
sβ∗
k
rk∗ (0). This says that, according to the a.c. (∗) property, we can define a graph
that extends p(0)∩δk∗ ∪rk∗ (0)∪
β∈ak∗
sβ∗
k ∩δk∗ . Let s be such a graph. We define
q , a candidate for a condition in Pα. Let q (0) = s and for β
k≤k∗
dom(rk)∪ak∗
let q (β) = p(β) δk∗
k≤k∗
rk(β). Note that q as a function is an element of
Nk∗ . We now claim that q is a condition in Pα. By induction on 1 β α we
prove that q β Pβ. For β = 1 this follows from the choice of s and for β limit
this is clear. Suppose that β = γ + 1 and γ is in dom(q ). If γ / dom(p) then
γ
k≤k∗
dom(rk) and q (γ) =
k≤k∗
rk(γ). This is well defined by the choice
of
rk’s. If γ dom(p) \
k≤k∗
dom(rk), then q (γ) = p(γ) δk∗ , so q β is a
condition because p satisfies (i) in the definition of completeness. Finally it may
happen that γ dom(p)
k≤k∗
dom(rk). In particular we have β ak∗ . Let k be
maximal such that γ γk. We claim that dom(fp(γ)) δk∗ = dom(fp(γ)) δm. If
this were not to be the case then at the stage k of the induction there would be an
unused ordinal in dom(fp(γ)), so γ = γk+1, a contradiction. Having established the
existence of q we now need to keep going to extend to a condition which extends
each rk. This is done in stages using the assumptions of completeness, for which
we refer the reader to the original article.
3.2
We now indicate how Lemma 3.2 implies that the forcing is ccc. We assume
that we have proved that the set of complete conditions is dense, so we only work
4In
[9], Lemma 1, it is shown that a better known P(3−) amalgamation property implies
a.c. (∗).
8 8
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