Page 76 - Textos de Matemática Vol. 38

68
PETER KOEPKE
Set
M = {I(x)|x ∈ }.
Proof . Consider y ∈ I(x) ∈ M. Choose v Rx x such that y = Ix(v). Then
(1) M is transitive. (v,Rx) ∈ and
y = Ix(v) = I((v,Rx)) ∈ M.
qed(1) (2) The function I : → M is surjective and preserves ≡ and =, and ◭ and ∈,
The above definitions imply: resp.:
∀x, y ∈ : ((x ≡ y ↔ I(x) = I(y)) ∧ (x ◭ y ↔ I(x) ∈ I(y))).
Hence
(3) M is a transitive ∈-model of the ZFC-axioms, i.e., M is an inner model. (4) S = {v ∈ M|v ⊆ Ord}.
Proof . Let v ∈ S. We build a point that will be interpreted as v. Choose an ordinal α such that v ⊆ α. Define a wellfounded relation Rx on α + 1 by
ξ Rx ζ iff (ξ < ζ < α or (ζ = α ∧ ξ ∈ v)).
Then x = (α,Rx) is a point. Let Ix(u) = {Ix(v)|v Rx u} be the recursive interpretation function for x. For ζ < α we have Ix(ζ) = ζ since we have inductively
And then
Ix(ζ) = {Ix(ξ)|ξ Rx ζ} = {ξ|ξ < ζ} = ζ. I(x)=Ix(α)={Ix(ξ)|ξRx α}={ξ|ξ∈v}=v.
Hence v = I(x) ∈ M.
The previous argument also shows that one may canonically represent an
ordinal ξ by the point (ξ, <↾ (ξ + 1)2):
I((ξ, <↾ (ξ + 1)2)) = ξ.
For the converse inclusion consider some v ∈ M, v ⊆ α ∈ Ord. Choose a point x ∈ such that I(x) = v. Since S satisfies the separation schema,
v = {ξ < α|ξ ∈ v} = {ξ < α|S (ξ, <↾ (ξ + 1)2) ◭ x} ∈ S.
qed(4)
The model M is unique since it is determined by its sets of ordinals (see [6], Theorem 13.28). ✷