Page 54 - Textos de Matemática Vol. 38

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PETER KOEPKE
This leads to
Definition 2.4. A set x is an ordinal number, Ord(x), if Trans(x) ∧ ∀y ∈
x Trans(y). Let
be the class of all ordinals.
Ord = {x|x is an ordinal number}
The class Ord contains the above natural numbers. We use small greek letters α, β, γ, . . . as variables ranging over ordinals. We write α < β instead of α ∈ β and α β instead of α < β ∨ α = β. Under appropriate set-theoretic axioms the class Ord is strongly well-ordered by the relation <. Let us recall the axiom of foundation which asserts the existence of ∈-minimal elements of sets:
∀x(∃yy ∈ x → ∃y(y ∈ x ∧ ¬∃z(z ∈ x ∧ z ∈ y))). Theorem 2.5. a) The class Ord is transitive.
b) Ord is linearly ordered by <. c) Ord is well-ordered by <, i.e.,
∀x ⊆ Ord(x ̸= ∅ → ∃α ∈ x∀β < αβ ̸∈ x).
Proof. a) Let x ∈ α ∈ Ord. Since α is an ordinal we have Trans(x). Consider y ∈ x. Since α is transitive we have x ∈ α and so Trans(x). Thus ∀y ∈ x Trans(y) and x ∈ Ord.
b) Let α, β, γ ∈ Ord and α < β < γ. Then α < γ by the transitivity of the ordinal γ.
Let α ∈ Ord. Then α ̸∈ α and so α ≮ α.
For the linearity of < assume that there are ordinals α, β such that
α ≮ β, α ̸= β, and β ≮ α.
By the axiom of foundation we can assume that α is minimal with that property, and that with respect to α the ordinal β is minimal with that property. We claim that α = β.
Let ξ ∈ α. By the minimality of α we have ξ < β or ξ = β or β < ξ. Assume ξ = β or β < ξ. Then β < α contradicting the minimal choice of α. Hence ξ ∈ β.
Conversely let ξ ∈ β. By the minimality of β we have ξ < α or ξ = α or α < ξ. Assume ξ = α. Then α < β, contradicting the choice of α and β. Assume α < ξ. Then again α < β, contradiction. Thus ξ ∈ α.
But α = β contradicts the choice of α and β.
c) follows directly from the axiom of foundation. ✷