Chapter 2. Ordinals, well-founded relations.∗
2.1. Well-founded Relations.
We start with some definitions and rapidly reach the notion of a well-ordered set.
Definition. For any X and any binary relation α and α − β = {α − γ | γ 0. Then there are unique σ, τ with τ < β such that α = β.σ + τ .
Proof. Let σ = { γ | β.s(γ) ≤ α } and τ = α − β.σ, = {δ | β.σ + δ < α }. Hence α = β.σ + τ .
In order to see that τ < β, observe that if not either β = τ , when α = β.σ + β = β.s(σ) or
β ∈ {δ | β.σ + δ < α } and, so, β.σ + β = β.s(σ) < α. So in any case β.s(σ) ≤ α and σ ∈ σ, a
contradiction!
Proposition. If x ⊆ On then α. x= { α.β | β ∈ x}. (Which means that . is continuous in the
second place/variable.)
Proof. α. x = {α.γ + δ | δ < α & (∃β ∈ x γ < β)} = { α.γ + δ | ∃β ∈ x (γ < β & δ < α)}
= { α.β | β ∈ x}.
Note. In the previous theorem σ = {γ | β.γ ≤ α }.
Ordinals arithmetic: 2.11 Exponentiation of ordinals.
Next we make another definition by recursion and define exponentiation.
Definition. Let α, β ∈ On. Then αβ = {0 } ∪ {αγ .δ + | γ < β & δ < α & < αγ }.
Proposition. Let α, β ∈ On. Then αβ ∈ On.
Lemma. ∀α, β, γ δ ∈ On we have (i) αβ .αγ = αβ+γ ; (ii) (αβ )γ = αβ.γ ; (iii) β < γ =⇒ αβ < αγ ;
and (iv) α ≤ δ =⇒ αβ ≤ δ β .
Proposition. Let α ∈ On. (i) α0 = 1; (ii) αs(β) = αβ .α for each β ∈ On; and (iii) if λ is a limit
ordinal, then αλ = {αγ | γ < λ}.
The proofs are similar to the proofs of the equivalents for ‘.’. You should check these if you desire.
ω
Note. ∀n ∈ ω nω = ω, but ω ω = ω, when ω.ω ω = ω ω . And if we define 0 = { ω, ω ω , ω ω , . . . },
then ω 0 = 0 .
When you come back to this section later you should remember that all of these facts are for
exponentiations of ordinals, that is very different from the theory of cardinal exponentiation that
we will see in Chapter (4).
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