By W. Weiss

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Let κ be an infinite cardinal. The formulas |κ| = |κ × {0}| and |κ × {0}| ≤ |κ × κ| imply that κ ≤ |κ × κ|. We now show that |κ × κ| ≤ κ. We use induction and assume that |λ × λ| = λ for each infinite cardinal λ < κ. We define an ordering on κ × κ by: max {α0 , β0 } < max {α1 , β1 }; α0 , β0 < α1 , β1 iff max {α0 , β0 } = max {α1 , β1 } ∧ α0 < α1 ; or, max {α0 , β0 } = max {α1 , β1 } ∧ α0 = α1 ∧ β0 < β1 . It is easy to check that < well orders κ × κ. Let θ = type κ × κ, < . It suffices to show that θ ≤ κ.

5. We say a relation R is total on C whenever ∀x ∈ C ∀y ∈ C [ x, y ∈ R ∨ y, x ∈ R ∨ x = y]. 6. We say R is extensional on C whenever ∀x ∈ C ∀y ∈ C [x = y ↔ ∀z ∈ C ( z, x ∈ R ↔ z, y ∈ R)]. 53 54 CHAPTER 6. RELATIONS AND ORDERINGS An example of a relation R on an ordinal is given by the membership relation: x, y ∈ R iff x ∈ y. R satisfies all the above properties. On the other hand, if α ∈ / ω, the reverse relation R given by x, y ∈ R iff y ∈ x is not well founded but nevertheless has all the other properties.

Using the penultimate lemma, let β0 = max {β : ω β ≤ α} and then let m0 = max {m ∈ ω : ω β0 m ≤ α} which must exist since ω β0 m ≤ α for all m ∈ ω would imply that ω β0 +1 ≤ α. 49 By the previous lemma, there is some α0 ∈ ON such that α = ω β0 m0 + α0 where the maximality of m0 ensures that α0 < ω β0 . Now let β1 = max {β : ω β ≤ α0 } so that β1 < β0 . Proceed to get m1 = max {m ∈ ω : ω β1 m ≤ α0 } and α1 < ω β1 such that α0 = ω β1 m1 + α1 . We continue in this manner as long as possible. We must have to stop after a finite number of steps or else β0 > β1 > β2 > .