By W. Weiss

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Z z = {κ : κ is a cardinal}. Theorem 25. For any infinite cardinal κ, |κ × κ| = κ. Proof. 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 .

Define f to be the function f = g ∪ { x, y }. It is straightforward to verify that f witnesses that x, y ∈ F . To prove (2), note that, by (1), for each x ∈ N there is n ∈ N and f : n → V such that F (x) = f (x) and, in fact, F |n = f . Hence, (∀m ∈ n) Φ(m, f |m, f (m), w). 38 CHAPTER 4. THE NATURAL NUMBERS We prove (3) by induction. Assume that (∀m ∈ n) H(m) = F (m) with intent to show that H(n) = F (n). We assume Φ(n, H|n, H(n), w) and by (2) we have Φ(n, F |n, F (n), w). By the hypothesis of the theorem applied to H|n = F |n we get H(n) = F (n).

In this case n is said to be the size of X. Otherwise, X is said to be infinite. Exercise 6. Use induction to prove the ”pigeon-hole principle”: for n ∈ N there is no injection f : (n + 1) → n. Conclude that a set X cannot have two different sizes. Do not believe this next result: Proposition. All natural numbers are equal. Proof. It is sufficient to show by induction on n ∈ N that if a ∈ N and b ∈ N and max (a, b) = n, then a = b. If n = 0 then a = 0 = b. Assume the inductive hypothesis for n and let a ∈ N and b ∈ N be such that max (a, b) = n + 1.