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X| ≥ |y| iff ∃ surjection f : x → y. Assume here that y = ∅. Theorem 24. (G. Cantor) ∀x |x| < |P(x)|. Proof. First note that if |x| ≥ |P(x)|, then there would be a surjection g : x → P(x). But this cannot happen, since {a ∈ x : a ∈ / g(a)} ∈ / g → (x). 61 For any ordinal α, we denote by α+ the least cardinal greater than α. This is well defined by Theorem 24. Exercise 21. Prove the following: 1. The supremum of a set of cardinals is a cardinal. 2. ¬∃z z = {κ : κ is a cardinal}. Theorem 25. For any infinite cardinal κ, |κ × κ| = κ.

Show that ω + 1 is not a cardinal and that, in fact, each other cardinal is a limit ordinal. Theorem 23. The following are equivalent. 1. κ is a cardinal. 2. , |κ| = κ. 3. (∀α < κ)(¬∃ injection f : κ → α). 59 60 CHAPTER 7. CARDINALITY Proof. We prove the negations of each are equivalent: ¬(1) (∀x)[(∃ bijection g : x → κ) → (∃α < κ)(∃ bijection h : x → α)] ¬(2) ∃α < κ ∃ bijection f : κ → α ¬(3) ∃α < κ ∃ injection f : κ → α ¬(2) ⇒ ¬(1) Just take h = f ◦ g. ¬(1) ⇒ ¬(3) Just consider x = κ. ¬(3) ⇒ ¬(2) Suppose α < κ and f : κ → α is an injection.

A relation R on X with the property that f (x) < f (y) whenever x, y ∈ R must be a well founded relation. In fact, this turns out to be a characterisation. Theorem 19. Let R be a relation on a set X. R is well founded iff there is an ordinal δ and a surjection f : X → δ such that f (x) < f (y) whenever x, y ∈ R. Proof. We treat only the forward implication. Using recursion on ON we define g : ON → P(X) by g(β) = x : x is a minimal element of X \ {g(α) : α < β} . From g we obtain f : X → ON by f (x) = the unique α ∈ ON with x ∈ g(α), if possible; 0, otherwise.

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