By W. Weiss
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Example text
X)(∃ <) [ X, < is a well ordered set]. Proof. We begin by using Theorem 5 to obtain a choice function f : P(X) \ {∅} → X such that for each nonempty A ⊆ X we have f (A) ∈ A. By recursion on ON we define g : ON → X ∪ {X} as: g(β) = f (X \ {g(α) : α < β}), if X \ {g(α) : α < β} = ∅; X, otherwise. 1) Now replace each x ∈ X ∩ran(g) by the unique ordinal β such that g(β) = x. The Axiom of Replacement gives the resulting set S ⊆ ON, where S = {β ∈ ON : g(β) ∈ X}. By Theorem 10 there is a δ ∈ ON \ S.
Assume (∃z)(z = ON). , an ordinal. This leads to the contradiction ON ∈ ON. Theorem 11. (Trichotomy of Ordinals) (∀α ∈ ON)(∀β ∈ ON)(α ∈ β ∨ β ∈ α ∨ α = β). Proof. The reader may check that a proof of this theorem can be obtained by replacing “N” with “ON” in the proof of Theorem 7. Because of this theorem, when α and β are ordinals, we often write α < β for α ∈ β. Since N ⊆ ON, it is natural to wonder whether N = ON. In fact, we know that “N = ON” can be neither proved nor disproved from the axioms that we have stated (provided, of course, that those axioms are actually consistent).
Theorem 12. (∃z)(z ∈ ON ∧ z = N). Proof. Since N ⊆ ON and N = ON, pick α ∈ ON \ N. We claim that for each n ∈ N we have n ∈ α; in fact, this follows immediately from the trichotomy of ordinals and the transitivity of N. Thus N = {x ∈ α : x ∈ N} and by Comprehension ∃z z = {x ∈ α : x ∈ N}. The fact that N ∈ ON now follows immediately from Theorem 6. , ω = N. Theorems 6 and 12 now show that the natural numbers are the smallest ordinals, which are immediately succeeded by ω, after which the rest follow.
An Introduction to Set Theory by W. Weiss
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