By W. Weiss

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**Extra resources for An Introduction to Set Theory**

**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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