Zermelo-Fraenkel Set Theory

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CHAPTER 4. ORDINALS 30 4.4 Fixed Point Hierarchies Compare Theorem 3.27 (p. 20). Definition 4.14 Let H be a monotone operator over the class U satisfying the property that X ∈ V ⇒ H(X) ∈ V (“if X is a set, then so is H(X)” — this is needed for the following hierarchy to exist). The least fixed point hierarchy asociated with H is the sequence {H↑α} α∈OR of stages H↑α, recursively defined by • H↑0 = ∅ • H↑(α + 1) = H(H↑α) • H↑γ = ⋃ ξ α, H↑β = H↑α), then the least such ordinal α is called the closure ordinal of the hierarchy. □ Let us call H ⋆-finite if, whenever a ∈ H(X), there is a subset x ⊂ X such that a ∈ H(x). Theorem 4.15 Suppose that H is ⋆-finite, monotone over U, and maps sets to sets. Then: 1. the least fixed point hierarchy associated with H is cumulative: α < β ⇒ H↑α ⊂ H↑β, 2. ⋃ α H↑α is the least fixed point H↑ of H, 3. if U is a set, then a closure ordinal exists. Proof. 1. Induction w.r.t. β. 2. Closure, H( ⋃ α H↑α) ⊂ ⋃ α H↑α: Assume that a ∈ H( ⋃ α H↑α). Since H is ⋆-finite, we have that a ∈ H(X) for some subset X ⊂ ⋃ α H↑α. So, ∀x ∈ X ∃α ∈ OR x ∈ H↑α. By Exercise 68, for some β ∈ OR, ∀x∈X ∃α∈β x ∈ H↑α. Thus, X ⊂ H↑β, and a ∈ H(X) ⊂ H(H↑β) = H↑(β + 1) ⊂ ⋃ α H↑α. Induction, H(X) ⊂ X ⇒ ⋃ α H↑α ⊂ X: Assume that H(X) ⊂ X. By induction on α it follows that H↑α ⊂ X. 3. If U is a set and a closure ordinal does not exist, then the least fixed point hierarchy (the map α ↦→ H↑α) constitutes an injection of OR into ℘(U). This contradicts Exercise 14 p. 9. □ By the argument for 4.15.3, the closure ordinal has power at most |℘(U)|. From the Axiom of Choice it even follows that the closure ordinal has power at most |U|: cf. Exercise 106 (p. 40). Exercises 69 ♣ Show that, for α ∈ OR, T↑α = α. (N.B.: T is the operator that generates OR.)
CHAPTER 4. ORDINALS 31 70 ♣ Show that the single recursion equation H↑α = ⋃ ξ
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Page 4 and 5: CONTENTS 2 7.6 Consistency of V = L
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CHAPTER 8. FORCING 80 By the same a
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CHAPTER 8. FORCING 84 Some examples
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CHAPTER 8. FORCING 88 Exercises 226
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CHAPTER 8. FORCING 102 Thus, ccc eq
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CHAPTER 8. FORCING 108 Proof. Choos
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CHAPTER 8. FORCING 110 Therefore, G
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Solutions to Exercises Chapter 2 5.
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B by x ≺ y ≡ the
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Zermelo-Fraenkel Set Theory

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