Heller M, Woodin W.H. (eds.) Infinity. New research frontiers (CUP, 2011)(ISBN 1107003873)(O)(327s)_MAml_

144 concept calculus: much better thanproof Let α, β α ) transitive. Hence, A ⊆ D α . We claim that A = {y: x > α+1 y}. To seethis, suppose x > α+1 y.Nowx > α y is impossible, given that x /∈ D α . Hence,y ∈ A. Conversely, suppose y ∈ A. Then x > α+1 y.For claim iii, let x ∈ D α . Let α ′ be least such that x ∈ D α ′. Then α ′ ≤ α andx ∈ D α ′\D α ′ −1. By ii, x = (α ′ − 1, {y : x> α′ y}). Set β = α ′ − 1.We prove claim iv by transfinite induction on α. Forα = 0, the statement isvacuously true given that D 0 =⊘. Suppose(∀x,y)(x > α y → (∃β α+1 y.Ifx> α y, then (∃β β+1 y.Ifx> β y, then x> α y. Otherwise, let x = (β,A),y ∈ A, where A is (D β , > β ) transitive. By iii, β α y.(∀x,y)(x,y ∈ D α (x> λ y → x> α y)).Lemma 6.2For all α, (D α ,> α ) is irreflexive and transitive.proof We prove this by transfinite induction on α. Clearly, (D 0 , > 0 )isirreflexive and transitive. Suppose (D α , > α ) is irreflexive and transitive. Supposex > α+1 x. By Lemma 5.1, x ∈ D α , and so x > α x. This violates the irreflexivityof > α .To see that (D α+1 , > α+1 ) is transitive, let x > α+1 y, y > α+1 z, where x,y,z ∈D α+1 . Then y ∈ D α ,z∈ D α .Ifx ∈ D α , then x> α z, x > α+1 z. Suppose x ∈D α+1 \D α . Let x = (α, A), where A is (D α ,> α ) transitive. Then y ∈ A, y > α z.Hence, z ∈ A, and so x > α+1 z. Hence, (D α+1 , > α+1 ) is irreflexive and transitive.Suppose for all β β ) is irreflexive and transitive. Then obviously(D λ , > λ ) is irreflexive. Let x > λ y ∧ y > λ z, where x,y,z ∈ D λ .Let x,y,z ∈ D β ,β β y ∧ y> β z. Therefore, x> β z. Hence,(D λ ,> λ ) is transitive.Definition Fix S to be a nonempty set of limit ordinals, with no greatest element,whose union is the limit ordinal λ. We define M[S] to be the following