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

interpretation of z in b + sde 153proof ≤ PWO is obviously reflexive. It is transitive by using composition ofbinary “relations.” It is connected by Lemma 8.11.If R< PWO R, then we have two different comparison relations between R, R,which is impossible. Hence, < PWO is irreflexive. S uppose R< PWO S ∧ S< PWOT . Then R< PWO T by composition of “relations.”Suppose R ≤ PWO S.If¬R < PWO S, then S ≤ PWO R, and so R ≤ PWO S.Conversely, obviously R< PWO S and R = PWO S separately imply R ≤ PWO S.Suppose R = PWO S. Then the comparison relation between R, S is from R,and the comparison relation between S, R is from S. If either comparison relationis not onto, then by composition, we get a comparison relation from R onto Sthat is not onto, or a comparison relation from S onto R that is not onto. Thiscontradicts R = PWO S and the uniqueness of comparison relations. Hence, bothcomparison relations are onto. Conversely, if the comparison relation between R,S is from R onto S, then R ≤ PWO S ∧ S ≤ PWO R.Suppose R< PWO S. If the comparison relation between R, S is onto S, thenS ≤ PWO R, which is a contradiction. If the comparison relation between R, S isfrom S, then S ≤ PWO R, which is a contradiction.Suppose the comparison relation between R, S is from R onto some S| < x.Then R ≤ PWO S.IfS ≤ PWO R, then the comparison relation between R, S iseither from R onto S or from S onto some R|< x. This violates the uniquenessof the comparison relation.Let R, S be prewellorderings. Then R ≤ PWO S ∨ S ≤ PWO R. Hence, R< PWOS ∨ S< PWO R ∨ R = PWO S. If more than one disjunct holds, then we violate theirreflexivity of < PWO .Lemma 8.12 Let R be a nonempty 6-ary “relation” of full codes forprewellorderings. There is a prewellordering S with a full code in R such thatevery prewellordering T with a full code in R has S ≤ PWO T .proof Let S be a prewellordering with a code in R. Look at the “set” B ofall x ∈ fld(S) such that the comparison relation between some S ′ with full codein R, and S, maps S ′ onto S|< x.IfB is empty, then S is as required. SupposeB is nonempty. Let y be an S least element of B. Let T have full code in R,where the comparison relation between T , S maps T onto S|< y. T hen T is asrequired.Definition A finite wellordering is a prewellordering whose equality relation isidentity, where every nonempty “subset” of the field has a greatest element.Definition A finite “set” is the field of some finite wellordering R.Definition We write x ≤ fin y if and only if x, y are finite “sets” such that thereis a one-one function from x into y.x < fin y ↔ x ≤ fin y ∧¬y ≤ fin x. x = finy ↔ x ≤ fin y ∧ y≤ fin x.Lemma 8.13 Every one-one function from a finite “set” into itself is onto. Letx, y be finite “sets.” Then x< fin y ↔ there is a one-one function from x into ythat is not onto.