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

154 concept calculus: much better thanproof Let R be a finite wellordering. We first prove that for all x ∈ A, everyf: {y: yRx}→{y: yRx} is onto. Let x be an R least counterexample. Letf: {y: yRx}→{y: yRx} be one-one and not onto. Then f is not empty. Let ube the R greatest element of dom(f ). We can obviously adjust f so that it is oneoneand not onto, and f (u) = u. Now delete f (u) = u, to obtain g: {y: yRu}→{y: yRu}, which is one-one and not onto. This is a contradiction.Suppose x< fin y. Let f: x → ybe one-one. If f is onto, then by taking theinverse function, y≤ fin x, which is impossible.Suppose f: x → y is one-one but not onto. Then x ≤ fin y.Ify≤ fin x, then bycomposition, we obtain a one-one function from x into x that is not onto. Thiscontradicts the first claim.Lemma 8.14 Let R, S be finite wellorderings. Then fld(R) ≤ fin fls(S)↔ R ≤ PWOS, fld(R) < fin fls(S)↔ R< PWO S, fld(R) = fin fls(S)↔ R = PWO S.IfA,B are finite“sets” and A is a proper “subset” of B, then A < fin B.proof Let R, S be finite wellorderings. Suppose R< PWO S. Let f be thecomparison map from R onto S|