前スレ 808より (参考)(再掲) 631より en.wikipedia.org/wiki/Well-ordering_theorem Well-ordering theorem Proof from axiom of choice The well-ordering theorem follows from the axiom of choice as follows.[9] Let the set we are trying to well-order be A, and let f be a choice function for the family of non-empty subsets of A. For every ordinal α, define an element aα that is in A by setting aα= f(A∖{aξ∣ξ<α}) if this complement A∖{aξ∣ξ<α} is nonempty, or leave aα undefined if it is. That is, aα is chosen from the set of elements of A that have not yet been assigned a place in the ordering (or undefined if the entirety of A has been successfully enumerated). Then the order < on A defined by aα<aβ if and only if α<β (in the usual well-order of the ordinals) is a well-order of A as desired, of order type sup{α∣aα is defined}. Notes 9＾ Jech, Thomas (2002). Set Theory (Third Millennium Edition). Springer. p. 48. ISBN 978-3-540-44085-7. (引用終り)
Thomas Jechの 証明 再録（前スレ 848より） P48 Theorem 5.1 (Zermelo’s Well-Ordering Theorem) Every set can be well-orderd. Proof： Let A be a set. To well-order A, it suffices to construct a transfinite one-to-one sequence （aα: α < θ) that enumerates A. That we can do by induction, using a choice fiunction f for the family S of all nonempty subsets of A. We let for every α aα=f(A-{aξ：ξ<α}) if A-{aξ：ξ<α} is nonempty. Let θ be the least ordinal such that A = {αξ: ξ < θ}. Clearly,（aα：α< θ) enumerates A. ■ (引用終り)
