証明のポイントは、 ”For every ordinal α, define an element aα that is in A by setting aα=f(A∖{aξ∣ξ<α}) ” の部分です。aα=f(A∖{aξ∣ξ<α})の部分が、選択公理における選択関数を成す A∖{aξ∣ξ<α}が集合族で、選択関数の定義域ですね
(参考) en.wikipedia.org/wiki/Well-ordering_theorem Well-ordering theorem 整列可能定理 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}.■
