＜公開処刑 続く＞
（『 ZF上で実数は どこまで定義可能なのか？』に向けて と
　 (あほ二人の”アナグマの姿焼き") に向けてｗｗ ；ｐ）

>>498
(再掲)>>497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.　注）＊
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.
注）＊
That we can do by induction, using a choice fiunction f for the family S of all nonempty subsets of A.

という具合に、後付けで、簡単に ”注）＊” とでも やっておけば、それで済む話では？
要するに、
”the family S of all nonempty subsets of A.”は、ZFのべき集合公理から従う
Aのべき集合公理を、いつものようにP(A)と書く。P(A)は、空集合Φを含むので
the family S＝P(A)＼Φ と書ける
分出公理を使うと、Sの部分集合として
{A∖Φ,A∖{Φ,a1},A∖{Φ,a1,a2},A∖{Φ,a1,a2,a3},・・ A∖{aξ∣ξ<α}・・}
これから 集合族 が出来て
A∖Φ,A∖{Φ,a1},A∖{Φ,a1,a2},A∖{Φ,a1,a2,a3},・・ A∖{aξ∣ξ<α}・・
集合族は、順序数で添え字付けられている と考えることができる
この集合族に、選択関数を適用すれば良い

”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.”
で大概の人は分かる

初学者向けに（君のために ；ｐ）
”That we can do by induction, using a choice fiunction f for the family S of all nonempty subsets of A.”
と書けば、多少親切ってことかな ；ｐ）

つづく