>>496
＜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）

ふっふ、ほっほ

　>>486より 再度転記しよう
T Jech 著 · 1997 · The Third Millennium Edition, revised and ... 2002. (Springer monographs in mathematics)
Thomas Jechの 証明
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 everv α
aα=f(A-{aξ：ξ<α})
if A-{aξ：ξ<α} is nonempt.
Let θ be the least ordinal such that A = {αξ: ξ < θ}.
Clearly,（aα：α< θ) enumerates A. ■

　対比で(参考)>>310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)

さて
1）前段のT Jech 著 では
　”That we can do by induction, using a choice fiunction f for the family S of all nonempty subsets of A.”とあるが
　後段の それによる en.wikipedia では、この1行は 省かれている
2）また、en.wikipediaから、他国のwikipedia 記載ぶりを見てみると
　中国：en.wikipediaと同じ （仏、伊などは Zornの補題使用）
3）思うに、T Jech 著
　”That we can do by induction, using a choice fiunction f for the family S of all nonempty subsets of A.”
　は、単なるイクスキューズ（excuse）で
　A-{aξ：ξ<α}（＝A∖{aξ∣ξ<α}）は、全部”the family S of all nonempty subsets of A”の中にあって
　単に その部分集合を 使っていますと 言い訳と補強をしているだけのこと と言いたいんじゃね？ （無くても良いと多くの人は 判断している）