ガロア第一論文と乗数イデアル他関連資料スレ12 (903レス)
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(9): 現代数学の系譜 雑談 ◆yH25M02vWFhP 01/26(日)14:09 ID:57hfZFiX(6/17) AAS
”<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) rio2016.5ch.net/test/read.cgi/math/1736907570/”

>A(=A\Φ),A\{a0},A\{a0,a2},・・,A\{a0,a2,・・},・・
>を得るにはP(A)-Φを定義域とする選択関数が必要。

妄想沸いてるよw ;p)
下記 Jechの証明を2つ再録しよう

1)
 >>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. ■

2)
また
(再掲)>>504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)

どちらも、aα=f(A-{aξ:ξ<α}) あるいは aα= f(A∖{aξ∣ξ<α})
つまり、関数で書くと
・f:A-{aξ:ξ<α} → aα
・f:A∖{aξ∣ξ<α} → aα

"P(A)-Φを定義域とする選択関数が必要"?
妄想沸いてるよ w ;p)
定義域 A-{aξ:ξ<α} または {aξ∣ξ<α}■
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