スレタイ箱入り無数目を語る部屋27(あほ二人の”アナグマの姿焼き”ｗ)

[過去ﾛｸﾞ] スレタイ箱入り無数目を語る部屋27(あほ二人の”アナグマの姿焼き”ｗ) (1002ﾚｽ)
上下前次1-新
通常表示 512ﾊﾞｲﾄ分割ﾚｽ栞
抽出解除ﾚｽ栞

このｽﾚｯﾄﾞは過去ﾛｸﾞ倉庫に格納されています｡
次ｽﾚ検索歴削→次ｽﾚ栞削→次ｽﾚ過去ﾛｸﾞﾒﾆｭｰ

491: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2024/11/23(土) 17:38:18.95 ID:dngn2gaF

>>482
>>それぞれの集合から一つずつ元を選び出して新しい集合を作ることができるというものである
>自然言語で間に合わせようとするから間違える。定式化してごらん。

・下記”Well-ordering theorem”で 単語”choice”は、重要キーワードですよ
　Well-ordering theoremから axiom of choiceが導かれるが
　その証明のキーは ”An essential point of this proof is that it involves only a single arbitrary choice, that of R;”とありますね
・なお、下記の英 Axiom of choice で
　”A proof requiring the axiom of choice may establish the existence of an object without explicitly defining the object in the language of set theory.”
　”Similarly, although a subset of the real numbers that is not Lebesgue measurable can be proved to exist using the axiom of choice, it is consistent that no such set is definable.[8]”
　非可測集合：”it is consistent that no such set is definable.[8]”かｗ ；ｐ）
・”Because there is no canonical well-ordering of all sets, a construction that relies on a well-ordering may not produce a canonical result, even if a canonical result is desired (as is often the case in category theory).”
　これも要注目です。”may not produce a canonical result”ｗ ；ｐ）

要するに、 ”An essential point of this proof is that it involves only a single arbitrary choice, that of R;”
”a subset of the real numbers that is not Lebesgue measurable can be proved to exist using the axiom of choice, it is consistent that no such set is definable.[8]”
””Because there is no canonical well-ordering of all sets, a construction that relies on a well-ordering may not produce a canonical result”
ｗ ；ｐ）

(参考)
en.wikipedia.org/wiki/Well-ordering_theorem
Well-ordering theorem
Ernst Zermelo introduced the axiom of choice as an "unobjectionable logical principle" to prove the well-ordering theorem.[3]
Proof of axiom of choice
The axiom of choice can be proven from the well-ordering theorem as follows.
To make a choice function for a collection of non-empty sets,
E, take the union of the sets in E and call it X.
There exists a well-ordering of X; let R be such an ordering.
The function that to each set S of E associates the smallest element of S, as ordered by (the restriction to S
of) R, is a choice function for the collection E.

An essential point of this proof is that it involves only a single arbitrary choice, that of R; applying the well-ordering theorem to each member S of E separately would not work, since the theorem only asserts the existence of a well-ordering, and choosing for each S a well-ordering would require just as many choices as simply choosing an element from each S.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1731325608/491

上下前次1-新書関写板覧索設栞歴

ｽﾚ情報赤ﾚｽ抽出画像ﾚｽ抽出歴の未読ｽﾚ AAｻﾑﾈｲﾙ

ぬこの手ぬこTOP 0.042s