現代数学の系譜工学物理雑談古典ガロア理論も読む62

[過去ﾛｸﾞ] 現代数学の系譜工学物理雑談古典ガロア理論も読む62 (1002ﾚｽ)
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856: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2019/03/25(月) 20:57:51.83 ID:whxE2Y+u

>>854
まあ、新人ROMさんには、経緯が分らないだろうから、説明すると
Hart氏のPDFは、下記でわずか２ページで、game2はP2の後半に出てくる（下記の通り）
（なお、時枝記事は>>21ご参照）
スレ61 https://rio2016.5ch.net/test/read.cgi/math/1550409146/64
(抜粋)
時枝と選択公理の関係で、Sergiu Hart氏のPDFに下記がありましたね（＾＾
これを認めるなら、選択公理なしで、時枝類似の数当ては成立つ

この議論は、過去なんども同じ経緯を辿って
あげく、Sergiu Hart氏のgame2: を指摘すると、しっぽを撒いて逃げ行く

その繰り返しです（＾＾

スレ44 http://rio2016.2ch.net/test/read.cgi/math/1506848694/463 より
Sergiu Hart氏のPDF http://www.ma.huji.ac.il/hart/puzzle/choice.pdf

Sergiu Hart氏は、ここに
A similar result, but now without using the Axiom of Choice.^2
Consider the following two-person game game2:
^2 Due to Phil Reny.（＝Phil Reny氏より）

として、”without using the Axiom of Choice” ”game2”を提案しているよ
(引用終り)

因みに、game1 は、その前のページで、２箇所に出てくる
Consider the following two-person game game1:
Theorem 1 For every ε > 0 Player 2 has a mixed strategy in game1 guaran-
teeing him a win with probability at least 1 － ε.
Remark. The proof uses the Axiom of Choice.

Apply the Axiom of Choice to choose an element in each
equivalence class; let F(x) denote the chosen element in the equivalence class
of x (thus F : X → X satisfies x ～ x' iff F(x) = F(x')).
(引用終り)

なお、このPDFの表題が、”Choice Games”となっているのは、”The proof uses the Axiom of Choice”に由来しているのだろう

つづく

http://rio2016.5ch.net/test/read.cgi/math/1551963737/856

865: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2019/03/25(月) 23:36:53.22 ID:whxE2Y+u

>>856 補足

Sergiu Hart氏PDFより
game2 (without using the Axiom of Choice )
Theorem 2
Proof. ”The proof is the same as for Theorem 1, except that here we do not use the Axiom of Choice."

＜参考引用＞
Sergiu Hart氏のPDF http://www.ma.huji.ac.il/hart/puzzle/choice.pdf
(抜粋)
Consider the following two-person game game1:

Theorem 1 For every ε > 0 Player 2 has a mixed strategy in game1 guaranteeing him a win with probability at least 1 － ε.
Remark. The proof uses the Axiom of Choice.

Proof. Fix an integer K. We will construct K pure strategies of Player 2 such
that against every sequence x of Player 1 at least K － 1 of these strategies yield a win for Player 2.
略

A similar result, but now without using the Axiom of Choice.^2
Consider the following two-person game game2:
^2 Due to Phil Reny.（＝Phil Reny氏より）

Theorem 2 For every ε > 0 Player 2 has a mixed strategy in game2 guaranteeing him a win with probability at least 1 － ε.

Proof. The proof is the same as for Theorem 1, except that here we do not use the Axiom of Choice.
Because there are only countably many sequences x ∈ {0, ・・・, 9}^N that Player 1 may choose
(namely, those x that become eventually periodic),
we can order them － say x^(1), x^(2), ・・・, x^(m), ... － and then choose in
each equivalence class the element with minimal index (thus F(x) = x^(m) iff m
is the minimal natural number such that*4 x ～ x^(m)).

*4 Explicit strategies σj may also be constructed, based on R^j being the index where the
sequence y^j becomes periodic.
(引用終り)

http://rio2016.5ch.net/test/read.cgi/math/1551963737/865

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