そのためもっと非自明なアイデアが必要なのであるが，それを説明する前に層を 圏論的な言葉で解釈しておこう．位相空間 X に対し，圏 OpenX を次のような圏 とする： ・ 対象は X の開集合． ・ V , U を X の開集合とするとき，V から U への射は包含写像 V ,?→ U． （Vが U に含まれるなら射は唯一，そうでないなら射はない．） このとき，X 上の前層は OpenX から Ab への反変関手に他ならず，前層の間の射 とは反変関手間の射に他ならない．前層が層になるための条件について考えよう．
定義 1.6 連続写像 f : Y ?→ X が局所同相であるとは，任意の y ∈ Y に対し y の開近傍 V , f(y) ∈ X の開近傍 U が存在して，f が V から U への同相写像を誘導すること をいう． 圏 LIsomX を次のように定める： ・ 対象は局所同相な連続写像 f : Y ?→ X（誤解のないときには単に Y とも表す）． ・ f : Y ?→ X から f′: Y′ ?→ X への射は，連続写像 g : Y ?→ Y′ で f′?g = f を満たすもの． LIsomX における射の族 (gi: Yi ?→ Y )i∈I（Y , Yi は LIsomX の対象）が被覆で あるとは，Y =∪i∈Ig(Yi) となることをいう．
＜Prussのmathoverflowでの発言について＞ （サルの>>718>>722より） (引用開始) スレ主がコピペしたPrussの発言に対しPrussとDenisが議論をしている。その中で、Prussは「勝率99/100以上」を認めてるよ（＾＾； For each fixed opponent strategy, if i is chosen uniformly independently of that strategy (where the "independently" here isn't in the probabilistic sense), we win with probability at least (n−1)/n. That's right.
あほサルの勝手読み Pruss氏の発言のその後があるだろう？ ”That's right. But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". ? Alexander Pruss Dec 19 '13 at 15:05 ”と 要するに、力点は、But以下の文にあるってことと 前文の”if i is chosen uniformly independently of that strategy”の部分が未証明だってことよｗ
https://books.google.co.jp/books?id=RXBoDwAAQBAJ&;pg=PA77&lpg=PA77&dq=%22conglomerability%22+assumption+math&source=bl&ots=8Ol1uFrjJQ&sig=ACfU3U1bAurNGJm5872wDblskzsSgsU0iA&hl=ja&sa=X&ved=2ahUKEwioiPyV_IPiAhXHxrwKHUeaArUQ6AEwCXoECEoQAQ#v=onepage&q=%22conglomerability%22%20assumption%20math&f=false Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018 P75 (抜粋) 2.5.3 COUNTABLE ADDITITVITY AND CONGLOMERABILITY (引用終り) 以上 0775１３２人目の素数さん2019/08/14(水) 14:21:44.78ID:MPteNw3f>>773 >前文の”if i is chosen uniformly independently of that strategy”の部分が未証明だってことよｗ おまえ中身なんにも分からずに接続詞が醸し出す雰囲気だけで言ってるだろｗ おまえは 「ランダム選択される i を事前に予想できないことが未証明」 って言ってるんだよｗ バカですか？ｗ そんでなんで俺がおまえの発言内容をおまえに説明してやらにゃあかんの？ｗ バカ過ぎて話にならないｗ
>Denisが、確率論が分かってないし、本を書くネタをmathoverflowで書くには余白と時間が限られているってことよ おまえの妄想に過ぎないｗ 証拠がひとつも無いｗ おまえすぐ妄想語るなｗ それ治療してもらえよｗ 0776１３２人目の素数さん2019/08/14(水) 14:22:19.83ID:c6g6R1pg>>773 ”But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy””
>>775 ”if i is chosen uniformly independently of that strategy” のindex iについてｗｗ
これ下記の”Alexander Pruss Dec 19 '13 at 15:05 ”の抜粋なｗ 以下の応答を嫁めｗ（＾＾
要するに、力点は、But以下の文にあるってことと 前文の”if i is chosen uniformly independently of that strategy”の部分が未証明だってことよｗ
(参考) https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 (抜粋) sked Dec 9 '13 at 16:16 Denis The Modification I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up. a
answered Dec 11 '13 at 21:07 Alexander Pruss Let's go back to the riddle. Suppose u￣ is chosen randomly. The most natural option is that it is a nontrivial i.i.d. sequence (uk), independent of the random index i which is uniformly distributed over [100]={0,...,99}. In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
Denis Dec 17 '13 at 15:21 Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice.
Alexander Pruss Dec 19 '13 at 15:05 What we have then is this: For each fixed opponent strategy, if i is chosen uniformly independently of that strategy (where the "independently" here isn't in the probabilistic sense), we win with probability at least (n-1)/n. That's right. But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy".
answered Dec 9 '13 at 17:37 In order for such a question to make sense, it is necessary to put a probability measure on the space of functions f:N→R. Note that to execute your proposed strategy, we only need a uniform measure on {1,…,N}, but to make sense of the phrase it fails with probability at most 1/N, we need a measure on the space of all outcomes. The answer will be different depending on what probability space is chosen of course.
If it were somehow possible to put a 'uniform' measure on the space of all outcomes, then indeed one could guess correctly with arbitrarily high precision, but such a measure doesn't exist. (引用終り) 以上 0783現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/14(水) 15:31:37.76ID:rg2Nhb+h>>781 タイポ校正
sked Dec 9 '13 at 16:16 Denis The Modification I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up. a ↓ asked Dec 9 '13 at 16:16 Denis The Modification I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up. 分かると思うが念のため askedのaが取り残されて分離されていたってことね（＾＾； 0784１３２人目の素数さん2019/08/14(水) 15:35:22.09ID:c6g6R1pg>>781-783 必死なのは、ゴキブリ、貴様だ
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 (抜粋) asked Dec 9 '13 at 16:16 Denis I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up.
Alexander Pruss Let's go back to the riddle. Suppose u￣ is chosen randomly. The most natural option is that it is a nontrivial i.i.d. sequence (uk), independent of the random index i which is uniformly distributed over [100]={0,...,99}. In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
Denis Dec 17 '13 at 15:21 Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice.
Tony Huynh Dec 9 '13 at 17:37 In order for such a question to make sense, it is necessary to put a probability measure on the space of functions f:N→R. Note that to execute your proposed strategy, we only need a uniform measure on {1,…,N}, but to make sense of the phrase it fails with probability at most 1/N, we need a measure on the space of all outcomes. The answer will be different depending on what probability space is chosen of course.
If it were somehow possible to put a 'uniform' measure on the space of all outcomes, then indeed one could guess correctly with arbitrarily high precision, but such a measure doesn't exist. (引用終り) 0788１３２人目の素数さん2019/08/14(水) 16:48:21.03ID:c6g6R1pg>>787 お前はすでに死んでいるｗ
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 (抜粋) asked Dec 9 '13 at 16:16 Denis I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up.
Alexander Pruss Let's go back to the riddle. Suppose u￣ is chosen randomly. The most natural option is that it is a nontrivial i.i.d. sequence (uk), independent of the random index i which is uniformly distributed over [100]={0,...,99}. In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
Denis Dec 17 '13 at 15:21 Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice.
Tony Huynh Dec 9 '13 at 17:37 In order for such a question to make sense, it is necessary to put a probability measure on the space of functions f:N→R. Note that to execute your proposed strategy, we only need a uniform measure on {1,…,N}, but to make sense of the phrase it fails with probability at most 1/N, we need a measure on the space of all outcomes. The answer will be different depending on what probability space is chosen of course.
If it were somehow possible to put a 'uniform' measure on the space of all outcomes, then indeed one could guess correctly with arbitrarily high precision, but such a measure doesn't exist. (引用終り) 0792１３２人目の素数さん2019/08/14(水) 17:05:25.60ID:c6g6R1pg>>791 自分でも理解できてないコピペは無駄
以前は、「時枝不成立」と言えば、いろんな”人”から時枝擁護の発言があった いまは、おそらくサル二匹だ 愉快だねｗ（＾＾ 0817１３２人目の素数さん2019/08/14(水) 21:12:30.77ID:MPteNw3f 惨めだねえ 0818現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/14(水) 21:30:46.54ID:Qpe2jc/f ありがとう お前がな（＾＾ 0819現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/14(水) 21:31:44.10ID:Qpe2jc/f 時枝不成立を理解するには、確率論と確率過程論の知識が必要だが ３年経っても、、確率論と確率過程論が理解できないサルｗ（＾＾ 0820現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/14(水) 21:33:48.79ID:Qpe2jc/f おサル I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1
おれ but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up.
ｗ（＾＾；
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 (抜粋) asked Dec 9 '13 at 16:16 Denis I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up. 0821哀れな素人2019/08/14(水) 22:14:45.00ID:Sxu+TbrS>>814 どこに英語の話題が出て来たのか(笑 イミフなレスしか書けない能無し(笑
