時枝記事の類似は、2013年12月09日にmathoverflowで、議論されている 二人の数学Dr Alexander Pruss 氏と Tony Huynh氏と、それ以外に質問者Denis氏(彼はコンピュータサインスの人)の周囲の人("other people argue it's not ok") たちは、「時枝の議論は測度論的に不成立」と言っている
answered Dec 11 '13 at 21:07 Math Dr. Alexander Pruss 氏 ・・・But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i ・・・Intuitively this seems a really dumb strategy.
answered Dec 9 '13 at 17:37 Math Dr. Tony Huynh氏 ・・・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. 0029現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/26(日) 10:37:29.90ID:uQ4z/5zX>>28 可測非可測の話で、ヴィタリ集合は時枝でも取り上げられている が、確率論ではもう一つ、「全事象の確率は1であるというコルモゴロフの確率の公理に反する」確率分布の話がある
Frechet filterの英wikipedia記事と ”Examples On the set N of natural numbers, the set of infinite intervals B = { (n,∞) : n ∈ N} is a Frechet filter base, i.e., the Frechet filter on N consists of all supersets of elements of B.” あと、MathWorld ”Cofinite Filter If S is an infinite set, then the collection F_S={ A ⊆ S:S-A is finite} is a filter called the cofinite (or Frechet) filter on S.”
(参考) https://en.wikipedia.org/wiki/Fr%C3%A9chet_filter Frechet filter (抜粋) In mathematics, the Frechet filter, also called the cofinite filter, on a set is a special subset of the set's power set. A member of this power set is in the Frechet filter if and only if its complement in the set is finite. This is of interest in topology, where filters originated, and relates to order and lattice theory because a set's power set is a partially ordered set (and more specifically, a lattice) under set inclusion.
The Frechet filter is named after the French mathematician Maurice Frechet (1878-1973), who worked in topology. It is alternatively called a cofinite filter because its members are exactly the cofinite sets in a power set.
Contents 1 Definition 2 Properties 3 Examples 4 See also 5 References
Examples On the set N of natural numbers, the set of infinite intervals B = { (n,∞) : n ∈ N} is a Frechet filter base, i.e., the Frechet filter on N consists of all supersets of elements of B.[citation needed]
External links ・Weisstein, Eric W. "Cofinite Filter". MathWorld. https://mathworld.wolfram.com/CofiniteFilter.html Cofinite Filter If S is an infinite set, then the collection F_S={ A ⊆ S:S-A is finite} is a filter called the cofinite (or Frechet) filter on S. 0040132人目の素数さん2020/07/27(月) 15:07:40.84ID:dppBRBhf>>39 > supersets
https://mathworld.wolfram.com/Superset.html mathworld.wolfram Superset A set containing all elements of a smaller set. If B is a subset of A, then A is a superset of B, written A superset= B. If A is a proper superset of B, this is written A superset B. 0041132人目の素数さん2020/07/27(月) 15:12:55.97ID:dppBRBhf>>40 文字化け訂正
If B is a subset of A, then A is a superset of B, written A superset= B. If A is a proper superset of B, this is written A superset B. ↓ If B is a subset of A, then A is a superset of B, written A ⊇ B. If A is a proper superset of B, this is written A ⊃ B. 0042現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/27(月) 21:41:57.91ID:slbIBvLt>>39 補足 https://arxiv.org/pdf/1212.5740.pdf Filters and Ultrafilters in Real Analysis 2012 Max Garcia Mathematics Department California Polytechnic State University
Abstract We study free filters and their maximal extensions on the set of natural numbers. We characterize the limit of a sequence of real numbers in terms of the Fr´echet filter, which involves only one quantifier as opposed to the three non-commuting quantifiers in the usual definition. We construct the field of real non-standard numbers and study their properties. We characterize the limit of a sequence of real numbers in terms of non-standard numbers which only requires a single quantifier as well. We are trying to make the point that the involvement of filters and/or non-standard numbers leads to a reduction in the number of quantifiers and hence, simplification, compared to the more traditional ε, δ-definition of limits in real analysis.
Contents Introduction . . 1 1 Filters, Free Filters and Ultrafilters 3 1.1 Filters and Ultrafilters . . .. 3 1.2 Existence of Free Ultrafilters . . . . . . 5 1.3 Characterization of the Ultrafilter . . . . . . 6 2 The Fr´echet Filter in Real Analysis 8 2.1 Fr´echet Filter . . . . . . . . . 8 2.2 Reduction in the Number of Quantifiers . . .. . . 10 2.3 Fr´echet filter in Real Analysis . . . . . . . 11 2.4 Remarks Regarding the Fr´echet Filter . . . . . 12 3 Non-standard Analysis 14 3.1 Construction of the Hyperreals *R . . . . . 14 3.2 Finite, Infinitesimal, and Infinitely Large Numbers . . . . . . . 16 3.3 Extending Sets and Functions in *R . . . . . . . . . . . . . . . 20 3.4 Non-Standard Characterization of Limits in R . . . . . . . . . 23 A The Free Ultrafilter as an Additive Measure 25 0043現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/27(月) 21:45:28.54ID:slbIBvLt>>42
これは、 フレシェ・フィルターなどを使う”non-standard numbers”、いわゆる超準解析についての論文ですね 0044132人目の素数さん2020/07/27(月) 22:54:10.00ID:Bn7Io8Ul>>37 >それって、時枝記事について、何も言ってないに等しいぞ! 当たり前だろw 同値関係を別の方法で再定義するってだけなんだからw 解答できなかったからって発狂すんなよw 0045現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/27(月) 22:57:10.15ID:slbIBvLt>>44 だから? なんだって? 0046現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/27(月) 23:02:30.88ID:slbIBvLt>>42 例えば ”We are trying to make the point that the involvement of filters and/or non-standard numbers leads to a reduction in the number of quantifiers and hence, simplification, compared to the more traditional ε, δ-definition of limits in real analysis.” ってあるよね
つまり、 ”traditional ε, δ-definition of limits in real analysis” に対して、Frechet Filter とか、 Ultrafiltersとかを使って、 ”Non-Standard Characterization of Limits in R”(いわゆる超準解析) を展開することを論じている
”Examples On the set N of natural numbers, the set of infinite intervals B = { (n,∞) : n ∈ N} is a Frechet filter base, i.e., the Frechet filter on N consists of all supersets of elements of B.[citation needed]”
このExampleは、時枝無関係でしょ つまり、 ”On the set N of natural numbers, the set of infinite intervals B = { (n,∞) : n ∈ N} is a Frechet filter base,” って、繰返すが、時枝無関係の標準的な、自然数N上のフレシェ・フィルターの例じゃんか?(^^
追加 ”On the set N of natural numbers, the set of infinite intervals B = { (n,∞) : n ∈ N} is a Frechet filter base,” って、”the set of infinite intervals B = { (n,∞) : n ∈ N}”って、フレシェ・フィルターに”∞”使われていますよwww(^^ 当然だけどな 超準(ノンスタ)だから、 (>>42より) ”3.2 Finite, Infinitesimal, and Infinitely Large Numbers . . . . . . . 16” ですからね、Infinitely Large Numberも扱いますよねwww(^^ 0051132人目の素数さん2020/07/27(月) 23:50:50.06ID:Bn7Io8Ul>>49 白紙答案の瀬田、相変わらず勝手に妄想して勝手に発狂してるw 誰がフレシェフィルタ使えば箱入り無数目が証明できると言ったんだ? 妄想バカに数学は無理w 0052132人目の素数さん2020/07/28(火) 00:00:02.70ID:96c6EGvu>>50 >”On the set N of natural numbers, the set of infinite intervals B = { (n,∞) : n ∈ N} is a Frechet filter base,” >って、”the set of infinite intervals B = { (n,∞) : n ∈ N}”って、フレシェ・フィルターに”∞”使われていますよwww(^^ >当然だけどな >超準(ノンスタ)だから、 バカだねえw Bのどの元にも∞は属さないよw おまえ()の意味わからんの?w なにがノンスタだよバカw
下記PDFで ”This new system would be constructed in a manner similar to Cauchy’s construction of the real numbers” ”Let us consider the factor ring R~^N = R^N/ 〜Fr where 〜Fr is the equivalence relation defined by (an)〜Fr(bn) if and only if {n : an = bn} ∈ Fr. This is no different to saying that (an) is equivalent to (bn) if and only if an = bn for all sufficiently large n. ”
ここに、Frは、フレシェ・フィルターです。 なるほど、なるほど、フレシェ・フィルターを使って、”similar to Cauchy’s construction of the real numbers”をやる ”where 〜Fr is the equivalence relation defined by (an)〜Fr(bn) if and only if {n : an = bn} ∈ Fr.” 数列のシッポの同値を使ってね
そうすると、”Non-standard Analysis 3.1 Construction of the Hyperreals *R ” が出る!
https://arxiv.org/pdf/1212.5740.pdf Filters and Ultrafilters in Real Analysis 2012 Max Garcia Mathematics Department California Polytechnic State University (抜粋) P12 2.4 Remarks Regarding the Fr´echet Filter
This new system would be constructed in a manner similar to Cauchy’s construction of the real numbers from rational sequences. The elements in this new system would be equivalence classes of real numbered sequences, which take into account sequence convergence (divergence) as well as the rate of convergence (divergence). Ideally, the resulting system will contain elements that can be used to characterize convergence in such a manner that we can do away with the limits of standard analysis or the set constructions from the Fr´echet approach. Let us consider the factor ring R~^N = R^N/ 〜Fr where 〜Fr is the equivalence relation defined by (an)〜Fr(bn) if and only if {n : an = bn} ∈ Fr.
This is no different to saying that (an) is equivalent to (bn) if and only if an = bn for all sufficiently large n. Thus the elements in our new system are equivalence classes of real sequences, denoted by <an>. We now define the relevant operations and order of our new system.
P14 Chapter 3 Non-standard Analysis 3.1 Construction of the Hyperreals *R (引用終り) 以上 0056132人目の素数さん2020/07/28(火) 11:09:00.61ID:U9fCF8yb>>54
ところで、このPDF https://arxiv.org/pdf/1212.5740.pdf Filters and Ultrafilters in Real Analysis 2012 Max Garcia Mathematics Department California Polytechnic State University
に、フレシェ・フィルター Fr を使って ”where 〜Fr is the equivalence relation defined by (an)〜Fr(bn) if and only if {n : an = bn} ∈ Fr.” 数列のシッポの同値ってやってますよね
で? なにか、新しいこと言えるの?
言えることがあれば、この論文読んで 言ってみてよ
無いわな! 読んだ限りではw
時枝記事について なにも新しいことは、言えないよね!! wwww(゜ロ゜; 0057132人目の素数さん2020/07/28(火) 13:39:21.62ID:U9fCF8yb>>55 ”This is no different to saying that (an) is equivalent to (bn) if and only if an = bn for all sufficiently large n. Thus the elements in our new system are equivalence classes of real sequences, denoted by <an>. We now define the relevant operations and order of our new system.”
1.Fr フレシェ・フィルター 使って、 ”to saying that (an) is equivalent to (bn) if and only if an = bn for all sufficiently large n. ” つまりは、十分大きなnの先で一致する数列、(an) と (bn) との同値(equivalent)が定義できる 2.で? Fr フレシェ・フィルター って、(an) と (bn) とか、具体的な数列には無関係なんですよね (>>50 "On the set N of natural numbers, the set of infinite intervals B = { (n,∞) : n ∈ N} is a Frechet filter base," とか PDF P8の”2.1.1 Definition (Fr´echet Filter). ”の通り) 3.だから、Fr フレシェ・フィルターを使ったところで、数列 (an) の具体的な各値 an については、何も言えませんね 4.一方時枝は、数列 (an) で、ある自然数数 ここではmとして、mより大きな数列 (an) の数値が分かれば その値から、am (あるいは i <m なる ai )の値が分かるという主張 5.つまりは、数列のシッポのある後半の部分の数を知ると、それより前(数列の先頭に近い)am ないし i <m なる ai の値を、確率99/100%で的中できるという主張 6.それって、明らかにムリゲーでしょw。なぜなら、数列 (an) のシッポとそれより前の am ないし i <m なる ai の値 は、無関係なんだから 7.そして、それは、大学の確率教程のIID(独立同分布)を知っていれば、反例になることはすぐ分かる 大学の確率教程のIID(独立同分布)を使って、確率変数 X1,X2,・・・Xn,・・・なる可算無限数列を作れば コイントスなら確率1/2、サイコロなら確率1/6 なととなって、確率99/100%なんて、どこからも出てこない 8.だから、数学的には上記7項で終わっている 数学的に面白いのは、「なぜ、当たるように見えるの?」「なぜみんな引っ掛かるの?」という部分なのです
answered Dec 11 '13 at 21:07 Math Dr. Alexander Pruss 氏 ・・・But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i ・・・Intuitively this seems a really dumb strategy.
answered Dec 9 '13 at 17:37 Math Dr. Tony Huynh氏 ・・・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 0059132人目の素数さん2020/07/28(火) 13:51:55.05ID:U9fCF8yb>>57 タイポ訂正
4.一方時枝は、数列 (an) で、ある自然数数 ここではmとして、mより大きな数列 (an) の数値が分かれば その値から、am (あるいは i <m なる ai )の値が分かるという主張 ↓ 4.一方時枝は、数列 (an) で、ある自然数 ここではmとして、mより大きな数列 (an) の数値が分かれば その値から、am (あるいは i <m なる ai )の値が分かるという主張
(>>28より再録) https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13 (抜粋) answered Dec 11 '13 at 21:07 Math Dr. Alexander Pruss 氏 ”The probabilistic reasoning depends on a conglomerability assumption, namely that given a fixed sequence u^→ , the probability of guessing correctly is (n?1)/n, then for a randomly selected sequence, the probability of guessing correctly is (n?1)/n. But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i, and we have no reason to think that the conglomerability assumption is appropriate.” と書いてある ”The probabilistic reasoning depends on a conglomerability assumption”つまり、確率的理由付けは、”conglomerability assumption”が成り立っている必要があるという
この”conglomerability”は、mathoverflow中にも説明がある。 また、本があるよ。下記の”Infinity, Causation, and Paradox Alexander R. Pruss”P75-77とかに詳しい説明がある (下記のGoogleのビューで、かなり読めるよ)
因みに、Alexander Prussは、数学Drで、いま大学教授(Professor of Philosophy) https://en.wikipedia.org/wiki/Alexander_Pruss Alexander Pruss (抜粋) Professor of Philosophy and the Co-Director of Graduate Studies in Philosophy at Baylor University in Waco, Texas. Biography Pruss graduated from the University of Western Ontario in 1991 with a Bachelor of Science degree in Mathematics and Physics. After earning a Ph.D. in Mathematics at the University of British Columbia in 1996 and publishing several papers in Proceedings of the American Mathematical Society and other mathematical journals,[4] he began graduate work in philosophy at the University of Pittsburgh. (引用終り) 以上 0067132人目の素数さん2020/07/31(金) 11:40:59.68ID:Trt2z5f1>>65 補足
これも、時枝記事の確率トリックのタネの一つだろう 当たりは存在するが、確率計算としては、0 ないし、むしろ「確率計算はできない(確率の公理に反する)」と言った方がいいかもしれない 0070132人目の素数さん2020/07/31(金) 12:14:18.90ID:Trt2z5f1>>69 時枝でいえば、決定番号は存在するが 決定番号を使った 確率計算は、できない(確率の公理に反する) ってことです 0071132人目の素数さん2020/07/31(金) 13:18:06.86ID:Trt2z5f1 (>>28より再録) https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13 (抜粋) answered Dec 9 '13 at 17:37 Math Dr. Tony Huynh氏 ・・・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)
Math Dr. Tony Huynh氏も分かっている ”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.”
つまり ”If it were somehow possible to put a 'uniform' measure on the space of all outcomes”が実現できれば なのだが 'uniform' measure=一様分布 (「一様分布」は、>>67の非正則事前分布の説明に出てくるね)
Math Dr. Tony Huynh氏も分かっているね 時枝における、「確率測度として成り立っていない!」は、ヴィタリ集合的なものではなく、 (全事象の積分ないし和が無限大に発散する)「非正則分布になる」ので、 ”全事象の確率は1であるというコルモゴロフの確率の公理”をうまく満たすことができない ってこと
Math Dr. Tony Huynh氏も分かっているねぇ〜(^^ 0072132人目の素数さん2020/07/31(金) 16:32:53.41ID:rnzodbOa>>68 なんでコソコソとsageてんの?
妄想はやめて記事を正しく読んで下さいねー 0074132人目の素数さん2020/07/31(金) 16:58:07.77ID:rnzodbOa>>65 数学の道を諦めて哲学の教授になられたPrussさんも確率99/100以上が正しいことを認めてますよー 「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. Alexander Pruss Dec 19 '13 at 15:05」
>answered Dec 11 '13 at 21:07 Math Dr. Alexander Pruss 氏 より後の日付なので、間違いに気付かれたようですねー 0075132人目の素数さん2020/07/31(金) 17:07:55.76ID:rnzodbOa>>65 もし不成立の補強としてPrussさんの投稿を引用したいなら、成立を明確に認めたDec 19 '13 at 15:05より後の投稿にして下さいねー 間違いに気付かれる前の投稿を引用しても無意味ですよー 0076132人目の素数さん2020/07/31(金) 17:24:16.99ID:rnzodbOa>>66 >因みに、Alexander Prussは、数学Drで、いま大学教授(Professor of Philosophy) あなたDrとか大学教授とか権威に弱いですねー モンティホール問題を沢山の数学者は間違えましたよー 「高度な知識を持つ数学者は勘違いしない」の反例ですねー 0077現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/31(金) 20:57:45.48ID:W/05pVKh>>74 あなた、それ不正確引用ですよ というか、意図してゴマカシていますね
<正確な引用> https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13 より (引用開始) 「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". ? Alexander Pruss Dec 19 '13 at 15:05 」 (引用終り)
いいですか あなたは、”But・・・”の前段の文だけを引用しましたね それは全くのゴマカシです
当然、Math Dr. Alexander Pruss 氏の主張の力点は、後段の But 以下の文 But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". にあります
QED (^^; 0078現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/31(金) 21:06:20.70ID:W/05pVKh>>77 文字化け訂正
「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". ? 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". - Alexander Pruss Dec 19 '13 at 15:05」
「not only ? but also」と「as well as」はほとんど同じ意味を持つ2つですが、使い方が違うため混同しやすいです。
この2つは必ずと言っても良い程、毎年どこかしらの試験の文法問題で出題されます。
今回は2つの違いをまとめましたので、確認してみてください!
【目次】
@not only A but (also) B (AだけでなくBも)
AA as well as B(Aももちろんだが、Bも)
Bnot only ? but alsoとas well as 0083132人目の素数さん2020/08/01(土) 12:57:59.88ID:zi34a+DT>>82 え??? >But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". って >@not only A but (also) B (AだけでなくBも) の構文じゃないんだけど・・・脳みそ腐ってるんすかー?
で、英文法がどうのはまったくどうでも良くて、さっさと「Prussの主張の力点」とやらの内容を書いてくれよ おまえが言い出したんだろ? >当然、Math Dr. Alexander Pruss 氏の主張の力点は、後段の But 以下の文 >But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". >にあります と
(>>28より再録) https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13 (抜粋) answered Dec 9 '13 at 17:37 Math Dr. Tony Huynh氏 ・・・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)
Math Dr. Tony Huynh氏も分かっている ”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.”
つまり ”If it were somehow possible to put a 'uniform' measure on the space of all outcomes”が実現できれば なのだが 'uniform' measure=一様分布 (「一様分布」は、>>67の非正則事前分布の説明に出てくるね)
Math Dr. Tony Huynh氏も分かっているね 時枝における、「確率測度として成り立っていない!」は、ヴィタリ集合的なものではなく、 (全事象の積分ないし和が無限大に発散する)「非正則分布になる」ので、 ”全事象の確率は1であるというコルモゴロフの確率の公理”をうまく満たすことができない ってこと
Math Dr. Tony Huynh氏も分かっているねぇ〜(^^ 以上 0094現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/08/02(日) 16:52:24.28ID:NrBYtRST>>92 タイポ訂正