20/03/20 11:34:24.47 +qJdNaLm.net
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0.99999……は１ではない その７
ｽﾚﾘﾝｸ(math板:79番)-80
(抜粋)
79 2020/03/20(金) ID:WMaa4Quj
conglomerabilityの定義を理解した上でPrussの論文を読み直せば、
自説がPrussによって真正面から否定されてると理解できます
80 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/20(金) ID:+qJdNaLm
おサルさん、DR Pruss氏は、mathoverflowの彼の回答の前段で、conglomerabilityを出しているが
（下記引用ご参照）
最後は、”the function is measurable”が不成立だから、”dumb（ダメダメ） strategy”と言っているよ
（下記の通り）
QED
（＾＾；
（参考）
URLﾘﾝｸ(mathoverflow.net)
Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13
DR Pruss氏
(抜粋)
By a conglomerability assumption, we could then conclude that P(X<=Y)=0, which would be absurd as the same reasoning would also show that P(Y<=X)=0.
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.
That's a fine argument assuming the function is measurable. But what if it's not?
So there is an extension P′ of P such that P′-almost surely the dumb strategy works. Just let P′ be an extension on which the set of representatives has measure 1 and note that the dumb strategy works on the set of representatives.
URLﾘﾝｸ(www.mdpi.com)
Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis
by Paul Bartha
Symmetry 2011, 3(3), 636-652;