現代数学の系譜 カントル 超限集合論2at MATH
現代数学の系譜 カントル 超限集合論2
- 暇つぶし2ch600:kewise 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. http://www.mdpi.com/2073-8994/3/3/636 Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis by Paul Bartha Symmetry 2011, 3(3), 636-652;
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