[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48 (625ﾚｽ)

現代数学の系譜 工学物理雑談 古典ガロア理論も読む48 http://rio2016.5ch.net/test/read.cgi/math/1513201859/

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118: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2017/12/16(土) 15:04:54.86 ID:/2xvBEHK >>116 >"c points"の意味が分らん（＾＾ 違うかも知れないが、検索ヒットと他にめぼしいヒットがないので下記を貼る （下記だと、cは連続濃度の意味だね） https://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable (抜粋) Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable? edited Nov 29 '12 at 22:06 19 answered Jul 16 '12 The answer to your second question (assuming the axiom of choice, to dodge Asaf's comment) is that 2^R/Σ has dimension 2^c, where c=2^?0 is the cardinality of the continuum. The main ingredient of the proof is a partition of [0,1] into c subsets, each of which intersects every uncountable closed subset of [0,1]. To get such a partition, first note that there are only c closed subsets of [0,1], so you can list them in a sequence of length (the initial ordinal of cardinality) c in such a way that each closed set is listed c times. Second, recall that every uncountable closed subset of [0,1] has cardinality c. Finally, do a transfinite inductive construction of c sets in c steps as follows: At any step, if the closed set at that position in your list is C and if this is its α-th occurrence in the list, then put an element of C into the α-th of the sets under construction, being careful to use an element of C that hasn't already been put into another of the sets under construction. You can be this careful, because fewer than c points have been put into any of your sets in the fewer than c preceding stages, while C has c points to choose from. At the end, if some points in [0,1] remain unassigned to any of the sets under construction, put them into some of these sets arbitrarily, to get a partition of [0,1]. つづく http://rio2016.5ch.net/test/read.cgi/math/1513201859/118

119: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2017/12/16(土) 15:05:17.21 ID:/2xvBEHK >>118 つづき Once you have this partition, notice that every piece has outer measure 1, because otherwise it would be disjoint from some closed set that has positive measure and is therefore uncountable. This implies that, among the 2c2c sets that you can form as unions of your partition's pieces, only φ and [0,1] can be measurable. In particular, no finite, nonempty, symmetric difference of these pieces is measurable. That is, they represent linearly independent elements of 2^R/Σ. (引用終り) http://rio2016.5ch.net/test/read.cgi/math/1513201859/119

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