[過去ﾛｸﾞ] 現代数学の系譜　カントル　超限集合論 (1002ﾚｽ)

現代数学の系譜　カントル　超限集合論 http://rio2016.5ch.net/test/read.cgi/math/1570237031/

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541: 現代数学の系譜 雑談 ◆e.a0E5TtKE [] 2019/11/30(土) 20:55:23.24 ID:4Ujjq2jv >>540 つづき ?As we know, the set of all objects (if it exists) enjoys paradoxical properties: unlike a theorem known to G. Cantor, the power of this set would not be inferior to that of the class of all its subassemblies. It is the same of the class composed of all the sets containing a single element; therefore, K classes do not check, Cantor's theorem. ?Taking this fact into account, one could question the very existence of classes K. By modifying Mr. Sierpinski's definition so as to remove that drawback, I get the following definition: The set M is finite, when the class of all its subsets (not empty) is the only class satisfying the conditions: 1. its elements are subsets (not empty) of M; 2. any set containing a single element of M belongs to this class; 3. if A and B are two sets belonging to this class, their set -sorn A + B also belongs to it. つづく http://rio2016.5ch.net/test/read.cgi/math/1570237031/541

542: 現代数学の系譜 雑談 ◆e.a0E5TtKE [] 2019/11/30(土) 20:55:39.85 ID:4Ujjq2jv >>541 つづき ?We can show that a finite set according to this definition is also in the ordinary sense and reciprocally. In other words: for a set to be finite according to the proposed definition, it is necessary and sufficient that the number of its elements can be expressed by a natural number (the notion of natural number being assumed to be known). ?Indeed, let M be a set whose number of elements can be expressed by a natural number; let Z be any class satisfying the conditions 1-3. We will show that every subset of M belongs to Z. This is - under condition 2 - subsets composed of a single element; at the same time, if this is so subsets containing n elements, it is the same - according to 3 - of those which contain n + 1. Since the number of elements of each subset of M is expressed by a natural number, it follows by induction that Z contains all the subsets of M. Therefore, since the class Z is necessarily identical to that of all the subsets of M, it is the only class satisfying the conditions 1-3. Thus, any set whose number of elements can be expressed by a natural number is a finite set in our sense. ?Suppose, on the other hand, that the number of elements of a set gives M does not let itself be expressed by a natural number. Let Z be the class of all the subsets of M whose number of elements can be expressed by a natural number. This class obviously satisfies conditions 1-3; at the same time, according to the hypothesis, M does not belong to Z and, consequently, Z is not identical to the class of all the subsets of M; therefore, the class of all subsets of M is not the only class satisfying the conditions 1-3 and M is not finite in our sense, c. q. f. d. (引用終り) 以上 http://rio2016.5ch.net/test/read.cgi/math/1570237031/542

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