19/11/30 20:55:39.85 4Ujjq2jv.net
>>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.
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