現代数学の系譜工学物理雑談古典ガロア理論も読む61

現代数学の系譜工学物理雑談古典ガロア理論も読む61at MATH

現代数学の系譜工学物理雑談古典ガロア理論も読む61 - 暇つぶし2ch737:licitly, although it had not yet been formally stated. For example, after having established that the set X contains only non-empty sets, a mathematician might have said "let F(s) be one of the members of s for all s in X." In general, it is impossible to prove that F exists without the axiom of choice, but this seems to have gone unnoticed until Zermelo. Not every situation requires the axiom of choice. For finite sets X, the axiom of choice follows from the other axioms of set theory. In that case it is equivalent to saying that if we have several (a finite number of) boxes, each containing at least one item, then we can choose exactly one item from each box. Clearly we can do this: We start at the first box, choose an item; go to the second box, choose an item; and so on. The number of boxes is finite, so eventually our choice procedure comes to an end. The result is an explicit choice function: a function that takes the first box to the first element we chose, the second box to the second element we chose, and so on. (A formal proof for all finite sets would use the principle of mathematical induction to prove "for every natural number k, every family of k nonempty sets has a choice function.") This method cannot, however, be used to show that every countable family of nonempty sets has a choice function, as is asserted by the axiom of countable choice. If the method is applied to an infinite sequence (Xi : i∈ω) of nonempty sets, a function is obtained at each finite stage, but there is no stage at which a choice function for the entire family is constructed, and no "limiting" choice function can be constructed, in general, in ZF without the axiom of choice. (引用終り)

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