you mean finite set? ZFC is not finite.



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送交者: steven 于 2005-6-17, 12:28:30:

回答: Yes 由 xinku 于 2005-6-17, 07:57:47:

The question is whether M is an uncountable set. Since the set of all languages U is a continuum, and M is a subset of U. According to Church-Turing thesis, M must be countable, hence there must exist a function f such that f(z) = m where z is an element in ZFC, and m is an element in M, for every z, and m. In other sense, ZFC is exactly M.



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