From: hardy@math.mit.edu (Michael Hardy) Subject: Re: Some well-defined sets ? Date: 16 May 2000 01:08:29 GMT Newsgroups: sci.math.research Summary: [missing] Nam Nguyen (nam.nguyen@calgary.qcdata.com) wrote: > I'm trying to construct some sets but don't quite know for sure if the > construction is correct, under ZFC (and Peano's axioms of > natural number). > > Let S be a well-defined non-empty set in ZFC and let P(S) means > the Power set of S, let's construct the following series of sets: > > P0(S) = S > P1(S) = P(S) > P2(S) = P(P(S)) > .. > .. > Pn(S) = P(Pn-1(S)) > .. > .. > Then, let: > > U = Union of all Pn(S) where n is in N, the set of all > Natural numbers. > > Are both U and P(U) well defined sets ? Since you're mentioning ZFC, I suppose the question would be not just whether it's well-defined, but whether the existence of such a set is entailed by ZFC. It is. The power-set axiom says the power set of any set is itself a set. Without that axiom, it would be consistent that the power set of any infinite set is a proper class rather than a set. The axiom of infinity entails the existence of a set whose members are precisely the natural numbers. The axiom of replacement implies that since {0, 1, 2, 3, ....} is a set, so is {P0(S), P1(S), P2(S), .....}, provided the mapping n |----> Pn(S) is definable within the language. I think it should be an easy, if tedious, exercise, to show that it is so definable. Finally, the union axiom says the union of any set is a set. (I suppose most mathematicians would say the union of any "set of sets" is a set. But in ZFC all sets are sets of sets.) And, since U is a set, so is P(U), by the power set axiom. Mike Hardy -- Michael Hardy hardy@math.mit.edu