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
<http://www-math.mit.edu/~hardy>