Re: Resolving the paradoxes of set theory

From: George Greene (greeneg_at_cs.unc.edu)
Date: 10/26/04 

Date: 26 Oct 2004 11:34:48 -0700

victorw@rogers.com (alephnull) wrote in message
> > > The power set is
> > > riddled with problems. For example
> > >
> > > a e A <-> {a} e P(A) is true.
> >
> > right.
> >
> > > A e A <-> {A} e P(A) is, however, false
> >
> > No, it isn't, not in ZFC anyway.
> > In ZFC, both sides of this biconditional
> > are always false, so the biconditional is
> > always true. This is not an example of any kind
> > of problem with powersets. This is just the way it is.
> > It is reasonable and normal. There is NOTHING "problem"atic about it.
>
> (to defeat logic you must use logic - sigh)
> A e A is only false because of the axiom of regularity

Right.

>
> > > Without the axiom of regularity (which presumably blocks the
> > > substitution a = A)
> >
> > No,it doesn't. When you substitute A for a in that
> > schema, you get AeA, and that's just false, it's not
> > blocked.
> >
> > > the power set would lead to inconsistency.
> >
> > No, it wouldn't. If A could in fact be an element of A,
> > THEN {A} e P(A) would become true, so the biconditional
> > would still be true, so there would be no inconsistency.
>
> But this would have other consequences - #P(A) would be 2^n + 1

That is incoherent: *what* is n, in the above??
I *think* you meant that n=#(A).
But if n is natural (which it presumably is, since
you called it n), then A is a finite set.
And in THAT case, whether AeA OR NOT is COMPLETELY
irrelevant! If #(A)=n then #(p(A))= 2^n AND NOT 2^n + 1,
EVEN WHEN AeA !

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