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On Sat, 26 Aug 2006, Rupert wrote:
> Rupert wrote:
> > MoeBlee wrote:
> > >
> > > In ZR(-I) (that's Z set theory with the axiom of regularity but not the
> > > axiom of infinity), prove:
> > >
> > > bXc = b -> b=0
> > >
> > > where 'X' stands for the Cartesian product.
> > >
> > > You may not use the 'infinite sequence of nested members' version of
> > > regularity. But the following previous theorems are available (where
> > > 'e' stands for 'is a member of') along with the usual "first chapter"
> > > theorems of set theory:
> > >
> > > ~SeS
> > >
> > > ~SeTeS
> > >
> > > ~SeTeVeS
> > >
> > > S subset SXS -> S=0
> > >
> >
> > If bXc=b, then (bXc)Xc=b, so b=bXc=bX(cXc), and ran(b)=c=cXc=0, so b=0.
>
> This argument of mine is flawed since it wrongly assumed
> (bXc)Xc=bX(cXc).
>
> But the following argument works.
>
> Suppose bXc=b and b!=0. Let x be in b. Say that yRx if there exists a z
> in c such that x=<y,z>, or there exists a w in b and a z in c such that
> y={w} and x=<w,z>. Even without infinity, we can define the class of
> objects that bear the transitive closure of R to x, though we cannot
> prove it is a set. However, since it is contained in b union P(b), it
> is a set. Now we can easily show that it has no minimal element,
> contradiction.
>
InterestingproofwhichwhenI'vetimetoeditittomakeitreadable,I'llbesuretojointhediscussion.
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