Re: bXc = b -> b=0 (with regularity but not infinity)

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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