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

["MoeBlee"]

Rupert wrote:
> This argument of mine is flawed since it wrongly assumed
> (bXc)Xc=bX(cXc).

I was wondering about that.

> 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.

I'll print that one out and think it through.

Thanks,

MoeBlee