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

Rupert, I don't understand your proof. It might take a bit of work for
you to explain it to me, so I understand that such work may not be
enjoyable for you; but if, on the the other hand, you don't mind
working it through with me, I do thank you for your time and effort.

Rupert wrote:
Suppose bXc=b and b!=0. Let x be in b.

Okay, I take that as existential instantiation to the variable x. I
only belabor that point because it bears upon why I don't understand
the rest of the proof.

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>.

Is x here under a universal quantifier or is it unquantified so that it
is the x we said is in b?

If x is under a universal quantifier, then I see that R is defined as a
predicate (but not yet proven to be a set) for all x and y, but without
any mention of the x that we said is in b. But if x is the x we said is
in b, then isn't R a 1-place predicate thus not a relation?

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),

I know now what the transitive closure of a relation on a set is. So
you mean that the transitive closure of R is a subset of buPb? But I'm
still confused about what R is and what set it is a relation on, so I
don't know how to see it's a subset of buPb.

it
is a set. Now we can easily show that it has no minimal element,
contradiction.

Thanks,

MoeBlee

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