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MoeBlee wrote:
> 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?
>
x is in b. It would have been clearer if I'd said, say that yRx if x is
in b and 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>. I should add that I'm
using the Kuratowski definition of the ordered pair whereby
<a,b>={{a},{a,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?
>
This is supposed to hold for any x in b. It would have been clearer if
I hadn't started by saying "Let x be in b". I really mean to say, for
any x in b and any y at all, say that yRx iff...
> > 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?
The transitive closure of R is contained in (b U P(b))Xb.
> 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.
>
If y bears the transitive closure of R to x, then x is in b, and either
y is in b, or there exists a w in b such that y={w}. So y is in b U
P(b). So the transitive closure of R is contained in (b U P(b)) X b.
> > it
> > is a set. Now we can easily show that it has no minimal element,
> > contradiction.
>
> Thanks,
>
> MoeBlee
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