🚀 go-pugleaf

In article <Bx693z.H37@cantua.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>Indeed, ZF intrinsically suggests it
>*without* the bother of a foundation axiom, by limiting the set formation
>possibilities so that self-membership is at least not immediately achieved

What is the reasoning here?  That a theory T "intrinsically suggests"
not-P for any P not in T?  If neither P nor not-P are in T, which of
not-P or not-not-P (= P) does T "intrinsically suggest?"

>The slick form of AF, that every nonempty set has a member disjoint from it,

Where is this form given?  And why doesn't it rule out Omega = {Omega}?
(Just asking, I'm not expert on this stuff.)

>But I'm open on well-foundedness: for all I know, there may well be sets like
>A = { 1 , { 2 , { 3 , { 4 , {.....}}}}} . Such a set seems very well defined on
>the face of it, so it *could* exist, for all I know. No self-membership there.
>
>REQUEST: If someone can produce a way of getting a self-membered set from the
>set A above, using the other ZF constructions, I would *love* to see it !!!!

If you are comfortable with A, you should not mind
B = { 1 , { 1 , { 1 , { 1 , {.....}}}}} .

But by AFA, B = {1, B}, making B a member of itself.

As soon as you weaken FA to AFA, the possibility of self-membership
arises.

>That is another reason I went to the trouble of posting my first article; it
>seems to me to make a very clear criterion for locating the essential
>impredicativity of Russell, which shows up most clearly in the *set* form:
>                 { x | P(x) }  ,
>the essential impredicativity comes from the bad range of x, not the form of P.
>
>Doesn't *anyone* else think so ?   :-(

I do.  But I forgot to say in my earlier message that the main point of
my positive reformulation of Russell's paradox was to remove this
impredicativity from the argument.  Witness Y was constructed as the
set of all *elements of X* not members of themselves.  This established
a positive result without any impredicativity.  Russell's theorem is
not so much a corollary of this theorem as an interpretation of it, in
that "there is no set of all sets" formalizes as "every set is
nonuniversal", or "every set has a nonmember".

>That's partly why I don't think that Aczel's
>set theory is really basic, delightful though it is. It can't look like
>fundamental set theory if a convincing show of consistency has to route
>through ZF's consistency.

ZF+FA is even stronger than ZF+AFA.  So by that reasoning ZF+FA can't
look like like a fundamental set theory either.  Yet you seem to prefer
ZF+FA to ZF+AFA.

>I hope you are taking your own rubbishing
>(concerning Z vs ZF) like a man !  It's a tough life on the net !

Being mistaken at the top of your voice does have its drawbacks.
Yesterday definitely wasn't my day, I got something really wrong on
sci.math involving measure theory, much more wrong than Z vs. ZF, which
I maintain is a highly debatable matter, the near-universal acceptance
of the cumulative hierarchy as a model of ZF notwithstanding.
-- 
Vaughan Pratt				There's no truth in logic, son.