Re: tedious sledding re set existence in FOL

george wrote:

> If you start out with the usual definition of set equality
> in ZF, namely, a=b <=df=> Ax[xea<->xeb], but you then relax
> to doing this in FOL withOUT equality, then you DON'T KNOW
> whether this does or doesn't also entail that a and b are members
> of the same sets. It turns out that it doesn't, but it's NOT like
> anybody
> around HERE has recently EXHIBITED the relevant counter-
> model. It is NOT at all clear how sets with the same members
> could satisfy all the other axioms of ZFC and still wind up being
> members of different sets. I mean, we know it's possible beacause
> Fraenkel and Torkel said so, but some intuition as to how such a
> state of affairs might actually come about REMAINS absent.

Is this related to Dana Scott's result? One book (I only have the
photocopied page) describes the situation as "Extensionality assures us
that a set is completely determined by its elements. From a casual
acquaintance with this axiom one might assume that extensionality is a
substitution principle having more to do with logic than set theory.
This suggests that if equality were taken as a primitive notion then
perhaps this axiom could be dispensed with. Dana Scott [Essays on the
Foundations of Mathematics. 1962, pp. 115-131] however, has proved that
this cannot be done without weakening the system. Thus, even if we were
to take equality as a primitive logical notion it would still be
necessary to add an extensionality axiom."

So Scott proved the independence of the axiom of extensionality (I'm
guessing no matter which version - either the xez <-> yez version or
the zex <-> zey version - since the situation is symmetrical if one is
the definition and the other is the axiom), right? What is Fraenkel's
result?

Thanks,

MoeBlee

.

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