Re: Russell's paradox and ZF(C)

On Dec 30, 6:15 am, Zaljo...@xxxxxxxxx wrote:
Hi all,

It is often told that ZF(C) avoids Russell's paradox, because of its
limited comprehension schema (Separation). This is a falsy!

The correct statment is that Separation prevents the exitence of the
set of all sets that are not in themselfs USING the negation Russell's
paradox (which is already a theorem of FOL), and that is not
prevention of Russell's paradox due to separation per se.

What prevents the existence of Russell's paradox without using the
negation of the paradox itself is actually Regularity and Pairing.

Because the set of all sets V will be prevented since it would be in
itself by definition and from pairing we can construct {V} which
violates Regularity.

Without pairing, Regularity canNOT do the job, since V would satisfy
regularity even if it is in itself.

Since V is prevented from existence by Regularity and Pairing, and
since regularity imply that Ax( ~yey <-> y=y ) , then the set of all
sets that are not in themselfs is prevented from existence, therefore
Russell's paradox doesn't raise.

ZF(C) wihtout regularity doesn't really avoid Russell's paradox due to
its axiomatization per se, it rather uses the negation of Russell's
paradox with separation to avoid the existence of the set of all sets
that are not in themselfs. So separation is not really a solution to
Russell's paradox. Separation is a schema that confirm with the
negation of Russell's paradox! But since it doesn't prevent the
paradox by itself and using theorem in FOL other than the negation of
Russell's paradox , then it remain questionable weather Separation
provides us with the limitation on comprehension sufficient to avoid
the paradox.

To clarify matters just suppose that the negation of Russell's paradox
is not a theorem of FOL, would separation then prevent the existence
of the set of all sets that are not in themselfs? definitely NOT!

while Regularity and Pairing would prevent the existence of such set
even if the negation of Russell's paradox is not a theorem of FOL with
identity

Also Separation with Regularity can prevent the paradox, since
{V} is provable from separation and that violate regularity!

In Quine's set theory, matters are different, the paradox is prevented
due to the axiomatization itself and the multisorted FOL that it uses,
so it is a more solid way of preventing Russell's paradox.

Also my theory: DST (dual set theory) is presumed (by myself) to
prevent Russell's paradox due to its axiomatization and its language
without refering to the negation of the paradox itself, so if DST is
consistent, then it affords a solution to Russell's paradox
(in addition to Burali-Forti and Cantor's paradox) that is simpler
than Quine's.

Zuhair

.

Relevant Pages

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  • Re: Russells paradox and ZF(C)
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