Re: Russell's paradox and ZF(C)

On Dec 30, 10:25 am, george <gree...@xxxxxxxxxx> wrote:
On Dec 30, 9:15 am, Zaljo...@xxxxxxxxx wrote:

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

You really don't know much about ZFC.
If you did then you would know that pairing is not even an
axiom of ZFC.   Once you have the F, you have Replacement.
Once you have Replacement, you can DERIVE pairing (and
separation, too, for that matter) as a theorem.
Basically you just create a 2-element set from the empty set
by iterating the powerset axiom, and then you can replace those
2 elements with anything.

All what you said I know it very well, and I discussed it in a prior
topic were G.Frege thankfully supplied different versions of axiom of
Replacement, and Moe Blee already presented the proof of pairing
depending on the stronge version of Replacement.
Not only that, I myself in many posts have made theories that are
equivalent to ZF that has pairing as a theorem.

The problem is that you didn't know what I mean by my statement
(.... have pairing), you noticed I didn't say (.. have pairing as an
axiom)
I only said (...have pairing) and I mean by that we have pairing as a
theorem ( according to some versions of ZF) or as axiom, it doesn't
matter, the priniciple is that we should have pairing.

Also separation can do the job but only in conjunction with Regularity
of course.
But pairing with Regularity is enough. Separation alone without
regularity cannot really prevent the paradox, it only uses the
negation of the paradox in preventing constructing a set of all sets
that are not in themselfs ( which is not necessarily the universal set
in ZF-Regularity ) and similar sets that if we suppose there existence
then separation with the negation of choice will prove them
paradoxical.

Zuhair

Zuhair
.

Relevant Pages

  • Re: Russells paradox and ZF(C)
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