Re: ZF and Russell paradox.

In article <1180559568.691156.189580@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
zuhair <zaljohar@xxxxxxxxx> wrote:

On May 30, 6:22 am, stevendaryl3...@xxxxxxxxx (Daryl McCullough)

It's not that separation *prevents* Russell's paradox,
it's that Russell's paradox relies on unlimited comprehension
(being able to form a set from an arbitrary formula
Phi(x)). ZF doesn't have unlimited comprehension. Separation
is a limited form of comprehension, carefully constructed
so that the Russell class cannot be formed using it.

This discrimination between what is a class and what is a set is only
artificial, we can dispense with it all together.

But then one no longer has ZF.

You seem to think that comprehension in ZF is Separation or
Replacement. That is incorrect. The original naive comprehension is
broken to axioms of pairing, union , power, separation, replacement
and infinity.

It is the axiom of regularity (or foundation) in ZF that bars RP.

The problem with Z or ZF is that we don't know the properties of the
universe of discourse. and what decides that a set exist in Z or ZF
universe of discourse or not, these axioms of Z or ZF only prove a
small portion of sets that exists in V, there migh be a lot of sets in
V of Z or ZF that these axioms don't prove. one of these might lead
to RP if we use Separation or Replacement, since we don't know all of
them then we cannot exclude RP

We can exclude anything that does not satisfy regularity, and RP does
not.

Let V be the universe of discourse for ZF.
Now if VeV then Russell's paradox 'RP' will be proved as a theorem
using separation itself.
if ~VeV , the theory is incomplete and doesn't tell us
if RP is a theorem or not.

Unless the "universe of discourse" can be proven to be a set, it is not
"in" ZF at all.
.

Relevant Pages

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