Re: ZF and Russell paradox.

On May 31, 7:12 am, zuhair <zaljo...@xxxxxxxxx> wrote:
On May 30, 6:22 am, stevendaryl3...@xxxxxxxxx (Daryl McCullough)
wrote:





zuhair says...

Yes, I know all of that. Not having naive comprehension doesn't mean
anything really,

On the contrary, it means that in ZF there is a distinction
between classes and sets. Not every class forms a set. A
class is any collection of sets definable by a formula:
{ x | Phi(x) }. To say that such a class forms a set
is to say that the corresponding instance of comprehension
holds:

Comp(Phi): Ey Ax (x e y) <-> Phi(x)

(There is some set y containing all sets x such that Phi(x) holds).

In ZF, not every formula Phi can be used to form a set. In particular
the Russell class defined by R = { x | ~ (x e x) } is not a set. To
prove that it is not a set, you have to prove the negation of the
instance of the comprehension schema for the formula ~ (x e x). The
corresponding instance of comprehension is

Comp(~(x e x)) : Ey Ax (x e y) <-> ~(x e x)

The negation is

~Comp(~(x e x)) : Ay Ex ~(x e y) <-> ~(x e x)

But ZF proves that no set is an element of itself.
So ~(x e x) is always true. So ~Comp(~(x e x))
simplifies to

Ay Ex ~(x e y)

which is an theorem of ZF (the existence of the empty set).
So it is provable in ZF that there is no set corresponding
to the Russell class. So Russell's paradox cannot be formed
in ZF.

still there is a possibility that Z or ZF encounter
Russell's paradox,

No, that really isn't a possibility. It is possible
that ZF is inconsistent for some other reason, but
the Russell route for proving it inconsistent does
not work.

(Okay, obviously if ZF is inconsistent then it
can prove anything, and in particular it can
prove that the Russell set R exists. However, the
logic would be: Because ZF is inconsistent, it
can prove that the Russell set exists, rather than
because ZF can prove the Russell set exists, ZF is
inconsistent.)

Separation can hardly be imagined to be preventive
from such paradox.

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.

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.

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.

This is inevitable. Let T be any consistent recursively enumerable
theory extending ZF. Then T does not prove that there exists a set
which is a model of T. You should study the theory of large cardinals.
All of your "multi-layered set theories" can easily be interpreted in
ZFC+"there exists an inaccessible cardinal". You are basically re-
inventing the concept of an inaccessible cardinal. There are many much
more powerful large-cardinal axioms.

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

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.

You cannot disprove nor prove RP.

There is a lot of Vague aspects of ZF so as to make it so suspecious.

There is no axiom in ZF that tell's us that

Ax( xeV <-> P ). were P is a certain formula that defines the
membership of x in V.
were V is the universe of discourse of ZF.

without such a universal axiom of membership in V of ZF
one can hardly be firm that this theory do avoid RP.





--
Daryl McCullough
Ithaca, NY- Hide quoted text -

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