Re: Dependant Choice.

On May 21, 3:06 am, zuhair <zaljo...@xxxxxxxxx> wrote:
Well, that's precisely the problem. In the language of Z-I we can only
talk about sets. You can't quantify over class variables. I've been
trying to get this across to you for quite a while now.- Hide quoted text -

- Show quoted text -

But you didn't say anything about the ulternative schema that I have
suggested and I will write it again below:

Axiom of Countability:if F is a formula in which x is not free then
all closures of
AyE!zF(y,z) -> Am(Ex(mex & Ay(yex->Az(F(y,z)->zex)))).
are axioms.

I think that does the job perfectly.

And I do think that the following theory is Z-R itself.

1) Extensionality 2) Separation 3) Countability 4) Union
5) Power

were Countability refers to the schema I have just wrote above.


No, sorry, actually this theory is a bit more powerful than Z-R. For
example you can prove the existence of {w,P(w),P(P(w),...}, which you
can't prove in Z-R.

Also I think that If I add Size Comprehension and Regularity as below,
I think we can get ZF.

6) Size Comprehension: If F is a formula in which x is not free then
all closures of
Ex( Ay(yex<->F)<->Ez(z supernumerous to x ) ).
are axioms.

7)Regularity.

Axiom of Size comprehension parallels the idea in NBG that a set
should have another set that it should be subnumerous to.

Size Comprehension avoids Russell's paradox :

Let F<->~yey
Then ~Ez( z supernumerous to x)
Proof:
Since from Regularity Ax ~xex
then Ay(yex<->~yey)<->x={y|y=y},
since AzAy(yez->y=y)
then z subset_of x
And no subset is supernumerous to its set
Thus ~Ez( z supernumerous to x)
we have ~Ay(yex<->~yey) as a theorem from FOL.
Then Size comprehension would be Ex(false<->false)
which is trivially true.

Regarding the case when F<->yey
we have from regularity Ax ~xex
Then it is clear that Ay(yex<->yey)<->x={}
and from Separation and Countability
we have {{}} as a set , which is supernumerous
to { } , So size comprehension is true.

I don't know if I should add another axiom stating that

8)Size limitation: AxEy( y supernumerous to x ).

I think with these 8 axioms we will have a theory that is equivalent
to ZF.

It is easy to prove that This theory ->ZF
Since Replacement can be proved easily in this theory.
And Infinity and Pairing follows from Countability.
But I don't know how to prove that ZF -> This theory.

So what I am sure of is that theory proves ZF, and what I am not sure
of is weather ZF prove this theory or not.

Zuhair


.

Relevant Pages

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