[FOM] Cardinality beyond Scott's:

[Zuhair Abdul Ghafoor Al-Johar]

Dear tf:

There is no intention at all to define H(x) as a set for every set x.

This definition doesn't require that at all! The only thing needed for 
this definition to work is to have:

"For every set x, there exist ordinals d,i such that 
x subnumerous to Pi(H(d))".

Let's denote the above requirement as "Z".

So H(d) need to exist as a set for "some" ordinal d, and not
necessarily for every ordinal d.

We know already that this is a theorem of ZF-
since H(0)=0. Also H(1)=1.

Now Z is strictly weaker than Regularity!

Regularity can be written in the following manner:

"For every set x, there exist an ordinal i such that 
x subset of Pi(0)". 

Which is a special case of Coret's assumption, which can be written as:

"For every set x, there exist an ordinal i such that 
x subnumerous to Pi(0)". 

Which is a special case of Z.

So Z is weaker (perhaps strictly) than Coret's assumption which is 
strictly weaker than Regularity, so Z is strictly weaker than Regularity.

So we neither need foundation nor Scott's trick!

Zuhair


On Jan 23 2010, T.Forster wrote:

>How do you plan to prove that H(x) exists, for all x? My guess is that
>you will find yourself using replacement, and then you can do Scott's
>trick cardinals. (I'm assumimg you have foundation in either case)