[FOM] Cardinality beyond Scott's:
Zuhair Abdul Ghafoor Al-Johar
zaljohar at yahoo.com
Sat Jan 23 21:21:40 EST 2010
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)
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