[FOM] Cardinality Beyond Regularity and Choice!
Zuhair Abdul Ghafoor Al-Johar
zaljohar at yahoo.com
Thu Dec 17 16:09:32 EST 2009
The full definition of recursive singletons is
x is recursive singleton iff
x is singleton &
for all y ( y e TC(x) -> y is singleton )&
for all y ( y e TC(x) -> ~ y e TC(y) )
This is the full definition of these recursive singletons
Now the lemma in ZF minus Regularity that
For all x , for all y
y e TC(x) if and only if there is a finite sequence
(x_0,x_1,x_2,...,x_n) were x_0 e x and
x_(i+1) e x_i for every i=0,1,2,...,n-1 and
y=x_n.
Now this Lemma will prevent us from having an uncountable number
of recursive singletons. Wouldn't it?
A generalization of the above would be to define x-recursive sets as
A is x-recursive iff
A equinumerous to x &
for all y ( y e TC(A) -> y equinumerous to x )&
for all y ( y e TC(A) -> ~ y e TC(y) )
Possibly one might as well define cardinality as:
Cardinality(x) is the class of all x-recursive sets.
We can unite these Recursive Cardinals with Scott's Cardinals to arrive
at a general definition of Cardinality of all sets with and without
Regularity and Choice.
Of course "for every set x there is x-recursive set" must be axiomatized
and added to ZF minus Regularity, since I don't think it can be proved in
ZF minus Regularity alone.
If I am correct, then this would be another example of defining
cardinality outside of Corets' assumption.
Zuhair
>On Dec 15 2009, T.Forster wrote:
>I would be amazed if gauntt's proof could not be made to work with these
>new objects. And incidentally i don't see why there shouldn't be
>uncountably many of them....
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