Prev: GOOD BYE AND HAPPY HOLIDAYS FROM INVERSE 19
Next: Big Rudin: International Edition?
From: zuhair on 18 Dec 2009 06:36
Hi all,

Define(x-recursive):-

For all A,x

A is x-recursive <->
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) ).

in other words

A is x-recursive <->
A equinumerous to x &
A hereditarily equinumerous to x&
for all y ( y e TC(A) -> ~ y e TC(y) ).

Now we can define Cardinality in ZF minus Regularity as:

Cardinality(x) is the class of all x-recursive sets.

So for example the Cardinality of the empty set { }
is the class of all sets equinumerous to { }
that are hereditarily equinumerous to { }
and since we have Extensionality then this would be
the class { { } }

So Card({}) = { {} }

Now the cardinality of any singleton set {y} would be
the class of all recursive singletons, i.e
the class of all singleton hereditarily singleton sets.

The cardinality of any doubleton sets {y,z} were ~y=z
is the class of all recursive doubletons, i.e
the class of all doubleton hereditarily doubleton sets.

same thing applied for every set x.

Now we may call these cardinals as :

"Recursive Cardinals".

Now what is the proof that these cardinals would be *set*s
and not proper classes, working in ZF minus Regularity.

The proof is the following Lemma in ZF minus Regularity:

For all x , for all y

y e TC(x) if and only if there exist a finite sequence
<x0,x1,x2,...,xn> were x0 e x and
xi+1 e xi for every i=0,1,2,...,n-1 and y=xn.

Now this lemma would guarantee that for every finite x
we cannot have an uncountable x-recursive sets, so
we can only have countably many x recursive sets
when x is finite, and even if it is said that we can have
uncountably many of these finite recursive sets,
still this would be subnumerous to power Omega.

Though I am not sure of that, the class of all
x-recursive sets for a particular x
is subnumerous to x^Aleph_0.

The most important thing, is that because of the Lemma
above, the class of all x-recursive sets for a particular x would be
a set because the process is exhaustive!, we cannot go
on having increasing numbers of these x-recursive sets
in such a manner as to have a proper class of these
x-recursive sets for a particular x.


While with the Ur-elements and with Quine atoms, we can
have proper classes of them, in such a manner that
for each x there would exist a proper class of sets of Quine atoms
or of Ur-elements that are equinumerous to x.

The reason is that with Ur-elements and Quine atoms there is
nothing to check on their identity, i.e. their identity is not
checked by their membership (the members they contain)
they are self identity objects so one can proliferate
them by will to the size of proper classes.

For example we can have the axiom

For all x Exist u
(for all y (y e u -> y is a Ur-element) & u equinumerous to x)

we can have the same axiom with Quine atoms.

For all x Exist u
(for all y (y e u -> y is a Quine atom) & u equinumerous to x)

BUT, we cannot have the same axiom with recursive sets.

i.e. the following is refuted in ZF:

For all m For all x Exist u
(for all y (y e u -> y is m-recursive) & u equinumerous to x)

That's why we can use these recursive sets to define cardinals.

Of course all of the above needs the axiom:

For all x Exist y ( y is x-recursive ).

I don't think there is a way in ZF minus Regularity to construct
these sets without the above axiom.

This axiom would be the Anti-Foundation axiom, i.e. it would
replace Regularity.

The above is an example of actually of defining Cardinality without
Regularity and Choice, and without Coret's assumption also.

Of course these recursive cardinals are shunned in ZF.

However to define cardinality that works both in ZF and
ZF minus Regularity one can actually have a joint definition of Scott
cardinals and the above.

Cardinality(x) is the class of all sets Equinumerous to x
of the least possible rank, or those that are x-recursive.

In ZF this would be reduced to the Scott cardinals.
In ZF minus Regularity outside Coret's assumption
this will be reduced the the recursive cardinals,
In ZF minus Regularity satisfying Coret's assumption
the cardinal would be the union of a Scott cardinal
and a recursive cardinal.

Zuhair
---------------------------------------------------------------
" And treat the sickness by what caused it in the first place"

An ancient saying!
First  |  Prev  | 
Pages: 1 2
Prev: GOOD BYE AND HAPPY HOLIDAYS FROM INVERSE 19
Next: Big Rudin: International Edition?