Re: Equivalent to Choice?
zuhair
For all A Exist x For all y
(y e x<->(~ y strictly supernumerous to A &
for all z (z in TC(y) -> ~ z strictly supernumerous to A)))
Zuhair
Butch Malahide
It seems clear that the statement implies AC, so I guess the real
question is whether it's implied by AC. Have I got that right?
Rupert
For me it was the other way round; I thought it was obvious that AC
implied the statement but the converse was not obvious to me.
Here is my proof that AC implies the statement.
Assume AC. Then A is equipollent to a cardinal kappa.
Let x=HC(kappa+). (This can easily be proved to exist if kappa+ is a
regular cardinal, which again follows from AC.)
Then x has the desired property, assuming Trichotomy of Cardinals,
which follows from AC.
zuhair
No it is the other way around. AC implies this statement.
But does this statement imply AC?
Zuhair
zuhair
> On Dec 14, 7:00 pm, Butch Malahide <fred.gal...@gmail.com> wrote:
>
> > On Dec 13, 7:50 pm, zuhair <zaljo...@gmail.com> wrote:
>
> > > Is the following equivalent to AC?
>
> > > For all A Exist x For all y
> > > (y e x<->(~ y strictly supernumerous to A &
> > > for all z (z in TC(y) -> ~ z strictly supernumerous to A)))
>
> > It seems clear that the statement implies AC, so I guess the real
> > question is whether it's implied by AC. Have I got that right?
>
> For me it was the other way round; I thought it was obvious that AC
> implied the statement but the converse was not obvious to me.
Yes, that what was in my mind.
My thought about that issue is the following:
if not every set is well order-able ( i.e. ~AC)
then for any non well order-able set x, we'll have a proper class
of ordinals incomparable to it.
For example take P(w) lets assume it is not well order-able, then
all ordinals from Aleph_1 and above would be incomparable
to P(w) and of course all would be in the set x above, since
all are not strictly supernumerous to P(w), thus x would
be a proper class, since we can put it in one-one correspondence
with the proper class ORD of all set ordinals.
So the above statement Must imply AC, and since AC implies it,
so it is equivalent to AC.
Zuhair
Butch Malahide
But the latter implication is trivial, as you yourself have observed.
To put it even more plainly: If AC is false, then some set A is not
well-orderable. Then no ordinal can be strictly (or nonstrictly for
that matter) "supernumerous" to A. Hence, if y is an ordinal (or for
that matter a set of ordinals), then y satisfies your condition
(~ y strictly supernumerous to A &
for all z (z in TC(y) -> ~ z strictly supernumerous to A)).
Thus x must contain all ordinals (even all sets of ordinals) as
elements, and so is a proper class.
Rupert
Yes. I believe you.
The Hartogs ordinal of P(w) is not necessarily aleph-1, but apart from
that the argument seems fine.
David Libert
> On Dec 14, 5:21=A0am, Rupert <rupertmccal...@yahoo.com> wrote:
>> On Dec 14, 7:00=A0pm, Butch Malahide <fred.gal...@gmail.com> wrote:
>>
>> > On Dec 13, 7:50=A0pm, zuhair <zaljo...@gmail.com> wrote:
>>
>> > > Is the following equivalent to AC?
>>
>> > > For all A Exist x For all y
>> > > (y e x<->(~ y strictly supernumerous to A &
> o A)))
>>
>> > It seems clear that the statement implies AC, so I guess the real
>> > question is whether it's implied by AC. Have I got that right?
>>
>> For me it was the other way round; I thought it was obvious that AC
>> implied the statement but the converse was not obvious to me.
>
> Yes, that what was in my mind.
>
> My thought about that issue is the following:
>
> if not every set is well order-able ( i.e. ~AC)
> then for any non well order-able set x, we'll have a proper class
> of ordinals incomparable to it.
>
> For example take P(w) lets assume it is not well order-able, then
> all ordinals from Aleph_1 and above would be incomparable
> to P(w) and of course all would be in the set x above, since
> all are not strictly supernumerous to P(w), thus x would
> be a proper class, since we can put it in one-one correspondence
> with the proper class ORD of all set ordinals.
>
> So the above statement Must imply AC, and since AC implies it,
> so it is equivalent to AC.
>
> Zuhair
>>
>> Here is my proof that AC implies the statement.
>>
>> Assume AC. Then A is equipollent to a cardinal kappa.
>>
>> regular cardinal, which again follows from AC.)
>>
>> Then x has the desired property, assuming Trichotomy of Cardinals,
>> which follows from AC.
>
Call Zuhair's statement (*) .
So ZF including regularity proves (*) <-> AC, as discussed above.
The proof of the -> direction, above, making the proper class of
von Neumann ordinals each not hereditarily supernumerous to a
non-well-orderable set, did not use regularity.
So -> is provable even over ZF with regularity dropped.
The other direction, <- , from AC , used H_kappa.
The usual ZFC proof that H_kappa exists uses regularity. Even Jech's
proof that H_happa exists from ZF only uses regularity.
Zuhair and I posted about ZFC - regularity models with a proper class of
Quine atoms or a proper class of singleton towers respectively. Those were
referenced in
[1] David Libert "The size of proper classes?"
sci.logic, sci.math Dec 3, 2009
http://groups.google.com/group/sci.logic/msg/dc2877f6b28621b2
[1] said ZFC. My base article behind [1] said start with ZF. But if
you start with ZFC it works the same and produces a ZFC - reguarity model
with a proper class of singleton towers as [1] said.
Zuhair's base article behind [1] already mentioned ZFC - regularity models
with a proper class of Quine atoms.
Those were counter models to H_kappa being a set. In fact H_2 is not a set.
But they are also countermodels to (*) . Any A woth #A > 1 is a
counterexample.
So these show over ZF without regualrity, AC -> (*)
is not provable if ZF is consistent.
Zuhair pointed out that his axiom of strong extensionality proves there are
not 2 Quine atoms. In fact by applying the axiom to {} and a QUine atom
it even proves there are no Quine atoms.
So the first ~ (*) model doesn't satisfy strong extensionality.
But the other model with singleton towers satisfies that no set is a
member of itself.
If no set is a member of itself, then extensiinality <->
strong extensionality. And the model satisfied usual extensionality.
So the singleton towers model shows that if we drop regularity and even
add in strong extensionality, we still can't prove AC -> (*) .
--
David Libert ah...@FreeNet.Carleton.CA
zuhair
Although I am not sure, but I think the towers of singletons might
not
cause any problem, but just in case it does, then we must shun them
from existence by the axiom:
For all x (x is singleton -> ~ For all y (y e TC(x) -> y is
singleton)).
in this way we stop these towers.
All these objects "distinct empty objects, Quine atoms, singleton
towers"
all of them are not classes in my opinion, and Extensionality must be
axiomatized in such a manner as to shun them from being confused with
sets. Or we can have strong Extensionality and the above axiom.
I shall point in a separate topic why should we shun these objects
from existence really, that is if we want to define cardinality in a
successful manner.
Zuhair