Re: A new definition of Cardinality.
zuhair
As far as I know, all the definitions of cardinality are limited in a
way or another, lets take them one after the other:
1) Von Neumann's Cardinals:
A cardinal is the least of all equinumerous ordinals.
2) Frege-Russell Cardinals:
A cardinal is an equivalence class of sets under equivalence relation
"bijection".
3) Scott-Potter Cardinals:
A cardinal is a class of all equinumerous sets from a common level.
Now lets come to discuss each one of them:
1) Von Neumann's cardinals has the limitation of being dependent on
choice, without choice one cannot know what is the cardinality of
Power(omega) for example.Accordingly in any theory which do not have
the axiom of choice among its axioms most of its sets would be of
indeterminable cardinality, which is a big draw back.
2) Frege-Russell cardinals contradict Z set theory, since their
existence would imply the existence of the set of all sets, which is
in contradiction with Z.
However in NBG and MK class theories, we can define
Frege-Russell cardinals, but by then they would be proper classes and
not sets, which is a great draw back, since proper classes cannot be
members of other classes, and they are hard to work with.
In NF and related theories, Frege-Russell cardinals are sets, but
these theories generally depend on the concept of stratification
of formulas, which is a complex concept, and even finite
axiomatization of NF and NFU and related theories is a complicated
approach, and at the end it also resort to stratification for most of
its inferences. All that make these cardinals undesirable.
3) Scott-Potter Cardinals: depend on the concept of "level" which
depends on the concept of type (Scott) and the iterative concept
(Potter), both concepts of which are complex and difficult to work
with, besides they are not the basic
concepts we use to compare set sizes.
I would like to suggest the following definition:
4) The cardinality of any set x is: The class of all sets
that are equinumerous to x were every member of their transitive
closure is strictly subnumerous to x.
So for any set x, any y is a member of the cardinality of x,
if and only if, y is equinumerous to x and every member of the
transitive closure of y is strictly subnumerous to x.
In symbols:
Define(cardinality(x)):-
z=cardinality(x) <->
for all y (y e z <->
(y equi-numerous to x &
for all m (m e Tc(y)->m strictly subnumerous to x)))
Were Tc(y) stands for the 'transitive closure of y' defined
in the standard manner.
Tc(y)=U{y,Uy,UUy,UUUy,......}
We can actually better define these cardinals through defining the
concept of "hereditary sets"
Define(hereditary):
x is hereditary <->
for all y (y e Tc(x) -> y strictly subnumerous to x)
So a cardinal can be defined in the following manner:
A Cardinal is an equivalence class of hereditary sets under
equivalence relation "bijection".
Or simply
A Cardinal is a class of all equinumerous hereditary sets.
So cardinality of x would be written shortly as:
Cardinality(x) = {y| y is hereditary & y equinumerous to x}
Now it can be proven in ZF that those cardinals would be 'sets', so
they are not proper classes! which makes them easy to handle.
These cardinals don't require choice.
They don't require complex concepts like "stratification,type,
iteration"
They simply depend on the basic concept used to compare set sizes,
which is the presence or absence of injections between the compared
sets.
To me this definition seems to be simpler, more general, and it works
with or without choice, with or without regularity.
So at the end I shall write the definition of cardinal again:
A Cardinal is an equivalence class of hereditary sets under
equivalence relation "bijection".
x is hereditary <->
for all y (y e Tc(x) -> y strictly subnumerous to x)
Cardinality(x) = {y| y is hereditary & y equinumerous to x}
Zuhair
zuhair
> Hi all,
>
> As far as I know, all the definitions of cardinality are limited in a
> way or another, lets take them one after the other:
>
> 1) Von Neumann's Cardinals:
>
> A cardinal is the least of all equinumerous ordinals.
>
> 2) Frege-Russell Cardinals:
>
> A cardinal is an equivalence class of sets under equivalence relation
> "bijection".
>
> 3) Scott-Potter Cardinals:
>
> A cardinal is a class of all equinumerous sets from a common level.
>
> Now lets come to discuss each one of them:
>
> 1) Von Neumann's cardinals has the limitation of being dependent on
> choice, without choice one cannot know what is the cardinality of
> Power(omega) for example.Accordingly in any theory which do not have
> the axiom of choice among its axioms most of its sets would be of
> indeterminable cardinality, which is a big draw back.
In addition to that, the concept of Cardinality has nothing to do
whatsoever with the concept of "order", so defining cardinals as
subclass of ordinals seems to be strange, though practical if choice
is assumed.
Rupert
But the question arises: can you prove in ZF that every set has a
cardinality, on this definition? Not quite obvious to me just at the
moment...
I have an idea that it is known that you need choice or regularity in
order to define cardinals...
zuhair
Well, to say the truth I am not sure either. But from my prior
discussions
I got the impression that working in ZF minus Regularity minus Choice,
it is provable that:
for any set there exist a set of exactly all sets that are
hereditarily strictly sub-numerous to it.
I remember the prove needed some form of induction or so.
Then from Separation on that set it is easy to define the cardinality
of x.
Zuhair
zuhair
zuhair
>
> Zuhair
The above link actually speaks of ZFC, but I don't know really if this
actually depends on choice or not? or weather regularity is needed or
not?
It seems that I got the wrong impression then.
Zuhair
zuhair
I think you mean in theories like ZF or Scott-Potters', however
in NF and related theories, Neither choice nor regularity
are need to define cardinals, and I do have
the vague sense that this definition of mine here
do not depend on them either?!
Zuhair
Rupert
That is certainly not a problem in ZFC but I certainly don't see right
now how to do it without regularity or choice. You should look into
that one carefully, I should think.
Jesse F. Hughes
> As far as I know, all the definitions of cardinality are limited in a
> way or another, lets take them one after the other:
>
[...]
> 2) Frege-Russell Cardinals:
>
> A cardinal is an equivalence class of sets under equivalence relation
> "bijection".
[...]
> 2) Frege-Russell cardinals contradict Z set theory, since their
> existence would imply the existence of the set of all sets, which is
> in contradiction with Z.
>
> However in NBG and MK class theories, we can define
> Frege-Russell cardinals, but by then they would be proper classes and
> not sets, which is a great draw back, since proper classes cannot be
> members of other classes, and they are hard to work with.
[...]
> I would like to suggest the following definition:
>
[...]
> A Cardinal is an equivalence class of hereditary sets under
> equivalence relation "bijection".
>
> x is hereditary <->
> for all y (y e Tc(x) -> y strictly subnumerous to x)
>
> Cardinality(x) = {y| y is hereditary & y equinumerous to x}
How is this any better than the Frege-Russell definition? It's much
more complicated and doesn't avoid the issue you called the difficult
point, does it?
Consider the set {0,1}. By your definition,
Card({0}) = { y | y is hereditary and y is a pair }
= { y | y is a pair and (Az)(z in Tc(y) -> z is empty or a
singleton) }
Let s be any set. The set {0,{s}} is in Card({0}), thus Card({0}) is
a proper class.
So what advantage have you here?
--
"Humanity is still a primitive species. I seem to have been born out
of my time, maybe centuries ahead, and I guess I'll just have to get
used to it. In ways, it's not so bad. Mostly it's boring though."
-- James S. Harris has problems beyond you and me.
David Libert
>> transitive closure of y is strictly subnumerous to x.
>>
>> In symbols:
>>
>> Define(cardinality(x)):-
>>
>> =A0for all y (y e z <->
>> (y equi-numerous to x &
>>
>> Were Tc(y) stands for the 'transitive closure of y' defined
>> in the standard manner.
>>
>>
>> We can actually better define these cardinals through defining the
>> concept of "hereditary sets"
>>
>> Define(hereditary):
>> =A0for all y (y e Tc(x) -> y strictly subnumerous to x)
>>
>> So a cardinal can be defined in the following manner:
>>
>> A Cardinal is an equivalence class of hereditary sets under
>> equivalence relation "bijection".
>>
>> Or simply
>>
>> A Cardinal is a class of all equinumerous hereditary sets.
>>
>> So cardinality of x would be written shortly as:
>>
>>
>> Now it can be proven in ZF that those cardinals would be 'sets', so
>> they are not proper classes! which makes them easy to handle.
>>
>> These cardinals don't require choice.
>>
>> They don't require complex concepts like "stratification,type,
>> iteration"
>>
>> They simply depend on the basic concept used to compare set sizes,
>> which is the presence or absence of injections between the compared
>> sets.
>>
>> To me this definition seems to be simpler, more general, and it works
>> with or without choice, with or without regularity.
>>
>> So at the end I shall write the definition of cardinal again:
>>
>> A Cardinal is an equivalence class of hereditary sets under
>> equivalence relation "bijection".
>>
>> x is hereditary <->
>>
>> Cardinality(x) =3D {y| y is hereditary & y equinumerous to x}
>>
>> Zuhair
>
> But the question arises: can you prove in ZF that every set has a
> cardinality, on this definition? Not quite obvious to me just at the
> moment...
>
> I have an idea that it is known that you need choice or regularity in
> order to define cardinals...
In ZF using regularity and no AC, we can prove every set has a cardinality.
Namely all such sets will be included in the set of all sets hereditarily <=
the cardinal, and from the thread Zuhair refences in this thread, Aatu noted
this is a ZF theorem. The proof uses both regularity and replacement.
Many common theorems of math don't require replacement.
But if we drop regularity it is not provable, even if we add AC back in fact.
For example, in a ZF model construct an inner model as follows. Declare each
set of form <alpha, n>, wherre alpha is an ordinal > 0 and n in omega,
to be an atom.
These atoms will look like sets in the final model, so we stipulate
membership facts for them.
Make each <alpha, n> become the singleton of <alpha, n+1>.
Namely we declare <beta, m> member <alpha, n> <->
m = n+1 and beta = alpha .
Now grow this into an inner ZF model without regularity though.
Namely declare certain sets of form <0, s> to be in the model
iteratively, by transfinitie recursion.
At stage alpha+1 we add atoms <alpha, n> and also all <0, s>
where s is a subest of stage alpha and s is not the singleton of
an atom. Dedine t member <0, s> iff t epsilon s .
Unions of stages at limit ordinals.
Singleton atoms are respresented by other atoms so we didn't add such
<0, s>.
<0, s> versus <alpha, n> alpha > 0 avoids confusion atoms versus
other sets so keeps the defintiion of membership well devined.
We also define <0, s> never member of atoms.
Avoiding singleton atoms in <0, s> avoided some coutnerexamples
to exensionality.
The atoms avoid contradicting etensionanlty by having other atoms
below them to stay distinct as members. Since the omega chain
never hits {} there is no day of reckoning.
So we end up with a proper class (by alpha varying) of singletons
that are hereditarily singeltons.
So there is no set to be cardinal 1.
This all relates to an axiom:
http://en.wikipedia.org/wiki/Aczel's_anti-foundation_axiom
That says that the shape of the graph of interated members determines the set.
(Well is existence and uniqueness).
My model above fails the uniqueness clause of that axiom.
What if we ad that axiom?
With AC also, we do get back only a set of solutions.
Without AC we don't though. Start with a ~AC ground model with <Ui | i in omega>
and <Wi | i in omega> so that each Ui bijective with each Wj,
yet there is no omega sequence of bijectioins f_i : Ui >->> Wi.
Basically: lots of bijections around so with AC picking bijectiins you could
make such an f_i sequence, but not really being able to is an AC failure.
Form a non-reguilarity models similoar to above, making a descnding chain of
Ui or Wi memebers.
Ok its a little more complicated, because these aren't singelons. Make the
iterated members be finsite equence of respective members in the Ui or Wi.
The point being: with regualrity, when you keep make isomorphic copies
all the way down to {}, the {} 's are =, and then iteratively up the ranks
the isomprphoc copoes of actually ='s become =.
Without reguilarity you never hit bottom, so extensionaloty can keep failing
based on lower failures.
I did that already in ther singleton example.
But the singleton example, Aczel's axiom could recognize the common shape and
so enforce = that way.
So I am making a more complicated example to fool Aczel's axiom. At each level
I am putting in isomorphisms to make it heriedary, but using a AC failure to
block the global comparison that would let Aczel's axiom kick in.
Anyway, cpmpying a version of the singletons along lines as above, which
I haven;t done in detail byu not spelling out all defintions, obtain
Ui and Wi based sets so U0 W0 that was each hereditarily <= a copy of U0,
but that are not recognized by Aczel's axiom as having the same shape.
So we can still get 2 non-= correspondings to the U0 copy.
Now do a fancier version, with a proper class of copies according to
the starting model, instead of only 2.
So again, mess up a set cardinal like in the singleton example, all
slipping by the guard of Aczel's axiom.
Ya, so construct exverything so Aczel's axiom was true also in the model.
(Gee, my hands are getting tired from waving so much :) ).
Anyway, with AC. and Aczel's, we can get back to up to iso only
a set of examples, so then get back a set.
--
David Libert ah...@FreeNet.Carleton.CA
zuhair
I think you missed the point of transitive closure here, you said let
s be any set right, then as you know not any s have all members in its
transitive closure
being a pair. so the set {0,{s}} might not be in Cardinality of {0,1}
by they way I think you made a typo since you right Card({0})
I think you mean Card({0,1}) since as you know
{0} is not equi-numerous to {0,{s}} for all s.
Card({0})= { y | y is hereditary and y equinumerous to {0} }
Zuhair
zuhair
The sketch of the proof goes like that:
First we prove that for every set x there exist a set of exactly all
sets that are
hereditarily strictly subnumerous to it,lets denote that later set by
H_(<x)
so we have: for every x there exist H_(<x)
H_(<x) = {y | y strictly subnumerous to x and
for all z (z e Tc(y) -> z strictly subnumerous to
x)}
as a theorem of ZF.
The next step is to take Power(H_(<x)) which is a set of course,
Now using separation on Power(H_(<x)) with the formula " y
equinumerous to x "
then we get :
for every set x there exist c such that
c= { y | y e Power(H_(<x)) and y equinumerous to x }
and of course c is the cardinal I that I defined.
So in ZF (with or without choice) one can prove the existence of these
cardinals for every set, however as David Libert points out it seems
that this depends on Regularity, but as he said it doesn't depend on
choice.
Zuhair
Jesse F. Hughes
Yes, you're right.
Given a set S, is there a simple argument that your definition Card(S)
is a set and not a proper class? Seems plausible, but I'm not sure
how to prove it.
> by they way I think you made a typo since you right Card({0})
That's right too. I changed from {0} to {0,1} midstream, but
screwed up.
> I think you mean Card({0,1}) since as you know
> {0} is not equi-numerous to {0,{s}} for all s.
>
> Card({0})= { y | y is hereditary and y equinumerous to {0} }
--
Jesse F. Hughes
"Sigh. Back to figuring out how to solve the factoring problem and
ending the world as we know it." -- James S. Harris
zuhair
He did?
The proof uses both regularity and replacement.
It would be of great help if you post the proof please. But definitely
replacement is needed no doubt.
zuhair
The sketch of the proof goes like that:
First we prove that for every set x there exist a set of exactly all
sets that are hereditarily strictly subnumerous to it,lets denote that
later set by
H_(<x)
so we have: for every x there exist H_(<x)
H_(<x) = {y | y strictly subnumerous to x and
for all z (z e Tc(y) -> z strictly subnumerous to x)}
as a theorem of ZF.
this is the lemma behind this definition, however Aatu presented a
proof
of it in the link I wrote in one of the posts to this topic, so you
can refer to it,
still I need the details of weather it requires choice or not, I mean
I want the complete proof.
The next step is to take Power(H_(<x)) which is a set of course,
Now using separation on Power(H_(<x)) with the formula
" y equinumerous to x "
then we get :
for every set x there exist c such that
c= { y | y e Power(H_(<x)) and y equinumerous to x }
and of course c is the cardinal I that I defined.
So in ZF (with or without choice) one can prove the existence of these
cardinals for every set, however as David Libert points out it seems
that this depends on Regularity, but as he said it doesn't depend on
choice, but I will know that for sure if he post the proof of the
lemma
above.
Zuhair
George Greene
>
> As far as I know, all the definitions of cardinality are limited in a
> way or another, lets take them one after the other:
>
> 1) Von Neumann's Cardinals:
>
> A cardinal is the least of all equinumerous ordinals.
> Now lets come to discuss each one of them:
>
> 1) Von Neumann's cardinals has the limitation of being dependent on
> choice, without choice one cannot know what is the cardinality of
> Power(omega) for example.Accordingly in any theory which do not have
> the axiom of choice among its axioms most of its sets would be of
> indeterminable cardinality, which is a big draw back.
Even WITH choice, the ZFC definition of cardinality STILL has a
serious drawback. You STILL CANNOT determine the cardinality of
p(w) under ZFC. The collection of all countable ordinals (aleph-one)
cannot possibly be bigger than p(w), but it could be equally large--
the cardinality of p(w) could be aleph-one, aleph-2, or any finite
aleph.
Rupert
Well, 'tis surely a theorem of ZFC, but yeah, why would it be a
theorem of ZF? Let's get clear about that first.
And you were saying you could do it in ZF-regularity.
But, yeah, I would like to know how you do it in ZF.
zuhair
That is my personal guess, I don't have a proof of that yet.
By the way can you help me regarding this issue, and post the complete
proof of this theorem in ZFC, I think this would be of great help.
Zuhair
Rupert
>
> - Show quoted text -
If I assume the axiom of choice, then every set can be well-ordered,
and is equipollent to some cardinal kappa, a cardinal being a von
Neumann ordinal not equipollent to any smaller ordinal.
Now let R(0) be the empty set, and for any R(alpha), let R(alpha+1) be
the set of all subsets of R(alpha) of cardinality less than kappa, and
for any limit ordinal beta, let R(beta) be the union of all R(alpha)
for alpha<beta. Iterate this up to R(kappa). In ZF I can prove this is
a set, using transfinite recursion and suchlike, and this is the set
you're after.
Fair enough?
But without the axiom of choice I know longer know that the set can be
well-ordered... and so it is not really clear what sort of transfinite
recursion to use.
Do you get that? Do you want me to explain in more detail?
George Greene
You can't do it in ZFC either.
ZFC DOES NOT assign a cardinality (in the usual sense)
to p(w).
George Greene
> If I assume the axiom of choice, then every set can be well-ordered,
Right.
> and is equipollent to some cardinal kappa,
WRONG.
Ex[whatever]
does NOT mean that there exists a UNIQUE x such that whatever!
There are MANY DIFFERENT well-orderings of p(w) under ZFC and
they correspond TO DIFFERENT ordinals!
> a cardinal being a von
> Neumann ordinal not equipollent to any smaller ordinal.
ZFC does NOT assign a cardinality to p(w) or any other uncountable set
except maybe the alephs. Under ZFC, beth-1 (=p(w)=p(aleph-0)
=2^beth-0)
"can" take ANY aleph-n (where n is natural) as a cardinality!
Remember, there was that whole conunundrum about
THE CONTINUUM HYPOTHESIS ?!?!?!!?
zuhair
Dear George Greene, please answer the following questions:
Is the following a theorem of ZFC or not?
For all s Exist x for all y
( y e x <-> (y strictly subnumerous to s and
for all z (z e Tc(y) -> z strictly subnumerous to
s)).
were Tc(y) stands for the transitive closure of y.
in case it is a theorem of ZFC, then is it a theorem of ZF? i.e.
without choice,
and if it is a theorem of ZF, then is it a theorem of ZF minus
regularity?
Simple questions that demand simple answer and proofs of these
answers!
Zuhair
David Libert
> On Nov 24, 6:15=A0pm, zuhair <zaljo...@gmail.com> wrote:
>> On Nov 24, 12:55=A0am, Rupert <rupertmccal...@yahoo.com> wrote:
>>
>>
>>
>>
>>
>> > On Nov 23, 10:37=A0pm, zuhair <zaljo...@gmail.com> wrote:
>>
>> > > On Nov 22, 9:20=A0pm, Rupert <rupertmccal...@yahoo.com> wrote:
>>
>> > > > On Nov 23, 11:37=A0am, zuhair <zaljo...@gmail.com> wrote:
>>
>> > > > > On Nov 22, 7:16=A0pm, Rupert <rupertmccal...@yahoo.com> wrote:
>>
>> > > > > > On Nov 23, 10:01=A0am, zuhair <zaljo...@gmail.com> wrote:
[Deletions]
>> > > The sketch of the proof goes like that:
>>
>> > > First we prove that for every set x there exist a set of exactly all
>> > > sets that are
>> > > hereditarily strictly subnumerous to it,lets denote that later set by
>> > > H_(<x)
>>
>> > > so we have: for every x there exist H_(<x)
>>
>> > > =A0H_(<x) =3D {y | y strictly subnumerous to x and
>> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 for all z (z e Tc(y) -> z str=
In my Nov 23 article in this thread
http://groups.google.com/group/sci.math/msg/721cb8170033cf84
I wrote (quoting Rupert and indirectly Zuhair):
>>> So at the end I shall write the definition of cardinal again:
>
>>> A Cardinal is an equivalence class of hereditary sets under
>>> equivalence relation "bijection".
>
>>> x is hereditary <->
>>> =A0for all y (y e Tc(x) -> y strictly subnumerous to x)
>
>>> Cardinality(x) =3D {y| y is hereditary & y equinumerous to x}
>
>>> Zuhair
>
>> But the question arises: can you prove in ZF that every set has a
>> cardinality, on this definition? Not quite obvious to me just at the
>> moment...
>
>> I have an idea that it is known that you need choice or regularity in
>> order to define cardinals...
>
> In ZF using regularity and no AC, we can prove every set has a cardinality.
>Namely all such sets will be included in the set of all sets hereditarily <=
>the cardinal, and from the thread Zuhair refences in this thread, Aatu noted
>this is a ZF theorem. The proof uses both regularity and replacement.
I was being careless in that writing. I was thinking of the same proof Rupert
wrote as quoted earlier above. In general I knew this as a ZFC theorem, but
I reviewed that proof, and everything seemed so explicitly definable in that
construction I thought AC had not been invoked.
I checked that Aatu was talking about the same proof, and since I had just
convinced myself AC was not used I wrote the older thread was also about a
ZF proof.
But Rupert above points out, at the outset we well-order x to find a cardinal
kappa to work with, thereby using AC as I overlooked.
So I agree, as Rupert and Zuhair wrote, in ZFC and using regularity, replacement
and AC, we can prove every set has a set of heritarily smaller sets (as opposed
to a proper class), and Zuhair's definition of cardinal is a set and not a proper
class.
The first quoted article above raised the question of proving this without using
regularity. I think I already got a countermodel to this in my mentioned Nov 23
article
http://groups.google.com/group/sci.math/msg/721cb8170033cf84
namely a inner model of ZFC with regularity removed with a proper class of
infinite descending singletons.
We all agree, these proofs are doing definitions by transfinite recursion,
and so use replacement. In fact I recall I had never written but in my
own thinking produced a model of ZC (Zermelo's theory with AC : no
replacement) in which the heritarily finite sets are a proper class.
So regularity and replacement are essential: have counterexample models
without.
What about AC? We know a proof using AC. Can we find a proof not using AC,
or can we find a ~AC counterexample model?
I can't do either yet, but I will report on some related partial answers
I know or have read about.
First off, a point about the AC based proof quoted above. I repeat the
quote:
> If I assume the axiom of choice, then every set can be well-ordered,
> and is equipollent to some cardinal kappa, a cardinal being a von
> Neumann ordinal not equipollent to any smaller ordinal.
>
> Now let R(0) be the empty set, and for any R(alpha), let R(alpha+1) be
> the set of all subsets of R(alpha) of cardinality less than kappa, and
> for any limit ordinal beta, let R(beta) be the union of all R(alpha)
> for alpha<beta. Iterate this up to R(kappa). In ZF I can prove this is
> a set, using transfinite recursion and suchlike, and this is the set
> you're after.
kappa is the cardinality of the set in question. The ordinary AC proof
actually iterates out to R(kappa+), ie kappa+ the successor cardinal
of kappa.
With AC, every successor cardinal is regular, that is had cofinality
= to itself.
This means any subset X of R(kappa+) must actually be a subset of
some R(alpha), some alpha dependnig on X < kappa+.
So R(kappa+) really closed off the growth and gets the closure
properties we wanted.
The ordinary proof that successor cardinals are regular uses AC. So
this is another way the proof depends on AC, apart from the original
production of kappa Ruppert mentioned.
I recall Moti Gitik from high large cardinal assumptions produced
a model of ZF in which every infinite von Meumann cardinal has
vountable cofinality, and indeed every infinite set is a union
of countably many subsets each of strictly smaller cardinality than
the big set.
I have a useful link listing home pages of set theorists:
http://www.math.ufl.edu/~jal/set_theory.html
In there I found Moti Gitik's home page:
http://www.math.tau.ac.il/~gitik/
He has many online papers there, but I didn't find that one.
Anyway it sure raises a question, whether the R(alpha) construction
above will ever close off in Gitik's model.
Here is a bogus arguement attempt to show it doesn't close off.
I will attempt a proof by transfinite induction that every
R(alpha+1) intriduces new sets.
R(0) is {{}}. So R(1) has a new member not in R(0) : {{{}}} .
For alpha a limit ordinal: find and alpha_n n in omega
sequence cofinal on alpha, by the proeprty of Gitik's model.
By induction hypothesis each R(alpha_n + 1) introduces a new sets
not previous in the lower R(beta).
So form the countable set of such elements, one for each n.
This is a new countable set not in any previous stage, so in
R(alpha + 1).
Next suppose alpha is itself a successor ordinal. By induction
hypothesis, it has a new member. So {this new member} is
a new member of R(alpha + 1) .
So this would appear to finish the proof.
But it is actually a bogus proof, because I had to pick one new
member for each R(alpha_n + 1). If we had countable choice I could,
but the whole point is Gitik's model is a ~AC model.
In fact come to think of it, from an omega sequence cofinal in
aleph_1 as Gitik's model had, if you had countable choice you
could pick an omega enumeration of each piece, and use that
to recover an analogue of Cantor's omega x omega ~ omege proof,
making aleph_1 countable.
So indeed Gitik's model doesn't satisfy countable choice.
You might be tempted to make a definable version of the omega
sequence of new R(apha_n + 1) by picking the ones previously
defined by transfinite recursion.
But that still depends on picking for each limit ordinal the
omega sequenxce of ordinals cofinal in it, which again is
only obvious with AC.
Ok so its not clear whether the R(alpha) construction closes
off in Gitik's model.
But this question depends on the details of how AC fails in
his model. So it will at least be a tricky querstion.
So these are all the issues that arise in considering
that proof in ~AC.
Here is a new point though. Rupert pointed out, to find
the set of sets hereitarily of card <= x, the usual proof
well-orders x, hence invoking AC.
I above pointed out the closure of the contruction becomes
another point.
But I can modify the proof to get around the well-ordering
of x issue.
Namely redo the construction, so at successor stages
R(alpha + 1) adds in all subsets of R(alpha) of cardinality
< #x, ie just use #x directyl in the definition.
We still iterate over ordinals.
ZF proves for any set x there is a set of what I will call
the surjective Hartog ordinal. The usual definition of the
Hartog ordinal of a set x is the least ordinal which does
not inject into x. ZF proves the so called Hartog ordinal
exists and is a set.
I take instead a surjecive version: the least ordinal
alpha such that x does not surject onto alpha.
ZF proves there is always such a set sized ordinal. Namely
to see this, any surjection of x onto alpha induces an
equivalence realtiion on x : x elements are equivalent
if they are sent to the same ordinal.
So any ordinal surejcted by x is isomorphic to a well-ordering
on some subset of P(x), ie pull the wellordering onn the ordinal
back to the equivalence class preimages.
So all the ordinals less than the surjective Hartog ordinal of
x are ismimorphic to various well-orderings on a single set
P(x).
So the collection of all well-orderings on P(x) has natural
well ordeing greater than all these, so its von Neuman ordinal
can't be surjected by x, so there is a least such.
So let kappa be the surjective Hartog ordinal of x.
Suppose there were some lambda > kappa with
cofinality(lambda) > kappa.
If AC: cofinality(kappa+) = kappa+, so we could take
lambda = kappa+.
Anyway assuming there is some such lambda, consider
R(lambda).
Suppse there was an #x sized subset X of R(lambda).
Map each element of y of x to the least alpha < lambda
containing the X element corresponding to y in the #x
enumeration of X.
So this map surjects x onto its range below lambda.
Since cofinality(lambda) > kappa, the range of this
map must be bounded below lambda.
So X is a member of the successor of this bound, and
hence in R(lambda).
So this R(lambda) hasd the desired closure properties.
So it all reduces to finding lambda with
cofinality(lambda) > kappa.
So the AC dependence based on well-ordering x can be
replaced, but the cofinality question remains.
There is Gitik's difficult model, but as noted above
Gitik's model makes it hard for us to prove the construction
closes off. But we don't have a proof it doesn't, and
we don't have a proof the desired heritary set doesn't
exist.
Which in turn brings me to the final point.
I looked up
http://en.wikipedia.org/wiki/Hereditarily_countable_set
and they say all H_kappa, hereditarily less than kappa
sets can be proven to exist in ZF.
They reference
http://www.jstor.org/pss/2273380
which is JSTOR for a 1982 article by Thomas Jech in Journal
of Symbolic Logic. "On Hereditarily Countable Sets"
vol 47, number 1, March 1982 .
Jech names the class of hereditarily countable sets HC.
Jech notes without AC it is not even obvious HC is a
set (as opposed to a proper class).
This is like the sort of questions I was raising with
Gitik's model.
But Jech's paper proves in ZF that it is a set.
I don't have insider access to JSTOR, so I can only
see the first page which states results.
That page says this was received by JSL Oct 15, 1979,
so I suppose this was new reseach then. All the old stuff
we were discussing the basic AC proofs would have gone
back to early 20th century. So this is a much bigger deal.
Also, Gitik's model is sure suggestive of a proper class
being possible, even for HC. So somehow Jech's proof is
going to show that doesn't happen.
The Wikipedia page doesn't give details why H_kappa.
I wonder if that is just generalizing Jech's HC.
Anyway, one point. The H_kappa always add sets
< kappa in cardinality, so sets injecting into a
well-order and hence well-orderable.
So for general non-well-orderable x in ~AC models, we
don't get x back into these H_kappa.
But I showed above how the H_kappa building techniques
can also sometimes apply to other x.
So if we understand more the H_kappa case, the general
x case might still go either way or might still be hard
to answer even knowing H_kappa.
We don't even know HC yet. Well we know the claim, but
not the proof, since I can't read past the opening
page.
I just thought to check Jech's pages from
http://www.math.ufl.edu/~jal/set_theory.html
I found 2, but a quick look didn't find that paper:
--
David Libert ah...@FreeNet.Carleton.CA
Aatu Koskensilta
> Here is a new point though. Rupert pointed out, to find the set of
> sets hereitarily of card <= x, the usual proof well-orders x, hence
> invoking AC.
As you note, this is not where AC comes into play, since in the
inductive definition we don't need a (von Neumann) cardinal the size of
x because we can just include sets < |x| directly. The problem is, again
as you note, proving the existence of a closure ordinal for the
inductive definition; or, to put it slightly differently, showing that
for every set x there's a V_alpha such that if all members of TC(A) for
a set A are of cardinality < |x|, A is in V_alpha.
PS. I have e-mailed you a copy of the Jech paper. I haven't myself
studied the argument for the existence of the set of hereditarily
countable sets in absence of choice, by establishing that the rank of a
hereditarily countable set <= omega_2, and so can say nothing about
whether it may be generalised beyond the countable case.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
zuhair
> ah...@FreeNet.Carleton.CA (David Libert) writes:
> > Here is a new point though. Rupert pointed out, to find the set of
> > sets hereitarily of card <= x, the usual proof well-orders x, hence
> > invoking AC.
>
> As you note, this is not where AC comes into play, since in the
> inductive definition we don't need a (von Neumann) cardinal the size of
> x because we can just include sets < |x| directly. The problem is, again
> as you note, proving the existence of a closure ordinal for the
> inductive definition; or, to put it slightly differently, showing that
> for every set x there's a V_alpha such that if all members of TC(A) for
> a set A are of cardinality < |x|, A is in V_alpha.
>
> PS. I have e-mailed you a copy of the Jech paper. I haven't myself
> studied the argument for the existence of the set of hereditarily
> countable sets in absence of choice, by establishing that the rank of a
> hereditarily countable set <= omega_2, and so can say nothing about
> whether it may be generalised beyond the countable case.
Can you e-mail me a copy of this paper.
Zuhair
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechen kann, darüber muss man schweigen"
zuhair
... Perhaps I misunderstand your definition, but with
Foundation one can prove that every non-empty set contains
0 (= empty set} in its transitive closure. If I am correct,
then your definition does not work....
Can anybody explain that?
Zuhair
Tim Little
> There are MANY DIFFERENT well-orderings of p(w) under ZFC and they
> correspond TO DIFFERENT ordinals!
But not to different cardinals.
- Tim
zuhair
> On 2009-11-24, George Greene <gree...@email.unc.edu> wrote:
>
> > There are MANY DIFFERENT well-orderings of p(w) under ZFC and they
> > correspond TO DIFFERENT ordinals!
>
> But not to different cardinals.
>
> - Tim
Yes to different *cardinals*, since Von Neumanns' cardinals are
special type of ordinals, in ZFC there many different well ordering of
p(w) and they correspond to different VON NEUMANN CARDINALS, it can be
aleph_1, aleph_2,..., aleph_n were n is is a natural number, all these
are CARDINALS, not just ordinals!
Zuhair
Tim Little
> Yes to different *cardinals*, since Von Neumanns' cardinals are
> special type of ordinals, in ZFC there many different well ordering
> of p(w) and they correspond to different VON NEUMANN CARDINALS
ZFC proves that only one well-ordering of p(w) has the order-type of a
von Neumann cardinal.
> it can be aleph_1, aleph_2,..., aleph_n were n is is a natural
> number, all these are CARDINALS, not just ordinals!
Those are only labels for different properties for card(p(w)) in
different models. In any given model, only one of the labels
describes card(p(w)), and the rest do not.
This does not mean that cardinality is undefined, any more than ring
theory leaves the concept of "ideals" undefined because (1+1) = R in
some models while (1+1) =/= R in others.
- Tim
zuhair
The authority himself explained that. There was a misreading of the
original definition.
Zuhair
Marc Alcobé García
> PS. I have e-mailed you a copy of the Jech paper. I haven't myself
> studied the argument for the existence of the set of hereditarily
> countable sets in absence of choice, by establishing that the rank of a
> hereditarily countable set <= omega_2, and so can say nothing about
> whether it may be generalised beyond the countable case.
Sorry, may I ask which of Jech's papers is here being referred to?
Is it available from http://www.math.psu.edu/jech/preprints/papers.html?
Thank you in advance.
Aatu Koskensilta
> Sorry, may I ask which of Jech's papers is here being referred to?
"On hereditarily countable sets".
> Is it available from http://www.math.psu.edu/jech/preprints/papers.html?
Apparently not.
George Greene
> Dear George Greene, please answer the following questions:
>
> Is the following a theorem of ZFC or not?
>
> For all s Exist x for all y
> ( y e x <-> (y strictly subnumerous to s and
> for all z (z e Tc(y) -> z strictly subnumerous to
> s)).
>
> were Tc(y) stands for the transitive closure of y.
Why do you expect me to care about what subnumerous
and transitive closure have to do with anything?
I am asking a much simpler question about the cardinality
of p(w)!
> in case it is a theorem of ZFC,
The question, rather, is whether
X is the cardinality of p(w)
is the cardinality of p(w)
is a theorem of ZFC, if X is any ordinal you can think of.
This IS NOT THE SAME question as whether
Ex[x is the cardinality of p(w)]
is a theorem of ZFC.
zuhair
Cardinals that I defined would not always be "sets" out of
Regularity, in other words the set-hood of these cardinals depends on
weather we work under Regularity or not, if we assume Regularity then
all these cardinals are sets. However without Regularity they can be
proper classes:
Below was the reply that I got regarding this issue:
----------------------
Without regularity (= Axiom of Foundation), one can construct in ZF
(or ZFC) a model of ZF (resp. ZFC) where there is a PROPER CLASS of
sets x = {x}. Call such crazy sets "nodes" for short.
Consider {x, y}, where both elements are nodes and different. Then,
by your definition, (x, y} would be a member of the cardinality of 2.
But there is a proper class of such pairs. So your cardinal would
not be a set.
----------------------
However the question weather the theorem that every set has a
cardinality (as defined in this thread) is dependent on choice, is
still not settled yet!
Zuhair
Tim Little
> The question, rather, is whether
> X is the cardinality of p(w)
That depends upon what *exactly* you mean by "ordinal you can think
of". The ordinal beth_1 would appear to suffice.
- Tim
Rupert
> On Nov 24, 3:23 am, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > If I assume the axiom of choice, then every set can be well-ordered,
>
> Right.
>
> > and is equipollent to some cardinal kappa,
>
> WRONG.
>
Nonsense. Of course it is correct. If a set can be well-ordered, then
there exists some cardinal kappa such that the set is equipollent to
kappa.
> Ex[whatever]
> does NOT mean that there exists a UNIQUE x such that whatever!
No, and I did not claim uniqueness, but as a matter of fact in this
case it is extremely easy to prove uniqueness as well.
> There are MANY DIFFERENT well-orderings of p(w) under ZFC and
> they correspond TO DIFFERENT ordinals!
>
But I spoke of a cardinal, an ordinal which is not equipollent to any
smaller ordinal. If a set can be well-ordered, then there exists an
ordinal (not unique) to which it is equipollent, and there exists a
unique *cardinal* to which it is equipollent. A cardinal being an
ordinal not equipollent to any smaller ordinal.
> > a cardinal being a von
> > Neumann ordinal not equipollent to any smaller ordinal.
>
> ZFC does NOT assign a cardinality to p(w) or any other uncountable set
> except maybe the alephs. Under ZFC, beth-1 (=p(w)=p(aleph-0)
> =2^beth-0)
> "can" take ANY aleph-n (where n is natural) as a cardinality!
>
So what? ZFC proves that P(w) has a cardinality, namely beth-1. We
cannot decide in ZFC which aleph it is.
> Remember, there was that whole conunundrum about
> THE CONTINUUM HYPOTHESIS ?!?!?!!?
That's completely irrelevant here.
George Greene
> On Nov 25, 2:20 am, George Greene <gree...@email.unc.edu> wrote:
>
> > On Nov 24, 3:23 am, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > If I assume the axiom of choice, then every set can be well-ordered,
>
> > Right.
>
> > > and is equipollent to some cardinal kappa,
>
> > WRONG.
>
> Nonsense. Of course it is correct. If a set can be well-ordered, then
> there exists some cardinal kappa such that the set is equipollent to
> kappa.
So FUCKING what??
There are models of ZFC with the set equipollent TO DIFFERENT kappas!
ZFC does NOT prove that the set has any PARTICULAR cardinal as
a cardinality!
>
> > Ex[whatever]
> > does NOT mean that there exists a UNIQUE x such that whatever!
>
> No, and I did not claim uniqueness, but as a matter of fact in this
> case it is extremely easy to prove uniqueness as well.
So FUCKING what??
OF COURSE the ordinal in question is unique in every INDIVIDUAL
model! So "the cardinal is unique" is a theorem. Yet DESPITE this,
the cardinal IS NOT unique because DIFFERENT models have DIFFERENT
cardinals playing this role!
> > There are MANY DIFFERENT well-orderings of p(w) under ZFC and
> > they correspond TO DIFFERENT ordinals!
>
> But I spoke of a cardinal, an ordinal which is not equipollent to any
> smaller ordinal. If a set can be well-ordered, then there exists an
> ordinal (not unique) to which it is equipollent, and there exists a
> unique *cardinal* to which it is equipollent.
NO THERE *DOESN'*, DUMBASS! (DAMN the STUPID is getting
THICK in here)! In each INDIVIDUAL model there exists such an
ordinal, but this ordinal can be ANY INITIAL ORDINAL YOU LIKE (or
any sufficiently small uncountable one, in the case of p(w), in
particular
it can be aleph-n for any natural n).
> A cardinal being
Oh, Shut The Fuck Up.
We know what a cardinal is.
You are the one who has not figured out that the continuum hypothesis
is totally relevant to the issue.
Zuhair started this.
He was saying that traditionally, AC was used to show that everything
had
a cardinality, and he was trying to show it some other way.
My point was that even WITH AC, you STILL don't know the cardinality
of anything uncountable except the alephs themselves.
As I said,
> > ZFC does NOT assign a cardinality to p(w) or any other uncountable set
> > except maybe the alephs. Under ZFC, beth-1 (=p(w)=p(aleph-0)
> > =2^beth-0)
> > "can" take ANY aleph-n (where n is natural) as a cardinality!
>
> So what?
SO YOU *DON'T KNOW* WHAT *THE FUCK* the cardinality of
p(w) is, THAT'S what, AND NEITHER DOES ZFC!!
> ZFC proves that P(w) has a cardinality, namely beth-1.
GODDAMN, RUPERT, YOU ARE *FUCKING STUPID*!!!
beth-1 (have you ever READ the definition?!?!?!?!?!?!?!?!)
IS *DEFINED AS* "the cardinality of p(w)", whatEVER-THE-FUCK
that may be!! That DOES NOT tell you WHAT it is!! It JUST *names* it!
And that name names something DIFFERENT in DIFFERENT models of ZFC!!
There is a model of ZFC where beth-1=aleph-1 and another where it
equals aleph-2009 !!
Neither this nor any other aleph is THE ACTUAL VALUE of beth-1!!!
Rupert
> On Nov 25, 7:04 pm, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > On Nov 25, 2:20 am, George Greene <gree...@email.unc.edu> wrote:
>
> > > On Nov 24, 3:23 am, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > > If I assume the axiom of choice, then every set can be well-ordered,
>
> > > Right.
>
> > > > and is equipollent to some cardinal kappa,
>
> > > WRONG.
>
> > Nonsense. Of course it is correct. If a set can be well-ordered, then
> > there exists some cardinal kappa such that the set is equipollent to
> > kappa.
>
> So FUCKING what??
So what I said was correct, and the reason why you are shouting and
swearing is unclear.
> There are models of ZFC with the set equipollent TO DIFFERENT kappas!
Well, that statement could use a bit more precision... But, yeah,
there are models of ZFC where beth-1 is equal to different alephs...
So fucking what, if I may borrow your turn of phrase?
> ZFC does NOT prove that the set has any PARTICULAR cardinal as
> a cardinality!
>
ZFC proves that each set has a unique cardinality, as I claimed.
Yaaaawn....
>
>
> > > Ex[whatever]
> > > does NOT mean that there exists a UNIQUE x such that whatever!
>
> > No, and I did not claim uniqueness, but as a matter of fact in this
> > case it is extremely easy to prove uniqueness as well.
>
> So FUCKING what??
> OF COURSE the ordinal in question is unique in every INDIVIDUAL
> model! So "the cardinal is unique" is a theorem.
Which was exactly what I claimed, so maybe you should acknowledge this
and stop shouting and swearing...
> Yet DESPITE this,
> the cardinal IS NOT unique because DIFFERENT models have DIFFERENT
> cardinals playing this role!
>
"The cardinal is unique" is a theorem, yet the cardinal is not unique.
Well, *there's* a thought. You claim one of the axioms is false, do
you?
You are deeply confused...
> > > There are MANY DIFFERENT well-orderings of p(w) under ZFC and
> > > they correspond TO DIFFERENT ordinals!
>
> > But I spoke of a cardinal, an ordinal which is not equipollent to any
> > smaller ordinal. If a set can be well-ordered, then there exists an
> > ordinal (not unique) to which it is equipollent, and there exists a
> > unique *cardinal* to which it is equipollent.
>
> NO THERE *DOESN'*, DUMBASS! (DAMN the STUPID is getting
> THICK in here)!
Oh go screw yourself, you tiresome pointless person.
<snip>
David Libert
[Deletion]
> PS. I have e-mailed you a copy of the Jech paper. I haven't myself
> studied the argument for the existence of the set of hereditarily
> countable sets in absence of choice, by establishing that the rank of a
> hereditarily countable set <= omega_2, and so can say nothing about
> whether it may be generalised beyond the countable case.
>
> --
> Aatu Koskensilta (aatu.kos...@uta.fi)
>
> "Wovon man nicht sprechen kann, dar�ber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Thanks. Jech's proof in ZF that HC is a set (not a proper class)
is in the first 2 pages of the paper.
I think Jech's proof straightdforwardly generalizes to H_kappa
for von Neumann cardinals kappa, showing from ZF that these are sets,
namely replacing omega in Jech's proof by kappa. (Well with
adjustments: HC is heritarily <= omaga and H_kappa is
heritarily < kappa : <= versus < ).
I also think Jech's proof generalizes to H_(< #x) for
non-well-orderable x, by an adjusted as above replacement
of omega by the surjective Hartog ordinal of x.
If this is correct, it would show Zuhair's definition of cardinal
is a set for arbtrary x as the theorem of ZF, depending on
regularity and replacement but no AC needed.
--
David Libert ah...@FreeNet.Carleton.CA
zuhair
> Aatu Koskensilta (aatu.koskensi...@uta.fi) writes:
>
> [Deletion]
>
> > PS. I have e-mailed you a copy of the Jech paper. I haven't myself
> > studied the argument for the existence of the set of hereditarily
> > countable sets in absence of choice, by establishing that the rank of a
> > hereditarily countable set <= omega_2, and so can say nothing about
> > whether it may be generalised beyond the countable case.
>
> > --
>
> > "Wovon man nicht sprechen kann, darüber muss man schweigen"
> > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
>
> Thanks. Jech's proof in ZF that HC is a set (not a proper class)
> is in the first 2 pages of the paper.
>
> I think Jech's proof straightdforwardly generalizes to H_kappa
> for von Neumann cardinals kappa, showing from ZF that these are sets,
> namely replacing omega in Jech's proof by kappa. (Well with
> adjustments: HC is heritarily <= omaga and H_kappa is
> heritarily < kappa : <= versus < ).
>
> I also think Jech's proof generalizes to H_(< #x) for
> non-well-orderable x, by an adjusted as above replacement
> of omega by the surjective Hartog ordinal of x.
>
> If this is correct, it would show Zuhair's definition of cardinal
> is a set for arbtrary x as the theorem of ZF, depending on
> regularity and replacement but no AC needed.
Yeah, I need to look into that carefully. It appears to agree with my
guess that these cardinals I defined are not choice dependent.
Thanks David.
Zuhair
>
> --
> David Libert ah...@FreeNet.Carleton.CA
Jesse F. Hughes
> NO THERE *DOESN'*, DUMBASS! (DAMN the STUPID is getting
> THICK in here)!
If I were sitting where you are, I'm sure I'd think the stupid was
pretty thick too.
Rupert was correct. You act as if he wasn't, but he was, and this is
one of those weird situations not solved by the caps lock key.
--
Jesse F. Hughes
"Really, I'm not out to destroy Microsoft. That will just be a
completely unintentional side effect." -- Linus Torvalds
zuhair
>
> [Deletion]
>
> > PS. I have e-mailed you a copy of the Jech paper. I haven't myself
> > studied the argument for the existence of the set of hereditarily
> > countable sets in absence of choice, by establishing that the rank of a
> > hereditarily countable set <= omega_2, and so can say nothing about
> > whether it may be generalised beyond the countable case.
>
> > --
>
> > "Wovon man nicht sprechen kann, darüber muss man schweigen"
> > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
>
> Thanks. Jech's proof in ZF that HC is a set (not a proper class)
> is in the first 2 pages of the paper.
>
> I think Jech's proof straightdforwardly generalizes to H_kappa
> for von Neumann cardinals kappa, showing from ZF that these are sets,
> namely replacing omega in Jech's proof by kappa. (Well with
> adjustments: HC is heritarily <= omaga and H_kappa is
> heritarily < kappa : <= versus < ).
>
> I also think Jech's proof generalizes to H_(< #x) for
> non-well-orderable x, by an adjusted as above replacement
> of omega by the surjective Hartog ordinal of x.
if there is a surjective ordinal on x, then x must be well orderable.
Zuhair
K_h
"zuhair" <zalj...@gmail.com> wrote in message
news:32349214-92a2-4803...@p33g2000vbn.googlegroups.com...
On Nov 24, 12:55 am, Rupert <rupertmccal...@yahoo.com>
wrote:
> On Nov 23, 10:37 pm, zuhair <zaljo...@gmail.com> wrote:
>
> > > > > > I would like to suggest the following
> > > > > > definition:
>
> > > > > > 4) The cardinality of any set x is: The class of
> > > > > > all sets
> > > > > > that are equinumerous to x were every member of
> > > > > > their transitive
> > > > > > closure is strictly subnumerous to x.
Here is a fast and easy way to define cardinals. Define the
surreal numbers and their order using Conway's simple
construction convention and definition (these only require a
few lines). Call surreals of the form {x|} "left surreals".
Then define the cardinality of a set to be the least left
surreal equinumerous to it.
You're done!
One reason this definition is so effective is because
infinite left surreals are equinumerous to all infinite left
surreals less than themselves. Additionally, one can
distinguish between finite and infinite left surreals by
whether or not they can be equinumerous to proper subclasses
of themselves.
http://planetmath.org/encyclopedia/OmnificIntegers.html
Rupert
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
This is not true. Let us assume that ZF is consistent. Then ZF can
prove that there is a surjection from aleph-one to the reals, but ZF
cannot prove that the reals can be well-ordered.
zuhair
I was speaking of the opposite. If there is a surjection from the
reals to an ordinal, then the reals are well orderable?
Zuhair
Rupert
>
> - Show quoted text -
No that is not true, either. It can be proved in ZF that there is a
surjection from the reals to omega. But it cannot be proved in ZF that
the reals can be well-ordered unless ZF is inconsistent.
zuhair
Oh yes. But in this way can't we prove in ZF that there would be an
injection from the reals to aleph-one.
Zuhair
zuhair
Yea, of course, I was confused. You are right.
Zuhair
Rupert
>
> - Show quoted text -
Yes, correct. But we can't prove that in ZFC either actually, because
the continuum may not be that small.
But given any set we can take the first ordinal greater than all the
ordinals injectable into it. In ZFC that will be the successor of the
cardinality of the set and the set will be injectable into it, but in
ZF the set is not necessarily injectable into it, as you observe.
Ostap S. B. M. Bender Jr.
> Hi all,
>
> As far as I know, all the definitions of cardinality are limited in a
> way or another, lets take them one after the other:
>
> 1) Von Neumann's Cardinals:
>
> A cardinal is the least of all equinumerous ordinals.
>
> 2) Frege-Russell Cardinals:
>
> A cardinal is an equivalence class of sets under equivalence relation
> "bijection".
>
> 3) Scott-Potter Cardinals:
>
> A cardinal is a class of all equinumerous sets from a common level.
>
> Now lets come to discuss each one of them:
>
> 1) Von Neumann's cardinals has the limitation of being dependent on
> choice, without choice one cannot know what is the cardinality of
> Power(omega) for example.Accordingly in any theory which do not have
> the axiom of choice among its axioms most of its sets would be of
> indeterminable cardinality, which is a big draw back.
>
> 2) Frege-Russell cardinals contradict Z set theory, since their
> existence would imply the existence of the set of all sets, which is
> in contradiction with Z.
>
> However in NBG and MK class theories, we can define
> Frege-Russell cardinals, but by then they would be proper classes and
> not sets, which is a great draw back, since proper classes cannot be
> members of other classes, and they are hard to work with.
>
> In NF and related theories, Frege-Russell cardinals are sets, but
> these theories generally depend on the concept of stratification
> of formulas, which is a complex concept, and even finite
> axiomatization of NF and NFU and related theories is a complicated
> approach, and at the end it also resort to stratification for most of
> its inferences. All that make these cardinals undesirable.
>
> 3) Scott-Potter Cardinals: depend on the concept of "level" which
> depends on the concept of type (Scott) and the iterative concept
> (Potter), both concepts of which are complex and difficult to work
> with, besides they are not the basic
> concepts we use to compare set sizes.
>
> I would like to suggest the following definition:
>
> 4) The cardinality of any set x is: The class of all sets
> that are equinumerous to x were every member of their transitive
> closure is strictly subnumerous to x.
>
What is"equinumerous"? How do you define that? "Of the same
cardinality"? :-)
What is the "transitive closure" of a set?
Here is a set: {Apple, PC, Pear, Banana}. What is its "transitive
closure"?
>
> So for any set x, any y is a member of the cardinality of x,
> if and only if, y is equinumerous to x and every member of the
> transitive closure of y is strictly subnumerous to x.
>
> In symbols:
>
> Define(cardinality(x)):-
>
> z=cardinality(x) <->
> for all y (y e z <->
> (y equi-numerous to x &
> for all m (m e Tc(y)->m strictly subnumerous to x)))
>
> Were Tc(y) stands for the 'transitive closure of y' defined
> in the standard manner.
>
> Tc(y)=U{y,Uy,UUy,UUUy,......}
>
> We can actually better define these cardinals through defining the
> concept of "hereditary sets"
>
> Define(hereditary):
> x is hereditary <->
> for all y (y e Tc(x) -> y strictly subnumerous to x)
>
> So a cardinal can be defined in the following manner:
>
> A Cardinal is an equivalence class of hereditary sets under
> equivalence relation "bijection".
>
> Or simply
>
> A Cardinal is a class of all equinumerous hereditary sets.
>
> So cardinality of x would be written shortly as:
>
> Cardinality(x) = {y| y is hereditary & y equinumerous to x}
>
> Now it can be proven in ZF that those cardinals would be 'sets', so
> they are not proper classes! which makes them easy to handle.
>
> These cardinals don't require choice.
>
> They don't require complex concepts like "stratification,type,
> iteration"
>
> They simply depend on the basic concept used to compare set sizes,
> which is the presence or absence of injections between the compared
> sets.
>
> To me this definition seems to be simpler, more general, and it works
> with or without choice, with or without regularity.
>
> So at the end I shall write the definition of cardinal again:
>
> A Cardinal is an equivalence class of hereditary sets under
> equivalence relation "bijection".
>
> x is hereditary <->
> for all y (y e Tc(x) -> y strictly subnumerous to x)
>
> Cardinality(x) = {y| y is hereditary & y equinumerous to x}
>
> Zuhair
zuhair
equinumerous bes the existence of a bijection.
x equinumerous to y <-> Exist a bijection between x and y.
the definition of "bijection" do not mention 'cardinality', so it
is not circular.
>
> What is the "transitive closure" of a set?
>
> Here is a set: {Apple, PC, Pear, Banana}. What is its "transitive
> closure"?
That is not a set in ZF, unless you add ur-elements to it, and you
should
add the constants apple,PC,Pear,Banana.
However all of our discussion is about ZF without ur-elements .
About the transitive closure of your set, this largely depends on how
you will integrate ur-element to a set theory like ZF, but anyhow
in most treatments of ur-elements , the transitive closure of your set
would be the set itself. Now if you chose the approach of
stipulating that every ur-element is an empty object (with
modification
of Extensionality to exclude them) then all your elements would be
subnumerous to any cardinality, and your set will enter as a member
into the fifth cardinal (i.e Natural number 4).
Regards
Zuhair
Jesse F. Hughes
> What is the "transitive closure" of a set?
He defines that right below!
>>
>> So for any set x, any y is a member of the cardinality of x,
>> if and only if, y is equinumerous to x and every member of the
>> transitive closure of y is strictly subnumerous to x.
>>
>> In symbols:
>>
>> Define(cardinality(x)):-
>>
>> z=cardinality(x) <->
>> for all y (y e z <->
>> (y equi-numerous to x &
>> for all m (m e Tc(y)->m strictly subnumerous to x)))
>>
>> Were Tc(y) stands for the 'transitive closure of y' defined
>> in the standard manner.
>>
>> Tc(y)=U{y,Uy,UUy,UUUy,......}
>
> Here is a set: {Apple, PC, Pear, Banana}. What is its "transitive
> closure"?
>
If Apple, PC, Pear and Banana are urelements, then
Tc({Apple, PC, Pear, Banana}) = {Apple, PC, Pear, Banana}.
I'm not sure that Zuhair's theory has urelements, so we can't answer
your question unless you tell us which sets Apple, PC, Pear and Banana
denote.
--
Jesse F. Hughes
Me: "Quincy, there's only *one* Truth, isn't there?"
Quincy (age 4): "Yeah, and it's *mine*."
-- A lesson in postmodernism goes awry.
George Greene
> Rupert was correct.
He is not correct, and you aren't either.
> You act as if he wasn't,
This is no act. You are just completely full of shit, and so is he.
Calling something beth-1 does not tell you what ordinal it is.
Proving that some initial ordinal (somewhere) must be equipollent to
p(w) does NOT tell you WHICH initial ordinal it might be.
Far more relevantly proving (as ZFC does, but as both you and Rupert
are totally INcorrect about) that different models put DIFFERENT
initial
ordinals into this role really does prove that the question simply has
no answer.
It proves a very fundamental mode of ignorance/indeterminacy on the
part of the
theory ITSELF about the question.
For Rupert to have said that "it proves it's beth-1"
is probably the single stupidest thing HE HAS EVER said.
This is one of those weird situations that is not solved by support
from Jesse Hughes.
Jesse F. Hughes
> On Nov 26, 9:23 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Rupert was correct.
>
> He is not correct, and you aren't either.
Every set is equipollent to some cardinal kappa. This is a simple
theorem of ZFC.
>
>> You act as if he wasn't,
>
> This is no act. You are just completely full of shit, and so is he.
>
> Calling something beth-1 does not tell you what ordinal it is.
> Proving that some initial ordinal (somewhere) must be equipollent to
> p(w) does NOT tell you WHICH initial ordinal it might be.
Look, you said Rupert was wrong when he claimed that every set is
equipollent to some cardinal kappa. Now you seem to agree that this
is so, and are instead disputing something else entirely.
--
Jesse F. Hughes
"The Hammer has arrived."
-- James S. Harris, Feb. 14 2006
George Greene
> Every set is equipollent to some cardinal kappa. This is a simple
> theorem of ZFC.
Therefore, NObody is disputing THAT.
So why are you even belaboring it?
This is called STUPIDITY.
ON YOUR part.
> Look, you said Rupert was wrong when he claimed that every set is
> equipollent to some cardinal kappa.
You are A DAMN LIAR.
I said NO such thing.
I said he was wrong about ZFC determining the cardinality of p(w).
And ZFC in fact does NOT determine that p(w) is equipollent to any
PARTICULAR kappa!
> Now you seem to agree that this
> is so, and are instead disputing something else entirely.
Nothing has changed.
I have always been disputing the SAME thing.
Rupert is the one who chose to respond to the dispute by saying
"it does so too have a cardinality -- beth-1".
THAT WAS STUPID.
Your failure to even properly identify the issue under discussion
isn't much smarter.
Jesse F. Hughes
> On Nov 30, 8:59 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Every set is equipollent to some cardinal kappa. This is a simple
>> theorem of ZFC.
>
> Therefore, NObody is disputing THAT.
Oh. I guess I can't read.
Here's the bit that gave me this odd notion that you doubted this
simple theorem.
,----
| > If I assume the axiom of choice, then every set can be well-ordered,
|
| Right.
|
| > and is equipollent to some cardinal kappa,
|
| WRONG.
`----
> So why are you even belaboring it?
> This is called STUPIDITY.
> ON YOUR part.
Yes, I must be stupid for misunderstanding that very clear exchange.
Rupert said that ZFC proves every set is equipollent to some cardinal
kappa and you said he was wrong. I thought that meant you believed he
was wrong.
I don't know where I got this silly notion.
>> Look, you said Rupert was wrong when he claimed that every set is
>> equipollent to some cardinal kappa.
>
> You are A DAMN LIAR.
> I said NO such thing.
Right. I probably just forged the above exchange.
Note: you continued that little exchange as follows:
,----
| Ex[whatever]
| does NOT mean that there exists a UNIQUE x such that whatever!
| There are MANY DIFFERENT well-orderings of p(w) under ZFC and
| they correspond TO DIFFERENT ordinals!
`----
But, of course, this is a non-sequitur. ZFC proves that there is a
unique cardinal equipollent to P(w).
> I said he was wrong about ZFC determining the cardinality of p(w).
> And ZFC in fact does NOT determine that p(w) is equipollent to any
> PARTICULAR kappa!
That may be what you meant, but it sure as hell isn't what you said.
More to the point, you mean that ZFC does not prove |P(w)|=aleph_alpha
for any ordinal alpha. That is clearly true.
But it ain't my fault if what you meant is not what you said.
>> Now you seem to agree that this
>> is so, and are instead disputing something else entirely.
>
> Nothing has changed.
> I have always been disputing the SAME thing.
> Rupert is the one who chose to respond to the dispute by saying
> "it does so too have a cardinality -- beth-1".
> THAT WAS STUPID.
Well, in a perfectly literal sense, Rupert's claim is true and correct
(and not terribly informative). P(w) has a cardinality and that
cardinality is beth-1 -- by definition of beth-1. Unfortunately, that
does not tell us which aleph is the cardinality of P(w), but no one
has disputed this obvious fact.
> Your failure to even properly identify the issue under discussion
> isn't much smarter.
I can see that your writing skills are lacking, but let's try your
reading comprehension. Pretend, for a moment, that these were not
your words, but were written by someone else. How would you interpret
them?
,----
| On Nov 24, 3:23 am, Rupert <rupertmccal...@yahoo.com> wrote:
|
| > If I assume the axiom of choice, then every set can be well-ordered,
|
| Right.
|
|
| > and is equipollent to some cardinal kappa,
|
| WRONG.
|
| Ex[whatever]
| does NOT mean that there exists a UNIQUE x such that whatever!
| There are MANY DIFFERENT well-orderings of p(w) under ZFC and
| they correspond TO DIFFERENT ordinals!
`----
--
"And that's what we do. We put in more troops to get to a position
where we can be in some other place. The question is, who ought to
make that decision? The Congress or the commanders? And as you know,
my position is clear. I'm the commander guy." --- George W. Bush
Jesse F. Hughes
> On Nov 30, 8:59 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Every set is equipollent to some cardinal kappa. This is a simple
>> theorem of ZFC.
>
> Therefore, NObody is disputing THAT.
> So why are you even belaboring it?
> This is called STUPIDITY.
> ON YOUR part.
Tell me, George, do you find it difficult to respond with such passion
and vitriol when you know that you're wrong? Or does being right
matter much less than being loud?
Honestly, something ain't right with you.
--
Jesse F. Hughes
"Have we learned nothing, nothing, from the downfall of Vanilla Ice?"
-- Time Magazine columnist Lev Grossman on
James Frey's /A Million Pieces/.
zuhair
In my search for making a better definition of Cardinality and during
my discussions with Prof. Dana S. Scott, some issues raised that makes
me ask the following four questions:
Question 1: is the following a theorem of ZF?
For all x Exist y ( y is transitive & y equinumerous to x )
were
y equinumerous to x <-> Exist f (f:x-->y, f is bijective).
Question 2: If the above is not a theorem of ZF, then is its negation
a theorem of ZF?
Question 3: is the following a theorem of ZF?
For all m Exist x
( x is transitive &
for all y ((y is transitive & y supernumerous to m) -> x subnumerous
to y))
were
y supernumerous to m <-> Exist f (f:m-->y, f is injective)
x subnumerous to y <-> Exist f (f:x-->y, f is injective)
Question 4: if the above is not a theorem of ZF, then is its negation
a theorem of ZF?
Zuhair
zuhair
Actually the important question is precisely the following:
We know from ZF that for every set x there exist a transitive closure
set TC(x).
Now does ZF prove or refute the following?
For all x Exist y
(y equinumerous to x &
~ Exist z (z equinumerous to x & TC(z) strictly subnumerous to TC
(y)))
were
x subnumerous to y <-> Exist f (f:x-->y, f is injective)
x equinumerous to y <-> Exist f (f:x-->y, f is bijective)
x strictly subnumerous to y <->
(x subnumerous to y & ~ x equinumerous to y)
This question is very important!
Zuhair
K_h
>4 important questions?
>
> In my search for making a better definition of Cardinality
> and during
Here is a fast and easy way to define cardinals -- I think.
Define the surreal numbers and their order using Conway's
simple construction convention and definition (these only
require a few lines). Call surreals of the form {x|} "left
surreals". Then define the cardinality of a set to be the
least left surreal equinumerous to it.
You're done!
One reason this definition should be a good one is because
infinite left surreals are equinumerous to all infinite left
surreals less than themselves. Additionally, one can
distinguish between finite and infinite left surreals by
whether or not they can be equinumerous to proper subclasses
of themselves. I don't see any problem with this approach
but perhaps I missed something.
Best,
REF:
http://planetmath.org/encyclopedia/OmnificIntegers.html
George Greene
> Oh. I guess I can't read.
Not much guessing required.
> Here's the bit that gave me this odd notion that you doubted this
> simple theorem.
>
> ,----
> | > If I assume the axiom of choice, then every set can be well-ordered,
> |
> | Right.
> |
> | > and is equipollent to some cardinal kappa,
> |
> | WRONG.
> `----
>
> > So why are you even belaboring it?
> > This is called STUPIDITY.
> > ON YOUR part.
This is just plain dishonest. These two things ARE NOT ADJACENT
in the original context. You have snipped without even acknowleding
RELEVANT context that you have removed!!
> Yes, I must be stupid for misunderstanding that very clear exchange.
No, you must be stupid FOR REMOVING THE CONTEXT AND LYING
about what was being said. WHAT ACTUALLY OCCURRED after I said
"WRONG" was:
>> Ex[whatever]
>> does NOT mean that there exists a UNIQUE x such that whatever!
>> There are MANY DIFFERENT well-orderings of p(w) under ZFC and
>> they correspond TO DIFFERENT ordinals!
This IS relevant, it DOES matter, and it DOES PROVE that Rupert was
wrong.
I was not disputing ANY of the THEOREMS!
YES, IT IS a theorem that any set is equipollent to some initial
ordinal.
YES, IT IS a theorem that any set (including p(w)) therefore has a
cardinality.
These truths DO *NOT MATTER*!!
They are NOT SUFFICIENT FOR ANYone to allege that ZFC *assigns*
or *determines* a cardinality FOR ANY uncountable set!
DESPITE these theorems, the ACTUAL PARTICULAR cardinality of
ANY uncountable set IS UNdetermined by ZFC!
This is A VERY BAD state of affairs!
I mean, if ZFC would just ADMIT that it couldn't determine these
cardinalities,
that would be BETTER. But INSTEAD it CLAIMS to fix them and then
FAILS to!
In any case, my point is, I DID state this, and YOU DID *CUT* it while
TRYING
to claim that you were saying something rational about the argument!
MORE to the point, I PREFIGURED all of this: I STARTED OUT WITH
calm reasonable claims that you are NOT BOTHERING to acquaint
YOURself with before intervening! E.g.,
>> Even WITH choice, the ZFC definition of cardinality STILL has a
>> serious drawback. You STILL CANNOT determine the cardinality of
>> p(w) under ZFC. The collection of all countable ordinals (aleph-one)
>> cannot possibly be bigger than p(w), but it could be equally large--
>> the cardinality of p(w) could be aleph-one, aleph-2, or any finite
>> aleph.
THIS IS THE ONLY sense in which I EVER said that Rupert was "WRONG",
and it WAS PRE-established by the context.
> Rupert said that ZFC proves every set is equipollent to some cardinal
> kappa and you said he was wrong. I thought that meant you believed he
> was wrong.
He IS Wrong, DUMBASS, IN THE CONTEXT of the argument,
WHICH I had pre-established and which YOU had not bothered to
acquaint yourself with. You can point out truthfully that a theory
proves a theorem AND STILL BE WRONG about what THIS entails
or implies ABOUT THE RELEVANCE OF THE THEORY *TO THE ISSUE*
under discussion!
> I don't know where I got this silly notion.
I know; I just don't approve.
> >> Look, you said Rupert was wrong when he claimed that every set is
> >> equipollent to some cardinal kappa.
>
> > You are A DAMN LIAR.
> > I said NO such thing.
>
> Right. I probably just forged the above exchange.
No, you just mangled and misquoted it.
> Note: you continued that little exchange as follows:
Well, better late than never.
It would have been nice for you to follow with it WHERE IT ACTUALLY
FOLLOWED,
instead OF MISleading as above.
>
> ,----
> | Ex[whatever]
> | does NOT mean that there exists a UNIQUE x such that whatever!
> | There are MANY DIFFERENT well-orderings of p(w) under ZFC and
> | they correspond TO DIFFERENT ordinals!
> `----
>
> But, of course, this is a non-sequitur.
And you, of course, are a lying dipshit.
> ZFC proves that there is a
> unique cardinal equipollent to P(w).
AND THAT is a non-sequitur.
Or, IF it follows, it follows in the same vein of spectacular
failure as before. DESPITE the fact that ZFC proves this,
THERE ISN'T a unique cardinal equipollent to p(w).
To the extent that it can talk about models of itself if you introduce
"There exists a model of ZFC" as a hypothesis, ZFC proves ABOUT
itself that it DOESN'T prove this, because it (under the assumption
that
a model exists) proves the existence of models with DIFFERENT
cardinals AS OPPOSED to any unique one IN THIS ROLE!
The uniqueness is PER MODEL! There are many DIFFERENT models
with DIFFERENT initial ordinals in this role! THAT IS WHAT MATTERS.
Rupert's rebuttal was "beth-1 exists", WHICH WAS STUPID.
Your choosing his side is even stupider. Your calling my
clarification OF WHAT I WAS SAYING "a non-sequitur" is just
unforgivable.
> > I said he was wrong about ZFC determining the cardinality of p(w).
> > And ZFC in fact does NOT determine that p(w) is equipollent to any
> > PARTICULAR kappa!
>
> That may be what you meant, but it sure as hell isn't what you said.
It SURE AS HELL IS what I *STARTED OUT* by saying.
I'm sorry if YOU weren't YET paying attention!
> More to the point, you mean that ZFC does not prove |P(w)|=aleph_alpha
> for any ordinal alpha.
This IS NOT "more to the point", dumbass, SINCE *I ALREADY SAID THAT*
BEFORE you even STARTED paying attention!
> That is clearly true.
SHIT. This IS NOT clearly true.
ANYbody TO WHOM this was clearly true WOULD NOT
be alleging in public that ZFC assigned *a* cardinality to p(w)!
They ESPECIALLY would not be reinforcing the MIS-guided notion
that "beth-1 is THE cardinality that ZFC assigns to p(w)"!
This fact WAS NOT CLEAR TO RUPERT and Rupert is an expert!!
This fact IS NOT CLEAR TO YOU DESPITE your claims to be agreeing
with it! If you ACTUALLY understood it then you would be picking MY
side
in this argument AND NOT the one you are picking!
This is NOT "clearly" true. This is CLOSER to "clearly"
true AFTER you get into an ARGUMENT
with ME about it. BEFORE then, it is "clear" that
ZFC *does* provide a cardinality for this set
BECAUSE IT PROVES THEOREMS THAT SAY IT DOES!!
The fact that these theorems ARE NOT SUFFICIENT to determine
the issue IS COUNTER-intuitive AND NOT clear UNTIL AFTER
you THINK about it for a while!
> But it ain't my fault if what you meant is not what you said.
It IS, however, your fault that you hadn't even read prior
relevant messages before picking a side, or that you decided
to talk about the caps lock key rather than THE ISSUE.
All that IS YOUR FAULT.
George Greene
> George Greene <gree...@email.unc.edu> writes:
> > On Nov 30, 8:59 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Every set is equipollent to some cardinal kappa. This is a simple
> >> theorem of ZFC.
>
> > Therefore, NObody is disputing THAT.
> > So why are you even belaboring it?
> > This is called STUPIDITY.
> > ON YOUR part.
>
> Tell me, George, do you find it difficult to respond with such passion
> and vitriol when you know that you're wrong? Or does being right
> matter much less than being loud?
>
> Honestly, something ain't right with you.
Like I GIVE A SHIT what YOU think about THAT.
What is or isn't right with me IS NOT THE ISSUE!
THE ISSUE IS WHAT ZFC SAYS ABOUT THE CARDINALITY OF p(w)!!
THAT is what was being discussed! THAT WAS ALL that was being
discussed!
EVERYBODY ELSE IS WRONG to the extent that they are trying to talk
about
ANYthing else WITH ME HERE IN THIS conversation!
Here is what I said AT THE BEGINNING:
>>On Nov 24, 12:55 am, Rupert <rupertmccal...@yahoo.com> wrote:
>> But, yeah, I would like to know how you do it in ZF.
> You can't do it in ZFC either.
> ZFC DOES NOT assign a cardinality (in the usual sense)
> to p(w).
THIS IS ALL that is going on!
NOTHING else is being discussed!
YOUR ATTEMPT to make this about me and who's loud
*IS*FUCKING*STUPID*.
It is ANTI-intellectual, ANTI-logical, and COMPLETELY OUT OF PLACE
HERE!
More to the point, I BEGAN BY CLARIFYING that
I was talking about ASSIGNING something in a PARTICULAR SENSE.
I PRE-advertised that something as lame as a non-constructive
existence
axiom MIGHT NOT be sufficient! I did NOT claim to be DISagreeing with
proved theorems and I AM NOT WRONG ABOUT ANY of this!!
Jesse F. Hughes
> On Nov 30, 1:40 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Oh. I guess I can't read.
>
> Not much guessing required.
>
>
>> Here's the bit that gave me this odd notion that you doubted this
>> simple theorem.
>>
>> ,----
>> | > If I assume the axiom of choice, then every set can be well-ordered,
>> |
>> | Right.
>> |
>> | > and is equipollent to some cardinal kappa,
>> |
>> | WRONG.
>> `----
>>
>> > So why are you even belaboring it?
>> > This is called STUPIDITY.
>> > ON YOUR part.
>
> This is just plain dishonest. These two things ARE NOT ADJACENT
> in the original context. You have snipped without even acknowleding
> RELEVANT context that you have removed!!
What context have I removed? Those quotations *are* adjacent in the
post on my server. And I included the paragraph following "WRONG."
in the post you're replying to, so you surely aren't complaining about
that snippage.
>> Yes, I must be stupid for misunderstanding that very clear exchange.
>
> No, you must be stupid FOR REMOVING THE CONTEXT AND LYING
> about what was being said. WHAT ACTUALLY OCCURRED after I said
> "WRONG" was:
>>> Ex[whatever]
>>> does NOT mean that there exists a UNIQUE x such that whatever!
>>> There are MANY DIFFERENT well-orderings of p(w) under ZFC and
>>> they correspond TO DIFFERENT ordinals!
>
> This IS relevant, it DOES matter, and it DOES PROVE that Rupert was
> wrong.
No, it doesn't. There are many different well-orderings of P(w),
true. And yet ZFC proves that P(w) is equinumerous to exactly one
cardinal.
Imagine! Just as Rupert said!
> I was not disputing ANY of the THEOREMS!
> YES, IT IS a theorem that any set is equipollent to some initial
> ordinal.
> YES, IT IS a theorem that any set (including p(w)) therefore has a
> cardinality.
> These truths DO *NOT MATTER*!!
They are all that Rupert asserted. Right there, where you responded
"WRONG." Can't you read?
> They are NOT SUFFICIENT FOR ANYone to allege that ZFC *assigns*
> or *determines* a cardinality FOR ANY uncountable set!
Rupert's post did not use the terms "assigns" or "determines". What
the heck are you replying to? Rupert's post has message id
<9be040de-1f99-4f15...@z35g2000prh.googlegroups.com>.
Why don't you read it again and tell me where you think Rupert claimed
something inconsistent with the independence of GCH.
Or don't. Whatever.
> DESPITE these theorems, the ACTUAL PARTICULAR cardinality of
> ANY uncountable set IS UNdetermined by ZFC!
>
> This is A VERY BAD state of affairs!
> I mean, if ZFC would just ADMIT that it couldn't determine these
> cardinalities,
> that would be BETTER. But INSTEAD it CLAIMS to fix them and then
> FAILS to!
>
> In any case, my point is, I DID state this, and YOU DID *CUT* it while
> TRYING
> to claim that you were saying something rational about the argument!
What I cut was irrelevant. All you said is that different
well-orderings of P(w) correspond to different bijections from P(w) to
some ordinal. Yes, but so what? Different well-orderings of w also
correspond to different bijections from w to some ordinal. This
trivial fact that there are different ways to well-order P(w) really
doesn't have anything much to do with CH.
>
> MORE to the point, I PREFIGURED all of this: I STARTED OUT WITH
> calm reasonable claims that you are NOT BOTHERING to acquaint
> YOURself with before intervening! E.g.,
>>> Even WITH choice, the ZFC definition of cardinality STILL has a
>>> serious drawback. You STILL CANNOT determine the cardinality of
>>> p(w) under ZFC. The collection of all countable ordinals (aleph-one)
>>> cannot possibly be bigger than p(w), but it could be equally large--
>>> the cardinality of p(w) could be aleph-one, aleph-2, or any finite
>>> aleph.
>
> THIS IS THE ONLY sense in which I EVER said that Rupert was "WRONG",
> and it WAS PRE-established by the context.
You're right, I didn't read any of your contributions prior to your
asinine reply to Rupert. But none of that matters. Rupert said
something true, and you replied with "WRONG." Followed by an
irrelevant observation about different well-orderings of P(w).
Rest snipped, because I'm afraid that my laptop will run out of
capital letters if I continue this tedious conversation.
You were wrong, George, and all your shouting doesn't change that.
Complain all you want about my reading comprehension. Rupert made a
very clear and correct claim and you read something else and *even
then* couldn't argue your point coherently.
--
Jesse F. Hughes
"Ultimately, I can bring the entire mathematical establishment to its
knees... Live in a fantasy world if you wish, but to me that's just
an expression of your intellectual inferiority." --James Harris