The size of proper classes?
zuhair
Can there exist a proper class x that is not supernumerous to the
class of all ordinals that are sets?
x supernumerous to y <-> Exist f (f:y-->x, f is injective)
I always had the idea that the class of all ordinals that are sets, is
the smallest proper class, i.e. there do not exist a proper class that
is strictly subnumerous to it, but can there exist a proper class that
is incomparable to it, i.e. there do not exist any injection between
it and that proper class.
If so can one give an example of such a proper class?
Zuhair
Herman Rubin
> Can there exist a proper class x that is not supernumerous to the
>class of all ordinals that are sets?
>x supernumerous to y <-> Exist f (f:y-->x, f is injective)
There are certainly Fraenkel-Mostowski models in which this
is false, and I believe Cohen models as well.
Fraenkel-Mostowski models are not models of ZF, but of ZFU;
the models needed are models of NBG, but Fraenkel-Mostowski
models can be extended.
>I always had the idea that the class of all ordinals that are sets, is
>the smallest proper class, i.e. there do not exist a proper class that
>is strictly subnumerous to it, but can there exist a proper class that
>is incomparable to it, i.e. there do not exist any injection between
>it and that proper class.
>If so can one give an example of such a proper class?
Not necessarily. The strongest class form of the Axiom of
Choice has all proper classes equinumerous to the class of
all ordinal numbers. See the book _Equivalents of the
Axiom of Choice II_ by Herman Rubin and Jean E. Rubin. The
construction in Godel's book, _Consistencey of the
Continuum Hypothesis_, constructs and inner model of NBG in
which it is true that the class of ordinal numbers is
equinumerous with the universe.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
zuhair
> In article <b1d5c9ac-d3ca-42a2-9d9c-e2decdf0c...@m35g2000vbi.googlegroups.com>,
>
> zuhair <zaljo...@gmail.com> wrote:
> >Working in NBG\MK minus choice
> > Can there exist a proper class x that is not supernumerous to the
> >class of all ordinals that are sets?
> >x supernumerous to y <-> Exist f (f:y-->x, f is injective)
>
> There are certainly Fraenkel-Mostowski models in which this
> is false, and I believe Cohen models as well.
> Fraenkel-Mostowski models are not models of ZF, but of ZFU;
> the models needed are models of NBG, but Fraenkel-Mostowski
> models can be extended.
If I didn't misunderstand you, what you are saying is the following:
There cannot exist a proper class x that is not supernumerous to the
class of all ordinals, in other words what you are saying is: the
Frankel-Mostowski models prove that every proper class is
supernumerous to the class of all ordinals that are sets, i.e for any
class x to be a proper class then there must exist an injection from
the class of all ordinals that are sets to the class x.
Is that what you are saying?
>
> >I always had the idea that the class of all ordinals that are sets, is
> >the smallest proper class, i.e. there do not exist a proper class that
> >is strictly subnumerous to it, but can there exist a proper class that
> >is incomparable to it, i.e. there do not exist any injection between
> >it and that proper class.
> >If so can one give an example of such a proper class?
>
> Not necessarily. The strongest class form of the Axiom of
> Choice has all proper classes equinumerous to the class of
> all ordinal numbers. See the book _Equivalents of the
> Axiom of Choice II_ by Herman Rubin and Jean E. Rubin. The
> construction in Godel's book, _Consistencey of the
> Continuum Hypothesis_, constructs and inner model of NBG in
> which it is true that the class of ordinal numbers is
> equinumerous with the universe.
This is a little bit vague, what was you referring to when you said
"Not necessarily"?Did you mean that we can have a proper class that is
strictly subnumerous to the class of all ordinals that are sets? or
can there exist a proper class that is
not comparable to the class of all ordinals that are sets? these
points are not clear from your answer.
Thanks for the references.
Zuhair
zuhair
> In article <b1d5c9ac-d3ca-42a2-9d9c-e2decdf0c...@m35g2000vbi.googlegroups.com>,
>
Let me re ask my question using more precise terminology, and working
in ZF instead of NBG\MK.
Is the following a theorem schema of ZF(without choice)?
If phi(y) is a formula in which at least y is free, and in which x is
not free, then all closures of
~ for all d
( d is ordinal ->
Exist x ( for all y ( y e x -> phi(y) ) and
d equinumerous to x ) )
-> Exist x for all y ( y e x <-> phi(y) )
are theorems.
I think the idea behind the question is very clear, if we cannot put
all ordinals that are sets into one-one relation with sets fulfilling
the predicate phi (this is equivalent to saying that we cannot have
an injection from the class of all ordinals that are sets to the
class of all sets fulfilling the predicate phi), then the predicate
phi defines a set, i.e. the class of exactly all sets for which the
predicate phi holds is a set.
Now is that true in ZF(without choice) ?
is that true in ZF without choice and without regularity?
Zuhair
George Greene
> Choice has all proper classes equinumerous to the class of
> all ordinal numbers.
What about the just-plain set/class distinction ITSELF,
in the form of a limitation-of-size principle, i.e., the claim
(or definition, even) that a class is proper if and only if it is
"equinumerous" to the class of all sets? And Is Therefore
"supernumerous" to each individual set? Is *that* involved with
or dependent on Choice?
George Greene
> Working in NBG\MK minus choice
>
> Can there exist a proper class x that is not supernumerous to the
> class of all ordinals that are sets?
Can there be proper-class ordinals??
"all ordinals that are sets" is almost surely redundant --
ordinals HAVE to be sets!
Proper classes, in ADDITION to being "big enough", ALSO
have to NOT be members of other classes! Every ordinal is a member of
its successor class, so it cannot be a proper class.
zuhair
> On Dec 2, 7:53 am, zuhair <zaljo...@gmail.com> wrote:
>
> > Working in NBG\MK minus choice
>
> > Can there exist a proper class x that is not supernumerous to the
> > class of all ordinals that are sets?
>
> While proper classes may be first-class objects in NBG,
> they are not so in ZF or other more usual theories.
> Therefore, EVEN ASKING whether they are or aren't "supernumerous"
> is problematic, INside the theory, since INside the theory, they are
> not
> even in the domain of discourse.
That is not correct. Proper classes exist in the domain of discourse
of NBG and MK, and we can even define cardinalities of these proper
classes, the relation "supernumerous" is well defined for proper
classes, and there is not problem with that whatsoever, so your
remarks above are all mistaken, when you are speaking of NBG\MK.
However they correct when you are speaking about theories that permits
'sets' only, i.e. non ur-elements(classes) that are members of other
classes. so in these theories yes you are correct those objects are
not in the domain of discourse, but yet we can have alternative
methods of dealing with them (please see my second reply to Herman
Rubin, in order to have an example of such equivalent treatements in
ZF and similar "set" theories).
These
>
> Even when they are in the domain, as in NBG, the set-class
> distinction is sharp AND SIMPLE, which is why long replies to this
> (even when they are expert and correct) are STILL misguided.
> You need to know the FUNDAMENTAL point about sizes of
> proper classes: The size of EVERY proper class is the SAME size
> as the size of the class of all sets.
No that is not correct. Not all models of NBG stipulate that, you are
taking about a model of NBG\MK that allows global choice, that is not
the only model, there are other models in which proper classes can be
defined as being supernumerous to
the class of all ordinals that are sets which is the smallest possible
proper class.
If you read my post carefully I said NBG(minus choice) , what I mean
by that
NBG that do not contain what you've just mentioned (your are talking
about NBG\MK in which the size limitation axiom stipulate that a set
is a class that is strictly subnumerous to V (the class of all sets))
there are many variants of this size limitation axioms, some of them
even don't mention V at all.
In a model of NBG\MK minus choice, you cannot have all proper classes
equal in size (i.e.equinumerous), but in any model of NBG\MK the least
(in size) possible proper class is the class of all ordinals that are
sets, so you can stipulate that a set is a class that is not
supernumerous (equal or bigger) than the class of all ordinals that
are sets, and in this way you leave room for V to be bigger (strictly
supernumerous) to the class of all ordinals that are sets, thus choice
would not be entailed in this model, there are various other
treatments of "size limitation" axiom that allow different comparisons
than those that I have mentioned.
> THAT IS the set/class distinction in these contexts: "subnumerous to"
> vs. "equinumerous to" the class of all sets.
No not necessarily, you are talking about one particular model of NBG
\MK that entail global choice, which is not the only model, actually
this model is too strong to the degree of making it undesirable, we
need the weaker models not this strong model.
>
> I can see a motivation for your question coming out of the fact that
> it
> initially appears that "most" sets are NOT ordinals, yet the class of
> all
> ordinals cannot be a set. So it is perhaps counter-intuitive that
> something
> consisting of as small and special a sliver of the class of all sets
> as JUST
> the ordinals could be the same size as the class of all sets.
> Especially since, without choice, you may not be able to show the
> existence of the actual bijection confirming the equipollence.
> But then again, you may, in SOME contexts anyway.
Yes, that is correct somehow. As I said above not in all models of NBG
\MK you have the size of the class of all sets (usually denoted as V)
equal to the size of the class of all ordinals that are sets, as I
said this concept is too strong.
The main idea is actually the following:
a) the class of all ordinals that are sets is the smallest possible
proper class in all models of NBG\MK
b) What is strictly subnumerous to the class of all ordinals that are
sets, is a set
and what is not comparable to the class of all ordinals that are
sets, is a set.
I am sure of the first half of b), but I am not sure of the second
statement in b).
Regards
Zuhair
George Greene
> On Dec 2, 8:40 pm, George Greene <gree...@email.unc.edu> wrote:
>
> > On Dec 2, 7:53 am, zuhair <zaljo...@gmail.com> wrote:
>
> > > Working in NBG\MK minus choice
>
> > > Can there exist a proper class x that is not supernumerous to the
> > > class of all ordinals that are sets?
>
> > This is NOT a well-formed question.
You were right.
I was wrong.
I retracted the message.
Unfortunately I was not in a position to retract it immediately
and you rebutted before I could retract it.
Aatu Koskensilta
> What about the just-plain set/class distinction ITSELF, in the form of
> a limitation-of-size principle, i.e., the claim (or definition, even)
> that a class is proper if and only if it is "equinumerous" to the
> class of all sets?
This principle, claim, definition (obviously) implies global
choice. Just pause to think about it for a moment, bearing in mind the
class of ordinals is a proper class.
If we're willing to take the consistency of ZFC + "there is an
inaccessible" as given, it's easy to see neither NBG nor MK proves all
proper classes are the same size. We need but recall that <V_kappa,
V_kappa+1> is a model of MK for an inaccessible kappa, and that it's
consistent with ZFC that GCH fails arbitrarily badly at kappa. With some
tweaking (e.g. by considering variants of ZFC with a restricted powerset
axiom) we can eliminate, in so far as consistency strength is
considered, the inessential inaccessible.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
George Greene
Actually I DID retract this before you replied to it.
I wrote it at 8:40 and got forcibly logged off the computer I wrote it
on,
as I was realizing that it was wrong. I had to go to another computer
and
retracted it around 8:55.
But you had probably already started replying to it and finished
drafting
your reply around 9:15.
Plus it takes a while for retractions to propagate, and not all
servers honor them.
I was really hoping I would get to it before you did.
zuhair
Yea, that is another error also, many people think that all ordinals
are sets, which is a wrong concept, to show this error we need to
recall the definition of ordinal, I shall use a definition that is
equivalent to the standard one, but to me it is simpler than the
standard one.
X is ordinal <->[ X is transitive & for all m ( m in X -> m is
transitive )]
were X is transitive <-> For all m,n ((n e m & m e X) -> n e X)
or simply a transitive set is a set were every member of it is a
subclass of it.
Now this definition is equivalent to the standard definition provided
we have axiom of Regularity ( Foundation ).
However without axiom of Regularity the definition would be
X is ordinal <->
[X is transitive & for all m ( m in X -> m is transitive ) &
For all Y ((Y subclass of X & ~Y=0) -> Exist z ( z e Y & z disjoint
Y ))].
were z disjoint Y <-> ~ Exist c ( c e z & c e Y )
Now this definition do not require Regularity and is equivalent to the
standard definition in all versions of ZF, NBG\MK.
To simplify matters lets assume Regularity and work with the first
definition.
So simply this definition says that an ordinal is a transitive class
of transitive sets.
So you see there is nothing in this definition that imply that an
ordinal is a set.
Now lets take the class D of all ordinals that are sets, now D itself
would be a transitive class of transitive sets i.e. an ordinal, but D
is not a set, thus there
exist an ordinal that is not a set, i.e. a proper class, and this
ordinal is
the class of all ordinals that are sets.
However sometimes one can say the class of all "ordinals", but in
reality what is meant is the class of all "ordinals that are sets",
but since it is trivial to
say "that are sets" then one might as well omit it from the expression
for purposes
of convenience since it is understood that what is meant is "ordinals
that are sets", but if one want to be precise then one should mention
"ordinals that are sets".
Zuhair
zuhair
Oh, sorry then, well I hope it clarifies issues to others.
Zuhair
George Greene
> > the claim (or definition, even)
> > that a class is proper if and only if it is "equinumerous" to the
> > class of all sets?
H.Rubin had already said,
> The strongest class form of the Axiom of
> Choice has all proper classes equinumerous to the class of
> all ordinal numbers.
I was trying to ask about something possibly weaker,
namely, about all proper classes being equipollent to the
class of all sets, as OPPOSED to the class of all ordinals.
Since HR is presumably expert/correct, and since the class of
all ordinals is a proper subclass of the class of all sets, mustn't
we agree with him that "all proper classes are equipollent to the
class
of all ordinals" is the "Strongest class form" of AC, while
"all proper classes are equipollent to the class of all sets" is a
slightly "weaker class form" of AC?
If that is so, then
On Dec 2, 10:06 pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> writing
> This principle, claim, definition (obviously) implies global choice.
is interesting, because even the "slightly weaker" claim STILL has to
be
not only Stronger than Choice, but Stronger than Global Choice.
So is the ordinal version (being StrongEST) Even Stronger Than That?
zuhair
there is another point here also, not all class theories define proper
classes as not being members of other classes! for example Ackermann's
class theory would become inconsistent if we assume that proper
classes are not members of other classes.
However NBG\MK do define proper classes as being not members of other
classes, as you mentioned ,that is correct.
If we take ZF + inaccessibles , then these inaccessible sets are
"sets" i.e. are members of other classes, but yet they have the size
of a proper class in NBG\MK
with no inaccessibles.
However what join all of these theories, is that proper classes have a
size criterion and they are always supernumerous (not necessarily
strictly) to the class of all ordinals that are sets (were 'set' is
defined according to the relevant theory).
Regards
Zuhair
zuhair
> Re
>
> > > the claim (or definition, even)
> > > that a class is proper if and only if it is "equinumerous" to the
> > > class of all sets?
>
> H.Rubin had already said,
>
> > The strongest class form of the Axiom of
> > Choice has all proper classes equinumerous to the class of
> > all ordinal numbers.
>
> I was trying to ask about something possibly weaker,
> namely, about all proper classes being equipollent to the
> class of all sets, as OPPOSED to the class of all ordinals.
They look equivalent to me! don't they?
zuhair
Not it imply choice, i.e Choice would be a theorem, actually global
choice.
How I see matters, is that there is no difference between saying
that "x is a proper class iff x is equinumerous with the class of all
sets", and between saying "x is a proper class iff x is equinumerous
with the class of all ordinal numbers (here ordinal numbers mean
ordinals that are sets)", I am speaking of course about NBG\MK
approach. However perhaps I am mistaken.
Zuhair
Aatu Koskensilta
> How I see matters, is that there is no difference between saying that
> "x is a proper class iff x is equinumerous with the class of all
> sets", and between saying "x is a proper class iff x is equinumerous
> with the class of all ordinal numbers (here ordinal numbers mean
> ordinals that are sets)", I am speaking of course about NBG\MK
> approach. However perhaps I am mistaken.
You're not.
Aatu Koskensilta
> If we take ZF + inaccessibles , then these inaccessible sets are
> "sets" i.e. are members of other classes, but yet they have the size
> of a proper class in NBG\MK with no inaccessibles.
This explanation or claim doesn't really make much sense.
Herman Rubin
zuhair <zalj...@gmail.com> wrote:
>On Dec 2, 1:35=A0pm, hru...@odds.stat.purdue.edu (Herman Rubin) wrote:
>> In article <b1d5c9ac-d3ca-42a2-9d9c-e2decdf0c...@m35g2000vbi.googlegroups=
>.com>,
>> zuhair =A0<zaljo...@gmail.com> wrote:
>> >Working in NBG\MK minus choice
>> > Can there exist a proper class x that is not supernumerous to the
>> >class of all ordinals that are sets?
>> >x supernumerous to y <-> Exist f (f:y-->x, f is injective)
>> There are certainly Fraenkel-Mostowski models in which this
>> is false, and I believe Cohen models as well.
>> Fraenkel-Mostowski models are not models of ZF, but of ZFU;
>> the models needed are models of NBG, but Fraenkel-Mostowski
>> models can be extended.
>If I didn't misunderstand you, what you are saying is the following:
>There cannot exist a proper class x that is not supernumerous to the
>class of all ordinals, in other words what you are saying is: the
>Frankel-Mostowski models prove that every proper class is
>supernumerous to the class of all ordinals that are sets, i.e for any
>class x to be a proper class then there must exist an injection from
>the class of all ordinals that are sets to the class x.
>Is that what you are saying?
>> >I always had the idea that the class of all ordinals that are sets, is
>> >the smallest proper class, i.e. there do not exist a proper class that
>> >is strictly subnumerous to it, but can there exist a proper class that
>> >is incomparable to it, i.e. there do not exist any injection between
>> >it and that proper class.
>> >If so can one give an example of such a proper class?
>> Not necessarily. =A0The strongest class form of the Axiom of
>> Choice has all proper classes equinumerous to the class of
>> all ordinal numbers. =A0See the book _Equivalents of the
>> Axiom of Choice II_ by Herman Rubin and Jean E. Rubin. =A0The
>> construction in Godel's book, _Consistencey of the
>> Continuum Hypothesis_, constructs and inner model of NBG in
>> which it is true that the class of ordinal numbers is
>> equinumerous with the universe.
>This is a little bit vague, what was you referring to when you said
>"Not necessarily"?Did you mean that we can have a proper class that is
>strictly subnumerous to the class of all ordinals that are sets? or
>can there exist a proper class that is
>not comparable to the class of all ordinals that are sets? these
>points are not clear from your answer.
>Thanks for the references.
>Zuhair
There exist models where all proper classes have
the same cardinality; i.e., the universe is
equinumerous with the class of ordinal numbers.
There exist models where there are proper classes
which are neither larger nor smaller than the class
of all ordinal numbers.
I am almost certain that there exist models with
proper classes strictly larger than the class of
all ordinal numbers, and all comparable.
Aatu Koskensilta
> I am almost certain that there exist models with proper classes
> strictly larger than the class of all ordinal numbers, and all
> comparable.
Using the consistency of ZFC + "there is an inaccessible" (and standard
well known independence results) it's easy to show there are such
models. As noted, by some tweaking we can remove the inessential
inaccessible, bringing us back to ZFC and MK in terms of consistency
strength.
George Greene
> I am almost certain that there exist models with
> proper classes strictly larger than the class of
> all ordinal numbers, and all comparable.
This invites thought about THREE different kinds of ordinals:
ordinals that are sets, ordinals that are proper classes, and ordinals
that are ordinal "numbers". Maybe the 1st and 3rd are the same,
but the 2nd is different?
George Greene
> hru...@odds.stat.purdue.edu (Herman Rubin) writes:
> > I am almost certain that there exist models with proper classes
> > strictly larger than the class of all ordinal numbers, and all
> > comparable.
So the class of all ordinal numbers, in this context, is proper, but
is comparably smaller than larger proper classes?
And in this context, there can be ordinal proper classes as well?
And the proper class of all set ordinals is one of them?
And this proper-class ordinal also has proper-class successors that
are also ordinals? Do these proper-class ordinals eventually get
big enough to be equipollent to the largest proper classes?
Zdislav V. Kovarik
On Wed, 2 Dec 2009, Herman Rubin wrote:
> In article <b1d5c9ac-d3ca-42a2...@m35g2000vbi.googlegroups.com>,
> zuhair <zalj...@gmail.com> wrote:
> >Working in NBG\MK minus choice
>
> > Can there exist a proper class x that is not supernumerous to the
> >class of all ordinals that are sets?
>
[...]
After an explicit permutation, we obtain the question
"The proper size of a class?"
And my answer is ... about 25 students. Beyond that, I cannot
even remember the faces and names, much less approach them individually.
Cheers,
ZVK(Slavek).
George Greene
> > there is no difference between saying that
> > "x is a proper class iff x is equinumerous with the class of all
> > sets", and between saying "x is a proper class iff x is equinumerous
> > with the class of all ordinal numbers (here ordinal numbers mean
> > ordinals that are sets)",
How is THAT supposed to be proved without choice??
MOST sets are NOT ordinals!!
zuhair
Ok, let me try to simplify matters.
(1) x is a proper class iff x is equinumerous with the class of all
sets"
(2)x is a proper class iff x is equinumerous with the class of all
ordinal numbers (here ordinal numbers mean ordinals that are sets)
Lets simplify things and take the MK definition of sets and proper
classes
x is a set <-> Exist y ( x e y )
x is a proper class <-> ~ Exist y ( x e y )
so a proper class cannot be a member of any class.
Now lets take (1) what (1) is saying is that x would be a proper class
which mean that there cannot exist any class that has x as a member,
if and only if x is equinumerous to the class of all sets, right.
Now in NBG\MK it can be proved easily that the class D of all ordinals
that are sets would be a proper class, the proof doesn't need choice
at all, you can prove it yourself and see if choice is needed!
(hint:Burali-Forti).
So D is a proper class, which mean that D must be equinumerous to V
were V is the class of all sets.
so (1) -> (2) right!
Now lets examine the opposite direction,
what (2) is saying is that every proper class must be equinumerous to
D, right.
Now we know that V is a proper class! also the proof of that doesn't
require choice at all, try to prove it yourself! (hint:Russell
paradox)
so from (2) we have:
V is equinumerous to D
so (2) -> (1)
thus we have (2) <-> (1).
QED
The essence of the matter is, that the above doesn't require choice,
but it would lead to choice, actually it will lead to global choice,
in other words "global choice" would be a theorem, not the opposite.
Zuhair
zuhair
> In article <d2a856ff-ae3a-4dbd-9799-f3a5637d5...@e31g2000vbm.googlegroups.com>,
Yes, it is these later models that I am seeking,
Can you please specify at least one of them
or point to a reference about these models.
In a separate topic of this Usenet, I recently
spoke about Z+size limitation, I think this
would be an example of such a theory.
The importance of that issue is connected
to the proof that the cardinals that I defined
would be sets! they are superior to the ordinary
cardinals in that they do not require choice
like Von Neumann Cardinals
and in that they do not always require regularity
as Scott's cardinals.
Cardinality(x) is the equivalence class of all sets
having every member of their transitive closures
strictly subnumerous to x, under equivalence
relation "bijection".
Zuhair
zuhair
> > I am almost certain that there exist models with proper classes
> > strictly larger than the class of all ordinal numbers, and all
> > comparable.
>
> Using the consistency of ZFC + "there is an inaccessible" (and standard
> well known independence results) it's easy to show there are such
> models. As noted, by some tweaking we can remove the inessential
> inaccessible, bringing us back to ZFC and MK in terms of consistency
> strength.
Ok, then, that is important. Then this mean that adding the following
size limitation schema to Z set theory, bears no problem.
Axiom schema of size limitation:
If phi(y) is a formula in which at least y is free, and in which x is
not free, then all closures of
~ for all d
( d is ordinal ->
Exist x ( for all y ( y e x -> phi(y) ) and
d equinumerous to x ) )
-> Exist x for all y ( y e x <-> phi(y) )
are axioms.
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
zuhair
> zuhair <zaljo...@gmail.com> writes:
> > If we take ZF + inaccessibles , then these inaccessible sets are
> > "sets" i.e. are members of other classes, but yet they have the size
> > of a proper class in NBG\MK with no inaccessibles.
>
> This explanation or claim doesn't really make much sense.
Agreed.
Zuhair
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
David Libert
[Deletion]
> In a separate topic of this Usenet, I recently
> spoke about Z+size limitation, I think this
> would be an example of such a theory.
>
> The importance of that issue is connected
> to the proof that the cardinals that I defined
> would be sets! they are superior to the ordinary
> cardinals in that they do not require choice
> like Von Neumann Cardinals
> and in that they do not always require regularity
> as Scott's cardinals.
You and I each posted articles models of ZFC without
regularity, in which there are a proper class of sets
with all members of the traansitive closure being singletons.
My article was
[1] David Libert "A new definition of Cardinality"
sci.logic, sci.math Nov 23
http://groups.google.com/group/sci.math/msg/721cb8170033cf84
Yours was
[2] Zuhair "A new definition of Cardinality"
sci.logic, sci.math Nov 25
http://groups.google.com/group/sci.math/msg/68fa234768c92dcc
Your Z+size limitation is so similar to ZF insofar as I have unde4rstood
it, I think similar models would apply if the basic theory is ok.
So these seem to show me that regularity is required.
> Cardinality(x) is the equivalence class of all sets
> having every member of their transitive closures
> strictly subnumerous to x, under equivalence
> relation "bijection".
>
> Zuhair
This is the same as your first definition, which was given in the
thread of [1] & [2].
I discussed this and also your second definition in
[3] David Libert "The magic of Hereditarily Hereditary Cardinals"
sci.logic, sci.math Nov 29
http://groups.google.com/group/sci.math/msg/1b40b261aeff6e96
In [3] I defined signatures for such definitions. The correct signature
for your definition above is = < .
In [3] I incorrectly wrote this as <= < , and noted that correction
in a followup to [3].
So regarding = < definition as you have above, in [3] I argued that
for x = A amorphous, Cardinality(x) = {} .
I still think that was correct.
In [3] I went on to note Cardinality({}) = {} .
I now think this was wrong. I think Cardinality({}) = {{}} .
So the first overlap I noted of A and {} getting same Cardinality
is retracted.
But in [3] I also noted we could also get a different Cohen model
with A1, A2 non-isomorphic amorphous sets, and so have
Cardinality(A1) = {}
Cardinality(A2) = {}
still making a problem for this definition.
I still think that last point is correct, making a problem for this
definition.
--
David Libert ah...@FreeNet.Carleton.CA
zuhair
> zuhair (zaljo...@gmail.com) writes:
>
Let me clarify matters:
This definition of Cardinality which is the same one I posted in topic
"a new definition of Cardinality", works with Regularity OR Choice. So
if we don't have Regularity but we have Choice then the definition
work! and also if we don't have Choice but we have Regularity also the
definition work, however the definition fail if we neither have
Regularity nor Choice, i.e. it fails for example in "ZF minus
Regularity" and it is this later condition that you are speaking of,
which is true.
So in ZF, the definition work
in ZFC minus Regularity, the definition work
in ZFC, the definition work
in ZF minus Regularity, the definition fail.
I am not sure if Scott Cardinals have the same property, they might!
but I got they idea that they don't work outside Regularity , anyhow ,
but this definition is simpler anyway.
I hope I clarified this point.
Zuhair
Herman Rubin
George Greene <gre...@email.unc.edu> wrote:
On this, you are correct. Ordinal numbers are ordinals
which are sets.
Aatu Koskensilta
> I am not sure if Scott Cardinals have the same property, they might!
> but I got they idea that they don't work outside Regularity , anyhow ,
> but this definition is simpler anyway.
In absence of regularity we must amend the usual definition slightly:
card(A) = the set of all well-founded sets the same size
as A of least possible rank
Unless every set is the same size as a well-founded set it may happen
that card(A) = card(B) = {} even though A and B are not equipollent. I
don't know of any definition of cardinals that works in absence of both
choice and regularity without any additional axioms.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
Herman Rubin
George Greene <gre...@email.unc.edu> wrote:
>On Dec 3, 12:16=A0pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>> hru...@odds.stat.purdue.edu (Herman Rubin) writes:
>> > I am almost certain that there exist models with proper classes
>> > strictly larger than the class of all ordinal numbers, and all
>> > comparable.
>So the class of all ordinal numbers, in this context, is proper, but
>is comparably smaller than larger proper classes?
Possibly, and possibly not. If V = L, as in Godel's model
which proves the consistency of the generalized continuum
hypothesis, the answer is not; the entire universe has the
same "cardinality" as the class of all ordinal numbers.
>And in this context, there can be ordinal proper classes as well?
In NBG, there must be.
>And the proper class of all set ordinals is one of them?
Yes.
>And this proper-class ordinal also has proper-class successors that
>are also ordinals?
Yes.
Do these proper-class ordinals eventually get
>big enough to be equipollent to the largest proper classes?
Every constructible ordinal class has the same cardinality as
the class of ordinal numbers. One cannot talk about the class
of these in NBG, but might be able to in Morse-Kelley. The
largest proper class is the universe in any case, and the
same question remains.
zuhair
> zuhair <zaljo...@gmail.com> writes:
> > I am not sure if Scott Cardinals have the same property, they might!
> > but I got they idea that they don't work outside Regularity , anyhow ,
> > but this definition is simpler anyway.
>
> In absence of regularity we must amend the usual definition slightly:
>
> card(A) = the set of all well-founded sets the same size
> as A of least possible rank
>
> Unless every set is the same size as a well-founded set it may happen
> that card(A) = card(B) = {} even though A and B are not equipollent. I
> don't know of any definition of cardinals that works in absence of both
> choice and regularity without any additional axioms.
Well the problem against that is that it is not a theorem of ZF that
every set is the same size of a well founded set. So this last
definition of yours is not enough by itself for cardinals to exist for
every set. In absence of Regularity The class V of all well founded
sets may not be able to supply cardinals for *all* sets in the
universe.
However, that was not my point. The cardinal you've mentioned above is
the Scott cardinal with slight modification, however this cardinal
doesn't work out of Regularity even if choice is assumed (I guess),
since a Von Neumann cardinal is not necessarily of the least possible
rank! What I am trying to say
is that the cardinal you've mentioned seem, to me at least, to be not
definable
in ZFC minus Regularity for every set.
While the Cardinal that I've defined, assuming size limitation axiom
schema that I've mentioned in the second of my replies to Herman
Rubin, this cardinal would work
i.e we can define cardinality of every set even if Regularity is not
assumed but provided that choice is assumed, also it can work in
absence of choice but then Regularity should be assumed.
IF add the following axiom to Z+size limitation
For all x Exist y ( y is well founded & x equinumerous to y )
which is Jean Coret axiom,
And if a modification of my definition of cardinality that is similar
to the one you've done with Scott's cardinals is done, then we can
have the following theorem
of Z+size limitation+Coret.
for every set x , cardinality(x) is a set.
even in absence of Regularity. While modified Scott cardinals cannot
achieve that.
The main draw back of the cardinals that I've defined is that the set-
hood of them
is difficult if not impossible to prove or refute in ZF, the set-hood
of those cardinals seems to be independent of ZF, that's why I
resorted to Z+ size limitation, to settle this matter.
Zuhair
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
zuhair
> In article <09620f7e-a9c0-442a-a476-ee5114e4c...@p19g2000vbq.googlegroups.com>,
> George Greene <gree...@email.unc.edu> wrote:
>
> >On Dec 3, 12:16=A0pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> >> hru...@odds.stat.purdue.edu (Herman Rubin) writes:
> >> > I am almost certain that there exist models with proper classes
> >> > strictly larger than the class of all ordinal numbers, and all
> >> > comparable.
> >So the class of all ordinal numbers, in this context, is proper, but
> >is comparably smaller than larger proper classes?
>
> Possibly, and possibly not. If V = L, as in Godel's model
> which proves the consistency of the generalized continuum
> hypothesis, the answer is not; the entire universe has the
> same "cardinality" as the class of all ordinal numbers.
>
> >And in this context, there can be ordinal proper classes as well?
>
> In NBG, there must be.
>
> >And the proper class of all set ordinals is one of them?
>
> Yes.
>
> >And this proper-class ordinal also has proper-class successors that
> >are also ordinals?
>
> Yes.
How is that possible? a proper class is a class that is not a member
of any class, on the other hand the definition of ordinal successor of
the ordinal x for example
is x U {x}, i.e. if y is the ordinal successor of the ordinal x, then
x must be a member of y, now if x is a proper class ordinal, then it
cannot have a successor, that is of course in NBG\MK.
However in Ackermann's class theory, every ordinal weather it is a set
or a proper class, must have a successor, since Ackermann's theory
becomes inconsistent if there is no ordinal successor for an ordinal.
So I think the above reply made by Herman Rubin is mistaken.
Aatu Koskensilta
> Well the problem against that is that it is not a theorem of ZF that
> every set is the same size of a well founded set.
The problem against what? I was replying to your comment that you "got
they idea that they don't work outside Regularity" about Scott
cardinals, and noted that this is indeed so, unless we adopt some
additional axiom.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
zuhair
> In article <09620f7e-a9c0-442a-a476-ee5114e4c...@p19g2000vbq.googlegroups.com>,
>
> >On Dec 3, 12:16=A0pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> >> hru...@odds.stat.purdue.edu (Herman Rubin) writes:
> >> > I am almost certain that there exist models with proper classes
> >> > strictly larger than the class of all ordinal numbers, and all
> >> > comparable.
> >So the class of all ordinal numbers, in this context, is proper, but
> >is comparably smaller than larger proper classes?
>
> Possibly, and possibly not. If V = L, as in Godel's model
> which proves the consistency of the generalized continuum
> hypothesis, the answer is not; the entire universe has the
> same "cardinality" as the class of all ordinal numbers.
>
> >And in this context, there can be ordinal proper classes as well?
>
> In NBG, there must be.
>
> >And the proper class of all set ordinals is one of them?
>
> Yes.
>
> >And this proper-class ordinal also has proper-class successors that
> >are also ordinals?
>
> Yes.
How is that possible? a proper class is a class that is not a member
of any class, on the other hand the definition of ordinal successor of
the ordinal x for example
is x U {x}, i.e. if y is the ordinal successor of the ordinal x, then
x must be a member of y, now if x is a proper class ordinal, then it
cannot have a successor, otherwise x would be a proper class that is a
member of a class, which is contradictive.That is the case of course
in NBG\MK.
However in Ackermann's class theory, every ordinal weather it is a set
or a proper class, must have a successor, since Ackermann's theory
becomes inconsistent if there is no ordinal successor for an ordinal.
But in Ackermann's proper classes are not defined as not being members
of other classes, they are defined as not being sets.
So I think the above reply made by Herman Rubin is mistaken, if we are
working in NBG\MK.
Aatu Koskensilta
> How is that possible? a proper class is a class that is not a member
> of any class, on the other hand the definition of ordinal successor of
> the ordinal x for example is x U {x}, i.e. if y is the ordinal
> successor of the ordinal x, then x must be a member of y, now if x is
> a proper class ordinal, then it cannot have a successor, that is of
> course in NBG\MK.
Presumably Herman had in mind class sized well-orderings of order type
greater than On, the class of ordinals. In NBG and MK we can talk of On
+ 1, On * 2, and so on, understood in this sense.
zuhair
> zuhair <zaljo...@gmail.com> writes:
> > Well the problem against that is that it is not a theorem of ZF that
> > every set is the same size of a well founded set.
>
> The problem against what? I was replying to your comment that you "got
> they idea that they don't work outside Regularity" about Scott
> cardinals, and noted that this is indeed so, unless we adopt some
> additional axiom.
Yes, I agree with your agreement with me about "Regularity", but I
want to make sure of one point though, does Scott cardinals work
outside Regularity if we assume choice. my guess is that they don't,
because a Von Neumann cardinal is not necessarily of the least rank,
does that make sense?
However the problem that I was saying is regarding having well founded
cardinals for all sets in the universe, I was speaking somewhat of a
more general matter, were even if we amend the definition of Scott
cardinals as you did, yet this amendment that you've done will not be
enough to make those well founded Scott cardinals defined for every
set of the universe in the desirable manner that we want from
cardinals, unless we assume Jean Coret's assumption that every set is
equinumerous to some well founded set, then we can have empty
cardinals for non equinumerous sets, thus departing from the desirable
quality we want cardinals to fulfill.
To speak more precisely your amendment will actually define cardinals
for every set, and those cardinal will be sets always!, but the
problem in having them non empty.
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
Aatu Koskensilta
> Yes, I agree with your agreement with me about "Regularity", but I
> want to make sure of one point though, does Scott cardinals work
> outside Regularity if we assume choice. my guess is that they don't,
> because a Von Neumann cardinal is not necessarily of the least rank,
> does that make sense?
The rank of a von Neumann ordinal alpha is alpha. I'm not sure exactly
what you're asking, but Scott cardinals certainly work when we have
choice, since given choice every set is the same size as an ordinal, and
all ordinals are well-founded. If we have choice, it is more natural to
define the cardinal of a set to be the initial ordinal equipollent to
it.
> To speak more precisely your amendment will actually define cardinals
> for every set, and those cardinal will be sets always!, but the
> problem in having them non empty.
Yes, I explicitly noted this in my post: unless we assume (or prove)
that every set is the same size as a well-founded set we can't use
Scott's trick to define cardinals.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
zuhair
> zuhair <zaljo...@gmail.com> writes:
> > Yes, I agree with your agreement with me about "Regularity", but I
> > want to make sure of one point though, does Scott cardinals work
> > outside Regularity if we assume choice. my guess is that they don't,
> > because a Von Neumann cardinal is not necessarily of the least rank,
> > does that make sense?
>
> The rank of a von Neumann ordinal alpha is alpha. I'm not sure exactly
> what you're asking, but Scott cardinals certainly work when we have
> choice, since given choice every set is the same size as an ordinal, and
> all ordinals are well-founded.
Oh yes, that is correct. so the cardinals you've mentioned do really
work
with Choice or Regularity, i.e. we need one of them at least.
Also the cardinals that I've defined do that, but the difference is
that
the set-hood of the cardinals that I've defined might be independent
of ZF, requiring more assumptions like size limitation, etc..., on the
other hand they might be proved in ZF to be sets? the matter is not
settled anyhow.
Thanks Aatu.
If we have choice, it is more natural to
> define the cardinal of a set to be the initial ordinal equipollent to
> it.
>
> > To speak more precisely your amendment will actually define cardinals
> > for every set, and those cardinal will be sets always!, but the
> > problem in having them non empty.
>
> Yes, I explicitly noted this in my post: unless we assume (or prove)
> that every set is the same size as a well-founded set we can't use
> Scott's trick to define cardinals.
outside Regularity and Choice, yes, that is true.
Zuhair
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
George Greene
> However NBG\MK do define proper classes as being not members of other
> classes, as you mentioned ,that is correct.
That is Not equivialent to defining proper classes as being
equinumerous with the class of all sets, or with the class of all
ordinals.
Under this definition,the class of all ordinals has to be proper
(Burali-Forti),
but it is not clear why it has to be equinumerous with the class of
all sets,
even though that too must also be proper given regualarity.
Aatu Koskensilta
> Under this definition,the class of all ordinals has to be proper
> (Burali-Forti), but it is not clear why it has to be equinumerous with
> the class of all sets, even though that too must also be proper given
> regualarity.
From the principle that a proper class is not a member of any class it
indeed doesn't follow that a proper class must be the same size as the
universe. What's the relevance?
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
George Greene
> Ok, let me try to simplify matters.
I am trying, too; I think mine will be simpler.
> Now in NBG\MK it can be proved easily that the class D of all ordinals
> that are sets would be a proper class, the proof doesn't need choice
> at all, you can prove it yourself and see if choice is needed!
> (hint:Burali-Forti).
Exactly. Not only do you not need choice, you do not even need a
definition of the set/proper-class distinction at all. The class of
all ordinal sets has
to be proper; it cannot be a set, because if it were, it would be an
ordinal
set and therefore be a member of itself -- which, even in the ABSENCE
of
regularity, would be the Burali-Forti paradox -- ordinals have to be
well-
founded EVEN when other sets don't.
> So D is a proper class,
Right. But D is in some sense the SMALLEST POSSIBLE proper class.
> Now we know that V is a proper class! also the proof of that doesn't
> require choice at all, try to prove it yourself! (hint:Russell
> paradox)
Well, yes, the class of ALL sets ALSO has to be proper.
But again, this is THE LARGEST possible proper class -- you
can't have any classes bigger than V, so if every class has to
be as big as V in order to be proper, as was said by your
> (1) x is a proper class iff x is equinumerous with the class of all sets" ,
then you are saying that all proper classes must be as LARGE as
possible.
This is NOT *intuitively* equivalent to saying that they all must be
as SMALL as possible, which is what was said by your
> (2)x is a proper class iff x is equinumerous with the class of all
> ordinal numbers (here ordinal numbers mean ordinals that are sets)
> so (2) -> (1)
>
> thus we have (2) <-> (1).
Indeed we do, yet it still seems paradoxical; the tension
is resolved by seeing that if you only have ONE size of proper class,
then obviously, that is equivalent to any other possible way of having
ONE size of proper class. Requiring that they all be small or that
they
all be big -- raising the floor to the ceiling or lowering the
ceiling to the floor --
winds up not mattering; you just have 1 size of proper classes
(and a lot of smaller sizes of sets) EITHER WAY.
The counter-intuitivity here is coming from using iff as opposed to if
in (1) and (2). What is intuitively necessary is that if you are not
at least as
big as the class of all ordinals, then you are not big enough to be
proper.
The back-arrow of (1) is NOT intuitive -- that is a LIMITATION to a
context of special models
where even the "big" proper classes -- like the class of all sets --
are ALSO small.
Conversely, in (2), what is intuitively necessary is that if you are
as big as the
class of all sets, then of course you are too big to be a set -- the
back-arrow,
that even if you are NOT (originally, necessarily) that big, but are
nevertheless proper,
then you must also be BIG & proper (after all), is, again, a
limitation
to a special case.
You point out that even prior to (1) and (2), we had, as theorems,
without choice or without even caring about the definition of the
set-class distinction, just purely as proven consequences
of the paradoxes,
(1P) the class of all ordinal sets is proper, and
(2P) the class of all sets is proper.
But it does not automatically follow that all proper classes are
always
as small as they are in (1) or as large as they are in (2).
Ross A. Finlayson
Burali-Forti says "ordinals" is irregular.
"The" physical universe of physical objects identified as mathematical
objects sees more than any finite complexity in relations, i.e. there
are infinitely many, compounded in autocorrelating that maps to
powerset of those same objects, identity indicates a simple physical
realization of counterexample to non-identity of physical universe to
itself and its self-same reconsidered powerset of things.
"Classes" as in "ZFC with Classes" where NBG is a schema-bound ZFC
(implicitly infinitely axiomatized) are a guilty subterfuge to
conscientious set-theoretic logicians acknowledging existence.
Ordinals are ubiquitous in identification to sets under ordinal N with
2^N ordinals.
Regards,
Ross F.
zuhair
Let us agree on two important issues here.
(1) The class of all ordinals that are sets 'D' is the smallest
possible proper class.
(2) The class of all sets 'V' is the Largest possible proper class.
Now when anybody say that *ALL* proper classes are equal in size to
the smallest possible proper class, then simply what he is saying is
that *ALL* proper classes are of the same size.
Now when another man say that *ALL* proper classes are equal in size
to the largest possible proper class, then simply what he is saying is
that *ALL* proper classes are of the same size.
So it doesn't really matter how you say it, you want to say all proper
classes are equal to the smallest possible proper class, or you want
to say all proper classes are equal to the Largest possible proper
class, in both versions all what you are saying is "ALL proper classes
are of the same size", so it doesn't really matter, the direction of
saying it is irrelevant, at the end you are saying the same thing but
in different ways. All what are you trying to say is Size D = Size V,
you can say D is as big as V, or if you want to say V is as small as
D, it doesn't matter, it is just two ways of saying the same thing
which is V is equal in size to D.
Of course in other inner models of MK, one can have a different axiom
of size limitation to the effect that size D not equal size V, i.e
size V strictly supernumerous to Size D, yes one can do that, there
are various models.
But the following if you add any one of the following size limitation
axioms to MK minus size limitation, then you will get the same result
(1) x e V <-> x strictly subnumerous to V
(2) ( ~ x e V & ~ y e V ) <-> x equinumerous to y
(3) ~ x e V <-> x equinumerous to D
(4) ~ x e V <-> x equinumerous to V
Zuhair
Daryl McCullough
>> Now in NBG\MK it can be proved easily that the class D of all ordinals
>> that are sets would be a proper class
[stuff deleted]
>Right. But D is in some sense the SMALLEST POSSIBLE proper class.
What's the argument for that? Is there a theorem in NBG to the effect
that, if C is a proper class than C is larger than D (in cardinality)?
--
Daryl McCullough
Ithaca, NY
zuhair
If "larger" mean 'bigger or equal', then must be. There is no way you
can have a proper class that strictly subnumerous to D.
Zuhair
Aatu Koskensilta
> What's the argument for that? Is there a theorem in NBG to the effect
> that, if C is a proper class than C is larger than D (in cardinality)?
With choice, yes. As an historical aside, Cantor thought he had found a
proof of the well-ordering theorem in the observation that every
"inconsistent multiplicity" must contain a submultiplicity equivalent to
the class of ordinals. For those of a more serious turn of mind, this is
evidence that Cantor "implicitly" accepted a form of
replacement. Incidentally, Jourdain produced a proof of the
well-ordering theorem that Zermelo found very unsatisfactory. Strangely,
this proof was exactly what we find in textbooks today: just pair the
elements of A with ordinals until you ran out of elements; since A is a
set you won't exhaust all the ordinals, and the process will terminate
at some set-sized ordinal alpha. Zermelo's less waffling proof is very
nice but, for some reason, not usually found in textbooks.
Aatu Koskensilta
You are assuming that "not bigger or equal" implies "strictly
subnumerous". This is an unwarranted assumption in absence of choice.
zuhair
> zuhair <zaljo...@gmail.com> writes:
> > On Dec 5, 7:03 am, stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
>
> >> What's the argument for that? Is there a theorem in NBG to the effect
> >> that, if C is a proper class than C is larger than D (in
> >> cardinality)?
>
> > If "larger" mean 'bigger or equal', then must be. There is no way you
> > can have a proper class that strictly subnumerous to D.
>
> You are assuming that "not bigger or equal" implies "strictly
> subnumerous". This is an unwarranted assumption in absence of choice.
Hmmm..., you are right.
What I wanted to say is that no proper class can be strictly
subnumerous to the class D of all ordinals that are sets . Of course
it can be incomparable with D or equinumerous to D or strictly
supernumerous to D, but it cannot be strictly subnumerous to D.
Zuhair
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
George Greene
Not strictly larger.
Larger-than-or-equal-to.
There is a theorem that if a proper class is injectible into you, then
you
are proper too. You are of course larger-than-or-equipollent-to any
class
that is injectible into you.
Obviously, mere subclasshood is not going to be the relevant sense of
smaller;
you can take 0 out of any proper-class-ordinal and get something that
is still
a proper class but is a proper subclass of the class of all ordinals.
But all the general rules are sort of out the window here since
we don't even have a fixed definition of the set-class distinction.
We are talking in general terms.
There was supposed to be 1 definition of the set-class distinction
whereby
being proper was equivalent to not being a member of any other class.
We are obviously NOT using that definition since that would outlaw
proper-class
ordinals (every ordinal is a member of its successor).
The single most interesting aspect of this subthread is that
the proper-class-hood of these two classes is INdependent of the
definition of the set-class distinction.
zuhair
I should say something about that, the correct statement is:
Every ordinal is a member of its successor if and only if there exist
a successor.
Let us assume Regularity.
we can define ordinal in the following manner:
An ordinal is a transtive class of transitive sets.
Now lets take for example MK class theory with the following size
limitation axiom.
For all x ( x e V <-> x strictly subnumerous to V )
were V is the class of all sets,
were a set is defined in MK as a class that is a member of some class,
while a proper class is defined as a class that is not a member of any
class.
Now according to MK with the above version of size limitation, the
class D which is the class of all ordinals that are sets, would
definitely be an ordinal, and as we know from Burali-Forti paradox D
cannot be a set, so it is a proper class, so D is an proper class
ordinal(i.e. an ordinal that is a proper class), this is to
differentiate it from "a set ordinal (i.e. an ordinal that is a set)
this is very easy to prove actually, but D has no successor according
to this theory!
in MK a proper class ordinal cannot have a successor, for a simple
reason, the successor of an ordinal x must contain x as a member in
it, thus if
x is a proper class ordinal, then obviousely it cannot have a
successor, since otherwise this would mean that x is a proper class
that is contained in a class (its successor), and that contradicts the
very definition of proper classes as not being members of any class.
Apparently you are not working under the set-proper class distinction,
that's why you said there can exist proper class ordinals that are
members of their successor, that reminds me of course of Ackermann's
class theory, were this is the case, and superisingly Ackermann's
class theory become inconsistent if we say that there exist an ordinal
that has no successor, but this case is applicable to Ackermann's
class theory, and not to NBG\MK.
However even if you remove this set-proper class distinction I refered
to upwards, still you cannot prove that every ordinal has a successor,
this will largely depend on the inner model of the theory you are
working in.
Zuhair
Aatu Koskensilta
> But all the general rules are sort of out the window here since we
> don't even have a fixed definition of the set-class distinction. We
> are talking in general terms. There was supposed to be 1 definition
> of the set-class distinction whereby being proper was equivalent to
> not being a member of any other class. We are obviously NOT using
> that definition since that would outlaw proper-class ordinals (every
> ordinal is a member of its successor).
You have been misled by Herman Rubin's comments. In all the theories
considered in these threads a class is proper iff it is not an element
of any class. In particular, there are no proper-class ordinals in your
sense, although there are well-orderings of the universe of order-type
greater than that of the class of ordinals.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
zuhair
> George Greene <gree...@email.unc.edu> writes:
> > But all the general rules are sort of out the window here since we
> > don't even have a fixed definition of the set-class distinction. We
> > are talking in general terms. There was supposed to be 1 definition
> > of the set-class distinction whereby being proper was equivalent to
> > not being a member of any other class. We are obviously NOT using
> > that definition since that would outlaw proper-class ordinals (every
> > ordinal is a member of its successor).
>
> You have been misled by Herman Rubin's comments. In all the theories
> considered in these threads a class is proper iff it is not an element
> of any class. In particular, there are no proper-class ordinals in your
> sense, although there are well-orderings of the universe of order-type
> greater than that of the class of ordinals.
This might be confusing Aatu since the phrase "in your sense" is
vague.
The class of all ordinals that are sets is definitely an ordinal, and
it is an ordinal that is a proper class, and in MK\NBG it doesn't have
a successor!
However it seems that George is thinking that the term
"ordinal" implies that their should be a successor, i.e. I think
George think if we say for example x is ordinal (weather it is 'set
ordinal' or 'proper class ordinal')
then there must exist a class y such that y is the successor of x,
i.e.
y=xUnion{x}, I think this is the source of confusion in George's
notes, so accordingly George seem to think that there is a class of
proper class ordinals, one succeeding the other like the case with
sets.
In most theories set ordinals do have successor, but a proper class
ordinal might or might not have a successor, in Ackermann's all
ordinals must have successors weather they are proper classes or sets,
while in NBG\MK all ordinals that are sets have successor, but there
exist only ONE proper class ordinal and that is
the class of all set ordinals, and it doesn't have any successor, and
yet it is ordinal!
I already clarified that there is nothing in the definition of
"ordinal" that implies that
there must exist a successor for each ordinal. Assuming Regularity an
ordinal can be defined as a transitive class of transitive sets,
that's all, nothing in this defintion per se entails that every
ordinal must have a successor, the existence of successors for
ordinals weather they are sets or proper classes depends on the theory
one is working with.
Zuhair
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechan kann, darüber muss man schweigen"
George Greene
> considered in these threads a class is proper iff it is not an element
> of any class. In particular, there are no proper-class ordinals in your
> sense, although there are well-orderings of the universe of order-type
> greater than that of the class of ordinals.
Thank you kindly for the clarification.
Zuhair was wanting to define a proper-class ordinal as
any transitive class of transitive sets. That would make the
class of ordinals a proper-class ordinal, but it is hard to see how
anything smaller than that would ever get to be a proper-class
ordinal,
since "smaller-than" FOR ordinals entails membership.
George Greene
> in Ackermann's all
> ordinals must have successors weather they are proper classes or sets,
So in that case, you are using a definition of the set/class
distinction that
is something OTHER than "not being a member of another class" --
unless
Ackermann's is using a different definition of successor.
> while in NBG\MK all ordinals that are sets have successor, but there
> exist only ONE proper class ordinal and that is
> the class of all set ordinals, and it doesn't have any successor, and
> yet it is ordinal!
Thank you for the clarification.
If it is the only one then that is a lot easier to tolerate.
We would prefer
talking about it as a unique and special thing
to
talking about the existence of proper-class-ordinalS, PLURAL
(sincethere is only one).
zuhair
I Know you are replying to Aato, but I should clarify some matters.
Let us be precise here so that we don't have any misunderstandings.
First the defintion of "ordinal" any "ordinal" not only an ordinal
that is a proper class, any ordinal, OK, is assuming Regularity is:
An ordinal is a transtive *class* of transitive sets.
This is proved to be equivalent to the standard definition of
*ordinal* if as I said we assume Regularity.
Without assuming Regularity this definition would be modified to
An ordinal is a transtive *class* of transitive sets, in which every
non empty subclass of it must contain a member that is disjoint of it.
Now this is also proved to be equivalent to the standard definition of
ordinal.
Now if that class was a set, then we have a "set ordinal"
If that class was a proper class, then we have a "proper class
ordinal".
The second point, You said: That would make the
class of ordinals a proper-class ordinal
That is not the precise statement, since we don't have a class of all
ordinals, otherwise this will lead to Burali-Forti paradox.
What we have is:
The class of all set ordinals. or more clearily
The class of all ordinals that are sets.
And this class is a proper class, and it is an ordinal. and in NBG\MK
it is the ONLY proper class ordinal, that's why I used the symbole D
to symbolize it, although the standard symbol for it is "ORD".
YOU said:
but it is hard to see how
anything smaller than that would ever get to be a proper-class
ordinal,
since "smaller-than" FOR ordinals entails membership
I think what you mean by smaller is "strictly subnumerous", if so then
you are correct, every ordinal that is strictly subnumerous to the
proper class ordinal ORD
is a set, there is no debate about that, and you are right when you
said
"smaller than" for ordinals entails membership, that is correct of
course.
Just to clarify points.
Zuhair
zuhair
Yes, in NBG\MK and I think also related models of ZF like ZFU and the
alike,
there would exist ONLY ONE proper class ordinal, and that is the class
of all set ordinals, which is unique, and it is denoted by the symbol
"ORD" although
I like to use the symbol D for short.
However in Ackermann's class theory, we do have multiple proper class
ordinals, each one succceed the other, exactly as it with sets, but
the definition of proper classes in Ackermann's is not the same as
that in NBG\MK\ZF related models.
In Ackermann's the definition of a proper class is: a class that is
not a set.
and a set is a primitive concept in Ackermann's class theory.
However we mainly use NBG\MK\ZF related models, and in those theories
we do only have ONE proper class ordinal.
NOTE ( I am not sure if there are ZF related models were we have many
proper class ordinals like the case with Ackermann's, but I greatly
doubt that).
Zuhair
Herman Rubin
George Greene <gre...@email.unc.edu> wrote:
Any subclass of an ordinal is order-isomorphic to an
ordinal. Therefore, and proper class which is a subclass
of the class of all ordinal numbers (ordinals which are
sets) must be order-isomorphic to the class of all
ordinals, and hence equinumerous.
zuhair
> In article <f8f5b867-fddd-4c83-bd15-857e3f7be...@k17g2000yqh.googlegroups.com>,
> George Greene <gree...@email.unc.edu> wrote:
>
> >On Dec 7, 1:28=A0am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> >> You have been misled by Herman Rubin's comments. In all the theories
> >> considered in these threads a class is proper iff it is not an element
> >> of any class. In particular, there are no proper-class ordinals in your
> >> sense, although there are well-orderings of the universe of order-type
> >> greater than that of the class of ordinals.
>
> <Thank you kindly for the clarification.
> <Zuhair was wanting to define a proper-class ordinal as
> <any transitive class of transitive sets. That would make the
> <class of ordinals a proper-class ordinal, but it is hard to see how
> <anything smaller than that would ever get to be a proper-class
> <ordinal,
> <since "smaller-than" FOR ordinals entails membership.
>
> Any subclass of an ordinal is order-isomorphic to an
> ordinal. Therefore, and proper class which is a subclass
> of the class of all ordinal numbers (ordinals which are
> sets) must be order-isomorphic to the class of all
> ordinals, and hence equinumerous.
Yes, but I think that was not what George Greene had in his mind (I
guess)
he was speaking of a *proper class* that is smaller (I thought smaller
means strictly subnumerous, but perhaps he meant ORDINALLY smaller
than) than
the class ORD of all ordinals that are sets, that is also an ordinal,
but this thing
do not exist, you can have a *proper* subclass of ORD that is well
orderable of course, and also that is equinumerous to the proper class
ORD, but still this would NOT be an ordinal, because it will not be
transitive! an obvious example is the class of all non empty ordinals,
it is a proper class, it is well orderable, it is equinumerous to to
ORD, yet still it is not an ordinal!
We cannot have an ordinal smaller than ORD and at the same time be a
proper class, this is just impossible, because if we suppose that x
for example is Ordinally smaller than ORD, then x is a member of ORD,
because this is the definition of Odinal smaller than, now if x is a
proper class, then it cannot be a member of any class, A
contradiction. Thus there cannot exist an ordinal that is Ordinally
smaller than ORD and is a proper class, also there cannot exist an
ordinal that is strictly subnumerous to ORD and is a proper class.
Zuhair