The size of proper classes?
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From: zuhair on 2 Dec 2009 16:14 On Dec 2, 1:35 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > In article <b1d5c9ac-d3ca-42a2-9d9c-e2decdf0c...(a)m35g2000vbi.googlegroups..com>, > > zuhair <zaljo...(a)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. > > >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...(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 Let me re ask my question using more precise terminology: 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
From: zuhair on 2 Dec 2009 16:17 On Dec 2, 1:35 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > In article <b1d5c9ac-d3ca-42a2-9d9c-e2decdf0c...(a)m35g2000vbi.googlegroups..com>, > > zuhair <zaljo...(a)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. > > >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...(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 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
From: zuhair on 2 Dec 2009 16:18 On Dec 2, 1:35 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > In article <b1d5c9ac-d3ca-42a2-9d9c-e2decdf0c...(a)m35g2000vbi.googlegroups..com>, > > zuhair <zaljo...(a)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. > > >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...(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 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
From: George Greene on 2 Dec 2009 20:40 On Dec 2, 7:53 am, zuhair <zaljo...(a)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. 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. 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. THAT IS the set/class distinction in these contexts: "subnumerous to" vs. "equinumerous to" the class of all sets. 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.
From: George Greene on 2 Dec 2009 20:58
On Dec 2, 1:35 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > The strongest class form of the Axiom of > 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? |