The size of proper classes?
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From: George Greene on 3 Dec 2009 15:37 On Dec 3, 12:00 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > 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?
From: George Greene on 3 Dec 2009 15:44 On Dec 3, 12:16 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > hru...(a)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?
From: Zdislav V. Kovarik on 3 Dec 2009 18:54 On Wed, 2 Dec 2009, Herman Rubin wrote: > In article <b1d5c9ac-d3ca-42a2-9d9c-e2decdf0c3b2(a)m35g2000vbi.googlegroups.com>, > zuhair <zaljohar(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? > [...] 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).
From: George Greene on 3 Dec 2009 19:44 > zuhair <zaljo...(a)gmail.com> writes: > > 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!!
From: zuhair on 3 Dec 2009 20:04
On Dec 3, 7:44 pm, George Greene <gree...(a)email.unc.edu> wrote: > > zuhair <zaljo...(a)gmail.com> writes: > > > 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!! 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 |