The size of proper classes?
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From: Herman Rubin on 4 Dec 2009 12:37 In article <db48db02-289d-4148-952b-d564da55ee17(a)o31g2000vbi.googlegroups.com>, George Greene <greeneg(a)email.unc.edu> wrote: >On Dec 3, 12:00=A0pm, 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? On this, you are correct. Ordinal numbers are ordinals which are sets. -- 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 hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: Aatu Koskensilta on 4 Dec 2009 12:43 zuhair <zaljohar(a)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. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Herman Rubin on 4 Dec 2009 12:45 In article <09620f7e-a9c0-442a-a476-ee5114e4cab6(a)p19g2000vbq.googlegroups.com>, George Greene <greeneg(a)email.unc.edu> wrote: >On Dec 3, 12:16=A0pm, 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? 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. -- 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 hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: zuhair on 4 Dec 2009 15:10 On Dec 4, 12:43 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > zuhair <zaljo...(a)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...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: zuhair on 4 Dec 2009 15:18
On Dec 4, 12:45 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > In article <09620f7e-a9c0-442a-a476-ee5114e4c...(a)p19g2000vbq.googlegroups..com>, > George Greene <gree...(a)email.unc.edu> wrote: > > >On Dec 3, 12:16=A0pm, 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? > > 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. > > 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. > -- > 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 |