The size of proper classes?
in [Math]
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From: zuhair on 9 Dec 2009 19:02 On Dec 9, 12:28 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > In article <f8f5b867-fddd-4c83-bd15-857e3f7be...(a)k17g2000yqh.googlegroups..com>, > George Greene <gree...(a)email.unc.edu> wrote: > > >On Dec 7, 1:28=A0am, Aatu Koskensilta <aatu.koskensi...(a)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 > -- > 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 |