The size of proper classes?
in [Math]
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From: zuhair on 2 Dec 2009 22:37 On Dec 2, 9:03 pm, George Greene <gree...(a)email.unc.edu> wrote: > 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? > > 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. 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
From: zuhair on 2 Dec 2009 22:38 On Dec 2, 10:12 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Dec 2, 9:16 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > On Dec 2, 8:40 pm, George Greene <gree...(a)email.unc.edu> wrote: > > 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. Oh, sorry then, well I hope it clarifies issues to others. Zuhair
From: George Greene on 2 Dec 2009 22:42 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. 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...(a)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?
From: zuhair on 2 Dec 2009 22:45 On Dec 2, 9:03 pm, George Greene <gree...(a)email.unc.edu> wrote: > 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? > > 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. 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
From: zuhair on 2 Dec 2009 22:50
On Dec 2, 10:42 pm, George Greene <gree...(a)email.unc.edu> wrote: > 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? > 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...(a)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? |