The size of proper classes?
in [Math]
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From: zuhair on 4 Dec 2009 16:09 On Dec 4, 3:40 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > zuhair <zaljo...(a)gmail.com> writes: > > Yes, I agree with your agreement with me about "Regularity", but I > > want to make sure of one point though, does Scott cardinals work > > outside Regularity if we assume choice. my guess is that they don't, > > because a Von Neumann cardinal is not necessarily of the least rank, > > does that make sense? > > The rank of a von Neumann ordinal alpha is alpha. I'm not sure exactly > what you're asking, but Scott cardinals certainly work when we have > choice, since given choice every set is the same size as an ordinal, and > all ordinals are well-founded. Oh yes, that is correct. so the cardinals you've mentioned do really work with Choice or Regularity, i.e. we need one of them at least. Also the cardinals that I've defined do that, but the difference is that the set-hood of the cardinals that I've defined might be independent of ZF, requiring more assumptions like size limitation, etc..., on the other hand they might be proved in ZF to be sets? the matter is not settled anyhow. Thanks Aatu. If we have choice, it is more natural to > define the cardinal of a set to be the initial ordinal equipollent to > it. > > > To speak more precisely your amendment will actually define cardinals > > for every set, and those cardinal will be sets always!, but the > > problem in having them non empty. > > Yes, I explicitly noted this in my post: unless we assume (or prove) > that every set is the same size as a well-founded set we can't use > Scott's trick to define cardinals. outside Regularity and Choice, yes, that is true. Zuhair > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: George Greene on 4 Dec 2009 20:35 On Dec 2, 10:45 pm, zuhair <zaljo...(a)gmail.com> wrote: > However NBG\MK do define proper classes as being not members of other > classes, as you mentioned ,that is correct. That is Not equivialent to defining proper classes as being equinumerous with the class of all sets, or with the class of all ordinals. Under this definition,the class of all ordinals has to be proper (Burali-Forti), but it is not clear why it has to be equinumerous with the class of all sets, even though that too must also be proper given regualarity.
From: Aatu Koskensilta on 4 Dec 2009 20:55 George Greene <greeneg(a)email.unc.edu> writes: > Under this definition,the class of all ordinals has to be proper > (Burali-Forti), but it is not clear why it has to be equinumerous with > the class of all sets, even though that too must also be proper given > regualarity. From the principle that a proper class is not a member of any class it indeed doesn't follow that a proper class must be the same size as the universe. What's the relevance? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: George Greene on 4 Dec 2009 21:49 On Dec 3, 8:04 pm, zuhair <zaljo...(a)gmail.com> wrote: > Ok, let me try to simplify matters. I am trying, too; I think mine will be simpler. > 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). Exactly. Not only do you not need choice, you do not even need a definition of the set/proper-class distinction at all. The class of all ordinal sets has to be proper; it cannot be a set, because if it were, it would be an ordinal set and therefore be a member of itself -- which, even in the ABSENCE of regularity, would be the Burali-Forti paradox -- ordinals have to be well- founded EVEN when other sets don't. > So D is a proper class, Right. But D is in some sense the SMALLEST POSSIBLE proper class. > 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) Well, yes, the class of ALL sets ALSO has to be proper. But again, this is THE LARGEST possible proper class -- you can't have any classes bigger than V, so if every class has to be as big as V in order to be proper, as was said by your > (1) x is a proper class iff x is equinumerous with the class of all sets" , then you are saying that all proper classes must be as LARGE as possible. This is NOT *intuitively* equivalent to saying that they all must be as SMALL as possible, which is what was said by your > (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) > so (2) -> (1) > > thus we have (2) <-> (1). Indeed we do, yet it still seems paradoxical; the tension is resolved by seeing that if you only have ONE size of proper class, then obviously, that is equivalent to any other possible way of having ONE size of proper class. Requiring that they all be small or that they all be big -- raising the floor to the ceiling or lowering the ceiling to the floor -- winds up not mattering; you just have 1 size of proper classes (and a lot of smaller sizes of sets) EITHER WAY. The counter-intuitivity here is coming from using iff as opposed to if in (1) and (2). What is intuitively necessary is that if you are not at least as big as the class of all ordinals, then you are not big enough to be proper. The back-arrow of (1) is NOT intuitive -- that is a LIMITATION to a context of special models where even the "big" proper classes -- like the class of all sets -- are ALSO small. Conversely, in (2), what is intuitively necessary is that if you are as big as the class of all sets, then of course you are too big to be a set -- the back-arrow, that even if you are NOT (originally, necessarily) that big, but are nevertheless proper, then you must also be BIG & proper (after all), is, again, a limitation to a special case. You point out that even prior to (1) and (2), we had, as theorems, without choice or without even caring about the definition of the set-class distinction, just purely as proven consequences of the paradoxes, (1P) the class of all ordinal sets is proper, and (2P) the class of all sets is proper. But it does not automatically follow that all proper classes are always as small as they are in (1) or as large as they are in (2).
From: Ross A. Finlayson on 4 Dec 2009 22:15
On Dec 4, 6:49 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Dec 3, 8:04 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > Ok, let me try to simplify matters. > > I am trying, too; I think mine will be simpler. > > > 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). > > Exactly. Not only do you not need choice, you do not even need a > definition of the set/proper-class distinction at all. The class of > all ordinal sets has > to be proper; it cannot be a set, because if it were, it would be an > ordinal > set and therefore be a member of itself -- which, even in the ABSENCE > of > regularity, would be the Burali-Forti paradox -- ordinals have to be > well- > founded EVEN when other sets don't. > > > So D is a proper class, > > Right. But D is in some sense the SMALLEST POSSIBLE proper class. > > > 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) > > Well, yes, the class of ALL sets ALSO has to be proper. > But again, this is THE LARGEST possible proper class -- you > can't have any classes bigger than V, so if every class has to > be as big as V in order to be proper, as was said by your > > > (1) x is a proper class iff x is equinumerous with the class of all sets" , > > then you are saying that all proper classes must be as LARGE as > possible. > This is NOT *intuitively* equivalent to saying that they all must be > as SMALL as possible, which is what was said by your > > > (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) > > so (2) -> (1) > > > thus we have (2) <-> (1). > > Indeed we do, yet it still seems paradoxical; the tension > is resolved by seeing that if you only have ONE size of proper class, > then obviously, that is equivalent to any other possible way of having > ONE size of proper class. Requiring that they all be small or that > they > all be big -- raising the floor to the ceiling or lowering the > ceiling to the floor -- > winds up not mattering; you just have 1 size of proper classes > (and a lot of smaller sizes of sets) EITHER WAY. > > The counter-intuitivity here is coming from using iff as opposed to if > in (1) and (2). What is intuitively necessary is that if you are not > at least as > big as the class of all ordinals, then you are not big enough to be > proper. > The back-arrow of (1) is NOT intuitive -- that is a LIMITATION to a > context of special models > where even the "big" proper classes -- like the class of all sets -- > are ALSO small. > Conversely, in (2), what is intuitively necessary is that if you are > as big as the > class of all sets, then of course you are too big to be a set -- the > back-arrow, > that even if you are NOT (originally, necessarily) that big, but are > nevertheless proper, > then you must also be BIG & proper (after all), is, again, a > limitation > to a special case. > > You point out that even prior to (1) and (2), we had, as theorems, > without choice or without even caring about the definition of the > set-class distinction, just purely as proven consequences > of the paradoxes, > (1P) the class of all ordinal sets is proper, and > (2P) the class of all sets is proper. > > But it does not automatically follow that all proper classes are > always > as small as they are in (1) or as large as they are in (2). Burali-Forti says "ordinals" is irregular. "The" physical universe of physical objects identified as mathematical objects sees more than any finite complexity in relations, i.e. there are infinitely many, compounded in autocorrelating that maps to powerset of those same objects, identity indicates a simple physical realization of counterexample to non-identity of physical universe to itself and its self-same reconsidered powerset of things. "Classes" as in "ZFC with Classes" where NBG is a schema-bound ZFC (implicitly infinitely axiomatized) are a guilty subterfuge to conscientious set-theoretic logicians acknowledging existence. Ordinals are ubiquitous in identification to sets under ordinal N with 2^N ordinals. Regards, Ross F. |