The size of proper classes?
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From: zuhair on 4 Dec 2009 22:35 On Dec 4, 9:49 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). Let us agree on two important issues here. (1) The class of all ordinals that are sets 'D' is the smallest possible proper class. (2) The class of all sets 'V' is the Largest possible proper class. Now when anybody say that *ALL* proper classes are equal in size to the smallest possible proper class, then simply what he is saying is that *ALL* proper classes are of the same size. Now when another man say that *ALL* proper classes are equal in size to the largest possible proper class, then simply what he is saying is that *ALL* proper classes are of the same size. So it doesn't really matter how you say it, you want to say all proper classes are equal to the smallest possible proper class, or you want to say all proper classes are equal to the Largest possible proper class, in both versions all what you are saying is "ALL proper classes are of the same size", so it doesn't really matter, the direction of saying it is irrelevant, at the end you are saying the same thing but in different ways. All what are you trying to say is Size D = Size V, you can say D is as big as V, or if you want to say V is as small as D, it doesn't matter, it is just two ways of saying the same thing which is V is equal in size to D. Of course in other inner models of MK, one can have a different axiom of size limitation to the effect that size D not equal size V, i.e size V strictly supernumerous to Size D, yes one can do that, there are various models. But the following if you add any one of the following size limitation axioms to MK minus size limitation, then you will get the same result (1) x e V <-> x strictly subnumerous to V (2) ( ~ x e V & ~ y e V ) <-> x equinumerous to y (3) ~ x e V <-> x equinumerous to D (4) ~ x e V <-> x equinumerous to V Zuhair
From: Daryl McCullough on 5 Dec 2009 07:03 George Greene says... > >On Dec 3, 8:04=A0pm, zuhair <zaljo...(a)gmail.com> wrote: >> Now in NBG\MK it can be proved easily that the class D of all ordinals >> that are sets would be a proper class [stuff deleted] >Right. But D is in some sense the SMALLEST POSSIBLE proper class. What's the argument for that? Is there a theorem in NBG to the effect that, if C is a proper class than C is larger than D (in cardinality)? -- Daryl McCullough Ithaca, NY
From: zuhair on 5 Dec 2009 07:13 On Dec 5, 7:03 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > George Greene says... > > > > >On Dec 3, 8:04=A0pm, zuhair <zaljo...(a)gmail.com> wrote: > >> Now in NBG\MK it can be proved easily that the class D of all ordinals > >> that are sets would be a proper class > > [stuff deleted] > > >Right. But D is in some sense the SMALLEST POSSIBLE proper class. > > What's the argument for that? Is there a theorem in NBG to the effect > that, if C is a proper class than C is larger than D (in cardinality)? If "larger" mean 'bigger or equal', then must be. There is no way you can have a proper class that strictly subnumerous to D. Zuhair > > -- > Daryl McCullough > Ithaca, NY
From: Aatu Koskensilta on 5 Dec 2009 07:42 stevendaryl3016(a)yahoo.com (Daryl McCullough) writes: > What's the argument for that? Is there a theorem in NBG to the effect > that, if C is a proper class than C is larger than D (in cardinality)? With choice, yes. As an historical aside, Cantor thought he had found a proof of the well-ordering theorem in the observation that every "inconsistent multiplicity" must contain a submultiplicity equivalent to the class of ordinals. For those of a more serious turn of mind, this is evidence that Cantor "implicitly" accepted a form of replacement. Incidentally, Jourdain produced a proof of the well-ordering theorem that Zermelo found very unsatisfactory. Strangely, this proof was exactly what we find in textbooks today: just pair the elements of A with ordinals until you ran out of elements; since A is a set you won't exhaust all the ordinals, and the process will terminate at some set-sized ordinal alpha. Zermelo's less waffling proof is very nice but, for some reason, not usually found in textbooks. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 5 Dec 2009 07:44
zuhair <zaljohar(a)gmail.com> writes: > On Dec 5, 7:03�am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> What's the argument for that? Is there a theorem in NBG to the effect >> that, if C is a proper class than C is larger than D (in >> cardinality)? > > If "larger" mean 'bigger or equal', then must be. There is no way you > can have a proper class that strictly subnumerous to D. You are assuming that "not bigger or equal" implies "strictly subnumerous". This is an unwarranted assumption in absence of choice. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |