The size of proper classes?
in [Math]
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From: zuhair on 6 Dec 2009 09:13 On Dec 5, 7:44 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > zuhair <zaljo...(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. Hmmm..., you are right. What I wanted to say is that no proper class can be strictly subnumerous to the class D of all ordinals that are sets . Of course it can be incomparable with D or equinumerous to D or strictly supernumerous to D, but it cannot be strictly subnumerous to D. 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 6 Dec 2009 15:46 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)? Not strictly larger. Larger-than-or-equal-to. There is a theorem that if a proper class is injectible into you, then you are proper too. You are of course larger-than-or-equipollent-to any class that is injectible into you. Obviously, mere subclasshood is not going to be the relevant sense of smaller; you can take 0 out of any proper-class-ordinal and get something that is still a proper class but is a proper subclass of the class of all ordinals. But all the general rules are sort of out the window here since we don't even have a fixed definition of the set-class distinction. We are talking in general terms. There was supposed to be 1 definition of the set-class distinction whereby being proper was equivalent to not being a member of any other class. We are obviously NOT using that definition since that would outlaw proper-class ordinals (every ordinal is a member of its successor). The single most interesting aspect of this subthread is that the proper-class-hood of these two classes is INdependent of the definition of the set-class distinction.
From: zuhair on 6 Dec 2009 23:50 On Dec 6, 3:46 pm, George Greene <gree...(a)email.unc.edu> wrote: > 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)? > > Not strictly larger. > Larger-than-or-equal-to. > There is a theorem that if a proper class is injectible into you, then > you > are proper too. You are of course larger-than-or-equipollent-to any > class > that is injectible into you. > > Obviously, mere subclasshood is not going to be the relevant sense of > smaller; > you can take 0 out of any proper-class-ordinal and get something that > is still > a proper class but is a proper subclass of the class of all ordinals. > > But all the general rules are sort of out the window here since > we don't even have a fixed definition of the set-class distinction. > We are talking in general terms. > There was supposed to be 1 definition of the set-class distinction > whereby > being proper was equivalent to not being a member of any other class. > We are obviously NOT using that definition since that would outlaw > proper-class > ordinals (every ordinal is a member of its successor). I should say something about that, the correct statement is: Every ordinal is a member of its successor if and only if there exist a successor. Let us assume Regularity. we can define ordinal in the following manner: An ordinal is a transtive class of transitive sets. Now lets take for example MK class theory with the following size limitation axiom. For all x ( x e V <-> x strictly subnumerous to V ) were V is the class of all sets, were a set is defined in MK as a class that is a member of some class, while a proper class is defined as a class that is not a member of any class. Now according to MK with the above version of size limitation, the class D which is the class of all ordinals that are sets, would definitely be an ordinal, and as we know from Burali-Forti paradox D cannot be a set, so it is a proper class, so D is an proper class ordinal(i.e. an ordinal that is a proper class), this is to differentiate it from "a set ordinal (i.e. an ordinal that is a set) this is very easy to prove actually, but D has no successor according to this theory! in MK a proper class ordinal cannot have a successor, for a simple reason, the successor of an ordinal x must contain x as a member in it, thus if x is a proper class ordinal, then obviousely it cannot have a successor, since otherwise this would mean that x is a proper class that is contained in a class (its successor), and that contradicts the very definition of proper classes as not being members of any class. Apparently you are not working under the set-proper class distinction, that's why you said there can exist proper class ordinals that are members of their successor, that reminds me of course of Ackermann's class theory, were this is the case, and superisingly Ackermann's class theory become inconsistent if we say that there exist an ordinal that has no successor, but this case is applicable to Ackermann's class theory, and not to NBG\MK. However even if you remove this set-proper class distinction I refered to upwards, still you cannot prove that every ordinal has a successor, this will largely depend on the inner model of the theory you are working in. Zuhair > > The single most interesting aspect of this subthread is that > the proper-class-hood of these two classes is INdependent of the > definition of the set-class distinction.
From: Aatu Koskensilta on 7 Dec 2009 01:28 George Greene <greeneg(a)email.unc.edu> writes: > But all the general rules are sort of out the window here since we > don't even have a fixed definition of the set-class distinction. We > are talking in general terms. There was supposed to be 1 definition > of the set-class distinction whereby being proper was equivalent to > not being a member of any other class. We are obviously NOT using > that definition since that would outlaw proper-class ordinals (every > ordinal is a member of its successor). 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. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: zuhair on 7 Dec 2009 07:13
On Dec 7, 1:28 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > George Greene <gree...(a)email.unc.edu> writes: > > But all the general rules are sort of out the window here since we > > don't even have a fixed definition of the set-class distinction. We > > are talking in general terms. There was supposed to be 1 definition > > of the set-class distinction whereby being proper was equivalent to > > not being a member of any other class. We are obviously NOT using > > that definition since that would outlaw proper-class ordinals (every > > ordinal is a member of its successor). > > 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. This might be confusing Aatu since the phrase "in your sense" is vague. The class of all ordinals that are sets is definitely an ordinal, and it is an ordinal that is a proper class, and in MK\NBG it doesn't have a successor! However it seems that George is thinking that the term "ordinal" implies that their should be a successor, i.e. I think George think if we say for example x is ordinal (weather it is 'set ordinal' or 'proper class ordinal') then there must exist a class y such that y is the successor of x, i.e. y=xUnion{x}, I think this is the source of confusion in George's notes, so accordingly George seem to think that there is a class of proper class ordinals, one succeeding the other like the case with sets. In most theories set ordinals do have successor, but a proper class ordinal might or might not have a successor, in Ackermann's all ordinals must have successors weather they are proper classes or sets, while in NBG\MK all ordinals that are sets have successor, but there exist only ONE proper class ordinal and that is the class of all set ordinals, and it doesn't have any successor, and yet it is ordinal! I already clarified that there is nothing in the definition of "ordinal" that implies that there must exist a successor for each ordinal. Assuming Regularity an ordinal can be defined as a transitive class of transitive sets, that's all, nothing in this defintion per se entails that every ordinal must have a successor, the existence of successors for ordinals weather they are sets or proper classes depends on the theory one is working with. Zuhair > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |