A new definition of Cardinality.
in [Logic]
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From: George Greene on 25 Nov 2009 16:02 On Nov 24, 3:10 pm, zuhair <zaljo...(a)gmail.com> wrote: > Dear George Greene, please answer the following questions: > > Is the following a theorem of ZFC or not? > > For all s Exist x for all y > ( y e x <-> (y strictly subnumerous to s and > for all z (z e Tc(y) -> z strictly subnumerous to > s)). > > were Tc(y) stands for the transitive closure of y. Why do you expect me to care about what subnumerous and transitive closure have to do with anything? I am asking a much simpler question about the cardinality of p(w)! > in case it is a theorem of ZFC, The question, rather, is whether X is the cardinality of p(w) is the cardinality of p(w) is a theorem of ZFC, if X is any ordinal you can think of. This IS NOT THE SAME question as whether Ex[x is the cardinality of p(w)] is a theorem of ZFC.
From: zuhair on 25 Nov 2009 16:34 Let me pass some news here. Cardinals that I defined would not always be "sets" out of Regularity, in other words the set-hood of these cardinals depends on weather we work under Regularity or not, if we assume Regularity then all these cardinals are sets. However without Regularity they can be proper classes: Below was the reply that I got regarding this issue: ---------------------- Without regularity (= Axiom of Foundation), one can construct in ZF (or ZFC) a model of ZF (resp. ZFC) where there is a PROPER CLASS of sets x = {x}. Call such crazy sets "nodes" for short. Consider {x, y}, where both elements are nodes and different. Then, by your definition, (x, y} would be a member of the cardinality of 2. But there is a proper class of such pairs. So your cardinal would not be a set. ---------------------- However the question weather the theorem that every set has a cardinality (as defined in this thread) is dependent on choice, is still not settled yet! Zuhair
From: Tim Little on 25 Nov 2009 18:57 On 2009-11-25, George Greene <greeneg(a)email.unc.edu> wrote: > The question, rather, is whether > X is the cardinality of p(w) > is a theorem of ZFC, if X is any ordinal you can think of. That depends upon what *exactly* you mean by "ordinal you can think of". The ordinal beth_1 would appear to suffice. - Tim
From: Rupert on 25 Nov 2009 19:04 On Nov 25, 2:20 am, George Greene <gree...(a)email.unc.edu> wrote: > On Nov 24, 3:23 am, Rupert <rupertmccal...(a)yahoo.com> wrote: > > > If I assume the axiom of choice, then every set can be well-ordered, > > Right. > > > and is equipollent to some cardinal kappa, > > WRONG. > Nonsense. Of course it is correct. If a set can be well-ordered, then there exists some cardinal kappa such that the set is equipollent to kappa. > Ex[whatever] > does NOT mean that there exists a UNIQUE x such that whatever! No, and I did not claim uniqueness, but as a matter of fact in this case it is extremely easy to prove uniqueness as well. > There are MANY DIFFERENT well-orderings of p(w) under ZFC and > they correspond TO DIFFERENT ordinals! > But I spoke of a cardinal, an ordinal which is not equipollent to any smaller ordinal. If a set can be well-ordered, then there exists an ordinal (not unique) to which it is equipollent, and there exists a unique *cardinal* to which it is equipollent. A cardinal being an ordinal not equipollent to any smaller ordinal. > > a cardinal being a von > > Neumann ordinal not equipollent to any smaller ordinal. > > ZFC does NOT assign a cardinality to p(w) or any other uncountable set > except maybe the alephs. Under ZFC, beth-1 (=p(w)=p(aleph-0) > =2^beth-0) > "can" take ANY aleph-n (where n is natural) as a cardinality! > So what? ZFC proves that P(w) has a cardinality, namely beth-1. We cannot decide in ZFC which aleph it is. > Remember, there was that whole conunundrum about > THE CONTINUUM HYPOTHESIS ?!?!?!!? That's completely irrelevant here.
From: David Libert on 26 Nov 2009 03:22
Aatu Koskensilta (aatu.koskensilta(a)uta.fi) writes: [Deletion] > PS. I have e-mailed you a copy of the Jech paper. I haven't myself > studied the argument for the existence of the set of hereditarily > countable sets in absence of choice, by establishing that the rank of a > hereditarily countable set <= omega_2, and so can say nothing about > whether it may be generalised beyond the countable case. > > -- > Aatu Koskensilta (aatu.koskensilta(a)uta.fi) > > "Wovon man nicht sprechen kann, dar�ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus Thanks. Jech's proof in ZF that HC is a set (not a proper class) is in the first 2 pages of the paper. I think Jech's proof straightdforwardly generalizes to H_kappa for von Neumann cardinals kappa, showing from ZF that these are sets, namely replacing omega in Jech's proof by kappa. (Well with adjustments: HC is heritarily <= omaga and H_kappa is heritarily < kappa : <= versus < ). I also think Jech's proof generalizes to H_(< #x) for non-well-orderable x, by an adjusted as above replacement of omega by the surjective Hartog ordinal of x. If this is correct, it would show Zuhair's definition of cardinal is a set for arbtrary x as the theorem of ZF, depending on regularity and replacement but no AC needed. -- David Libert ah170(a)FreeNet.Carleton.CA |