How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: MoeBlee on 5 Jul 2010 15:03 On Jul 5, 11:56 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > MoeBlee wrote: > > Yes. Aatu claims ZFC is consistent. He's written many posts on the > > subject. > > Did Aatu "claim" or present _a formal proof_ that ZFC is consistent? Aatu has answered this over and over already! > > For even more on the subject, see Franzen's book > > Sorry. I always respect what TF had to offer. But I could not ask a > dead man any question, which I'm sure I'd like to ask in the subject. Would you please read the book? It's easy reading, not too technical. And instead of saying that there's questions he can't answer because he's dead, just read the book to see what questions he DOES address anyway. THEN perhaps you might have additional questions. MoeBlee
From: MoeBlee on 5 Jul 2010 16:16 On Jul 5, 1:06 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > I take it that you meant to answer you don't know the answer to this > question. (But you have to let me know if this is the case). A question such as yours presupposes a framework from which the question is asked. Without myself adopting whatever presuppositions you might or might not have, I gave you a detailed and to the point answer to your question from within my own modest framework. On the other hand, I'm not interested in a mode that is less an explanation of certain notions and has more the character of a deposition. MoeBlee
From: MoeBlee on 5 Jul 2010 16:53 On Jul 5, 1:46 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > MoeBlee wrote: > > On Jul 5, 1:06 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > > >> I take it that you meant to answer you don't know the answer to this > >> question. (But you have to let me know if this is the case). > > > A question such as yours presupposes a framework from which the > > question is asked. > > Agree. And usually when it's left unsaid it's assumed to be FOL= > framework of formal axiom systems (which are syntactical). This > is what I presupposed. (I thought we understood that when we > talked - many times - about formal proofs, FOL, etc... No?) No, I mean even more general presuppositions (and, to be fair to you, I don't expect you'd know that since I didn't specify what kind of presuppositions I meant). Never mind though. It's not worth the ordeal now of going into yet another issue as to what kind of presuppositions I have in mind; I meant it merely in the sense of a GENERAL disclaimer. ASIDE from that, I've failed in virtually every attempt to communicate with you on virtually every matter, informal or informal, I've discussed with you. I need to give up. MoeBlee
From: Transfer Principle on 5 Jul 2010 18:35 On Jul 3, 5:45 pm, "K_h" <KHol...(a)SX729.com> wrote: > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > In NAFL, N is a proper class and there is no > > such thing as P(N). You cannot quantify over all > > the infinite sub-classes of N to get P(N), > > because quantification over proper classes is not > > permitted. > Why is N a proper class in NAFL and what is the basis to even have proper classes > in NAFL? The whole reason to bifurcate collections into sets and proper classes > is so that mathematics is self-consistent. That is, the problems presented by > Cantor's theorem (for example the Cantor and Burali-Forti paradoxes) can only be > solved by dividing collections into two types: sets and proper classes. (The > problems presented by Russell's paradox are solved by the axiom of foundation, no > collection can be a member of itself.) So it's because of Cantor's theorem that > such a bifurcation is necessary. If Cantor's theorem is false in NALF then on > what basis does NALF bifurcate collections into the two types? As far as I can > tell, NAFL has no basis to do so and so N should be a set. As a set, it is easy > to show that |N|<|P(N)|. So K_h acknowledges that because of two paradoxes, Cantor and Burali-Forti, we have theories with proper classes -- in particular, we have the proper classes V and On, respectively. But now let's take a look at N. We know that by the von Neumann definition of natural number, a natural number is a finite ordinal. So now we ask, how do we know that N isn't equal to the class On of all ordinals? In standard theory, the Axiom of Infinity proves that omega is a set, and an ordinal, but an infinite ordinal. Thus we have (omega)eOn and yet ~(omega)eN. Therefore ~N=On. But without the Axiom of Infinity, we can't prove that omega is a set. And according to Srinivasan, the theory NBG-Infinity, with NAFL as the underlying logic, proves that N=On and so N is a proper class, indeed the proper class of all ordinals. If we use FOL instead of NAFL, we can't prove that N=On unless we add the additional axiom D=0 (since On intersect D is exactly N). This raises another question. I believe in the following analogy: ZF-Infinity(+~Infinity) is to ZF as ZF(+~Inaccessible) is to ZF +Inaccessible Indeed, it is often stated that in a way, Infinity is the first large cardinal axiom. We can replace all references to ZF in the above with NBG so that we can discuss proper classes (or we can adopt MoeBlee's trick and treat class symbols as syntactic sugar for expressions that omit references to classes). And so now let's define a class M to be the class of all "accessible" ordinals (i.e., less than the first inaccessible ordinal). We note that in NBG+~Inaccessible, all ordinals are accessible and so M is exactly the class On of all ordinals, hence M is a proper class. But in NBG+Inaccessible, M is exactly the first of the inaccessible ordinals, and so M is actually a _set_. So, when K_h writes: > As far as I can > tell, NAFL has no basis to do so and so N should be a set. As a set, it is easy > to show that |N|<|P(N)|. I can just as easily write: "As far as I can tell, NAFL (or even FOL) has no basis to do so and so M should be a set. As a set, it is easy to show that |M|<|P(M)|." thus forcing K_h to accept inaccessibles! In other words, for me to require K_h to accept inaccessibles is analogous to requiring a finitist like Srivinasan to accept infinite sets. > We must show that |N|<|P(N)|. For this it must demonstrated that |N|<=|P(N)| and > |N|=/=|P(N)| are both true. To establish |N|<=|P(N)|, note that for all n in N > the function f(n)={n} does the job. To establish |N|=/=|P(N)|, a proof by > contradiction is needed. Subsets of N, like {2,5} and {1,3,5,7,9,...}, can be > written like {_,_,2,_,_,5,_,_,_,...} and {_,1,_,3,_,5,_,7,_,9,_,...} where the > underscore denotes the natural numbers that are missing from the subset. Since > one subset of N is N itself, N={0,1,2,3,4,5,6,7,8,9,...} is the only subset > without underscore characters and the empty set will have nothing but underscore > characters. Assume there is a bijection, f, between N and P(N) as follows: Does this proof work if N=On? I doubt it. Otherwise, why can't we substitute M for N above in the proof? Sure, K_h (assuming that he's working in a consistent ZF/NBG) can't refute that M=On, but neither does Srinivasan refute that N=On. > It is clear that the subset S of N is not in the list because it is different > from every set in the list according to the axiom of extensionality. Recall that > the axiom of extensionality says that two sets are equal iff they contain the > same elements. The set S={0,3,4,...} is clearly a subset of N S is clearly a sub_class_ of N. If N is a proper class, then it's possible for S to be a proper class as well, but only _sets_ can be elements of other sets. Therefore, the proof appears to be invalid unless we can prove that N is a set -- and in Srinivasan's theory, N isn't a set. Although the following isn't directly related to Srinivasan's theory, the following might be relevant in other threads, including Herc's: K_h states that the main reason to make a distinction between sets and proper classes is Cantor's Theorem. Thus, in a theory which refutes Cantor's Theorem, such as NFU, we expect there to be no distinction between sets and proper classes. Indeed, we see that this is exactly the case -- NFU proves that even V is a set. Notice that Infinity isn't an axiom of NFU (but in practice, set theorists work in NFU+Infinity or even NFU+Infinity+AC). But suppose we had the theory NFU+"N is a set" (with "N is a set" being syntactic sugar for a valid formula in NFU). Is K_h's proof now a valid proof that card(N) < card(P(N)), and thus, N must be a Cantorian set? I suspect that the answer might be "yes" -- which is why even though NFU proves the existence of non-Cantorian sets, the set N must still be Cantorian no matter what. This is why, in the Herc threads, I'm trying to find out whether Herc opposes Cantor's Theorem per se (so that NFU satisfies his desiderata) or the idea of cardinals larger than card(N) (so that NFU still won't satisfy Herc)? If the latter, then we'd need a theory other than NFU, one which proves N to be non-Cantorian, in other to meet Herc's desiderata. > > Again, you are arguing backwards. In NAFL, > > truth as provability is a fundamental logical > > requirement (as I have illustrated above while > > discussing the law of non-contradiction). > > Therefore the incompleteness theorems do not > > hold for NAFL theories (e.g. F, the theory of > > finite sets). > > You haven't demonstrated what you say you have. All formal systems are limited > in what they can mechanically prove but those limitations are not limitations on > existence. > > _- Hide quoted text - > > - Show quoted text -
From: Transfer Principle on 5 Jul 2010 18:54
On Jul 5, 3:35 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > axiom D=0 (since On intersect D is exactly N). Correction: On\D = N -- that is, On set-subtraction D equals N. |