How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Logic]
|
Prev: equivalence
Next: How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
From: MoeBlee on 1 Jul 2010 12:32 On Jul 1, 11:00 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > how about a high > level summary of the proof and how ZFC axioms are needed to formalize > that? That's been posted already! (No, I'm not gonna get post numbers for you; It's enough that at least a few posters have been providing such summaries all along.) As to the axioms used, the axioms of Z-R suffice. Whether there is a proper subset of those axioms that suffice, I don't opine. MoeBlee
From: R. Srinivasan on 1 Jul 2010 12:49 On Jul 1, 5:01 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > "R. Srinivasan" wrote: > > On a more intuitive level, how can we fault the existence of the > > infinite set N? If you consider the statement "All natural numbers are > > not upper bounds for N" and ask "how many natural numbers are exceeded > > by some element of N?", the answer has to be "infinitely many". > > The answer is finitely many. Let's suppose N starts with 0 (if you take > it as starting with 1 my comment need only a slight adjustment). Let's > take the "some" element to be n. How many natural numbers are exceeded > by n? Answer n. E.g. take n = 4: > > 0, 1, 2, 3, *4*, 5, 6, ... > > and count the numbers exceeded by 4: 0 < 4, 1 < 4, 2 < 4, 3 < 4, and > nothing else is exceeded by 4, so that makes 4. > > Btw, your "All natural numbers are not upper bounds for N" is more > idiomatically expressed as "No natural numbers are upper bounds for N", > which I agree with. > I didn't ask how many natural numbers are exceeded by some specific, fixed n. I am asking you to count the number of natural numbers that are not upper bounds for N. Since all natural numbers fit this description, the answer has to be "infinitely many". The assertion that infinitely many natural numbers have been exceeded within N is paradoxical, as I pointed out. The root cause of the paradox is the existence of infinitely many finite natural numbers. RS
From: Frederick Williams on 1 Jul 2010 12:59 "R. Srinivasan" wrote: > > > I didn't ask how many natural numbers are exceeded by some specific, > fixed n. You asked "how many natural numbers are exceeded by some element of N?" But never mind... > I am asking you to count the number of natural numbers that > are not upper bounds for N. Since all natural numbers fit this > description, the answer has to be "infinitely many". Agreed. > The assertion > that infinitely many natural numbers have been exceeded within N is There is nothing in N that exceeds infinitely many natural numbers. > paradoxical, as I pointed out. The root cause of the paradox is the > existence of infinitely many finite natural numbers. > > RS -- I can't go on, I'll go on.
From: MoeBlee on 1 Jul 2010 13:01 On Jul 1, 11:07 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jul 1, 11:53 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 30, 11:17 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > The problem isn't to conclude that a model exists, using ZFC. The > > > problem is to prove that PA is consistent, using ZFC. > > > I've gone over this already with you. ZFC proves that if a theory has > > a model then the theory is consistent. It's an extremely simple > > exercise I proposed you might think about and complete in about two > > minutes. > > Then there's no reason for not giving it. Bravado is no substitute > for Mathematics. The reason for giving it as an exercise is to get you STARTED THINKING. WHAT bravado? Come on, really. A theory is a set of sentences (all in a language) closed under entailment. If a theory T is inconsistent, then there is a sentence P such that both P and ~P are in T. But P is true in model M iff ~P is false in model M, and no sentence is both true and false in a given model M (by the definition-by-recursion function that maps sentences to true or (exclusive or) to false per a model). So if a theory is inconsistent, then the theory has no model (lest there be a sentence P that is both true and false in the model, which is impossible). Note: This does not preclude that there are models ('structures' if you prefer) for the LANGUAGE of an inconsistent theory. For any language, there are many models for that language. But if a theory is inconsistent, then there is no model in which all of the sentences of the theory are true. Now, I PROMISE myself. No more explanation of this for you. If you don't understand or have some question or objection about it. Then just study the matter. I suggest Enderton's book for this particular matter. > Didn't you read my response? Hinman doesn't refer to ZFC's axioms at > all in his proof. The axioms are used in the various steps leading up to the proof. That's how mathematics works. A proof of a theorem may rely on previously proven theorems. > He even admits that. What specific quote do you have in mind? He doesn't need ZFC. ZF is sufficient (less is sufficient too). One disclaimer: I've given you the Hinman reference since you've asked for a reference. I have not scrutinized his particular proof, since this is something I proved myself long before I got Hinman's book. Nevertheless, it is a reference as you asked for one, and you may elect or not to read all the steps in the book that lead to his exposition of why ZF proves PA is consistent. MoeBlee
From: R. Srinivasan on 1 Jul 2010 13:11
On Jul 1, 9:59 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > "R. Srinivasan" wrote: > > > I didn't ask how many natural numbers are exceeded by some specific, > > fixed n. > > You asked "how many natural numbers are exceeded by some element of N?" > True. In the sense that An Em m > n For each n, there is some m that exceeds it. So how many such n's are there that are exceeded *within* N? Certainly more than finitely many. > > But never mind... > > > I am asking you to count the number of natural numbers that > > are not upper bounds for N. Since all natural numbers fit this > > description, the answer has to be "infinitely many". > > Agreed. > > > The assertion > > that infinitely many natural numbers have been exceeded within N ... > > There is nothing in N that exceeds infinitely many natural numbers. > True. But the assertion that "more than finitely many natural numbers are not upper bounds for N" translates precisely to "Infinitely many natural numbers are exceeded within N". This is the intuition behind the existence of nonstandard natural numbers, by the way. RS |