How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Charlie-Boo on 1 Jul 2010 15:25 On Jul 1, 12:23 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > Of course infinity = PA = arithmetic. > > Of course. PA doesn't mention infinity. Just because these chaps > > 0, 0', 0'', 0''', ... > > exist, doesn't mean that this chap > > {0, 0', 0'', 0''', ...} > > exists. Which--speaking loosely--is why ZF needs an axiom of infinity > and PA doesn't. No, PA and ZFC both have Peano's Axioms. The only difference is the universal set, which is N in PA and sets in ZFC. That's why it is expressed differently - different alphabets as well. C-B > -- > I can't go on, I'll go on.
From: Charlie-Boo on 1 Jul 2010 15:30 On Jul 1, 12:32 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > 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. The point is that ZFC would have to have an axiom other than Infinity (i.e. PA) that is necessary to prove PA consistent in order for it to be impossible in PA and possible in ZFC. But again you don't show that. And of course there are still no references to anyone doing it. Just 2 or 3 BS references that I exposed as being bogus - par for academia. C-B > MoeBlee
From: Charlie-Boo on 1 Jul 2010 15:45 On Jul 1, 1:01 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > 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. My fault that you don't answer the question? Did you know that a common tactic of abusive people is to blame the abused for making the abuser abuse them? It's not a theorem until proven. I have nothing to do with that. C-B > 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: Frederick Williams on 1 Jul 2010 15:50 Charlie-Boo wrote: > No, PA and ZFC both have Peano's Axioms. What formulations of first order PA and ZFC have _any_ of their non-logical axioms in common? None. Why? Well, for starters, all of the proper axioms of ZFC have the predicate symbol $\in$ in them, none of PA's axioms do because $\in$ is not even in the language of PA > The only difference is the > universal set, which is N in PA There are no sets, universal or otherwise, in first order PA. > and sets in ZFC. That's why it is > expressed differently - different alphabets as well. -- I can't go on, I'll go on.
From: Frederick Williams on 1 Jul 2010 16:02
Charlie-Boo wrote: > The axiom of infinity was added to give ZFC the capabilities of PA. No, the axiom of infinity is there so that ZFC may talk about infinite sets: omega and a whole load of others. Set theory begins with Cantor and has some famous difficulties that arise in connection with infinite sets. Zermelo's purpose was to axiomatize away the difficulties but keep the infinite sets. > Now, "to give" here refers to the state of mind of its author, so > let's not be psychotic. Let me say that it seems that's all it gives > you - that's all it's used for. So mathematicians have nothing to say about uncountable sets? > But all of you ZFC-ites can say if it > has been used for anything else. -- I can't go on, I'll go on. |