How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Tim Little on 26 Jun 2010 22:51 On 2010-06-26, R. Srinivasan <sradhakr(a)in.ibm.com> wrote: > The theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > exist"). Actually ~Inf does not assert "Infinite sets do not exist". It only asserts "there does not exist a successor-closed set containing the empty set". It may turn out to prove an equivalent statement under the rest of the axioms, but ~Inf does not actually mean "infinite sets do not exist". > This proof obviously implies that "There does not exist a model for > PA", for a model of PA must have an infinite set as its universe Even if your intepretation of ~Inf were correct, all it would prove is that ZF-Inf+~Inf does not model PA. > Now I am sure a lot of people are going to jump up and down and > protest at this interpretation. But it is logical. No, it is not. It exhibits a fairly elementary failure of logic. The statement "X models PA" implies "PA has a model". However, "X does not model PA" does *not* imply "PA has no model". - Tim
From: Charlie-Boo on 26 Jun 2010 23:18 On Jun 25, 2:16 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Thu, 24 Jun 2010 21:21:05 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > >> On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > >> > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy > >> > to prove that PA is consistent (its axioms and rules preserve > >> > truth) yet (by Godel-2) PA can't do such a simple proof as that. > > >> So what? ZFC can prove it. > >> ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove > >> that PA is consistent. This is why you can't say that "ZFC/PA > >> doesn't prove PA is consistent." "ZFC/PA" is just a meaningless > >> locution in any case. > >> ZFC is one thing. PA is another. > > > PA is a subset of ZFC, > > No it isn't. (Curious that you continue to assert otherwise.) What does that assertion mean, actually? > > so I emphasize that by calling it ZFC/PA (it makes more sense to > > distinguish the two anyway.) This is besides the point. Who has > > proved PA consistent using ZFC? > > It is an elementary exercise. A proof can be found in almost any > reasonably thorough text in set theory. You know, you've always been a perfect target. You are one of those who try being condescending AND rely on the "vague = undefined = unrefutable" principle, while I am one who loves to define the undefined and thus refute the unrefutable (see past posts.) Ok, so if one went through a bunch of well-referenced texts, the great majority of them would contain a proof of PA written in ZFC - or however you think it should be phrased? And being elementary, it must be small and completely given? And being elementary, it should be in almost all texts. How about one that is on-line (from all those many instances of its being published)? Or a couple of your favs? C-B
From: George Greene on 27 Jun 2010 00:04 On Jun 26, 11:18 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > Ok, so if one went through a bunch of well-referenced texts, the great > majority of them would contain a proof of PA written in ZFC Not a PROOF OF PA -- look at your OWN subject line! It would contain a proof that *PA is Consistent*. That's what you SAID you wanted! Consistent does NOT mean the same thing as "true". Individual STATEMENTS get proven. PA is not just one sentence. It is a WHOLE THEORY of sentences that ALL follow from its axioms. Proving a theory consistent is NOT the same KIND of thing as just proving ONE sentence TRUE. > And being elementary, it must be small and completely given? And > being elementary, it should be in almost all texts. > > How about one that is on-line (from all those many instances of its > being published)? > > Or a couple of your favs? Please. You can google it yourdamnself. But it is not even complicated. You do, however, need to understand some background. One way to prove that a theory is consistent is to prove that it HAS A MODEL. Inconsistent theories do NOT have models. If a theory is inconsistent then there is no way to completely interpret all the functions, predicates and constants that are named in its axioms, that will make each individual AXIOM come up true. If the theory is consistent, then ZFC is a good tool to use for CONSTRUCTING A MODEL for it. That's one of the most common uses to which ZFC-sets IN GENERAL are put. You can represent (or interpret) functions as sets of ordered pairs. You can represent (or interpret) predicates as sets of their argument- lists. Or, given a set-representation of some constant in the theory (e.g., if you interpret PA's zero as ZFC's empty set), you can define set-functions in ZFC and interpret the functions from PA's axioms accordingly (e.g., if a number x from PA has been interpreted as a set X in ZFC, then the successor function s(.) from PA can be interpreted as a function S(.) in ZFC by S(X) = X U {X}. PA's addition and multiplication can be interpreted using disjoint union and cartesian products. This is not exactly short but it is very easy to understand, if you find a decent treatment. The point is, the mere existence of these interpretations (along with a proof that each axiom of PA comes out true under them) constitutes a proof of the consistency of PA.
From: George Greene on 27 Jun 2010 00:19 On Jun 26, 3:56 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > Do you know how N is represented in ZFC? By constructing SHUT UP. YOU DON'T know, NOT EVEN after having read "Set Theory for the Working Mathematician". If you WANT to know, ASK. THEN somebody will tell you that von Neumann figured out that the right way to represent natural numbers as ZF sets was to represent (interpret) every natural number as the set of all smaller natural numbers (or their sets). Since there are no natural numbers smaller than 0, 0 is interpreted as the empty set (which I will spell O just to highlight the difference). Thereafter, in accordance with the above, 1 is interpreted as {O}, 2 is interpreted as { O,{O} } (i.e, as {0,1} ) and 3 is interpreted as {0,1,2}, ad inf. Since you canNOT count to infinity, the completed infinity of all of N *could*never*get* constructed this way; therefore ZF has AN AXIOM OF INFINITY that postulates the existence of N. In other words, N is really NOT constructed at all: it is just blatantly asserted to exist JUST BECAUSE WE FEEL LIKE IT.
From: George Greene on 27 Jun 2010 00:21
On Jun 25, 5:14 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > So why not have a try at it? You'll find all the details you need in any > decent text. This is a little sparse. I really don't think that the model existence theorem is going to leap out at him here. |