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 27 Jun 2010 12:11 On Jun 25, 5:14 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > Who has proved PA consistent using ZFC? If it were possible then I > > assume someone would have done it. It certainly would be a very > > educational exercise. > > So why not have a try at it? > You'll find all the details you need in any > decent text. What is the latest text with it - you know, modern treatment? Lots of new texts coming out all the time, aren't there? C-B > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Charlie-Boo on 27 Jun 2010 12:40 On Jun 27, 5:18 am, William Hale <h...(a)tulane.edu> wrote: > In article > <ff54cc7d-b23f-4a45-9040-0459145ff...(a)j8g2000yqd.googlegroups.com>, Charlie-Boo <shymath...(a)gmail.com> wrote: > > [cut] > > > If ZFC can't calculate what PA can, how can anyone say that ZFC is a > > good basis for doing mathematics - PA is used by lots of > > mathematicians. > > PA is not used by any mathematicians to do algebra, number theory, real > analysis, complex analysis, topology, or differential geometry. These > mathematicians represent most mathematicians. They use ZFC as their > axiomatic system. PA is not used but ZFC is? But ZFC invokes the Peano Axioms carte blanche to represent N - so PA is used by ZFC and thus by all of these Mathematicians. Good point! C-B
From: Charlie-Boo on 27 Jun 2010 12:50 On Jun 27, 8:21 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > On Jun 25, 4:58 pm, Frederick Williams <frederick.willia...(a)tesco.net> > > wrote: > > > Charlie-Boo wrote: > > > > point. Who has proved PA consistent using ZFC? > > > > Also, see Gentzen > > > and Ackermann. Gentzen's proof used far less than full ZFC. > > > References please. On-line?? Thanks! > > Gentzen: Mathematische Annalen, vol. 112, pp 493-565 > and Forschungen zur Logik und zur Grundlegung der exakten > Wissenschaften no. 4, pp 19-44. Gentzen's consistency proof "reduces" the consistency of mathematics, not to something that could be proved. Wikipedia http://en.wikipedia.org/wiki/Gentzen%27s_consistency_proof Wiki doesnt say anything about ZF in its write-up of Gentzens proof of the consistency of PA! What happened?? C-B > Ackermann: Mathematische Annalen, vol 117, pp 162-194. > > For Gentzen in English see his collected papers edited by Szabo. $449.94 - and worth every penny of it! http://www.amazon.com/collected-Gerhard-Gentzen-foundations-mathematics/dp/072042254X/ref=sr_1_2?ie=UTF8&s=books&qid=1277657239&sr=1-2 > For an > account of Ackermann's proof see Wang, Logic, Computers and Sets, Ch > XIV. > > > > You may wish to know that ZFC with the axiom of infinity replaced > > by > > > its negation is a model of PA and vice versa. > > > Wow, that sounds cool. I'll have to think anout that one. Where can > > I read about it? > > I wish I could remember. Chris Menzel will tell us shortly. > > -- > I can't go on, I'll go on.
From: Charlie-Boo on 27 Jun 2010 12:57 On Jun 27, 5:44 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 27, 11:24 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > > > > On Jun 26, 7:51 pm, Tim Little <t...(a)little-possums.net> wrote: > > > > On 2010-06-26, R. Srinivasan <sradh...(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". > > > This has come up time and time again. I myself have claimed that > > the theory ZF-Infinity+~Infinity proves that every set is finite, > > and someone (usually MoeBlee or Rupert) points out that this > > theory only proves that there's no _successor-inductive_ set > > containing 0, not that there is no infinite set. > > There is a simple way to sidestep this controversy. Suitably extend > the language of ZF-Inf to admit the set D, where > > D = {x: An (x not in P_n(0))} Good idea. > Here 0 is the null set, n ranges over the non-negative integers and > P_n(0) is the power set operation iterated n times on 0 with P_0(0) = > P(0). > > Note that by definition, D does not include any hereditarily finite > set, but it will contain every other set. Good point. > Consider the theory F = ZF-Inf+{D=0} It is clear that F will only > admit models with hereditarily finite sets. Use the theory F instead > of ZF-Inf+~Inf in my post. Obviously. C-B > > > > > > And every time this comes up, I want to say _fine_ -- so if > > ZF-Inf+~Inf _doesn't_ prove that every set is finite, then there > > should exist a model M of ZF-Inf+~Inf in which "there is an > > infinite set" is true, even though "there exists a set containing > > 0 that is successor-inductive" is clearly false (assuming, of > > course, that ZF is itself consistent), just as the fact that ZFC > > doesn't prove CH implies that there is a model of ZFC in which > > CH is false (once again, assuming that ZF is itself consistent). > > > Yet no one seems to accept the existence of this model M. > > > Either this model M exists, or ZF-Inf+~Inf really does prove that > > every set is finite. There are no other possibilities. > > > So let's settle this once and for all. Assuming that ZF is > > consistent, I ask: > > > 1. Is there a proof in ZF-Inf+~Inf that every set is finite? > > 2. Does there exist a model M of ZF-Inf+~Inf in which "there > > is an infinite set" is true? > > > Notice that exactly one of these questions has a "yes" answer > > and exactly one has a "no" answer. (Actually, come to think > > of it, since the base theory is ZF and not ZFC, it's possible > > that the answer to 1. is "yes" if by "finite" we mean one > > type of finite, say Dedekind finite, and "no" if we mean some > > other type of finite. In this case, I'd like to know which > > types of finite produce a "yes" answer.) > > > If 1. is "yes," then I hope that I will never again see a post > > claiming that ZF-Inf+~Inf doesn't prove that every set is in > > fact finite. In fact, I'll go as far as to suggest that if 1. > > is "yes," then those who claim that ZF-Inf+~Inf doesn't prove > > that every set is finite deserve to be called five-letter > > insults -- if posters are going to call those who deny the > > proof of Cantor's Theorem by five-letter insults, then those > > who deny the proof of "every set is finite" in ZF-Inf+~Inf > > also ought to be called the same. > > > If 2. is "yes," then what I'd like to know is how can I take > > _advantage_ of this fact? Suppose I want to consider a theory, > > based on ZF-Inf, which actually proves that an infinite set > > exists, yet also proves that no successor-inductive set > > containing 0 exists. > > I am also very much interested in knowing the outcome of this > controversy. A quick look at Wikipedia, > > http://en.wikipedia.org/wiki/Axiom_of_infinity, > > throws up the following; > > \begin{quote} > Indeed, using the Von Neumann universe, we can make a model of the > axioms [of ZF] where the axiom of infinity is replaced by its > negation. It is V_\omega \!, the class of hereditarily finite sets, > with the inherited element relation. > \end{quote} I couldn't've said it better myself. > > > > > > > In the current Tony Orlow thread, there is a discussion about > > whether TO is defining N+ to be a successor-inductive set. It > > is possible that the theory that I mentioned above might be > > useful to discussing TO's ideas. > > > But of course, we can't proceed until we know, once and for > > all, whether ZF-Inf+~Inf proves every set to be finite or not.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 27 Jun 2010 13:03
On Jun 27, 9:12 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Jun 26, 10:41 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> Charlie-Boo <shymath...(a)gmail.com> writes: > >> > On Jun 25, 10:21 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> >> People say that atoms are made up of subatomic particles. But you > >> >> can't make atoms up from protons, because they repeal each other. So > >> >> why would people think this? > > >> >> This is a great argument, because the class of protons is a subset of > >> >> the class of subatomic particles, just as the theorems of PA are a > >> >> subset of the theorems of ZFC (with suitable extension of the language > >> >> of ZFC). > > >> > "with suitable extension of ZFC" > > >> > Yikes! > > >> Yes. The usual language of ZFC does not have a successor function > >> symbol, while the language of PA does. Thus, we must extend *the > >> language* of ZFC and also add a defining axiom for the successor > >> function. > > > "add an axiom" > > > Yikes! Yikes! > > You might want to learn about conservative extensions of a theory. Any > time you add a function symbol to a language, you must also add a > defining axiom to the theory if you want the function to be defined. You should already have that axiom as a theorem. You are adding Peano's Axioms, one at a time (as opposed to the standard way of all at once when a set for N is defined.) > In "good" cases, one can prove that the extension is conservative. > > Utterly standard. What difference would that make? C-B > Wikipedia has pages on both "Conservative extensions" and "Extensions by > definitions". > > -- > "Witty adolescent banter relies highly on the use of 'whatever.' > Anyone out of high school forced to watch more than an hour of > 'Laguna Beach' might possibly feel the urge to beat themselves about > the head with a large stick." -- NY Times on an MTV reality show- Hide quoted text - > > - Show quoted text - |