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 12:50 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 -
From: Charlie-Boo on 27 Jun 2010 13:14 On Jun 27, 12:50 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > 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. Wikipediahttp://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-mathemati... Hey Frederick, I bet you $449.94 that Gentzen's book doesn't contain a proof that PA is consistent, carried out in ZFC. You on? I'll pay via PayPal or Western Union, and will absorb any delivery charges. Or do you say things that you don't believe? C-B > > > > 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.- Hide quoted text - > > - Show quoted text -
From: William Hale on 27 Jun 2010 13:22
In article <dbebec73-757b-4de5-ae60-d5f6a6ab8830(a)t10g2000yqg.googlegroups.com>, Charlie-Boo <shymathguy(a)gmail.com> wrote: > 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. ZFC does not invoke the Peano Axioms to represent N. Textbooks may mention the Peano Axioms when they show how ZFC incorporates what it deems to be the natural numbers, but this mention of PA is only for putting things in a historical perspective or to give some motivation for what is going on in ZFC, but it is not necessary for ZFC to have any mention of PA in order to develop natural numbers. A textbook could mention how Euclid developed natural numbers (for historical purposes or to show similarities etc), but this does not mean that ZFC invokes the five Euclidean postulates of geometry. > Good point! > > C-B |