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 28 Jun 2010 03:28 On 2010-06-27, Charlie-Boo <shymathguy(a)gmail.com> wrote: > How do you define "PA is a subset of ZF"? "Every sequence of symbols in the set of theorems of PA is a sequence of symbols in the set of theorems of ZF". This statement is obviously false, since there are symbols in the theorems of PA that are not even in the language of ZF. > Can't ZF prove what PA can? Strictly speaking, ZF itself cannot. In an informal sense, it can. You can extend ZF by definitions into a superset (let's call it ZF* following convention earlier in the thread) of PA, which does prove what PA can. In practice almost nobody uses ZF itself, but if you're going to use the terminology you should get it right. - Tim
From: Tim Little on 28 Jun 2010 03:39 On 2010-06-27, Charlie-Boo <shymathguy(a)gmail.com> wrote: > PA is not used but ZFC is? But ZFC invokes the Peano Axioms carte > blanche to represent N ZFC does not invoke the Peano Axioms at all. In fact to say that a formal theory "invokes" anything is at the very least somewhat odd. Mathematicians working with ZFC may invoke axioms and inference rules of ZFC to prove theorems such as the existence and uniqueness of a given set they denote "N". They may then define operations such as "S", "+" and "*" on its elements, and prove that these operations are well-defined. Where in that do they "invoke the Peano Axioms carte blanche"? - Tim
From: Charlie-Boo on 28 Jun 2010 07:56 On Jun 27, 6:31 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Sun, 27 Jun 2010 10:50:16 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > ... > > ZFC declares that there is a set that satisfies Peanos Axioms and > > defines N to be that set. > > You have it completely backwards. The set is defined first, independent > of any mention of PA. So? Doesn't that set satisfy Peano's Axioms? Whether it says it or not doesn't change the question. > One can then prove (in ZF) that that set (and > indeed, infinitely many others) can serve as the domain of a model of > PA. And then doesn't it define N to be that set? C-B
From: Charlie-Boo on 28 Jun 2010 07:57 On Jun 27, 9:40 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Sun, 27 Jun 2010 17:14:02 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > >> > > ZF, for example, proves Con(PA) which, of course, PA does not > >> > > (assuming its consistency). > > > > > You have to assume PA is consistent? > > > > You have to ask? > > > Yes or no, please. > > Yes. (You *really* had to ask?) Do you think PA is consistent and why? C-B
From: Charlie-Boo on 28 Jun 2010 08:01
On Jun 28, 3:28 am, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-27, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > How do you define "PA is a subset of ZF"? > > "Every sequence of symbols in the set of theorems of PA is a sequence > of symbols in the set of theorems of ZF". Narrowminded. The first reference that I grabbed (quoted) says ZFC proves everything PA does. If I have two systems and they both prove P(N) when N is a constant and P expresses an r.e. set and P(N) is true, then aren't they equivalent? > This statement is obviously false, since there are symbols in the > theorems of PA that are not even in the language of ZF. > > > Can't ZF prove what PA can? > > Strictly speaking, ZF itself cannot. In an informal sense, it can. > You can extend ZF by definitions into a superset (let's call it ZF* > following convention earlier in the thread) of PA, which does prove > what PA can. > > In practice almost nobody uses ZF itself, but if you're going to use > the terminology you should get it right. Which terminology is not used right and what should it be? Do you reject my reference? C-B > - Tim |