How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: MoeBlee on 1 Jul 2010 13:16 On Jul 1, 11:17 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jul 1, 12:02 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > I've given you a reference. > > Yes, and as a Conservative you know the creed that to name a book is > to defeat your enemy - especially if it is very hard to obtain. (And > to make sure, refuse to quote from it!) WHAT are you talking about? I'm not a conservative. I didn't say I "defeated" you by naming a book. You asked for a reference and I gave you one. You asked for reference. I'm not a bookseller who knows what's hard or easy to obtain. I just happened to notice that Hinman is at least one book that provides a reference for this matter. And it seems you got the book quickly enough anyway. And whatever I said about quoting from it, it's understandable that I'd rather let you read for yourself since (1) It's hard to quote in ASCII all his speical symbols, plus give all the special definitions of his notation. (2) The whole page itself is not comprehensible if one has not read and understoof the steps leading up that page, some of which go back perhaps hundreds of pages previous in the book. (3) I didn't even say that I'm an expert in Hinman's own exposition. Indeed, I only gave it as a reference if you wish to study for yourself. Personally, I've studied mostly for other books, and from that study am able to prove Z |- Con(PA) myself. (4) I don't have unlimited time to tutor you in this subject. I gave you a reference; I didn't thereby also promise to help you read it. You seem to THRIVE on making these conversations as unproductive as possible. I need to stop (I'm BEGGING myself to stop). MoeBlee
From: MoeBlee on 1 Jul 2010 13:21 On Jul 1, 12:11 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > the assertion that "more than finitely many natural numbers > are not upper bounds for N" translates precisely to "Infinitely many > natural numbers are exceeded within N". Okay. > This is the intuition behind the existence of nonstandard natural > numbers, by the way. There are lots of routes to nonstandard models. I don't know how you determined the above is "the intuition". (And that is not a request that you post a bunch more of your claims about NAFL.) One simple route to non-standard models is the compactness theorem. MoeBlee
From: R. Srinivasan on 1 Jul 2010 14:03 On Jul 1, 10:21 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jul 1, 12:11 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > the assertion that "more than finitely many natural numbers > > are not upper bounds for N" translates precisely to "Infinitely many > > natural numbers are exceeded within N". > > Okay. > > > This is the intuition behind the existence of nonstandard natural > > numbers, by the way. > > There are lots of routes to nonstandard models. I don't know how you > determined the above is "the intuition". (And that is not a request > that you post a bunch more of your claims about NAFL.) > > One simple route to non-standard models is the compactness theorem. > Of course. The route I mentioned is via Edward Nelson's Principle of Idealization in his Internal Set Theory. RS
From: billh04 on 1 Jul 2010 14:24 On Jul 1, 10:47 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jul 1, 2:18 am, billh04 <h...(a)tulane.edu> wrote: > > > > > On Jul 1, 1:03 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > On Jul 1, 1:20 am, billh04 <h...(a)tulane.edu> wrote: > > > > > On Jun 30, 11:34 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > > > On Jun 30, 5:23 am, Frederick Williams <frederick.willia...(a)tesco.net> > > > > > wrote: > > > > > > > Charlie-Boo wrote: > > > > > > > The best way to explain that a ZFC axiom is not used is to give the > > > > > > > proof without using any ZFC axioms - good luck! > > > > > > > > How would you prove the PA axioms in ZFC, then? You keep saying it > > > > > > > isn't from an axiom but can't say how it is done - so how do you know > > > > > > > it isn't? > > > > > > > As Tim Little says elsewhere in the thread, you define S, +, and * in > > > > > > the language of ZFC and then prove the counterparts of PA's axioms as > > > > > > theorems. It's done in Suppes (probably without C). > > > > > > Which requires the ZFC equivalent of Peano's Axioms (the axiom of > > > > > infinity.) > > > > > I'm not sure that I understand what you are saying. > > > > > Let me expand what you are saying a little bit: > > > > > === Your statement ======== > > > > To prove the counterparts of PA's axioms as theorems requires the ZFC > > > > equivalent of Peano's Axioms (the axiom of infinity.) > > > > ======================= > > > > > Assuming this is correct, let me expand it a little bit more: > > > > > === Your statement ======== > > > > To prove the counterparts of PA's axioms as theorems requires a ZFC > > > > axiom (or seveal ZFC axioms) equivalent to Peano's Axioms (the axiom > > > > of infinity.) > > > > ======================= > > > > > If this is what you mean, then why do you think so? > > > > > Here is our statement, similar to yours, but not as strong: > > > > > === Our statement ======== > > > > To prove the counterparts of PA's axioms as theorems requires a ZFC > > > > axiom (or seveal ZFC axioms) that imply Peano's Axioms. > > > > ======================= > > > > > For example, we need to prove the following theorem in ZFC: > > > > > Theorem. (all x in N)(all y in N)(if S x = S y then x = y) > > > > > Here N is a set (defined as mentioned before). > > > > Here x and y are sets since everything in ZFC is a set. > > > > Here S is a set (the successor function). > > > > Thus, the theorem is a statement about sets. > > > > ZFC can proves statements about sets using the ZFC axioms. > > > > > Why do you think this theorem cannot be proved using the ZFC axioms? > > > > If it could be done without the axiom of infinity, then the axiom of > > > infinity wouldn't be needed in ZFC. Maybe it is, but I doubt that. > > > But the axiom of infinity is a ZFC axiom. So, I am allowed to use it > > in proving a theorem in ZFC. > > > It seems that you are just saying that we are cheating when we say > > that we can prove statements in ZFC that correspond to the statements > > of the axioms of PA > > No no no. This whole thing got started when I said that ZFC just > consists of PA plus some dinky little axioms about what sets exist, > and that the latter does not contribute to the proof so ZFC can't > prove PA consistent any more than PA can. I don't understand how this relates to the points that I was trying to make in my post. > > Then those who like to carpet bomb attacked the question of how ZFC > manages to include PA, whether it actually "includes" PA, the > definition of "includes", etc. etc. > > My mistake was to respond. It is of no consequence. > > C-B > > > because the axioms of ZFC imply the statements of > > the axioms of PA. Whether we are cheating or not, we don't deny that, > > and in fact that is pretty much what we are saying. We are not trying > > to hide that the axioms of ZFC imply the statements of the axioms of > > PA. That is what we are claiming. > > > > C-B > > > > - Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -
From: Charlie-Boo on 1 Jul 2010 15:20
On Jul 1, 4:37 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > On Jun 30, 5:23 am, Frederick Williams <frederick.willia...(a)tesco.net> > > wrote: > > > Charlie-Boo wrote: > > > > The best way to explain that a ZFC axiom is not used is to give the > > > > proof without using any ZFC axioms - good luck! > > > > > How would you prove the PA axioms in ZFC, then? You keep saying it > > > > isn't from an axiom but can't say how it is done - so how do you know > > > > it isn't? > > > > As Tim Little says elsewhere in the thread, you define S, +, and * in > > > the language of ZFC and then prove the counterparts of PA's axioms as > > > theorems. It's done in Suppes (probably without C). > > > Which requires the ZFC equivalent of Peano's Axioms (the axiom of > > infinity.) > > You seem to be saying that the axiom of infinity, suitably translated, > is a PA axiom, or maybe even all of the PA axioms collectively. The axiom of infinity was added to give ZFC the capabilities of PA. 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. But all of you ZFC-ites can say if it has been used for anything else. C-B > -- > I can't go on, I'll go on.- Hide quoted text - > > - Show quoted text - |