How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: billh04 on 1 Jul 2010 02:18 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 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 -
From: Ki Song on 1 Jul 2010 02:34 Perhaps an analogy is in order! I feel like what Charlie-Boo is asking people to do is analogous to asking someone to perform the addition: Sqrt{2}+Sqrt{3}, with a decimal precision of 10^(10^(10^100)) place. On Jul 1, 2:04 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > To see it means to have actually created it, and its actual creation > > would answer the very interesting question of whether ZFC can prove > > that PA is consistent even though PA can't. > > We already know the answer. The axioms used in the proof are a > restricted form of comprehension, infinity, union and pairing. Spelling > out the formalization of "PA is consistent" in the language of set > theory would be a tedious, trivial and pointless undertaking, of no > apparent mathematical interest whatsoever. > > > It would substantiate your assertions. Is that of value? > > Well, no. I can't think of any pressing reason I should want to convince > you of anything. > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Frederick Williams on 1 Jul 2010 04:37 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. -- I can't go on, I'll go on.
From: R. Srinivasan on 1 Jul 2010 07:00 On Jul 1, 3:15 am, "K_h" <KHol...(a)SX729.com> wrote: > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > news:46d58d89-34b1-40a9-a5a8-1ee250ba57e3(a)e5g2000yqn.googlegroups.com... > On Jun 29, 8:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 29, 2:09 am, "R. Srinivasan" <sradh...(a)in.ibm.com> > > wrote: > > > > ZF-"Inf'+"~Inf" > > > > That theory entails that every object is finite. And > > > there is no > > > definition of any infinite object possible in that > > > theory. > > > OK. Here I want ~Inf to be stated in the form that you > > mentioned, that is, every set is hereditarily finite. > > Why do you think the axiom of infinity is false? What is > the basis for your belief in ~Inf? To me it is > self-evident that all the naturals exist. > First of all I happen to work in a logic (NAFL) where I have a *proof* of ~Inf. Essentially, if you define truth (as provability) such that all vestiges of Platonism are thrown out, infinite sets will not survive. However, there can and do exist infinite classes, like N, the class of all natural numbers. But quantification over classes is not allowed and classes can only be defined by construction -- there is no "arbitrary" infinite class in NAFL theories. Despite these seemingly severe restrictions, I show that it is possible to define a method for real analysis in NAFL based on translating Euclidean geometry into a theory of finite sets with classes. I also show that the paradoxes of classical real analysis, like Zeno's paradox, Banach-Tarski paradox, etc. will be eliminated in such a system of real analysis. On a more intuitive level, how can we fault the existence of the infinite set N? If you consider the statement "All natural numbers are not upper bounds for N" and ask "how many natural numbers are exceeded by some element of N?", the answer has to be "infinitely many". Yet if infinitely many natural numbers are exceeded *within* N, it seems that the only way out is that N must contain an infinitely large number. This is precisely the intuition that leads to nonstandard models of arithmetic, where there are nonstandard integers that exceed every "standard" natural. To call such numbers "finite" is grotesque, to say the least. Yet that is the only way to save the consistency of classical Peano Arithmetic. We have to sacrifice our well-known and well-accepted intuition of what "finite" means, which is something I am not willing to do. If the above considerations do not already leave a bad taste in the mouth, consider the definition of N as a set. It is an essentially impredicative definition. Here I am talking about the simple basic definition of N, which uses universal quantifiers in an essential way. These quantifiers quantify over an universe that already contain N. That such a definition is "harmless" is a commonly stated assertion. If you think carefully, such a defense of circularity is based on Platonism, namely, that N "really" exists, and our attempted definition only tries to access something that is already "out there" in the universe of sets. Note that we do not have this problem with finite sets, even if these are defined using quantifiers. Because we can always define them predicatively by listing their elements. Here is a post (by Brian Hart) in the FOM newsgroup that says Platonism is essential to defend the impredicative methods used in modern logic, physics, mathematics: http://www.cs.nyu.edu/pipermail/fom/2010-May/014713.html \begin{quote} If one axiomatizes the logical universe (the one containing strictly logical objects such as proper and hyper-classes) impredicativity is a requirement as these objects cannot be defined non-circularly. \end{quote} This post seemed to be the first sensible one in an FOM thread where dozens of badly-off-the-mark posts had appeared earlier. Guess what? Precisely after this post appeared, the "moderator" of FOM, Martin Davis, decided to call off the discussion: http://www.cs.nyu.edu/pipermail/fom/2010-May/014716.html \begin{quote} This discussion has long since reached the point of diminishing returns. Hereafter only messages on this topic judged to be of very special interest will be posted. \end{quote} FOM is supposed to be the newsgroup where all the elite logicians and philosophers ponder the foundations of mathematics and logic. And you can see the narrow-minded intolerance that prevails at that level just by looking at this thread and the manner in which it was closed, without allowing any reply to the post of Brian Hart I quoted above. Regds, RS
From: Frederick Williams on 1 Jul 2010 08:01
"R. Srinivasan" wrote: > On a more intuitive level, how can we fault the existence of the > infinite set N? If you consider the statement "All natural numbers are > not upper bounds for N" and ask "how many natural numbers are exceeded > by some element of N?", the answer has to be "infinitely many". The answer is finitely many. Let's suppose N starts with 0 (if you take it as starting with 1 my comment need only a slight adjustment). Let's take the "some" element to be n. How many natural numbers are exceeded by n? Answer n. E.g. take n = 4: 0, 1, 2, 3, *4*, 5, 6, ... and count the numbers exceeded by 4: 0 < 4, 1 < 4, 2 < 4, 3 < 4, and nothing else is exceeded by 4, so that makes 4. Btw, your "All natural numbers are not upper bounds for N" is more idiomatically expressed as "No natural numbers are upper bounds for N", which I agree with. > Yet if > infinitely many natural numbers are exceeded *within* N, it seems that > the only way out is that N must contain an infinitely large number. -- I can't go on, I'll go on. |