How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: George Greene on 10 Jul 2010 11:55 On Jul 9, 11:32 am, Marshall <marshall.spi...(a)gmail.com> wrote: > An interesting read! And far less stupid than I had feared. > But ultimately I found it unconvincing. Of course you did! "In this talk, I give reasons for taking seriously the possibility that contemporary mathematics, including Peano Arithmetic, may be inconsistent". Well, No, He Doesn't, at least not about the PA part. The rest of it is (obviously!) different! The PROOFS OF CONSISTENCY of PA may not be finitistic, but PA itself IS (or at least it is innocent of infinity in terms of what it can prove); in order to prove the consistency of T, you basically have to be "stronger than" T (counterexamples dismissed with "basically"; they weren't basic enough).
From: George Greene on 10 Jul 2010 12:42 On Jul 9, 11:32 am, Marshall <marshall.spi...(a)gmail.com> wrote: > > No need to. The link is right here: > > >http://www.math.princeton.edu/~nelson/papers/hm.pdf We could have another thread on this paper. Nelson is not going to stoop to sci.logic, but the point is, what he is saying is sufficiently crankISH that the people who usually disagree with the cranks might be able to disagree with Nelson (or agree critically) in a way that might be intellectually productive (unlike what we are doing with the cranks), about issues that may actually be legitimate (which has hitherto been obscured for us, by the crankiness of the cranks).
From: George Greene on 10 Jul 2010 14:02 On Jul 9, 11:34 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jul 8, 10:00 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > > On Jul 8, 6:15 pm, Marshall <marshall.spi...(a)gmail.com> wrote:. > > > > And I [don't] trust your ability to faithfully represent > > > others' views > > > No need to. The link is right here: > > >http://www.math.princeton.edu/~nelson/papers/hm.pdf > > An interesting object lesson there. Cranks can't even begin to > appreciate Nelson's paper I don't appreciate it either. This all appears to depend on two things: > One might be tempted to try to do so by induction, > but the induction axioms of arithmetic were postulated for formulas ' > of the specified language of arithmetic and is a counting number is not > in that language. He thinks it is somehow legal to augment theories where on axiom is SPECIFIED VIA AN AXIOM-SCHEMA with new predicates that DO NOT FALL UNDER the schema! He acts as though the original specification of the theory was a fixed infinite collection of axioms and that the schematic letter in the original specification was not important or relevant! In point of actual fact, if you wanted to be doing anything consistent with the spirit of the original theory, if you were to introduce a new predicate or functor into the signature of the language, then that predicate or functor would HAVE to fall under the schema. Theories where the schema applies to some formulas and not all are obviously going to be a completely different beast. There is a monotonicity property about what happens when you add axioms that makes this sort of tolerable, but again, that was about adding axioms under the assumption that THE LANGUAGE WAS FIXED! With Nelson, you don't JUST add an axiom: you add something new to the language and then REdefine the SCHEMA to be INcomplete, ex post facto! I don't think it's worthy of any consideration beyond incompleteness. He has another claim that evaluation of "super"exponentiation is "circular" and that it might therefore somehow be consistent to assume that some primitive recursive function is not total. I don't find any of this to be adequately explicated so far. If anybody can give me reference to anybody besides Nelson regarding the difficulty with evaluating"super"exponentiation, I would try.
From: R. Srinivasan on 12 Jul 2010 16:49
On Jul 7, 6:29 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jul 6, 3:29 pm, "K_h" <KHol...(a)SX729.com> wrote: > > > "Transfer Principle" <lwal...(a)lausd.net> wrote in message > > > And according to Srinivasan, the theory NBG-Infinity, with NAFL as > > > the underlying logic, proves that N=On and so N is a proper class, > > How so? Why wouldn't NBG-infinity be agnostic on this issue? > > Normally, in FOL, it would be. Apparently, there is something > peculiar about NAFL such that one can prove that N isn't a > set in NAFL. Srinivasan, the inventor of NAFL, can explain > further how NAFL proves that N is a proper class. > > I have been busy with my day job, so I could not respond earlier. ******************************************************************************** WARNING: This is a fairly long post in which I outline my response to issues raised by K_h and Transfer Principle ******************************************************************************** I will give an outline here of the NAFL proof that infinite sets cannot exist.. First we need the following result: Let P be a proposition that is undecidable in a consistent NAFL theory T (i.e., neither P nor ~P is provable in T). Then T does not prove ~(P&~P). (*) The argument for (*) is as follows. A purported refutation of P&~P in T, when T does not refute either P or ~P, would have to be via one of the following arguments: 1. P -> ~~P (i.e., "If P, then it is not the case that ~P") 2. If ~P, then ~P (i..e., "If ~P, then it is not the case that P") 3. Suppose P&~P. Then we may deduce an arbitrary proposition from this hypothesis, which is absurd. Therefore ~(P&~P). The first two arguments, based on classical logic, are refuted via the axiomatic nature of truth in NAFL theories. Thus 1. would be refutation of P&~P in the theory T+P, and 2. would be a refutation of P&~P in the theory T+~P. Hence neither of these arguments can be made within the NAFL theory T. Here, as I noted in a previous post, NAFL must treat "If P...." (or "If ~P ....") as axiomatic declarations of P (or ~P), for there is no Platonic truth in NAFL. Now consider 3. This would essentially be the intuitionistic argument for ~(P&~P). This is refuted in NAFL by noting first of all that an arbitrary proposition is not deducible from P&~P in NAFL. Also note the following interesting fallacy in 3. It rejects the hypothesis P&~P via the observation that an arbitrary proposition can be deduced from it, which leads to an absurdity. Then the conclusion ~(P&~P) seemingly follows. However, this conclusion, namely, ~(P&~P), has also been deduced from an argument that starts with the hypothesis P&~P, as is clear from 3. Since anything can be deduced from the hypothesis P&~P, why should we believe that 3. is a valid derivation of ~(P&~P)? NAFL does not permit the hypothesis P&~P in any valid proof, and so 3. cannot be a valid derivation of ~(P&~P) in NAFL. From (*) and the completeness theorem (which is accepted as a valid metamathematical result in NAFL), it follows that: If P is a proposition that is undecidable in a consistent NAFL theory T, then there must exist a non-classical model M for T in which P&~P is the case. (**) In this non-classical model M, 'P' is interpreted as "~P is not provable in T" and '~P' is interpreted as "P is not provable in T". Thus one can see that 'P&~P' is indeed true in this non-classical sense in M. Since truth is provability in classical NAFL models of T (in which either P or ~P would be true via provability in T+P or T +~P), one can also interpret the non-classical model as expressing that "P (~P) is neither true nor false in a classical sense". Armed with these results, let us now get to the business of refuting the existence of infinite sets in NAFL. Consider the NAFL theory F of finite sets (the classical equivalent would be NBG-Infinity). Consider the class D defined by D= {x: An(x not in P_n(0))} (***) Here P_n(0) is the power set operation iterated n times on the null set 0 and P_0(0)=P(0). As mentioned in a previous post, D will exclude all hereditarily finite sets by definition and include all other sets. The theory F, in contrast to its classical equivalent (NBG-Infinity) must prove that D=0. The proof of this result is as follows. To get a contradiction, suppose that D=0 is undecidable in F. Then by (**) there must exist a non-classical model M for F in which (D=0 & D=/=0) is the case. But such a non-classical model for F cannot exist, for (D=0 & D=/=0) would contradict the axiom of extensionality, which is an axiom of F. Therefore we conclude that D=0 cannot be undecidable in F. We also know that there exists a model for F in which D=0 is true. It follows that D=0 must be provable in F, and hence infinite sets cannot exist in any consistent NAFL theory. There are many subtleties that I have avoided mentioning for the sake of brevity. For example, note that I have used the following result: If a proposition P is not undecidable in the NAFL theory F, then it must be either provable or refutable in F. This is not an intuitionistically valid result, but holds in NAFL, where the notion of undecidability (or provability) of a proposition P in a theory F is taken as a Platonic truth (i.e., every valid proposition in the language of a consistent NAFL theory F is intrinsically either provable or refutable or undecidable in F, independent of any proof of this claim). It follows that the notion of undecidability (or provability) of a proposition P in the language of F is not formalizable within F, because truth for formal propositions in the language of F must be equated with provability in NAFL. Essentially the same argument shows that If F is a consistent NAFL theory, then there cannot exist any F- undecidable propositions (****) Note that (****) is the direct opposite of Godel's theorems. In a sense this is not surprising, for NAFL contradicts Godel by equating truth with provability. It follows from (****) that there cannot exist any nonstandard models for F. In particular, there cannot exist any nonstandard integers in NAFL, and we can see in hindsight that the definition of D in (***) via the hierarchy P_n(0) is perfectly valid. There would be an ambiguity with the definition of P_n(0) if and only if we had any doubt about what "finite" means in "n is finite" and this is the case classically. The issue of defining "finite" in a classical context has been addressed by David Libert in a separate post. In summary, I believe that NAFL has emerged as a serious alternative and challenger to classical/intuitionistic/constructive logic. NAFL attempts to address shortcomings in these logics. > > > > Let's review some basics about proper classes. Logically, a proper class has a > > "cardinal" or "ordinal" size of absolute infinity. (I use the equivocal quotes > > there because it's technically an abuse of those terms.) So the idea is that a > > proper class has the largest size possible and so there cannot be another class > > with an "ordinal" or "cardinal" size bigger than that. But you CAN get a bigger > > "ordinal" object than N. For instance, consider the sum of (1/2)^n for all n in > > {0,1,2,3,...}. Each natural number n provides one term in the sum: > > 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... > > When this is summed for all naturals we get 2. Can we now add more? Yes we can: > > we can add the number 3 to 2 to get 5 and then we can add the number 4 to 5 to > > get 9. So it makes sense to say that the number 3 is the w term in the sum and > > the number 4 is the w+1 term in the sum. Ordinals simply mean do this first, do > > that second, etc, in a given order -- e.g. addition. That's the motivation > > underlying the ordinals. If N were a proper class then its "ordinal" size would > > have to be absolute infinite and, in this example, the number 4 would correspond > > to an "ordinal" term greater than absolute infinite! That is plainly a > > contradiction and this simple fact is one reason why N really must be a set. > > Interesting. Still, I'm curious as to how a _finitist_ like > Srinivasan would react to an _infinite_ series such as: > > 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... > > Presumably, a finitist would be opposed to the mere > existence of infinite series, and so would be unmoved > by the use of infinite series to prove that N is a set. > > In the NAFL theory F of finite sets, infinite proper classes do exist provably. But these are essentially constant symbols and quantification over proper classes is not allowed. At first sight, this appears to be a severe restriction and seemingly rules out any possibility of doing real analysis in NAFL. However, this is not the case. The theory F of finite sets can define natural numbers and the NAFL counterpart of Peano Arithmetic (which I will denote as NPA).. Also F can define negative integers and a theory of rational numbers Q. Sequences of rationals may be represented as infinite (proper) classes. However, there is no such thing as an "equivalence class" of such sequences of rationals, and so the question arises as to how these rational sequences can be used to define real numbers. The short answer to this question is that we do not really define real numbers in NAFL. Instead their existence is taken for granted as "points" on the Euclidean real line. Indeed the diagrammatic constructs of Euclidean geometry are taken as apriori existent in the human mind and not capable of any further justification.It turns out that we may *translate* these diagrammatic concepts into the NAFL theory F to get a NAFL version of real analysis. This has been done in the paper: http://arxiv.org/abs/math.LO/0506475 Real numbers are encoded into F as Cauchy sequences of rationals. Thus the square root of 2 can be represented by any such Cauchy sequence converging to sqrt(2). Such a Cauchy sequence is an infinite (proper) class in F. We may define addition and multiplication of reals in a straightforward manner. Intervals of reals can also be represented by sequences of rationals. For example, the interval [0,1] is represented by a sequence of rationals having precisely [0,1] as its limit points. It follows that open intervals of reals cannot exist and indeed, it is impossible to define any such open interval geometrically (or diagrammatically). Cantor's diagonal argument will not go through in the NAFL version of real analysis because of the ban on quantification over proper classes. Further dy/dx *is* 0/0, where the zeros in the numerator and denominator are sequences representing real numbers. The objection that "0/0 i s undefined" is answered with the rebuttal that we merely *define* 0/0 in a particular way whenever we deduce an expression for dy/dx. It turns out that the NAFL version of real analysis rules out the paradoxes of classical real analysis, in particular those of measure theory, like the Banach-Tarski paradox. Also Zeno's paradoxes cannot be formulated in the NAFL version of real analysis. Because the NAFL theory F does not tolerate any undecidable propositions, (see ****) non-Euclidean geometries do not exist in NAFL. In summary, there is no need for infinite sets to do real analysis in NAFL. This post is already too long, so I will stop here. RS > > > (As an aside, note that there are sci.math posters, > including Tony Orlow and tommy1729, who don't > accept the existence of the ordinal omega+1 as > used in K_h's post, yet both do accept that N is a > a set and that P(N) is a larger set. Go figure.) > > > > K_h states that the main reason to make a distinction between > > > sets and proper classes is Cantor's Theorem. Thus, in a theory > > > which refutes Cantor's Theorem, such as NFU, we expect there > > > to be no distinction between sets and proper classes. Indeed, > > > we see that this is exactly the case -- NFU proves that even V > > > is a set. > > Why do you think NFU refutes Cantor's theorem instead of being agnostic on it? I > > am just curious -- whatever you believe is your own business. > > Let's find out more about NFU: > > http://math.boisestate.edu/~holmes/holmes/nf.html > > "Russell's paradox: "x is not an element of x" is not a stratified > predicate. But the universe, V = { x | x = x }, does exist in NF and > its > subsystems known to be consistent (see below). > > "Cantor's paradox of the largest cardinal: Cardinal numbers are > defined > in NF as equivalence classes of sets of the same size. The form of > Cantor's theorem which can be proven in Russell's type theory asserts > that the cardinality of the set of one-element subsets of A is less > than > the cardinality of the power set of A. Note that the usual form > (|A| < |P(A)|) doesn't even make sense in type theory. It makes sense > in NF, but it isn't true in all cases: for example, it wouldn't do to > have > |V| < |P(V)|, and indeed this is not the case, though the set 1 of all > one-element subsets of V is smaller than V (the obvious bijection > x |-> {x} has an unstratified definition!). > > "Sets with the property that the set P1(A) of one-element subsets of A > is the same size as A are called "Cantorian" sets; sets with the > stronger > property that the restriction to A of the singleton "map" x |-> {x} > exists are > called "strongly Cantorian" sets." > > Let's see what NFU has to say about the set N of natural numbers: > > "The set of natural numbers is provably Cantorian; the assumption that > it is > strongly Cantorian, known as Rosser's Axiom of Counting, is known to > strengthen NF (if it is consistent -- one of the few independence > results not > obtained by permutation methods) and known to be consistent with NFU. > Cantorian and strongly Cantorian sets clearly satisfy Cantor's theorem > in > its usual unstratified form (|A| = |P1(A)| < |P(A)|)." > > So N is Cantorian, and therefore card(N) < card(P(N)), just as it is > in > standard theory. > > Even though I was referring to posters other than Srinivasan when I > started discussing NFU, the following line caught my eye: > > "Ali Enayat and Solovay have some new results relating the strength of > NFU + "the universe is finite" and some of its extensions to systems > of arithmetic." > > NFU + "the universe is finite"? Unfortunately, the link doesn't > discuss > what any of these results are exactly. |