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 6 Jul 2010 13:34 On Jul 5, 5:08 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > MoeBlee <jazzm...(a)hotmail.com> writes: > > And, in that regard, I have said all along that there is no finitistic > > proof of the consistency of PA. I base that on the second > > incompleteness theorem. > > The first incompleteness theorem already suffices. This elementary point > is often overlooked, and has recently been stressed by Richard Zach. I > said a bit about this in an earlier post, not that it matters much in > the grand scheme of things. I'm sorry I overlooked that post. I always welcome these kinds of sharpenings. I'll try to go back to your post and I'll keep an eye out for Zach's remarks too (would you suggest where I can find them?). > > (I haven't personally verified every detail in a proof of the second > > incompleteness theorem, so my remarks are to the extent we can be > > confident that the second incompleteness theorem does indeed withstand > > complete scrutiny, as it is reported in the literature that it does.) > > Come now, you're surely not hedging your bets and saying you're not > absolutely certain the second incompleteness theorem, I'm on some tough deadlines right now, so I can only give a short answer that deserves some later elaboration. For me, it's not a matter of 'absoluteness' or 'certainty' per se, except as I would need to describe the sense in which I would mean them. My remarks do not at all reflect any special, significant, or important doubts that such and such a formal proof (such as of second incompleteness) does indeed exist. Perhaps later I'll explain later more specifically what I have in mind and why I gave such a disclaimer as above. Anyway, the purpose of my remarks above, the reason for the disclaimer I gave, is not to raise any real question as to to whether second incompleteness is indeed a theorem. > which has > withstood the most exacting and extreme scrutiny humanly possible, far, > far beyond what any run of the mill result in mathematics -- the > classification of finite groups, the Heine-Borel theorem, most of stuff > in analysis, Wiles's proof of Fermat's last theorem, the fundamental > theorem of arithmetic ... -- usually has to endure to prove its good > name, holds. I make the following comment NOT toward any purpose (which I do not have) of claiming need for skepticism regarding second incompleteness. That said, as far as I can tell, the literature on second incompleteness does not seem to me to be as thorough in proof details as you seem to convey. Most textbooks admit to providing only an outline and to omit certain "tedious" sections. As far as I know, there are not many full expositions other than Hilbert & Bernays, which still awaits translation into English. Even a tome such as Hinman's (which tends toward fussy details) provides only such an outline. Pace, again, my point is not to say that this requires important doubts as to theoremhood, but rather just to say that I don't think that, as a matter of literary fact, second incompleteness is accessible to a student's verification as are many other resuts in mathematics. (That there are results that may be even LESS accessible I do not contest.) > (I apologise for the clumsiness in the construction of the > previous sentence. My only excuse is that I'm again suffering of severe > lack of sleep. Aatu, I really hope you can find an ongoing solution for your insomnia. I know what suffering it can be. Maybe exercise? I don't know. But I hope the best for you. P.S. I've been reading over your other post in which you proposed an exercise, though I haven't written a post in response. MoeBlee
From: MoeBlee on 6 Jul 2010 16:47 On Jul 5, 6:46 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Aatu Koskensilta wrote: > > It seems your posts have nevertheless led Nam [...] to incorrect > > ideas about my views. This is not any fault of yours. > > Huh? You didn't say anything in that conversation and silently > let him "represent" your views and if his "representing" goes > wrong that would be my fault, no explanation needed? > > Where are your (and his) sense of credibility and responsibility? vomitous: see above. MoeBlee
From: K_h on 6 Jul 2010 18:29 "Transfer Principle" <lwalke3(a)lausd.net> wrote in message news:e272b12a-3214-487c-beef-ebe1ac54dc74(a)y4g2000yqy.googlegroups.com... > On Jul 3, 5:45 pm, "K_h" <KHol...(a)SX729.com> wrote: >> "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message >> > In NAFL, N is a proper class and there is no >> > such thing as P(N). You cannot quantify over all >> > the infinite sub-classes of N to get P(N), >> > because quantification over proper classes is not >> > permitted. >> Why is N a proper class in NAFL and what is the basis to even have >> proper classes >> in NAFL? The whole reason to bifurcate collections into sets and >> proper classes >> is so that mathematics is self-consistent. That is, the problems >> presented by >> Cantor's theorem (for example the Cantor and Burali-Forti paradoxes) >> can only be >> solved by dividing collections into two types: sets and proper >> classes. (The >> problems presented by Russell's paradox are solved by the axiom of >> foundation, no >> collection can be a member of itself.) So it's because of Cantor's >> theorem that >> such a bifurcation is necessary. If Cantor's theorem is false in >> NALF then on >> what basis does NALF bifurcate collections into the two types? As >> far as I can >> tell, NAFL has no basis to do so and so N should be a set. As a set, >> it is easy >> to show that |N|<|P(N)|. > > But now let's take a look at N. We know that by the von Neumann > definition of natural number, a natural number is a finite ordinal. So > now we ask, how do we know that N isn't equal to the class On of > all ordinals? Yo, transfer dude. If I may kindly ask, why would it be? > 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? > thus forcing K_h to accept inaccessibles! In other words, for me to > require K_h to accept inaccessibles is analogous to requiring a > finitist like Srivinasan to accept infinite sets. Okay. I accept inaccessibles; Srivinason is free to reject infinite sets if he likes. >> It is clear that the subset S of N is not in the list because it is >> different >> from every set in the list according to the axiom of extensionality. >> Recall that >> the axiom of extensionality says that two sets are equal iff they >> contain the >> same elements. The set S={0,3,4,...} is clearly a subset of N > > S is clearly a sub_class_ of N. If N is a proper class, then it's > possible for S to be a proper class as well, but only _sets_ can > be elements of other sets. Therefore, the proof appears to be > invalid unless we can prove that N is a set -- and in Srinivasan's > theory, N isn't a set. 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. > 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. _
From: Transfer Principle on 6 Jul 2010 21:29 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. > 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. (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.
From: Marshall on 7 Jul 2010 22:12
On Jul 7, 7:05 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Aatu Koskensilta wrote: > > Nam Nguyen <namducngu...(a)shaw.ca> writes: > > >> MoeBlee wrote: > > >>> And, in that regard, I have said all along that there is no > >>> finitistic proof of the consistency of PA. > >> So you've agreed that there's no formal proof for PA's consistency and that > >> if you go by formal proof only then you don't have knowledge of PA's > >> consistency. > > > There are many formal proofs of the consistency of PA. None of them are > > finitistic. > > So it doesn't seem you used the phrase "formal proof" in the standard > way that textbook (e.g. Shoenfield's) would use. In that standard > usage, a formal proof is a (finite) syntactical proof of a FOL formal > system theorem. > > Given that standard definition of "formal proof", would you agree with > my statement above that: > > >> there's no formal proof for PA's consistency and that if you go by > >> formal proof only then you don't have knowledge of PA's consistency.. > > ? > > -- > ---------------------------------------------------- > There is no remainder in the mathematics of infinity. > > NYOGEN SENZAKI > ---------------------------------------------------- Give it up, loon. PA is provably consistent. Deal with it. Marshall |