Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Shmuel (Seymour J.) Metz on 20 Jul 2005 07:59 In <dbjqog$tg7$1(a)xenon.Stanford.EDU>, on 07/19/2005 at 09:19 PM, amorgan(a)xenon.Stanford.EDU (Alan Morgan) said: >No, you didn't. You proved this for finite sets only. You claimed, >without proof, that this result applied to infinite sets. The result does apply to infinite sets. There is no largest natural in an infinite set of naturals, and hence any statement about the size of "the largest natural" is vacuously true. Unfortunately, tony doesn't understand that proving flying pigs are reptiles is not useful when there are no flying pigs. -- Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel> Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap(a)library.lspace.org
From: Tony Orlow on 20 Jul 2005 10:01 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> > >> > David Kastrup said: > >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> >> > >> >> > Alec McKenzie said: > >> >> >> "Stephen J. Herschkorn" <sjherschko(a)netscape.net> wrote: > >> >> >> > >> >> >> > Can anti-Cantorians identify correctly a flaw in the proof > >> >> >> > that there exists no enumeration of the subsets of the > >> >> >> > natural numbers? > >> >> >> > >> >> >> In my view the answer to that question a definite "No, they > >> >> >> can't". > >> >> >> > >> >> >> However, the fact that no flaw has yet been correctly > >> >> >> identified does not lead to a certainty that such a flaw > >> >> >> cannot exist. Yet that is just what pro-Cantorians appear to > >> >> >> be asserting, with no justification that I can see. > >> >> >> > >> >> > Even though every subset of the natural numbers can be > >> >> > represented by a binary number where the first bit denotes > >> >> > membership of the first element, the second bit denotes > >> >> > membership of the second element, etc? > >> >> > >> >> Well, what number will then represent the set of numbers dividable by > >> >> three? > >> > 100100...100100100100 > >> > >> > Of course, you will argue that this infinite value is not a > >> > natural number, since all naturals are finite, but that is > >> > clearly incorrect, as it is impossible to have an infinite set of > >> > values all differing by a constant finite amount from their > >> > neighbors, and not have an overall infinite difference between > >> > some pair of them, indicating that at least one of them is > >> > infinite. > >> > >> You have not shown such a thing, and of course it would be > >> inconsistent with the Peano axioms defining the naturals. > > > > That is simply not true. > > Sulking won't help. Don't be a jerk. > > > There is nothing in Peano's axioms that states explicitly that all > > natural numbers are finite. > > It is an immediate consequence. Read on. > > > The fifth axiom, defining inductive proof, is used to prove this > > theorem, but it is a misapplication of the method. > > An axiom is not a "misapplication". No, the proof is a misapplication of the axiom. The inductive proof method works well for constant equalities, but in the case of proving that all naturals are finite, you are incrementing the value at each of an infinite number of steps and maintaining that the value remains finite, because you are looking only at individual steps. When you maintain something is "finite", that is to say it is less than any infinite number, but if it grows by a constant finite amount at each of an infinite number of steps, then it has grown by an infinite amount. Basically, adding X repeatedly Y times is the same as adding Y repeatedly X times, so adding 1 and inifnite number of times is the same as adding infinity once. Now, I have heard the argument that inductive proof does not prove things for an infinite number of steps, but only a finite number, but if this is the case, then it does not prove anything for an entire infinite set of natural numbers. Either you agree that there are an infinite number of steps involved, or that the set of naturals is finite, or that inductive proof does not prove a property true for all natural numbers as Peano stated. > > > I offered, and you saw, a deductive proof that proves that the > > largest natural in a set must be at least as large as the set > > size. > > But the set of naturals does not have a largest element. So you can > prove any property you want about it, like it having webbed feet and > dancing swing polka with a vengeance. Doesn't make a difference, > since no such beast exists anyway. At each step in the proof, the set has a largest element which is precisely the same as the size of the set. This constant equality holds for all such sets defined by ANY natural number in the set. Most generally, it is impossible to have ANY set of natural numbers, finite or infinite, which has a number of elements that is greater than its largest element value. This should be clear to anyone who thinks about it. It is impossible to have an infinite set of strings, of which digital numbers are a type, without having either an infinite base, or an infinite number of digits. You have not refuted that proof either. Therefore, all the hand-waving in the world has not convinced me that I am in error at all. Besides, as Petry said, those of us on this side of the fence, while our perspectives may vary considerably, actually share a good deal of overlap in our objections. For instance, WM may be a neo-finitist, and I may be a nested-infinity thinker, but we certainly agree on the basic finite relationship between nodes and paths in a binary tree, finite or infinite, while Cantorians devise proofs that can show one or the other to be "uncountable", depending on their moods. > > > So, which inductive proof do you believe? You cannot add 1 an > > infinite number of times to your maximal element, > > There is no maximal element, and the Peano axioms don't define > "infinite number of times" or similar processes. If Peano's fifth is correct, and inductive proof applies to the entire infinite set in a stepwise manner, then there are indeed an infinite number of steps implied. It's an "immediate consequence", as you said above. > > > without it achieving an infinite value. I provided two other proofs > > that an infinite set of naturals must include infinite values, which > > were dismissed, but never refuted, or any flaw pointed out. > > You are delusional. Insults are really a waste of time. Grow up. > > > I will have my web pages published before too long, so I am not > > getting into a mosh pit with you again right now. Just be aware that > > anti-Cantorians are sick of being called crackpots, and the day will > > soon come when the crankiest Cantorians will eat their words, and > > this rot will be extricated from mathematics. > > Oh good grief. Successor in interest to JSH, are we? I don't know really what James Harris' point was. I think you fellows ran him out of here before I had a chance to see his points. Maybe they were junk. I don't know. But, I am tired of being told things about infinity and infinite sets that make no sense, and are proven using assumptions that are unfounded, and told that the Banach-Tarski result is a "paradox" and not a proof by contradiction. There really seems to be a bad influence going on that allows people to think they have an infinite language, when they only allow finite strings, or that the number of paths in a binary tree, which is always half of the number of branches, is suddenly infinitely larger than the number of branches, when those numbers become infinite. I simply can't understand how mathematics has come to accept such illogical results as "counterintuitive" rather than incorrect. > > >> >> Let's take the number representing the set of numbers dividable > >> >> by three. Is this number dividable by three? > >> > > >> > Why does it have to be? > >> > >> It does not have to be. But if it is a natural number, it either > >> is dividable by three, or it isn't. You claim that it is a natural > >> number. So what is it? Is it dividable by three, or isn't it? > >> > >> It must be one, mustn't it? > > > > The number is 100100...00100100. It's certainly even, and a multiple > > of 4. Is it divisible by 3? That can only be determined in a binary > > system with a finite number of digits, as far as I can > > tell. > > Oh, certainly not. Natural numbers are defined by the Peano axioms. > Now let us define the set of all numbers with a well-defined remainder > from division by 3: > > a) 0 has a well-defined remainder of 0 > > b) If n has a well-defined remainder of 0,1,2 respectively, > S(n) has a well-defined remainder of 1,2,0 respectively. > > c) different numbers n with well-defined remainder have different > successors S(n) with well-defined remainder > > d) 0 is not the successor of any natural with well-defined remainder > > e) if a set contains 0, and for each of its elements x contains S(x), > then this set contains all numbers with well-defined remainders. > > Shiver me timbers, looks just like we have the Peano axioms here. So > all natural numbers have a well-defined remainder from division by 3. > Looks like 100100...00100100 is not a natural number. > > If it is, just point out which of the above laws is wrong. Well, it would appear that, despite the insistence of Cantorians that all we know about finite sets goes out the window for infinite sets, even such basic obvious facts as proper subsets having a smaller size, they would like to insist that infinite numbers behave precisely the same way as finite numbers. Does this seem a little unfair to you? Why do you insist that eveything one can do with a finite number should work exactly the same for an infinite number? The determination of divisibility by a number which is prime relative to the number base depends on a termination to the process of division. > > > Infinite whole numbers aren't always as convenient as finite ones, > > but they still must exist for the set to be infinite, > > Says you. Says the nature of digital numbers, and says the Peano axioms, which define an infinite number of successors, each a constant finite quantity greater than the last. It's an "immediate consequence". The sum of an infinity of 1's is 1 infinity. > > > and they still can be used to represent infinite subsets of the > > naturals. > > Unfortunately not subsets in- or excluding themselves at will. Whether the number that is represented by a binary string which represents a subset is a member of that subset is irrelevant. There is no will involved, but a simple method for representing set membership as a number. So, what are you saying here? > > > If you divide this number by 3 (11) you find it is divisible or not, > > depending on whether you have an odd or even number of 100's in your > > infinite string. Of course, this question is not really answerable, > > so I don't have an answer for you. What do you think? What is > > aleph_0 mod 3? > > Oh, I never claimed that aleph_0 was a member of the natural numbers, > so I don't need to make claims about aleph_0 mod 3. But the answer to this question depends on the number of digits in the infinite number. If that number is a multiple of 6, then the number representing the set is divisible by 3, but that isn't really decideable, is it? Perhaps not everything that can be determined regarding a finite number can be determined for an infinite number. If you can allow discrepancies between you finite and infinite sets, I am not going to apologize for not being able to perform a mod operation on an infinite number. It is still a fact that they must exist for the set to be infinite, says me. > > -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 10:08 Robert Low said: > Tony Orlow (aeo6) wrote: > > Stephen J. Herschkorn said: > >>To those who insist there is a smallest positive real number? > > 000...000.000...001 > > And how many 0's are there after that decimal point? N? log_2(N)? N-1? > N+1? (Whatever the hell any of those answers mean...) > :D Gee that depends on the number base. Let's say it's binary, and let's say there are N whole numbers. Then there are log2(N) digits to the left of the binary point, and the same to the right. As long as it's infinite, it doesn't really matter does it? Perhaps we should say there are as many digits as the solution to the harmonic series, which is considerably smaller than N. That would do as well. There are a lot of infinities. These kind of power set relations can be nested indefinitely. Once we enumerate all the subsets of N in terms of binary numbers, we can enumerate the power set of those, the set of all sets of subsets of N, in the same way. Similarly, it seems to me, that infinitiies and zeroes can be nested in ways that allow for infinitesimals and infinite layers of infinities. -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 10:12 Alan Morgan said: > In article <MPG.1d473242aff1e303989f30(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >That is simply not true. There is nothing in Peano's axioms that states > >explicitly that all natural numbers are finite. The fifth axiom, defining > >inductive proof, is used to prove this theorem, but it is a misapplication of > >the method. I offered, and you saw, a deductive proof that proves that the > >largest natural in a set must be at least as large as the set size. > > No, you didn't. You proved this for finite sets only. You claimed, without > proof, that this result applied to infinite sets. > > Alan > I proved it for all n in N, which I think you agree is an infinite number of naturals. I think you were perhaps one of those claiming that inductive proof only works for finite iterations, but then it wouldn't work for the infinite set of naturals, now, would it? I am not going in these stupid circles with you. Again this goes back to the Cantorian mantra, "no largest finite", which is in direct contradiction with your omega, a smallest infinite, rgarding which the exact same logic applies. Neither one exists, and it is really pointless to spedn your time trying to find the "line" between finite and infinite. -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 10:18
David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> > >> > David Kastrup said: > >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> >> > >> >> > There is no reason to expect the natural number corresponding to > >> >> > the subset to be a member of that subset. > >> >> > >> >> There is no such expectation. The only expectation is that _every_ > >> >> natural number is _either_ a member of its corresponding subset, or > >> >> not. > >> > Okay. > >> >> _Depending_ on that, the constructed subset will either _not_ or > >> >> _do_ contain the number, respectively. > >> > Redundant, but okay. > >> > >> Sure it is redundant. But you should be the last person to complain > >> about your need to get this hammered bit by bit into your skull. > >> > >> >> This constructed subset then does not correspond to _any_ natural > >> >> number. > >> > >> > This is where it goes wrong. > >> > >> Uh yes. That's the conclusion: that the assumption of the mapping > >> goes wrong, since it fails to cover the constructed subset. > >> > >> > The consructed subset corresponds to the natural number denoted by > >> > the binary string constructed from the right to the left, where each > >> > successive bit is a 1 if the successive natural is a member, and 0 > >> > if it is not. > >> > >> You are assuming a particular mapping, the proof holds for _any_ > >> mapping. So let's see where your mapping takes us. Since 2^n>n > >> always, the subset we get is simply N itself, at least as long as we > >> assume that N only contains finite numbers. However, that is an > >> assumption that you are not willing to make, so let is call this > >> particular subset of N by the name Q. > >> > >> So what natural number corresponds to Q according to your logic? I'll > >> take a daring guess at your muddled thoughts and suppose 111...111 or > >> something like that. Is 111...111 itself a member of Q? > > > Well, that's an interesting question. yes, it is 111...111, and > > 111...111 is also a member of N, but those two are a little > > different. The first is a map of membership, and has N number of > > 1's. The second, is a digital number, and presumably denotes > > N-1. Therefore, it has log2(N) 1's, rather than N. To represent 2^N > > reals, you need N digits, but to represent N naturals, you only need > > log2(N) digits. > > So is would not seem like the "111...111 that is a map of membership" > could itself be contained in the "111...111 that is a map of > membership" since then you'd need "2^N digits" instead of just the "N" > you have. > > Of course, this is complete and utter bullshit, since N is not a > natural number to start with, but even if we dive into your utter > bullshit with a vengeance, it does not hold water. Nice language. I'm going to tell your mother. You are correct that the mapping string for the complete set cannot represent a number within the set, as it is 2^x-1, where x is the number of elements in the set. This is true for finite AND infintie sets. Otherwise, what does Cantor's diagonal proof for the reals prove? Are the reals not a larger set than the naturals? > > >> > It is so basic, I cannot even quite see what erroneous assumption > >> > you are making. It seems like you are assuming subset number X > >> > must contain X as a member. If so, how do you justify this > >> > assumption? > >> > >> I don't. But I assume that for every X, subset number X must > >> either contain number X as a member, or doesn't. And if it does, > >> subset Q will not contain number X as a member, and if it doesn't, > >> subset Q will contain number X as a member, and so subset Q differs > >> from every subset X at the position of number X. > > > > This is essentially the diagonal argument in disguise. You are > > assuming you have the same number of numbers as digits, which is > > clearly a false assumption when using any base greater than 1. > > Fine. So you agree that there can't be a surjection of the naturals > onto the subsets of N. Thanks for ceding that point. Now was this so > hard? I don't think I ceded anything there. I am refuting the validity of the diagonal argument's conclusion that one CANNOT list or order the reals. > > -- Smiles, Tony |