Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 20 Jul 2005 10:28 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: > >> >> > >> >> > David Kastrup said: > >> >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: > >> >> >> > >> >> >> > David Kastrup <dak(a)gnu.org> wrote: > >> >> >> > > >> >> >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: > >> >> >> >> > It has been known for a proof to be put forward, and > >> >> >> >> > fully accepted by the mathematical community, with a > >> >> >> >> > fatal flaw only spotted years later. > >> >> >> >> > >> >> >> >> In a concise 7 line proof? Bloody likely. > >> >> >> > > >> >> >> > I doubt it had seven lines, but I really don't know how many. > >> >> >> > Probably many more than seven. > >> >> >> > >> >> >> It was seven lines in my posting. You probably skipped over it. It > >> >> >> is a really simple and concise proof. Here it is again, for the > >> >> >> reading impaired, this time with a bit less text: > >> >> >> > >> >> >> Assume a complete mapping n->S(n) where S(n) is supposed to cover all > >> >> >> subsets of N. Now consider the set P={k| k not in S(k)}. Clearly, > >> >> >> for every n only one of S(n) or P contains n as an element, and so P > >> >> >> is different from all S(n), proving the assumption wrong. > >> >> > >> >> > I still do not get this. You have a set of naturals {0,1,2,3...}, > >> >> > and a set of binary numbers {0,1,10,11,100,101,110,111,....}. Surely > >> >> > there is a bijection between these two sets. > >> >> > >> >> Fine. > >> >> > >> >> > So, what is the problem if one interprets the binary numbers (with > >> >> > implied leading zeroes) as being a map of each subset, where each > >> >> > successive bit represents membership in thesubset by each successive > >> >> > natural number? > >> >> > >> >> Ok, so let's construct P. It is actually easy enough, since n<2^n, > >> >> and so P=N. So what number corresponds to N itself in your mapping? > >> >> > >> > An infinite string of 1's: 1111.....1111. This is (2^aleph_0)-1. > >> > > >> > Infinite whole numbers are required for an infinite set of whole > >> > numbers. > >> > >> And what number corresponds to the subset of N without the number > >> 1111.....1111 in it? > >> > >> > > What is this, 20 questions? It's 01111....1111. Duh! > > And now if you also take out 01111....1111, what do you get then? > > Well, no you've removed an element from the middle (top of the first half), and the representation of that becomes difficult for infinite digital numbers. With the understanding that we have equal numbers of digits to either side, the answer would be 0111...1101...1111. If you're trying to point out that there is an infinite regress and that we get into sticky business with infinite whole numbers, I readily concede that that is true. There is no need to play rope-a- dope. Many things that are easy with finite numbers are not with infinite numbers, but many things that are true of finite sets are dismissed by Cantor when it comes to infinite sets. -- Smiles, Tony
From: David Kastrup on 20 Jul 2005 10:41 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: >> >> >> >> > David Kastrup said: >> >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> >> >> >> >> > 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. That's your job description. >> > 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, What is supposed to be a "constant equality"? > but in the case of proving that all naturals are finite, you are > incrementing the value at each of an infinite number of steps Nonsense. Complete bullshit. I do nothing at all like that. There are no steps whatsoever involved here, and certainly not an infinite number of them. n is finite for n=0, and it is shown that for any finite n, n+1 is finite. That is not an infinite number of steps, but at best 2 steps. > and maintaining that the value remains finite, because you are > looking only at individual steps. That's what induction is about. Showing f(0), and proving f(n)->f(n+1). Two steps, and this covers the naturals. If you want to crosscheck more than that, feel free to do so, but the axioms don't require doing that. That's their whole point: not having to check things piece by piece. > 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. It is irrelevant. The axioms don't talk about steps, and certainly not an infinite number of them. You are babbling what you consider intuitive, but it bears no relation to the math. > 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. Derive that from the axioms. If you can't, it is irrelevant. > 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. There is a strictly limited number of steps involved in an induction proof, and it provides proof for all natural numbers, which happen to form an infinite set. >> > 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. Fine, and so you prove something for a set that has a largest element. The set of natural numbers is no such set, and so your proof does not hold for it. > This constant equality holds for all such sets defined by ANY > natural number in the set. Certainly. Unfortunately, the set of natural numbers is no such set (since there is always a larger number than any given number in it), and so your proof does not apply to it. > 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. If it has a largest element value. The set of natural numbers has no largest element, and so your proof does not apply to it. > 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. Oh, there is an infinite number of digits, but every single string only occupies a finite number of them. >> > 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, There is no "stepwise manner" in the axioms. > then there are indeed an infinite number of steps implied. It's an > "immediate consequence", as you said above. It is an immediate consequence of the axioms that the described set is infinite, since the successor relation gives a bijection to a proper subset. But there are no steps involved at all. You just make them up for the sake of your personal intuition. They are a personal crutch of your own, and cause you to make mistakes that are not inherent in the axioms. >> > 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, Then get a fscking set theory book and learn. That's what everybody else does. > and are proven using assumptions that are unfounded, Axioms are not "unfounded", and not assumptions. > 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. Then look up and understand the definition of an infinite set. It is _the_ decisive mark of an infinite set that it does no longer obey the pigeon hole principle. > I simply can't understand how mathematics has come to accept such > illogical results as "counterintuitive" rather than incorrect. Then get yourself books and learn. >> >> >> >> 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. You got it all confused again. There are no infinite natural numbers, so the "Cantorians" don't insist on anything for those non-existing entities. > 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? There is no infinite natural number. The finite natural numbers are defined by the Peano axioms, and yes, everything that works with finite naturals works with all of them. That is the whole point of defining them with those axioms: to have a body of numbers governed by the same rules. If you don't like it, you can invent your own axioms and numbers, but then you can't cry foul that your laws don't hold with the numbers the others are talking about. > The determination of divisibility by a number which is prime > relative to the number base depends on a termination to the process > of division. Good. And since all natural numbers are finite, every division among those naturals is guaranteed to finish. >> > 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, No. They don't. They only define the existence of a single successor for every number. The infiniteness of the set is a _consequence_ of the axioms, but it is not defined in them. > 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. The Peano axioms are not concerned about sums, neither with a finite nor an infinite number of terms. They only have a nonspecified "successor" relation without inherent arithmetic properties. >> > 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. Not if you are establishing a bijection. >> > 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. Since there are no infinite numbers, I need not answer the question. You claim that there are, so it is up to you to provide an analysis of your claims. > 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. Which is why infinite numbers have not been admitted into the naturals. It would be impractical to have a set of numbers that are not governed by the same laws. > 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. Sulking won't help. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 20 Jul 2005 10:51 Stephen Montgomery-Smith said: > Tony Orlow (aeo6) wrote: > > > I was asked that before, and never got around to fully analyzing the axioms for > > lack of time, but the diagonal proof suffers from the fatal flaw of assuming > > that the diaginal traversal actually covers all the numbers in the list. Any > > complete list of digital numbers of a given length, even a given infinite > > length, is exponentially longer in members than wide in terms of the digits in > > each member. Therefore, the diagonal traversal only shows that the anti- > > diagonal does not exist in the first aleph_0 terms. Of course, the entire list > > is presumed to be aleph_1 long, being a list of the reals, and the antidiagonal > > simply exists on the list, below the line of diagonal traversal. Cantorians > > seem to think infinity is simply infinity, even during the course of a proof > > that that is not the case. > > I got it!!! > > The usual proof starts - suppose that there is a complete countable list > of real numbers. But your rebuttal is amazing in its simplicity - > suppose that there isn't. > > Have you considered the usual proof that there are infinitely many prime > numbers? I think your method might also work to reveal the flaw there > as well. > I love the way mathematikers love to spew insults instead of replies. The conclusion of the proof is that there isn't such a list, which is what I disagree with, so your statement is either totally confused, or deliberate obfuscation, which is more likely. Whatever you "got", you didn't catch it from me. -- Smiles, Tony
From: David Kastrup on 20 Jul 2005 10:53 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: >> >> >> >> > David Kastrup said: >> >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> >> >> >> >> > David Kastrup said: >> >> >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: >> >> >> >> >> >> >> >> > David Kastrup <dak(a)gnu.org> wrote: >> >> >> >> > >> >> >> >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: >> >> >> >> >> > It has been known for a proof to be put forward, and >> >> >> >> >> > fully accepted by the mathematical community, with a >> >> >> >> >> > fatal flaw only spotted years later. >> >> >> >> >> >> >> >> >> >> In a concise 7 line proof? Bloody likely. >> >> >> >> > >> >> >> >> > I doubt it had seven lines, but I really don't know how many. >> >> >> >> > Probably many more than seven. >> >> >> >> >> >> >> >> It was seven lines in my posting. You probably skipped over it. It >> >> >> >> is a really simple and concise proof. Here it is again, for the >> >> >> >> reading impaired, this time with a bit less text: >> >> >> >> >> >> >> >> Assume a complete mapping n->S(n) where S(n) is supposed >> >> >> >> to cover all subsets of N. Now consider the set P={k| k >> >> >> >> not in S(k)}. Clearly, for every n only one of S(n) or P >> >> >> >> contains n as an element, and so P is different from all >> >> >> >> S(n), proving the assumption wrong. >> >> >> >> >> >> > I still do not get this. You have a set of naturals >> >> >> > {0,1,2,3...}, and a set of binary numbers >> >> >> > {0,1,10,11,100,101,110,111,....}. Surely there is a >> >> >> > bijection between these two sets. >> >> >> >> >> >> Fine. >> >> >> >> >> >> > So, what is the problem if one interprets the binary >> >> >> > numbers (with implied leading zeroes) as being a map of >> >> >> > each subset, where each successive bit represents >> >> >> > membership in thesubset by each successive natural number? >> >> >> >> >> >> Ok, so let's construct P. It is actually easy enough, since >> >> >> n<2^n, and so P=N. So what number corresponds to N itself in >> >> >> your mapping? >> >> >> >> >> > An infinite string of 1's: 1111.....1111. This is (2^aleph_0)-1. >> >> > >> >> > Infinite whole numbers are required for an infinite set of whole >> >> > numbers. >> >> >> >> And what number corresponds to the subset of N without the >> >> number 1111.....1111 in it? >> >> >> >> >> > What is this, 20 questions? It's 01111....1111. Duh! >> >> And now if you also take out 01111....1111, what do you get then? >> >> > Well, no you've removed an element from the middle (top of the first > half), and the representation of that becomes difficult for infinite > digital numbers. In short, your nonsense breaks down. > With the understanding that we have equal numbers of digits to > either side, Equal number of digits? What is that supposed to mean when we are talking about an infinite number of them? > the answer would be 0111...1101...1111. Oh, and if we remove 0111...1101...1111 from that set? Don't you realize that the absurdity of your increasingly silly pseudo-constructions is unlimited? > If you're trying to point out that there is an infinite regress and > that we get into sticky business with infinite whole numbers, I > readily concede that that is true. There is no need to play rope-a- > dope. Many things that are easy with finite numbers are not with > infinite numbers, Which is the reason that mathematics has excluded them from the naturals, by decree of the Peano axioms. If you want to deal with infinite strings, use real numbers behind the decimal point, or p-adics before it. But neither obey the Peano axioms. > but many things that are true of finite sets are dismissed by Cantor > when it comes to infinite sets. Because they don't hold. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 20 Jul 2005 10:57
Barb Knox said: > In article <MPG.1d4726e11766660c989f2f(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > [snip] > > >Infinite whole numbers are required for an infinite set of whole numbers. > > Good grief -- shake the anti-Cantorian tree a little and out drops a > Phillite. Here's a clue: ALL whole numbers are finite. Here's a > (2nd-order) proof outline, using mathematical induction (which I > assume/hope you accept): > 0 is finite. > If k is finite then k+1 is finite. > Therefore all natural numbers are finite. > > That's the standard inductive proof that is always used, and in fact, the ONLY proof I have ever seen of this "fact". Is there any other? I have three proofs that contradict this one. Do you have any others that support it? Inductive proof proves properties true for the entire set of naturals, right? That entire set is infinite right? Therfore, the number of times you are adding 1 and saying, "yep, still finite", is infinite, right? So, you have some way of adding an infinite number of 1's and getting a finite result? You might want to discuss this with your colleagues specializing in infinite series. There is a very simple rules that says no infinite series can converge to a finite value unless the terms of the series have a limit of zero as n approaches infinity. Does this constant term, 1, have a limit of zero? No it doesn't, and the infinite series of constant 1's cannot converge, but diverges to infinity. Can you actually deny this? If so, then Poincare was right. -- Smiles, Tony |