Obections to Cantor's Theory (Wikipedia article)
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From: Daryl McCullough on 26 Jul 2005 07:37 Han de Bruijn says... >Yeah. Sure. It's all so rigorous that they don't even have a clue how >to handle an elementary limit which is the easiest one I've ever done. If you are talking about the proof that (number of evens)/(number of naturals) = 1/2, then what is being rejected is your *terminology*. The claim limit as n --> infinity (cardinality of the positive evens less than n)/n = 1/2 is not in dispute. The interpretation of that claim as a fact about the *size* of the set of evens *is* in dispute. Let A_n be the set of all numbers between 0 and 2*n. Clearly, limit n --> infinity A_n/n = 2 but that doesn't mean that A_n approaches a set that is twice as large as the naturals. Your notion of "set size" is inconsistent. -- Daryl McCullough Ithaca, NY
From: Dik T. Winter on 26 Jul 2005 08:13 In article <MPG.1d4f1bfe605de2db989f80(a)newsstand.cit.cornell.edu> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Dik T. Winter said: .... > > those digits. A short table to show the idea: > > 1 = 1 > > 2 = 2 > > 3 = 11 > > 4 = 12 > > 5 = 21 > > 6 = 22 > > 7 = 111 > > 8 = 112 > > 9 = 121 > > 10 = 122 > > etc. > > In this representation each finite natural number is represented by a > > single finite string. Now how many of such finite strings are there, > > given that the stringlength is unbounded? > > This is precisely the kind of question which cannot be answered. Why can it no be answered? It is the same question as the question about the number of binary strings that have a leading 1 in a finite position. > If you allow infinite strings and values, then I say there are N numbers, > by definition. Which definition? > Now, this is a little different from normal digital number > systems. At first I said "two symbols, S=2", but the problem here is that > the strings have different lengths and cannot be simply extended using > leading zeroes and using S=3, because the zeroes are ONLY allowed to the > the left of the 1's and 2's, so we cannot have some constant infinite L. Indeed, there is no constant L. > So, we can say there are 2^L strings for each L, and therefore > sum(x=1->n: 2^x) strings less than or equal to length x, or 2^(n+1)-2. Yes, right. > Therefore, if 2^(n+1)-2 > =N, then 2^(n+1)=N+2, n+1=log2(N+2), and > n=log2(N+2)-1 digits required for N numbers. What is N here? And why do you switch from >= to =? > In other words, it has one > fewer digits than would be required to make a binary number that is 2 > greater in value. I do not understand what you are telling here. I would formulate it as follows: For all numbers except those of the form (2^k - 1) you need one fewer digit in the dyadic system compared to the binary system. For numbers of the form (2^k - 1) the number of needed digits is the same. > > How would your "number" "N" be represented in > > that representation? > > The largest number would be 2222...222 obviously. Since the digit places > represent powers of two, as in binary, this would appear to be twice as > large as 111...111 in binary, but as noted above, with one less digit we > can make a number that is two greater. This last remark is false (if I do understand its meaning correctly). In the dyadic system you need 4 digits to represent the numbers 15 to 30. In the binary system you need 5 digits to represent the numbers 16 to 31. > When we have one less binary digit, > we are essentially dividing by 2, which we can do by shifting each digit > to the right, or by dividing each digit by 2, since each digit is a 2 in > this case. Indeed in the dyadic system when all digits are 2 we divide by 2 by replacing each digit by 1. But shifting to the right is *not* the same as (truncating) division by 2. That is only the case if the least significant digit is 2. > Then we get 111...111 in normal binary, which is what we expect. > This is the normal number of digits, which is one less than would be > required to represent a number that is 2 greater, since that would require > the additon of another digit. > > That was a hard one, but not impossible. Somthing a bit harder. In the binary system a number is divisible by 2 if and only if the least significant digit is 0 (easy induction proof). In the dyadic system a number is divisible by 2 if and only if the least significant digit is 2 (also an easy induction proof). Now according to the above, N is divisible by 2 if it is written in the dyadic system, but is not divisible by 2 if it is written in the binary system. Now what? > > You are missing the point. Each set "larger" in some sense than the > > naturals as mathematicians think about them is (by definition) > > uncountable. > > Sure, unless you consider sets like the halves or square roots to be larger > than N, which I do. besides, the reals are only "uncountable", without > bijection with the naturals, only because the naturals are not allowed to be > infinite. Yes. So what is arbitrary about the "conflation" of "larger and infinite" with "uncountable", when it just follows? In *your* theory it would be arbitrary, and actually unfounded. Do not confuse the terms when they are applied in *your* theory to the terms when they are applied in standard mathematics. > > Wrong, the diagonal proof proves the same thing as the proof that the > > powerset of a set is larger than the base set. The reals can be put in > > a 1-1 relation with the powerset of the natural numbers. (But only when > > you consider natural numbers in the sense of the mathematicians.) > > And if you allow infinite naturals, then the naturals can also be put in 1-1 > correspondence with the power set of the naturals via binary strings. Can it? How come when you earlier stated: > > > You know, if the only conclusion drawn from Cantor's proofs was that > > > a power set is necessarily larger than its base set, I would have > > > absolutely no problem. You have absolutely no problem with this conclusion, nevertheless you also state that you can prove in some way that the sets have equal size. I think this is a contradiction. The point is, when you allow "infinite" naturals, then those naturals can be put in 1-1 correspondence with the power set of the *finite* naturals. There is no problem with that. But they can not be put in 1-1 correspondence with the power set of the enlarged set of naturals. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Han de Bruijn on 26 Jul 2005 08:48 Daryl McCullough wrote: > Let A_n be the set of all numbers between 0 and 2*n. Clearly, > > limit n --> infinity A_n/n = 2 > > but that doesn't mean that A_n approaches a set that is twice > as large as the naturals. Finally! That's *exactly* what boggles my mind! > Your notion of "set size" is inconsistent. Maybe. But it means that there is NO smooth transition from the finite to Cantor's "infinite". And this is precisely what anti-Cantorians find unacceptable. IMHO an "infinite" set cannot consistently have a "size". Han de Bruijn
From: Tony Orlow on 26 Jul 2005 09:37 imaginatorium(a)despammed.com said: > Tony Orlow (aeo6) wrote: > > imaginatorium(a)despammed.com said: > > > > > > > > > Tony Orlow (aeo6) wrote: > > > > Daryl McCullough said: > > > > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > > > > > > > > >>> There is nothing in Peano's axioms that states explicitly that all > > > > > >>> natural numbers are finite. > > > > > > > > > > Let's get more specific, and consider the sets S_n = the set of all > > > > > natural numbers less than n. Are you claiming that there is a natural > > > > > number n such that S_n is not finite? > > > > Actually, no, I am saying that for all finite n, S_n is finite, and that, if > > > > all n in N are finite, then so is N. Conversely, since S_n is finite for finite > > > > n, if S_n is infinite then n is infinite. > > > > > > > > > > What definition of finite are you using? > > > > Less than any infinite number. Not infinite. With a known end, or bound. I am > > > > sure you know what I mean. > > > > > > Hmm. Been here before... > > > > > > Does "with a known end" mean that you can name the known end of the set > > > of pofnats, which you tell us is finite? And tell us what its successor > > > is? And explain how exactly there is no contradiction with the Peano > > > axioms? > > The contradiction come from your decalration that they are all finite. > > Hmm. This is getting tricky. Look, for the purposes of argument, I am > accepting the notional existence of your 'set' of 'natural numbers', > the Tonats, some of which are infinite, and some of which are finite. I > am then *defining* a different set, a subset of the Tonats, which I > call the pofnats, and I am defining this set to include only those of > the Tonats that are finite. That means that when I say "the pofnats", > this expression could be replaced by the expression "the subset of the > Tonats containing only those which are unambiguously finite". This > replacement is totally impractical, but does not change the meaning at > all. So when you say the problem is in my "declaration", I'm not sure > what you mean. Can I not select those of the Tonats that are finite? Sure you can talk about such a set. You just can't draw very much in the way of conlusions about its size or upper bound. > > OK, I know what your getout is here: to you being "finite" is something > like being "small" or being "interesting". 139 is ever so slightly less > finite than 3 is. Well, this is why people keep asking you what you > mean. They can't understand your definition of "finite" if it means > that some numbers are a bit less finite than others. When people say > they can't understand your terminology, please have the decency to > understand that this means that they really *can't* understand what you > are saying. I'll try, but there are so many times when it seems my words are deliberately misrepresented, and grammer is misconstrued, and the point obfuscated, and that it is a simple defenseive maneuver. If people just say "define!", then it seems like not-picking nonsense. If they as a specific question, then I can clarify, but it's hard to answer questions that aren't asked, and I am sick of answering the same questions repeatedly due to forgetfulness or something. So.... "Finite" means it has an end, whether its a set, quantity, process, structure, or whatever. Infinite means it doesn't. In an infinite loop as implied by inductive proof, which defines the entire unending set, you may only be adding a finite number at each step, but over an infinite number of steps, it becomes an infinite addition. When I say the number is getting "less finite", that simply means bigger, or closer to infinity. A finite number of finite steps won't get you to infinity, but an infinite number of finite steps will, as will a finite number of infinite steps. A finite times an infinite is infinite. The best way i can think of defining "finite" is to say it has an end. > > But this "gradated finitude": really, how does it work? It sounds like > you could rephrase "finite" as "close to 0". In some (rather woolly) > sense, 10^10^100 is "close to zero", when we think about the interval > [10^10^100, 47^47^470]. Yet 13 is clearly closer to zero than > 10^10^100. How could this have anything to do with "having an end"? The > interval [0, 13] has a left end at 0 and a right end at 13. The > interval [10^10^100, 47^47^470] has a left end at 10^10^100 and a right > end at 47^47^470. Who but Mueck could think these ends are in any way > less endular than any other end? Less endular? Hopefully, my statement above clarified my ill-worded "less finite". Yes, the finites are around zero, and the infinites are around infinity. So, as a number increases, it is on the way from zero towards infinity. That's how I see it. Infinity is the inverse of zero. That's how people found it to begin with. > > * Sorry, I'm probably using 'interval' nonstandardly. By [n, m], n and > m pofnats, m>n, I mean the set {n, n+1, ... m-1, m} > > So if you want to use "gradated finitude", we really need a definition, > or a first draft of a definition. I keep suggesting that a reasonable, > if informal, definition of a finite set, that does not rely on any > formal set theory, is that a set is finite if it can be counted against > a ditty, and the ditty stops. If the ditty does not stop, the set is > infinite. Any claim that there can be some half-way house at which the > ditty 50% stops is quite beyond my comprehension, without a very large > amount of explanation. Well, you just said it. The finite ditty "stops". It has an end. As you count upward, at each step you are a little farther along on the trip to infinity, even if you are still infinitely far from it in finite-land. It's not that the ditty "50% stops", but that you have traveled some of the way through the ditty already, and are no longer at the beginning, at zero. > > > > I have > > already repeatedly pointed out the flaw in that proof. Go ask your infinite- > > series friends. They'll tell you. > > You keep wittering about infinite series, and posting links to a > Mathworld page on tests for convergence. The relevance of this is hard > to determine. In particular, there is no "proof" at all, in what I just > said. I selected a set, the pofnats; you appear to accept that this set > exists; you claim it is finite; you suggest this means it has a last > member; I ask how this is compatible with the Peano axioms, which say > that any pofnat has a successor, since this appears to imply that the > successor is also a pofnat. This appears to me to be not a proof, but > simply a contradiction, and I ask how you resolve it. No, I NEVER claimed it has a last member. Stop listening to Virgil's lies about what I have said. Any synopsis on his part is bogus. I have repeatedly said there is no greatest finite number. That notwithstanding, if all whole numbers are finite, then the number of them is also finite, since each is 1 greater than the last, and an infinite number of them, according to an infinite series of 1's, which do NOT have a limit of zero, must diverge and have an infinite sum. It's like saying I can move a peg 1 hole to the left, and infinite number of times, and it will still be a finite number of holes from the starting point. It just ain't true. > > You also appear to wish that there are - somewhere in the world of > normal maths - people (the "infinite series" crowd) who could explain > all these errors in "Cantorianism". Why don't you find one? Try your > local college, or whatever. After all, if you are going to be able to > "destroy" set theory, you can expect to be at least as famous as JSH > appeared to expect to be, and that's Quite Famous. Yeah maybe after I'm dead, or finally give up trying. This isn't my only gig. It's just a major issue that I see needs fixing. I am tired of being incorrectly corrected and accused of being dense when I correct in turn. Is this an ego thing? No, I walked away from cardinality 25 years ago and said "let them have that, I don't want it", and studied other ideas. It's not until I see this being argued here with such fervor, and see the level of confusion over these things, and start to understand how pervasive this is, that I feel motivated to do something about it. I really don't have time for this to be honest, but I get a little obsessive when it comes to studies of infinity and zero. I think it's really important mathematically and spiritually to grasp these things properly, and I have a hard time when I see people barking down others that are, as far as I can see, absolutely correct in their point. Oh well, I'm out for the rest of the day. > > Brian Chandler > http://imaginatorium.org > > -- Smiles, Tony
From: David Kastrup on 26 Jul 2005 09:51
Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > Daryl McCullough wrote: > >> Let A_n be the set of all numbers between 0 and 2*n. Clearly, >> limit n --> infinity A_n/n = 2 >> but that doesn't mean that A_n approaches a set that is twice >> as large as the naturals. > > Finally! That's *exactly* what boggles my mind! > >> Your notion of "set size" is inconsistent. > > Maybe. But it means that there is NO smooth transition from the > finite to Cantor's "infinite". Of course not. > And this is precisely what anti-Cantorians find unacceptable. IMHO > an "infinite" set cannot consistently have a "size". This is a _perfectly_ valid point of view (as opposed to the views of those you sympathize with which are wildly inconsistent). After all, talking about the size of something that maps 1:1 to a proper subset of itself sounds disingenious. So far, so well. But it also turns out that among the infinite sets (which can map 1:1 to a proper subset of themselves if they want to) there are some that you can't map 1:1 on others, no matter how much you try. And that makes it convenient to assign labels to their equivalence classes (those that _can_ be mapped 1:1), and try to establish some sort of order between those classes as long as this does not lead to inconsistencies. Whether or not you use the words "size" or "number" "cardinality" or "surjection equivalence class" or whatever else, there still appears to be a criterion according to which you can introduce another tiny bit of order. It is clear that if you view this as a form of "size", then the usual arithmetic conventions don't work with it. So you might prefer not to label it in this manner at all. But that more or less is a matter of notation instead of substance. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |