Obections to Cantor's Theory (Wikipedia article)
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From: malbrain on 26 Jul 2005 13:56 Tony Orlow (aeo6) wrote: > Robert Kolker said: > > Han de Bruijn wrote: > > > > > > True. That's why: > > > > > > A little bit of Physics would be NO Idleness in Mathematics > > > > You are pissing and moaning that a peach is not a pear. Mathematics is > > deductive. Physics is empirical. Mathmematics is about the relation of > > ideas. Physicis is about how the real world works. Two completely > > different things. > > > > Bob Kolker > > > Mathematics is NOT purely deductive. Before you can apply your rules to the > facts at hand, you have to develop the rules. Right. Mathematics has a history as all crafts do. I'm hopeful that we can interest one of the mathematicians on the list to enlighten us on the history of the axiom of induction -- from what it came from. karl m
From: Tony Orlow on 26 Jul 2005 14:02 Dik T. Winter said: > 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. Yeah, it's finite, but that's all that can be said. > > > If you allow infinite strings and values, then I say there are N numbers, > > by definition. > > Which definition? The definition of N as the number of positive integers, the unit discrete infinity. > > > 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. Indeed..... > > > 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 =? I think that '>' was accidentally moved from the following line or something. It shouldn't be there. N is still the number of positive integers, and n is the number of digits required to represent that many numbers in this system. > > > 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. Well, now you have to determine what fraction of them are of each form, how many digits are required for those numbers in binary, and what the resulting sum is. I guarantee you get the same result, if you do it right. > > > > 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. Okay, I left out a detail here, which is important on the finite scale, but not for the ultimate conclusion about the infinite set. Log2(x) does not produce integers generally, but we want an integer result. We round up the result to the nearest integer using ceiling(). So, if we require n=log2(x+2)-1 digits to represent x numbers, we can see that for 2 numbers we require ceiling(log2(4))- 1, or 1 digit. For 30 numbers we need ceiling(log2(32))-1=5-1=4 digits. For 31 we need ceiling(log2(33))-1=6-1=5 digits. The ceiling() function is important for correctly measuring the digits required for finite initial segments of the set. > > > 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. That's not true. If each digit place represents a power of the base, then shifting everything to the right one digit divides the denoted value of that digit by the number base. Here, each place denotes a power of 2, and moving all digits to the right one place divides the entire number by two, perhaps with remainder. > > > 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? That IS an interesting question. I have a piece of brain working part-time on the concept of divisibility of infinite numbers, and haven't a conclusion yet on that. In some sense, N seems prime to me, and yet in another, that seems almost impossible. I don't have an answer to thta question, but I think it may turn out to be unanswerable. At some point soon I'll do more playing with infinite division. Maybe I'll have an answer for you. > > > > 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. Okay, I guess. It just seems like counting is a dismal apporach to infinity, since it never gets there, and that the only reason reals are considered less countable in digital format is because naturals aren't allowed to be infinite. Powersets of naturals seem perfectly enumerable to me. > > > > 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. Uh, no. You misunderstand. Bijection is useful if you keep in mind your mapping, and use it to determine relative infinities. In itself, it does not prove equal size. That is a serious conflation right there. While it may be true that bijections between finite sets prove equal size, the very definition of an infinite set is that it can have a bijection with a proper subset. To declare the set the same size as a proper subset is a lot of what offends the sensibilities of anti-Cantorians, and to declare that this is the case for infinite sets, rather than to admit that bijections of themselves do NOT prove equal size in that case, is a baffling choice on the part of the mathematical community. > > 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. Why not? The "enlarged set" goes on forever, and the set of subsets goes on forever, and can be put into a linear order based on membership of elements, which can be seen to correspond with the naturals that go on forever. Where do you see a problem? > -- Smiles, Tony
From: Tony Orlow on 26 Jul 2005 14:11 imaginatorium(a)despammed.com said: > > > Tony Orlow (aeo6) wrote: > > Martin Shobe said: > <big snip> > > > > BTW, there is a caveat on convergence. You have to assume the > > > standard topology. In other topologies, that sequence can converge. > > > You mean with a ring? That's really not what we're talking about, unless you > > agree that the number line is a circle, and even then it's not relevant. In > > pure quantitative terms, a sum of infinite 1's is infinite. > > Tony, could you please clarify: when you use the word "ring", what do > you mean? > > (a) The algebraic structure known by mathematicians as a ring > (b) Something else (in which case please call it a T-ring) > (c) You're sure it is (a), but cannot actually sketch the axioms for a > ring (a) > > If you select (c), please confirm you really meant (a) by sketching the > axioms. > If you select (a), please suggest why you think the name "ring" is > used. > > (Since I really have no idea, I'd be interested in informed comments on > the last question.) > > Brian Chandler > http://imaginatorium.org > > > > > > > So, please make up your mind. Do we increment to get a successor an infinite > > number of times, or only a finite number of times, to get N? > > > > > > Martin > > > > > > > > > > -- > > Smiles, > > > > Tony > > I am not an expert in rings, nor am I going to sketch the axioms that you know better than I, nor does it help the conversation when you snip the original statement was repsonding to, which had soemthing to do with numbers being their own multiples of more than 1, or something. It didn't make sense in the context of normal quantitative addition. It was a vague guess as to what the idea was. The sum of an infinite number of 1's is infinite. That's all. -- Smiles, Tony
From: Tony Orlow on 26 Jul 2005 14:25 imaginatorium(a)despammed.com said: > > > Tony Orlow (aeo6) wrote: > > Daryl McCullough said: > > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > > >> "Bigger" in the sense of no surjection from the "smaller set to the > > > >> "larger", is one thing, "bigger" in the sense of having the "smaller" > > > >> set as a proper subset is different. While these two measures happen to > > > >> coincide for finite sets, they do not coincide for infinite sets, as the > > > >> definition of infinite for sets should hint to you. > > > >> > > > >gee, they coincide for finite sets AND infinite sets, under Bigulosity > > > >but I don't suppose you consider that extra consistency any sort of > > > >progress. > > > > > > No, that's no progress at all. You can prove that for finite sets > > > if there is a bijection between set A and set B, then they have > > > the same Bigulosity. But that fails for infinite sets. So Bigulosity > > > is an inconsistent notion of "size". > > > > > > -- > > > Daryl McCullough > > > Ithaca, NY > > > > > > > > Huh? Bigulosity doesn't rely on bijections. That why it DOES work. > > Remind us how you determine the bigulosity of the set > { 1, 1/2, 1/4, 1/8,...} > > Brian Chandler > http://imaginatorium.org > > Okay, I don't know what I was saying. It relies on bijections, but doesn't consider any bijection to mean equality. The mapping functions are used to determine relative size when it comes to numeric sets. Sorry. -- Smiles, Tony
From: Randy Poe on 26 Jul 2005 14:27
malbrain(a)yahoo.com wrote: > Randy Poe wrote: > > malbrain(a)yahoo.com wrote: > > > David Kastrup wrote: > > > > Of course not. But the basis for choosing an axiom is irrelevant to > > > > the application of the axiom. And this axiom was chosen exactly in a > > > > manner that does not require recursive application. > > > > > > When working a proof one chooses one's axioms based on the desired > > > result. How can you say that the choice is irrelevant to the > > > application? > > > > Eh? > > > > Can you give an example that you think illustrates this > > "choosing one's axioms based on the desired result"? Do > > you think different number theories decide which of the > > Peano axioms to accept and which not, or add additional > > axioms of their own, based on a "desired result"? > > Perhaps terminology is misleading us. I'm talking about applying > axioms in a proof, and you seem to be talking about including them into > the system in the first place. That's why we're asking for the history > of the axiom of infinity v. axiom of induction. > > > If you want to make a contribution to the same theory > > that everybody else is working on, you start from the > > same axioms. If you choose different axioms, you aren't > > working on the same system. > > We're discussing the system of natural numbers. > karl m |