Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 27 Jul 2005 16:30 In article <MPG.1d518c097a07ce5989fa7(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Daryl McCullough said: > > Tony Orlow writes: > > > > >Barb, you're not saying anything new. I have heard it all before. I am not > > >drawing my conclusions in any such confused way > > > > Uh, yes you are. Why don't you write down what you consider to be > > valid axioms for working with infinite sets, and then try to write > > a formal proof of your claim > > > > If a set S of strings is infinite, then S contains some infinite > > strings. > > > I guess you can't understand a proof if it has natural language. Your > statement > is not quite right. This is what I have proven: > > Given an alphabet of finite size S, an infinite set of unique strings made > from > this alphabet must contain infinite strings. > > Here's my two axioms: > > (1) N=S^L, where N is the size of the set of ALL strings of length L, > constructed from an alphabet of size S. N is the upper limit on the number of > strings from a set of S symbols and a maximum length of L. > > (2) A^B is finite if and only if A and B are finite. > > Given (2), S^L is finite iff S and L are finite. > Given finite S, S^L is finite iff L is finite. > Given (1), N is finite iff L is finite. > Conversely, N is infinite iff L is infinite. > > ... > > Well, I just got back from getting lunch, during which it occurred to me that > it would be good to put this proof in inductive format, and watch you try to > discredit it while hanging on to your proof that all naturals are finite. > This > proof relies on some assumptions. Let me know if you disagree with any of > them: > > (1) All naturals are finite. (you already "proved" this one) > (2) The number of strings of length L that can be constructed from a set of > symbols of size S (set size, not symbol size, remember (sigh)), is S^L. > (3) For finite A and B, both A^B and A+B are finite. > > Are we good to go? Okay. > > Proof that f(n), the number of strings in the set of all strings up to and > including length n in N, on a finite alphabet of size S, is finite: > > 1. For n=1, we have S^1 strings of length 1, for a total of S strings less > than > or equal to 1 in length. f(1)=S is finite, as stated. > > 2. For a finite number of srings of length n or less, we can add S^(n+1) > strings of length n+1, to get f(n+1),the number of strings of length n+1 or > less. S is finite and n+1 (a natural number) is finite, so S^(n+1) is finite. > S^(n+1) is finite and f(n) is finite, so f(n+1)=f(n)+S^(n+1) is finite. > > 3. Therefore, for all n in N, f(n), the number of strings up to and including > n > symbols, which are created from a finite alphabet, is finite. There is no n > in > N, in other words, which maximum length of string will allow you to have an > infinite set of strings. That requires that if a set of finite strings is finite that some finite string in it cannot be made longer by appending another character to its end. Which is clearly ridiculous. > > I eagerly await any response on this, but I won't hold my breath. Oh do. It might force some blood to your brain.
From: Daryl McCullough on 27 Jul 2005 16:18 Tony Orlow (aeo6) wrote: > >Virgil said: >> Okay, TO! Why is the limit on the size of finite string lengths smaller >> than the limit on size of finite sets of finite strings? > >Maybe you should ask Stephen, since I never made any such claim. That's because you never explore the *consequences* of the things you do claim. That's the difference between you and a competent mathematician. >He just made my point anyway, when he said "How many binary strings >are there then? 1 + 2 + 4 + ... + 2^L, which we all know is >2^(L+1)-1, which is finite". If you have >finite lengths only, then you have a finite set. Thanks, Stephen. Stephen was pointing out how nonsensical your claim was. You are saying that there is a maximum string length L. The number of strings of that length or smaller is M (where M = 2^(L+1)-1). M is bigger than L. But it's still finite, right? So why can't you strings of length M? -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 27 Jul 2005 16:38 Dik T. Winter said: > In article <MPG.1d5044b418d0c79d989f94(a)newsstand.cit.cornell.edu> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > Dik T. Winter said: > ... > > > 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 last 2 should have been 1.) > > > 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. > > Lessee. In the dyadic system 8 = "112". Shifting one position to the right > gives "11" = 3. I would not call that dividing by 2. Uh, wouldn't it give 11.2, or 3 and 2/2, or 4? Yep, it's still dividing by 2. Whew! > > > To > > declare the set the same size as a proper subset is a lot of what offends > > the sensibilities of anti-Cantorians, > > Not all anti-Cantorians. There are also those who state that oo+1 = oo, > which is in essence the same. Those are pre-cantorians. I am post-cantorian. :D > > > 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. > > Well, bijections provide an excellent way to compare sets, because it is > an equivalence relation, and so they define equivalence classes. The denote equivalence for finite sets, and some kind of relation for infinite sets, but not necessarily equivalence. > > > > 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? > > Pray explain your mapping, it is not clear. > Define each subset as an infinite binary string, where the first bit denotes membership of the first natural, the second bit denotes membership of the second natural, etc. Each subset int he pwoer set of the naturals is thus represented by a unique infinite binary string. Each finite binary string is accepted as corresponding to a finite natural, but the infinite strings, with infinite values, are rejected on the basis that natural numbers cannot have infinite values. If they were allowed to have infinite values, the the set of all infinite binary strings which represents the power set of the naturals could be put in bijection with the set of natural numbers, and you would have a bijection between the naturals and their power set. -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 16:40 Dik T. Winter said: > In article <MPG.1d5025572ee38731989f8e(a)newsstand.cit.cornell.edu> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > Dik T. Winter said: > ... > > > > > Back on your horse again. Tell me about the binary numbers > > > > > (extended to the left with 0's) where the leftmost 1 is in a > > > > > finite position. Are all those numbers finite? Are there > > > > > only finitely many of them? > > > > > > > > > yes and yes > > > > > > I think you should apply for the reward for solving Collatz' problem. > > > > > I just looked that up. Does this have anything at all to do with that > > problem? > > Your assertion at least makes Collatz' problem decidable. > How so? Maybe we should collaborate. ;) -- Smiles, Tony
From: Daryl McCullough on 27 Jul 2005 16:24
Tony Orlow (aeo6) wrote: >imaginatorium(a)despammed.com said: >> You claim, simultaneously that: >> >> (a) There is no largest pofnat. >> (b) There are only a finite number of pofnats. >> >> You suddenly became very explicit (elsewhere in the thread) and appear >> to agree that to say a set is *finite* means that it can be counted >> against a ditty, and the ditty stops. >> >> OK, so arrange the pofnats in normal ascending order, and count them >> against a ditty. When the ditty stops, you have reached the last >> pofnat, which is therefore the largest. This is a consequence of (b), >> and contradicts (a). > >How about you tell ME where the ditty stops, and the values for the naturals >end, if you're so sure they are all finite? You are asking Brian to explain *your* theory to you? Brian isn't claiming that there are only finitely many finite naturals, you are. -- Daryl McCullough Ithaca, NY |