Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 27 Jul 2005 15:42 In article <MPG.1d516f5910dfd0c3989fa3(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Robert Low said: > > Tony Orlow (aeo6) wrote: > > > > > Barb, you're not saying anything new. I have heard it all before. I am > > > not > > > drawing my conclusions in any such confused way, > > > > Differently confused, then. It's hard to tell. > > > > > If I > > > thought the way you are suggesting, then wouldn't I also be claiming that > > > the > > > reals in [0,1) must have infinite values too? > > > > So, consider the rationals in [0,1). Each of them is (by definition) > > of the form p/q, where p is a natural number, q is a natural number > > other than 0, and p and q have no common factors. > > > > Are there rationals in [0,1) where p and/or q have to be infinite? > > > Interesting question. In order to have an infinite number of them, yes, you > would need either an infinite number of numerators or an infinite number of > denominators, both of which are whole numbers. So, you would require infinite > values in either numerator or denominator, in order to achieve an infinite > set. While the question might be interesting to TO, the answer by TO is simply silly. > Now, of course, you would counter that, in order for the value to be in the > interval [0,1) and have either an infinite numerator or denominator, it must > have both, which is true, and that a ratio between two infinite values cannot > be between 0 and 1, but I really have no problem declaring that N/(N+1) is in > that range, even though N is infinite. I mean, it is trivial to prove > inductively that for all n in N, 0 <= n/(n+1) < 1, and for me that includes > infinite n. That means that it is only true for one person, at least with infinite n, and false for everybody else. Given that a,b,c and d are disctinct "infinite naturals", how does one determine which of a/b < c/d, a/b = c/d or a/b > c/d is true? Given two "infinite naturals" how does one determine their greatest common factor? Their product? Given any "infinite integer" how does one find its factorization into primes? Can it have infinitely many prime factors? Can it have infinite prime factors, and if so, how does one determine if an infinite natural is prime or composite? Is the reciprocal of an infinite natural infinitesimal? TO's assertion that there exist infinite naturals creates a lot of questions that he cannot answer satisfactorily.
From: Tony Orlow on 27 Jul 2005 15:45 Martin Shobe said: > On Tue, 26 Jul 2005 14:46:42 -0400, Tony Orlow (aeo6) > <aeo6(a)cornell.edu> wrote: > > >Martin Shobe said: > >> On Mon, 25 Jul 2005 15:52:08 -0400, Tony Orlow (aeo6) > >> <aeo6(a)cornell.edu> wrote: > >> > >> >Martin Shobe said: > >> >> On Wed, 20 Jul 2005 10:57:58 -0400, Tony Orlow (aeo6) > >> >> <aeo6(a)cornell.edu> wrote: > >> >> > >> >> >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? > >> >> > >> >> Yep. > >> >> > >> >> >That entire set is infinite right? > >> >> > >> >> Yep. > >> >> > >> >> >Therfore, the number of times you are adding > >> >> >1 and saying, "yep, still finite", is infinite, right? > >> >> > >> >> Yep. But be careful here, at *every* stage of this process, we have > >> >> still only done it a finite number of times. > >> >Uhhh.... Look at what you just agreed to. The number of times you are adding 1 > >> >is infinite. But, now you say it is always a finite number of times? make up > >> >your mind. > >> > >> It's quote made up. I have chosen to follow the route you so > >> cavalierly ignored. > >?????? Huh? > > There is no finite upper bound on how many times I can apply it. > However, each application occurs after only a finite number of > previous applications. Do you, or do you not, have an infinite set? Do you, or do you not, egenrate one member per iteration. How many iterations are requires to produce an infinite set, one at a time? (sigh) > > >> >> > So, you have some way of > >> >> >adding an infinite number of 1's and getting a finite result? > >> >> > >> >> Nope. You weren't careful. > >> >You contradicted yourself. Do you apply successor and increment the value a > >> >finite number of times, or an infinite number of times? Be careful. > >> > >> There is no end to how much I can apply successor. However, I can > >> only apply it a finite number of times. No contradiction here. > >Except for the fact that somehow you got an infinite set ina finite number of > >steps, producing 1 element at a time. How does that work? > > Do you remember that proof of mine about the existence of a smallest > cardinal? The first few lines of the proof get you one of the more > common constructions of the naturals. I get it in finitely many steps > by showing that there is a set at least that big, and throwing out the > things I don't want. I don't recall that in detail. By smallest cardinal, I assume you mean smallest infinite cardinal? That doesn't exist, unless you allow bullshit like omega-1 =omega. Sorry > > >> >> > 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? > >> >> > >> >> Nope. > >> >> > >> >> > No it doesn't, and the > >> >> >infinite series of constant 1's cannot converge, but diverges to infinity. > >> >> > >> >> Yep. > >> >> > >> >> > Can > >> >> >you actually deny this? If so, then Poincare was right. > >> >> > >> >> 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? > >> > >> No. One example would be to use the trivial topology. In this > >> topology, every sequence converges to every value. > >I don't know what that means, but it sure sounds useless. > > Well, it is trivial... > > > I am talking about > >actual quantities in the normal linear "topology". If I am talking about > >infinite series, i wonder why you are changing the subject. Gee, I wonder.... > > That's why I put in the BTW. I clearly warned you that this wasn't > the main issue. I am glad it wasn't meant to be pertinent. > > Martin > > -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 15:50 imaginatorium(a)despammed.com said: > Tony Orlow (aeo6) wrote: > > imaginatorium(a)despammed.com said: > > > Tony Orlow (aeo6) wrote: > > > > > > > I don't see where you pointed out any specific flaw, except to rant about your > > > > largest finite number again. > > > > > > No, well, I give up. Just for my curiosity, though, I still cannot > > > understand your point when you complain about "ranting about my[sic] > > > largest finite number". It has been pointed out to you so many times - > > > with absolutely no effect - that the Peano axioms (or any similar more > > > informal notion of pofnats) imply that there cannot be a largest > > > pofnat. Just tell me: do you claim... > > > > > (sigh) > > > (1) There _is_ a largest pofnat. > > no > > > (2) There is no largest pofnat (but the contradictions with your ideas > > > escape you) > > yes, please explain the contradiction, without the mantra. I have heard Virgil > > claim that I think there is one, or that I MUST produce one, if I am to claim > > there are infinite whole numbers. I see no such need. I ahve agreed that one > > cannot count finitely from the finite to the infinite, and it has been agreed > > that one cannot count down from the infinite to the finite. The first fact does > > not mean the infinite whole don't exist, any more than the second means that > > finite wholes cannot exist. So, where is the contradiction? > > 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? As soon as you end your ditty on the element values, my ditty on the set size will be complete. Deal? > > > > > > (3) The answer to "Is there a largest pofnat?" is somehow neither 'Yes' > > > nor 'No'. > > No, the answer is no, just like the answer to "is there a smallest infinite > > number?" There is no distinct line between the finite and infinite. That line > > is infinitely wide, and requires an infinite difference to cross. > > I wouldn't call it a "line", personally, but roughly speaking this > appears to be a description of the set of surreals {0, 1, 2, 3, ...} U > {..., w-3, w-2, w-1, w} > (using union notation to prevent dotty confusion). It is similar, if not the same. I started reading about the surreals, and it seemed liek avery sensible approach, though I have not had time to read it all. Hopefully during vacation. > > Brian Chandler > > > http://imaginatorium.org > > > > > > > > > > -- > > Smiles, > > > > Tony > > -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 15:52 Virgil said: > In article <dc6fn8$51j$1(a)news.msu.edu>, stephen(a)nomail.com wrote: > > > In sci.math Daryl McCullough <stevendaryl3016(a)yahoo.com> wrote: > > > Tony Orlow (aeo6) wrote: > > >> > > >>Daryl McCullough said: > > > > >>> But for the set we are talking about, there *is* no L. We're talking > > >>> about the set of *all* finite strings. That's an infinite union: If > > >>> A_n = the set of all strings of length n, then the set of all possible > > >>> finite strings is the set > > >>> > > >>> A = union of all A_n > > >>> = { s | for some natural number n, s is in A_n } > > >>> > > >>> This set has strings of all possible lengths. So there is no L > > >>> such that size(A) = S^L. > > > > >>If those lengths cannot be infinite, then the set cannot be either. > > > > > Why do you believe that? > > > > >>Either you have an upper bound or you do not, and if there is no > > >>upper bound on the values of the members, then they may be infinite. > > > > > Why do you believe that? > > > > >>> You are assuming that every set of strings has a natural number L > > >>> such that every string has length L or less. That's false. > > >> > > >>I am saying that if L CANNOT be infinite > > > > > I'm saying that there *is* no L. So don't talk about the case > > > where L is infinite or the case where L is finite. I'm talking > > > about the case where there *is* no maximum size L. > > > > > Why do you think that there is a maximum size L? > > > > I doubt you will get any rational response. The idea > > that a set of finite objects must be finite is so engrained > > in some people's mind that they cannot see past it, despite > > all the obvious contradictions. > > > > For example, presumably there is some maximum length > > to the finite binary strings, which we will call L. > > How many binary strings are there then? 1 + 2 + 4 + ... + 2^L, > > which we all know is 2^(L+1)-1, which is finite, and is clearly > > larger than L (assuming L > 0). Now why someone would believe that > > there can exists 2^(L+1)-1 binary strings, but there cannot exist > > binary strings with length 2^(L+1)-1 is quite beyond me. They > > are both finite numbers. Why is the limit on finite string lengths > > smaller than the limit on finite sets of finite strings? > > > > Stephen > > Nice point! > > 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. 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. -- Smiles, Tony
From: malbrain on 27 Jul 2005 15:53
Tony Orlow (aeo6) wrote: > Martin Shobe said: > > There is no finite upper bound on how many times I can apply it. > > However, each application occurs after only a finite number of > > previous applications. > Do you, or do you not, have an infinite set? Do you, or do you not, egenrate > one member per iteration. How many iterations are requires to produce an > infinite set, one at a time? (sigh) Yes, you have an infinite set. Yes, you generate one member per iteration. We have no finite-natural-number for the total number of iterations involved in all the generations of all the finite-natural-numbers. karl m |