Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Virgil on 27 Jul 2005 23:57 In article <MPG.1d51ae843adae7ee989fc5(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > 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) One for each finite natural after the first(sigh).
From: cbrown on 27 Jul 2005 23:59 Poker Joker wrote: > "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message > news:dc9h1k08iu(a)drn.newsguy.com... <snip> > > Okay, maybe this will make it clear to you: There is > > exactly 1 natural number that is less than 1: namely 0. > > There is exactly 2 natural numbers that are less than 2: > > namely 0 and 1. There are exactly 3 natural numbers that > > are less than 3: 0, 1, and 2. > > > > You see the pattern? For any n, the number of naturals > > less than n is equal to n. > > > > Now, how many natural numbers are less than infinity? > > I disagree with grouping infinity in with the natural > numbers. That's what Cantorians do. Assuming that "Cantorians" means mathematicians: no, they don't. Daryl didn't say above that infinity was a natural number. In fact, he has already proven (elsewhere) that every natural number is finite. Cheers - Chas
From: Virgil on 28 Jul 2005 00:02 In article <MPG.1d51b00ca3d8b7ab989fc7(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > 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? Iteration on finite sets ends. With what finite natural does the add one iteration of the finite naturals end?
From: Virgil on 28 Jul 2005 00:06 In article <MPG.1d51b11579a055d1989fc8(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Chris Menzel said: > > On Tue, 26 Jul 2005 16:44:24 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > > > I am saying that if L CANNOT be infinite, then S^L CANNOT be > > > infinite, > > > > No one disagrees with that, for fixed S and L. > > > > > and the fact that so many find this impossible to understand > > > demonstrates that Poincare was right, and Cantorian transfinite > > > cardinality is a disease in mathematics. > > > > Google for a recent post by Keith Ramsay for a correction of this > > historical myth. > > > > > For finite S, S^L can ONLY be infinite with infinite L. Why is this so > > > hard to understand? If S and L are both finite, then S^L is finite, > > > isn't it? > > > > Yes, of course. But that's for a fixed L, say 17. But for any given > > nonempty set S of natural numbers, the set of *all* finite sequences of > > elements of S -- i.e., S^1 U S^2 U S^3 ... -- is infinite. That's the > > part you don't seem to get. > > > > > Not unless you allow L to go to infinity. That depends entirely on what one means by "go to" Letting L increase without finite limit is quite sufficient, one never needs to reach infinity, just reach for it. Like any "limit" process, one never arrives.
From: imaginatorium on 28 Jul 2005 00:07
Daryl McCullough wrote: > Tony Orlow (aeo6) wrote: > > >If I prove that there is no n in N for which the set of all strings of length > >less than or equal to n is infinite, then I have proven that no finite limit on > >the length of strings can produce an infinite set. > > Right. The set of all finite strings has no maximum string length. > If there was a maximum string length, then the set would be finite. > But since there is no maximum string length, the set is infinite. > > There is no number n such that all finite strings have length less > than or equal to n. Ha ha! Gotcha! If there is no such n, some of the "finite strings" must be infinite, then, mustn't they? This show will run, and run. I think Tony has now far exceeded Phil's staying power, but it makes no difference. Round and round in circles, even though cranks reciprocate. Note to Tony: First sentence above is a parody of crank-argument. Please do not claim that I said it, out of context. Note to me: Actually, I think the danger of young minds being confused by TO has passed. Must get on with something productive. Brian Chandler http://imaginatorium.org Incidentally, what *is* an "obection"? Sounds like a geometrical obscurity... |