Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 27 Jul 2005 15:20 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > You didn't like my extension of Peano's axioms? I haven't seen any > comment yet from anyone. I'll keep looking. I commented on it already. You get the union of two sets that are both isomorphic to the integers. So your axioms would not permit including the set of infinite bit streams. You are basically thrashing around wildly and without a clue what you are doing in the hope that something mathematical will result from it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 27 Jul 2005 15:26 stephen(a)nomail.com said: > 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. That is obviously not my position, as I have pointed out. I doubt that was the logic behind any of the "cranks'" arguments, since that would also mean that the reals in [0,1) include infinite values. Repeated statements like this are a clear indication that you are not comprehending the argument at all. > > 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? It's not, but as you just said, if the length of the strings is finite, then so is the number of strings. Thanks for the support. Everyone seems to be making my points for me today. Funny that! > > Stephen > > > -- > > Daryl McCullough > > Ithaca, NY > > -- Smiles, Tony
From: David Kastrup on 27 Jul 2005 15:24 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > At the point that the runtime became infinite, which is obviously > not an identifiable point. At what point DOES the runtime become > infinite? A million years? a billion? If you have infinite runtime, > starting with a finite amount of time, then you have infinite > numbers, starting with finite ones. You can't have it both ways. I prefer to use induction and get my proofs finished before the end of time. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: David Kastrup on 27 Jul 2005 15:24 malbrain(a)yahoo.com writes: > malbr...(a)yahoo.com wrote: >> Tony Orlow (aeo6) wrote: >> > Some finite, indeterminate number. You tell me the largest finite number, and >> > that's the set size. It doesn't exist? Well, then, I can't help you. >> >> >From websters 1913 dictionary: >> >> De*ter"mi*nate (?), a. [L. determinatus, p. p. of determinare. See >> Determine.] >> >> 1. Having defined limits; not uncertain or arbitrary; fixed; >> established; definite. >> >> >> Thus "indeterminate" is the exact opposite of fine. You can't have it >> both ways. > > Ooops. "indeterminate" is the exact opposite of finite. You've > uncovered a contradiction about the count of elements in an infinite > set that cannot be resolved from the definitions of finite and > indeterminate. karl m Uh, no. "Indeterminate" just means unspecified, not infinite. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: David Kastrup on 27 Jul 2005 15:26
Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Okay, given the standard Peano axioms, it would seem that one might > not be able to count to infinite numbers, because you can't define > the point where they beocme infinite. But any infinite set of whole > numbers MUST contain infinite values for the reasons I have put > forth. Whining does not make it so. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |