Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 27 Jul 2005 15:28 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Virgil said: >> In article <MPG.1d506ab0de98a030989f9e(a)newsstand.cit.cornell.edu>, >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >> >> > 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. >> >> Or they may not. >> >> >> > how do you have an infinite set of strings with only finite >> > lengths? >> >> The usual way, by not having any finite bound on their lengths. > Only an infinite bound? No bound. If you want to artificially introduce an infinite bound, that is fine as long as you don't confuse it with finite values. > Then they CAN be infinite? No, they can't, but there is no limit to how large they can become. There is always a larger, finite one. > You make no sense. To people with quantifier dyslexia that can't understand the difference between infinite and arbitrarily large. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Daryl McCullough on 27 Jul 2005 15:13 Tony Orlow (aeo6) says: >Daryl McCullough said: >> Let enum(e,s,t) mean "enumeration e produces element s at >> or before step t". Then >> >> 1. S is finite >> <-> exists enumeration e, exists step t, forall s in S, >> enum(e,s,t) >> >> 2. S is countable >> <-> exists enumeration e, forall s in S, exists step t, >> enum(e,s,t) >> >> Note the difference between 1. and 2. The difference is in >> the order of quantifiers. That difference is important. >That has nothing to do with it. Yes, it does. You don't understand quantifiers. >Do you produce the naturals through this inductive process >ala Peano, adding 1 each time to produce the next natural number? Yes. For every natural number n, there exists a step t such that n is produced at stage t. That's what countable means. >Is it an infinite set? Yes. For every step t, there exists a natural number n such that n is not yet produced by step t. That's what infinite means. >So, are you not adding 1 an infinite number of >time to generate an infinite number of naturals? Why is this so hard for you? No *single* number requires more than finitely many steps. But the entire infinite *collection* requires an infinite number of steps. >I can't argue this anymore. You could never argue it in the first place. -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 27 Jul 2005 15:35 Dave Rusin said: > Bark Knox: > > > Using your computational view, consider the following > > infinite loop (using some unbounded-precision arithmetic > > system similar to java.math.BigInteger): > > > > for (i = 0; ; i++) { > > println(i); > > } > > > > Now, although this is an INFINITE loop, every value printed will > > be FINITE. Right? > > > Tony Orlow: > > > In any case, sure, the program will spit out finite numbers, > > snce it is a finite machine running in finite time. > > If the machine had infinite capacity > > and infinite funtime, it could conceivably produce infinite results. > > [Heh,heh -- he said "funtime".] > > Barb Knox: > > > HOW?? No matter how long it runs, EVERY printed value is finite. > > WHEN EXACTLY do you think it would start producing "infinite" values > > Duh! Obviously in the year N = 1000...000 . I'm sure Tony will > agree you'll get finite numbers in finite years, and infinite numbers > in infinite years. > > It's better when you throw Moore's Law into the mix, though. > ("Processor speeds double every 18 months" or something like that.) > > If it takes t_k years to print out all the k-digit numbers, > then in that amount of time processor speeds will increase by > a factor of 2 ^ ((t_k)/(1.5)) . So if after those t_k years we > buy a new computer, the looping job will take less time per > iteration, by a factor of 2 ^ (-(t_k)/(1.5)), i.e. that previous job > would now only take (t_k) * 2 ^ (-(t_k)/(1.5)) years if we were to > redo it. But instead of redoing the last loops, we use the new machine to run > through the displaying all the (k+1)-digit numbers, of which there are > ten times as many so the job takes 10 times as long to run, i.e. > the run time is now > t_{k+1} = (t_k) * 10 * 2 ^ (-(t_k)/(1.5)) years > Interestingly, this makes t_{k+1} > t_k if t_k < 4.9829 but > t_{k+1} < t_k if t_k > 4.9829 -- that is, the new job takes > LESS time than the old one (unless the old one was really short). > You can prove in this way that it will never take more than > 8 years to print all the k-digit numbers, irrespective of what > natural number k is. (The proof uses induction ...) > Winking a little at the precise numbers, this says roughly that > we print all the numbers up to 10^k in about k decades. > > I mention all this because it leaves TO in the uncomfortable position > of having to say that the year (decade) and the number printed are both > finite "until they are both infinite" (or whatever he would say), > that is, you start printing infinite numbers exactly when the number > of decades passed is infinite; but by then ("decade N") the number being > printed is "10^N", meaning that numbers stay finite below 10^N, > thus proving "N = 10^N " in his language, which I thought was > an assertion he agreed couldn't be true. > > Of course those of us who believe there are no "infinite integers" > believe that there is no "infinite year" and no "infinite integers" > will ever be printed, and there is no contradiction. A much less > sucky position to be in. > > dave > Well, then, I guess the question was nonsensical to begin with, and you can never produce an infinite set either, for the same reason you can never produce infinite numbers. I see you like to get in on the fun, but your last post agreed with me up until you got nonsensical and spoke of "almost-not-quite-but- sometimes-maybe" something or other, instead of giving any actual valid objection to my argument. I guess you didn't have any, which I specifically asked for. Yes, life is just a gamne in the ivory tower of academia. How fun. I sure hope you're paid well for your contributions. They're priceless. -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 15:40 Chris Menzel said: > On Tue, 26 Jul 2005 16:39:58 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > > ... > > The big problem in transfinite cardinality is the assumption that a > > bijection necessarily indicates equal sizes for infinite sets, as it > > does for finite sets. > > It's not an assumption, it's a definition, one that you reject. That's > fine, but until you provide equally rigorous alternatives, it's the only > game in town. And frankly, you've got no business rejecting it just > because some of the consequences conflict with your intuitions. The > mathematics of the infinite is occasionally surprising, especially when > one doesn't have a complete understanding of the subject matter. > > > When the only way to form a bijection is with a mapping function, > > How else? > > > then that function needs to be taken into account. This nonsense > > about an infinite set of finite whole numbers is pretty bad too, but > > probably without any real consequences. > > You seem to agree that the set of whole numbers is infinite. But there > was an inductive argument a few posts back that all the whole numbers > are finite, and hence that the set of finite whole numbers is infinite. > There was some real mathematics there. Why have you not responded to a > mathematical proof that all the members of the infinite set of whole > numbers are finite? It would be your chance to show everyone where the > error in the argument lies. > > Chris Menzel > > I have repsonded to that proof over and over. Where were you when we were discussing the nature of inductive proof, and the implicit infinite loop in the construction that no one seems to have considered, and the fact that adding 1 an infinite number of times, as is done in the generation of the infinite set of naturals, produces an infinite sum? I tell you what. You refute my inductive proof that the set of naturals is finite, without refuting your own proof, and then we'll talk. Good luck. I don't expect any response, since you folks tend to ignore any proof you can't throw "largest finite" at. Your bag of tricks is really rather finite, and more and more transparent. -- Smiles, Tony
From: Daryl McCullough on 27 Jul 2005 15:22
Tony Orlow (aeo6) wrote: >Daryl McCullough said: >> But you claimed that the set of all finite naturals is a finite set. >> Every finite set of naturals has a largest element. >That's what they say, and I disagree. No, you sometimes agree, and sometimes disagree. You sometimes accept the axiom of induction, you sometimes reject it. You just say whatever pops into your mind. >The "largest element" is not a valid criterion for finiteness. Let Phi(n) be the statement If S is any set of naturals, and the size of S is n+1, then S has a largest element. Then you can easily prove forall n, Phi(n) by induction. >Besides, it is my position that the set of whole >numbers IS infinite, but contains infinite values. We're not talking about the set of whole numbers, we're talking about the set of *finite* whole numbers. Is that set finite, or not? If it is finite, then it has a largest element. -- Daryl McCullough Ithaca, NY |