Obections to Cantor's Theory (Wikipedia article)
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From: Dave Rusin on 26 Jul 2005 19:41 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
From: Martin Shobe on 26 Jul 2005 20:04 On Tue, 26 Jul 2005 13:20:30 -0600, Virgil <ITSnetNOTcom#virgil(a)COMCAST.com> wrote: >In article <MPG.1d5012d12d5990ae989f86(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >> Virgil said: > >> > The inductive axiom shortcuts that recursion, which is the point of the >> > inductive axiom. It says that if the recursive step can be proved in >> > general, then it never need be applied recursively. >> > >> > If TO wishes to reject the inductive axiom, only then can he argue >> > recursion. > >> What a load of bilge water! > >TO apparently is not familiar with the statement of the inductive axiom >if he denies so clearly what it actually says. >And as it is accepted in some form in ZF, ZFC and NBG, it requires no >proof. A bit of a nitpick here. In ZF, ZFC, and NGB, induction does require proof. It's not particularly difficult, but it still requires proof. (And you can't prove it by directly applying that fact that -> is transitive. In order to finish it that way would require that you perform induction on the number of ->). Martin
From: Chris Menzel on 26 Jul 2005 19:47 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
From: Martin Shobe on 26 Jul 2005 20:13 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. >> >> > 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. >> >> > 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. Martin
From: malbrain on 26 Jul 2005 20:15
Chris Menzel wrote: > On Tue, 26 Jul 2005 16:39:58 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > > ... > > 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. How does it follow that the count of finite whole numbers is infinite? How is this established by the Peano axioms? You have Tony agreeing to the axiom of infinity apriori, when this is not indicated. karl m |