Obections to Cantor's Theory (Wikipedia article)
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From: Martin Shobe on 25 Jul 2005 21:07 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. >> > 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. >> > 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. Martin
From: Virgil on 25 Jul 2005 21:34 In article <MPG.1d4f0a8eab9b915f989f7c(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Actually, whether you are talking about reals in [0,1) or n in N, or > ANY infinite set of elements represented as strings of symbols, in > order to have an infinite set, you either need an infinite set of > symbols, or infinitely long strings. To the right of the point, these > infinitely long digital strings represent finite values, whereas to > the left, they represent infinite values (mostly). This is a fact of > any symbolic systems. > > > > I'm sure I'm not the only one to suggest the following, but here > > goes: > > > > 1. Note that the successor function is a 1:1 function from the set > > of natural numbers to a proper subset. This establishes N as an > > infinite set, since it is in 1:1 correspondence with a proper > > subset. > > > > 2. Note that the induction axiom (one of the Peano axioms) states: > > > > If A is a subset of N that is closed under the successor > > function, and which contains 0, then A = N. > > > > The set of all finite natural numbers satisfies the requirements of > > the induction axiom. Therefore, the set of finite natural numbers > > is equal to the set of all natural numbers. In other words, every > > natural number is finite. > Take a look at the rewrite of Peano's axioms I posted earlier today. > > > > 3. Note that the set of finite natural numbers is (without the > > induction axiom) already an infinite set by virtue of (1) above: > > the successor of every finite natural number is another finite > > natural number, and the set of all successor numbers is a proper > > subset. > > > > Whatever set you take to be the set of natural numbers, if it > > contains infinite natural numbers, is not the same set that > > mathematicians are referring to when they use the term "natural > > numbers". > Okay. Let's just call them integers. > > > > Perhaps you mean to be claiming that there is no model of Peano > > arithmetic? > I am claiming that after infinite succession, we have changed values > by an infinite amount. > > > > I'll also note that your "infinite series" argument is not germane. > > The issue is not whether one *can* formulate infinite series, but > > whether one *must* formulate these. Arithmetic by itself does not > > mandate such, nor does algebra. In fact, you nod to that point by > > raising the notion of convergence, a notion that is decidedly > > non-algebraic, but instead topological, in nature. The very act of > > defining convergence and of assigning numerical values to the > > formal sums that can be written involves an expansion of the notion > > of summation; if it were not so, then there would be no choice in > > the matter of assigning the value of sums. Instead, there is a > > great deal of choice in the matter, viz. p-adic numbers, being > > completions of the rationals wrt norms other than the usual one. In > > addition, note that the phenomenon of conditional convergence > > carries with it the ability to assign the value of a sum to any > > real value whatsoever by merely rearranging terms. > That may all be true, but none of it makes it possible to increment a > value an infinite number of times and get a finite value. TO requires that we at some point must be able to increment a finite value once and get an infinite value. > You are > adding 1 an infinite number of times to produce the set. Definition: a natural is finite if the initial segment of naturals of which it is the largest is a finite set (has no injection into a proper subset). (1) The first natural is finite. (2) if the natural n is finite then the natural n+1 is finite. By the inductive axiom, it follows that ALL naturals are finite. If TO's version of natural numbers do not work this way then they are not the same as Peano's.
From: Virgil on 25 Jul 2005 22:01 In article <MPG.1d4f0cf3bdd3d39a989f7e(a)newsstand.cit.cornell.edu>, 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. > > > > > 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. When, and only when, TO can point to a particular incrementation as the one which goes from finite to infinite, will he have any case. > > > > > 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? If TO does not know what a topology is, he shold not b e talking about converence and divergence. His ignorance here, as elsewhere in mathematics, makes him act the fool. > That's really not what we're talking about, That's not what toplogy is about, ignoramus. > > So, please make up your mind. Do we increment to get a successor an > infinite number of times, or only a finite number of times, to get N? Only a finite number of times for any n in N, but without end, but also never reaching the "end" that is not there.
From: Virgil on 25 Jul 2005 22:04 In article <MPG.1d4f0d1369ffd55b989f7f(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1d4858812235c0b0989f4c(a)newsstand.cit.cornell.edu>, > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > David Kastrup said: > > > > > > What is supposed to be a "constant equality"? > > > > > > > An equality that holds true for n=1, and for n=n+1 given true for n. > > > > In mathematics, n=n+1 is always false. > > > excuse the typo. again Actually it is more likely TO's assumption that mathematics must act like a special version of C.
From: stephen on 25 Jul 2005 22:01
In sci.math malbrain(a)yahoo.com wrote: > Barb Knox wrote: >> In article <MPG.1d4ecd45545679a8989f6b(a)newsstand.cit.cornell.edu>, >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >> [snip] >> >> >keep in mind that >> >inductive proof IS an infinite loop, so that incrementing in the loop creates >> >infinite values, and the quality of finiteness is not maintained over those >> >infinite iterations of the loop. >> >> 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? > Not so fast. The behaviour of incrementing i after it reaches INT_MAX > is undefined. There is no INT_MAX for an unbounded-precision arithmetic system. Stephen |