Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 20 Jul 2005 14:05 Daryl McCullough said: > Tony Orlow <aeo6(a)cornell.edu> said: > > >> > but the diagonal proof suffers from the fatal flaw of assuming that > >> > the diaginal traversal actually covers all the numbers in the list. > >> > Any complete list of digital numbers of a given length, even a given > >> > infinite length, is exponentially longer in members than wide in terms > >> > of the digits in each member. Therefore, the diagonal traversal only > >> > shows that the anti- diagonal does not exist in the first aleph_0 > >> > terms. > > Yes, that's exactly what the diagonal argument proves. There is > no list of length aleph_0 that contains all real numbers. > > -- > Daryl McCullough > Ithaca, NY > > Okay, that I can agree with, at least in digital terms, which is what the proof relies on. Try it with Roman numerals! What it really shows is that digital systems with a given number of digits have more strings than digits. it is not necessary to have aleph_0 digits, if you allow for smaller infinities. You only need SOME infinite number of digits. For N naturals, you only need log(N) digits to produce a list N long. -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 14:09 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > 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? > > For each of its members. > > > That entire set is infinite right? Therfore, the number of times you > > are adding 1 and saying, "yep, still finite", is infinite, right? > > No. He is not adding 1 more than a single time, just to check that > for n finite (which means that the set 0..n obeys the pigeon-hole > principle) n+1 is still finite (the case of one additional pigeon-hole > can be reduced to the case n if you check for the hole in position > n+1 and in 0..n both before and after permutation). How do you know 6598367 is finite? Because you got it by adding 1 to 6598366, the 6598366th number. So how to you get the aleph_0th number? What IS that number? it's aleph_0. > > > So, you have some way of adding an infinite number of 1's and > > getting a finite result? > > No, he is just checking that the conditions for the fifth Peano axiom > hold. The whole point of the axioms is not to have to check an > infinite number of steps in order to get a statement about the members > of a particular infinite set, the naturals. And yet, one cannot apply an increment an infinite number of time without adding infinity. > > > 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? No it doesn't, and the infinite > > series of constant 1's cannot converge, but diverges to > > infinity. Can you actually deny this? If so, then Poincare was > > right. > > Nobody is bothered about adding numbers. The Peano axioms do not even > define addition. Just succession. Addition requires the added notion of measure, but that is inherent inn the natural numbers. Each one, as a real number, represents its distance from the origin, 0. > > -- Smiles, Tony
From: stephen on 20 Jul 2005 14:12 In sci.math malbrain(a)yahoo.com wrote: > Daryl McCullough wrote: >> Tony Orlow writes: >> > >> >Dik T. Winter said: >> >> >> Back on your horse again. Tell me about the binary numbers (extended to the >> >> left with 0's) where the leftmost 1 is in a finite position. Are all those >> >> numbers finite? Are there only finitely many of them? >> > >> >yes and yes >> >> What definition of "finite" are you using? > Main Entry: finite > Pronunciation: 'fI-"nIt > Function: adjective > Etymology: Middle English finit, from Latin finitus, past participle of > finire > 1 a : having definite or definable limits <finite number of > possibilities> b : having a limited nature or existence <finite beings> > This definition from webster should suffice. Definitions from webster rarely suffice for mathematical arguments. > Binary numbers with ones > in finite positions have a limited number of possibilities. > karl m What is that limit? How is it defined? Do you seriously believe that there are only a finite number of finite positions? A binary number with one's in finite positions can have an arbitray number of one's. There is no limit on the possibilities. The set of f finite binary strings is infinite. Stephen
From: Virgil on 20 Jul 2005 14:20 In article <MPG.1d48308522352190989f3d(a)newsstand.cit.cornell.edu>, 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? Unless TO has a definition of finiteness of naturals that makes the above proof invalid, one valid proof is enough. We have yet to see any of TO's alleged counter-proofs that are not fatally flawed. > > Inductive proof proves properties true for the entire set of naturals, right? Wrong! It proves things only for the MEMBERS of that set, not the set itself! Definitions (Cantor): (1) a set is finite if and only if there do not exist any injective mappings from the set to any proper subset (2) a set is infinite if and only if there exists any injection from the set to any proper subset. Clearly then, a set is finite if and only if it is not infinite. Definitions (Auxiliary): (3) a natural number, n, is finite if and only if the set of naturals up to it, {m in N: m <= n}, is finite (4) a natural number, n, is infinite if and only if the set of naturals up to it, {m in N: m <= n}, is infinite If these definitions are valid, then it is easy to prove buy induction that there are no such things as infinite naturals: (a) The first natural is finite, since there is clearly no injection from a one member set the empty set. (b) If any n in N is finite then n+1 is also finite. This is also while quite clear, though a comprehensive proof would involvev a lot of details. By the inductinve axiom, goven (a) and (b), EVERY MEMBER of N is finite, but that does not say that N is finite.
From: malbrain on 20 Jul 2005 14:21
step...(a)nomail.com wrote: > In sci.math malbrain(a)yahoo.com wrote: > > > > Daryl McCullough wrote: > >> Tony Orlow writes: > >> > > >> >Dik T. Winter said: > >> > >> >> Back on your horse again. Tell me about the binary numbers (extended to the > >> >> left with 0's) where the leftmost 1 is in a finite position. Are all those > >> >> numbers finite? Are there only finitely many of them? > >> > > >> >yes and yes > >> > >> What definition of "finite" are you using? > > > Main Entry: finite > > Pronunciation: 'fI-"nIt > > Function: adjective > > Etymology: Middle English finit, from Latin finitus, past participle of > > finire > > 1 a : having definite or definable limits <finite number of > > possibilities> b : having a limited nature or existence <finite beings> > > > This definition from webster should suffice. > > Definitions from webster rarely suffice for mathematical > arguments. They are illustrative of how the set of contradictions is resolved by the majority. > > > Binary numbers with ones > > in finite positions have a limited number of possibilities. > > > karl m > > What is that limit? How is it defined? Do you seriously > believe that there are only a finite number of finite positions? > > A binary number with one's in finite positions can have an arbitray > number of one's. There is no limit on the possibilities. > The set of f finite binary strings is infinite. Right. We're discussing the number of permutations of each of these strings taken individually. karl m |