Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 21 Jul 2005 01:05 In article <MPG.1d489bbdafd9082c989f5f(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: > > 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. > In this case the definition is fine. > > > > > Binary numbers with ones > > > in finite positions have a limited number of possibilities. > > > > > karl m > > > > What is that limit? How is it defined? > 2^n > > Do you seriously > > believe that there are only a finite number of finite positions > Of course. If there were an infinite number of positions, then some would be > infinitely far from zero, using positions as a unit of measure. > > > > 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 finite binary strings is infinite. > That is absolutely incorrect. Then there must be an absolutely largest number representable by such finite binary strings. TO must show us that number, or at least that it exists, or recant. > If the farthest out 1 is at finite position n, then there can only be > n 1's, no more, and the value can equal 2^(n+1)-1, no more (with ALL > 1's). the set of finite binary strings has a number of strings no > greater than 2^n, where n is the longest string. if n is finite, then > so is 2^n. There's the actual math on that, despite your "theory". But one can always prepend another 1 to the longest finite string without making it infinite.
From: Barb Knox on 21 Jul 2005 01:10 In article <85hdepqo4q.fsf(a)lola.goethe.zz>, David Kastrup <dak(a)gnu.org> wrote: >Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> David Kastrup said: [snip] >>> Good. And since all natural numbers are finite, every division >>> among those naturals is guaranteed to finish. >> >> For finite naturals, yes. > >Which are the only ones there are, since the Peano axioms defining >them leave no room for other naturals. At the risk of appearing to give aid and comfort to the crackpots, let me make the pedantic point that non-standard models of the (1st-order) Peano axioms DO contain numbers which are infinite. But of course these non-standard numbers still obey all the consequences of the axioms, so (e.g.) division is still well-defined on them. [snip] -- --------------------------- | BBB b \ Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | -----------------------------
From: Virgil on 21 Jul 2005 01:27 In article <MPG.1d489d7fea8af732989f60(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > 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. > I have explained the flaw in this proof, and it is met with confusion > because none of you seems to appreciate the recursive nature of > inductive proof. 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. > I am wasting my time with you, unless I write a > complete elementary textbook. Please do, we can use the laughs. > > > > We have yet to see any of TO's alleged counter-proofs that are not > > fatally flawed. > You have yet to point out any fatal flaw. That TO does not choose to acknowledge those flaws does not mean that they are not there. > The best you have done is repeat your mantra of "no largest finite" > on the inductive one, which is irrelevant. Except to the issue at hand. If there is no largest finite natural then the successor function on the naturals proves that the set of finite naturals is infinite in the sense of the Cantor definition of infinite. And then there is no need for any of TO's alleged "infinite naturals". > You have been mute on the information theory one, TO's "information theory" claim requires that at some point one can no longer add another character to a character string and still have a "finite" string. > and tried to claim there is no infinite sum of 1's in the infinite > series one. Which addition of one to a prior natural carries over from finite to infinite? > There is no fatal flaw that anyone has pointed out in my > valid proofs. Willful blindness is not an adequate argument. > Try addressing the situation, without making dishonest > statement repeatedly conscerning my position or your achievements in > refuting them. You're really a dishonest fellow, I must say. That is no more true than what TO mislabels proofs. > > > > > > 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! > And if a set is defined by each member with properties relating to > that member, then those are all properties of that member. You have > claimed repeatedly that I am making some sort of leap, and I have > corrected you on that, and you failed to reply to those corrections, > only to repeat your lies at a later time. Shut up and listen for a > change. Maybe you'll learn something new for a change. > > > > 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. > > > N is finite if every member of N is finite. Show me how you get > infinite S^L with finite S and L. 1^L + 2^L + 3^l + ... diverges, S^1 + S^2 + S^3 + ... diverges for all S > 1. Unless TO can show that each of these has a finite limit, he is refuted.
From: David Kastrup on 21 Jul 2005 01:36 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > N is finite if every member of N is finite. Show me how you get > infinite S^L with finite S and L. (silence) By not having a limit on S and L. They can get arbitrarily large values, but all of them are finite. There is no maximum. THERE IS NO MAXIMUM!!!!!!! And that's what causes a set consisting just of finite values to have an infinity of values. There is always one more value, even though all of them are finite. Here is a game for you: you name a finite number, and then I name a finite number. If I can't name a higher number, you win. You'll always lose, because the set we draw my values from is infinite and I can always add 1 to the number you are naming. Yet the numbers we name are all finite. We don't need infinitely large values for the natural numbers to be an infinite set. All we need is arbitrarily large values. "infinitely large" would be a property associated with a single member. "arbitrarily large" is a property associated with a collection of members that form an infinite subset. The Peano axioms don't give us "infinitely large", but they _do_ give us "arbitrarily large". -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: David Kastrup on 21 Jul 2005 01:44
"Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <MPG.1d483583ff4dfb97989f41(a)newsstand.cit.cornell.edu> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > ... > > > Now this case is quite dissimilar. Cantor's proof (about the > > > cardinality of powersets vs. the cardinality of the base sets) > > > is indeed simple and can be written in very few steps. > ... > > You know, if the only conclusion drawn from Cantor's proofs was > > that a power set is necessarily larger than its base set, I would > > have absolutely no problem. > > Actually that is the only conclusion. From his first version that worked just for digit strings instead of real numbers (real numbers with terminating expansion have an alternative digit strings). You get the (minusculy stronger) result for real numbers only after fixing the ambiguities. > Wrong, the diagonal proof proves the same thing as the proof that > the powerset of a set is larger than the base set. The reals can be > put in a 1-1 relation with the powerset of the natural numbers. > (But only when you consider natural numbers in the sense of the > mathematicians.) Almost. The difference is small, but there. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |