Obections to Cantor's Theory (Wikipedia article)
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From: malbrain on 26 Jul 2005 14:28 Tony Orlow (aeo6) wrote: > I am not an expert in rings, nor am I going to sketch the axioms that you know > better than I, nor does it help the conversation when you snip the original > statement was repsonding to, which had soemthing to do with numbers being their > own multiples of more than 1, or something. It didn't make sense in the context > of normal quantitative addition. It was a vague guess as to what the idea was. > > The sum of an infinite number of 1's is infinite. That's all. Here's what Pascal had to say about his triangle in the 1600s: "Even though this proposition may have an infinite number of cases, I shall give a very short proof of it assuming two lemmas. The first, which is self evident, is that the proposition is valid for the second row. The second is that if the proposition is valid for any row then it must necessarily be valid for the following row. From this it can be seen that it is necessarily valid for all rows; for it is valid for the second row by the first lemma; then by the second lemma it must be true for the third row, and hence for the fourth, and so on to infinity" karl m
From: Virgil on 26 Jul 2005 14:40 In article <MPG.1d500de16aaa9f18989f85(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > that is not at all my logic. Pay attention. Since we have never seen anything from TO that qualifies as logic, there has been nothing to pay attention to, at least of logic. TO claims that an iterative process, n -> n+1, that has no end must achieve actual infinity, but cannot say at which iteration such a transformation occurs. At the same time TO claims that the iterative process, 1/n -> 1/(n+1), never actually reaches zero.
From: Tony Orlow on 26 Jul 2005 14:46 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? > > >> > 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? > > >> > 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. 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.... > > Martin > > -- Smiles, Tony
From: Tony Orlow on 26 Jul 2005 14:48 malbrain(a)yahoo.com said: > Virgil wrote: > > In article <MPG.1d4ef2d3df7d5616989f71(a)newsstand.cit.cornell.edu>, > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > > > > Is there a 1-1 correspondence between digital strings and whole numbers? > > > > Depends on what one means strings and by whole numbers. TO includes > > among his "strings" things which have infinitely many characters > > including a first and last, yet are supposed to be serially ordered > > with, except for the first, each having both an immediate predecessor, > > and, except for the last, an immediate successor. > > > > > > Can you have an infinite set of strings without infinite lengths? > > > > Everyone but TO can. But TO handicaps himself. > > > > > > How can you have an infinite set of digital whole numbers, without having > > > infinitely long strings representing infinite values? > > > > By allowing those strings to get arbitrarily long (no limit on length). > > That's what Tony is illustrating. Each of his infinite strings has a > finite value. karl m > > Not if it has a non-zero digit infintiely far to the left of the digital point. Then the value is infinite. -- Smiles, Tony
From: Virgil on 26 Jul 2005 15:20
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. > Accepting the axiom as a general rule does not mean one has to > immediately forget about the lgical basis for the axiom. The > underlying reason, outside of the axiomatic system, that this axiom > holds true, is the transitive nature of logical implication, such > that (a->b ^ b->c) -> (a->c). The construction of inductive proof > produces an infinitely long chain of implications, WRONG! Arbitrarily long, but alway finite. The chain of implications is made of individual links, one link at a time. There is never a step that can append infinitely many links. throughout which > the transitive property holds, so that the implication then applies > to all members of the set. If you want to pretend that it's true > because Peano said so, then you are just not thinking. WE take it as true because it is one of the axioms of the system we are working in. If TO does not wish to work in that axiom system, he has no right to criticize what others find in that system. But even if we took TO's view, induction can never bridge the gap from a finite natural to an infinite one, since at each iteration, it can only go from finite to finite. > > > > > > I am wasting my time with you, unless I write a > > > complete elementary textbook. > > > > Please do, we can use the laughs. > You do enough laughing. TO gives us plenty of cause to laugh. > > > > > > > > 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. > That Virgil wants to claim they are there does not mean that this isn't just > more dishonest garbage. If Virgil actually had any objection to the symbolic > system one, then he would put it forth, and not shoot himself in the foot, as > he did, below. > > > > > 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. > Wouldn't it be nice if the "Cantor definition" agreed with ANYTHING else? It is sufficient to itself. That it does not agree with such idiocies as "bigulosity" is to be credited in its favor. > > > > And then there is no need for any of TO's alleged "infinite naturals". > I have shown that there is, despite your repeated assertions to the contrary. To has claimed there is, but his "showings" do not show. > > > > > 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. > You obviously have not been paying attention. I never said anything even > remotely like that. Learn to read. I didn't say you actually said that, but what you did say requires that there be a longest possible finite string. If there is no longest possible finite string, then there can easily be more than any given finite number of finite strings without any problems. And a set containing "more than any finite number" of objects is infinite by any definition. > > > > > There is no fatal flaw that anyone has pointed out in my > > > valid proofs. > > > > Willful blindness is not an adequate argument. > Neither are empty statements and claims to victory. > > > > > 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. > Yes, very good. Respond with more ad hominems. A clear sign of a weak > argument, or really, none at all. AS it was TO who brough up the matter of honesty above, he points that finger at himself! > > > > > > > > > > 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. > > > > I am tired of your foolishness, Virgil. Because my foolishness is more valid than TO's wisdom. You are creating infinities > by combining, first, all the strings of length L from a set of 1 > symbol, plus those from a set of 2 symbols, etc, up to an infinite > set of symbols, and in the second, combining the set of strings from > a set of symbols S of length 1, plus those of length 2, etc, up to > infinite lengths. WRONG! All the terms in each series above are finite terms. The partial sums increase without finite limit, creating a divergent series, so the sums would be infinite if it were possible to achieve them, thus that the "number" of possible naturals expressable by finite strings is larger than any finite natural number. |