Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 27 Jul 2005 18:22 In article <MPG.1d51952f7c8459e7989fb3(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Daryl McCullough said: > > Then how is bigulosity an improvement over cardinality? > Because Bigulosity takes into account the nature of the bijection in > order to determine a precise relative size of infinity. The question > of size for finite sets is trivial: count the elements. But mapping > functions can also be used to exactly measure the size of a finite > numeric set and an infinite numeric set equally well. Since injections and bijections ARE mapping functions, it appears as if bigulosity is no better than cardinality in drag. > It's a much > better measure than bijection which, as we all know, even works on > proper subsets, when they are infinite. A system that even makes > proper subsets the same size as their supersets seems to be broken, > to me. WEll a system that requires infinite naturals and declares that finite ordered sets need not have maximal members is quite definitely broken, to all of us.
From: malbrain on 27 Jul 2005 18:29 Dik T. Winter wrote: > In article <1122395188.769778.34530(a)z14g2000cwz.googlegroups.com> malbrain(a)yahoo.com writes: > > Dik T. Winter wrote: > > > In article <1122347583.518181.245300(a)g14g2000cwa.googlegroups.com> malbrain(a)yahoo.com writes: > > > > > > > > The C language is defined by the C standard, as defined by ISO. There > > > > are no "unbounded" standard types in the C language. karl m > > > > > > Who is talking about C? > > > > Of the billions of computer systems deployed since the micro-computer > > revolution, the overwhelming majority are programmed with C. > > That is not an answer. Well, the OBVIOUS answer to your question is, "I'm talking about C" However, I'm not that vulgar. I tend to translate discussions into C because I find it to be more universally understood than java. karl m
From: Virgil on 27 Jul 2005 18:30 In article <MPG.1d519866bed1b5fa989fb4(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > 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. > Okay, then you disagree that inductive proof proves a property true for an > infinite number of iterations, and therefore you must either reject the idea > that the naturals are an infinite set, or that inductive proof proves the > property for the entire set, as Peano stated. So, which is it? Or, would you > like to retract that spewage? > > > > 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. > If you aimply accept the axioms handed to you (see Daryl?) then you are not > thinking any mor than necessary, which is already abundantly obvious. > > > > 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 guess you never get an infinite set then, or the set is finite. make up > your > mind. > > > > > > > > > > > > > > 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. > Yeah, or those idiotic infinite series, or stupid symbolic systems, or even > its > own retarded axiom of induction. Shut up, Virgil. > > > > > > > > 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. > another repeated asserion. Ho hum. > > > > > > > > > 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. > "largest finite. largest finite" That's YOUR bag. In my thinking the lack of > a > largest finite doesn't prohibit infinite numbers any more than the lack of a > smallest infinite precludes the existence of finite numbers. That's just > stupid, to think that. So, keep your parroting stupidity to yourself. It > doesn't wash. > > > > And a set containing "more than any finite number" of objects is > > infinite by any definition. > Well, you seem to think the set of naturals is finite, anyway, since the > axiom > of induction states that it proves something for the entire set, and you > claim > induction only covers a finite number of iterations. Why don't you draw > yourself a picture or something. > > > > > > > > > > > > > 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! > I brought up the matter of honesty because you keep lying about me. That's > not > exactly an ad hominem attack, but a defensive statement. > > > > > > > > > > > > > > > > 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. > Yes, and the series are infinite, dolt. My point was that for finite S and L, > S^L is finite. If S or L goes to infinity, then it becomes infinite. Comment > on > my inductive proof that the set of naturals is finite, if you dare. Comment: TO's S^L argument is pure LSD. > > > > 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. > You're forgetting that S and L in those series go to infinity. How is it that S, L and S^L though all finite, can be of different sizes? Whenever S^L is larger that either S or L, why not take a new S and L each equal to the old S^L and recalculate a new S^L. Does this process ever have to stop? If at some step S^L ever reached an infinite value, for which finite S and L does S^L become infinite?
From: stephen on 27 Jul 2005 18:46 In sci.math Daryl McCullough <stevendaryl3016(a)yahoo.com> wrote: > Tony Orlow (aeo6) wrote: >> >>Virgil said: >>> Okay, TO! Why is the limit on the size of finite string lengths smaller >>> than the limit on size of finite sets of finite strings? >> >>Maybe you should ask Stephen, since I never made any such claim. > That's because you never explore the *consequences* of the things > you do claim. That's the difference between you and a competent > mathematician. >>He just made my point anyway, when he said "How many binary strings >>are there then? 1 + 2 + 4 + ... + 2^L, which we all know is >>2^(L+1)-1, which is finite". If you have >>finite lengths only, then you have a finite set. Thanks, Stephen. > Stephen was pointing out how nonsensical your claim was. You > are saying that there is a maximum string length L. The number > of strings of that length or smaller is M (where M = 2^(L+1)-1). > M is bigger than L. But it's still finite, right? So why can't > you strings of length M? Tony apparently does not understand the question. However I cannot think of any other way to rephrase it. Here is another inconsistency for Tony to mull over. Apparently there is a largest finite natural number (of course Tony denies this in every other post, but uses it in every "proof"). This base 10 representation of this number requires L digits. The base 2 representation of this number, which presumably exists because all natural numbers can be representd in any base, requires roughly 3L digits. Why do no 3L digit long base 10 numbers exists? Or do no 3L digit long numbers exist in any base, so that really larger numbers cannot be represented in small bases? The mysteries of Orlovian math. Stephen
From: malbrain on 27 Jul 2005 19:09
Tony Orlow (aeo6) wrote: > malbrain(a)yahoo.com said: > > Chris Menzel wrote: > > > On Tue, 26 Jul 2005 16:44:24 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > > > > I am saying that if L CANNOT be infinite, then S^L CANNOT be > > > > infinite, > > > > > > No one disagrees with that, for fixed S and L. > > > > > (...) > > > > > > > For finite S, S^L can ONLY be infinite with infinite L. Why is this so > > > > hard to understand? If S and L are both finite, then S^L is finite, > > > > isn't it? > > > > > > Yes, of course. But that's for a fixed L, say 17. But for any given > > > nonempty set S of natural numbers, the set of *all* finite sequences of > > > elements of S -- i.e., S^1 U S^2 U S^3 ... -- is infinite. That's the > > > part you don't seem to get. > > > > Obviously, he doesn't. Perhaps using the definition for all will help: > > The set made by taking each and every (finite) L sequence of elements > > of S is an infinite set. > > > > karl m > > > > > So, the sum of a finite number of finite terms is infinite. Sure. Sometimes. The sum of infinite 1/2^n is one, for example. I gave you this example yesterday in Barb's java program that generates all the natural numbers. Perhaps you weren't paying attention? karl m |