Obections to Cantor's Theory (Wikipedia article)
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From: imaginatorium on 27 Jul 2005 14:02 Tony Orlow (aeo6) wrote: > Daryl McCullough said: > > Tony Orlow writes: > > > > >Barb, you're not saying anything new. I have heard it all before. I am not > > >drawing my conclusions in any such confused way > > > > Uh, yes you are. Why don't you write down what you consider to be > > valid axioms for working with infinite sets, and then try to write > > a formal proof of your claim > > > > If a set S of strings is infinite, then S contains some infinite > > strings. > > > I guess you can't understand a proof if it has natural language. Your statement > is not quite right. This is what I have proven: > > Given an alphabet of finite size S, an infinite set of unique strings made from > this alphabet must contain infinite strings. Er, no it isn't. At least if the following proof is supposed to be the relevant one. > Here's my two axioms: > > (1) N=S^L, where N is the size of the set of ALL strings of length L, > constructed from an alphabet of size S. N is the upper limit on the number of > strings from a set of S symbols and a maximum length of L. > > (2) A^B is finite if and only if A and B are finite. (1) at least doesn't need to be an axiom, since it follows from elementary arithmetic. > Given (2), S^L is finite iff S and L are finite. > Given finite S, S^L is finite iff L is finite. > Given (1), N is finite iff L is finite. > Conversely, N is infinite iff L is infinite. > > ... > > Well, I just got back from getting lunch, during which it occurred to me that > it would be good to put this proof in inductive format, and watch you try to > discredit it while hanging on to your proof that all naturals are finite. This > proof relies on some assumptions. Let me know if you disagree with any of them: > > (1) All naturals are finite. (you already "proved" this one) > (2) The number of strings of length L that can be constructed from a set of > symbols of size S (set size, not symbol size, remember (sigh)), is S^L. > (3) For finite A and B, both A^B and A+B are finite. > > Are we good to go? Okay. > > Proof that f(n), the number of strings in the set of all strings up to and > including length n in N, on a finite alphabet of size S, is finite: > > 1. For n=1, we have S^1 strings of length 1, for a total of S strings less than > or equal to 1 in length. f(1)=S is finite, as stated. > > 2. For a finite number of srings of length n or less, we can add S^(n+1) > strings of length n+1, to get f(n+1),the number of strings of length n+1 or > less. S is finite and n+1 (a natural number) is finite, so S^(n+1) is finite. > S^(n+1) is finite and f(n) is finite, so f(n+1)=f(n)+S^(n+1) is finite. > > 3. Therefore, for all n in N, f(n), the number of strings up to and including n > symbols, which are created from a finite alphabet, is finite. There is no n in > N, in other words, which maximum length of string will allow you to have an > infinite set of strings. Absolutely. There is no n in N for which the set of strings of length up to n is infinite. Right. (You've done this before, though; you vacillate between correctly proving one thing, then restating it as something else, which is false.) If for any finite integer the strings are limited to this length, then the set is finite. If, however, the strings are not limited to the length of any single integer, but may have a length of any integer, then the set is infinite. As always, you have to be careful reading the quantification of the two parts of this sentence correctly: (a) For any string, there is an integer which is its length. (b) There does not exist an integer such that none of the strings exceeds it in length. Brian Chandler http://imaginatorium.org
From: Daryl McCullough on 27 Jul 2005 13:45 Tony Orlow (aeo6) wrote: >Obviously, i am rejecting some of your axioms and assumptions, so I am not >working with the same set as you are. Then write down what set you *are* using. -- Daryl McCullough Ithaca, NY
From: Robert Low on 27 Jul 2005 14:08 Daryl McCullough wrote: > Tony Orlow (aeo6) wrote: >>Obviously, i am rejecting some of your axioms and assumptions, so I am not >>working with the same set as you are. > Then write down what set you *are* using. His set of axioms and assumptions seems to be as well-formed as his set of finite natural numbers...
From: Tony Orlow on 27 Jul 2005 14:11 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. > > 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. God, you're confused. Do you really think you can get anywhere squirming like that? Sheesh! > -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 14:14
Virgil said: > In article <MPG.1d50173fe4d9949d989f8a(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > It seems like axioms are given a status that makes > > them unquestionable and almost incomprehensible, beyond the > > application of them. For instance, everyone's dismissal of the > > infinity inherent in the recursive nature of inductive proof is a > > sign that the axiom is not really understood, but accepted without > > question. For me, mathematics without meaning is unsatisfying, and > > symbolic manipulation without understanding is boring. With a closer > > examination of axioms and a willingness to revise them, I would hope > > that mathematicians could develop a fully integrated set of universal > > axioms which would cover all of math. But, that won't happen as long > > as axioms are taken without question, just because they "work". > > TO again conflates the issues. > > The choice of axioms to make up an axiom system is one thing, and the > composition of the axiom systems that TO finds so barren are actually > the result of an extremely long and arduous series of trial and error > refinings of a complexity well above TO's capacity for understanding. > > Once an axiom system is chosen, one must abide by those axioms if that > system is to be given a fair evealuation. > > Since TO will not abide by the axioms of any system containig the > equivalent of the Peano axioms, all his claims must refer to what occurs > in some other system on his own invention, for which he has not, and > perhaps cannot, give a comprehensive axiom system. > You didn't like my extension of Peano's axioms? I haven't seen any comment yet from anyone. I'll keep looking. -- Smiles, Tony |