Obections to Cantor's Theory (Wikipedia article)
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From: hale on 20 Jul 2005 17:47 Tony Orlow (aeo6) wrote: > malbrain(a)yahoo.com said: [cut] > > That's not what our axiom says. It says that induction covers all the > > natural numbers in a single step, a single leap-of-faith. > That's really not the case. It is a recursive proof where the property is > proven true for each element depending on its truth for the preceding element. > f(n)->f(n+1), for n=1 to oo. Otherwise, how do you think it proves things for > each and every n in N? Do you agree that one can prove the statement "If n is even, then n+1 is odd" without using induction? I would say that one could. In such a proof, one would start off with something like: "Suppose n is even. Then, there is an m such that n = 2 * n. etc" In such a proof, you would not be running through elements n = 1 to oo. Or, do you disagree with this? If you disagree, then we can reduce the problem to this case and eliminate the discussion about how induction works. Then, to answer your last question, I prove that f(n) is true for each and every n in N by invoking "axiom of induction" after I have proved f(1) is true and "f(n)->f(n+1)" is true (this without invoking induction). Induction is an axiom, not a recursive proof. The idea of recursing from n = 1 to oo is an intuitive justification of the axiom of induction. But, an axiom doesn't need a justification (in a certain sense). -- Bill Hale
From: Tony Orlow on 20 Jul 2005 18:30 Daryl McCullough said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > >So, inductive proof does not rely on proving for n+1 based on n? The infinite > >number of successive naturals for which you prove your property do not > >constitute an infinite number of implied steps in inductive proof? > > As has been pointed out before, an inductive proof does not > have an infinite number of steps. To prove > "for all natural numbers x, Phi(x)", you only > need to prove the following two statements: > > 1. Phi(0). > 2. for all natural numbers x, Phi(x) implies Phi(x+1). > > You seem to be thinking that proving statement 2 somehow > requires an infinite number of steps. If that's the case, > then statement 2 doesn't *have* a proof (because proofs > have to be finite). > > A proof of a universal statement is not the concatenation > of infinitely many singular proofs. > > -- > Daryl McCullough > Ithaca, NY > > It seems to me you are all being dense and objecting to something I am not saying. I hope that's not the case. The proof has a finite form, much like a recursive algorithm. A recursive algorithm will run forever if it doesn't have some stop condition, like running out of nodes in a tree path, which is bad for a computer program. But this recursive proof, like the recursive definition of the naturals, DOES go on forever. Your #2 above is the recursive part of the proof; it proves something true based on the truth of its predecessor. When you actually "run" the proof, #2 acts like a loop, iterating its way through all the members of the set, with no stop condition. The infinity of steps is implied in the link from one number to the next. It is stated as one statement, but that statement is applied to every member of the set, an infinite number of times. If that step involves an increment, since it is dealing with a property directly relating to the value of that number, which is incremented (notice the n+1?), then it is incremented an infinite number of times when the proof is applied. Yes, all programs are finite, but programs can also go into "infinite loops". I hope this clears up what seemed rather obvious to me, having said it about 100 times in this newsgroup. Probably not. I'm going away for four days, so maybe once I get the hell outta this office you chilluns can have fun discussing what an idiot I am or whatever floats your boat. -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 18:35 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 f finite binary strings is infinite. That is absolutely incorrect. 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". > > Stephen > -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 18:43 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. I am wasting my time with you, unless I write a complete elementary textbook. > > 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. The best you have done is repeat your mantra of "no largest finite" on the inductive one, which is irrelevant. You have been mute on the information theory one, and tried to claim there is no infinite sum of 1's in the infinite series one. There is no fatal flaw that anyone has pointed out in my valid proofs. 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. > > > > 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. (silence) -- Smiles, Tony
From: Daryl McCullough on 20 Jul 2005 18:29
Robert Low says... > >Daryl McCullough wrote: >It's pretty unclear in a lot of the thread when people >are talking about first order and when second order PA; >but I don't see how that helps. The only difference is >replacing the axiom scheme "If P is a one-place predicate, >then P(0) together with \forall n (P(n)->P(n+1)) entails >\forall n P(n)" with the axiom "If S \subset N, and >0 \in S, and S is closed under successor, then S=N". > >We still don't have an explicit mention of order, so >the axioms can't just *say* than N is well-ordered, >even if they imply it. I just meant that you can formalize and prove the claim "Every nonempty set of naturals has a smallest element" in 2nd order PA. -- Daryl McCullough Ithaca, NY |