Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 20 Jul 2005 14:46 In article <MPG.1d483583ff4dfb97989f41(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > 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. That fact is always the case and I don't > dispute it. Infinite sets of finite natural numbers, on the other > hand, are self contradictory, and the conflation of "larger and > infinite" with "uncountable" is arbitrary. But the inductive property of the naturals requires that all naturals be "finite". Not until TO can cross that pons asinorum will he make any progress. > The diagonal proof has hidden assumptions that mathematicians seem to > acknowledge and not question, and discussions of the relative > infinites of digital numbers, tree parts, subsets etc, seem to > purposely ignore the properties of the elements of which they are > constructed. > Being "anti-Cantorian" doesn't mean you disagree with > everything he said, just with the stuff that contradicts other math > and reality. But no one, and certainly not TO, has ever produced any such stuff. There have always been fatal flaws in the anti-Cantorian arguments, though they are usually invisible to their overly fond progenitors.
From: Virgil on 20 Jul 2005 14:54 In article <MPG.1d4837e2e5a2d827989f43(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1d472484f809e37c989f2c(a)newsstand.cit.cornell.edu>, > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > > > > The idea of uncountability as being equivalent to "larger than > > > the set of naturals" is unfounded. > > > > The meaning of "uncountable" is a matter of definition. One may, > > as TO > > does, object to claims of its being instanciated, but it is > > perfectly well-defined. > > > > > > > There is no reason to believe that larger sets cannot be > > > enumerated. > > > > What does "larger than" mean, then? > You ask this as if any set that is larger than any other set is > "uncountable". Do you consider the two terms, "larger" and > "uncountable" to be synonymous? It depends. I know of at least two distinct meanings which are not synonymous. Proper supersets are "larger" in one sense and if there is no surjection from one set to another, the other is larger is a somewhat different sense. > > > > There is no reason to believe that larger sets cannot be > > enumerated. To ennumerate a set means, by definition, to find a bijection from either N or some initial segment of N to that set. In this sense, the reals cannot be ennumerated. > > > > That TO does not understand the reasons does not invalidate them. > Why don't you try explaining them? I, and others, have tried, but have been defeated by TO's invincible ignorance. > > > > > the power set of the naturals can be enumerated and bijected with > > > the naturals, as I described in another post, as long as infinite > > > natural numbers are allowed. > > > > But to allow infinite naturals means that one must have a finite > > natural so large that adding one to it gives an infinite result, > > since every natural, except the first, is produced by adding 1 to a > > previous natural. > > > There is no way to specify any such number. Because there are no infinite naturals.
From: imaginatorium on 20 Jul 2005 14:54 Tony Orlow (aeo6) wrote: > Daryl McCullough said: > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > >>> I offered, and you saw, a deductive proof that proves that the > > >>> largest natural in a set must be at least as large as the set size. > > > > Yes. So we have: > > > > If S is a set of natural numbers, and S has a largest member N, > > then N >= the cardinality of S. > > > > How do you prove that every set of natural numbers has a largest > > member? > > > > -- > > Daryl McCullough > > Ithaca, NY > > [Annotating with * and # for clarity] > Essentially, the proof shows that no set can have a larger number[*] of naturals > in it than the values[#] of all the naturals in it. What does it mean to compare a number [*] with multiple values[#]? Let's consider the set {1, 2, 3}. You appear to be saying that the number [*] of naturals in this set (which I hope we agree is three) cannot be larger than the "values[#] of all the naturals in it". I don't quite understand this comparison - I don't think you intend to refer to the sum of the values[?], so perhaps you mean one at a time. Well, there are only three cases: let's look at them. First case: the element 1. Oh dear, 3>1, so plainly the set can have more naturals in it than this first value. Second case: the element 2. Oh dear, 3>2, so plainly the set can have more naturals in it than this second value. Third case: the element 3. Ah! 3 is not more than 3, so in this one case only, your claim is indeed true. One out of three. Not enough to pass, I'm afraid. But what about that one out of three: which one was it? Aha! It was 3, and 3 is the largest element in the set. So here's a revised claim: Essentially, the proof ought to show that a set _can_ have a larger number[*] of naturals in it than any value of a natural in it that is not the largest member. It seems to me this gets three out of three. (Infinity out of infinity, even ^_^) > I will devise a proof without > a largest element, if need be, but that "largest element" argument is a waste > of time. I still don't see how you can say that for all finite sets this is > true, but that one can get an infinite set, and still have all finite numbers. > If each finite n in N is the size of the set including all m<=n, then each of > them corresponds to a finite set. How do we get an infinite set, then, if m<=n > is finite for any finite n in N? Because this set goes on and on, from one finite number to the next, each one totally 100% finite, because it is one more than the previous one, and this set goes on and on and on, and this going on and on never stops, ever, never gets to an end because there isn't one, as Wolf Kirchmeir's grandchild could tell you. Brian Chandler http://imaginatorium.org
From: Daryl McCullough on 20 Jul 2005 14:35 Tony Orlow <aeo6(a)cornell.edu> said: >Daryl McCullough said: >> Yes, that's exactly what the diagonal argument proves. There is >> no list of length aleph_0 that contains all real numbers. >Okay, that I can agree with, at least in digital terms, which is what the proof >relies on. That's all that Cantor's proof shows, and it's all then anyone has ever claimed that it shows. >What it really shows is that digital systems with a given number of digits have >more strings than digits. it is not necessary to have aleph_0 digits, if you >allow for smaller infinities. You only need SOME infinite number of digits. That's not true. If S is an infinite set of strings, then there is a difference between (1) There is no finite bound on the lengths of strings in S. (2) There is a string in S that is infinite. If you wrote these out as logical statements, you would see that you are mixing up the order of quantifiers: (1) forall b, exists s in S, (if b is a finite bound, then length(s) > b) (2) exists s in S, forall b (if b is a finite bound, then length(s) > b) Statement (1) says that the *set* S has no finite bound. Statement (2) says that S contains an *element* that has no finite bound. Those are two different statements. -- Daryl McCullough Ithaca, NY
From: Virgil on 20 Jul 2005 14:57
In article <MPG.1d483916d7024f5b989f44(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1d4725fbdfa9bb24989f2d(a)newsstand.cit.cornell.edu>, > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > > > > There is a simple, demonstrably valid proof of Cantor's Theorem > > > > in ZF set theory. So you must think the proof is unsound. > > > > Which axiom of ZF do you believe to be false? > > > > > > > > Chris Menzel > > > > > > > > > > > I was asked that before, and never got around to fully analyzing > > > the axioms for lack of time, but the diagonal proof suffers from > > > the fatal flaw of assuming that the diaginal traversal actually > > > covers all the numbers in the list. > > > > If it misses any of them, it must miss a first one. Which is the > > first one misssed? > > > > If there is not a first one missed, then none are missed. > > > The first one missed is the first one directly below the diagonal > traversal. If any are missed then there must be an n in N corresponding to that "first one" missed. If TO cannot produce such a finite n from N, then no such n exists, and that anti-diagonal is complete. |