Obections to Cantor's Theory (Wikipedia article)
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From: malbrain on 27 Jul 2005 14:48 malbr...(a)yahoo.com wrote: > Tony Orlow (aeo6) wrote: > > Some finite, indeterminate number. You tell me the largest finite number, and > > that's the set size. It doesn't exist? Well, then, I can't help you. > > >From websters 1913 dictionary: > > De*ter"mi*nate (?), a. [L. determinatus, p. p. of determinare. See > Determine.] > > 1. Having defined limits; not uncertain or arbitrary; fixed; > established; definite. > > > Thus "indeterminate" is the exact opposite of fine. You can't have it > both ways. Ooops. "indeterminate" is the exact opposite of finite. You've uncovered a contradiction about the count of elements in an infinite set that cannot be resolved from the definitions of finite and indeterminate. karl m
From: Daryl McCullough on 27 Jul 2005 14:38 Tony Orlow writes: >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: Of course it is. There is no disagreement about that. Let A_n = the set of all bitstrings of length less than or equal n. Then by definition, the number of elements in A_n is f(n). And that's always finite if n is finite. Fine. The disagreement is over the set A = the set of *all* strings over the alphabet {0,1} that are finite in length You seem to be under the impression that there must be some L such that A = A_L But why do you think that? It's not true. There *is* no L such that A = A_L. There is no *finite* L such that A = A_L, and there is no *infinite* L such that A = A_L. -- Daryl McCullough Ithaca, NY
From: David Kastrup on 27 Jul 2005 14:55 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > No, I understand what you mean by "countable". My issue with your > "uncountability of the powerset of the naturals" is that you can > easily form a bijection between the members of the powerset and the > set of infinite bit strings denoting set membership in the subsets, Yes. > and between those bit strings and whole numbers, No. > and the only reason this bijection is rejected is because of the > refusal to allow infinite whole numbers in the set of whole numbers, They don't fit the axioms. But let's, for the sake of the argument, take a look at the set of infinite binary bit strings (which is a superset of the naturals). All positions of those bit strings _are_ already completely used up in the powerset mapping for just the naturals. Let's call the bit stream picking out all numbers dividable by k B_k, so B_1 = ...111111111, B_2=...10101010101, B_3=...1001001 and so on. Now tell us the number of the set {B_2, B_3}. It has 2 bits set. What is the distances between those 2 bits, and where is the rightmost bit? The infinite bitstreams you dragged in to be able to number the power set have exacerbated the problem: you can't indicate subsets containing them. The only subsets you can now index are subsets containing only finite integers. Since the additional members have not solved the problem of the powerset mapping, it is less complicated to not admit them in the first place. Then we only have a single level of problems. > despite the fact that an infinite set of whole numbers requires > infinite whole numbers, Whining does not make it so. It requires arbitrarily large numbers, but none of them need to be infinite. > so you don't consider the infinite bitstrings to represent whole > numbers. It's like all the wrong choices have been made, almost on > purpose, in order to make some grand distinction which really isn't > there. Well, it's like you try to fix something that you perceive as a problem by making it much worse. > If you allow infinite whole numbers, as is required, Whining won't make it so. > suddenly this whole "uncountability of the power set" vanishes in a > puff of smoke. But then you still can't account for the resulting powerset: you can only cover the powerset of the finite integers with your infinite bit strings. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 27 Jul 2005 14:59 Robert Low said: > Tony Orlow (aeo6) wrote: > > Robert Low said: > >>Tony Orlow (aeo6) wrote: > >>>There ARE no infinite numbers, only infinite sets, in your > >>>lexicon. > >>Not only do we know about infinite numbers, but there are > >>at least two different kinds of them: cardinal and ordinal ones. > > So, you think there can be infinite whole numbers, but that they are not in the > > infinite set of whole numbers? > > I didn't say that there were infinite *natural* numbers; > I said that there were at least two different types of > infinite number, namely cardinal and ordinal ones. The > finite ordinals are the natural numbers, and the set > of all finite ordinals is an infinite set each of whose > elements is itself finite. > > > Why am I being challenged on the very notion of > > such numbers, if their existence is already established? > > You are being challenged on the notion of 'infinite > natural numbers' as necessarily existing in the > set of all standard natural numbers, because that's poppycock. Okay, given the standard Peano axioms, it would seem that one might not be able to count to infinite numbers, because you can't define the point where they beocme infinite. But any infinite set of whole numbers MUST contain infinite values for the reasons I have put forth. Either one abandons the idea that the set of naturals is infinite, or one includes infinite numbers in the set. > > There *are* models of the (first order) Peano Axioms > which contain objects you could call infinite integers. > These models even have some of the properties you > seem to want to ascribe to the natural numbers. > These models are called non-standard precisely to distinguish > them from the usual, or standard model which does not > contain such integers. > > I suspect you'd find this stuff interesting if you > bothered to learn about it, and you might even come > to the conclusion that one of these non-standard models > is the 'right' model of the natural numbers. If so, you > should read up on non-standard analysis, and internal > set theory in particular. Thanks for the suggestion. I just printed out the page from MathWorld and will hopefully have time to read it soon. It is not that I am not willing to do research. I have done a lot while discussing things here and leanred a lot. I have limited time, though, and nothing has been put forth here that even remotely makes me question my conclusions. Rather, it confirms them all the more. So, I am happy to read up on other things. I already have a pile of things to read, so I can't be expected to read everything that is suggested in a particularly timely manner. > > But the debate between the mainstream approach and that > one is not that only one of the approaches is consistent, > rather that one approach is more useful than the other. > The non-standard one being the more useful, I would imagine. -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 15:03
Daryl McCullough said: > Tony Orlow (aeo6) wrote: > > >> (3) The answer to "Is there a largest pofnat?" is somehow neither > >>'Yes' nor 'No'. > >No, the answer is no. > > But you claimed that the set of all finite naturals is a finite set. > Every finite set of naturals has a largest element. > > -- > Daryl McCullough > Ithaca, NY > > That's what they say, and I disagree. The "largest element" is not a valid criterion for finiteness. Besides, it is my position that the set of whole numbers IS infinite, but contains infinite values. -- Smiles, Tony |