Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Robert Low on 25 Jul 2005 09:17 Han de Bruijn wrote: > Robert Low wrote: >> Han de Bruijn wrote: >>> Robert Low wrote: >>>> Han de Bruijn wrote: >>>>> Clearly, it has never crossed their minds that such a nice >>>>> relationship >>>>> between topology and calculus could possibly esists. >>>> Apart from trivialities like de Rham cohomology and >>>> the Atiyah-Singer Index theorem, anyway. > But, as I suspected, these don't compare with my remarkably _simple_ > result. Which nevertheless went unnoticed by mainstream mathematics. I was referring you to large bodies of mathematics which relate topology and calculus. If you want a simple one, then the *usual* integral to find the winding number is already well-known to compute the crossing number. But the complaint that something is a large body of well-developed mathematics is hardly strengthening the claim that mainstream mathematics is ignorant of nice relationships between calculus and topology.
From: Tony Orlow on 25 Jul 2005 09:26 Daryl McCullough said: > 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. Yes, I understand the difference between those two statements, and in this case the two are equivalent. If the length of strings is L and the symbol set has a finite size of S, then you have S^L strings, which is infinite IF AND ONLY IF L is infinite. Infinite Set <-> Infinite Element. This is not the case for the real numbers, because they do not have any minimum finite difference between them. That is why you can have [0,1), an infinite set with all finite numbers. You cannot have such a set with integers or strings. > > 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. I don't need a lesson in logic, thanks. I see now why WM is constantly accused of quantifier dyslexia. It's because folks here refuse to see what is abundantly clear and obvious, no matter how carefully it is explained, or how many times. It's Cantorian dyslexia. This point could not be more clear, and yet you seem almost to insist that S^L can be infinite with finite S and L, and that you can increment a value an infinite number of times and still have a finite value. Can this really be the position of the professional mathematical community? Do you really not understand what "if and only if" means? > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony
From: Tony Orlow on 25 Jul 2005 09:29 Daryl McCullough said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > >imaginatorium(a)despammed.com said: > >> Tony has no clue what mathematics is, nor how it is done, so he doesn't > >> normally bother with definitions. The closest we got from him for a > >> definition of "finite" was that a finite number is less than an > >> infinite one. And you can guess the "definition" of infinite. > >Well, that's about as close to a lie as one can get, eh? > > >I asked for a definition of infinite, and no one could give me a > >definition of that word. The best I could get was that an infinite > >set can have a bijection with a proper subset, which is hardly a > >definition of the word "infinite". > > On the contrary, that's a perfectly good definition of the concept > "infinite set". > > -- > Daryl McCullough > Ithaca, NY > > But not the word "infinite" on its own. Do you think the dictionary has the word "set" in the definition of "infinite"? Are sets the only way to think about infinity? Not hardly. -- Smiles, Tony
From: David Kastrup on 25 Jul 2005 09:31 Dave Seaman <dseaman(a)no.such.host> writes: > On Mon, 25 Jul 2005 10:45:20 +0200, David Kastrup wrote: > >> There is no room for a bit of error in the tenth place if you are >> factoring primes. > > I can factor primes without looking at any of their places. Give me half the bucks of Bill Gates, and I won't mind repeating his mathematical blunders. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 25 Jul 2005 09:38
imaginatorium(a)despammed.com said: > 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. Uh, no. Incorrect. The fact is that any set of whole numbers greater than or equal to 1 MUST have at least one element with a value at least equal to the size of the set. That is a fact and has been proven. > > 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. I bet if Wolf asked his grandchild what number you would get if you added 1 forever, he or she would say infinity. You, on the other hand, despite the knowledge of infinite series, seem to believe you can add an infinite number of 1's and NOT get infinity. Really quite fascinating, this theory of yours.... > > > Brian Chandler > http://imaginatorium.org > > -- Smiles, Tony |