Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 28 Jul 2005 00:10 In article <MPG.1d51b16564a56569989fc9(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > malbrain(a)yahoo.com said: > > Chris Menzel wrote: > > > On Tue, 26 Jul 2005 16:44:24 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > > > > I am saying that if L CANNOT be infinite, then S^L CANNOT be > > > > infinite, > > > > > > No one disagrees with that, for fixed S and L. > > > > > (...) > > > > > > > For finite S, S^L can ONLY be infinite with infinite L. Why is this so > > > > hard to understand? If S and L are both finite, then S^L is finite, > > > > isn't it? > > > > > > Yes, of course. But that's for a fixed L, say 17. But for any given > > > nonempty set S of natural numbers, the set of *all* finite sequences of > > > elements of S -- i.e., S^1 U S^2 U S^3 ... -- is infinite. That's the > > > part you don't seem to get. > > > > Obviously, he doesn't. Perhaps using the definition for all will help: > > The set made by taking each and every (finite) L sequence of elements > > of S is an infinite set. > > > > karl m > > > > > So, the sum of a finite number of finite terms is infinite. The sum of enough finite naturals can be made larger than any finite quantity. To ignore this as TO does, demonstrates once again his quantifier dyslexia.
From: Virgil on 28 Jul 2005 00:16 In article <MPG.1d51bac9dec3f89b989fcc(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > Well, bijections provide an excellent way to compare sets, because > > it is an equivalence relation, and so they define equivalence > > classes. > The denote equivalence for finite sets, and some kind of relation for > infinite sets, but not necessarily equivalence. Does TO then deny that bijectability is an equivalence relation? His ignorace is peeking through again.
From: imaginatorium on 28 Jul 2005 00:19 Daryl McCullough wrote: > Tony Orlow (aeo6) wrote: > > >> Who cares about S_oo? That's not one of the S_n. > > >You mean there aren't an infinite number of n in N? > > Okay, maybe this will make it clear to you: I'm not sure which is the more serious ailment: quantifier dyslexia, or terminally hopeless optimism. > There is > exactly 1 natural number that is less than 1: namely 0. > There is exactly 2 natural numbers that are less than 2: > namely 0 and 1. There are exactly 3 natural numbers that > are less than 3: 0, 1, and 2. > > You see the pattern? For any n, the number of naturals > less than n is equal to n. > > Now, how many natural numbers are less than infinity? Why, even I can do that one: "infinity", of course. Rename "infinity" as some fancy Hebrew name, and it becomes just what we were looking for - the indeterminate finite number that is the size of the finite numbers. I remember in my youth watching some TV drama about alchemy, and I remember that the crank^H^H^H^H^Halchemist kept talking about "imponderable fluids". Have you ever thought, I mean thought deeply, about an imponderable fluid? Brian Chandler http://imaginatorium.org
From: Virgil on 28 Jul 2005 00:29 In article <MPG.1d51c03096557080989fd1(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Randy Poe said: > > > > > > Tony Orlow (aeo6) wrote: > > > > > > Why do you keep saying that? It's provably false. The set of > > > > all finite strings is an infinite set. It's infinite by *your* > > > > definition of infinite, in the sense that it is "without end". > > > > The set of all finite strings is the union of > > > > > > > > S_1 = the set of strings of length 1 S_2 = the set of > > > > strings of length 2 S_3 = the set of strings of length 3 > > > > ... > > > > > > > > The collection of subsets S_n goes on without end. > > > So, each of these sets is finite right, given finite S and L? > > > There are an infinite number of such finite sets? Do they then > > > go, say, from S_1 to S_oo? And S_1 is the set of strings of > > > length 1, and S_2 is the set of strings of length 2, etc, so S_n > > > is the set of strings of length n? Okay. What length are the > > > strings in S_oo? > > > > Who cares about S_oo? That's not one of the S_n. > You mean there aren't an infinite number of n in N? No he means, as usual, that there are infinitely many finite n in N. > > > > The question being contested is, how many strings are there in the > > union S_1 U S_2 U S_3 U ... where the union is over all the S_n for > > finite natural numbers n? > 2^(n+1)-2, which is finite for finite n. Why not 2^(n+2)-2 ? Is n+1 not a finite value for n? Or 2^(n+3)-2, and so on. Where does it end? If TO cannot say where it ends, it doesn't have to end. > > > > You say that's a finite number. You say there's some finite L which > > is the size of the largest string in this collection. > For finite n, 2^(n+1)-2 is finite, no matter what n is. Finite n -> > finite set of strings. But a set of finite strings can be infinite in the sense of having ever larger but still finite members. It is like a divergent monotone increasing sequence of all finite values. It keeps reaching farther and farther without ever reaching actual infinity.
From: Virgil on 28 Jul 2005 00:31
In article <MPG.1d51c252ad53d1d4989fd2(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Robert Low said: > > Anyway, I'd still like to know how many elements there are in the > > set of all finite integers. When you can answer that, there might > > be some scope for further development. > > > LOL. You call it aleph_0, so why should I argue? It's no wonder you > are satisfied with set sizes that you can't do anything with. If > that's the size, though, it's a finite number. I suppose it's a > wonderful name for a number that can't be pinned down. That "number" does have the interesting property the any set of that size allows injections into proper subsets. |