Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 27 Jul 2005 16:57 In article <MPG.1d5190cb1cee0bd2989fae(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > malbrain(a)yahoo.com said: > > Randy Poe wrote: > > > Tony Orlow (aeo6) wrote: > > > > Well, if this is how mathematicians feel about their study, and claim > > > > it has no > > > > connection to reality, then Cantorian set theorists really have no > > > > reason to > > > > claim that anything they say is more correct than any competing claims. > > > > > > As usual, you are confused. Your arguments deal with conclusions > > > which you draw FROM THE SAME AXIOMS as the competing, mainstream > > > theorems you denigrate. > > > > Ok. Perhaps you might give us the pre-set-theory version of the axiom > > of induction. I presume it didn't just spring-up from nothing. > Zeno's paradox is certainly an early form of inductive reasoning. TO conflates mathematical induction with inductive reasoning. They are quite different.
From: Tony Orlow on 27 Jul 2005 17:01 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? Is every subscript finite? Are there only a finite number of such sets? > > 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. > > 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 if you believe there's a finite L, don't you think > there's a set S_L in this collection? And isn't there > a set S_L+1? There is no largest finite L, like there is no largest finite n. So? Shall I get the chicken blood and the rattle? > > - Randy > > -- Smiles, Tony
From: Daryl McCullough on 27 Jul 2005 16:42 Tony Orlow <aeo6(a)cornell.edu> said: >So, the sum of a finite number of finite terms is infinite. Sure. No. The sum of a finite number of finite terms is finite, and the sum of an infinite number of finite terms is infinite. -- Daryl McCullough Ithaca, NY
From: David Kastrup on 27 Jul 2005 17:06 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Obviously, i am rejecting some of your axioms and assumptions, so I > am not working with the same set as you are. Then you have no business using the terms defined by those axioms for your own means. You want to change a building by first pulling down the first two stories and putting something else in afterwards. You'll find that you can't hope for the upper stories to just stay put. You'll have to build your own house from the ground up. And then you don't get to redecorate the existing upper stories. > Mathematics as a whole should be viewed as a system of axioms that > should be consistent overall. Does a pretty good job at that. > That is the real litmus test of a given subsystem: external > consistency within the broader field of all math. "External consistency". What a beautiful oxymoron. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Robert Low on 27 Jul 2005 17:08
Tony Orlow (aeo6) wrote: > Daryl McCullough said: >>Tony, you are just winging it, making up your answers as you go, > If I am just winging it, I'm doing pretty good. <cough> > Do you have a specific > incosistency you'd like to mention? I don't see it. Apart from your insistence that there are only finitely many finite numbers, so that there's one which adding 1 to makes it infinite, but that's ok because you don't know which one it is? If I put my fingers in my ears and sing the theme tune to "The Banana Splits Show" real loudly, I can't hear people telling me to shut the hell up. It doesn't mean they aren't saying it. >> n is finite >> n+1 is infinite > Virgil has the same problem. In the case of the finite naturals, it simply does > not hold true. Give the proof and I'll take a look at it. Suppose that the set containing the natural numbers 1 to n is finite, but the set containing 1 to n+1 is infinite. Then there is an bijection, f, from the set {1...n+1} to some subset of itself. If n+1 is not in the range of f, then restricting f to {1...n} gives a bijection from {1...n} to a subset of itself (since the range of f does not contain f(n+1), which is in {1...n}). Hence {1...n} is infinite. Contradiction. If n+1 is in the range of f, then if f(n+1)=n+1, restricting f to {1...n} again gives a bijection from {1..n} to a subset of itself, so {1..n} is infinite. Contradition. If n+1 is in the range of f, but f(n+1) is not n+1, then f(n+1)=m for some m<=n, and f(k)=n+1 for some k<=n. Define g to be equal to f except that g(k)=m and g(n+1)=n+1. Then restricting g to {1..n} gives a bijection from {1..n} to some subset of itself, and so {1..n} is infinite. Contradiction. Not that this will convince you, of course. |