Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Daryl McCullough on 27 Jul 2005 13:04 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >Daryl McCullough said: >> 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? Right. For each n, S_n is finite. >There are an infinite number of such finite sets? Yes, there are infinitely many possible values for n. >Do they then go, say, from S_1 to S_oo? The sequence S_1, S_2, ... is an *infinite* sequence; it's a sequence that has no end; it has no last set. So no, the last set is not S_oo because there is no last set. >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? There *is* no S_oo. The possible values of n are the *finite* naturals. I'm telling you that S_1, S_2, ... is a sequence without end, with no last set, and you are asking me what the length of the strings in the last set. That doesn't make any sense. There is no last set. The sequence goes on forever. -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 27 Jul 2005 13:23 Daryl McCullough said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >It seems like axioms are given a status that makes > >them unquestionable and almost incomprehensible, beyond the application of > >them. > > They are not incomprehensible to the people who work with them. Right. That's why nobody here seems to have recognized the basis for the inductive axiom, and misses the fact that it constitutes an infinite recursion. Sure. You guys really understand your tools. > There purpose is to *clarify* what is going on, and they serve > that purporse well. Axioms allow for precise communication between > mathematical workers, and they allow for objective criteria for > when a proof is valid or not. Axioms need to be proven too, from outside the system where they are axioms, often inductively. > > In contrast, what you consider to be a proof seems to be purely > subjective. Yeah, sure, read the two I gave you today and respond specifically, please, without making foolish statements like that. > > >For instance, everyone's dismissal of the infinity inherent in the > >recursive nature of inductive proof is a sign that the axiom is not really > >understood, but accepted without question. > > Nothing in mathematics is excepted without question. Not by > mathematicians, anyway. That's not what they say all the time here. They typically consider axioms unquestionable truths, or at least act like they are. > > Yes, it is certainly the case that *if* you can prove by > induction "forall x, Phi(x)", *then* you can write a > corresponding recursive function that given a number n, > produces a proof of Phi(n). Nobody disputes that. What > people are disputing is your bizarre belief that proving > "forall x, Phi(x)" by induction means that you have proved > Phi(0), Phi(1), Phi(2), ... It means that you *can* prove > all those infinitely many statements, not that you *have*. It means there is an infinite chain of implication, defined recursively. To forget that is folly. > > >For me, mathematics without meaning is unsatisfying, and > >symbolic manipulation without understanding is boring. > > That's true for all mathematicians. The difference with you > is that you don't want to do all the work necessary to understand > real mathematics. Don't make dumb statements, Daryl. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 13:26 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > Various axioms have their various issues. The most pertinent to this > > discussion right now, it seems, is Peano's 5th. I don't disagree > > with the axiom or with the concept of inductive/recursive proof, > > There is no such thing as "recursive proof" in this context. If you don't see it, you're blind. > > > but in order to eb careful that what we are doing is correct, we > > need to keep in mind the original justifications for axioms when > > applying them. > > Wrong. An axiom needs to stand on its, absolutely. If it requires > additional considerations, it was ill-chosen. Fortunately, this does > not appear to be the case with the 5th Peano axiom. See, Daryl? Unquestioning acceptance without critical examination. Typical. > > > If you are applying a method such as inductive proof, with an > > inherent infinite loop, you cannot maintain finiteness through an > > infinity of iterations, each involving finite increase in value. > > Completely irrelevant chitchat to the 5th Peano axiom. It is not > bothered about "increase" in value, it is not bothered about > "maintaining finiteness", it is not bothered about "iterations" or an > "infinity" of them. The proof that all naturals are finite is not concerned with maintaining finiteness as values increase? Uh, sure, David. Whatever you say. Go back to bed. > > It works without you having to keep an eye on all that folderol. > > That's what makes it a good choice. Good because it's easy. I see. Go back to bed. > > -- Smiles, Tony
From: Robert Low on 27 Jul 2005 13:25 Tony Orlow (aeo6) wrote: > 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: But nobody has disagreed with that. The point of contention is whether the union over all n of S^n contains an infinite string; it doesn't, but you claim it does. When you answer the question: "How many elements does the set of all finite integers contain?" you will be enlightened, and all these mysteries will become clear to you.
From: Daryl McCullough on 27 Jul 2005 13:06
Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >If you place no restriction on the length of strings, then they can be >infinitely long. No, there is only one restriction on the length of strings, and that is that the length is finite. -- Daryl McCullough Ithaca, NY |