Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 26 Jul 2005 10:55 Han de Bruijn said: > Tony Orlow (aeo6) wrote: > > > 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". In fact I went to > > the etymology, which literally means "without end". Finite means with a known > > end or bound, and infinite means without end. Of course, I got all sorts of > > flack for my definition, from those that couldn't even suggest one outside of > > the set theory they were regurgitating. let's try to be straight here, and no > > more insulting than necessary, so it doesn't come back to bite us, why don't > > we? > > The infinite they define as "that an infinite set can have a bijection > with a proper subset" of itself is known as "_actual_ infinite", which > is rejected by most anti-Cantorians as sheer nonsense. Apparently, you > are rather talking about "_potential_" infinity: something finite that > becomes larger and larger. The indisputably useful concept of a "limit" > falls within the latter category. Everything that involves the infinite > and cannot be handled with limits is rather suspect IMHO. > > Han de Bruijn > > I agree. I suppose Dave Petry is right in ascribing some of the anti- Cantorianism to a sense of computation. Limits provide wonderfully precise ways of dealing with infinities and zeroes. I really have no objection to the current definitions of finite and infinite, using bijections with subsets. Part of what I consider ill-fated is this distinction between natural numbers and set sizes, since that is what a natural number essentially is. Still, I don't see much reason to find a better definition for infinite. It works. I think we all know what the difference is. -- Smiles, Tony
From: David Kastrup on 26 Jul 2005 10:55 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Virgil said: >> In article <MPG.1d489d7fea8af732989f60(a)newsstand.cit.cornell.edu>, >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >> >> > I have explained the flaw in this proof, and it is met with >> > confusion because none of you seems to appreciate the recursive >> > nature of inductive proof. >> >> The inductive axiom shortcuts that recursion, which is the point of >> the inductive axiom. It says that if the recursive step can be >> proved in general, then it never need be applied recursively. >> >> If TO wishes to reject the inductive axiom, only then can he argue >> recursion. > > What a load of bilge water! Accepting the axiom as a general rule > does not mean one has to immediately forget about the lgical basis > for the axiom. Of course not. But the basis for choosing an axiom is irrelevant to the application of the axiom. And this axiom was chosen exactly in a manner that does not require recursive application. Whether it was chosen because it is equivalent to arbitrarily deeply nested recursion or because Kronecker's neighbor had a particularly ugly elephant locked in his belfry is irrelevant to its application. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 26 Jul 2005 11:02 Han de Bruijn said: > David Kastrup wrote: > > > The rules are: if you don't like some conclusions, you have to change > > the axioms, and then you lose all other conclusions (many of them > > might be easy to reacquire, but that process is not automatic). > > But what if your method is not axiomatic ? I mean, in intuitionism, the > emphasis is not on formal reasoning and axions, but "constructiveness". > > Han de Bruijn > > I think that's why I have such trouble here. I don't work axiomatically, though I intend to work on that for the benefit of this issue. I generally construct a concept and analyze it for patterns, rather than apply rules in symbolic form. It's an inductive approach, rather than deductive, and it's equally important. Without inductive methods to produce your axioms and rules, you don't have anything to deduce. It's the old Yin-Yang of logic. -- Smiles, Tony
From: Daryl McCullough on 26 Jul 2005 10:46 Tony Orlow (aeo6) wrote: > >imaginatorium(a)despammed.com said: >> ...That means that when I say "the pofnats", >> this expression could be replaced by the expression "the subset of the >> Tonats containing only those which are unambiguously finite"... >> Can I not select those of the Tonats that are finite? > >Sure you can talk about such a set. You just can't draw very much >in the way of conlusions about its size or upper bound. On the contrary, you can easily prove of this set (call it A) 1. There is no n in A such that size(A) = n. 2. There is no n in A such that all n is the upper bound for A. >I'll try, but there are so many times when it seems my words are >deliberately misrepresented, and grammer is misconstrued, and the >point obfuscated, and that it is a simple defenseive maneuver. >If people just say "define!", then it seems like not-picking nonsense. No, it's not. The whole point of defining one's terms and writing down the axioms for using those terms is that then *anyone* can prove theorems about the subject, and *everyone* will agree that those are indeed theorems. In contrast, if (as you prefer) you never give definitions for your terms, and you never write down axioms for using those terms, then *nobody* except you is able to prove anything about your concepts. For example, you claim that if S is a set of bit-strings, and S is an infinite set, then S must contain at least one bit-string of infinite length. Nobody can prove such a claim except you. In contrast, with the usual definitions of "infinite set", "bit-string", and "length", plus some basic facts about naturals, *anybody* can prove the negation: That there exists a set S of bit strings such that (1) S is infinite, and (2) no string in S is infinite. By not giving precise definitions and axioms (and by precise, I mean a *mathematical* definition, in terms of functions, relations, membership, logical operators, etc.) you are basically limiting your mathematics so that it only is relevant to you. >"Finite" means it has an end, whether its a set, >quantity, process, structure, or whatever. >Infinite means it doesn't. Do you think that the set of real numbers between 0 and 1 (inclusive) is an infinite set? But it certainly has an "end". It has two ends: 0 and 1. So your definition "without end" is not a useful definition. A better definition might be: A set S is infinite if it is possible to hop from one element of S to another, and never stop, and never return to a place you've visited earlier. But mathematically, what does that mean? How do you characterize such a progression? Well, you can characterize it by two quantities s_0 = some starting point in the set S next-hop = a function that takes you from where you are now to your next point in S To say that the progression never ends and never repeats can be mathematically formalized this way: Let S1 be the set { s_0, next-hop(s_0), next-hop(next-hop(s_0)), ... } Then next-hop is a function from S to S with the properties that 1. By definition of S', if x is in S', then next-hop(x) is in S'. 2. for any x in S', next-hop(x) is not equal to s_0 3. for any x and y in S', if x is not equal to y, then next-hop(x) is not equal to next-hop(y). But those three conditions are *exactly* what mathematicians mean when they say that next-hop is a bijection between S' and the set S' - { s_0 }. So the mathematical definition, that a set is infinite if and only if there exists a bijection between that set and a proper subset *exactly* captures the intuitive notion of the possibility of going on "without end". -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 26 Jul 2005 11:06
Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >Right. The Peano axioms don't fully address this issue, hence the confusion. Yes, they do. (well, as well as it *can* be addressed in first-order logic). >However, you cannot argue that any particular S^L can be infinite with >finite S and L, can you? No. There are only finitely many strings of size 1, finitely many strings of size 2, finitely many strings of size 3, etc. But if you place no restriction on the size of the strings, then there are infinitely many strings (each of which is finite in length). >If S is finite (2 for binary, etc) then we must allow infinite >strings, in order to have infinite numbers of strings. 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. -- Daryl McCullough Ithaca, NY |