Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 27 Jul 2005 12:08 Daryl McCullough said: > 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. That's funny, because I proved that, if N starts with 1, then at any point in the generation of the set, the set size is also equal to the maximal element. > 2. There is no n in A such that all n is the upper bound for A. You mean ther eis no single n which is the upper bound for the set? True. The only upper bound for the set of finite whole numbers is whatever smallest infinity one can concoct. > > >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. I understand that, but as far as defining "finite", "infinite", "string", and "set", when I am using these terms, as far as I can tell, by their normal definitions, it sounds like a deliberate attempt to wear me down. This is already kind of draining, battling with this pack. I have started on a set of axioms and definitions, and will keep your suggestions in mind, but don't have the system ready for delivery yet. Sorry. In the meantime, if it sounds like I am using a word in some non-standard way, then ask. That would help me realize what I may have to clarify. I am not used to putting things in such rigorous verbal format. It's much easier for me to draw pictures. :) > > 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. I gave you an axiom which is easily demonstrable inductively, and with which no one has taken issue (I hope): With a set of symbols of size S, one can create a set of unique strings of length L, whose maximum size is N=S^L. Among the Axioms of Finiteness (which probably exist already, somewhere): For A and B finite (non-zero) and infinite, A^B or A*B is either infinite or finite according to the following table- A Finite Infinite ----------------------------------- B | Finite | Finite Infinite Infinite | Infinite Infinite Note that this table is equivalent to the OR truth table, when you replace "finite" with 0 and "infinite" with 1. In other words, the result is infinite if either A OR B is infinite, and is finite if A AND B are finite. When we combine these two "axioms", we see that N=S^L is only infinite for either infinite S or L. Therefore, reapplying what our symbols mean to "reality", we can only have an infinite set of strings if we either have an infinite set of symbols with which to build them, or allow infinite length strings. Now we introduce the digital number systems, defined as usual, with S equal to the radix, a finite whole number greater than 0, the symbols in the set representing the whole numbers from 0 to the radix minus 1, and each string position representing a power of the radix that increases going leftward, with a digital point directly to the right of the digit place representing the zero power of the radix. The value represented by the digital number, V, is given by V=sum(i=-oo->+oo: a_i * b^i), where a_i is the digit at digit place i, and b is the radix. That sounds about right, doesn't it? Did I miss anything? Now, if i is infinite and positive for a given digit, then it is infinitely far to the left of the point, and represents the radix to an infinite power, right? So, given our axiom of finiteness, we can say that a finite radix to an infinite power gives an infinite result, right? And, if a_i is non-zero, then we can also use our axiom of finiteness to say that this infinite b^i multiplied by the finite non-zero a_i, yields an infinite result for V, right? Finally, I think we can agree that, if all digits in infinite digit places are zero, then it is equivalent to having a finite number of digits, not an infinite number, so if we require an infinite number of digits, that means we require an infinite number of non-zero digits. So, we have three axioms (actually it needs to be about 10, right, to cover the digital numbers ?): 1. N=S^L 2. A^B and A*B are only finite for finite A and B, and otherwise infinite. 3. A digital number with nonzero digits in digit places infinitely to the left of the point represents an infinite value. So, infinite(N)->infinite(S) OR infinite(L) by 1 and 2 NOT(infinite(S))->infinite(L) by definition of radix infinite(L)->infinite(V) by formula for V Hopefully this is step-by-step enough to get SOMEBODY to understand this line of reasoning. > > 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. Those are bounds on the values, and endpoints to the line segment, which is a finite space. Are you telling me you don't already know what part of that set is infinite, and doesn't end? It is not infinitely long, to be sure, but it is infinitely divisible, being a real interval. The process that never ends is the nested subdivision of the line segment. It's a process infinity, more than a quantity infinity. But, you knew that already, didn't you? ;) Nice try. > > So your definition "without end" is not a useful definition. Sure it is. It can be applied to all infinities, each in their own way. > 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 }. You basically just replaced successor with next-hop, for the Peano deinition of the naturals. > > 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". Sure. I don't diagree that that definition works for defining infinite sets. My problem with bijections is the notion that being able to form them between two sets automatically means those sets are equal in size, when they are infinite. The nature of the bijection must be taken into account, whether symbolic, numeric, or in any other sort of unending system. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony
From: Tony Orlow on 27 Jul 2005 12:17 David Kastrup said: > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > > > David Kastrup wrote: > > > >> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > >> > >>>And this is precisely what anti-Cantorians find unacceptable. IMHO > >>>an "infinite" set cannot consistently have a "size". > > > >> This is a _perfectly_ valid point of view (as opposed to the views > >> of those you sympathize with which are wildly inconsistent). > > > > Uh, uh. How do you know with which I symphathize? Do you keep a record > > of my responses to everybody? > > Well, you use the word "anti-Cantorians" above. If this was not > intended to mean mostly a particular set of some outspoken people in > this Usenet group, it would appear that I misinterpreted this. > > > [ .. rest deleted .. ] > > > > OK. It seems that we finally have arrived somewhere. I have one > > final question, though. Is it "legal" (according to mainstream > > mathematics) to call a set "countable" if it can be brought in 1:1 > > correspondence with the naturals? > > That's the usual usage of the word. > > > And uncountable otherwise? > > Well, in one direction: {1} can't be brought into 1:1 correspondence > with the naturals, either. But if you can't map the naturals to cover > the set exhaustively, than the usual term would be "uncountable". > > It may be that there is a bit of excluded middle: I think finite sets > are usually classed neither as countable nor uncountable. Not sure > about that, though. > > > Perhaps it's trivial for anyone, but I'm a bit at lost (due to those > > heated "Cantor's diagonal argument" threads I think :-) > > Tony Orlow appears to assign some other meaning to those words, but he > has not yet come up with a useful definition. Apart from him, I think > most of the participants in the group seem to agree on the meaning. > > No, I understand what you mean by "countable". My issue with your "uncountability of the powerset of the naturals" is that you can easily form a bijection between the members of the powerset and the set of infinite bit strings denoting set membership in the subsets, and between those bit strings and whole numbers, and the only reason this bijection is rejected is because of the refusal to allow infinite whole numbers in the set of whole numbers, despite the fact that an infinite set of whole numbers requires infinite whole numbers, so you don't consider the infinite bitstrings to represent whole numbers. It's like all the wrong choices have been made, almost on purpose, in order to make some grand distinction which really isn't there. If you allow infinite whole numbers, as is required, suddenly this whole "uncountability of the power set" vanishes in a puff of smoke. -- Smiles, Tony
From: Daryl McCullough on 27 Jul 2005 12:00 Tony Orlow (aeo6) wrote: >Robert Low said: >> Are there rationals in [0,1) where p and/or q have to be infinite? >> >Interesting question. In order to have an infinite number of them, yes, you >would need either an infinite number of numerators or an infinite number of >denominators, both of which are whole numbers. Tony, you are just winging it, making up your answers as you go, without any regard to whether they are consistent with what you've already said. That's the point of rigorous definitions and axioms--they keep the mathematician honest. He can't just make stuff up, he has to follow the rules---even if they are rules that the mathematician made up himself. Your basic claim is that if S is a set of naturals, then either S is finite, or S contains an infinite natural. Why in the world do you claim that? It's provably false. Let's go through the steps, and you can say which ones you disagree with: 1. First, do you agree that if S is a finite set, then the number of elements in S must be some (finite) natural number? If that's not the case, then I don't know what you mean by finite (and I suspect that neither do you). 2. Second, do you agree that if n is any natural number, and S has exactly n+1 elements, then S has a largest element? I don't know how you could believe otherwise. It's provable by induction on n. Maybe you want to say that induction is only sometimes true and sometimes false, depending on how you feel at the moment? So, from 1 and 2, it follows that Let FN = the set of finite natural numbers. If FN is finite, then FN has a largest element. So your claim that FN is finite implies that there is some number n that is the largest finite natural number. That means that n is finite n+1 is infinite By any sane definition of "infinite" I would think that would be impossible. -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 27 Jul 2005 12:25 Daryl McCullough said: > 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 you place no restriction on the length of strings, then they can be infinitely long. If they cannot be infinitely long, then they are restricted to finite lengths, and the set size is restricted to a finite size. Set theorists need to get back to using real numbers more often. > > >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. 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? > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony
From: Randy Poe on 27 Jul 2005 12:34
Han de Bruijn wrote: > > Randy Poe wrote: > > > >> Where do you believe set theory comes up in physics? > > > > If you believe that music is physics I don't, and I'm more than passing familiar with both. > > ... I have an application where set > > theory is employed But that doesn't mean that the sets are real, or that the chords are really sets. All it means is that you have created an abstraction of chords in terms of sets. And your set description of chords says nothing whatsoever about the physics of music. Can you tell me why {C,E,G} is a chord and {C,C#,D,D#} is not? What does set theory contribute to that? > >> By the way, I don't think your music idea is valid beyond your > >> trivial example. Show me how it tells me the right chord > >> progression to go from, oh, C Maj to F# Maj to G minor for > >> instance. > > > > Oh well. C Maj = {C,E,G,B} , F# Maj = {F#,A#,C#,F} , G minor = {G,A#,D} > > Hence F# Maj \ C Maj = {F#,A#,C#,F} On , C Maj \ F# Maj = {C,E,G,B} Off. > > (You've defined disjoint sets.) Now wait and listen. Then: > > G minor \ F# Maj = {G,D} On , F# Maj \ G minor = {F#,C#,F} Off. Why is that the "right" one? How about this one? {C,E,G,B} = C Maj {C,F#,G,B} {C,C#,F#,G} {A#,C#,F#,G} {A#,C#,F#,F} = F# Maj, according to you By the way, the set description of chords also does not distinguish inversions, yet inversions are crucial to "pleasing" chord changes. - Randy |