Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 20 Jul 2005 15:01 In article <MPG.1d4839e77d7d2c0a989f45(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > If TO's assumprtions were actually the case, there would have to be a > > finite natural so large that adding 1 to it would produce an infinite > > natural. But TO cannot produce either a largest finite nor a smallest > > infinite, so the set of all finite naturals is already big enough. > > > We have been through all this before. You lay these requirement on me, but > when > I say you cannot have a smallest infinite omega, or omega-1 would be finite, > you say omega-1=omega. I do not say anything at all about omega. I merely go by the Peano properties, which prohibit TO's vision of "infinite naturals". So, I can play your stupid game, and say that alpha is > the largest finite, but alpha+1=alpha. Tada! I am as senseless as you! Isn't > that great? Let me know when you want to get off the "largest finite" kick. I > won't be responding to it any more. It is not me who keeps requiring a largest finite, but TO, by his delusional insistence that there must be non-finite naturals within the Peano system.
From: Daryl McCullough on 20 Jul 2005 14:51 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >imaginatorium(a)despammed.com said: >> Tony has no clue what mathematics is, nor how it is done, so he doesn't >> normally bother with definitions. The closest we got from him for a >> definition of "finite" was that a finite number is less than an >> infinite one. And you can guess the "definition" of infinite. >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". On the contrary, that's a perfectly good definition of the concept "infinite set". -- Daryl McCullough Ithaca, NY
From: stephen on 20 Jul 2005 15:19 In sci.math Daryl McCullough <stevendaryl3016(a)yahoo.com> wrote: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >>imaginatorium(a)despammed.com said: >>> Tony has no clue what mathematics is, nor how it is done, so he doesn't >>> normally bother with definitions. The closest we got from him for a >>> definition of "finite" was that a finite number is less than an >>> infinite one. And you can guess the "definition" of infinite. >>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". > On the contrary, that's a perfectly good definition of the concept > "infinite set". These seems to be another common misconception among the anti-Cantorians that words cannot have specific meanings in specific contexts. Somehow they think an all-encompassing definition of 'infinite' must be provided before someone can say what an infinite set is. I am not sure what they mental hangup is. I wonder how any of them would ever learn a foreign language. What is so hard about "A set is infinite if there exists a bijection between the set and a proper subset of the set." ? There is no need to get all mystical and metaphysical just because the word 'infinite' shows up. The word 'infinite' may have other definitions in other contexts, but that is true of all words, and is irrelevant in a discussion of infinite sets. Stephen
From: David Kastrup on 20 Jul 2005 15:21 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > David Kastrup said: >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> > Don't be a jerk. >> >> That's your job description. > I didn't accuse anyone of sulking, especially when they were doing > anything but that. Typical Cantorian ad hominem. Thanks for proving my point by that ad hominem. >> > No, the proof is a misapplication of the axiom. The inductive proof >> > method works well for constant equalities, >> >> What is supposed to be a "constant equality"? > > An equality that holds true for n=1, and for n=n+1 given true for n. For n=n+1. Do you even read the nonsense you write? > An equality, as opposed to an inequality, or a vague property such > as "finite", which is really an inequality that is constantly > decreasing. This is hand-waving hogwash. Please come up with a proper definition of "constant equality" if you want to argue using it. >> > but in the case of proving that all naturals are finite, you are >> > incrementing the value at each of an infinite number of steps >> >> Nonsense. Complete bullshit. I do nothing at all like that. >> There are no steps whatsoever involved here, and certainly not an >> infinite number of them. n is finite for n=0, and it is shown that >> for any finite n, n+1 is finite. > So, inductive proof does not rely on proving for n+1 based on n? It relies on exactly that. But you need not check this explicitly for any n except 0. It is sufficient to show the implication from n to n+1, and the case for 0. Those two steps are all that is required. > The infinite number of successive naturals for which you prove your > property do not constitute an infinite number of implied steps in > inductive proof? No. The inductive proof consists of two steps. Once you have covered them, the proof is established for _all_ n. That's what induction is all about. > Because you only require three lines for the construction of the > proof, you think there are only three steps implied? Yes, exactly that. Exactly that is the purpose of defining the naturals by a small set of axioms: being able to deduce general properties without having to check each number individually. > Talk about nonsense, and complete bullshit. It is called generalization. >> That is not an infinite number of steps, but at best 2 steps. >> >> > and maintaining that the value remains finite, because you are >> > looking only at individual steps. >> >> That's what induction is about. Showing f(0), and proving >> f(n)->f(n+1). Two steps, and this covers the naturals. If you >> want to crosscheck more than that, feel free to do so, but the >> axioms don't require doing that. That's their whole point: not >> having to check things piece by piece. > The truth for n+1 depends on the truth for n, for all n in N, from 1 > to oo, an infinite number of times, proving the property true for an > infinite number of numbers. It does not depend "an infinite number of times". It is just one general dependency. And yes, as a general dependency it obviously holds for an infinite amount of numbers. But that does not mean that I have to check each number individually. > If the inductive portion of the proof, where truth is derived for > n+1 from truth for n, increments a value, f(n)->f(n+1) is a static relation between two different statements depending on n. It does not increment anything. >> It is irrelevant. The axioms don't talk about steps, and certainly >> not an infinite number of them. You are babbling what you consider >> intuitive, but it bears no relation to the math. > the first step is 1, and subsequent steps are denoted by n, and > there are an infinite number, as many as you have n in N. The Peano axioms don't involve "steps" or any progress or whatever. It is utterly irrelevant in what order you determine f(6)->f(7), f(3)->f(4). If you have ascertained f(0), and the validity of f(n)->f(n+1) by whatever means and in whatever order, the validity of f(n) for all n is established. No need for an infinite number of steps. The whole idea of proof by induction is to prove with two steps a statement valid for all elements of N (which happens to be an infinite set). >> > Basically, adding X repeatedly Y times is the same as adding Y >> > repeatedly X times, so adding 1 and inifnite number of times is the >> > same as adding infinity once. >> >> Derive that from the axioms. If you can't, it is irrelevant. > So you contend that X*Y<>Y*X? Interesting. Wrong. I contend that the Peano axioms neither define nor use multiplication, and so multiplication is irrelevant in contexts where just the Peano axioms are used. > I am sure you must have an axiom in your bag for the commutative > property of multiplication? Check the side pocket.... Indeed, for _arithmetic_ on natural and whole numbers, there is another bag of axioms. You can't expect to be talking about multiplication before you have even defined the term. >> There is a strictly limited number of steps involved in an >> induction proof, and it provides proof for all natural numbers, >> which happen to form an infinite set. > And I guess only three of them are involved in the proof? How do you > prove it for number x? Does that not depend on x-1? The number of > inductive steps implied in proving a property true for all n in N is > infinite. Here is an example for a proof by induction: Proposition: The number of arrangements of k items is k!, where 0!=1, and (n+1)!=(n+1)n!. Proof: There is exactly one way to arrange 0 items, and 0! = 1 by definition. Now let us arrange n+1 items. This can be done by arranging n items (which offers us n! possibilities by the induction premise). After arranging the n items, there are n+1 possibilities to place the last item: for n=0, there is 1 possibility, and for larger n we can place the remaining item to the left of the n items, or to the right of the n items, or in any of the n-1 positions in between. So we have (n+1)n! possibilities for the combined placement. Finished. I don't need to check for n=7. This is already covered by the proof. >> > At each step in the proof, the set has a largest element which is >> > precisely the same as the size of the set. >> >> Fine, and so you prove something for a set that has a largest >> element. The set of natural numbers is no such set, and so your proof >> does not hold for it. > > What the proof shows is that, no matter how large the set gets, it > never has more members than the number represented by its largest > member. For every set that is characterized by a largest member. The set of natural numbers is no such set, and so your proof does not hold for it. > Therefore, if the elements are all finite, then the set size cannot > be infinite, If the set has a largest member. The set of natural numbers has no largest number, and so your reasoning does not hold for it. > since that would be larger than the value of every member in the > set. I am sure I can work out a proof without any reference to > largest element. Do so and then come back. >> > This constant equality holds for all such sets defined by ANY >> > natural number in the set. >> >> Certainly. Unfortunately, the set of natural numbers is no such set >> (since there is always a larger number than any given number in it), >> and so your proof does not apply to it. > > It applies to ALL n in N. Sorry. Take your complaint to Peano. Sure does. And that means that it applies to all sets defined by some n in N as its last element. The set of natural numbers is no such set, since it has no last element. Tough. >> > Most generally, it is impossible to have ANY set of natural >> > numbers, finite or infinite, which has a number of elements that >> > is greater than its largest element value. >> >> If it has a largest element value. The set of natural numbers has >> no largest element, and so your proof does not apply to it. > Yes, it does. No set of naturals can contain a greater number of > elements than all element numbers in the set. What is this? Proof by whining? The set of positive unit fractions has 0 as its lower bound, yet 0 is not a member of that set. >> > This should be clear to anyone who thinks about it. It is >> > impossible to have an infinite set of strings, of which digital >> > numbers are a type, without having either an infinite base, or an >> > infinite number of digits. >> >> Oh, there is an infinite number of digits, but every single string >> only occupies a finite number of them. > > You mean it only has non-zero values in a finite number of digits? Every single number has non-zero values only in a finite number of digits. But there is no finite number of digits that would contain _all_ numbers. > That's the same as only having a finite number of digits, in which > case one can only have a finite number of unique digital numbers. Each number has a finite number of digits, but there is no fixed finite number of digits that would contain all numbers. >> >> > So, which inductive proof do you believe? You cannot add 1 an >> >> > infinite number of times to your maximal element, >> >> >> >> There is no maximal element, and the Peano axioms don't define >> >> "infinite number of times" or similar processes. >> >> > If Peano's fifth is correct, and inductive proof applies to the >> > entire infinite set in a stepwise manner, >> >> There is no "stepwise manner" in the axioms. > true for n+1, given true for n, for each n in N. steps, an infinite number of > them. Where is there a "step" in the fifth axiom? You are fantasizing your own rules. >> > then there are indeed an infinite number of steps implied. It's an >> > "immediate consequence", as you said above. >> >> It is an immediate consequence of the axioms that the described set is >> infinite, since the successor relation gives a bijection to a proper >> subset. But there are no steps involved at all. You just make them >> up for the sake of your personal intuition. They are a personal >> crutch of your own, and cause you to make mistakes that are not >> inherent in the axioms. > Uh huh. And the inductive construction of the naturals using the > successor operator at each step also involves no steps. There are no steps, so it does not make sense to talk about "at each step". The fifth axiom relies on "for every n". It is irrelevant whether you establish that in any order, forwards, backwards, even numbers first, then odds, or simultaneously, so it is nonsensical to talk of steps. There is no prescribed order in the axioms. > Maybe you could use a stepwise crutch so you don't have to crawl on > your belly. It is amazing that you cannot see that incrementing an > infinite number of times to generate an infinite set of naturals > does not result in infinite values. Poincare was right. But there is no incrementing in the axioms. If there is, point it out. And certainly not a "number of times". >> > and are proven using assumptions that are unfounded, >> >> Axioms are not "unfounded", and not assumptions. > Excuse me, but axioms are just that, for the most part. They are > statements assumed to be true for the sake of argument and > proof. But they are not assumed anew for every argument and proof. They are cornerstones of the mathematics built upon them. You can choose them arbitrarily, but there is a cost of throwing them away afterwards, and this cost gets higher as mathematics progresses. Sometimes it is still worth to pay the price for abandoning an established axiom. But some bumbling bozo not understanding their implications is not a worthwhile reason. >> > and told that the Banach-Tarski result is a "paradox" and not a >> > proof by contradiction. There really seems to be a bad influence >> > going on that allows people to think they have an infinite language, >> > when they only allow finite strings, or that the number of paths in >> > a binary tree, which is always half of the number of branches, is >> > suddenly infinitely larger than the number of branches, when those >> > numbers become infinite. >> >> Then look up and understand the definition of an infinite set. It is >> _the_ decisive mark of an infinite set that it does no longer obey the >> pigeon hole principle. > That's one conception, and one I reject. 1 is 1, whether it's in N, > or in {1,2,3}. It is not a "conception", it is the bloody _definition_. The word is no longer free for the taking. The very least you have to do is to _define_ it if you are going to use it with a different than the established meaning. Everything else is nonsensical. >> > I simply can't understand how mathematics has come to accept such >> > illogical results as "counterintuitive" rather than incorrect. >> >> Then get yourself books and learn. > > Yes, spend years trying to qwrap my head around concepts that are > clearly wrong. What a grand use of my time. Then just shut up. There is no sense in spewing off about something which you do not even plan to understand. > I am afraid I WILL have to tear it apart bit by bit to convince > anyone, but even the most obvious objections to your set of naturals > are rejected with insults, and no real refutation except your own > established definitions. Uh, that's what the definitions are for. Establishing what one is talking about. If you want to talk about something different, you need to start with defining things differently. If you don't, and then people prove you wrong _by_ _the_ _definitions_, then you are just wrong, period. >> You got it all confused again. There are no infinite natural >> numbers, so the "Cantorians" don't insist on anything for those >> non-existing entities. > You are doing exactly that when I suggest the inclusion of infinite > naturals. You are saying they don't behave the same way, so they > are different, and not in the set. Quite so, since the set is defined by the axioms. > I might as well say that infinite sets are not sets at all, because > my concept of a set is that every set is finite. Yes, and this would be a perfectly valid argument. Once you define a set as something finite, it is clear from the Peano axioms that the natural numbers can't form such a set, since the successor relation clearly violates the pigeon hole principle that a finite set obeys. But as long as we are talking about "natural numbers" and "sets" and "infinite sets" without redefining them, we are talking about established concepts. > Don't throw your definitions at me as if they are proofs. They're > not. But a proof that can't be traced back to the definitions is not sound. You can't prove things if you don't even know what you are talking about. >> > Does this seem a little unfair to you? Why do you insist that >> > eveything one can do with a finite number should work exactly the >> > same for an infinite number? >> >> There is no infinite natural number. The finite natural numbers >> are defined by the Peano axioms, and yes, everything that works >> with finite naturals works with all of them. That is the whole >> point of defining them with those axioms: to have a body of numbers >> governed by the same rules. > > And why don't you want all sets to have the same rules? if infinite > sets can have slightly different rules, why can't infinite numbers? Oh, they can. But they are not part of the naturals, because the definition of the naturals happens to rule them out. So you need to find a different context to talk about infinite numbers rather than them being natural numbers. >> If you don't like it, you can invent your own axioms and numbers, >> but then you can't cry foul that your laws don't hold with the >> numbers the others are talking about. > > Sure, except my numbers are a proper superset of your, so if > anything holds for all of them, then it holds for yours too. Feel free to come up with axioms and definitions that meet that criterion. Until you do, you'll look less stupid if you shut up. >> > The determination of divisibility by a number which is prime >> > relative to the number base depends on a termination to the >> > process of division. >> >> Good. And since all natural numbers are finite, every division >> among those naturals is guaranteed to finish. > > For finite naturals, yes. Which are the only ones there are, since the Peano axioms defining them leave no room for other naturals. >> > Whether the number that is represented by a binary string which >> > represents a subset is a member of that subset is irrelevant. >> >> Not if you are establishing a bijection. > What? Now it's my turn to say "bullshit". There is no requirement > that the subset of N associated with the number x must contain x as > an element. No, but it must either contain or not contain x as an element, and that is all that is required to establish that you can't biject between set and powerset. >> >> > If you divide this number by 3 (11) you find it is divisible >> >> > or not, depending on whether you have an odd or even number of >> >> > 100's in your infinite string. Of course, this question is not >> >> > really answerable, so I don't have an answer for you. What do >> >> > you think? What is aleph_0 mod 3? >> >> >> >> Oh, I never claimed that aleph_0 was a member of the natural >> >> numbers, so I don't need to make claims about aleph_0 mod 3. >> >> > But the answer to this question depends on the number of digits >> > in the infinite number. >> >> Since there are no infinite numbers, I need not answer the >> question. You claim that there are, so it is up to you to provide >> an analysis of your claims. > > My claim is not that all arithmetic holds for infinite whole > numbers, but that infinite whole numbers are required in any > infinite set of whole numbers. They aren't. You require an unlimited amount of finite numbers, but each one of them is finite, even though there is no maximum to them. > The arithmetic is a sidebar, and if you want to claim that I have to > work all that out to assert their existence, you're full of > ptooey. That is an entirely diferent question. The divisability is a sideeffect of the Peano axioms. They leave no room for numbers suddenly stopping to be either divisable or not divisable by 3. >> Which is why infinite numbers have not been admitted into the >> naturals. It would be impractical to have a set of numbers that >> are not governed by the same laws. > They certainly need to be handled differently, Yes, and thus they are not natural numbers. Since they can't be handled with the Peano axioms. > but that's what Cantorians say, unapologetically, about infinite > sets. Sure. There is nothing wrong with claiming infinite numbers or whatsoever, as long as you don't make the mistake of assuming that the Peano axioms or anything else derived from them would extend to those numbers. It is perfectly fine to talk about such entities as long as one finds axioms that fit them. The Peano axioms don't, and so you can't call them natural numbers. > When asked why a certain property is not true of infinite sets, the > typical answer is, "because not everything that holds for finite > sets must hold for infinite sets", which doesn't answer the question > at all. So, you get no apologies from me for including infinite > values in the set of whole numbers (okay not naturals, see?), in > order for that set to be infinite. Unfortunately, the moniker "whole numbers" is already taken as well. Call them TO numbers, and devise a coherent set of axioms for them, and nobody will complain. > By your definition of naturals as all finite, and the basic math > governing quantities and strings, I can only conclude that you have > a finite set. Wrong. But since you have been hit with the proof for that about five dozen times, it appears useless to do it again. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 20 Jul 2005 15:36
Daryl McCullough said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > >You ask this as if any set that is larger than any other set is > >"uncountable". Do you consider the two terms, "larger" and > >"uncountable" to be synonymous? > > Not quite. "Uncountable set" means "set with a larger cardinality > than the set of naturals". Yes, that seems to be the Cantorian definition. Does it actually follow that an infinite set larger than the naturals can't be enumerated? Personally, I don't see the connection at all, and view it as a conflation. > > Look, Tony. Your objections to standard mathematics all seem > to revolve around disagreements about the meanings of words. > Words such as "infinite", "finite", "larger", etc. *Real* > mathematics doesn't depend on word choice. That's interesting, since the mathematiicians here seem to like to haggle over words that they themselves can't define, like "infinite". Do YOU have a definition for "infinite", which can eb applied equally well to sets and numbers? > > Can you express what you are trying to say without using > any of those controversial words? Normal mathematics can. > The use of the word "larger" to mean "having a greater > cardinality" is just terminology. All of mathematics would > go through just as well without *ever* using the word "larger". > You could just as well use the word "more bloppity": > > By definition, a set S is said to be more bloppity than a set R > if there is a 1-1 function from R to S, but there is no 1-1 function > from S to R. > > Instead of using the term "size" to refer to sets, we could > refer to the "bloppitude". I use Bigulosity, to distinguish my measures from cardinality. > > Instead of using the words "infinite", we could use the term > "mega-bloppity". Try "unending". Or, as I requested, give ANY synonym, or definition of the word itself. > > Nothing of any importance about mathematics would change > if we substituted different words for the basic concepts. Then you shouldn't be having a word problem with me, right? > > In contrast, your arguments are about nothing *but* terminology. > To me, that shows that there is no actual content to your arguments. > An actual mathematical argument does not depend on word choice. My arguments have NOTHING to do with terminology. If I say that the set of evens is smaller than the set of naturals, I think you all know what I mean and why I am saying it. Cardinality is supposed to be a measure of set size, but when it comes right down to it, mathematicians back away from that declaration, for good reason. It's one way to handle sets, but the results are at best unituitive, and at worst, contradictory to the rest of mathematics. > > As a challenge, see if you can express your claims about > infinite sets, or infinite naturals, or set size, or whatever, > *without* using the words "infinite", "larger", "size", etc. Yeah sure, and you describe your fluffy pink flying elephant without using the words "fluffy", "pink", "elephant" or "flying". Doesn't this suggestion sound rather absurd to you? How do you expect me to talk about infinity or infinite sets without using the word "infinite"? > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony |