Obections to Cantor's Theory (Wikipedia article)
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From: Chris Menzel on 25 Jul 2005 12:38 On Mon, 25 Jul 2005 11:27:28 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > ...You cannot form an infinite number of strings with a finite > alphabet, without strings of infinite length. Oh jeez, you're as clueless about formal languages as you are about set theory, not that this is a surprise. Do you know that what you are saying here is contradicted in the early chapters of any book on formal languages? Really! Go have a look! Here, let me save you some trouble: On page 1 of their famous standard text *Introduction to Automata Theory, Languages and Computation*, Hopcroft and Ullman define a "string" to be *finite* sequence of symbols. (Note: they don't think the notion of string can't be generalized to the infinite, it's just that in their text they are only interested in the finite ones.) Then, on the very next page, they point out that "The set of palindromes (strings that read the same forward and backward) over the alphabet {0,1} is an infinite language." See? Do you realize what a fool you are making of yourself by making assertions about set theory, transfinite arithmetic, formal languages, computability, etc. that show you don't understand even the most elementary concepts, or grasp the most elementary theorems, of set theory, transfinite arithmetic, formal languages, computability, etc? C'mon, stop embarrassing yourself. Go *learn* some mathematics before you start spouting off about it. It's painful to watch. Chris Menzel
From: Chris Menzel on 25 Jul 2005 12:39 On 25 Jul 2005 15:54:51 GMT, Chris Menzel <cmenzel(a)remove-this.tamu.edu> said: > ... > Really, do you mean that you think my comment above has any such > implication? TO, in the comment to which I was responding, essentially > asserted that he believed CH, though with typical cluelessness he was > unable to recognize the fact. I was simply it pointing out to him... "it pointing" -> "pointing it"
From: MoeBlee on 25 Jul 2005 13:09 Tony Orlow: What is your logistic system, your primitive terms, and your axioms? MoeBlee
From: David Kastrup on 25 Jul 2005 13:12 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > David Kastrup said: >> >> >> > 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? > oops. sue me in typo court. >> >> > 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. > A formula using one of these "=". Where I come from, we call it an > equal sign, and it means "is the same as". So that is a constant equality. And what is a non-constant equality? And a constantly decreasing inequality? This is all a bunch of nonsense. > I am sure you have come across these from time to time. Do you > honestly not know the difference between an equality and an > inequality? If you do, can you tell which one "is finite" might be > most like? "is finite" is neither an equality, nor an inequality, but a property. And there are no "decreasing inequalities" and no non-constant equalities. >> >> > 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. > Yes, I am well aware of the construction of inductive proof. Are you > aware of WHY it is understood to work? Because it is the most basic application of the fifth Peano axiom. > Do you understand the concept of a recursive stepwise process with a > starting point, but no ending point? The lack of endpoint is what we > call infinite. Anyway, you are putting the cart before the horse. The _idea_ of a process that one can repeat indefinitely certainly underlies the selection of the fifth Peano axiom. But that's it. The axiom stands on its own. The idea might well be still around, but it is not relevant. > If it ends, it's finite. Inductive proof does not end, Inductive proof consists of two steps, and that is that. And this is _exactly_ the reason why the set of natural numbers is defined in terms of the induction property instead of some indefinite process: because mathematical laws that would require you to check an indefinite number of conditions would be mostly useless as long as you can't systematize the process. The induction axiom provides the necessary shortcut. It _replaces_ a non-terminating process, but does neither require nor perform it. >> > 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. > > It is called bullshit. It's like you really don't understand what > inductive proof is doing. It is a complete and exhaustive replacement for the much worse idea of checking every value. And since it is so much more useful, it is _induction_ that has become the _axiom_, not some fuzzy unfinished process. > You memorized all three steps and now you think you're smart. But, > if you can't see that it proves the property true for each n in N, It proves the property true for each n in N, period. > in turn, There is no order implied. > recursively, There is no stepwise process implied. > then I can't imagine what you think is going on. One checks the preconditions for applying the fifth axiom to a property of integers, and if they are met, one has shown the property to hold for all integers. That is what is going on. If you feel like it, you are, of course, free to check a few consecutive cases manually. But that is not a part of the proof. > This is simple denial in the face of an obvious contradiction > between counting an infinite number of times and still having only > finite values. There is nothing like "counting an infinite number of times". > Now you want to say induction has no "steps", when the step that > gets repeated an infinite number of times is the meat of the proof, No step gets repeated an infinite number of times in induction. Induction checks two conditions, and if they are met, the property holds for all integers. It is a bloody axiom. Any process you want to invent is a construct of your own, and is not relevant to the proof technique. > and in the proof of finiteness, that step involves adding 1. What is > in the water at these institutions, anyway? Ask your guard if the taste offends you. >> > 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. > No, of course not. Otherwise it wouldn't be a proof, but > fact-checking. See, you get it if you really try. > Still, you need to be careful about your property in this case, and > keep in mind that inductive proof IS an infinite loop, Too bad. You lost it again. > so that incrementing in the loop creates infinite values, and the > quality of finiteness is not maintained over those infinite > iterations of the loop. Please point out where a "loop" or "process" is in the following axiom: A set containing 0, and for each of its elements k also containing its successor S(k), contains all natural numbers. I can switch the order around without changing the logic: A set containing for each of its elements k also the successor S(k), contains all natural numbers as long as it contains 0. See? I can start the statement with the successor stuff without even having specified a "starting point" or whatever. >> 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 Peano axioms define an infinite set as a self-referential > recursive loop. No. They might have been inspired by such an idea, but they most certainly and definitely don't define any such thing. See above. > Naturals are defined as the successor of the previously-generated > natural, Wrong. There is one Peano axiom that says "if n is a natural, S(n) is also a natural". But this single axiom alone does not define the natural numbers. It takes all five axioms together to do that. > You say we can go in any order, but that's not what axioms say. The axioms don't prescribe an order at all, and so there is no order. > They explicitly talk about the successor to each natural. The successor relation has nothing to do with a succession in time. > You're just spouting defensive nonsense here. That is a nice characterization of what _you_ do. Please come up with any temporal order or process defined in the Peano axiom. And no, that one call a defining relation "successor" is just picturesque language for the sake of children. It is a name, nothing else. And unless I am mistaken, it was never written out in the original formulation of the axioms, anyway, but was just S(n). >> 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. > > So, you can add 1 an infinite number of times and get a finite > value. Also very interesting. You are babbling incoherent nonsense without any apparent relation to what you pretend to be answering. Smokescreen. > Whatever math doesn't suit your case is simply irrelvant. Whatever math is not present in the axioms and definitions is irrelevant. Yes. > Time to grow up, Dave. A fine resolution. Go to, go to. >> 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. > We both know what multiplication is. Apparently you don't know it. It is not defined in the context of the basic five Peano axioms. You need more axioms (usually associated with the whole numbers) to get it into the picture. Since we are not talking about anything but natural numbers, multiplication does not apply. > Don't you ever get tired of playing dumb? Don't you ever get tired of blaming your deficiencies on others? >> >> 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. > Very good. You used an equality as a property. That works every > time. In fact, I don't think I can think of any other case where > this type of proof doesn't work, except for the proof of finiteness > of naturals. It works there as well, since it is based on an axiom. > Then again, "finite" not be a very good mathematical term to use in > such a proof. Go via finite sets. Theorem: If f(0) = {}, the empty set, and f(n+1) = f(n) U {n+1} for all n, then f(n) for all n can't be put into 1:1 correspondence with a proper subset of it. Proof 1: we prove that f(n) has n members by induction. f(0) has 0 members. f(n+1) has f(n)+1 = n+1 members (this is a shortcut since we have to show that n+1 is not already in the set, but that also can be done by induction). So we need to put n+1 members into correspondent with n members. This is impossible for all n. So the set f(n) is a finite set for all n, and we call all n by assocation also finite numbers, since they indicate the cardinality of a finite set. >> >> > 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. > "No largest finite. Shake rattle" The proof holds for all n in N. For all sets that end with an n in N. N itself is no such set. > Your refusal to even think about the proof is silly. There is no way > the size of the set can be larger than every element within it. Just sulking does not make it so. > Just think about it a minute and stop playing lawyer-math. Math is about laws, much more rigidly than the legal profession could ever make it. > However large your set, you cannot fill it with only values that are > strictly smaller than that size, or you run out. Just sulking does not make it so. Math requires proof, not conviction. >> > 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. > > Your reasoning is simply not happening at all. What is this supposed to be? A paraphrase of "LA LA LA CAN'T HEAR YOU"? >> > 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. > I'll work on it. Fine. Until then, shut up. >> >> > 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. > > At every step in the proof, the new element is the largest element > and equal to the sizeof the set. You are confusing steps in the proof with some weird iteration scheme again. Anyway, you are always proving something for a finite set. N is no finite set. > At no time can you ever have the set be bigger than all elements, Because you are always talking about finite sets here. N is no finite set. Your proof does not apply to it. > but it is always equal to the last element added. N is no set with a last element. So your proof does not apply to it. > Whatever the size of your set, that IS the upper bound on element > values. This is too obvious. For sets that end at a given n. N is no such set. > Stop playing dumb, with your "largest element" excuse. It is not an excuse. It is what is relevant. Even though you pretend not to get it. >> >> > 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. > What is this? Proof by non-sequitur? This has nothing to do with > what we're talking about. Correct. But since you don't get what we're talking about, maybe you would have understood a similar setting. >> >> > 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. > Prove it. Assume that a finite number of digits is sufficient to contain all natural numbers. A finite number is characterized by not being able to be put into bijection with a proper subset of itself. A finite number k of digits can assume a number of states 10^k. Now we use the successor relation S(n) to map all possible numbers to a different number. Different numbers have different successors, so we again get 10^k different numbers. However, 0 is not the successor of any number, so our 10^k states have been mapped to just 10^k-1 states if the finite number k would have been sufficient to represent 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. > Then what makes you think they are all finite? How do you have an > infinite set of finite length strings on a finite alphabet, when the > formula is N=S^L. No one seems to want to address this issue. How do > you get infinite N with finite S and L? You don't. Probably nobody wants to "address this issue" because you are not bothering to define your variables. Guessing their meaning, I'd say your mistake is to assume L is limited. L can take on only finite values, but arbitrarily large ones. In consequence, N takes on arbitrarily large finite values, depending on the value of L. >> >> > 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. > The recursive statement, the meat of the proof, where it is > demonstrated that f (n)->f(n+1). That is a single step you need to show for all n (with no implied order). Not an infinite number of them. >> >> > 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. > Gee, what was that "n" doing in there again? What IS that thing > anyway? Maybe we should go ask Gumby and Pokey.... For all n. No order implied, no steps implied. >> > 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". > > What did "n+1" mean again? I fell down and banged my head, and now I > don't remember..... n->n+1. That is the only step involved here. There is no infinite number of steps or anything like that. >> >> > 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. > > Don't be so hard on yourself. You can probably be made to > understand, if you try hard. Well, I am not the one fantasizing about infinite steps and recursion and all other manner of junk that is not in the axioms. >> >> > 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. > > What's nonsensical is pretending you can talk about infinity solely > in terms of counting and get anywhere, and then pretending you got > somewhere, but using axiomatic sleight of hand. You are again babbling incoherent nonsense. > It is certainly possible for some to think about infinite sets > without resorting to bijections. But then they are not thinking about what is called "infinite set" in mathematics. > That is not the only way to conceive of them, and your insistence > that it is, is what needs correction here. Cantor does not have a > monopoly on infinity. Oh, he does in a manner: his definition was adopted as the most useful one. You are, of course, free to disagree, but then you bloody have to properly define your different terms instead of using established words to mean something different and complaining that people don't follow your usage. >> > 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. > > Apparently I understand it better than you experts, since I see > clearly where is makes foolish assumptions, Look, you can't even bloody see where you make foolish assumptions when one rubs your nose into it twenty times in a row. So you are probably the least qualified person on Earth to spot fallacious assumptions in anybody else's reasoning. > and you can't even follow connections to infinite series or symbolic > systems, with simple formulas provided. You are putting the cart before the horse. The Peano axioms are not bothered about infinite series or symbolic systems. If infinite series or symbolic systems had a problem with the Peano axioms (which they don't), it would be entirely the problem of the infinite series or symbolic systems, and they would have to sort that out. > DO you go and read lots of books on Hitler so you can be sure nazism > is wrong? The only reason I am slogging tis out with you all is that > it's become clear to me just how entrenched and nonsensical > cardinality is, and what its effects are on the mind. It's not > good. Since it has not yet registered on your mind, I'd call that sour grapes. >> > 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 can't prove things by definition. I did above, a few times. > If I am talking about something different, and you throw definitions > I am not using If you are not using them, then you have no business complaining about their consequences. That's what others are talking about when they are using those words. If you want to talk about something different, invent your own bloody terminology. > as proof that I'm wrong, then I tell you to get your head out of > your axioms and listen to the point I am making. I am afraid that the axioms are more relevant to the matter than any point you try making. >> > 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. > You mean like infinite recursion. Read up a little. Not an axiom. And if you want to make yourself ridiculous in yet another area: I have been working as a programmer since the 70s. >> >> > 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. > > Fine. Call them whole numbers. If you insist that all naturals are > finite, that they each differ by at least 1 from all others, and > that there is an infinite set, then i don't believe in natural > numbers, and will henceforth not use the term. Can you live with > infinite WHOLE numbers? You'd have to define them first. >> >> 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. > > Unfortunately for you, I don't mind looking stupid, expecially when > that's in the eye of the beholder, so I don't think I'll shut > up. Why don't you try it, and maybe you'll look like less of a > crybaby. So, you want to talk about some numbers that are not supposed to be the natural numbers and which you are not going to define. And that is going to prove exaxtly what to whom? >> >> > 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. > > Given any subset denoted by a binary string, and any element in the > set of all whole numbers, one can easily determine whether the > subset contains the whole number denoted by the binary > string. Simply look at the bit at the position corresponding to that > element and see if it's a 1. If this is your criterion, it is met. Uh, no. There is no number mapping to the set of numbers dividable by 3. You seem to have a bad memory. >> >> >> > 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. > > Which is entirely impossible if they each differ from each other by > at least some finite amount. Sulking won't make it so. > Then, you have an infinite sum of finite differences, for an overall > infinite range of values, which means some values muct be infinite. You are babbling again. Try to form coherent words. >> >> 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. > > They are whole numbers that are required for the infinite > set. Peano's axioms can easily be inverted to produce them. You are babbling again. >> > 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. > > They do if inverted. One can start at infinity and count down quite > easily. You never get to finite values that way. Besides, you'd have to prove that the "inverted axioms" are equivalent, and I don't see how you will invert both "if n is a natural number, S(n) is a natural number" as well as "0 is not equal to S(n) for any natural number n". >> > 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. > > Can I quote you on that? Let's see how your prediction pans out: > > > 1. oo and 0 are numbers > > 2. If x is a number, the successor of x is a number. > > 3. oo and 0 are not successors of any number. > > 4. Two numbers of which the successors are equal are themselves equal. > > 5. (induction axiom.) If a set T of numbers contains oo and 0, and also the > successor of every number in T, then every number is in T. > > > Any complaints? Well, then your numbers basically consist of the union of the disjoint sets {0, S(0), S(S(0))...} and {oo, S(oo), S(S(oo))...} with both of the sets being isomorphic to N. It is a pretty pointless structure, but not yet invalid. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: David Kastrup on 25 Jul 2005 13:16
Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > David Kastrup said: >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> > How do you know 6598367 is finite? Because you got it by adding 1 >> > to 6598366, the 6598366th number. >> >> Certainly not. Don't tell me that you have spent the effort of >> counting to 6000000 manually just to be able to talk about that >> number. >> >> I know it is finite because it has a form (a finite number of >> digits) that can be shown to always refer to a finite number. I >> see that the number has 7 digits, and I know it is finite. >> >> That is, I know it is finite because it obeys certain laws, and >> those laws can be shown to hold for all natural numbers, by >> induction. I don't need to carry out any individual steps to make >> use of that knowledge. > Then why do you ignore those laws of representation, which are based > on N=S^L, and pretend you can have an infinite set of such finite > strings? Because L can become arbitrarily large, there is no fixed S^L. > Why do you feel entitled to violate established laws of math? You > cannot form an infinite number of strings with a finite alphabet, > without strings of infinite length. This should be obvious. Without strings of _arbitrary_ length. If you don't get the difference between "infinite length" and "arbitrary length", you will never understand this. >> >> > So how to you get the aleph_0th number? What IS that number? it's >> > aleph_0. >> >> But that number does not obey a form I can recognize as belonging to a >> finite number. And indeed, taking a look at its properties, it >> becomes clear that it can't be a finite number. > Well, no, it's infinite. Aleph_0 is basically 111...111+1. This is a > clear contradiction within set theory, this assumption of all finite > values. Try my new axioms. You can count from both sides at once. No, I can't. oo is not the successor of any number according to your axioms, so I can't go backwards from there. > The Peano axioms do just that. they start at 0 (or 1) and define > successors, which are interpreted The axioms don't interpret. > to be the previous value plus 1. That's called an increment, and if > you do it x times, It is your problem if you want to do anything in a particular order or a number of times. The axioms are not bothered about that. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |