Cantor Confusion
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From: Virgil on 30 Dec 2006 19:29 In article <1167515365.408977.36260(a)k21g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1167492220.138771.111750(a)h40g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > cbrown(a)cbrownsystems.com schrieb: > > > > > > > Have you given up on your "rational relation" proof that |N| = |R|? > > > > > > No, why should I? It is correct. (A series with a first but no last > > > term can be reversed to have a last but no first term - without oosing > > > its value.) > > > > How does one "sum" a series with no first term? There is no place to > > start! > > There is a place to start, namely at the end, because there is a last > term. Did you ever hear of an order omega^star (w*). Why should it be > impossible to reflect a sequence or the terms of a series and > nevertheless maintain the value? If you also agree to sum from the end term, there is no problem, but there is no point either. > > > > > > But it is easier to see, and should be visible even for such as you, > > > that the tree built from the union of all finite trees (with > > > representations of rational paths) is the same as the complete infinite > > > tree. > > > > In the union of all finite trees, there are no infinite paths at all, > > No? Where do they end? You should be able to answer this question and > to prove your answer. In any union of finite trees, every path is a member of some finite tree, and it ends where the paths in that tree end. QED. > Hint: In the union of all finite initial segments of N there is an > infinite initial segment, namely N (so some people claim at least). According to that definition there must be a member of a union which is not a member of any set in the union. That is definitely non-standard set theory. > > > since to be represented in the union, a path would have to be in one of > > the finite trees, ergo a finite path. > > > > In the same way, the union of any family of sets of finite naturals does > > not contain anything but finite naturals. > > The question is not whether the edges have a finite distance from the > root, but if there are infintely many of them. If you can find a finite tree containing an infinitely long path only then. (Compare the irrational > numbers with only finite indexes but, as you claim, infinitely many of > them.) > > Regards, WM
From: Dik T. Winter on 30 Dec 2006 23:47 In article <1167493176.710056.286330(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > At what point is 1/3 represented? > > In the binary tree it is not represented, but it could be represented > by using a ternary tree. Now you are arguing differently from before. When I stated that 1/3 was not in your tree, you argued that it was in your tree. Do you now think differently? > There is no representation > of 1/3, unless it were in the infinite. But if you insist on an > existing decimal representation of 1/3 in the infinite, then also the > union of all finite binary trees contains such an existing binary > representation of 1/3. I still do not understand what subtleties you are making. In the *complete* binary tree 1/3 does exist. But that is *not* the union of your rational trees. > > > Take the union of all these rational trees. It contains all paths > > > representing rational numbers. > > > > No. It contains all paths representing rational numbers for which the > > denominator is a power of two. > > Every decimal number gives a power of 10. A distraction again. Pray keep to the discussion. > > > What distinguishes this union of rational trees from the complete tree? > > > > That some rational numbers are missing. If the union contains all > > rational numbers, there must be one of the rational trees that contains > > (for instance) 1/3. There is none. > > My question is only the following: Do you think that the union of all > rational binary trees is the complete infinite binary tree or not? It is not. But I already did answer that when I answered your question what the distinction was between the union of rational trees and the complete tree. So why do you ask again? I state they are distinct, and indicate in what way they are distinct, and you ask whether I think they are distinct? > > Where is 1/3? At what point does it get into the union of rational > > trees? > > > My question is only the following: Do you think that the union of all > rational binary trees is the complete infinite binary tree or not? How many times must I answer that question? > > > That is the reason why the cardinal number of the paths in both trees > > > cannot be different. > > > > That is the reason why the cardinal number of the paths in both trees > > *is* different. > > > With a different number and configuration of nodes (or edges) or with > exactly the same? You lost me here. We were talking about the union of finite trees and the complete tree. > > > But if we subtract the rational paths from the complete tree, then > > > nothing remains but the empty set which obviously is the set of all > > > paths representig irrational numbers. > > > > > > This shows that irrational numbers are nothing (but a chimera). > > > > And apparently a lot of rational numbers are nothing? > > A lot of binary representations, yes. But every rational number has > some representation - in contrast to irrational numbers. You are wrong. sqrt(2) has a pretty good representation: "sqrt(2)". -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 31 Dec 2006 00:06 In article <1167514645.819727.261070(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Why don't you simply give the formula for the definable well ordering? > > > The formula could be evaluated, and, step by step, we would get a list > > > of all reals. > > > > Wrong. It would not give a list. It would give a well-ordering. Do you > > not know the difference? > > Every non empty set of a well-ordering has to have a first element. If > all reals are well-ordered, then all reals are first elements in some > sets. What is the relevance? If I order the naturals as: {1, 3, 5, 7, ..., 2, 4, 6, 8, ...} I have a well-ordering but not a list. A well-ordering is *not* a list. > > And I doubt whether you could apply the formula > > one by one to each real. > > I doubt that too, in fact it is impossible in ZFC. Therefore, there is > no well-ordering of all the reals. You keep to state that, without proof. For a well-ordering it is not necessary to be able to state the elements one by one. > > > Because non-distinguishable sets are not different sets. You just > > > proved that there is no element of a path which distinguished it from > > > every other path. > > > Right. But that does *not* mean that it can not be distinguished from > > every other path. > > If every edge of a path is shared by another path, how would you > distinguish them? What is "them" here? Give me two paths and I can give you an edge that distinguishes them (or at least an algorithm that shows it). > > Well, I understand why he [Leibniz] is your favourite. > > He is not my special favourite. But he was very wise and he said: The > arithmetic of the infinite has to be based upon the arithmetics of the > finite. Yes, and he was not even able to define the things he was working with. He simply gave some rules of calculation that did work, without any justification at all. > > Again, a finitistic view. There are algorithms that do calculate that > > number. [pi*10^10^100] > > Is it even or odd? Is that relevant? If you *really* want to know it, I can start a program that attempts to give the result. But you have to wait some time. > > That they can not be implemented in a finite world is of no > > concern to the mathematician. On the other hand, there *is* a natural > > number that is equal to that value. > > Is it even or odd? Why bother? > > But apparently you also eschew the proof that li(x) and pi(x) cross > > each other infinitely often, because it is impossible to even calculate > > the first cross-over. > > I do not "eschew" the proof for li(x) and pi(x). These "numbers" seem > to eschew existence. Yes, a purely finitistic view. Pray develop mathematics based on your view. But do not think that current mathematics will be any help. How (in your finitistic view) do you allow that lim{n->oo} 1/n = 0 ? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: cbrown on 31 Dec 2006 00:41 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > I did not calculate |N|/|R| but edges per path. > > > > Wonderful. But what relevance does the calculation of "edges per path" > > have to do with the /actual question/: does there exist a surjective > > function from the set N to the set R? > > If there are not less edges than paths, then a surjection is possible, > i.e., the possibility of a surjection is proved. If x and y are real numbers, then x >= y has a well-understood and agreed upon meaning. But when x and y are set cardinalities, only if by "x is not less than y" you mean "there is a surjection from x onto y" does it make sense to most hearers to say "x >= y". The statement "There is some x such that 2*x = 3" is false if by x and y we mean naturals; and yet true if by x and y we mean complex numbers. You are confusing two different meanings of "the size of". It is rarely asked in these threads, but what is the use of the usual definition of cardinality, anyway? Let H be the additive group of rationals with denominator of the form 2^i * 7^j for naturals i, j; and K be the additive group of rationals with denominator of the form 3^m * 5^n for naturals m, n. How should we calculate "lim" |H|/|K|? Is H "not less than" K because of the value of this limit? If not, why not? > It is really the same > as with the proof of a well-ordering of the real numbers. No, it is completely different. There is no contradiction with the usual definitions of N and R when we simply assume a well-ordering of the real numbers. Many do not accept this assumption, but this is usually on philosophical, not logical, grounds. On the other hand, simply assuming that |N| = |R| results in a contradiction - again, with the /usual meanings/ of N, R, and |N| = |R|. You seem to be confused by the use of the symbols "<", "=", and ">". Let us ignore these symbols then for a moment. Instead, let Mesmo(A,B) be the proposition "there exists a bijection between A and B". Let "Umaum(A, B)" be the proposition "there exists an injection from A to B". Let "Em(A, B)" be the proposition "there exists a surjection from A onto B". I will grant you that Em(A, B) and Em(B, A) implies Mesmo(A, B); and similarly that Umaum(A, B) and Umaum(B, A) implies Mesmo(A, B) (i.e., Cantor-Bernstein-Schroeder). We have previously agreed that Mesmo(N, edges), and Mesmo(paths, R). Can you prove your claim that Mesmo(N, R) using your "rational relation" argument? Why does the fact that lim sum 1/2^n = 1 imply Em(N, R)? Or don't you speak Portuguese? > And, in fact, > if the real numbers could be well-ordered, then the paths could be > well-ordered and a bijection between paths and edges (which already are > well-ordered) could be completed (because it has been proven that there > are not more paths than edges). For the thousandth time, you claim you have a proof. But when you are questioned, you respond with a chain of disconnected claims which all either assume what you want to prove, or are completely incoherent. > This is the reason why my proof is > important. Do you now see what relevance the calculation of "edges per > path" has to do with the /actual question/: does there exist a > surjective function from the set N to the set R? No. You do not recognize that you are comparing apples and oranges. "3/2 < 2" and "|N| < |R|" have /completely different meanings/. The first is a linear order applied to a field. The second is an ordering based on surjections / injections. Sometimes by "bread" I mean an edible product made of ground wheat. At other times, by "bread" I mean money. Sometimes by "Bread", I mean the cheesy pop group of the 70's. That doesn't imply that I claim that the band that brought us "If a picture paints a thousand words... " is composed of edible baked wheat products that can be exchanged for goods and services. > > > > That is what is meant by "|N| >= |R|". The use of ">=" here has a > > /different meaning/ than "1 + 1/2 + ... >= 1", a difference which seems > > to elude you. > > > > In the latter case, we are not asking "is there a surjective function > > from 1 + 1/2 + ... onto 1?", because that would be simply nonsense. > > Only if one does not know the meaning of the arithmetic of rational > numbers and limits. A truly fantastic claim! Then please give a rule for determining a surjection from sqrt(2) to 1 that is neither nonsense nor dependent on first /assuming/ sqrt(2) >= 1. I begin to wonder if you have even forgotten what "f is a surjection from A onto B" means. You sometimes appear to say whatever words come into your head, without regard to the meaning of your assertions. > Sums of rational numbers are nothing else but sums > of integers, given in the unit of the greatest denominator. In case of > limits this kind of arithmetic is extended, but the basis is > calculating with natural numbers which are to be compared by mappings, > i.e., the arithmetic of ordinal numbers. 3 is not identical to 6. 7 is not identical to 14. And yet we say 3/7 is identical to 6/14. In fact, there are at least as many names for the number "3/7" as there are natural numbers. This is the same as the number of names for the number "2/9"; which is to say there is a bijection between the names for "3/7" and the names of "2/9". Your thoughts? > > > Similarly, to try to show that |N| >= |R| by calculating the limit of > > edges per path is equally nonsense - it requires misreading the meaning > > of ">=". > > You seem to be not very well informed about arithmetic in general, be > it rational or ordinal. BTW, have you learned meanwhile that 2^omega is > a countable ordinal? What does that have to do with showing that Mesmo(N, R) follows from lim (2 - 1/2^n) = 2? Why do you feel that acceptance or rejection of the premise "2^omega is countable" plays any role at all in your argument? Cheers - Chas
From: mueckenh on 2 Jan 2007 06:35
Franziska Neugebauer schrieb: > > Different from what you believe is behind these "numbers". > > The contradiction of your answers remains in effect because what "I > believe" is not part of your contradiction. What you believe comes in when we talk about the object which you call an irrational number. > > Do you want to withdraw your answer to question 3.? > > >> You have to decide which version of "ideas" shall be valid. Without > >> that decision you contradict yourself. > > > > Common irrational numbers are identical with my "ideas" > > Repetition of acknowledgement of your answer to question 2. No reason not to repeat it. > > > But the common interpretation of common irrational numbers is false, > > in particular the assertion of unlimited facility of approimation. > > The contradiction of your answers remains in effect because "common > interpretations" are not part of your contradiction. Where is a contradiction? > > Do you want to withdraw your answer to question 3.? Why? Regards, WM |