An uncountable countable set
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From: Dik T. Winter on 12 Jul 2006 10:53 In article <1152707786.981875.231380(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > When Cantor's proof was published, there was not such an axiom. > > > > Indeed, at that time it was not yet an axiom, but the thought was present > > that infinite sets did exist. > > There was not such an assumption in general. Infinite sets did not > exist (as they do not exist yet), but only the potential infinite was > accepted. Cantor was little understood and was blamed to do philosophy > or theology but not mathematics. "... die Gesamtheit aller endlichen Zahlen 1, 2, 3, ..., v, ...", Cantor. > > > If an axiom states the existence of 100 natural numbers below 20, it > > > has to be abolished in order to save mathematics. The axiom of > > > infinity, interpreted as you do, is such an axiom. But it was built to > > > model Cantor's worldview. > > > > The first of these two axioms would lead to an inconsistent system. The > > axiom of infinity does not. It leads only to an inconsistency with *your* > > view. > > It leads to the inconsistency that you must demand 0.111... *- (0.1 + > 0.11 + 0.111 + ...) = 0.000... but 0.111... not being in the sum. As you have not defined the meaning of the second term, this makes no sense. Offhand I would say it goes 0.1, 0.21, 0.321, 0.4321, 0.54321, 0.654321, 0.7654321, 0.87654321, 0.987654321, 1.0987654321, and I see no obvious limit emerging when we do the infinite sum. > It > leads to a binary tree having infinitely many path mapped upon the very > same edge though no path can split without two new edges. Care to explain? I do not understand. Given a balanced binary tree with infinite depth, obviously there go infinitely many paths through each edge. That is in itself not inconsistent. > And you must > refuse to calculate with fractions in mappings. Here I have no idea what this means. > That is inconsistency > enough. As I do not understand anything you have written here, I still see none. -- 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 12 Jul 2006 11:13 In article <1152708585.848976.307310(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > The limit is the stable state where no further transpositions are > > > applied, because there are no two elements remaining, which were not > > > ordered by magnitude. > > > > That makes no sense as a definition, as you will never reach that stable > > state. > > That is not my problem but the problem of non-existing infinite sets. That *is* your problem, because it makes it a non-definition. So you can not talk about the "limit" of your infinite sequence of transpositions. > > But it is for the rationals. And I think it can also be found for the > > algebraic numbers, but it will take a bit more care. > > Is it for rationals *without repetition*? How does it read? Yes, it is. I gave one in this newsgroup a few days ago for the rationals from 0 to 1. > > What do you mean with a sentence like "in which line of Cantor's list a > > certain digit of the diagonal number will be placed"? > > How can you find out whether a certain real of Cantors list has the > number 5782712398413208741? Why is that an interesting question? If it does occur you can find it, if it does not occur you can not prove that. But the question remains, why is that an interesting question? > > > > But what is known is that the n-th digital > > > > The mapping from the naturals to the list gives the n-th element. It is > > in the *definition* of the list. > > Do you want to define a proof or to give a proof? Eh? Do you know what a list is? > > > > A mapping that well-orders > > > > the rationals is *not* a sequential process. The determination of the > > > > n-th digit of the diagonal from a given list is *not* a sequential > > > > process > > > > > > Wrong. You cannot know line number n without knowing line number n-1. > > > > The mapping gives that. > > > > > But even if you were right, your argument would be void. If infinity > > > would exist, then an infinite set could be exhausted by a definition > > > like that of Cantor or that of mine. > > > > Makes no sense. The natural numbers can not be exhausted by a sequential > > process. > > Therefore you resist to count lines. But in case of proof (not > definition of proof) you are forced to count all of them. No. You show that it is valid for *any* n, and hence it is valid for *all* n. > > Nobody has a problem with that, as it is well known that > > what is the case in the limit is not necessarily what is the case outside > > the limit. > > Why don't you apply this knowledge in case of Canor's list? In the > limit n --> oo it may well be that 4 cannot be distinguished from 5. Makes no sense. > > > > But it is just that part that is interesting. Try the same with a > > > > sequence of algebraic numbers. You need to prove that what you get is > > > > also an algebraic number (and you cannot). So for algebraic numbers > > > > the proof fails. For real numbers the proof goes through, because you > > > > can prove that the resulting number is also a real number. > > > > > > I proved in my special list even that the diagonal number is a > > > rational. > > > > I wonder whether it was a proof or just some handwaving. > > 0.0 > 0.1 > 0.11 > 0.111 > ... > replace 0 by 1. As far as I see the diagonal starts with 1.000... Am I right? And indeed, in that case it is a rational. But you can not prove that it is a rational for *all* lists of rationals. Only for particular lists can you prove that it is a rational. > > I am lost. Mathematically a sequence of symbols like 0.999... makes no > > sense (as a number) unless there is a definition. > > 10 * 0.999... = 9.999... > 9.999... - 9 = 0.999... That is not a definition. How do you define the multiplication of strings of digits? Subtraction? Still parts of the definitions are missing. > That does not need special definition, after * and - are defined and > can be applied digit by digit. So you give a definition for * and - on infinite strings. Ah well, I am wondering when you are done with 10 * 0.999... And how about 20 * 0.999... ? > 1/n becomes arbitrarily small for large enough n. Without further > definition but that of /. > > > > The sequence makes > > perfect sense as a sequence. But as a sequence I can only state that > > 1.000... is not equal to 0.999.... Only when we want to interprete them > > as numbers we need a definition, and with the common definition those two > > are the same *as numbers*. > > Do you really believe another definition would be possible in |R? Probably not. But the fact remains that you need definitions, and you will find that when you do it properly, for that you need limits. > > > I did not say that Cantor's strings were binary numbers. I said that > > > they might be *interpreted* as binary (dyadische, dual) > > > representations, safely. Cantor and Zermelo at least did so. > > > > You can do so *if* you take dual representations in account. But at least > > with Cantor I do not find any evidence that he *did* interprete them as > > binary numbers. > > No, he did not. But when extending his proof to functions, he applied > numbers 0 and 1, which obviously can also serve as m and w in the > orignal version. You stated "binary representations". Now I am a bit at a loss, what you mean with that. Representations of *what*? However, for binary representations of something arbitrary you need not use the digits 0 and 1, you can also use 1 and 2, or 0 and 5, or p and q. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 12 Jul 2006 11:32 David Hartley schrieb: > Given a sequence of total-orderings <_n of a set S we can define a > partial ordering <_L on S by > > For any x,y in S, x <_L y iff there exists n in N st. > for every m > n, x <_m y > > This ordering will not necessarily be total, but if it is, it seems > reasonable to say that the sequence has a limit, and that it is <_L. > > Using this definition, Daryl McCullough, Dik Winter and I have all given > examples of sequences of well-orderings whose limit is not a > well-ordering. WM's original proposal was rather muddled, but his > present version works. He *has* described how to construct a sequence of > orderings of the positive rationals, starting with that induced by an > arbitary enumeration, whose limit is the usual ordering. (In fact, the > same technique will give a sequence whose limit is any given total > ordering, and a similar, although a bit more complicated, argument shows > that there are sequences starting with any given total ordering whose > limit is any given ordering isomorphic to N.) > > What he hasn't proved is his claim that, since each ordering in the > sequence is order-isomorphic to the naturals, so is the limit. When > challenged on this, he refuses to acknowledge the problem, just claiming > that it is "clearly" true. Unless he attempts to prove this there is > little point in arguing with him; you can't point out the errors in a > non-existent proof. If you do continue, at least attack the part of his > argument which fails, not the part that he has - just about - got right. I really can't see your problem. Every rational q_n_0 has an index n_0 due to the initial well-ordering. This initial index will never get lost (contrary to the current indices) because this initial index is all that identifies this rational. Regards, WM
From: mueckenh on 12 Jul 2006 11:35 Virgil schrieb: > In article <1152639759.882706.252490(a)s13g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > The first edge is mapped on the first path. > > > > > > Which of the infinitely many paths through that edge is the "first" one? > > > > > That at the root. > > All paths start at the root. There these all are one path. You may call it bunch of paths. > > > > > > > > > > If this splits in two, 1/2 of the first edge in addition to the new > > > > edge is mapped on the new paths. > > > > > > You are not allowed to split edges. Besides which, the first edge has > > > already been entirely used up. > > > > It is inherited by the childs of the first path. > > Paths do not have children, only nodes have children. Call it as you like. The first bunch splits in two bunches. These two inherit half of the first edge each. > > > > > > > > > If the new paths split, 1/4 of the first edge and 1/2 of the > > > > second and all of the third are mapped on each path. Do you know what a > > > > geometric series is? What do you object to this mapping? > > > > > > If one can split edges into infinitely many peices one can equally split > > > paths. it is only unsplit edges and unsplit paths that are to be matched > > > up in injections, surjection or bijections.. > > > > You dislike fractions in calculations? Several thousand years ago > > humans discovered that fraction can be useful. Why should they not be > > useful in set theory? > > Some things are indivisible. An edge must start at one (parent) node and > end at another (child) node. Any proper fraction of one won't reach. We do not consider its function here, but its value of 1 edge. > > In the set of naturals {} is the first. > What does half of {} mean? Or any other fractional part of {}? In the set of naturals 1 is the first. The first edge is one edge, not {} edge. However, one edge is the first one, and this first one is divided in two halves. Regards, WM
From: mueckenh on 12 Jul 2006 11:36
Virgil schrieb: > In article <1152639642.547075.258360(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > All positions which can be enumerated are in the list, by definition. > > > > Hence, the decimal representation of 1/9 does not exist. > > > > > > Then how is it that so many people use it? > > > > Because they do only believe they used it. > > For numbers, such things only exist by belief in them anyway, so if > enough people believe in 0.111..., then it does exist. And presently enough people believe in infinity. Regards, WM |