An uncountable countable set
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From: Virgil on 12 Jul 2006 14:51 In article <1152718980.956031.285220(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > > You are using "to index" simultaneously for two different meanings. The > > first is "to index a specific position" the second is undefined. What > > does "0.1111 is indexed by the list number 4 = 0.1111" exaclty mean? > > The unary number n indexes every position 1,2,3,...,n. Actually a unary numeral, or any other numeral for a natural number, can only index one position. The MEMBERS of a natural, however expressed, can index up to that natural. The members of N can index all natural positions.
From: mueckenh on 12 Jul 2006 16:17 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > >> > To prove the existence and uniqueness of (|Q_+, <)? > >> > >> I want to know what the application of all of your conditional > >> transpositions to the ordered set *means*. I ask for a definition of > >> the *result*. And it would be nice if you show if it exists and is > >> unique. This is not (yet) proven "by the axioms". > > > > Many things cannot be proven by ZFC+FOPL. > > But you still have not proven your claim, that it is - at least - > meaningful to write about application of *all* conditional > transpositions to a given sequence. Until then it is not meaningful. The transpositions are a set of order omega. The existence of this set is guaranteed by the axiom of infinity. That the transpositions are conditional does not disturb the proof. Cantor's diagonal argument is also conditional. That the conditions have to be executed one after the other does not disturb the proof. Cantor's diagonal argument requires counting which is a sequential act too. > > > ZFC is not all, as even set-theorists increasingly recognize. > > You may chose any other axiomatic system. But please take care that it > is not contradictory by design again. > > > (|Q_+, <) is the result, because if there was any q_i not yet ordered > > by magnitude then it would become ordered by magnitude by my > > definition (not by yours). > > You have still not shown, what the application of *all* of your > conditional transpositions means. Writing about ordings of q_i > requires a meaningful (q_i), which you still have not proven to exist. The first well-order taken from Cantor's. The others remain well-orders because: wenn in einer wohlgeordneten Menge irgend zwei Elemente m und m' ihre Plätze in der gesamten Rangordnung wechseln, so wird dadurch der Typus nicht verändert, also auch nicht die "Anzahl" oder "Ordnungszahl". Daraus folgt, daß solche Umformungen einer wohlgeordneten Menge die Anzahl derselben ungeändert lassen, welche sich auf eine endliche oder unendliche Folge von Transpositionen von je zwei Elementen zurückführen lassen, d. h. alle solche Änderungen, welche durch Permutation der Elemente entstehen. So I am on the safe side with respect to each and every objection which could be raised by jealous people. Regards, WM
From: mueckenh on 12 Jul 2006 16:23 Dik T. Winter schrieb: > The representation 0.111... *does* exist. There is a specific definition > for it (as I have already given), and an axiom through which you may > prove that it does exist. The axiom says "there is an infinite set". It does not say that 0.111... does belong to that set, in particular because all numbers which in fact do belong to the set are different from 0.111... . > What is wrong about it? Take some particular p and take n any value larger > than or equal to p. We see that An[p] = 1 = K[p]. So what is wrong? If you take some particular p, this p is a natural, otherwise you could not take it. Therefore you cannot show that there are digits in 0.111... which are not indexed by naturals. And in fact, such digits do not exist, because they would be undefined. But as K is not in the list of natural numbers, it must differ from the numbers in the list. And this difference cannot be accomplished other than by K having more digits than any number in the list. Again, this is impossible. Hence we have a contradiction. > > 0.111... - 0.111...1 = 0.000...0111... > > > > such that there are more 1's in K than in any An. > > The last is true. But it is also true that for all p there is an An. > Or please exhibit a p for which it is not wrong. This equation 0.111... - 0.111...1 = 0.000...0111... says that for any An = 0.111...1 there is a p = n+1 which cannot be indexed. > > All digits which are indexed by smaller list numbers can be indexed by > > one larger list number. Therefore always only one is required. > > Care to explain the implication you mention above? How does K differ from all natural unary numbers together? > > > > > You do not need the first two list numbers 0.1 and 0.11, because the > > > > third alone is sufficient: 0.111 does index the first three digits p = > > > > 1 to 3 of 0.111... > > > > > > I am a bit at a loss here what you mean with "indexing". > > Again care to explain? How can a single number index three digits? instead of "to indeex" you can also say "to cover", i.e. to have more 1's than (or at least as many as) the number indexed or covered. > > No, indeed. So you have to prove it. > > > So let us find a resolution of this dilemma by mathematics. > > > > Define "*-" by > > a_i *- b_i = a_i - b_i if a_i > b_i else a_i *- b_i = 0 > > (subtraction down to zero but without negative numbers) > > > > Consider 0.111... and the list numbers as sequences or as vectors, such > > that *- can be applied to each term separately. > > > > If you are right, you must maintain the result: > > > > 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000... > > I can make no sense of the second term. What do you *mean* with > (0.1 + 0.11 + 0.111 + ...) The sum of all list numbers by digit, as shown in the following example: 0.1 +0.11 +0.111 _____ =0.321 Regards, WM
From: mueckenh on 12 Jul 2006 16:26 Dik T. Winter schrieb: > 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. just because of views like that. > > > > > 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. The numbers are to be understood as sequences or vectors, so your last sum is 0.(10),9,8,7,6,5,4,3,2,1. The next would be 0. (11),(10),9,8,7,6,5,4,3,2,1 and so on. The magnitudes of the numbers do not matter if they are larger than zero, because 1 *- omega = 1 *- 1 = 0. My question is whether 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000...? > > And you must > > refuse to calculate with fractions in mappings. > > Here I have no idea what this means. The binary tree: All paths, not yet disperged, start through the root edge a. (Let us denote them as a single path as long as they are together.) Map this root edge a on this single path. In the next level the path splits in two paths. Map half of edge a on each of them. The right one passing through edge b, gets b mapped on it and it inherits half of a. After splitting again, each of the paths gets the next edge, say c, half of b and quarter of a. | a o / \ b o o / \ / \c By this consideration we see that it is impossible for a path to carry a load of less than one edge at any level of the tree. Contrary there is always a load of nearly two edges, and in the limit of level infinity each path carries a load of precisely two edges. It is very simple to observe that no path can split without two additional edges available. In addition we find simple relations valid for all levels, such that no deviationa can occur: Number of paths at level n: P = 2^n Numer of edges up to level n: E = -1 + 2(n+1). There is no chance to get E < P. Regards, WM
From: mueckenh on 12 Jul 2006 16:33
Dik T. Winter schrieb: > > > > 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? I use only the digits behind the point: So the diagonal is 0.111... = 1/9. > > > > 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... ? = 2* 10 * 0.999... = 20 > > > 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. > If there is no other outcome possible, I don't need a further definition. 0.999... = 1 follows from the definition of (+,-,*,/) in the reral numbers. Regards, WM |