An uncountable countable set
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From: Franziska Neugebauer on 11 Jul 2006 14:23 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> Every "list number" (= sequence member) is a sequence of digits >> indexed by every natural number. > > Not as an unary representation. Every unary representation *is* by definition a sequence of digits. Indexed are *the* positions in that sequence. > The list number 0.1111 is indexed by the list number 4 = 0.1111 What exactly does "indexed" mean here? > but not by the list number 3 = 0.111, because 3 cannot index the > fourth 1 of 0.1111. 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? >> > I say: What can be indexed by different list numbers can also be >> > indexed by one alone. >> >> Meaning? Proof? > > The first 1 of 0.1111 can be indexed by 0.1, the second 1 of 0.1111 > can be indexed by 2 = 0.11 etc. But all 1's of 0.1111 can be indexed > by 4 = 0.1111 simultaneously. What does "being index simultaneously" mean? Can't see that. >> > If you do not believe that, >> >> I do not know what you mean. >> >> > then you should be able to show an example where more >> > than one list number is required. Of course it must be a finite >> > example, because there are only finite list numbers. And note: all >> > list numbers are unary representations of natural numbers. >> >> But not all unary sequences are representations of natural numbers. >> 0,111... is a sequence which does not represent a natural. > > And it cannot be completely indexed by natural numbers. What does "completely indexed" mean? F. N. -- xyz
From: Virgil on 11 Jul 2006 17:25 In article <1152638595.246620.165700(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > > It can only be proved > > > after having counted the digits from 1 to n without leaving out a > > > single one. > > > > Nonsense. The Cantor rule generates a digit to go in the nth decimal > > place of the number being created without any reference to any other > > decimal place. > > > You cannot identify any place without counting from 1 to that place. The axiom of infinity guarantees us all those "places" without counting anything.
From: Virgil on 11 Jul 2006 17:27 In article <1152638676.290941.149680(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > In ZFC the elements of a set do exist and omega does exist too. I did > not *claim* that omega does not exist, but I *proved* that as a fact. > > Regards, WM WM has as yet to show that he can prove anything. All of his attempts so far have been failures in everyone's eyes but his own. And those of us who have proved at least something are well aware of how deceptive one's own eyes can be in such matters.
From: Virgil on 11 Jul 2006 17:31 In article <1152638895.480011.23630(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > 0,111... is a sequence which does not represent a natural. > > And it cannot be completely indexed by natural numbers. If such an obviously well ordered set has digits not indexed by members of N it must have a first such digit Which digit is that first non-indexed digit, "mueckenh"? Can't answer? Thought not!
From: Virgil on 11 Jul 2006 17:58
In article <1152639144.800362.201000(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > But it is impossible to count over all natural numbers. Therefore the > > > prescription replace a_n by b_n does not cover all natural numbers, it > > > is void. > > > > It is only impossible to do in a way which Cantor did not attempt to do > > it, by iteration of individual cases. It is quite possible, and even > > trivial, to do it the way Cantor did it, by a general rule which covers > > all cases simultaneously. > > There is no simultaneity in counting. No one was doing it that way. > n is not defined without counting > from 1 to n. All n are "defined" by the axiom of infinity. > The whole set can never be exhausted by counting, > therefore Cantor has no advantage over me. But as Cantor does not have to count, and "Mueckenh" does, Cantor wins. > > > > According to "mueckenh"'s way of thinking, it would be impossible to > > prove that all natural numbers are greater than -1, since we would have > > to do it one natural last time and would never finish. It would be > > equally impossible in "mueckenh"'s way of thinking o prove that the sum > > of two consecutive naturals is odd. > > Both proofs, and many similar ones, are trivial if one is not > > constrained to do a separate proof for each individual case, but are > > allowed to do general cases. > > Of course these proofs can be done. But they do not prove that all > natural numbers form a set with an actually infinite cardinal number. They show that the cardinality is not finite. > Of course you can define "replace 4 by 5", but in order to go through > the list and produce the diagonal number from the single digits, you > have to count. That "mueckenh" has to count to get anywhere does not limit anyone else. > I see that the diagonal proof will produce digits which differ from > the diagonal digits of the list numbers. But I see no evidence that the > diagonal proof will at any point produce even half of the digits > required for an irrational number. There is no requirement that the number be irrational, merely that it not be in the list presented. If all the numbers in that list are irrational, the diagonal could very well be rational. > I can disprove Cantor's claim by showing that the diagonal is not > different from any list number because the list is by definition > complete. Let's see you do that little thing, then. In that sense I can equaly "define" a circle to be square, but that does not make it so. > > > > You claim that the final result is in natural order and simultaneously > > well ordered. Between the first and second in the well ordering their > > average value is a rational that in natural ordering must be between > > them, but is not. This, among many other contradictions, disproves your > > claim. > > In the well order of the rationals, all rationals have finitely many > predecessors while in the order by magnitude each one has infinitely > many predecessors and as many successors. Hence, a well-ordering of the > rationals is impossible. Well orderings of the rationals are actual. Here is one: Each rational has a unique representation as i/n where (1) i is an integer, (2) n is a non-zero natural (3) for non-zero i, i and n are relatively prime (4) for i = 0, n = 1 We then define f: Q --> N by (a) if i < 0 then f(i/n) = 2^(-i) * 3^n (b) if i >=0 then f(i/n) = 5^i * 7^n This function injects Q into N with image, say, Im(f). Since Im(f) is a subset of N, it is naturally well ordered by the well ordering of N, and induces a well ordering on Q by q1 < q2 in this new ordering if and only if f(q1) < f(q2). This is a simple well ordering of Q. >This, among many other contradictions, > disproves Cantor's > claim. Only in your dreams. > You have not understood the scheme. Look into my replies to Franziska. > My scheme gives always the correct well-ordering - for all sets which > exist. Only for finite sequences. It fails for any infinite sequence of umberswhich requires more than a finite number of transpositions for its success. This includes any well ordering of the rationals. |