An uncountable countable set
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From: Dik T. Winter on 22 Jul 2006 22:03 In article <1153482158.663876.226910(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > What your > > example shows is that an infinite sequence of transpositions can destroy > > well-orderedness. What is the difference? The only conclusion is that > > Cantor's statement is incorrect or that the reading of Cantor's statement > > is incorrect. Take your pick. > > Your hinting at any doubt about the clear meaning of Cantor's statement > is outrageous. Perhaps. In that case his statement was clearly wrong. > Wrong. My arguing is directed against the complete existence of the set > of all naturals, the set of all digits of a real number, the actual > infinity, the first transfinite number. That all is purest Cantor. As I said before. You are arguing agains the axiom of infinity. But your inconsitency proofs show nothing, because in part of your proofs you always use the negation of that axion. -- 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 22 Jul 2006 22:14 In article <1153482677.919976.211910(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > What has your statement in parentheses to do with the statement I gave? > > There is no argument at all about K being different from all natural > > numbers. It is just a plain statement: > > For all p, there is an n such that An[p] = K[p] > > you claim that is false, and again without direct proof. Do you disagree > > with the statement: > > For all p, Ap[p] = K[p]? > > I know that Ap[p] = 1 = K[p]. Now chose in the above n = p and you get > > the new statement. Why would this second statement be false? > > You said "for all p". If p is a natural number, then the segment K[p] K[p] is a single digit. > is the unary representation of a natural number and, hence, is covered > by a natural number. Remember: > > For each natural (i.e. finite) number p we know: to index and to cover > are equivalent. > p covers K[p] <==> p indexes p. > > Would K have only natural indexes p, then it would be covered by at > least one natural number. Again, a conclusion without proof. Consider the digits of K indexed by natural numbers p. Further consider the statements: For all p, Ap[p] = K[p] and For all p, Ap[p+1] != K[p+1] Which of these is false, and why? Pray provide a proof. -- 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 23 Jul 2006 13:06 Dik T. Winter schrieb: > In article <1153478099.434931.273560(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > Where in the guidelines, and where in the > > > consitution, is it layed down that zero is considered a natural number? > > > I have access to both, so pray give pointers. > > > > I do not remember the source, but it was a trustworthy one. > > Yes, so what am I to say about it? I simply do not believe it. I would be glad if you were right. But I don't believe it. Perhaps someone else might be able to give a hint? > > > > When you can derive, > > > within that system, contradicting conclusions, your axioms are inconsistent. > > > In principle, no other knowledge than the axioms and the definitions is > > > needed. > > > > Within this system I defined a *+ sum and showed that > > > > The *- sum of > > > > 0.1 > > 0.11 > > 0.111 > > ... > > 0.111... > > Nope. You did *not* define how to do a sum of order w, nor did you define > how to do a sum of order w+1. I said if one 1 is present, then the *+sum is 1. Hence the order is absolutely unimportant. This includes: The *+sum of any order is 1 if at least one 1 is present. Nevertheless, i am sure you will claim, the sum be undefined. What is your reason for such an obviously false claim? > And when I tried to interprete it you said > I was wrong. So *define* the meaning of "...", and not through examples. > > > This is a contradiction, because by tertium non datur only one case can > > apply. One of these results is wrong if 0.111... represents a number. > > You are again making no sense. Using your *+ notation: > 0.10 *+ 0.01 = 0.11 > (sum not in the list) > 0.10 *+ 0.01 *+ 0.11 = 0.11 > (sum in the list) This is the same invalid argument Franziska uses to post. It does not hold for unary representations. Therefore I used such representations: If n = 1 then m = 1 forall m < n. > And do not come with the objection that I use 0.01. Because all of > 0.10, 0.01 and 0.11 can represent a number (1/5, 1/25 and 6/25 to give > an example). But they cannot represent naturals in the unary form which I defined with purpose. Regards, WM
From: mueckenh on 23 Jul 2006 13:12 Dik T. Winter schrieb: > > The numbers count the lines. As long as there are only finite numbers, > > the list is finite. > > Why? There are infinitely many finite numbers. Just that is in question. More precisely: Are there actually or only potentially infinitely many numbers. > You are trying to establish > that the Axiom of Infinity leads to a contradiction. But your very statement > "As long as there are only finite numbers, the list is finite" is in itself > in contradiction to the axiom. The statement is just opinion, without proof. > So there is not yet a contradiction. Please show a proof of your statement. The proof is given by the list: 1 0.1 2 0.11 3 0.111 .... ... w 0.111... The set of unary numbers cannot result in the *+sum 0.111... . (I defined: the *+ sum is 1 if there is at least one 1 in it.) With respect to the left column this means the cardinality of 1,2,3,... cannot be w without w in it. > > > I thought we were talking about natural numbers. I have not yet seen a > > > definition that calls the last one a natural number. > > > > I never said so. > > So, why do you think that it is in the list, when the list is a list of > natural numbers in unary notation? It is not *in* the list, but has been added at that position which is due to its 1's, namely omega. This is justified because all numbers are listed at that position which is due to the 1's in their unary representations and because omega is a number (though not a natural) which covers only *one* place in the sequence of ordinals. > But they give different information. When the "notations" are unary > representations of natural numbers, their *+ sum is not the representation > of natural numbers, and so is not in the list. When the "notations" > are decimal (or ternary, or whatever) representations of rational numbers > their *+ sum is a representation of a rational number that is not in the > list. So in both cases, the *+ sum is not in the list. In both cases the *+ sum is a representation of a number which is different from all list numbers (naturals or terminating rationals). > > In both cases, when you insert the *+ sum early in the list you get either > (case 1) a non-natural number in the list, so it makes no sense, or > (case 2) a different rational in the list, and in this case the *+ sum > is that rational. But in the latter case I see no problem. In the list there are only terminating rationals. 0.111... has omega digits. Therefore we see perfect parallels. > > > This is only possible, if it was in the list before already. > > Wrong. It is because all elements are idempotent under your *+ operation. > Union {a} with {b} is the same as union {a} with {b} and {a, b}. Again you forgot the special from of unary representations. Is the special property of unary representation so difficult to see, so hard to remember? Regards, WM
From: Virgil on 23 Jul 2006 13:13
In article <1153674388.435551.29190(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Nope. You did *not* define how to do a sum of order w, nor did you define > > how to do a sum of order w+1. > > I said if one 1 is present, then the *+sum is 1. Hence the order is > absolutely unimportant. This includes: The *+sum of any order is 1 if > at least one 1 is present. Then you are saying that 0.1 *+ 0.11 = 1. In fact what "mueckenh" is saying is that the "*+" sum of any set of ordinals, when represented in his loony notation, is their LUB, expressed in that same loony notation.. |