An uncountable countable set
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From: Dik T. Winter on 21 Jul 2006 21:03 In article <1153478871.303029.5360(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > But it is obvious for *all* *finite* *unary* number. And only finite > unary numbers are involved in the second list below: > > The *- sum of > > 0.1 > 0.11 > 0.111 > ... > 0.111... 0.111... is not a finite unary number. Rather, I would say it is not a unary number at all. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 21 Jul 2006 22:06 In article <J2s449.F6K(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > 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. And do > *not* come up with the objection that I should show that it is not there, > unless you want me to post all the guidelines plus the constition to let > you show it is not there. > > > > 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. And when I tried to interprete it you said > I was wrong. So *define* the meaning of "...", and not through examples. If one identifies "mueckenh"'s 0.1, 0.11, 0.111, ..., and 0.111... in the obvious way with ordinal numbers (with 0.111... being the first transfinite ordinal) then "mueckenh"'s *+ operation is merely the least upper bound operation, LUB, and any bounded set of ordinals has a least upper bound ordinal > > > This is a contradiction, because by tertium non datur only one case can > > apply. Actually 2 cases apply, the LUB of S = {0.1, 0.11, 0.111, ...} as a set of ordinal numbers is the same as the LUB of (S union {0.111...}), and in both cses is 0.111... . > One of these results is wrong if 0.111... represents a number. Not if it represents an ordinal. > > 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) > > And here I use how you defined *+ addition, and used finite addition > only (the only thing you have ever defined). > > 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).
From: Virgil on 21 Jul 2006 22:10 In article <J2s5LM.J4J(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article <1153478871.303029.5360(a)i3g2000cwc.googlegroups.com> > mueckenh(a)rz.fh-augsburg.de writes: > > But it is obvious for *all* *finite* *unary* number. And only finite > > unary numbers are involved in the second list below: > > > > The *- sum of > > > > 0.1 > > 0.11 > > 0.111 > > ... > > 0.111... > > 0.111... is not a finite unary number. Rather, I would say it is not a > unary number at all. All the problems go away if we regard all of "mueckenh"'s strings as representing ordinal numbers and his *+ operation as the least user bound operation, LUB, on bounded sets of ordinals.
From: Franziska Neugebauer on 22 Jul 2006 04:24 Virgil wrote: > In article <J2s5LM.J4J(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > wrote: > >> In article <1153478871.303029.5360(a)i3g2000cwc.googlegroups.com> >> mueckenh(a)rz.fh-augsburg.de writes: >> > But it is obvious for *all* *finite* *unary* number. And only >> > finite unary numbers are involved in the second list below: >> > >> > The *- sum of >> > >> > 0.1 >> > 0.11 >> > 0.111 >> > ... >> > 0.111... >> >> 0.111... is not a finite unary number. Rather, I would say it is not >> a unary number at all. With the definition of the unary notation (essentially meaning an infinite sequence of zeros and ones) unary(omega) may be defined as the sequence (a_i = 1)_{i e omega}. Of course it does not represent a natural but omega. Whether we should /call/ it unary _number_ is not worth arguing. > All the problems go away if we regard all of "mueckenh"'s strings as > representing ordinal numbers That is what unaries are made for. The real problems of the "game" WM is playing are: 1. Using undefined notation. 2. Uncapability or unwillingness to argue within the logical framework. 3. Intentional unwillingness do learn what certain notions mean (i.e. definition, proof, contradiction). 4. Utilizing non-existing entities like mckenheimian element/ member/sequence in the discourse. (i.e. larget element of an infinite set, last member of an infinite sequence, sequence containing its LUP at the last position). 5. Taking the unaries for the origin of numbers, i. e. confounding numbers with their representations. > and his *+ operation as the least user bound operation, LUB, on > bounded sets of ordinals. The core of his *-+-operation is a function {0, 1} x {0, 1} |-> {0, 1} which we usually call boolean or. The *-addition of two unaries is defined as the memberwise *-sum. The *-sum of an infinite sequence of unaries (aka "list") is indeed a kind of LUB. F. N. -- xyz
From: Dik T. Winter on 22 Jul 2006 21:59
[reformatting to standard discussion style] In article <1153479321.840071.258130(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter wrote: > > You again refrain from answering questions. The definition of 0.999... > > is as follows: > > 0.999... = lim{n -> oo} sum{k = 1 .. n} 10^(-k) > > You said that definition is silly. What is silly about it? > > Sorry, could you help me with "silly". My find function shows with > "silly" only the sentence: "It is only in modern set theory that such > silly things as the rejection of the binary tree occur." You actually said that it made no sense. But I will try to find the correct quote. > > what indices in the limit above are undefined? > > Those which cannot be enumerated by natural numbers. What indices in the above limit can not be enumerated by natural numbers? > > Because, and I state it again, there is a difference between sequential > > processes and simultaneous processes. > > I know that in the infinite binary tree *every* split is realized by > the presence of two edges and would be impossible without them. > > | > o > /\ > > Nothing can be more simultaneous than this knowledge which is present > without considering any sequential process and which concerns the whole > tree. So, I can perform the 102349176-th split without knowledge about previous splits? Or was that not the 102349176-th split? When you do your splitting you are performing a sequential process. Further splits depend on earlier splits. > Dik T. Winter wrote: > > Four your count of edges and > > paths you need limits, but those limits do not exist. > > Oh, the limit of 1 + 1/2 + 1/4 + ... = 2 is not existing? Strange. Is that the count of edges or the count of paths? > There is no question that every path can be interpreted as the > representation of a real number. And now "those limits do not exist"? > Very strange. You are confused. The count of paths have no limit. But if you wish to interprete the count of paths as a real number, be my guest, and tell me how to represent the count of paths as a real number. But whatever, if you were ever to complete your picture, both the number of edges and the number of paths would become uncountable (you did confuse me a bit by first talking about nodes and now about edges). It is the number of nodes that remains countable. The number of edges after adding the n-th splitting is 2n+1, the number of paths is 2n, and the number of nodes is n. > > Again, you need something like a limit here, because adding nodes is a > > sequential process. > > Adding lines in Cantor's list is not a sequential process? Perhaps, but the list is given by some (unidentified process). Without the given list, there is even no start on the determination of the diagonal. The proof clearly states "given a list", so the assumption is there that there is a list. > They are > given by the "higher Being which rules matheology". Nope, the assumption is that you are able to give a list. And the proof shows that you are not able to. > But adding levels > of the binary tree is a sequential process. Are you really in earnest > about that question? Yes. I state that adding levels (or individual nodes, but I grant that all nodes of the same level can be added simultaneously), is a sequential process. The definition of the list may or may not be a sequential process, but that is irrelevant. Given a list, the definition of the diagonal number is not sequential. > Take the nodes as digits. There is no question that every path can be > interpreted as the representation of a real number. Oh, yes. And every edge can be interpreted as such (with quite a bit of duplication). And every node can be interpreted as a rational number with a finite binary representation. (Obviously I ignore the dual representations here.) > Dik T. Winter wrote: > > Also, it is true for each natural that it cannot index all 1's of 0.1111. > > That means that for all integers it is impossible to index all 1's of > > 0.1111. On the other hand, with the *set* if integers it is possible > > to index all 1's of 0.1111, and also of 0.111... . That is, when with > > "to index" we mean "point to an individual element" as in K[p]. The > > reason is that each natural can index one and only one digit. > > > On the other hand, you on occasion use index to mean something completely > > different. For that I could use the word "cover" where A covers B if > > in all positions where B has a 1, A also has a 1. Now in the finite > > case we find that the digits of 0.1111 are covered by all naturals > > greater than or equal to 4. But the digits of 0.111... are not covered > > with any natural. > > Why then do you expect that all digits are indexed "in the infinite > case"? You apparently are not aware of common mathematical terminology. When something happens for natural n, and we can define a limit when n grows without bound, it is common to state that that limit is the infinite case in informal discourse. As you are not very formal (you do not provide definitions), I thought this was an informal discourse. > Dik T. Winter wrote: > > But nobody has stated that that is possible, it is > > only you who *claims* that that should be possible, without giving any > > reasonable proof of it. > > There is no difference! If a natural indexes the digit n, then it > covers all digits m =< n. This is guaranteed by the construction of the > list of unary representations - for *all finite numbers*!!! How can you > believe you could make someone believe this situation would change? Well, it does not change, and I never said so. When you have a list A1 to An, the *+ sum will be An. There is no dispute about that. > Therefore I say sometimes each natural n can index all digits m =< n > which it covers. Whether you call this to cover or to index is > unimportant. Both definitions imply each othe as long as *finite > numbers* are concerned Yes, for each finite number n An covers all of A1 to An. No dispute about that (and I never disputed that). In shorthand: For each n, An = (*+){i = 1 .. n} Ai. True, without a prob |