Cantor Confusion
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From: mueckenh on 9 Jan 2007 09:25 William Hughes schrieb: > > > > But it is supposed to exists as the complete, finished diagonal. > > [The claim is that the line L_D does not exist. > Consider each of your comments in light of this claim.] > > No. No such claim was made. The claim was that L_D does not > exist. Knowing whether a complete finished diagonal > exists will not tell you whether L_D exists. You are wrong. If the complete diagonal exists, then the complete line L_D exists. If the complete diagonal does not exist, then the complete line need not exist. That is just my arguing. Of course an actually existing complete line is nonsense, but it results from the assumption of a complete diagonal, which, unfortunately, is not considered nonsense by set theorists. > > This comment says nothing about whether L_D exists. > > > > > > L_D does not exist > > > so the union of line ends including the end of L_D does not exist. > > > > Therefre a complete, finished diagonal does not exist. > > Irrelevent. The diagonal is not a line. > > This statement says nothing about whether L_D exists. The diagonal consists of line ends, of finished lines. > > > > Every end of a line belongs to a line. However, these lines > can be different for different line ends Your assertion is wrong. There are never different lines required for different line ends. Different line ends are always elements of one single line. > > L_D does not exist. Of course does it not exist, because th diagonal does not exist. To spell it out clearly and herewith closing this discussion: Your error is to assume that more than one line could be necessary to supply two or more different natural numbers, elements (or indexes) of the diagonal. That is wrong. Regards, WM
From: mueckenh on 9 Jan 2007 09:37 Franziska Neugebauer schrieb: > >> L_D does not exist > >> so the union of line ends including the end of L_D does not exist. > > > > Therefre a complete, finished diagonal does not exist. > > As far as I can see is the diagonal "only potentially infinite". Nobody > claimed that it is "complete". Nonetheless, the conclusion that a > single line L_D does not exist remains valid. If the diagonal is not complete, then it has a largest element. A line containing this element (and all smaller ones) does exist. If it turns out, that the diaogonal has a larger element, then it turns out that a line containing this (and all smaller ones) does also exist. In any case, a line containig all elements of the diagonal does exist. > > >> > Fine. There are no infinite sets. There is only potential > >> > infinity. Every line including the diagonal > >> > >> The diagonal is not a line. The diagonal is a potentially infinite > >> set. > > > > Correct. Therefore it does not contain all natural numbers (as > > indexes) because then it would be actually infinite. > > Please pay attention to your wording! The diagonal _contains_ numbers > (values). Not "as indexes", but as value of the diagonal _at_ a certain > index. So your statement is rather meaningless. To define what you are talking about: The diagonal of the EIT contains only digits 1. 1 11 111 .... But these digits can be indexded by natural numbers 1,2,3,... The diagonal therefore, contains natural numbers as indexes as I said. Regards, WM
From: mueckenh on 9 Jan 2007 09:47 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > >> > The finite union of two or more finite trees is a finite tree. An > >> > infinite union of finite trees is the infinite tree. > >> > >> What is an infinite union? (Please give a definition) > > > > An infinite union of trees is the union of all (or nearly all) finite > > trees which reach from level 0 to level n where n eps N. > > 1. Please give a definition of "union of all finite trees". Definitin: Denote the nodes of the tree by (0,0) (1,0) (1,1) (2,0a) (2,1a) (2,0b) (2,1b) .... (n,0a) (n,1a) ... The union of all trees up to the n-levels tree is {(0,0)} U {(1,0), (1,1)} U .. U {(n,0a) (n,1a) ...} Example: The union of the one-level tree and the two-levels tree is {(0,0), (1,0), (1,1)} U {(0,0), (1,0), (1,1), (2,0a), (2,1a), (2,0b), (2,1b)} > 2. Please give a definition of "union nearly all finite trees". Union of all trees with a finite number of exceptions. Regard, WM
From: William Hughes on 9 Jan 2007 09:49 mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > But it is supposed to exists as the complete, finished diagonal. > > > > [The claim is that the line L_D does not exist. > > Consider each of your comments in light of this claim.] > > > > No. No such claim was made. The claim was that L_D does not > > exist. Knowing whether a complete finished diagonal > > exists will not tell you whether L_D exists. > > You are wrong. If the complete diagonal exists, then the complete line > L_D exists. >If the complete diagonal does not exist, then the complete > line need not exist. That is just my arguing. Of course an actually > existing complete line is nonsense, but it results from the assumption > of a complete diagonal, which, unfortunately, is not considered > nonsense by set theorists. > > > > This comment says nothing about whether L_D exists. > > > > > > > > > L_D does not exist > > > > so the union of line ends including the end of L_D does not exist. > > > > > > Therefre a complete, finished diagonal does not exist. > > > > Irrelevent. The diagonal is not a line. > > > > This statement says nothing about whether L_D exists. > > The diagonal consists of line ends, of finished lines. But this does not make the diagonal a line, so the question of whether the diagonal exists is not relevant. > > > > > > > > Every end of a line belongs to a line. However, these lines > > can be different for different line ends > > Your assertion is wrong. There are never different lines required for > different line ends. Different line ends are always elements of one > single line. The assertion was that the lines can be different. This is trivially true. Your statement that "Different line ends are always elements of one single line" is equivalent to the statement that L_D exists. We know this is false. "Different line ends are always elements of one single line" is not true for the potentially infinite set of all "Different line ends". > > > > L_D does not exist. > > Of course does it not exist, because th diagonal does not exist. > Make up your mind. Your repeated claim is that L_D does exist. > To spell it out clearly and herewith closing this discussion: > Your error is to assume that more than one line could be necessary to > supply two or more different natural numbers, elements (or indexes) of > the diagonal. That is wrong. No. It is clear that given any two natural numbers, or indeed any set of natural numbers you can write down, there exists a single line which contains all the natural numbers you wrote down. But this line *can* depend on which set of natural numbers you write down. So this fact can be true whether or not L_D exists. This fact cannot be used to show that L_D exists. L_D does not exist. - William Hughes
From: Franziska Neugebauer on 9 Jan 2007 09:51
mueckenh(a)rz.fh-augsburg.de wrote: > Of course an actually existing complete line is nonsense, > but it results from the assumption of a complete diagonal, So you _now_ want switch over to discuss the complete (not only "potential") existence presuming a complete (not only "potential") existing diagonal? F. N. -- xyz |