Cantor Confusion
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From: William Hughes on 15 Nov 2006 19:35 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > If we add one element to each of the columns then > > > > > > the supremum of the lengths of the initial segments of the columns > > > > > > changes to omega +1. If we add one elelment to each of the lines > > > > > > the supremum of the lengths of the lines does not change, it > > > > > > remains omega. > > > > > > > > > > Fine. This new matrix has a diagonal, if the old matrix had one. > > > > > > > > Let he original matrix be A. > > > > > > > 1 > > > 12 > > > 123 > > > ... > > > > > > > Let B be the matrix we get by adding one element to each > > > > of the columns of A. This matrix does not have a diagonal. > > > > > > Every matrix has a diagonal. The question is only whether all lines and > > > columns are crossed by the diagonal. Not every matrix has a bijection > > > between lines and columns. > > > > OK. We say that every matrix has a diagonal, but > > only a matrix with the same number of lines and columns > > has a diagonal which crosses every line and every column. > > > > The matrix A has the same number of lines (omega each > > column has omega elements) and columns (omega the supremum > > of {n | n in N}). > > You say so. But if so, then there must be a bijection such that a > column is mapped on a line for every column and every line. All finite > values are bijected without problem n <--> n. But there remains a taken > supremum to be bijected with a not taken supremum. That is impossible. No. The taken supremum has nothing to do with the bijection. Let the set of lines be S, and the lengths of the lines be T, and the columns beU and the lengths of the columns be V. You are confusing the elements of S with the number of elements in S. Likewise the elements of T with the number of elements in T. The bijection is between S and U. There are omega lines, so S has omega elements, and omega columns, so U has omega elements. The number of lines is the supremum of the lengths of the columns. This supremum is taken. The number of columns is the supremum of the lengths of the columns. This supremum is not taken. There is no bijection between the sets T and V > > > > The matrix B does not have the same number of lines (omega+1 each > > column has omega+1 elements) and columns (omega the supremum > > of {n+1 | n in N}). > > of {n | n in N}). > > > The matrix D does not have the same number of lines (omega+1 each > > column has omega+1 elements) and columns (omega the supremum > > of {n+1 | n in N}). > > If there was a bijection between columns and lines in A, then there is > a bijection of columns and lines in B. No B has the same number of columns as A. B has one more line than A. You cannot extend the bijection from A to a bijection for B. > > > > > You want to say that A has a diagonal, but addition of one element to > > > each line, each column and the diagonal eliminates this property? > > > > > > Yes, because adding one element to each column changes the number > > of lines, but adding one element to each line does not > > change the number of columns. > > Then there must have been a difference in columns and lines in A > already Yes, but there was no difference in the number of columns and the number of lines. Two different sets can have the same number of elements. As before let the set of lines be S, and the set of columns be U. Then the elements of S are different than the elements of U ..However, the number of elements of S is the same as the number of elements of U. > > > > > The original matrix has the same number of > > lines and columns, however,the lines and columns are different. > That is impossible, because we have only the ordered set N to construct > columns and lines of the matrix. There is exactly one ordered set of > infinitely many initial subsets: > > 1 > 2,1 > 3,2,1 > ... > > They are the lines. The first column is the union of them. Correct. The first column is the union of all of them. The nth column is the union of all of them minus the set {1,2,3,...,n-1} The nth line is the union of the first n of them Both lines and columns are put together from subsets of the one set of infinitely many initial sequences however they are put together from different subsets. > Adding one > element w to each of them has the same effect. > > Impossible. Both sets are the natural numbers. Further we have the > segments of the first column to be the same as the lines. This > > 1 > 2 > 3 > ... > n > and n, ..., 3.2.1 > Yes but the first column is the union of all of the initial segments. No line is. > > Doing the same thing > > to the lines and columns does not have the same result. > > That's what I demonstrated. The reason is that there cannot be a > bijection. No. There is a bijection. Doing different things ("does not have the same result") means that the bijection cannot be extended. > > > > > > > > > > > Only for matrixes B and D is there a last line. > > > > > > > > So if there is a diagonal, there is no last line. > > > > > > How about adding the same elements not at the ends but at the > > > beginnings of the lines and the columns? > > > > Adding an element to the beginning of an infinite sequence does > > nothing. So this is the same as adding an element to the > > end of each line, but not adding an element at the end of each > > column. > > Adding an element at the beginning of every column does the same as > adding an element at the beginning of each line. But at the ends, there > are differences? There should be a reasonable reason for this different > behaviour. There is. Columns. There is a beginning but no end. You can add an element at the beginning because the beginning exists. You cannot add the element at the end because no end exists. You have to add the element "after" the end (hence transfinite). There is a difference between adding and element *at* the beginning and *after* the end. Lines. There is both a beginning and an end. You can add an elment *at* the beginning or *at* the end. > > > > > > > No. If you add one line, but you do not add one column you > > do n
From: Dik T. Winter on 15 Nov 2006 19:32 In article <1163603386.223046.315750(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > Why? Yes, I know why. You think that the complete set of natural numbers > > does contain an un-natural number. > > It must, but cannot. As an un-natural number cannot be a natural > number, this set N cannot exist. Again, you state only it must. > > I have asked you for a proof based on > > the basic principles. But you simply refuse to do so. You think it is > > the case, so it must be the case. So, please, a proof that N contains > > something that is not a natural number based on the standard axioms of > > set theory. Once you have done so you will have shown an inconsistency > > because it is easy to proof that there is no such number. > > Here is one of several proofs: > > Take the set of natural numbers in form of a list or matrix: > > 1 > 11 > 111 > ... > > This matrix has length omega and width omega. And its diagonal has > length omega. No line has length omega. Therefore the width is larger > than any line. And the diagonal is longer than any line. This is > impossible. No, that is very possible. > impossible. In order to see that, add one element to every column and > to every line. Now the order type of any column is omega + 1, the > length of the matrix has order type omega + 1, the order type of any > line is n+1 < omega, and the width of the matrix has order type omega. Yes, doing that changes the matrix from a square matrix to a non-square matrix. The height is omega + 1, the width is omega. > The diagonal is a bijection between columns and lines. It des no exist. In a non-square matrix the diagonal does not cross either all lines or all columns. If a non-square matrix has more lines than colunms, there is no bijection between the elements of the diagonal and the lines. This does *not* make the diagonal non-existent, only that the elements are either not in bijection with the lines or not in bijection with the columns. > This shows that the diagonal in the original matrix did not exist > either, unless there was a line of order type omega, i.e., an > un-natural number. Nonsense. If we fill up your matrix with zero's to make it a true matrix (and not some triangle), we can index the elements of the original by a pair of indices [i,j] where both i and j are natural numbers, i is the row number and j the column number. The i-th row consists of the elements a[i,j] with j any natural number, and the j-th column consists of the elements a[i,j] with i any natural number, and the diagonal consists of the elements a[i,i] with i any natural number. When you add one line, you add the elements a[w,j] with w = omega and j any natural number. Obviously it does not cross the diagonal. -- 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 15 Nov 2006 19:57 In article <1163604980.169629.197680(a)f16g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > But you do not allow elliptic or hyperbolic > > geometry? If not, why not? > > I do not forbid it. It is clear that already the simple geometry on a > sphere does not yield two parallels. Euclid simply did not consider > such kind of plane. In another system we have other axioms (or better > fundamental truths). Well, the same in set theory. In one form it has AC as a fundamental truth, in another version it is not a fundamental truth. What is the essential difference between the cases? > > > > I do not ask what you think. Reread my question. What is true about a > > set of numbers in nature or reality? > > There are so many truths. Take order, 1 < 11, commutativity of addition > and multiplication, n + m = m + n. These things do not become invalid > or to be proved only because matrix multiplication or quaternions were > invented. Again, no answer. What is true about a set of numbers in nature or reality? > I think there is a great difference. It is not necessary to call > negative solutions "false" solutions as even Descartes did, (because it > was customary at his time. Although this custom was justified as long > as only positive numbers were called numbers.) But it is necessary to > distinguish between negative and positive numbers or real and complex > numbers or Euclidean and non Euclidean spaces. And you were vehemently objecting at calling the irrational numbers numbers. What is the essential difference between all those cases? I once asked you for a definition of "number", and you did never supply a proper definition. Now you say yourself that the definition has been changed in the course of time. I may note that there is no trichonomy between the complex numbers, it is not possible to consistently order them. > Then it will not be > astonishing that some new axioms arise or some existing axioms have to > be modified. Going from addition to subtraction, we find that > commutativity is no longer valid. And the natural numbers are not > closed under subtraction. But these adaptions of axioms (or I should > better say: of fundamental truths) has to be justified and not only to > be decided by an arbitrary choice. But in the cases I know the choices are justified. Perhaps you think the justification is insufficient, but that can only be a matter of opinion. Interestingly, Hardy (whose work largely builds on work by Kronecker and Dedekind: algebraic number theory, I think Kronecker was the driving force in that theory) once said he was proud that there was no practial application for his work. He would turn in his grave if he saw how it is used nowadays (cryptology). So perhaps one day we will find practical use of the p-adics... Or for set theory without AC, but with an alternative axiom... In the geometry case the parallel postulate was felt to be unnatural, and so Saccheri attempted to proof that geometry with the negation of that axiom would be inconsistent. He thought he had shown it, but actually the inconsistency proof was wrong and he had created Lobachevsky geometry before its actual creation. But even at those times there were not yet practical applications of those geometries. It took some time before it was realised that the geometry on a sphere was non-Euclidean if one applied a specific definition of "straight line" on the sphere. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 15 Nov 2006 20:00 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > Franziska Neugebauer wrote: > > > William Hughes wrote: > > > > > > >> You want to say that A has a diagonal, but addition of one element to > > > >> each line, each column and the diagonal eliminates this property? > > > > > > > > Yes, because adding one element to each column changes the number > > > > of lines, but adding one element to each line does not > > > > change the number of columns. > > > > > > What did I tell you!? > > > > Yes, but the discussion has reached the point where > > the distinction between adding an element to an infinite > > sequence and adding an element to a finite sequence > > can be made clear. > > If there is a bijection between lines and columns, then line n has > exactly the same properties as the initial segment of the first column: > 1,2,3,...,n <--> 1,2,3,...,n. The first column is an infinite sequence. Equivalently, it does not have a last element. Each line is a finite sequence. Equivalently, each line does have a last element. If you add an elment to an ordered sequence of natural numbers without a last element you add a transfinite number. If you add an element to an ordered sequence of natural numbers with a last element you add a natural number. This is the difference between lines (sequences with a last element) and columns (sequences without a last element). - William Hughes
From: Dik T. Winter on 15 Nov 2006 20:17
In article <1163605613.609012.99490(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1163510733.868272.250410(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > They have no finite binary representation. But if they can appear in a > > > list, then they can appear in a tree. > > > > Yes, of course they can appear in a tree. But they do not appear in *this* > > tree. 1/3 can appear in a ternary tree. > > So you say that 1/3 does not have a binary representation? No, that is specifically *not* what I say. 1/3 does not have a *finite* binary representation. > So you say that pi has no representation at all in a fixed base > (n-adic or n-ary), and, therefore, cannot appear in Cantor's list? No *finite* representation. But "Cantor's list" does allow *infinite* representations. More specific, his diagonal proof was about lists where each element consisted of an *infinite* sequence of symbols. > > > Even of [0, 1] because 0.111... = 1. > > > If you have this opinion, I will happily agree, but then you must also > > > apply it to every binary representation of the reals. The tree is > > > nothing other than such a representation, a special one. > > > > And that is wrong. The tree contains only the set of finite binary > > representations. > > i.e., Cantor's original list contains only the set of finite sequences > of m and w. I agree. No "i.e." and wrong. The list was a list of *infinite* sequences. > > > You just stated that there are no infinite strings. A agree. > > > > Where did I state that? > > Above you said "The tree contains only the set of finite binary > representations." Yup. There is *no* node 1/3. > If there are infinite strings elsewhere, then they > are in my tree too (in form of paths). But you had said that it were the nodes that represented the numbers in the tree, so the paths are irrelevant. There are infinite paths in your tree, but they do not contain a node that represents (for instance) 1/3. So, if the nodes represent numbers (as you have said), 1/3 is not in your tree. You are not clear about what the numbers in your tree are. Are they the nodes? Are they the paths? Sometimes you say one thing other times you say something different. So to get proper understanding. What are the things that represent numbers? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |