Cantor Confusion
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From: William Hughes on 8 Nov 2006 05:25 mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > Oh no? The lengths of the columns of the list > > > > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > > > > are not omega? I was under the impression that there are aleph_0 lines > > > with less than aleph_0 columns. > > > > Be careful with your quantifiers. The fact that every line has less > > than aleph_0 columns does not mean there are less than aleph_0 > > columns, it just means that no line has all columns. > > It just means that not all columns are there. There is no actual > infinity. > Interesting is that all columns have aleph_0 lines. So all lines are > there. > > You should at least be able to see the difference between the actuality > of the aleph_0 lines which are present, which are realized, and the > aleph_0 columns which are not realized and therefore, are not present. > > > > There are aleph_0 columns. > > Each line has less than aleph_0 columns. > > No line has all columns > > Every column is in some line. > > This is potential infinity. No line has aleph_0 columns. > There is another kind of infinity: Every column has aleph_0 lines? > Got it? No. There are aleph_0 lines and aleph_0 columns. Both infinities are exactly the same. The fact that no line contains all the columns does not change the number of columns. The fact that no line contains all the columns and the first column contains all the lines does not change either the number of columns or the number of lines. > > > > > > > > You see not difference between the infinite man 1's of 0.111... and the > > > finite many 1's in every line? > > > > I see a difference. However, I do not see a contradiction. 111... > > is not one of the lines and has a property that none of the lines > > share. > > 111... can be mapped on a line, say the first line > Then the first line > has aleph_0 1's. As no line has so much 1's, now the first line is > longer than all other lines. > > > > > > > > > > > > > 0.1 > > > > > > > 0.11 > > > > > > > 0.111 > > > > > > > ... > > > > > > > > > > > > > > you remember? Without an infinite number in the list there is no > > > > > > > infinite diagonal defined. > > > > > > > > > > > > I state (you know that) that the diagonal: 0.111... > > > > > > > > > > Yes, again mixing up maximum and supremum. > > > > > > > > No, not mixing at all. Supremum only. > > > > > > Then 0.111... is not different from the finite sequences of 1's? > > > > Yes is it different. So what? The supremem of A may not be a member > > of A. The supremum of A can be different from every element of A. > > But why is it not different from the columns? It is different from the set of columns in any line. It is not different from the set of columns. The set of columns not the same as the columns in any line. > > > Again this is true but there is no contradiction. The number of > > columns > > is not the maximum of the column lengths but the supremum. > > This supremum is a thing which is not realized by the number of columns > but which is realized by the numer of lines (if the notion of finished > infinity is correct). What ma ybe the reason for this difference? > Construction. So what? The supremum of a set A may or may not be a member of A. In the case of the set of column lengths, C, the supremum is a member of C. In the case of the set of line lengths, L, the supremum is not a member of L. However, the supremum of L is equal to the supremum of C (two different sets can have the same supremum). > > Because, by construction, every column is infinite and no line is. > > Why do you insist this simple fact is a contradiction? > > Because in natural numbers we have n = n, e.i., the n-th number has > size n. If you transpose the matrix, then nothing can change. But in > our matrix, we have a change. This is not possible for natural numbers. > Yes we have a change. The sets C and L change. However, since the supremums of C and L are identical, the supremums do not change. > > > But you don't see a difference between a diagonal having aleph_0 1's > > > and the lines which it is constructed from which have less 1's? > > > > No. Because the diagonal is the supremum, and the supremum of a finite > > set does not have to be finite. > > Maybe. But a diagonal cannot exist in a domain where no lines reach. Correct, however there is no such domain. To say that the diagonal has aleph_0 1's does not mean that the diagonal has an index aleph_0. It does not. The diagonal only has finite indexes. Given any finite index there is a line that reaches that index. - William Hughes
From: mueckenh on 8 Nov 2006 05:28 David Marcus schrieb: > > If he considers omega to be not the maximum but only the supremum of > > the set of lines, then we agree that actual infinty does not exist. > > Omega is the least ordinal greater than every natural number. In that > sense, omega is the sup of the set of line numbers. How does this imply > that the set of line numbers does not exist? If the maximum of the set of enumerated lines does exist as aleph_0 and if the maximum of the set of numbers in a line does not exist, then the bijection from numbers enumerating the lines and numbers within the lines (which exists in case of finite numbers) cannot exist in case of the limit. > > Does "actual infinity does not exist" mean the same as when you say you > disagree that "there exists the set of lines"? It means the same as to say "finished infinity is nonsense" for he set of lines as well as for the set of numbers in a line. Regards, WM
From: mueckenh on 8 Nov 2006 05:32 David Marcus schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > > > For the diagonal (d_kk) of the list > > > > > > > > 0 > > > > 1 2 > > > > 3 4 5 > > > > 6 7 8 9 > > > > ... > > > > > > > > we have the following mappings: > > > > d_kk --> d_mk and d_kk --> d_km with k, m eps N. > > > > > > Normally, the word "mapping" means function. If you are trying to define > > > functions, it isn't clear to me what you mean since m appears on the > > > right, but not the left. > > > > > > Is d_mk the kth element in the mth row? > > > > Yes. That is the usual notation. > > > > > > It is curious that the set of terms of (d_km) has omega as a maximum > > > > for every fixed m eps N while the set of terms of (d_mk) has omega not > > > > as a maximum for every fixed m eps N. > > > > > > For m in N, you appear to be considering two sets: > > > > > > A = {d_km | k in N}, > > > B = {d_mk | k in N}. > > > > > > If we take m = 0, then > > > > > > A = {d_00, d_10, d_20, ...} = {0,1,3,6,...}, > > > > This set has omega elements. > > > > > B = {d_00, d_01, d_12, ...} = {0, undefined, undefined, ...}. > > > > B = {0} > > > > Take m = 2, for instance: > > > > Then B' = {3,4,5} > > > > > Regardless, no set of natural numbers has omega as a maximum. I've no > > > idea why you think your list shows that omega is the maximum of a set of > > > natural numbers. > > > > What do you think is the difference between the first column A and any > > line B of the above matrix, concerning the number of elements? Is there > > a difference? > > If we let B_j be the set of numbers in line j, and we start labeling the > lines at zero, then > > |A| = aleph_0, > |B_j| = j + 1, for j in N. > > So, for any natural number j, |B_j| < |A|. Now what? No try to set up a bijection beween the lines and columns which is necessary to prove that a matrix is a square matrix. Regards, WM
From: mueckenh on 8 Nov 2006 05:34 David Marcus schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > > David Marcus schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > Virgil schrieb: > > > > > > > > > In article <1162470874.593282.36250(a)b28g2000cwb.googlegroups.com>, > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > WM merely repeated his automatic error several more times here. > > > > > > > > > > WM claims that a list in which the nth listed element is a string of > > > > > length at least n characters cannot produce a diagonal of length > > > > > greater that any finite number of characters. > > > > > > > > > > > > > The diagonal needs an element from every line, the n-th element from > > > > the n-th line. Therefore it cannot be longer than every line. > > > > > > By "it cannot be longer than every line", do you mean its length can't > > > be greater than the sup of the lengths of the lines? > > > > Its length can't be the length of a column, i.e. omega, if the width of > > the matrix has only the supremum omega. > > All I asked was what "[the diagonal] cannot be longer than every line" > meant. Was what you wrote supposed to be an answer to this question? Please follow the discussion with those who understand. Perhaps you will understand later on too. Regards, WM
From: mueckenh on 8 Nov 2006 05:38
Randy Poe schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Virgil schrieb: > > > > > > > > > > If it cannot be a fraction because ZF does > > > > > > not yet know how to divide elements, > > > > > > > > > > In ZF we define various operations of division. As far as I know, there > > > > > is not a dvision operation for sets in general. > > > > > > > > Why then do you not understand how an edge of the binary tree can be > > > > divided? > > > > > > But once divided, the parts are no longer edges at all, so are useless. > > > > Once divided the parts of a cake are useless? > > An edge of a graph is not a cake. It is a connection between two > nodes. How do you divide a connection? What does half a connection > mean? > > Here is a simple graph: A--B. > > In this graph, A is adjacent to B. > > If I divide the connecting edge between A and B into thirds, > what is my resulting graph? What does that division mean? > Is A a neighbor of B or not? In my tree argument, the edges are not physically divided. The following edge 1 / 0 remains the connection between this 1 and this 0. It is only the "ownership" which is shared equally by all the paths going through this edge. Regards, WM |