Cantor Confusion
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From: mueckenh on 9 Nov 2006 07:04 MoeBlee schrieb: > As I said, I don't care about (3). My mathematical notions include the fact that the diagonal of a matrix cannot have more elements than every line. If in your opinion this position is not valid in ZFC or in your logic, then we should stop here. > > The diagonal is assumed to exist such that each of its digits exists, > > actually. This is established by the mapping on a column. > > No it is not. > > > But it cannot > > be established by the mapping on any line. You should recognize that > > the following bijection between columns and lines shows a > > contradiction, because one element is missing: > > > 1 <--> 1 > > 1,2 <--> 2 > > 1,2,3 <--> 3 > > ... > > 1,2,3,...n <--> n > > ... > > 1,2,3,... <--> omega > > 1, 2, 3 ... is NOT a line. But it is the whole column which would correspond to the line omega (which does not exist). > > And it is NOT a contradiction that 1, 2, 3, .. omega is equinumerous > with omega+1. Of course, the set 1,2,3... is equinumerous to the set 2,3,4,..., 1. Nevertheless the diagonal of a matrix cannot have more elements than every line. If in your opinion this position is not valid in ZFC or in your logic, then we should stop here. > > > I answered that already. And the cardinality of the diagonal is not > > > assumed, but is proven, to be omega. > > > > It is *assumed* by stating the axiom of infinity. Without this > > assumption the length was not omega. > > It is PROVEN from the axiom of infinity. But the existence of this axiom is assumed. > since I would not have bothered to even post > a proof about denumerable sequences and talk with you about it for so > long if I accepted any condition that I can't use the axiom of > infinity. That is *not my condition* but it is the result which follows from the fact that the diagonal cannot be longer than every line. If I have the choice either to accept the axiom of infinity with the condition that a diagonal can be longer than every line, or to drop both notions, then I choose the second. If you can live with the contrary, then try to do it. I wish you nice dreams. Regards, WM
From: mueckenh on 9 Nov 2006 07:13 MoeBlee schrieb: > About a couple of weeks ago you presented an argument about trees, and > as you presented your argument, as you described it, it was clearly > reasonable to regard you as intending that as a proof in set theory (it > would have been UN reasonable to think the contrary). But later you > switched, so that your argument was not to be taken as in set theory. Why don't you try to find out how it could be presented in set theory? Or why that cannot be done? > > Now, when I say that I am giving a proof in set theory, and after I > discuss that with you while having reminded you of that so many times, > you switch not your OWN argument this time, but you switch to make the > terms of MY argument subject to criticism for USING AN AXIOM OF SET > THEORY! You misunderstood. See below. > > I'm really very curious what satisfaction you get from such games you > play. > I hope that there are some lurkers who will learn what are the consequences of accepting ZFC. > There can't be any intellectual satisfaction in such mindless games. So > what is it get from them? I do not prohibit to use the axiom of infinity, but I show that the consequences of its use are absurd. It leads to such results as a diagonal which is longer than any line of a matrix or a vase which contains zero numbers at noon or Tristram Shandy getting ready with his diary. In the binary tree it leads the result that there are not less edges than paths. In my opinion that is enough to prove ZFC being wrong. If you prefer to remain within this wrong system, good luck. Regards, WM
From: William Hughes on 9 Nov 2006 07:55 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > It is necessary to distinguish between the lengths of the > > lines and columns and the indexes of the lines and columns. > > > > Let the set of line indexes be LI. > > Then LI is just |N. For every natural number n we have a line > > n. There is no last line so there is no line infinity. > > > > Let the set of column indexes be CI. > > Then CI is just |N. For every natural number > > n there is a column n. > > > > Let the set of column lengths be CL > > Then CL has only one element, aleph_0. > > The supremum of CL (also the maximum) is > > aleph_0. > > There are two sets the bijection of which has not yet been considered > but sould be: > A = The set of lengths of initial segments of the (first) column. > B = The set of lines. > Here we can set up a bijection, expressing the lines by the natural > indexes of the elements: > > 1 1 > 2 1,2 > 3 1,2,3 > ... ... > n 1,2,3,...,n > ... ... > omega 1,2,3,... > > which shows that omega or aleph_0 does not exist. No. This shows that there is no line omega. The number of lines is omega. This does not mean that there is a line omega. - William Hughes
From: William Hughes on 9 Nov 2006 08:01 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > The set of lengths of columns, C, consists of the single element > > > > omega. > > > > > > This is a maximum taken by every length of a column. > > > > > > > The set of widths of lines, W, is the set of all natural > > > > numbers. > > > > W has more elements than C, but every element of W is > > > > smaller than every element of C. > > > > > > Correct. > > > > > > > > The supremum of W is omega. The width of the > > > > matrix is the supremum of the set of widths of the lines, i.e. > > > > omega. Thus the width of the matrix is equal to the height of the > > > > matrix. > > > > > > But this supremum is not taken. > > > > > > The diagonal connects width and length by a bijection. The element d_nn > > > of the diagonal maps the n-th line on the first n elements of the first > > > column. As long as n is a natural number, this is no problem. Only for > > > aleph_0 the diagonal has to map a not existing maximum on an existing > > > maximum. This is a hard task. > > > > No. You are confusing the length of the columns with the set of line > > indexes contained in each column. These are not the same. > > Explain how in a square matrix (squarity proven by the existence of a > diagonal) the columns can be longer than the lines for *every* line > and every column. To be square we must have: for every line k, there exists a colmun k. This is true. Note that there is no line aleph_0 and no column aleph_0. > > > > The set of line indexes contained in the first column is |N. > > The length of the first column is the size of |N, aleph_0. > > > > The set of column indexes is |N. > > > > The bijection defined by the diagonal connects the line indexes > > with the column indexes. Finding a bijection between |N and |N > > is not hard. > > Of course not, because you can do that for finite indexes only. But > finding that there is a column with aleph_0 indexes while there is no > ine with aleph_0 indexes shows that the idea of the existence of > aleph_0 is false. Since there is neither a column nor a line with index aleph_0, there is no problem.. > > > > In other words, the fact that there are aleph_0 lines does not mean > > that there is a line with index aleph_0. Such a line would have > > to be the last line, however, there is no last line. > > If there were aleph_0 lines with aleph_0 a (non-natural) number, then > we would need a last line. No. This is simply false and the root of your problem. - William Hughes
From: William Hughes on 9 Nov 2006 08:13
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > 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 > > There are aleph_0 lines and aleph_0 columns. Both infinities > > are exactly the same. > > Why then is there a column with aleph_0 terms but no line with aleph_0 > terms? Is that really *exactly* the same? Well, probably in set theory > where one does not look too sharp. Because the set of all columns is different from the set of columns in any line. The first set has aleph_0 terms and is the same as the set of terms in the first column. Remember, the first column does not have an aleph_0 th term. > > > > 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. > > That is an assertion. But it is wrong. > > > > > > But why is it not different from the columns? > > > > It is different from the set of columns in any line. > > Of course it is different from the set of columns in any line. Because > this set is the set of terms of a line. > > > > It is not different from the set of columns. > > That is in question. For this purpose we should consider the terms in a > column and the tems in a line. There is always a difference. But > according to the symmetry of a square matrix, there should be no > difference. > No, there needs to be a bijection. There is. > > > 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 may be the reason for this difference? > > > > > > > Construction. So what? The supremum of a set A may or may > > not be a member of A. > > If it is not a member, then A' has one member less than the otherwise > same set A where it is a member. Therefore A and A' are different, > aren't they? > Completely irrelevent. The bijection is not between one set without supremum and this set plus supremem. The fact that there are aleph_0 terms in the first column, does not mean there is an aleph_0 th line. > > 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). > > Two sets which are distinct by one element, are two different sets. Completely irrelevent. The bijection is not between one set without supremum and this set plus supremem. The fact that there are aleph_0 terms in the first column, does not mean there is an aleph_0 th line. > > > > > > 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 after the change the lines have infinitely many indexes while the > columns have only finitely many. What about the iterpretation as > finitely many infinite numbers? > . > > > 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. > > But it has infinitey many indixes, while no line has infinitely may > indexes. So there is a lack of second indexes n in d_nn. No single line has infinitely many indexes. However, given any index there is always a line containing that index. Remember, the fact that there are infinitely many lines does not mean that ther is a line with index aleph_0. - William Hughes |