Cantor Confusion
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From: David Marcus on 10 Nov 2006 07:23 mueckenh(a)rz.fh-augsburg.de wrote: > Therefore we can compare the diagonal with every line. We find that the > diagonal is longer than every line. 1 1 1 1 1 1 1 1 1 1 .... The length of the diagonal is clearly the same as the length of the first column. And, the first column "goes on forever", while each line does not "go on forever". Therefore, for each line L, the length of the diagonal is longer than the length of line L. Hence, "the diagonal is longer than every line." What is wrong with my reasoning? > This is about the same as the vase > which is empty at noon or Tristram Shandy who completes his diary. > > No reason to bother about this. -- David Marcus
From: William Hughes on 10 Nov 2006 07:27 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > 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. > > To be square we must have both: For every line, there exists a column, > and there exist as many lines as columns. Consider Cantor's argument. > His list of reel numbers forms a matrix with omega lines and omega > columns. Both are maxima, because every reel number has omega digits. > If you drop the decimal points then there are no natural numbers but > infinite strings. This matrix has a diagonal with omega elements. The > list of natural numbers cannot have an infinite diagonal because of the > finiteness of every number. > > > > > > 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. > > False is the assertion that there can be omega lines actually existing > without an index omega actually existing. You may call the potential > infinity omega. This notion, however, would only mean that the process > of constructing natural numbers will never end. No process that creates an infinite set one element at a time can end. This is the definition of infinity. However, the fact that you cannot create a set one element at a time does not mean that the set does not exist. - William Hughes
From: William Hughes on 10 Nov 2006 07:28 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > > The length is not only a supremum but exists as a maximum (if all > > > natural numbers exist). > > > > > > > No. > > > > Consider the two sets. > > > > A={1,2,3,...} > > B = {1,2,3,...,omega} > > > > Both sets have omega as supremum. Set A does not contain its supremum, > > set B > > does. In the first case the supremem is not a maximum, in the second > > case > > it is. So it is not true that the supremum must also be a maximum. > > That is correct for two independent sets. But it is wrong for two sets > which are connected by a bijection d_nn like the diagonal of a matrix. > Every element d_nn maps a line with n elements to an initial segment of > the column with n elements. > 1 1 > 2 12 > 3 123 > ... > n 123...n > ... > > Continuing we collect less than omega lines. No. Continuing we find that every line index is finite. Since it is possible to have an infinite set each of whose elements are finite this is not a contradiction. - William Hughes
From: David Marcus on 10 Nov 2006 07:32 mueckenh(a)rz.fh-augsburg.de wrote: > Daniel Grubb schrieb: > > > >> Since the length of the diagonal equals the length of the first column, > > >> you are also saying that there is some line whose length is greater than > > >> or equal to the length of the first column. Is that correct? > > > > >The diagonal of actually infinite length omega cannot exist without the > > >existence of a line of actually infinite length omega. > > > > Why not? Do you have a proof of this claim? > > The diagonal of a matrix is defined as consisting of elements of this > matrix. For a diagonal longer than every line (or every column) this > is impossible. It is impossible for a finite square matrix. However, you are working with an infinite triangle. What is your reason for saying that the diagonal of an infinite triangle cannot be longer than each line? Consider this infinite triangle: 1 1 1 1 1 1 1 1 1 1 .... Each line contains a finite number of 1's, but the first column and the diagonal contain an infinite number of 1's. This is clear from the picture. Why should something that is clear from the picture be "impossible"? -- David Marcus
From: William Hughes on 10 Nov 2006 07:36
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > > > Every part of the discussion is crystal clear except for WM's > > > > maunderings which seem more and more to be deliberately obfuscating. > > > > > > > > If every line is finite, say, for example, the nth line is of length > > > > 2*n, and the diagonal is not finite , then the diagonal MUST be longer > > > > that every line. > > > > > > By definition the diagonal of a matrix cannot be longer than every line > > > of the matrix. > > > > No. The length of the diagonal is the supremum of the lengths of > > the lines. > > A diagonal consists of certain line elements. Therefore a supremum is > not sufficient. > Why is the supremum "not sufficient" to define the length of the diagonal. (Note, there is no line with an infinite index, so your answer should not depend on a line with an infinite index)? > > The length of the diagonal will be longer than every > > line if and only if there is no line with maximum length. > > Therefore also the diagonal cannot have maximum lengh omega > n. > > > As there is no last line, there is no line with maximum length. > > Exactly. And therefore there is no "number" omega The length of the set of natural numbers is the supremum of the lengths of the initial segments. This supremum is omega. However, this supremum is not a natural number. - William Hughes |