Cantor Confusion
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From: mueckenh on 10 Nov 2006 05:13 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. It would not allow to state omega > n for every n. Regards, WM
From: mueckenh on 10 Nov 2006 05:18 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. Regards, WM
From: mueckenh on 10 Nov 2006 05:21 William Hughes schrieb: > > 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. What makes up the difference to the fact that there are not aleph_0 terms in any line? Or is there no difference between aleph_0 terms and less than aleph_0 terms? > > No single line has infinitely many indexes. However, given any > index there is always a line containing that index. Uninteresting when considering the difference between aleph_0 indexes and less than aleph_0 indexes. > Remember, > the fact that there are infinitely many lines does not mean > that ther is a line with index aleph_0. The column has infinitely many indexes. No line has infinitely many indexes. No difference? Lines with finitely many indexes cannot exhaust the column with its infinitely many. We learn: Infinity cannot be exausted. Regards, WM
From: mueckenh on 10 Nov 2006 05:29 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. > 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. Regards, WM
From: mueckenh on 10 Nov 2006 05:35
MoeBlee schrieb: > > > > > > Here is my translation: It is allowed to understand the new number > > > > > > omega as limit to which the (natural) numbers n grow, if by that we > > > > > > understand nothing else than: omega shall be the first whole number > > > > > > which follows upon all numbers n, i.e., which is to be called larger > > > > > > than each of the numbers n. > > > > > > > > > > Okay. I don't see any problem with that. Would you please refresh the > > > > > context by saying what point it is that you draw from that quote? > > > > > > > > Cantor's first book (general paper): Grundlagen einer allgemeinen > > > > Mannigfaltigkeitslehre (Leipzig 1883). Foundations of a general set > > > > theory. > > > > § 11, showing how one is lead to the *definition* of the new > > > > numbers... (Es soll nun gezeigt werden, wie man zu den Definitionen der > > > > neuen Zahlen geführt wird und auf welche Weise sich die natürlichen > > > > Abschnitte in der absolut-unendlichen realen ganzen Zahlenfolge, welche > > > > ich Zahlenklassen nenne, ergeben.) > > > > > > I asked you what point you draw from the quote that you translated. > > > > Es soll nun gezeigt werden, wie man zu den Definitionen der neuen > > Zahlen geführt wird > > It is now to be shown how one is lead to the definition of the new > > numbers. And this definition, given by Cantor and translated by myself > > is given above. There is no point to draw but only to understand > > Cantor's *definition* (or not). > > Okay, so for you there is no point to draw about your quote other than > understanding Cantor's own writings. And, I'll add that in this > particular instance, Cantor, as you translated, is not in conflict with > the theorem of current set theory that omega is a limit ordinal and the > first oridinal that is greater than all natural numbers. That is what I said. omega is a limit. In modern set theory there are limits. Further this (Cantor's) definition supports my definition of definition. Regards, WM |