Cantor Confusion
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From: David Marcus on 10 Nov 2006 06:56 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > 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. > > Your assertion: "The number of lines is omega. This does not mean that > there is > a line omega" contains a self-contradiction. > > Consider the bijection between initial segments of columns and lines > like this typical example > > 1 > 2 > 3 > ... > n <--> 1,2,3,...n > > The lines are mapped on finite segments of the column only. There is no > line with a finite number of elements mapped on an initial segment with > omega elements. OK. > Therefore the number of lines is finite only, but not > infinite. That doesn't follow. How do you go from for n in N, let f(n) := {1,...,n} to there exists m in N such that for all n in N, n <= m? Please show the intermediate steps. Whatever your logic is, wouldn't the same logic let us go from for n in N, let g(n) := n to there exists m in N such that for all n in N, n <= m? > Therefore there are less than omega lines. > > As a consequence an actually existing number omega > n cannot exist. The phrases "actually existing" and "cannot exist" are not defined. -- David Marcus
From: David Marcus on 10 Nov 2006 07:08 mueckenh(a)rz.fh-augsburg.de wrote: > > MoeBlee schrieb: > > > > > > > > > Here is my translation: It is allowed to understand the new num= > ber > > > > > > > omega as limit to which the (natural) numbers n grow, if by tha= > t we > > > > > > > understand nothing else than: omega shall be the first whole nu= > mber > > > > > > > which follows upon all numbers n, i.e., which is to be called l= > arger > > > > > > > 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. > > > > > =A7 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 gef=FChrt wird und auf welche Weise sich die nat=FCrli= > chen > > > > > Abschnitte in der absolut-unendlichen realen ganzen Zahlenfolge, we= > lche > > > > > 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 gef=FChrt 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. Moe said that "omega is a limit ordinal". He did not say that "omega is a limit". The two statements are not the same. Do you really believe the two statements are the same? Or, are you trolling? > Further this (Cantor's) definition supports my definition of > definition. Regardless of whether Cantor meant his remark to be a definition, you can't change what the word "definition" means in modern mathematics. It is rude to use a common word with a personal meaning and not point this out. If someone asks you for a "definition", etiquette and honesty requires you to say, "I'm sorry, but I do not know what you mean." -- David Marcus
From: William Hughes on 10 Nov 2006 07:12 mueckenh(a)rz.fh-augsburg.de wrote: > 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? Each line has only a finite number of column indexes. There are an infinite number of lines. There are an inifinite number of column indexes. There is a difference between the set of column indexes in any one line and the set of line indexes. There is no difference between the set of column indexes and the set of line indexes. > Lines with finitely many indexes cannot exhaust the column with its > infinitely many An infinite number can. -William Hughes ..
From: David Marcus on 10 Nov 2006 07:19 mueckenh(a)rz.fh-augsburg.de wrote: > > MoeBlee schrieb: > > > > > This number does not exist. The diagonal has not omega positions. > > > > So I guess the answer to my question is that you are not prepared to > > show that my proof has anything more than first order logic applied to > > Z set theory nor are you prepared to demonstrate a contradiction in Z > > set theory. > > Your proof shows either a contradiction in Z or it shows that Z is > false. > > Z contains a contradiction to your proof, if one can show in Z that a > diagonal of a matrix has not more elements than every line. Z is false, > if one cannot show that. So, why further bother about Z? It appears you are mislead by your loose terminology. In a finite square matrix, the length of the diagonal equals the height and the width, e.g., 1 2 3 4 5 6 7 8 9 However, why should the diagonal of a triangle with infinite height be the same length as one of its lines? You have claimed this, but you have not explained why you believe this to be true. Looking at the following picture, it seems to be obviously false. 1 2 3 4 5 6 7 8 9 10 .... 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". What is wrong with my reasoning? -- David Marcus
From: William Hughes on 10 Nov 2006 07:24
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > 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. > > Your assertion: "The number of lines is omega. This does not mean that > there is > a line omega" contains a self-contradiction. > > Consider the bijection between initial segments of columns and lines > like this typical example > > 1 > 2 > 3 > ... > n <--> 1,2,3,...n > > The lines are mapped on finite segments of the column only. There is no > line with a finite number of elements mapped on an initial segment with > omega elements. Therefore the number of lines is finite only, but not > infinite. Therefore there are less than omega lines. No. What this means is that every line index is finite. It is possible to have an infinite set each of whose elements is finite. Therefore the fact that each of the line indexes is finite does not mean that there are less than omega lines. - William Hughes |