Cantor Confusion
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From: William Hughes on 16 Nov 2006 07:55 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > You say so. But if so, then there must be a bijection such that a > > > column is mapped on a line for every column and every line. All finite > > > values are bijected without problem n <--> n. But there remains a taken > > > supremum to be bijected with a not taken supremum. That is impossible. > > > > No. The taken supremum has nothing to do with the bijection. > > > Make it easier for you to understand it by considering the initial > segments of the first column. The initial segment > > 1 > 2 > 3 > ... > n > > stand in bijection with the line 1,2,3,...,n by the diagonal element > d_nn. > > If we add 1 element to every initial segment, then we obtain from the > complete column (which has not an omegath element but allegedly does > exists having omega elements) a segment of order type omega + 1. And it > there was a bijection in the original version, then there will be a > bijection in the extended version. Yes, from the complete column you do obtain a seqment of order type omega+1. This is a segment you did not have before. However, there is no complete line. So by adding an element to a line you do not add a segment that you did not have before. Let us restate this in terms of indexes (and index is just the last term of an initial segment that has a last term). Each column does not have an end. The only index at which we can add an element is a transfinite index . Each line does have an end. Therefore we can add an element at a natural number index. Let W be the set of line indexes before we add elements. Let X be the set of column indexes before we add elements. Let W' be the set of line indexes after we add elements. Let X' be the set of column indexes after we add elements. Let N be the set of all natural numbers. Then W=N and X=N. There is an obvious bijection between W and X. Now add elements. For each column the element gets added at the transfinite index omega. Thus X' is {X,omega} and is not the same as X. For each line the element gets added at a natural number index. Thus W' does not change. W'=W. Thus we cannot extend the bijection from W->X to a bijection from W'->X' - William Hughes
From: mueckenh on 16 Nov 2006 09:09 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > You say so. But if so, then there must be a bijection such that a > > > > column is mapped on a line for every column and every line. All finite > > > > values are bijected without problem n <--> n. But there remains a taken > > > > supremum to be bijected with a not taken supremum. That is impossible. > > > > > > No. The taken supremum has nothing to do with the bijection. > > > > > > Make it easier for you to understand it by considering the initial > > segments of the first column. The initial segment > > > > 1 > > 2 > > 3 > > ... > > n > > > > stand in bijection with the line 1,2,3,...,n by the diagonal element > > d_nn. > > > > If we add 1 element to every initial segment, then we obtain from the > > complete column (which has not an omegath element but allegedly does > > exists having omega elements) a segment of order type omega + 1. And it > > there was a bijection in the original version, then there will be a > > bijection in the extended version. > > > Yes, from the complete column you do obtain a seqment > of order type omega+1. This is a segment you did not > have before. However, there is no complete > line. So by adding an element to a line you do not add a segment > that you did not have before. Therefore there is no bijection between lines and initial segments of columns. There is no partner for the complete initial segment of the column. > > Let us restate this in terms of indexes > (and index is just the last term of an initial segment that > has a last term). And every such segment can only be bijected to another element that also has a last term in this matrix. And every such element remains in possesion of a last term after having added another last term. > > Each column does not have an end. The only index at which we can add > an element is a transfinite index . Each > line does have an end. Therefore we can add an element > at a natural number index. > > Let W be the set of line indexes before we add elements. > Let X be the set of column indexes before we add elements. > > Let W' be the set of line indexes after we add elements. > Let X' be the set of column indexes after we add elements. > > Let N be the set of all natural numbers. > > Then W=N and X=N. There is an obvious bijection between > W and X. No. That is obviously no so. You only think so. Because: If you add one element to N = N, then the result is the same. > > Now add elements. > > For each column the element gets added at the transfinite > index omega. Thus X' is {X,omega} and is not the same as X. > > For each line the element gets added at a natural number > index. Thus W' does not change. W'=W. You see, you have a problem to explain that W = N and X = N. It is in fact very easy to see. > > Thus we cannot extend the bijection from W->X to a bijection > from W'->X' Because there was nothing which could be extended. Regards, WM
From: Franziska Neugebauer on 16 Nov 2006 09:36 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > William Hughes schrieb: [...] >> Yes, from the complete column you do obtain a seqment >> of order type omega+1. This is a segment you did not >> have before. However, there is no complete >> line. So by adding an element to a line you do not add a segment >> that you did not have before. > > Therefore there is no bijection between lines and initial segments of > columns. There is no partner for the complete initial segment of the > column. http://mathworld.wolfram.com/InitialSegment.html What you call "complete initial segment" is not an *initial* segment but the whole set of lines. F. N. -- xyz
From: William Hughes on 16 Nov 2006 11:45 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > You say so. But if so, then there must be a bijection such that a > > > > > column is mapped on a line for every column and every line. All finite > > > > > values are bijected without problem n <--> n. But there remains a taken > > > > > supremum to be bijected with a not taken supremum. That is impossible. > > > > > > > > No. The taken supremum has nothing to do with the bijection. > > > > > > > > > Make it easier for you to understand it by considering the initial > > > segments of the first column. The initial segment > > > > > > 1 > > > 2 > > > 3 > > > ... > > > n > > > > > > stand in bijection with the line 1,2,3,...,n by the diagonal element > > > d_nn. > > > > > > If we add 1 element to every initial segment, then we obtain from the > > > complete column (which has not an omegath element but allegedly does > > > exists having omega elements) a segment of order type omega + 1. And it > > > there was a bijection in the original version, then there will be a > > > bijection in the extended version. > > > > > > Yes, from the complete column you do obtain a seqment > > of order type omega+1. This is a segment you did not > > have before. However, there is no complete > > line. So by adding an element to a line you do not add a segment > > that you did not have before. > > Therefore there is no bijection between lines and initial segments of > columns. There is no partner for the complete initial segment of the > column. The bijection is between the line indexes and the column indexes. The bijection is not between the initial segments of the first column and the lines. There is a bijection between the column indexes and the lines. There is no bijection between the line indexes and the initial segments of the first column. There is almost a bijection, but not quite. Each line index is associated with an initial segment of the form {1,2,3,...,n}, that is an initial segment that ends. Each initial segment that ends is associated to a line index. However, there is one initial segment, {1,2,3,...} that does not end. This initial segment is not associated with any line index. So the fact that there is no bijection between the initial segments of the first column and the lines, does not mean that there is no bijection between the line indexes and the column indexes. - William Hughes
From: Lester Zick on 16 Nov 2006 12:44
On Thu, 16 Nov 2006 01:35:12 -0500, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >Virgil wrote: >> In article <1163426009.651510.237050(a)h48g2000cwc.googlegroups.com>, >> mueckenh(a)rz.fh-augsburg.de wrote: >> > It is difficult to answer this question, because the expression "set" >> > is occupied in modern mathematics by collections of elements which are >> > actually there (you don't know what that means, imagine just a set as >> > you know it). Such infinite sets do not exist. >> >> While infinite collections in any physical sense are not possible, why >> are imaginary infinities, such as sets of numbers must be, unimaginable? Why are square circles unimaginable? >For that matter, we can always switch from Platonism to formalism and >declare the question of whether sets really exist to be a philosophical >question. So is the switch from platonism to formalism a philosophical question? ~v~~ |