Cantor Confusion
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From: mueckenh on 15 Nov 2006 04:44 William Hughes schrieb: > > The suprema are not the same. After havin added one term to each line, > > each line has the ordinal number n+1 < omega. And the width of the > > matrix has the ordinal number omega. > > So? Adding one element may or may not change the supremum. If the supremum is a maximum, it will be changed. If not, it will not be changed. > > The supremum of the lengths of the initial segments of > the columns is omega. The supremum of lengths of the lines is > omega. The supremums are the same. > > However, we can make the supremums different by adding elements. This shows that they were not quite the same before. > > If we add one element to each of the columns then > the supremum of the lengths of the initial segments of the columns > changes to omega +1. If we add one elelment to each of the lines > the supremum of the lengths of the lines does not change, it > remains omega. Correct. And if we add one element to the diagonal, then its supremum changes form omega to omega + 1. > > > > > > If two sets A and B > > > are not the same, adding one element to A may not > > > give the same result as adding one element to B. > > > > May be. But he diagonal bijects the columns to the lines. The element > > d_nn maps the n-th column to he n-th line. It should do so, at least. > > > > And it does. Are you really sure? > > There is no line with infinite index. > There is no column with infinite index > There is no element of the diagonal with infinite index > Now, after extending: There is an element of every column with transfinite index. There is an element of the diagonal with transfinite index. There is no element of a line with transfinite index. There is no bijection between lines and columns. Regards, WM
From: mueckenh on 15 Nov 2006 04:52 William Hughes schrieb: > mueck...(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > If we add one element to each of the columns then > > > the supremum of the lengths of the initial segments of the columns > > > changes to omega +1. If we add one elelment to each of the lines > > > the supremum of the lengths of the lines does not change, it > > > remains omega. > > > > Fine. This new matrix has a diagonal, if the old matrix had one. > > Let he original matrix be A. > 1 12 123 .... > Let B be the matrix we get by adding one element to each > of the columns of A. This matrix does not have a diagonal. Every matrix has a diagonal. The question is only whether all lines and columns are crossed by the diagonal. Not every matrix has a bijection between lines and columns. > > Let C be the matrix we get by adding one element to each of > the lines of A. This matrix has a diagonal. > > Let D be the matrix we get by adding one element to > each of the lines of B. This matrix does not have a diagonal. You want to say that A has a diagonal, but addition of one element to each line, each column and the diagonal eliminates this property? Of course you must say so because otherwise you had to confess that set theory is self contradictory. On the other hand you could argue with a split diagonal. Just a bit counter intuitive, but a hard-boiled set theorist will swallow that without feeling nausea. No, William, if a matrix has a diagonal crossing every line and every column, and if one element is added to every line and every column and to the diagonal itself, then this diagonal again crosses every line and every column. Briefly: If D(A_ij) then D(A_i+1,j+1) And if the property D(A_i+1,j+1) is false, then the property D(A_ij) is false. Just that is what I assert and what has only been rejected by you with the argument a diagonal could have more element than every line. Here you see that you are forced to postulate a square matrix changes into a non-square matrix when lines and rows are extended symmetrically. > > Only for matrixes B and D is there a last line. > > So if there is a diagonal, there is no last line. How about adding the same elements not at the ends but at the beginnings of the lines and the columns? In that case you claim the existence of a diagonal. No problem, nobody can check it (in particular because any infinity is a void claim). But if we find a way to circumvent this spiritual existence and to fasten it by adding probes at the end, then the diagonal mysteriously disappers? The diagonal is nothing but the means to biject lines to columns. If there is no diagonal crossing all lines and columns, this simply means that there is no bijection between lines and columns. And that is fact after increasing each one by one element as well as before. > > >It > > bijects the columns to the lines. The element d_nn maps the n-th column > > to the n-th line. It should do so, at least. > > > > > > And it does. > > > > Could you construct this bijection? It is easy to begin: > > > > Column 1 <--> Line 1 by the diagonal element d_11 > > Column 2 <--> Line 2 by the diagonal element d_22 > > Column 3 <--> Line 3 by the diagonal element d_33 > > ... > > Column ? <--> Line omega+1 by the diagonal element d_?? > > > > But t is difficult to end, although there is a last element in the set > > of lines > > No. If there is a diagonal there is no last line. A new theorem? Can it be applied generally, or only in this special case? If we may apply it generally, then let's consider Cantor's list. Cantor's list has a diagonal. If we add a last line, then the diagonal argument can no longer be applied? Then let's construct some Cantor list, take any line, for instance the first, and add it to the end of the list as a last line. a_21,a_22,a_23,... a_31,a_32,a_33,... a_41,a_42,a_43,... .... .... .... a_11,a_12,a_13,... And abracadabra, all uncountability proofs vanish. Regards, WM
From: Franziska Neugebauer on 15 Nov 2006 05:32 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: >> Let he original matrix be A. [...] > 1 > 12 > 123 > ... Something is missing here: 1uuu... 12uu... 123u... .... Usaually a matrix m(a,b) is a function of domain A x B (it is rectangular not a triangle) and some co-domain. Here A equals B equals omega by definition. A matrix with domain A x B is called (generalized) square iff A = B. "u" stands for undefined (empty), not set. The "triangle" you see is not the structure of the matrix but its occupancy. Prima facie "Adding x to each column" ("at the end") is not "possible", since the columns have no end (no last element). "Adding x to each column" can *meaningfully* only be regarded as "adding a new line" "behind all lines", so we obtain the new matrix 1uuu... 12uu... 123u... .... xxxx.... This operation changes the domain A into A + 1 (omgea + 1). The order type of domain A has changed. Since the domain has changed this new matrix has no longer the original structure. And since omega + 1 is not equal to omega, this matrix is no longer a square one. It is not possible to "add x to each column" and to retain the domain. "Adding x to each row" means simply changing the occupancy of the matrix without changing its domain: 1xuu... 12xu... 123x... .... There is no change in domain "necessary" to "perform" this operation. [...] >> Let C be the matrix we get by adding one element to each of >> the lines of A. This matrix has a diagonal. >> >> Let D be the matrix we get by adding one element to >> each of the lines [clolumns!] of B. This matrix does not have a >> diagonal. > > You want to say that A has a diagonal, but addition of one element to > each line, each column and the diagonal eliminates this property? Of > course you must say so because otherwise you had to confess that set > theory is self contradictory. On the other hand you could argue with a > split diagonal. Just a bit counter intuitive, but a hard-boiled set > theorist will swallow that without feeling nausea. > > No, William, if a matrix has a diagonal crossing every line and every > column, and if one element is added to every line and every column and > to the diagonal itself, then this diagonal again crosses every line > and every column. Briefly: As explained above: "Adding x to each row" is entirely different from "Adding x to each column". You are comparing apples and oranges. > If D(A_ij) then D(A_i+1,j+1) > > And if the property D(A_i+1,j+1) is false, then the property D(A_ij) > is false. > > Just that is what I assert and what has only been rejected by you with > the argument a diagonal could have more element than every line. Here > you see that you are forced to postulate a square matrix changes into > a non-square matrix when lines and rows are extended symmetrically. As explained above "adding x to each row" and "adding x to each column" are different operations. The first changes the occupancy the second the domain of the matrix. >> Only for matrixes B and D is there a last line. >> >> So if there is a diagonal, there is no last line. > > How about adding the same elements not at the ends but at the > beginnings of the lines and the columns? This simply means shifting the occupancy. Domains are retained. [...] F. N. -- xyz
From: William Hughes on 15 Nov 2006 06:13 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > The suprema are not the same. After havin added one term to each line, > > > each line has the ordinal number n+1 < omega. And the width of the > > > matrix has the ordinal number omega. > > > > So? Adding one element may or may not change the supremum. > > If the supremum is a maximum, it will be changed. If not, it will not > be changed. > > > > The supremum of the lengths of the initial segments of > > the columns is omega. The supremum of lengths of the lines is > > omega. The supremums are the same. > > > > However, we can make the supremums different by adding elements. > > This shows that they were not quite the same before. > > > > If we add one element to each of the columns then > > the supremum of the lengths of the initial segments of the columns > > changes to omega +1. If we add one elelment to each of the lines > > the supremum of the lengths of the lines does not change, it > > remains omega. > > Correct. And if we add one element to the diagonal, then its supremum > changes form omega to omega + 1 Correct, but to add one element to the diagonal we have to add one element to the lines and one element to the columns. If we just add one element to the lines we no longer have a diagonal. .. > > > > > > > > > > If two sets A and B > > > > are not the same, adding one element to A may not > > > > give the same result as adding one element to B. > > > > > > May be. But he diagonal bijects the columns to the lines. The element > > > d_nn maps the n-th column to he n-th line. It should do so, at least. > > > > > > > And it does. > > Are you really sure? > > > > There is no line with infinite index. > > There is no column with infinite index > > There is no element of the diagonal with infinite index > > > Now, after extending: > There is an element of every column with transfinite index. > There is an element of the diagonal with transfinite index. > There is no element of a line with transfinite index. > > There is no bijection between lines and columns. > No After extending There is an element of every column with transfinite index. There is no element of a line with transfinite index. There is no diagonal. - William Hughes
From: mueckenh on 15 Nov 2006 06:34
William Hughes schrieb: > No all the natural cardinal numbers have already been used up for > sets of the form {1,2,3,...,n}. Correct. > There are no natural numbers > left for the set {1,2,3,...}. Then, what does this set consist of, if all natural numbers have been used up by finite sets? > So it is not a question of maybe > not knowing what the natural cardinal number is. We know that > there is no natural cardinal number. Why then do you assert that this set exists? > Let N be the set of natural numbers. Then induction is valid > for all elements of the set N. "the set N" is not an element of N. > Induction > is not valid for "the set N". I do not want to prove anything for this chimera. All I say it that induction is valid for all natural numbers. Call this a collection or a sequence if the notation "set" leads you to astray. The "all natural numbers" (whatever you may call it) has not an actually infinite number of elements. > > Sets of the form {1,2,3,...,n} are not elements of N either. > However, we can associate > each set with its cardinal, a unique element of N, and then use > induction > on the elements of N to show something about the sets of this form. > However, > the set {1,2,3,...} does not have a cardinal which is an element > of N. So we cannot use induction to tell us anything about this set. You say: However we cannot use induction to tell us anything about this set so we cannot use induction to tell us anything about this set. Let the set be what it may. Talk about all the natural numbers. Regards, WM |