Cantor Confusion
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From: William Hughes on 15 Nov 2006 12:19 mueckenh(a)rz.fh-augsburg.de wrote: > 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? We are not talking about the elements of the set but the cardinal number of the set (the number of elements in the set). The cardinal number of the set {1,2,3,...} cannot be a natural cardinal number because all the natural cardinal numbers have been used up on other 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? I use the axiom of infinity. If you do not use this axiom, then you do not know whether this set exists. Fine, knock yourself out. However, you want to do more, you want to show that using the axiom of infinity leads to absurd results. To do this you have to use the axiom of infinity. > > > 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. If you use the axiom of infinity then the "all natural numbers" has an infinite number of elements. - William Hughes
From: Lester Zick on 15 Nov 2006 13:28 On Tue, 14 Nov 2006 17:51:18 -0700, Virgil <virgil(a)comcast.net> wrote: >In article <1163504837.420660.251800(a)h54g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >> Lester Zick schrieb: >> are. >> > > >> > >Try to construct as many numbers as you can using only 100 bits. Then >> > >increase to 10^10 bits, then increase to 10^100 bits. More is not >> > >available. I find this very convincing. >> > >> > I don't. It's a problematic argument at best. Based once again on a >> > hypothetical finitude of the "physical" universe whatever that means. >> >> Here I cannot understand you. The accessible universe is finite, >> allowing for not more than 10^100 bits (a closer estimation would be >> 10^205, but that is irrelevant). Now, to express a number requires at >> least one bit. What more is needed to see? > >WM is talking about accessibility of numbers, not their existence, which >is quite a different issue. Not so sure, Virgil. Arithmetic infinities like the square root of 2 are certainly accessible through rac construction techniques. Apart from that I don't know anyone who ever claimed that fractional expansions for irrationals were accessible to begin with. So if that's all WM is claiming it would seem to be a non starter to begin with. ~v~~
From: Lester Zick on 15 Nov 2006 13:52 On 15 Nov 2006 04:05:48 -0800, mueckenh(a)rz.fh-augsburg.de wrote: > >Lester Zick schrieb: > > >> >> I don't. It's a problematic argument at best. Based once again on a >> >> hypothetical finitude of the "physical" universe whatever that means. >> > >> >Here I cannot understand you. The accessible universe is finite, >> >> This is not true or at best problematic. > >It is enforced by the finity of the maximum signal velocity. It may well be that the accessible universe is finite. So what? That doesn't necessarily mean that irrationals themselves aren't accessible and cannot exist because they're inaccessible unless you're just talking about fractional expansions for them. Certainly arithmetic infinities such as the square root of 2 are very accessible through rac construction techniques. Also pi is accessible through circular arcs. So the problem seems like a non starter unless you're talking about arithmetic fractional approximations for irrationals alone. >> >allowing for not more than 10^100 bits (a closer estimation would be >> >10^205, but that is irrelevant). Now, to express a number requires at >> >least one bit. What more is needed to see? >> >> The issue seems to be whether reduction to a finite number of bits or >> whatever is determinate of the number. Let's assume we have only room >> for 2 bits or 4 bits or whatever. Is that determinate of the numbering >> capacity of our thoughts and numbers represented in our thoughts? I >> don't think so. > >If there were only 4 bits, then our brain would be included. It would >have less capacity than that of an earth worm. Our brain has about >10^11 neurons. This huge number lets us overlook that it is limited. If you assume as you appear to that the brain has no alternative but to designate numbers neuronically then I would agree. I don't know why you would think that or why you would think the brain is a bit TvN machine. That's absurd. The brain is a differential machine and functions tautologically. So all this talk about bits is irrlevant. >> I see infinitesimal subdivision able to express itself >> to any degree necessary for the computation of relationships between >> infinitesimal ratios. In other words instead of extending out numbers >> infinitely all we're doing is subdividing unity. And this requires no >> further finite space than unity regardless of precision or extension. > >If you get to the subdivision number [pi*10^10^100], you cannot >determine his number, neither by any brain nor by all ressources of he >universe. Pi is certainly accessible through circular arcs and their diameters. Are you suggesting circles and their diameters don't exist? It looks to me like you're confusing the accessibility of calculations for fractional expansions with the accessibility of the number itself. If one ignores geometry and makes arithmetic the sole mathematical paradigm of course your conclusion follows. That's the price to be paid for exclusive reliance on infinite set analytical techniques as exhaustive criteria for the properties of numbers. ~v~~
From: mueckenh on 15 Nov 2006 15:31 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > 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. > > OK. We say that every matrix has a diagonal, but > only a matrix with the same number of lines and columns > has a diagonal which crosses every line and every column. > > The matrix A has the same number of lines (omega each > column has omega elements) and columns (omega the supremum > of {n | n in N}). 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. > > The matrix B does not have the same number of lines (omega+1 each > column has omega+1 elements) and columns (omega the supremum > of {n+1 | n in N}). of {n | n in N}). > The matrix D does not have the same number of lines (omega+1 each > column has omega+1 elements) and columns (omega the supremum > of {n+1 | n in N}). If there was a bijection between columns and lines in A, then there is a bijection of columns and lines in B. > > > 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? > > > Yes, because adding one element to each column changes the number > of lines, but adding one element to each line does not > change the number of columns. Then there must have been a difference in columns and lines in A already. > > > The original matrix has the same number of > lines and columns, however,the lines and columns are different. That is impossible, because we have only the ordered set N to construct columns and lines of the matrix. There is exactly one ordered set of infinitely many initial subsets: 1 2,1 3,2,1 .... They are the lines. The first column is the union of them. Adding one element w to each of them has the same effect. Impossible. Both sets are the natural numbers. Further we have the segments of the first column to be the same as the lines. This 1 2 3 .... n and n, ..., 3.2.1 > Doing the same thing > to the lines and columns does not have the same result. That's what I demonstrated. The reason is that there cannot be a bijection. > > > > > > > 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? > > Adding an element to the beginning of an infinite sequence does > nothing. So this is the same as adding an element to the > end of each line, but not adding an element at the end of each > column. Adding an element at the beginning of every column does the same as adding an element at the beginning of each line. But at the ends, there are differences? There should be a reasonable reason for this different behaviour. > > > > No. If you add one line, but you do not add one column you > do not have a bijection. If it was a bijection before, then it must be a bijection after. Before there were so many elements in a column that increase of 1 leads to omega +1. In case of a bijecion, as much elements have been in the union of the set of lines. According to you the original matrix with nothing added supplies a bijection between lines and columns. Although there is no line with order type omega, you postulate that the order type of the set of lines is omega. If this were correct, then adding one element to each line would increase the order type of the set of lines to omega + 1. As this is not the case, your assertion is wrong. > > > > > > > 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? > > 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. > > This you cannot do. There is no end of the list. You can > add it after the list, but then you have a transfinite sequence > not a list. Cantor's argument applies to lists. It applies also to two lists, because even 10 lists are not sufficient to contain all reals. What you call a transfinite sequence is nothing but two list, one of them consisting of 1 line only. And according to your new law these two lists cannot have a common diagonal. Therefore one cannot prove that not all reals can be written in two lists. But I see you have specialized your law. Regards, WM
From: mueckenh on 15 Nov 2006 15:33
William Hughes schrieb: > > Wrong. If there is a bijection f between A and B, and we > add one elment to A to form A' and no elements to B > to form B', we cannot extend f to a bijection between A' and B' We have added one element to all entities which constitute the matrix. This matrix does not contain anything further which could be extended. So, if the result is different here, then there was no bíjection possible. Regards, WM |