Cantor Confusion
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From: Franziska Neugebauer on 11 Nov 2006 07:30 William Hughes wrote: > Franziska Neugebauer wrote: [...] > I am quite willing to go to a much more formal set > of terms and definitions, but I try to use the language > preferred by whomoever I am discussing something with > so if WM insists on using informal language that is what > I will use.. The nice thing about informal language is that it allows for talking at cross purposes. > Yes, my switch from "length of a line or a diagonal" > to "length of the natural numbers" was an attempt to > move back to slightly more formal language. Whether > this will work I do not know. > > Note. It has become clear that at the base > we have two statements. > > a: Any infinite set of numbers must contain > an infinite number > > b: It is possible to have an infinite set of numbers > that does not contain an infinite number. > > that cannot both be true. In WM's eyes there are no infinite sets "in reality" at all. For the sake of continuing the discussion he will not present that "argument" too early. Otherwise his mission ("prevent student from studying set theory") would be in danger ... > WM claims that a is true and that furthermore b leads to a > contradiction. However, all of the > contradictions he has been showing depend on assuming that > a is true. I would like to discuss statements a and b directly, > however, I do not think that WM will cooperate. True. F. N. -- xyz
From: mueckenh on 11 Nov 2006 08:46 Daniel Grubb schrieb: > >> >The diagonal of actually infinite length omega cannot exist without the > >> >existence of a line of actually infinite length omega. > >> > >> Why not? Do you have a proof of this claim? > > >The diagonal of a matrix is defined as consisting of elements of this > >matrix. For a diagonal longer than every line (or every column) this > >is impossible. > > Let me restate the situation: > > We have the 'matrix' > > 1 > 2 3 > 4 5 6 > 7 8 9 10 > etc > > and you argue that the length of the diagonal cannot be > infinite. > > Of course, the first problem is that this is not a square > matrix. If you want to make it into a square matrix, you can > make it into this matrix: > > 1 0 0 0 0 0 .... > 2 3 0 0 0 0 ... > 4 5 6 0 0 0 ... > .... > .... > > In which case, it is a square matrix and every line > has infinite length. So the diagonal, which has infinite > length, has the same length as every row and every column. There is no line without trailing zeros. The diagonal has no zeros. > > An alternative is to look at the incomplete, triangular array > that we started with and notice that in triagular arrays the > length of the diagonal is always at least as large as the > length of each row. In the situation where there are finitely > many rows, the length of the diagonal is the same as the number > of rows. > > The sticking point is whether the length of the diagonal is > the same as the length of some row even when there are infinitely > many rows. You seem to be claiming the answer is yes. As we have not the least idea of what "actual infinity" could be, we can do nothing but extrapolate from the finite domain. Would you believe that the diagonal of our infinite triangle can be longer than the first column? > While > it is clear that the length of the diagonal is at least the length of > any row, it seems clear that the length of the diagonal is actually > longer than the length of each row in this case since it is at > least as long as the next row down. Since there is always a > 'next row down', the length of the diagonal is longer than > the length of every row. For every line there is always a next line longer than the first. Nevertheless you do not believe that there are lines with infinite length. Regards, WM
From: mueckenh on 11 Nov 2006 08:47 Franziska Neugebauer schrieb: > David Marcus wrote: > > > [...] you are working with an infinite triangle [...] > > ,----[ http://en.wikipedia.org/wiki/Triangle ] > | A triangle is one of the basic shapes of geometry: a polygon with > | three vertices [...] > `---- > > Are there really three vertices in WM's "triangle"? If finished infinities like omega do exist, yes. Then you can add 1 term to the first (and any other) column and one term to the diagonal obtaining tree vertices: 1 22 333 .... omega...omega Addition of one term to every line, however, will not yield anything. > Since when do sets have a length? It can be defined: length of a set of natural numbers = cardinal number of that set. Regards, WM
From: mueckenh on 11 Nov 2006 08:50 David Marcus schrieb: > If you think that your "mathematical proof" is valid and you prove > something that is absurd, this only shows that your mathematical > reasoning leads to absurdities. Everyone who uses mathematical reasoning > consistent with ZFC cannot prove your absurdity. So, if you learn ZFC, > you will stop proving absurdities. Only then the absurdities would begin. But I would not notice. > > > We need no experts in trees. This tree does nothing else but represent > > the real numbers of an interval. Here it is in a form which can be > > understod by ever mathamatician. > > > > Consider a binary tree which has (no finite paths but only) infinite > > paths representing the real numbers between 0 and 1 as binary strings. > > The edges (like a, b, and c below) connect the nodes, i.e., the binary > > digits 0 or 1. > > > > > > > > 0. > > > > /a \ > > > > 0 1 > > > > /b \c / \ > > > > 0 1 0 1 > > > > .......................... > > > > > > > > The set of edges is countable, because we can enumerate them. Now we > > set up a relation between paths and edges. Relate edge a to all paths > > which begin with 0.0. Relate edge b to all paths which begin with 0.00 > > and relate edge c to all paths which begin with 0.01. Half of edge a is > > inherited by all paths which begin with 0.00, the other half of edge a > > is inherited by all paths which begin with 0.01. > > Half of the paths include edge a. One quarter of the paths include edge > b. > > Suppose we have a finite tree of height n. Then there are 2^n paths. > Consider a path (e1,e2,...,en). Edge e^j is contained in 2^-j of the > paths. We can define a real-value function on paths by g(e1,e2,...,en) = > sum_{j=1}^n 2^-j = 1 - 2^-n. Not sure what to do next... > > > Continuing in this > > manner in infinity, we see by the infinite recursion > > > > f(n+1) = 1 + f(n)/2 > > What is f(n)? f(n) is the number of edges related to the initial segment of one path which has passed through the first n edges. Regards, WM
From: mueckenh on 11 Nov 2006 08:52
William Hughes schrieb: > > A diagonal consists of certain line elements. Therefore a supremum is > > not sufficient. > > > > Why is the supremum "not sufficient" to define the length of the > diagonal. (Note, there is no line with an infinite index, so > your answer should not depend on a line with an infinite index)? Every column has a fixed number omega of terms. You can add 1 further term and you will get omega + 1 terms in every column. Every line has less than omega terms. If you add 1 term to every line, the number of terms in every line remains less than omega. What about the diagonal? It has to satisfy both conditions, which is impossible. Therefore your conviction, expressed in many postings (it is possible to have an infinite set each of whose elements are finite) is wrong. > > > > 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 > > The length of the set of natural numbers is the supremum of the > lengths of the initial segments. This supremum is omega. However, > this supremum is not a natural number. But it is an ordinal number which can be increased by 1. And if infinity in fact is finished, then it is a maximum. Regards, WM |