Cantor Confusion
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From: Virgil on 11 Nov 2006 20:28 In article <1163252760.153565.112090(a)h54g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daniel Grubb schrieb: > > > 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. So that the diagonal is different from every line, as required. > > > > 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. For "we" read "I", as there are lots of people who have quite concrete ideas of what any sort of infinity is. > > Would you believe that the diagonal of our infinite triangle can be > longer than the first column? Why? > > > 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. Not unless the construction under consideration provides lines of infinite length
From: Virgil on 11 Nov 2006 20:32 In article <1163252852.669083.310910(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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 Why is it not the height (or width, or area or volumn, or any other word indicating quantity) of the set rather than its "length"
From: Virgil on 11 Nov 2006 20:36 In article <1163253023.847019.319670(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. One often does not notice one's own absurdities, so it is quite possible that WM would not notice any such new ones more that he notices his many old ones. > > > > 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. Then f(n) = n, unless "related to" has some esoteric meaning as yet unexplained.
From: Virgil on 11 Nov 2006 20:44 In article <1163253136.983027.246700(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. The whole issue is that according to us the diagonal does NOT have to satisfy the condition of being a line in the listing. So that WM proves our point! > > Therefore your conviction, expressed in many postings (it is possible > to have an infinite set each of whose elements are finite) is wrong. Has WM just proved 2 + 2 = 1? He has certainly not proved anything else, at least not to any one else's satisfaction. > > But it is an ordinal number which can be increased by 1. And if > infinity in fact is finished, then it is a maximum. Of what? it is certainly a LUB or supremum in some senses, but how can a anything be the maximum of things of which it is not a member? What WM is claiming is equivalent to saying that 10 is the /maximum/ of one digit decimal integers. What corrupt definition of "maximum" is WM using?
From: William Hughes on 11 Nov 2006 20:47
David Marcus wrote: > William Hughes wrote: > > mueckenh(a)rz.fh-augsburg.de wrote: > > > David Marcus schrieb: > > > > William Hughes wrote: > > > > > > > > > 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. > > > > > > > > WM, please tell us if you agree or disagree with the statements a and b > > > > above. > > > > > > You will not understand it. But I'll try. > > > > Not a very clear attempt. > > It was clearer than most of WM's attempts. Now that's what I call damning with faint praise! > Be encouraging! Ahw! do I havta? Sigh! Ok. - William Hughes |