Cantor Confusion
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From: mueckenh on 23 Dec 2006 16:12 Dik T. Winter schrieb: > > The sum 1 + 1/2 + 1/4 + ... has no last term. Nevertheless we know, > > even in ZFC, that it is larger than 1 and less than 10. That has > > significance to our question. The series exactly maps the infinity of > > the paths. > > The sum as stated has *no* meaning in mathematics unless you are using > limits The bijection {1,2,3,...n} <--> {2,4,6,...2n} is a bijection for finite sets. In fact, it is false for infinite sets. The sum 1 + 1/2 + 1/4 + ..., however, is larger than 1, with any definition of limit. > > In the previous messages (and this one) you state something akin to: > lim{n->oo} ((# edges@n)/(# paths@n)) = > lim{n->oo} (# edges@n) / lim{n->oo} (# paths@n). > The first limit is indeed 2. But that does not mean that both limits > on the right hand side can be compared with that. The two limits do > not exist. So the theorem "the limit of a quotient is the quotient of > the limits" can not be applied because that has only been proven for > the finite case. The sum 1 + 1/2 + 1/4 + ... > 1 is sufficient for my proof. No division, no limits will save you from going down. Regards, WM
From: mueckenh on 23 Dec 2006 16:15 Newberry schrieb: > When I was thaught the diagonal argument in school my first reaction > was to construct a binary tree and start numbering the nodes from top > to bottom and from left to right. They laughed maliciously and said > "you left out all the infinite paths." I said "I exhausted the whole > tree." They said "no, you did not., there are finite subsets, co-finite > subsets, and infinite subsets." And I saw how dumb I was. No, not you were dumb but the laughers. > > Again, I do not know if this argument makes any sense. But it is a > verbalization of the fact that in ZFC the diagonal argument and > numbering of the tree nodes can co-exist without generating a formal > contradiction. It seems to me though that your "geometric" argument > (mapping the paths on the edges) shows the Cantorist verbalization to > be entirely implausible. It is implausible, because there, in fact, the limit is required for the diagonal element. And it is assumed erroneously. Also, for example {1,2,3,...n} <--> {2,4,6,...2n} is a bijection for finite sets only. There is no limit consideration by Cantor et al. at all. Regards, WM
From: mueckenh on 23 Dec 2006 16:17 Franziska Neugebauer schrieb: > > Recently we had the following claim (by Virgil) > > > > The number of equations > > > > 1+1+1+...+1 = n > > > > with finitely many 1 is infinite. Considering the fact that there are > > exactly as many 1 as different numbers, it is impossible that the > > number of numbers can be infinite while the sme number of 1 is finite. > > The number of sums (left to the "=") and the number of n's (right to the > "=") is not bounded. What precisely is your problem? The infinitely many sumands in 1 1+1 1+1+1 .... yield an infinitude of sums but not an infinitude of summands. How can finitely many summands in this linear order yield an infinitude of sums? Regards, WM
From: mueckenh on 23 Dec 2006 16:18 Franziska Neugebauer schrieb: > > In short: If any natural number n does exist, then there exists a line > > in which this and any smaller number does exist. > > This holds for every number n. > > Every natural n. > > > You know that the logical symbol for every is forall. > > I don't see how this could support your "famous" quantifier reversal. > Hence that does not prove the existence of a _single_ line (*). Show me a number which is not in a single line. Or show me two lines, each of which lacks a number the other one does not lack. Regards, WM
From: mueckenh on 23 Dec 2006 16:21
Gc schrieb: > mueckenh(a)rz.fh-augsburg.de kirjoitti: > > > Gc schrieb: > > > > > > Every two > > > paths separete on some finite level edge, but only when you got the > > > whole countably infinite path (the union of it`s all finite subpaths > > > starting from the beginning) all infinite long pathes separate from > > > each other. > > > > Of course. And therefore all the paths can be counted by the number of > > split positions. > > No. There is no point (except union of all the points) where a path > separates from all the other pathes and becomes a unique path. This union is covered by the limit n --> oo, i.e., by the infinite tree. Regards, WM |