Cantor Confusion
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From: Franziska Neugebauer on 18 Dec 2006 16:06 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: [...] >> Assume that there exists an L_D which contains all the numbers >> contained in the diagonal. L_D is bounded, > > If an unbounded diagonal exists, then obviously an unbounded line must > exist. > The conclusion is false, so the antecedent cannot be true. Non sequitur: Logical Implication p q p -> q F F T *1) F T T T F F *2) T T T *2) is the interpretation I would prefer. I may help you: If the moon exists today then obviously Germany must be a monarchy today. Following the M�ckenheim-reasoning one would "conclude" that the moon does not exist. The point is that you erroneously think by simply juxtaposing p, "->" and q you "prove truth" on the implication "p->q". F. N. -- xyz
From: Franziska Neugebauer on 18 Dec 2006 16:08 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: [...] >> L_D is a line and is therefore bounded. >> >> Contradiction. >> Hence, the _assumed_ L_D does not exist. >> > Assumed is the diagonal. Premise: "If the diagonal is unbounded." > L_D is a line and is therefore bounded. Yes. And because L_D is bounded it cannot "contain" every element of the diagonal. Exactly this constitutes the contradiction to your claim P1. F. N. -- xyz
From: Virgil on 18 Dec 2006 16:49 In article <1166474763.304897.177520(a)80g2000cwy.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > In article <1166451538.861033.303300(a)48g2000cwx.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Newberry schrieb: > > > > > > > Sorry that I joined a bit late. > > > Doesn't matter. > > > > > > >Are you saying that (in an infinite > > > > binary tree) the set of paths is uncountable but the set of edges is > > > > countable? > > > > > > The set of edges is obviously countable by, e.g., > > > > > > 1, 2, > > > 3,4,5,6, > > > 7,... > > > > > > As no path can separate from another one without the existence of two > > > more edges, the number of edges is an upper bound for the number of > > > paths. > > > > > > That is only the case in finite trees. > > It fails miserably in infinite binary trees in which no path has a last > > node or last edge. > > Do you agree that the number of edges is the upper bound of the number > of paths but disagree that the edges are countable? I do not agree that in an infinite binary tree, such as may be constructed in ZFC, the number of edges is an upper bound for then number of paths. It is my contention that the number of paths through every edge is in that case uncountable. > > I guess if the number of edges > number of paths and you accept the > diagonal argument then it follows that the edges are uncountable. I do > not know if there is any way to prove directly that the edges are > countable. Actually, it is not too hard to enumerate the edges. At the first level, issuing from the root node there are two edges, which can be numbered from "left" to "right" as 1 and 2. At the next level, issuing from child nodes of the root node, there are 4 edges numberable 3,4,5 and 6. At the next level 8, numberable from 7 through 14. And so on. General rule: at the nth level there are 2^n edges, which can be numbered from 2^n - 1 to 2*(2^n-1).
From: Jonathan Hoyle on 18 Dec 2006 16:51 > WM wants to divide something into a countably infinite number of equal > shares. That would be a neat, considering that it has been proven in Measure Theory than an interval (of finite but positive length) cannot be partitioned in countably infinitely many intervals each of the same size. > If s is the size of one share then WM wants to have s + s + s ...= 1. > > What is the numerical value of s, WM? > > Even in ZFC and NBG, we know better than that. Yup, what you have written is essentially a sketch of the proof. Good luck trying to wrangle out of this one, WM.
From: Jonathan Hoyle on 18 Dec 2006 16:53
> WM wants to divide something into a countably infinite number of equal > shares. That would be a neat trick, considering that it has been proven impossible by Measure Theory to partition an interval (of finite but positive length) into countably infinitely many sub-intervals each of the same length. > If s is the size of one share then WM wants to have s + s + s ...= 1. > > What is the numerical value of s, WM? > > Even in ZFC and NBG, we know better than that. Yup, what you have written is essentially a sketch of the proof. Good luck trying to wrangle out of this one, WM. |