Cantor Confusion
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From: mueckenh on 20 Dec 2006 12:33 Newberry schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Newberry schrieb: > > > > > > > >The edges have cardinal number aleph_0. > > > > > > Certainly. > > > > > > > A > > > > subset of the paths can be shown to be in bijection with the real > > > > numbers of the interval [0,1]. > > > > > > Which subset? Path as a set of intervals or path as a set of eges? How > > > does the above follow from this? > > > > The paths are binary representations of all real numbers of the > > interval [0,1]. Some rational reals have double representations, namely > > such ending by 000... and such ending by 111... like 1.000... and > > 0.111... . So there are not less real numbers in the interval [0, 1] > > than paths in the binary tree. We have > > > > aleph_0 = |{edges}| >= |{paths}| >= |{reals of [0,1]}| = 2^aleph_0 > > > > while according to set theory > > > > aleph_0 < 2^aleph_0 > > > > Regards, WM > > Well, a Cantorist would say that this part does not hold > > |{edges}| >= |{paths}| > > because you considered only the finite paths but not the infinite paths. In the infinite tree there are no finite paths. But if we calculate the limit of a finite tree, the result is the same. Meanwhile we have found out that set theorists must assume 1 + 1/2 + 1/4 + ... > 10 be possible unless we have some special definition of limits. I think his shows the whole insanity of their position well enough. Regards, WM
From: Virgil on 20 Dec 2006 14:47 In article <1166613023.292578.130190(a)a3g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Jonathan Hoyle schrieb: > > > > 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. > > Small wonder that it is impossible to divide something being by > something not being. > > > > > 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. > > Out of what? In principle I do not need any share which is less than > 1/1024 edge. (And I am ready to further negotiate.) WM does not seem to have any principles, at least no mathematical ones, if he claims that.
From: Virgil on 20 Dec 2006 14:55 In article <1166613226.916636.162010(a)i12g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Gc schrieb: > > > mueckenh(a)rz.fh-augsburg.de kirjoitti: > > > > > 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. > > > > The edges and paths are there completely from the beginning. > > In fact, if irrational numbers do exist, this must be the case. > > > 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. Non sequitur, unless one requires that all paths be finite. And ever then there are more paths than splitting positions. > > But we can look at the problem also from another point of view: > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > different pathes and E(n) = 2*2^n - 2 different edges. So we find > lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, > E(n)/P(n) >= 1 in any case. Hence, the assertion "E(n)/P(n) < 1 in an > actual infinite tree" has been disproved by simplest application of > mathematics. Except that there is no justification provided for claiming that what holds for finite trees must hold for infinite ones. And some of what olds for finite trees cannot hold for infinite ones. In a finite tree, every path has a leaf (terminal) node. By WM's form of argument, every infinite path would also have to have a leaf node.
From: Virgil on 20 Dec 2006 14:59 In article <1166613276.709962.230670(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166478692.220101.200880(a)f1g2000cwa.googlegroups.com>, > > "Jonathan Hoyle" <jonhoyle(a)mac.com> wrote: > > > > > > 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. > > > > WM usually ignores what he cannot find a way to wriggle out of. > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > different pathes and E(n) = 2*2^n - 2 different edges. So we find > lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, > E(n)/P(n) >= 1. Hence, the assertion "E(n)/P(n) < 1 in the actual > infinity" has been disproven by simplest application of mathematics. While one can show that, for finite n, E(n) <= P(n), one can also show that L(n) = P(n) where L(n) is the number of leaf nodes (termnal nodes), or, equivalently, terminal edges. So that by WM's own argument every unending path in an infinite tree must end.
From: Virgil on 20 Dec 2006 15:04
In article <1166613407.943112.211570(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David R Tribble schrieb: > > > Assuming that the largest natural that has been "brought into > > existence" is m, then we could easily define q as m+1. > > But then we have a contradiction: since m "exists", so does q, > > but this contradicts your statement that q is "not yet a number", > > i.e., q does not exist yet. m exists, but m+1 does not. So one > > of your assumptions about the "existing" naturals must be wrong. > > It is not my assumption that a non existing number does exist. Could > well be an axiom of set theory. On the contrary, it is precisely WM's assumption that for any list of naturals there is a largest, which requires that the successor of that largest be a non-existing number. |