Cantor Confusion
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From: mueckenh on 20 Dec 2006 06:10 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.) Regards, WM
From: mueckenh on 20 Dec 2006 06:13 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. 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. Regards, WM
From: mueckenh on 20 Dec 2006 06:14 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. Regards, WM
From: mueckenh on 20 Dec 2006 06:16 David R Tribble schrieb: > muecken h wrote: > > Not yet a number, unless you can specify it such that one (at least > > you) can decide whether q < n or q = n or q > n for any natural number > > n given to you. Only saying that it is not equal to any natural number > > given is not enough. > > I did not say that q is not equal to any given natural number. > I said that q is not equal to any natural number that has been > "brought into existence". Trichotomy is well defined for q. > For any natural n given, if n has been "brought into existence", > then q > n. > > Since you believe that only those numbers that have been "brought > into existence" exist, you should readily grasp the definition of q. Yes. You mean a number q which exists but is larger than any existing number. > > > 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. Regards, WM
From: mueckenh on 20 Dec 2006 06:20
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > > > Element n+1 can be shown to exist in L_D (which is obviously a line > > > > > > > > containing n+1). > > > > > > > > > > > > > > No. L_D is bounded. The largest element of L_D is n. > > > > > > > L_D does not contain n+1. > > > > > > > > > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > > > > contained in the diagonal. If your L_D does not contain them, then you > > > > > > have the wrong L_D. > > > > > > > > > > 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. > > > > > > The diagonal is the potentially infinite set of natural numbers. > > > This exists and is unbounded. > > > > Yes, but not in the way you understand "to exist". Potentially infinite > > means: always finite. The diagonal is always finiter though not > > bounded > > The only property of the diagonal that is used is that > it is unbounded (i.e. given a finite set of elements of the diagonal, > it is possible to add one element). That property is correct. > The existence of all the elements > of the diagonal is neither claimed nor used. The "infiniteness" > of the diagonal is neither claimed nor used. Fine. > > The contradiction (a line must be both bounded and unbounded) No line must be unbounded in case of a potentially infinite (i.e. always finite) diagonal. If you add one element to the diagonal then all its elements are in another line than before, but they are in a finite line. Therefor all finite elements of the diagonal are in one single line. Regards, WM |