Cantor Confusion
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From: Virgil on 21 Dec 2006 17:28 In article <1166737239.205667.205290(a)f1g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1166617872.309270.165750(a)79g2000cws.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > The inequality holds for all finite paths. Why it also should hold > > > > for > > > > infinite paths escapes me. > > > > [ This was about the number of edges in finite case n: 2^(n+1) - 2, > > and the number of paths in finite case n: 2^n.] > > > > > > Because we can prove that 2 - 1/2^n is always, i.e., for all finite > > > natural n, less than 100. > > > > Yes, so I did state. > > > > > And because even infinity does not supply > > > other than finite natural numbers. > > > > That has no significance to the question. You fail to see that in > > the finite case each path has a last edge, but in the infinite case > > *no* path has a last edge. > > The sum 1 + 1/2 + 1/4 + ... has no last term. It is not a sum, but is often used to represent an infinite sequence of partial sums. Such a sequence may have a defined limit value, but that limit is not "the sum" except by abuse of language. > Nevertheless we know, > even in ZFC, that it is larger than 1 and less than 10. We do not know yet that it exists at all, until we have defined what it would mean. And, as that is a matter of delta-epsilonics, it is not as trivial as WM implies.
From: Virgil on 21 Dec 2006 17:45 In article <1166737382.495151.299530(a)i12g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Newberry schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > > > Yes, but you arrived at |{edges}| >= |{paths}| by calculating > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > > In this limit all the n are finite. So a Cantorist would claim that you > > examined only the finite sub-paths, > > > Every real number (like, e.g., pi) has digits r_1, r_2, r_3, ..., only > at finite digit positions: 1,2, 3, ...(because an infinite digit > position could not be identified and the corresponding digit could not > be determined).. But in ZFC or NBG, there are infinitely many of those finite digit positions, one for each of the infinitely many finite natural numbers. > > > and not the infinite paths. I do > > not know if this Cantorist claim makes sense, > > It helps them to veil the contradictions appearing in set theory, > because they can easily jump from "every natural number is finite" to > "there are infinitely many natural numbers". Which is easily verified in ZFC and NBG. WM has no coherent system in which it is not verifiable. > > > but the question is > > whether you can generate a direct contradiction of the type P & ~P in > > ZFC. > > I do not think that this is of any interest after it can easily be > shown that the results of ZFC are inacceptable. Except that the only demonstration of such "inacceptability" would be the discovery of some internal contradiction within ZFC or NBG. And the only conflicts WM can up with are with statements outside of ZFC and NBG, which are irrelevant to what occurs within them. > Compare for instance > the resuls I have recently prepared in a discussion with Virgil: > > You accept a diagonal of a matrix longer than any line. I accept an infinite path in an infinite binary tree > You accept more separated paths than splitting positions (= origins of > separated paths) [in the binary tree]. If, by "splitting positions", WM means nodes, then in an infinite binary tree I do accept that there are "more: paths than nodes, in the Cantor sense that one can inject the set of nodes into the set of paths but not vice versa. Complete and detailed descriptions of the node to path injections have been posted. Wm has made various vague claims of a reverse mapping, but has never been explicit enough about it for any sound analysis to be performed. > You accept that 1 + 1/2 + 1/4 + ... > 10 is possible. Not so. What I do not accept is that 1 + 1/2 + 1/4 + ... has any more relevance that 1 + 1/2 + 1/3 + .... > And at last you accept the existence of an actually infinite finite > number. What "actually infinite finite number" is that, WM? ZFC allows, indeed requires something like actually infinite non-natural "numbers", but only WM claims that any of them are also finite.
From: Virgil on 21 Dec 2006 17:49 In article <1166737725.202451.87180(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > cbrown(a)cbrownsystems.com schrieb: > > > > > > > > > > > No. But I need no proof in order to see that > > > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > > > > > > > > > > Yes, it appears that you need no proof to make any of your assertions. > > > > You simply "see" them; or "feel" them. That's not what I would call > > > > "really correct mathematics". > > > > > > Call it whatever you want. > > > > I call it "mathiness". > > > > > I can calculate 1 + 1/2. > > > > Good for you! > > Bad for you, if you can't. > > > > > But considering the > > > fact that you would call the following result "correct mathematics", I > > > prefer to stay with my version. > > > > > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > > > different paths 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. Nevertheless "E(n)/P(n) < 1 in the actual infinity". > > > > It is /not/ a fact that I would call this "result" correct mathematics. > > I would call it yet another example of your devotion to "mathiness" > > whenever you claim to provide a "contradiction". > > > There can't be a contradiction in ZFC because everything looking like a > contradiction is mathiness. > > While "E(n)/P(n)" is certainly well-defined when both E(n) and P(n) are > > real numbers, it is not well-defined otherwise. You simply "feel" that > > Cantor's result must imply "E(n)/P(n) < 1 in the actual infinity"; > > without considering what that assertion /means/. > > > I know that 2^aleph0 > aleph0 is a result of Cantor's theory. If you > don't know that, you should try to inform you. > And I have proved that E(n) > P(n) for every n and for the limit n --> > oo. > > > You apply "mathiness" to the problem. You "know" (need no proof) that > > "2 < 3" implies that "2/3 < 1". > > I know that 2/3 < 1 is a convenient way to express 2 < 3. But if you > don't know that, then we can stay with 2 < 3 as well. > > > You "see" the assertion "|N| < |R|", > > This is a provable theorem in ZF > > > and because it "looks" like "2 < 3", you "feel" that you can therefore > > conclude "|N|/|R| < 1". > > > > Instead, the statements "|N|/|R| < 1" and "E(n)/P(n) < 1 in the actual > > infinity" are simply mathematical nonsense: neither true nor false. > > Are you unable to see that the formal use of fractions like |N|/|R| < 1 > is completely irrelevant but only a brief way to express |N| < |R| ? By definition, |N| < |R| means no more than that there are injections from set N to set R but none from set R to set N. How is E(n) < P(n) relevant to that definition?
From: Virgil on 21 Dec 2006 18:02 In article <1166737855.828116.90470(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > Please give a counter example for a finite number n. > > > > > > > > No such counterexample exists. The claim is that despite > > > > the fact that given any n we can find a single line, L(n), that works > > > > for this n, there is no single line, L_D, that will work for all n. > > > > > > This claim is obviously false. At least my counter claim cannot be > > > disproved by an example.> I have several time disproved it by one of WM's own examples, infinite binary trees: Any infinite path is a diagonal for some EIT. The finite subpaths staring at the root nod and ending at a node of the given path are the edges. Given any node there is a line containing it BUT There is no line containing every node. Note that WM has so far carefully avoided responding to this refutation. Any infinite path is a diagonal for some EIT. The finite subpaths staring at the root nod and ending at a node of the given path are the edges. Given any node there is a line containing it BUT There is no line containing every node. > > You are in error. You gave no example showing that not all natural > numbers can be in one line. Note that WM has so far carefully avoided responding to this refutation: In any infinite tree: Any infinite path is a diagonal for some EIT. The finite subpaths staring at the root nod and ending at a node of the given path are the edges. Given any node there is a line containing it BUT There is no line containing every node. > If all natural numbers would exist, then, of course, also this L_D > would exist. Not in the model of a path in an infinite tree. > If L_D does not exist, then we have the proof that not all > natural numbers do exist. WRONG! Not in the model of a path in an infinite tree. > > There exists a natural number m, such that for any n that can be shown > to exist, m >= n. if m can be shown to exist, so can m+1. So WM claims a natural m so large that it is greater than its successor. > It is a number which is at least as large as the largest existing > natural number. And must be larger than its successor.
From: Virgil on 21 Dec 2006 18:38
In article <1166737915.714901.106300(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > See my recent reply to William Hughes. > > > > Please copy and paste. I am not able to guess what you have in mind. > > > We talked about the potentially infinite set N. > > Which means that for any n we can find a single line L(n) that works > > for this n. It does not mean that that we can find a single > > line L_D, that works for all n. > > > > That is impossible, beause there is no "all n". There is an "all n" in ZFC or NBG. What system is WM proposing in which things are otherwise? > If "all n" were > somewhere, then we had no problem to put them in one single line. Only when that "line" is the diagonal. if every line, except the first with only one element, is a successor to another line, every line must have a last, but the diagonal need not have a last, at least not in ZFC or NBG. Due > to the finiteness of every n, every position of the line was finite. > And due to the fact that a number with only finite indexes is a natural > > number, the whole number was a natural number. > > 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. Except that I am not suggesting that we add the infinitely many equations to each other, which is the only way to get infinitely many 1's. Within each equation, where those actual additions occur, only finitely many 1's can occur. > > > That is obviously our old EIT. A proper EIT is like an infinite path in an infinite binary tree. Any path is itself like a diagonal with the finite suppaths from the root node to one of the odes in the "diagonal" being the lines. In this model, WM is totally wrong. > > > 1 > 11 > 111 > ... > > > The number of 1 is not *actually* infinite without one line being > actually infinite. That false belief is at the root of WM's perennial failures. > > > > > > Therefore your claim is void, unless you can find a linear set of > > > finite elements which forbids exchange of quantifiers. DONE! In model described above, an infinite path in an infinite tree: For every node(natural number) in the diagonal (infinite path) there is a line (finite path) containing that node BUT NOT There is a line (finite path) containing every node of that diagonal(infinite path). I.e., we have an infinite linear set of nodes, each a finite element in the infinite path, in which one cannot switch quantifiers. |