Cantor Confusion
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From: mueckenh on 21 Dec 2006 16:43 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).. > 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". > 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. 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. You accept more separated paths than splitting positions (= origins of separated paths) [in the binary tree]. You accept that 1 + 1/2 + 1/4 + ... > 10 is possible. And at last you accept the existence of an actually infinite finite number. Enough? Regards, WM
From: mueckenh on 21 Dec 2006 16:48 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| ? > > Equally, if we "stay with your version", "E(n)/P(n) >= 1 in the actual > infinity" is nonsense; just as "Triangle/Square <= 2" is nonsense. > > On the other hand, "down to level n", E(n) is finite for all finite n. > Nevertheless, "E(n) is not finite in the actual infinity". > > Do you consider the above also not "correct mathematics"? What is > "incorrect" about it? Your attidute which should make us believe you understood mathematics. Regards, WM
From: mueckenh on 21 Dec 2006 16:50 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.> > > see the example that disproves your counter claim below. > (by mistake you snipped it) You are in error. You gave no example showing that not all natural numbers can be in one line. > > > > > > > > Remember, even if there is not a last line, every n eps N is finite. > > > > > > 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". > > "It does not mean that that we can find a single > line L_D, that works for all n." written out in full is > > It does not mean that there exists an L_D with the property that: > if a natural number n can be shown to exist in the diagonal > n must be an element of L_D. If all natural numbers would exist, then, of course, also this L_D would exist. If L_D does not exist, then we have the proof that not all natural numbers do exist. > > There is no claim that "all n" exists. > > Restoring the example you snipped > (and expanding the language slightly) > > There exists a linearly ordered set of finite elements that > does not support quantifier reversal. > > Consider the (potentially infinite) set of natural numbers. > This is linearly ordered. Every element is finite. > > For any natural number n that can be shown to > exist, a natural number m(n), > such that m(n) > n can be shown to exist. > > However it is not true that: > > There exists a natural number m, such that for any > number n that can be shown to exist, m>n. There exists a natural number m, such that for any n that can be shown to exist, m >= n. It is a number which is at least as large as the largest existing natural number. Regards, WM
From: mueckenh on 21 Dec 2006 16:51 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". If "all n" were somewhere, then we had no problem to put them in one single line. 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. That is obviously our old EIT. 1 11 111 .... The number of 1 is not *actually* infinite without one line being actually infinite. > > Therefore your claim is void, unless you can find a linear set of > > finite elements which forbids exchange of quantifiers. > Consider the (potentially infinite) set of natural numbers. > This is linearly ordered. Every element is finite. > For any natural number n, there exists a natural number m(n), > such that m>n > However it is not true that: > There exists a natural number m, such that for any natural > number n, m>n. It is true for any set of natural numbers which can be written down. You cannot give a counter example. The only things you can do is either to present a line which does not contain all numbers, or to assume that a line contains all numbers and then to show that this is not the case. What a surprise! Or you can try to talk of dancers each of which has a women partner but not all have the same. 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. You know that the logical symbol for every is forall. Regards, WM
From: Virgil on 21 Dec 2006 17:22
In article <1166737132.026953.46380(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > Indeed. No axiom is necessarily true. > > How then should the truth of AC be proved? One does not prove axioms, one either assumes them or does not assume them. For a given axiom system, those statements in that system which have been proved from those axioms are called theorems, at least within that system. |