Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 17 Dec 2006 08:21 Dik T. Winter schrieb: > > The notion "rational relation" is a further development of the notion > > "relation". While a relation connects elements of a domain to elements > > of a range, the rational relation connects shares of the elements of > > the domain to (shares of the) elements of the range. A simple example: > > Again, what are "shares"? An edge is considered to be a set of shares. You can divide it into m equal shares. If you divide n edges into m equal shares each, then you have m * n shares and you can combine m abitrary shares to have one full edge. As an example use euro and cent. > > > A) Consider the relation: > > > > 1 ---> a > > 2 ---> b > > > > B) Now a rational relation could be: > > > > 1.1 ---> a > > 1.2 ---> b > > > > 2.1 ---> a > > 2.2 ---> b > > > > If 1.1 and 1.2 are the only shares 1, i.e., if 1.1 u 1.2 = 1 and if 2.1 > > and 2.2 are the only shares of 2, i.e., if 2.1 u 2.2 = 2, then one full > > element of the domain is available for every element of the range. > > Makes barely sense. What is 1.1 united 1.2? I have no definition, so I > have no idea what that means. I wrote "1.1 u 1.2 = 1". But perhaps it is bewildering to use the decimal point here. Let us denote the two equal shares of edge number 1 by 1#1 and 1#2. Consider you have two coins of 50 cent value each. Then you can buy something which costs 1 euro. And even going into your reasoning, I see > still nothing that allows the "then one full element of the domain is > available for every element of the range". I would like to ask for a > proof. You are still reasoning in the finite case. Whether it is > applicable in the infinite case remains a question. For this sake we have mathematics, I mean really correct mathematics, not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + .... = 2. > > > This proof is similar to the proof of a well-ordering of the real > > numbers. It cannot be done (and it can even be proved that it cannot be > > done) but it has been proved to exist. > > You are entirely wrong here. > (1) Axiom of choice valid: it can be done (and I think it has been done). LOL. Would you be so kind to tell me in which form this well-ordering has been done? Is a catalog available at some bibliothy? Or is there a formula stating which real comes first and which comes after that number aleph * omega^omega^omega? > (2) Axiom of choice not valid: you can prove that it can not be done. > But you confluate the two together and apparently do not see the > difference. In *mathematics*, either it can be done or it cannot be done. And if it has ben done, one need not forget it after abolishing AC. > > > The rational relation devised in the binary tree proves that a > > surjection of edges onto paths exists. In so far my assertion "there is > > a surjective mapping" was not quite wrong. > > Not so. Because you are using a lot of terms that lack definition. Revisit the 3rd class of the elementary school. There the teacher will teach you how money can be divided into smaller pieces and different pieces can be collected to yield a desired value, for instance euro 28.50 to order my book. If you found a company and every member gives 5 euro, then 6 members will suffice to buy the book including postage. Regards, WM
From: Virgil on 17 Dec 2006 16:05 In article <1166346410.767514.191120(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > We see it but do not believe that it does what you say it does. > > > > Your example assumes a bijection and constructs a "rational relation". > > I assume nothing but that every path gets a share of that edge through > which the path runs. As infinitely many paths run through each edge, whether one believes that to be a ccountable or an uncountable infinity, each edge must be divided into infinitely many parts. > You cannot deny that an edge must exist if a path runs through it. > I prove that there are enough shares to build 2 edges per path. WM claims to do a lot of things which he, in fact, never has done and cannot do. > > > > But in the absence of any original bijection there is no assurance of > > any final bijection by such construction. > > Without any original bijection we can conclude from > 1.1 ---> a > 1.2 ---> b > 2.1 ---> a > 2.2 ---> b > and the knowledge that > 1.1 U 1.2 = 1 and > 2.1 U 2.2 = 2 > than there are as many nunmbers as letters. Since every edge must be divided by the number of paths running through it, in order to give each of those paths a part of it, one has to be able to divide a finite path into infinitely many equal parts, which is impossible. > > And assuming what you are trying to prove is a no-no. > > no * no = yes In my logic, which is fairly standard Yes = 1, no = 0, so no * no = no + no = no. > > Regards, WM
From: Virgil on 17 Dec 2006 16:10 In article <1166346555.117491.40240(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > Do you believe that paths beginn to run separated wihout any edge > > > > > being > > > > > inolved? > > > > > > > > No. And so what? But only looking at the paths that way you never get > > > > the full infinite paths. > > > > > > If we look at least at one part of each path, then we look at all paths. > > > > Not so. To separate any infinite path from /all/ other infinite paths > > you must look at ALL its edges, not just some of them. > > > The first n edges play no role. The two edges proceeding from any node play a roll in separating one set of paths from another. But to separate one path from ALL others one must look at ALL its edges.
From: Virgil on 17 Dec 2006 16:23 In article <1166346721.953467.204620(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > > (It is contained in the union of all lines, but the > > > > > > > > > > union of all lines is not a line) > > > > > > > > > > > > > > > > > > That is a void assertion unless you can prove it by showing > > > > > > > > > that > > > > > > > > > element by which the union differes from all the lines. > > > > > > > > > > > > > > > > Not quite. In order to achieve that the diagoal is not in any > > > > > > > > linem all > > > > > > > > that is required is: > > > > > > > > Given any line there is an element of the diagonal not in > > > > > > > > THAT line. > > > > > > > > It is not requires that: > > > > > > > > There is an element of the diagonal that is not in any line. > > > > > > > > > > > > > > > > > > > > > For linear sets you cannot help yourself by stating that the > > > > > > > diagonal > > > > > > > differs form line A by element b and from line B by element a, > > > > > > > but a is > > > > > > > in A and b is in B. This outcome is wrong. > > > > > > > > > > > > > > Therefore your reasoning "there is an element of the diagonal not > > > > > > > in > > > > > > > THAT line. It is not required that: There is an element of the > > > > > > > diagonal > > > > > > > that is not in any line." is inapplicable for linear sets. You > > > > > > > see it > > > > > > > best if you try to give an example using a finite element a or b. > > > > > > > > > > > > > > > > > > In every finite example the line that contains > > > > > > the diagonal is the last line. > > > > > > > > > > Every example with natural numbers (finite lines) is a finite > > > > > example. > > > > > > > > > > > Your claim is that there is a line which contains the diagonal. > > > > > > > > > > Because a diagonal longer than any line is not a diagonal. > > > > > > > > > > > Call it L_D. Question: "Is L_D the last line?" > > > > > > > > > > There is no last line > > > > > > > > Then, there is a line that comes after L_D. > > > > > > > > Therefore :L_D does not contain every element > > > > that can be shown to exist in the diagonal. > > > > > > All elements that can be shown to exist in the diagonal can be shown to > > > exist in one single line. > > > > > > > Call it L_D > > > > L_D contains a largest element. n. > > > > L_D is not the last line, so there is > > a line with element n+1, > > > > Element n+1 can be shown to exist in the diagonal. > > > Element n+1 can be shown to exist in L_D (which is obviously a line > containing n+1). > > All elements that can be shown to exist in the diagonal can be shown to > exist in one single line. Which line is that? That presumes a last line, which presumption is unwarranted in ZFC and NBG and most other set theories. > > The easiest way to see that would be to name all elements of the > diagonal. Every element is a natural number and as such a set > containing all the smaller natural numbers. In fact it is irrelevant > whether you write > 1 > 2 > 3 > or > > 123. > > If the diagonal contains only finite segments, then each one is in a > line and vice versa each line is in the diagonal. Only if there is an > infinite segment in the diagonal, then it is not in a line. But in > order to have the infinite segment, you need an element which is not in > any line, i.e., an infinite element. The idiotic idea that an endless sequence must contain an infinite object is unwarranted. It is a delusion that such as WM feed their other delusions upon. Consider, for example, the sequence f(n) = sin(n radians). It has no last term but every term has a distinctly finite value falling between -1 and 1, and each term is distinct from all previous terms. Similarly, consider the sequence, g(n) = 1/2^n. According to WM there must be some n so large that either g(n+1) = g(n) or g(n+1) = 0.
From: Virgil on 17 Dec 2006 16:27
In article <1166359875.644197.31300(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Bob Kolker schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > Do you believe that today's standards are sound by tomorrow's > > > standards? > > > > Wait a day and we will find out. Hilbert's axiomatization of Euclidean > > Geometry done back in 1899 is as good today as it ever was. Aristotelean > > logic and semantics have held up very well over 2400 years. Too bad that > > Aristotelean physics fell way short. > > > > The difference between today's math and tomorrows math will be mostly in > > scope, not validity. The arithmetic argument that Euclid used to prove > > the infinitude of primes > > He proved that there are more primes than given. In fact, more than any finite number. > > > is as valid today as it was 2200 years ago. > > The difference between today's math and tomorrows math will be that the > physical and physiological foundations will be explored and taken into > account. WM would, if he had the power, allow only some of the words and none of the music. |