Cantor Confusion
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From: mueckenh on 22 Nov 2006 07:25 Virgil schrieb: > In article <1164143353.398396.215510(a)m73g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > > Perhaps D. Marcus' understanding in > > > <MPG.1fcbb0f0d873ccdd989977(a)news.rcn.com> > > > may help you to cope with my sentence under discussion. > > > > > You wrote: > > 1) "The cardinality of omega is |omega| not omega." > > And after learning from me that this is wrong, > > 2) "The cardinality of omega, also written as |omega|, is omega". > > > > Now I am interested whether or not this is a contradiction in your > > eyes. > > WM is guilty of much more serious trangressions of both logic and common > sense that is involved with whether omega and |omega| are the same. That is not the question any longer. The question is: Can a set theorist admit that she is in error? It seems impossible. They all are too well trained in defending ZFC. Regards, WM
From: mueckenh on 22 Nov 2006 07:28 Virgil schrieb: > In article <1164143519.034139.154960(a)h54g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Randy Poe schrieb: > > > > > > > What about all contructible numbers or all words? They cannot be mapped > > > > on N though they are a countable set. > > > > > > The set of all finite words, the set of polynomials, or any > > > countable set can be put in bijection with N. Where do you > > > get your view that it "can't be mapped to N"? > > > > The list of all finite words cannot be constructed. > > It can be constructed inductively, which means that in ZF or NBG, it can > be constructed. The natural numbers cannot be constructed inductively. Induction proofs do not cover N, I was told. > > > > The list of all > > constructible numbers cannot be constructed. > > > The list of all naturals can be constructed inductively, which means > that in ZF or NBG, it can be constructed. It can be proven inductively that all initial segments of the set of natural numbers are finite. So, if the induction proof concerns all n in N, then N is finite. Regards, WM
From: Franziska Neugebauer on 22 Nov 2006 07:35 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> You have not commented on the first part of my posting: > No. So you don't want to tell me whether I miscomprehended you or not? ,----[ <4562fc49$0$97261$892e7fe2(a)authen.yellow.readfreenews.net> ] | mueckenh(a)rz.fh-augsburg.de wrote: | | > Franziska Neugebauer schrieb: | >> > Name one single element of the diagonal which is not contained in | >> > a line (which contains this and all preceding elements). | >> | >> 1. Burden of proof weighs upon you who claimed "And there are lines | >> as long as the diagonal is.". The question remains: Which lines? | > | > The diagonal D is a subset of the union U of lines L_n. | > Therefore D - UL_n = empty set. | | Do I miscomprehend your sentence | | "And there are lines as long as the diagonal is."? | | Condider the following scenario: | | state 1: Given a parking space with two cars having their | wheels mounted. (each car has 4 wheels) | | state 2: Unmount all the wheels of both cars and put them on one | stack of wheels. (stack of wheels has 8 wheels) | | Proposition 1: "There are cars in state 1 which have as many wheels | as the stack of wheels in state 2 has." | | Proposition 2: "There is a car in state 1 which has as many | wheels as the stack of wheels in state 2 has." | | Now my interpretation: Claim 1 states that there are at least two (due | to the plural form "cars" and due to the "are" instead of "is") cars | each of which have the property of having as many wheels as the stack | of wheels has (8). That is wrong in my view. | | Claim 2 states that there is (at least) one car which has as many | wheels as the stack of wheels has (8). Which is wrong, too. | | Question 1: Do you agree with my interpretation? | | We frequently use formulas like Ex(p(x)) which is translated as "There | is an x having property p". | | Question 2: Is it correct that you mean by writing | "And there are lines as long as the diagonal is." you mean | "There is a line as long as the diagonal" is? I. e. in the sense of | Ex (p (x)) with x for line and p(x) "lenght of line equal length of | diagonal"? `---- >> >> Do you want to posit, that there is some line identical to the >> >> diagonal? That is not (yet) proven. >> > >> > The number of elements of the diagonal is larger than all n: >> > E omega A n : omega > n. >> >> 1. n is not suitably constrained. For any n > omega the formula >> is omega > n is wrong hence the formula is wrong. > > Please read yourself. In future I am not willing to explain everything > twice to you. > We discuss a diagonal with omega natural elements. Do you mean A n (n e omega -> n < omega) ? >> > The lines have only finitely many elements. >> > >> > In a linear order of finitely many elements n exactly the following >> > is implied: >> > If for every n there exists an line L(n) such that L(n) contains >> > all elements m <=n >> > then there exists a line L such that for every n, L contains all >> > elements m <=n . >> >> > This is valid for the finitely many elements of every line. (How >> > many lines there are is irrelevant.) >> >> Proof? > > It is a self evident truth for finitely many elements like the > elements of a line, because every finite set has a largest element. So you mean your implication is not universally valid? > There exists no counter example, and it is impossible to construct a > counter example. F. N. -- xyz
From: William Hughes on 22 Nov 2006 07:35 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > > If there is a bijection with all finite line ends, then there is a > > > bijection with one line. > > > > No. > > The diagonal contains only finite line ends. Every element > > d_nn, of the diagonal, is the end of the line which ends > > at the natural number n. > > > > The bijection with all finite line ends is: The line which ends > > in n > > maps to the element d_nn. > > > > There is no bijection with one line. Such a bijection would > > imply that there is a largest d_nn. > > No. It implies only that all d_nn are finite. If you disagree, please > give an example for two elements of the diagonal which cannot be found > in one single line. > It is trivially true that given any two elements of the diagonal that a line exists that contains both elements. It would be quantifier dyslexia to claim that there exists a line that contains any two elements of the diagonal. ["two" can of course be replace by any natural number.] Your claim is that there is a bijection with one line. Call this line Kumquat. Kumquat has a largest element. Since there is a bijection between the elements of Kumquat and the d_nn, there must be a largest d_nn. - William Hughes
From: William Hughes on 22 Nov 2006 07:46
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > No. You can develop "potential set theory". Just define > > an element of a potential set to be an element of any one > > of the arbitrary sets that can be produced. Now go through > > set theory and add the word "potential" in front of each > > occurence of the word set. There is no difference between > > saying "a potentially infinite set exists" and saying "an actually > > infinite set exists". > > A potentially infinite set has no cardinal number. It cannot be in > bijection with another infinite set because it does not exist > completely. > Piffle. You need to study your potential-set theory. We have defined what it means to be an element of a potential set. Therefore we can define bijections between potentially infinite sets. Therefore we can define cardinalities of potentially infinite sets. - William Hughes |