Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Virgil on 14 Dec 2006 19:47 In article <1166109473.022083.10120(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166044694.629951.178340(a)79g2000cws.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > You are aware of the number 2^omega being a countable one? If so, why > > > then do you speak of an uncountable set of shares? > > > > Because 2^omega, representing the set of functions from omega to {0,1}, > > is demonstrably not countable. > > > > > > Do you think that lim {n-->oo}1/2^n * 2^n consists of uncountably many > > > parts? > > > > I have no idea why WM thinks that relevant to anything under > > discussion, but it certainly does not effect the demonstrable > > uncountability of 2^omega. > > > > The ordinal 2^omega is a countable set. Even omega^omega is a countable > set. Not in ZFC or NBG. And until WM presents an axiom system in which it is true, not in ANY system. > You could learn this elementary knowledge from my book: > http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN > =3-8322-5587-7 Considering the illogic in WM's postings here, I will guess that there is nothing worth my attention in any book or paper WM writes, except possibly as entertainment.
From: Virgil on 14 Dec 2006 19:52 In article <1166109663.622495.208470(a)f1g2000cwa.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: > > > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > > A_1 = {1} > > > > > > > > > > A_2 = {1,2} > > > > > > > > > > A_3 = {1,2,3} > > > > > > > > > > > > > > > > > > > > B = {1,2,3} > > > > > > > > > > > > > > > > > > > > then B is contained in the last A_i. If there is no last > > > > > > > > > > A_I, then > > > > > > > > > > there is > > > > > > > > > > no A_i that contains B > > > > > > > > > > > > > > > > > > That has nothing to do with "last". > > > > > > > > > > > > > > > > If A_i contains B, then A_i contains any A_j. > > > > > > > > Therefore A_i is "last". > > > > > > > > > > > > > > > > > > > > No comment? > > > > > > > > > > If B contains any A_i then B is the last. > > > > > > > > B cannot be the last A_i, as B is not > > > > one of the A's (B corresponds to the diagnonal, > > > > the A_i correspond to the lines) > > > > > > In which element does it differ from every A_i? > > > > In no element. > > > > We know that B is contained in all of the A_i > > put together. > > How can you put all together, if there is no last one? You will never > be sure that you have all. > > > > The question is: "Is B contained in one > > of the A_i?". > > > > The answer is: "B is contained in one of the A_i > > if and only if there is a last A_i." > > > Otherwise B cannot be contained in all A_i put together. Sure it can. The union of a family of sets need not be a subset of any of them. > > > > >Don't forget, the A_i > > > form a linear set. It is not possible for B to differ from A_n in > > > element a but not in element b and from A_m in element b but not in > > > element a. > > > > So A_i will contain all the elements from all the other > > A_j, if and only if for every A_j, i>=j. > > > > > > Another way to put this is 'Can we replace "all of the A's" > > by ''a single A_i" ?'. The answer to this question is > > 'We can make the replacement, if and only if there is > > a last A_i.' > > We can make the replacement for all A_i which are smaller than a finite > A_j. > If there are only finite A_j, then we can make the replacement for all. And if, as is the case in ZFC and NBG, there are more that any finite number of them, then no A_j can contain all the A_i. WE have axiom systems that works our way. WM has not produced any axiom system which works his way. For those who insist on knowing what is being assumed before consenting to it, WM is out of the running.
From: Virgil on 14 Dec 2006 19:55 In article <1166109758.397975.173050(a)16g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Another way to put this is 'Can we replace "all of the A's" > > > by ''a single A_i" ?'. The answer to this question is > > > 'We can make the replacement, if and only if there is > > > a last A_i.' > > > > > > > > > - William Hughes > > > > WM assumes (sometimes covertly, sometimes explicitly) that there is a > > last A_i, which is, of course, false in ZFC or NBG and most other set > > theories in general use. > > We can make the replacement for all A_i which are smaller than a finite > A_j. > If there are only finite A_j, then we can make the replacement for all. But unless we assume that there can only be finitely many we cannot conclude that there is a last or largest conctaining all the others. When one accepts ZFC or NBG, all the assumptions have been laid out clearly, so one knows what is being assumed. WM has no such system of axioms, so we are just supposed to let his claim whatever he wants? I don't buy that.
From: Virgil on 14 Dec 2006 20:00 In article <1166109895.141235.41650(a)j72g2000cwa.googlegroups.com>, 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. I do not find that as any part of the definition of a "diagonal". > > > Call it L_D. Question: "Is L_D the last line?" > > There is no last line because there is no unique diagonal. There CAN be a diagonal even in the absence of a last line, in the right axiom systems. For example, in ZFC and NBG, one has the finite ordinals as successive lines and the first transfinite ordinal as diagonal.
From: Ralf Bader on 14 Dec 2006 21:18
Virgil wrote: > In article <1166109473.022083.10120(a)j72g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >> Virgil schrieb: >> >> > In article <1166044694.629951.178340(a)79g2000cws.googlegroups.com>, >> > mueckenh(a)rz.fh-augsburg.de wrote: >> > >> > >> > > You are aware of the number 2^omega being a countable one? If so, why >> > > then do you speak of an uncountable set of shares? >> > >> > Because 2^omega, representing the set of functions from omega to {0,1}, >> > is demonstrably not countable. >> > > >> > > Do you think that lim {n-->oo}1/2^n * 2^n consists of uncountably >> > > many parts? >> > >> > I have no idea why WM thinks that relevant to anything under >> > discussion, but it certainly does not effect the demonstrable >> > uncountability of 2^omega. >> > >> >> The ordinal 2^omega is a countable set. Even omega^omega is a countable >> set. > > Not in ZFC or NBG. > > And until WM presents an axiom system in which it is true, not in ANY > system. It seems you are not aware of the fact that ordinal exponentiaton and cardinal exponentiation are defined differently. The result of ordinal exponeniation should be an ordinal again. If the exponent is finite, one achieves this by lexicographic ordering of finite sequences. If the exponent is omega, one takes the union of all the sets obtained by exponentiating with finite exponents; so omega^omega comprises all sequences of naturals of arbitrary but finite length. This is a bit easier to well-order than map(omega,omega). Mückenheim, however, commits a similar but much more stupid error on p.5 of the remarkable flatulency he put on the arxive today: http://www.arxiv.org/pdf/math.GM/0403238 >> You could learn this elementary knowledge from my book: >> http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN >> =3-8322-5587-7 > > Considering the illogic in WM's postings here, I will guess that there > is nothing worth my attention in any book or paper WM writes, except > possibly as entertainment. I must admit that I'm not totally free from lust for trash, and Mückenheim serves this quite well (probably too well, and not only me). But of course I wouldn't pay a rusty nail for /that/. Ralf |