Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Dik T. Winter on 14 Dec 2006 08:14 In article <1166094987.634497.223120(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > When you say "You are aware of the number 2^omega being a countable > > one?", I would respond : that's what you're trying to /prove/. You > > can't just assert it that it's true for other reasons, and claim that > > therefore you have proven it using your "rational relation". > > No. According to ZFC the ordinal 2^omega is a countable set. Even > omega^omega^omega is a countable set. You two are talking at cross purposes. Cbrown is thinking about cardinal exponentiation, you are thinking about ordinal exponentiation. These two are *very* different. In cardinal exponentiation 2^aleph0 is the cardinality of the set of infinite sequences of the symbols 0 and 1. In ordinal exponentiation 2^omega is the ordinality of the set of infinite sequences of the symbols 0 and 1 where in each sequence there are only finitely many 1's, and that with a specific ordering. So in terms of the tree, only those paths are in the set 2^omega that go only finitely many times to the right. Or, saying it another way, only the rationals with a finite binary expansion are in it. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 14 Dec 2006 10:17 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. 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 Regards, WM
From: mueckenh on 14 Dec 2006 10:21 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. > > >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. Regards, WM
From: mueckenh on 14 Dec 2006 10:22 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. Regards, WM
From: mueckenh on 14 Dec 2006 10:24
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 because there is no unique diagonal. Regards, WM |