An uncountable countable set
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From: mueckenh on 9 Jul 2006 09:07 Dik T. Winter schrieb: > In article <pTsU8Z33pqrEFw3G(a)212648.invalid> David Hartley <me9(a)privacy.net> writes: > ... > > (WM's fallacy is his claim that the (order-preserving) indexation by the > > naturals carries over to the limit.) > > Indeed. The most basic standard fallacy he uses. Indeed. The most basic standard fallacy Cantor uses. Regards, WM
From: mueckenh on 9 Jul 2006 09:09 Dik T. Winter schrieb: > In article <J21Gtn.BCI(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > > Let me expand on this. > > > In article <1152282364.618937.306970(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > You start with the list of natural numbers in unary notation: > A0 = 0. > A1 = 0.1 > A2 = 0.11 > A3 = 0.111 > ... > calculate the diagonal number K: > K = 0.111... > and claim that K equals An for some value of n. I stated: either it does or it does not. > > Note the following: > > > It is not by pure accident that I chose 0.111... . This number has also > > > a meaning as decimal representation of 1/9. So we can do the > > > calculations. > > Indeed. Let's do the calculations, and let each An also have the meaning as > decimal representation, so > An = (1 - 10^(-n))/9 > and now you claim that for some n > (1 - 10^(-n))/9 > eqals K = 1/9. > > If that were true, there must be some n such that 10^(-n) equals 0. Indeed, if you consider 1.000... - 0.999... = 0 and if you claim that all positions are enumerated by natural numbers, then there must be natural number n with 10^(-n) = 0. Hence, this assumption is obviously wrong. > > > > may not, simply cannot occur. We get from (2) > > > A p E n : p = n > > > and by construction of n = 0.111...1 we obtain that all 1's in K > > > coincide with the 1's of n. > > > > Right. But this does *not* prove that there is an n that is equal to K. > > To reformulate clearly > > For all p there is an n such that An[p] = K[p] > > This was not entirely correct: > For all p there is an n0 such that for all n >= n0 An[p] = K[p]. > > this does *not* imply that > > There is an n such that for all p An[p] = K[p], > > which is what you are arguing. > > > > Pray explain how you come (in logical steps) from the first statement > > to the second. > > Note that when we expand the An's with an infinite (countable, mind) sequence > of digits 0 we can do something different. After the decimal point we change > each 0 to 1 and each 1 to 0. We get the following sequence: > A0 = 0.11111111... > A1 = 0.01111111... > A2 = 0.00111111... > and so > K = 0.000000... > Now we see: > For all p there is an n0 such that for all n >= n0 An[p] = K[p]. > and your claim is that from that it follows that > There is an n such that for all p An[p] = K[p], > But as An = 1/(9 . 10^n), this would imply that K (the limit of the sequence) > must be in the list and for some n, An must be 0. > > So, are you now arguing that the limit of a sequence is an element of that > sequence? Or what else? And if so, for what natural number (eh, finite...) > is 1/(9 . 10^n) equal to 0? Either K is in the list (which would contradict analysis in the same way as my original example), then there is an n with 10^(-n) = 0. Or K is not in the list. Then there must be a position which cannot be enumerated by natural numbers (the unary representations of them are now the zeros behind the decimal point). Then this position is undefined, because it cannot be found and, hence, does not exist. Regards, WM
From: mueckenh on 9 Jul 2006 09:11 Virgil schrieb: > What we have said, and what is quite true, is that for every natural > there is a larger natural. Of course. And therefore it is impossible to exhaust all of them or to find a set which is larger than all naturals together. > > > > Right! There are more than any finite number of finite naturals, indeed, > an endless supply of them which collectively form an infinite set. There is no "supply". The elements of a set do exist, instantaneously and immediately. Sets are static. Regards, WM
From: Franziska Neugebauer on 9 Jul 2006 10:56 mueckenh(a)rz.fh-augsburg.de wrote: > The set ordered by size and by natural indices is the limit of my > transpositions. 1. What is a limit of transpositions? 2. How do you prove its existence and uniqueness? F. N. -- xyz
From: Franziska Neugebauer on 9 Jul 2006 11:00
mueckenh(a)rz.fh-augsburg.de wrote: >> After each of your countably many transpositions, there is still a >> countably infinite subset of the rationals left which are not in >> natural order. That is a necesssary consequence of the entire set >> being originally well ordered. > > After each of his countably many exchanges there is still a > countably infinite subset of the list numbers left which are not yet > exchanged. 1. There is no "exchance". 2. d_i def= f (a_ii) is done A i e N. There are no "list numbers left". F. N. -- xyz |