An uncountable countable set
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From: Franziska Neugebauer on 9 Jul 2006 11:16 mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: >> In article <J21Gtn.BCI(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> >> writes: >> [...] >> 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. K is not in the list. > Then there must be a position which cannot be enumerated by natural > numbers non sequitur. All positions are indexed by definition of decimal representation. > (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. These positions are not undefined, they do not exist. By definition of decimal representation. F. N. -- xyz
From: Franziska Neugebauer on 9 Jul 2006 11:21 mueckenh(a)rz.fh-augsburg.de wrote: > 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. What exactly does "larger than all naturals together" mean? >> 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. Sudden insight? F. N. -- xyz
From: Virgil on 9 Jul 2006 14:14 In article <1152449428.459811.14210(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > Also without bound n is always finite, or it would not be a natural > > > number! > > > > Ok, good, so your sequence of transpositions will never give the standard > > order of the rationals (as you claimed). > > This is correct, but it is correct too in case of Cantor's list, even > if it is completed in zero time. "Mueckenh" seems to think that the lists themselves are Cantor's, but only the rules for finding elements not in those lists are Cantor's, and once such a rule is completely stated, it applies completely to any and all such lists forever. > > There *is* a limit, as I wrote already: > > The set ordered by size and by natural indices is the limit of my > transpositions. Then it is a limit both unachieved and unachievable. > The idea that Cantors diagonal was defined as a nunber accrding to > Cauchy AND that each digit was well distinguished from the digit of the > corresponding list-number is a wish, an imagination, a fairy tale. Only in "mueckenh"'s fairyland. In real life any one of the uncountably many "Cantor diagonal numbers" for a given list is as real a number as any real number in the list.
From: Virgil on 9 Jul 2006 14:18 In article <1152449701.757688.240470(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > But it is in the list, because this list > is the complete list of all natural numbers. What has this to do with the listability of the reals? "Mueckenh" is off in his own private world again.
From: Virgil on 9 Jul 2006 14:25
In article <1152449799.247262.45300(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Cantor has nowhere shown that his diagonal ever covers the whole list. More to the point, Cantor HAS shown that the whole list does not include any of his uncountably many "diagonals". > > > > 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. In Cantor's 'diagonal" proof, no numbers are transposed. It is merely shown that a rule of construction can be stated such that for an arbitrary list of reals and an arbitrary member of that list the constructed number is not equal to the selected number. From which one may deduce that a complete listing of reals is not possible. |