From: Sylvia Else on 22 Jun 2010 20:03 On 23/06/2010 6:38 AM, WM wrote: > On 22 Jun., 21:34, Virgil<Vir...(a)home.esc> wrote: > >>>> But (A0, A1, A2, ...) is obviously countable. Above you say it's "certainly >>>> not countable", but it is. >> >>> The set is certainly countable. But it cannot be written as a list >> >> But it HAS been written as a list (A0, A1, A2, ...), > > Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, > L0)? I can't see that you've proved that it doesn't. All you've proved is that it doesn't contain the anti-diagonal of (A0, A2, A2, ...). But that means nothing of significance, because your construction doesn't require or imply that it would so contain. Sylvia.
From: Virgil on 22 Jun 2010 20:56 In article <103cc56f-746e-482f-9315-0c9e8eaf45e1(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Jun., 21:05, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > > > Ln = > > > > > An > > > ... > > > A2 > > > A1 > > > A0 > > > L0 > > > > > where L0 is thelist of all rationals. > > > > > This list Ln contains a countable set of numbers > > > > ..correct, {An, ...A0, L0(0), L0(1), ...L0(n),...} > > is obviously countable. [L0(k) is the k'th element in list L0] > > > > > but the set of its > > > diagonals is not listable, because An is not in the list. > > > > The "set of its diagonals" = {An}. A list has just one diagonal. Every set > > of one element is listable. Like Sylvia I must be misunderstanding what you > > mean. (But I'm not misunderstanding what you actually say. :-) > > To spell it out clearly: The set of all diagonals (including or > exluding all rationals - that does not matter) > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > that are constrcuted according to my prescription cannot be listed > although it is countable. Except in WM's world, a set is listable (can be a bijective image of the naturals) if and only if it is countable (there is abijection with the naturals). WM is very careful NOT to give his own definition of listable and countable in which there are countable sets which are not listable. > > If we use Cantor's definiton of "countable", then the set > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > is uncountable. Cantor's definition says a set is countable if there is a surjection from the naturals onto that set. But such a surjection IS a list. > > If we use the definition that a subset of a countable set is > countable That is not a definition but a theorem, at least if countable includes all finite sets. , then the set > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > is countable. > > > > > > > > > This is wrong. An obvious listing is (A0, A1, ...) > > The set > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > cannot be listed. You just listed it, since every-double-open-ended list can easily be single-open-end listed. > > > > > there exists a countable set M, such that > > If L is a Cantor-list, then > > (anti-?)diagonal of L belongs to M. > > > > That is so obviously false that its banal. > > No it is not. It is outside of WM's world, such as in FOL+ZFC. > If there exists a Cantor-list, i.e., that what Cantor > really understood by the term list, then it is a list of *defined* > reals. Again, this holds only in WM's world, and not in such standard set theories as FOL+ZFC. And then its anti-diagonal is a defined real too. Then exists a > countable set M, namely the set of all defined reals, that is > countable. Nevertheless it cannot be listed. If it cannot be listed, how does WM know that it satisfies any standard definition of countability? Answer: He doesn't and it doesn't.
From: Virgil on 22 Jun 2010 21:03 In article <2000e81b-7c5a-41be-b6af-98f96f2fb630(a)w31g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Jun., 21:34, Virgil <Vir...(a)home.esc> wrote: > > > > > But (A0, A1, A2, ...) is obviously countable. Above you say it's > > > > "certainly > > > > not countable", but it is. > > > > > The set is certainly countable. But it cannot be written as a list > > > > But it HAS been written as a list (A0, A1, A2, ...), > > Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, > L0)? Since (..., An, ... A2, A1, A0, L0) does not appear to be a list, why should there be any antidiagonal for it? On the other hand, if (A0, A1, A2, ...) is a list of reals, then it will have an antidiagonal. >Then Cantor's argument fails. Does it not? Then a list can not > list all anti-diagonals that belong to the countabe set constructed in > my argument. Outside of WM's world a recursive construction such as he suggests does allow an anti-diagonal to any list infinite anti-diagonals.
From: Sylvia Else on 22 Jun 2010 21:33 On 23/06/2010 11:03 AM, Virgil wrote: > In article > <2000e81b-7c5a-41be-b6af-98f96f2fb630(a)w31g2000yqb.googlegroups.com>, > WM<mueckenh(a)rz.fh-augsburg.de> wrote: > >> On 22 Jun., 21:34, Virgil<Vir...(a)home.esc> wrote: >> >>>>> But (A0, A1, A2, ...) is obviously countable. Above you say it's >>>>> "certainly >>>>> not countable", but it is. >>> >>>> The set is certainly countable. But it cannot be written as a list >>> >>> But it HAS been written as a list (A0, A1, A2, ...), >> >> Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, >> L0)? > > Since (..., An, ... A2, A1, A0, L0) does not appear to be a list, why > should there be any antidiagonal for it? Ach! Let's scrap A0 - it's confusing. If we let L_n be the nth element in the list L0, and An the anti-diagonal of the An-1, An-2,...., A1, L_1, L_2, L_3,... then L_1 A1 L_2 A2 L_3 A3 L_4 .... is a list. I'm still thinking about that. Sylvia.
From: Sylvia Else on 22 Jun 2010 22:18
On 23/06/2010 11:33 AM, Sylvia Else wrote: > On 23/06/2010 11:03 AM, Virgil wrote: >> In article >> <2000e81b-7c5a-41be-b6af-98f96f2fb630(a)w31g2000yqb.googlegroups.com>, >> WM<mueckenh(a)rz.fh-augsburg.de> wrote: >> >>> On 22 Jun., 21:34, Virgil<Vir...(a)home.esc> wrote: >>> >>>>>> But (A0, A1, A2, ...) is obviously countable. Above you say it's >>>>>> "certainly >>>>>> not countable", but it is. >>>> >>>>> The set is certainly countable. But it cannot be written as a list >>>> >>>> But it HAS been written as a list (A0, A1, A2, ...), >>> >>> Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, >>> L0)? >> >> Since (..., An, ... A2, A1, A0, L0) does not appear to be a list, why >> should there be any antidiagonal for it? > > Ach! Let's scrap A0 - it's confusing. > > If we let L_n be the nth element in the list L0, and An the > anti-diagonal of the An-1, An-2,...., A1, L_1, L_2, L_3,... > > then > > L_1 > A1 > L_2 > A2 > L_3 > A3 > L_4 > ... > > is a list. I'm still thinking about that. > > Sylvia. Hmm... A1 is the antidiagonal of L1 L2 L3... A2 is the antidiagonal of L1 A1 L2 L3... A3 is the antidiagonal of L1 A1 L2 A2 L3 L4... Each An is thus constructed from a list that is different from the list into which it is inserted. So the construction does not lead to a list that should contain its own anti-diagonal, and it doesn't. Sylvia. |