On 22 Jun., 21:05, "Mike Terry" <news.dead.person.sto...@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.
If we use Cantor's definiton of "countable", then the set {..., An, ...A0, L0(0), L0(1), ...L0(n),...} is uncountable.
If we use the definition that a subset of a countable set is countable, 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.
> 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. 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. 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.
