In article <firstname.lastname@example.org>, email@example.com wrote:
> On Wednesday, 12 June 2013 18:06:40 UTC+2, Zeit Geist wrote: > > > > > > And for which k do you fail to well-order by size all rationals > > > > > which have (m + n) < k? > It will fail at no k, all k are natural > > > > > numbers. It will fail for the set of all natural numbers. > > My proof does not concern "the set of all natural numbers" whatever you may > understand by and expect from that phrase. My proof concerns "all natural > numbers" and "all rational numbers" without any exception.
and your "proofs" are mere 'poofs' outside the wild weird world of WMytheology > > If your "sets" are different or more, then I am not interested
Then we are not interested in your wild weird world of WMytheology either. --