On 20 Jun., 01:58, Tim Little <t...@little-possums.net> wrote: > On 2010-06-19, Virgil <Vir...@home.esc> wrote: > > > There is, however, some question in my mind about the existence of a > > list of all and ONLY computable reals. > > Why? The computable reals can be proven countable, as you already > know, so there is a bijection between N and the set of countable > reals.
That is not true. An exclusive list need not exist.
The reals in a certain Cantor-list are countable. And if you form the anti-diagonal of that list and add it (for instance at first position) to the list, this new list is also countable. Again consttruct the anti-diagonal and add it to the list. Continue. The situation remains the same for all anti-diagonals you might want to construct. Therefore all reals constructed in that way are countable. Nevertheless it is impossible to put all of them in one list, because there would be another resulting anti-diagonal.
Conclusion: It is impossible to obtain a bijection of all these reals with N although all "these" reals are countable with no doubt.