In article <7d59f93c-4bc8-43e2-9b40-485733a01305@z8g2000yqz.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> 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.
Equal cardinality of two sets requires, by definition, a bijection between them.
The naturals are, by definition of COUNTABLE cardinality so that any set of equally countable cardinality must have a bijection with the naturals.
Cantor has shown that the set of all reals cannot biject with the naturals, ergo, the set of all reals is NOT of countable cardinality.
Actually Cantor's original "diagonal" proof was based on the set of all functions from the naturals to a two element set, but several people have constructed bijections between that set of functions and the set of all reals. I have done it myself.
