"Tim Little" <tim@little-possums.net> wrote in message news:slrni1o6pf.jrj.tim@soprano.little-possums.net... > On 2010-06-18, Virgil <Virgil@home.esc> wrote: >> Given the axiom of choice, as in ZFC, any countable set must be, at >> least theoretically, listable, though such a listing need not be >> computable. > > The definition of countability is the existence of a bijection with a > subset of N. Since N can be well-ordered even in ZF, the theorem > (countable -> listable) is easily proven without AC. > > > - Tim
Well gee, so you say.
I agree that computable reals are countable. But I do not agree this means they can be listed. In fact, I can easily prove they are not. If you give me a purported list of all computable Reals I can use a diagonal argument to form a computable Real not on the list.
Clearly there are countable sets that cannot be listed.
Cantor proved that the Reals cannot be listed. His diagonal proof does *not* show they are uncountable, any more than doing exactly the same proof with computable Reals "proves" they are uncountable.
