On 15 Jun., 18:53, stevendaryl3...@yahoo.com (Daryl McCullough) wrote: > WM says... > > > > >On 15 Jun., 16:17, stevendaryl3...@yahoo.com (Daryl McCullough) wrote: > > >> In this sense, the antidiagonal of the list of all computable reals > >> is definable (but not computable). > > >That is nonsense. To define means to let someone know the defined. If > >he knows it, then he can compute it. > > That's just not true.
It is just true.
> For example, we can define a real r as follows: > > r = sum from n=0 to infinity of H(n) 2^{-n} > > where H(n) = 1 if Turing machine number n halts on input n, > H(n) = 0 otherwise. > > That's definable, but it is not computable.
Anyhow it is not a definition. It would be more useful to define some number by the legs of a crowd of unicorns touching the ground at a given time.
Nevertheless your "definition" belongs to a countable set, hence it is no example to save Cantors "proof".
Either all entries of the lines of the list are defined and the diagonal is defined (in the same language) too. Then the proof shows that the countable set of defined reals is uncountable. Or it does not show anything at all.
To switch "languages" is the most lame argument one could think of. The diagonal argument does not switch languages. And it cannot be applied at all because the list of all finite defiitions does not contain infinite entries. Those however are required for the diagonal argument.
