> 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.
Yes. If you provide a list of Reals, then the diagonal is computable and does not appear on the list.
> Then the proof shows > that the countable set of defined reals is uncountable.
No, it shows that all "definable" (computable) Reals cannot be explicitly listed. This is *not* the same as being uncountable.
> Or it does not > show anything at all. >
It shows that all "definable" (computable) Reals cannot be explicitly listed. This is a well known proof in set theory. This is *not* the same as being uncountable.
> 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. >
No, that paragraph above is close to gibberish. Cantor said and proved that any purported list of all Reals cannot contain all Reals. His proof is simple and clear, provides an explicit construction of at least one missing Real, and does contain or require any concepts of uncomputable numbers, or use of the Axiom of Choice.
Perhaps if you were to identify the step in Cantor's proof that you consider wrong, then we might gain some idea as to what you are actually objecting to?
