In article <email@example.com>, "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
> > 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.
Isn't there a missing "not" in that last sentence between "does" and "contain"?