>Hmmm. But you cannot provide me with a list of all Computable numbers, can >you? Which is what Cantor's diagonal proof requires.
Cantor's proof assumes the *existence* of such a list. It doesn't assume that you know how to compute it.
>Of course there exists a mapping from N to computable numbers. >But Cantor's proof requires more than that; it requires the >mapping to be recursively enumerable such that we can also >explicitly list them.
No, it doesn't. If f is any mapping from N to computable numbers, then antidiag(f) is a new real that is not in the image of f. It doesn't matter whether f is recursively enumerable or not.
>That a set cannot be listed is not the same as it is uncountable,
I would say it this way: That a set is not recursively enumerable does not imply that it is not enumerable.
The set of computable reals is enumerable, but not recursively enumerable. The set of *all* reals is not enumerable.
>You cannot give me a list of all Computable numbers, >because I can use a diagonal construction to form a >computable Real not on the list.
If the list is computable, that's correct. If the list is not computable, then that's not correct.
>Cantor's proof applied to Reals does *not* prove they are >uncountable;
It certainly does. If you assume that the set of reals is countable, then that means that there is a bijection f from the naturals to the reals. Cantor's proof shows that there is no such bijection.
>> If that doesn't answer your question, you'll have to clarify what you >> mean by it. >> > >What I mean is that Cantor proved you cannot provide a list of all Reals.
He proved that there is no bijection between the naturals and the reals. By definition, that means that the reals are uncountable.