On 2010-06-20, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: > Cantor's proof requires the purported list of all Reals to be > provided.
That is your fantasy, not part of Cantor's proof. Cantor proved a closed-form proposition, which means that there are no free variables that can be "required to be provided".
> No idea what you are talking about.
Then perhaps you should read a reasonable reference on predicate logic and mathematical proof.
> Perhaps if you could explain how my proof differs from Cantor's > proof, that would be a start.
I have done already, but perhaps I can explain it in different terms for you.
> You seem to accept Cantor's proof, but not mine. Yet they are almost > identical, the only difference being I have inserted the word > "computable" in front of "reals". > > I can't see how you can accept one and not the other.
Cantor's proof shows that for any mapping L:N->R, antidiag(L) is real and not in range(L). Cantor's proof does include a proof that antidiag(L) is real. It does not show that antidiag(L) is a computable real.
You cannot just drop "computable" in there and suppose that the logic works, just as you cannot drop "rational" in there and suppose that the logic works.
If you want to prove that for any mapping L:N->R, antidiag(L) is computable, you need to include a proof that antidiag(L) satisfies the mathematical definition for a computable real.
> It is extremely easy to compute. > > Take the nth decimal place of the nth term.
The nth decimal place of the nth term of what? The list L? That's fine if you permit infinite lists of reals as input to the algorithm, but the definition of "computable real" permits no such thing.
> I am using Cantor's proof exactly, except for inserting the word > "computable" before every use of the word "real". That is the whole > point.
That doesn't work because Cantor's work does not include any proof that antidiag(L) is computable: only that it is real. Hence there is a gaping hole in the validity of your modified text.
You will not be able to fill that hole, because there are well-defined functions L for which antidiag(L) is *not* a computable real. There are even explicitly-definable such lists, and furthermore there are such lists where there are only two values in the range: for example, let L(n) = 0 if the n'th digit of Omega is 0, L(n) = 1/9 otherwise.
Both members of range(L) are very obviously computable - they're even rational! But the decimal antidiagonal is not computable.