On 2010-06-20, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim@little-possums.net> wrote in message >> Where is the phrase "provided in advance" used in the proof? > > Because Cantor's proof requires us to know the nth digit of the nth > item on the list.
No, it just requires that the list L (a mathematical object) be of the type that an n'th digit of the n'th item *exists*.
> Indeed Cantor's proof starts with a list
No, Cantor's proof starts with a conditional assumption that some mathematical object *is* a list.
> If you don't what is on the list already, you cannot possibly prove > that some Real is missing.
Now it looks like your deficiency might be in knowing what a proof in predicate logic is. You don't need a separate proof for each list, each one based on the contents of that list. You have a single proof that there does not exist a mathematical entity satisfying the definition of a surjective mapping from natural numbers to reals.
> My proof is *exactly* the same, except for inserting the word > "computable" in front of Real.
And therefore deficient as making that textual substitution leaves a gap between Cantor's proof that antidiag(L) is a real, and your substituted claim that antidiag(L) is computable real.
> Give me a list of all computable numbers in the same form as > Cantor's proof requires a purported list of all Reals, ie a list > where the nth digit of the nth term is known.
The constructive portion Cantor's proof requires nothing more than a mapping from N to R. In particular, it does not require that the n'th digit of the n'th term in that mapping "is known". That requirement is entirely due to your imagination.
> So by definition if the list contains only computable Reals, we know > the nth digit of each term.
Careful with your quantifiers! For a list of computable reals, the following is true:
for all n, there exists Turing machine T such that T(n) computes the n'th digit of L(n),
but the following is false:
there exists Turing machine T such that for all n, T(n) computes the n'th digit of L(n).
