"Tim Little" <email@example.com> wrote in message news:firstname.lastname@example.org... > On 2010-06-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> Which of these two statements do you agree with: >> >> 1. You cannot form a list of computable Reals. >> >> 2. The computable Reals are countable. > > Answering the same questions for the fourth time now, 1 is false, 2 is > true. >
OK, how about you give me a list of computable Reals - in exactly the same format as Cantor requires, ie each item in the list can be evaluated to any desired degree of accuracy - and I will produce a Real not on the list.
> >> You have failed to explain why Cantor's diagonal proof "proves" the >> Reals are uncountable, but the same proof applied to computable >> Reals does *not* prove the Computable Reals are uncountable. > > I have, multiple times now. The antidiagonal of a list of computable > reals is not necessarily a computable real.
yes it is.
There is a very simple, explicit construction that will calculate the anti-diagonal number to any required degree of accuracy.
Take the list, determine the nth digit of the nth item, blah blah. Remember, just as in Cantor's proof, you have to provide the list first.
If you don't believe me, give me a list of all computable Reals in the same format as Cantor uses in his proof, and I will produce a computable number not on the list.
> Your insistence that it > must be is almost certainly because you don't know what a computable > real is. (There are other possibilities, even less flattering to your > math skills) > > In particular, the antidiagonal of a list of all computable reals is > never a computable real. >
I am still waiting to see your list of all computable Reals. In the same format as Cantor used in his proof; after all, I want the two proofs to be identical other than one considers all Reals and the other only computable Reals.