"Tim Little" <tim@little-possums.net> wrote in message news:slrni1m606.jrj.tim@soprano.little-possums.net... > On 2010-06-18, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> Give me a list of all computable Reals. > > I specified such a list previously in this thread. It is not a > computable list, but you didn't ask for a computable list - just a > list of computable reals. They are not the same thing. >
Cantor said that given a list of Reals, he can generate a Real not on the list.
P Webb says that given a list of computable Reals, he can form a computable Real not on the list.
The proof is identical.
But yet computable numbers are countable.
> >> I will use Cantor's diagonal argument to explicitly construct a Real >> which isn't on the list. This is diagonal number clearly computable > > In this case it clearly is not computable.
Give me a list which you claim contains all computable numbers, and I will explicitly compute a missing number.
> > >> as there is an explicit construction for it. > > No, there is only a finite algorithm *relative to* being provided with > the list initially.
Yes, exactly as in Cantor's proof that the Reals cannot be listed.
> There is no algorithm to produce the antidiagonal > starting without input,
Yes, exactly as per Cantor's proof that the Reals cannot be listed.
> which being computable would require. >
Cantor's proof requires the list to be specified in advance. So does mine. It is exactly the same proof.
And this is deliberate. The whole point is to demonstrate that Cantor's proof does *not* show the Reals are uncountable; it shows they cannot be listed. This only works because I am using *exactly* the same form of proof.
> Come back when you read what the term "computable number" means. >
You give me a list of computable numbers.
Every number on that list is computable.
Therefore the nth digit of every Real on the list is computable.
