Re: sketch of PROOF OF COUNTABLE REALS

rupertmccallum_at_yahoo.com
Date: 02/02/05 

Date: 1 Feb 2005 23:01:15 -0800

|-|erc wrote:
> <rupertmccallum@yahoo.com> wrote in message
> > To construct the anti-diagonal for the list of all computable
numbers,
> > you need to pick out the Turing machines that halt for every digit.
> > This cannot be done by a Turing machine. Hence the anti-diagonal is
not
> > computable.
> >
>
> This is proof that nobody here reads what they are replying to.
>
> "If the real is non-computable, then "
>
>
> 1/ But it is representable
>

I thought you were putting forth the list of computable reals as your
countable list of all real numbers. What list are you putting forward?
What does "representable" mean?

> 2/ That is not the problem of the people putting forward a countable
list.
>
> If the list has incomplete digits or not it still maps to every real.
> It is the onus of the people asserting diagnoalisation disproves that
assertion
> to find some rigging in which diagonalisation works.
>
> You can, just take the matrix at 'some' scale of completion,
substitute an 11th digit
> for NULL and the diagonalisation argument is viable at any stage.
>
> The claim is
>
> reals U sequences_with_blanks = computable list --- UTM(n ranges over
N, digit) mod 10
>
> We don't need a real list, we only need to verify the membership
relation is possible here :
>
> real e {UTM(index, digit)} = [T | F]
>
> As long as that is decidable we have the functionality of a real
list.
>
> Herc

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