Re: sketch of PROOF OF COUNTABLE REALS

From: |-|erc (h_at_r.c)
Date: 02/02/05 

Date: Wed, 2 Feb 2005 12:33:01 +1000

<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

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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