On Mon, Jun 1, 2020 at 3:12 AM 'Brent Meeker' via Everything List <
[email protected]> wrote:

> On 5/30/2020 10:44 PM, Bruce Kellett wrote:
>
> On Sun, May 31, 2020 at 2:26 AM Bruno Marchal <[email protected]> wrote:
>
>>
>> Let us write f_n for the function from N to N computed by nth expression.
>>
>> Now, the function g defined by g(n) = f_n(n) + 1 is computable, and is
>> defined on all N. So it is a computable function from N to N. It is
>> computable because it each f_n is computable, “+ 1” is computable, and, vy
>> our hypothesis it get all and only all computable functions from N to N.
>>
>> But then, g has have itself an expression in that universal language, of
>> course. There there is a number k such that g = f_k. OK?
>>
>> But then we get that g_k, applied to k has to give f_k(k), as g = f_k,
>> and f_k(k) + 1, by definition of g.
>>
>
>
> That is a fairly elementary blunder. g_k applied to k, g_k(k) = f_n(k)+1,
> by definition of g_k. You do not get to change the function from f_n to f_k
> in the expression. It is only the argument that changes: in other words,
> f_n(n) becomes f_n(k). So you are talking nonsense.
>
>
> No, I think that's OK.  It's a straight substitution n->k.  The trick is
> that g(n) is not some well defined specific function because n has infinite
> range.  So none of this works in a finite world.  But it's not surprising
> that there is incompleteness in an infinite theory.
>


Yes, I had misunderstood what g(n) was supposed to be -- it is simply a
representation of the diagonal elements of the array, plus 1. But Bruno's
attempt to use the diagonal argument here fails, because  he has to show
that f_n(n)+1 is not contained in the infinite list. He has failed to do
this.

Bruce

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