> On 31 May 2020, at 07:44, Bruce Kellett <bhkellet...@gmail.com> wrote:
>
> On Sun, May 31, 2020 at 2:26 AM Bruno Marchal <marc...@ulb.ac.be
> <mailto:marc...@ulb.ac.be>> 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
What is g_k? The only enumeration here is the f_k, then we have define a
precise, single, function g such that
g(n) = f_n(n) + 1. (f_n(n) is the diagonal term, you can see this by making
the table (the infinite matrice) with the number in the top row, and the f_i in
a column):
0 1 2 3 ...
f_0 f_0(0) f_0(1) f_0(2) f_0(3)
f_1 f_1(0) f_1(1) f_1(2) f_1(3)
f_2 f_2(0) f_2(1) f_2(2) f_2(3)
…
Here the underlining means “+1”.
> applied to k, g_k(k) = f_n(k)+1,
There are no g_k.
g is the function defined by diagonalisation. g(x) = f_x(x) + 1, that g(0) =
f_0(0) + 1, g(1) = f_1(1) + 1, g(2) = f_2(2) + 1, ...
> by definition of g_k.
The only enumeration was the enumeration of the functions f_k
> You do not get to change the function from f_n to f_k in the expression.
We do.
> It is only the argument that changes: in other words, f_n(n) becomes f_n(k).
This makes no sense. What is g(2) ? f_n(2) + 1 ? What is n then?
> So you are talking nonsense.
You miss the diagonal. Read again.
Bruno
>
> Bruce
>
> So f_k(k) = f_k(k) +1.
>
> And f_k(k) has to be number, given that we were enumerating functions from N
> to N. So we can subtract f_k(k) on both sides, and we get 0 = 1. CQFD.
>
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