On 5/30/2020 10:44 PM, Bruce Kellett wrote:
On Sun, May 31, 2020 at 2:26 AM Bruno Marchal <[email protected]
<mailto:[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.
Brent
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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