On 5/31/2020 3:49 PM, Bruce Kellett wrote:
On Mon, Jun 1, 2020 at 8:31 AM 'Brent Meeker' via Everything List
<everything-list@googlegroups.com
<mailto:everything-list@googlegroups.com>> wrote:
On 5/31/2020 3:23 PM, Bruce Kellett wrote:
On Mon, Jun 1, 2020 at 3:12 AM 'Brent Meeker' via Everything List
<everything-list@googlegroups.com
<mailto:everything-list@googlegroups.com>> wrote:
On 5/30/2020 10:44 PM, Bruce Kellett 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 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.
All computable functions are in the list ex hypothesi.
That is what the diagonal argument is all about: you hypothesize that
all bit strings (for example) are in your infinite list. Then you flip
the diagonal bit of each string and form a new string from all the
diagonal elements. And lo, that new string is not in the initial list.
Therefore your hypothesis that all bit strings are in the list is
disproven.
Bruno has attempted toride to glory on this argument, and has failed
miserably!
That's a general problem with reductio arguments. When you get to end
you don't know which premise was wrong. Bruno, isn't changing the
hypothetical list though, so he's saying the premise that you can order
the total functions is wrong. You can order the functions (say
lexigraphically) but you can't know which are total.
ISTM the result, that there's an incompleteness theorem for the set of
all functions, is quite intuitive. But Bruno seems to be saying this is
all finitist because he doesn't assum and axiom of infinity. Yet the
"diagonalization" doesn't work in a finite world.
Brent
Bruce
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