On Sun, May 31, 2020 at 5:21 PM Bruno Marchal <[email protected]> wrote:
> On 31 May 2020, at 07:44, Bruce Kellett <[email protected]> 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 > > > What is g_k? > That is your notation: "But then we get that g_k, applied to k has to give f_k(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. > You defined g_k!! g_k applied to k is f_k(k), and that is your error. 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, ... > But that is not what you said before. > 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? > n is the number of the function in the ordered list of all functions from N to N. Adding 1 to f_n(n) gives a different function. Diagonalization does not help you here. So g(n) = f_n(n)+1 is a different function. It is NOT f_n() with some different argument. So your attempt to make them the same function is invalid. Bruce So you are talking nonsense. > > > You miss the diagonal. Read again. > > Bruno > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLRDc72tk9MbP20wd_Ch5jxj1x2LSw4bAPaveen8-Dk6Zg%40mail.gmail.com.
