On Sun, May 31, 2020 at 5:21 PM Bruno Marchal <marc...@ulb.ac.be> wrote:
> 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> 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 everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLRDc72tk9MbP20wd_Ch5jxj1x2LSw4bAPaveen8-Dk6Zg%40mail.gmail.com.
