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 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. 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. > -- 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/CAFxXSLTO1wbdNLtXPKvxyfB--g-muHdq98TP5tCeAk0aF8H%3Dhw%40mail.gmail.com.
