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 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 [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLTO1wbdNLtXPKvxyfB--g-muHdq98TP5tCeAk0aF8H%3Dhw%40mail.gmail.com.
