> On 31 May 2020, at 10:08, Bruce Kellett <[email protected]> wrote: > > On Sun, May 31, 2020 at 5:21 PM Bruno Marchal <[email protected] > <mailto:[email protected]>> wrote: > On 31 May 2020, at 07:44, Bruce Kellett <[email protected] > <mailto:[email protected]>> wrote: >> On Sun, May 31, 2020 at 2:26 AM Bruno Marchal <[email protected] >> <mailto:[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),”
That was a typo error. You need to read “g applied to k”. Sorry. But you should have just ask “what is g_k?”. I mean it is not a blunder, but a typo error (probably among many). The typo error is detectable from just the few words which go after. > > > 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_k" just has no meaning at all. Nowhere is a sequence g_k defined. > > 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. It is. Read again, avoid the typo error when I apply g on its code k. > > >> 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. It defines g. g(n) = f_n(n) + 1 To compute g, enumerate the f_i up to f_n, and compute f_n on n, and add one. Of course this will lead to a contradiction, which shows that the bijection n —> f_n is not computable. (But that does not destroy CT, it only make the f_n mixed with partial (defined on a subset of N) computable function, and that mowing has to be non computable (if not we can filter out the partial and get a computable enumeration of the f_n, and get again the contradiction). > > So g(n) = f_n(n)+1 is a different function. Different from what? If it is different from all f_n, the language is no more universal. It means that if the language is universal, g is not computable, and what is not computable is not the f_n, nor “+ 1”, so it can only be the enumeration f_n itself. That is the point against conventionalism: computable entails the existence of non computable well defined entities. > It is NOT f_n() with some different argument. So your attempt to make them > the same function is invalid. ? Are you saying that you believe that we can enumerate computably the computable functions from N to N? If that is the case, you have missed the argument. Apology for the typo error. But now that it is has been corrected, you might read more cautiously the argument; I just do not understand you last remark. I will re-explain, if necessary. This post is the minimal think to proceed. It ex^lmains also why the universal dovetailer has to … dovetail, necessarily. Bruno > > 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] > <mailto:[email protected]>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLRDc72tk9MbP20wd_Ch5jxj1x2LSw4bAPaveen8-Dk6Zg%40mail.gmail.com > > <https://groups.google.com/d/msgid/everything-list/CAFxXSLRDc72tk9MbP20wd_Ch5jxj1x2LSw4bAPaveen8-Dk6Zg%40mail.gmail.com?utm_medium=email&utm_source=footer>. -- 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/71B8790D-17DC-46F3-BE38-1AABEEA90A9B%40ulb.ac.be.
