On Mon, Jun 1, 2020 at 9:05 AM 'Brent Meeker' via Everything List < everything-list@googlegroups.com> wrote:
> On 5/31/2020 3:49 PM, Bruce Kellett wrote: > > On Mon, Jun 1, 2020 at 8:31 AM 'Brent Meeker' via Everything List < > everything-list@googlegroups.com> wrote: > >> On 5/31/2020 3:23 PM, Bruce Kellett wrote: >> >> On Mon, Jun 1, 2020 at 3:12 AM 'Brent Meeker' via Everything List < >> everything-list@googlegroups.com> wrote: >> >>> On 5/30/2020 10:44 PM, Bruce Kellett 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 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. >>> >>> >>> No, I think that's OK. It's a straight substitution n->k. The trick is >>> that g(n) is not some well defined specific function because n has infinite >>> range. So none of this works in a finite world. But it's not surprising >>> that there is incompleteness in an infinite theory. >>> >> >> >> Yes, I had misunderstood what g(n) was supposed to be -- it is simply a >> representation of the diagonal elements of the array, plus 1. But Bruno's >> attempt to use the diagonal argument here fails, because he has to show >> that f_n(n)+1 is not contained in the infinite list. He has failed to do >> this. >> >> >> All computable functions are in the list ex hypothesi. >> > > > That is what the diagonal argument is all about: you hypothesize that all > bit strings (for example) are in your infinite list. Then you flip the > diagonal bit of each string and form a new string from all the diagonal > elements. And lo, that new string is not in the initial list. Therefore > your hypothesis that all bit strings are in the list is disproven. > > Bruno has attempted toride to glory on this argument, and has failed > miserably! > > > That's a general problem with reductio arguments. When you get to end you > don't know which premise was wrong. Bruno, isn't changing the hypothetical > list though, so he's saying the premise that you can order the total > functions is wrong. You can order the functions (say lexigraphically) but > you can't know which are total. > > ISTM the result, that there's an incompleteness theorem for the set of all > functions, is quite intuitive. But Bruno seems to be saying this is all > finitist because he doesn't assum and axiom of infinity. Yet the > "diagonalization" doesn't work in a finite world. > Take all bit strings of length N (finite) and apply the diagonal argument. The string resulting from putting all the flipped diagonal bits together is not in the original list, contradicting the assumption that the list is complete. Of course, the list of all strings of length N contains more than N elements, so the diagonal argument does not apply. The set of all strings of infinite length is certainly infinite, so one might work the diagonal argument there -- if one doesn't worry too much about cardinality issues...... I think Bruno should rephrase his argument -- it might be sensible, but as presented it was clearly invalid. Bruce -- 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/CAFxXSLRM0SwFN3ujhJ7Av0MAV5aR98Gma4uA6NWg-5D8DP%2B7RA%40mail.gmail.com.
