```Date: Mar 10, 2013 5:02 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 222 Back to the roots

On 9 Mrz., 23:53, William Hughes <wpihug...@gmail.com> wrote:> On Mar 9, 5:57 pm, WM <mueck...@rz.fh-augsburg.de> wrote:>> > What do you understand by the not changeable function (d)?>> If x is a potentially infinite sequence of 0's and 1's> then we say (x) is associated to x, if (x) is an algorithm> which given a natural number produces a finite string> of 0's and 1's, such that for every natural number n,> (x) produces the nth FIS of x.Ok. So it is clear that a finite number has to be given and a finitestring is produced.>> We will say x is coFIS to (y) iff>>      i.  We have (x) associated to x and>          (y) associated to y>>      ii.  For every n, (x) and (y) produce the same>           finite string."Every given n" is tantamount to "there is a last given n".This maximum is the same for line l_max and max FIS of d.>> Some comments.>> We can never have more than a limited number of the strings> produced by (x) existing at any time, since we can never> have more that a limited number of natural numbers.Correct.>> We cannot show that x is coFIS to (y) by using (x)> and (y) to produce "all possible strings" and> comparing them, since "all possible natural numbers"> does not exist.  However, we may be able to use induction> to show "For every n, (x) and (y) produce the same finite> string"Induction is fine, but also restricted to the (variable) maximum.>> To show that x is coFIS to (y), it is not enough> to show that  every existing FIS of x, is equal to an> existing FIS of y.More cannot be shown.>> d, the diagonal of the list L, is a potentially infinite> sequence of 0's and 1's with associated (d):> for any natural number n produce> a sequence of n ones.This sequence is identical to a line l_max of the list L, byconstruction of d_max.>> l_k, the k_th line of L is a potentially infinite> sequence of 0's and 1' with associated (l_k):> for any natural number n produce a sequence> of k 1's followed by a sequence of n-k 0's>> A line of L, l_k is findable iff the index> k is findable.>> Do you agree with the statement>>              There does not exist>              (in the sense of not findable)>              a natural number m such that>              the mth line of L is coFIS to>              (d)No. The construction of (d) implies that for every d_k there is a linel_k = d_1, ..., d_k, hence for every d_max there is a line l_max.For reference, the list L is given by11,21,2,3...Regards, WM
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