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 finite string 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, by construction 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 line l_k = d_1, ..., d_k, hence for every d_max there is a line l_max.
