On 10 Mrz., 20:20, William Hughes <wpihug...@gmail.com> wrote:
> > > So do you agree with the statement. > > > > If G is a set of lines of L with a findable > > > last element, then there is no line s of > > > G such that s is coFIS to (d) > > > Yes. How often will you ask? > > (d) is a prescription to find or to construct FIS d_1, ..., d_n. > > > Would you expect that > > "write 0. and then add the digit 1 with no end" is coFIS with a line > > of > > 0.1 > > 0.11 > > 0.111 > > ... > > No, the other way round.
There is no way. This is a sequence of less than 10 words: "write 0. and then add the digit 1 with no end". It is not coFIS with any line of the list. But it defines the lines of the list. > > Recall > > 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.
(x) and (y), if describing infinite sequences, are phrases of few words. They are probably not coFIS. > > The statement x is coFIS to (y) means approximately > that x and the potentially infinite sequence described > by (y) are COFIS. > > Do you agree with the statement > > For every n, the nth FIS of x is > contained in g iff > g is coFIS to (x)
Let us stay in the concrete example: L is the list
1 1,2 1,2,3 ... 1,2,3,...,max
and d is the diagonal 1,2,3,...,max.
For every n, the nth FIS of d is contained in the list L and, therefore, in the last, unfixable and unfindable, line 1,2,3,...,max. Since the last line and the sequence 1,2,3,...,max described by (d) are identical, every line l_n = 1,2,3,...,n of the list is contained in the last line and in the sequence described by d.
Is that a sufficient answer to your question? If not, don't hesitate to ask. But I would be glad, if you could stay with our example L and d.