Date: Mar 10, 2013 3:39 PM
Subject: Re: Matheology § 222 Back to the roots

On 10 Mrz., 20:20, William Hughes <> 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


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

Regards, WM