On 15 Feb., 23:58, William Hughes <wpihug...@gmail.com> wrote: > On Feb 15, 10:30 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 15 Feb., 00:44, William Hughes <wpihug...@gmail.com> wrote: > > > > > > Two potentially infinite sequences x and y are > > > > > equal iff for every natural number n, the > > > > > nth FIS of x is equal to the nth FIS of y > > > > So we note that it makes perfect sense to ask > > > if potentially infinite sequences x and y are equal, > > > and to answer that they can be equal if they are actually infinite. > > But this answer does not make sense. > > You cannot prove equality without having an end, a q.e.d.. > > A very strange statement. Anyway there is no reason to > claim equality. Let us define the term coFIS > > Two potentially infinite sequences x and y are said to be > coFIS iff for every natural number n, the > nth FIS of x is equal to the nth FIS of y. >
So for every natural number the list 1 12 123 ... is coFIS with its diagonal.
> We note that it makes perfect sense to ask > if potentially infinite sequences x and y are coFIS,
up to every n. Correct. And that is the case in the list.
> we have cases where they are not coFIS and cases > where they are coFIS.. We also note that no > concept of completed is needed, so coFIS can > be demonstrated by induction. In particular, you > do not need a last element to prove that x and y > are coFIS. > > So WMs statements are > > there is a line l such that d and l > are coFIS
Of course, for every n there is a line 1, 2, 3, ..., n that is coFIS to the diagonal 1, 2, 3, ..., n. And there is not more than every n.
> > there is no line l such that d and l > are coFIS
That would only be true if there was an n larger than every n, either in a line or in the diagonal. But that would not only require actual infinity but also the diagonal surpassing every line or some line surpassing the diagonal. Both is impossible by construction.
Do you know an n of the diagonal that is not in a line? Or vice versa?