|
|
Re: Matheology § 203
Posted:
Feb 3, 2013 2:51 AM
|
|
On 3 Feb., 00:22, William Hughes <wpihug...@gmail.com> wrote:
> On Feb 2, 11:58 pm, William Hughes <wpihug...@gmail.com> wrote:
>
>
>
>
>
> > On Feb 2, 11:42 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 2 Feb., 23:36, William Hughes <wpihug...@gmail.com> wrote:
>
> > > > On Feb 2, 11:15 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > > On 2 Feb., 20:11, William Hughes <wpihug...@gmail.com> wrote:
>
> > > > > > > > > > Can a potentially infinite list
> > > > > > > > > > of potentially infinite 0/1
> > > > > > > > > > sequences have the property that
> > > > > > > > > > if s is a potentially infinite 0/1
> > > > > > > > > > sequence, then s is a line of L
>
> > > > <snip>
> > > > > For every s: There is alsways a list that contains the first n bits of
> > > > > s.
>
> > > > Is there a single line which contains s
> > > > Yes or no
>
> > <snip>
>
> > > There is no complete s.
>
> > Then the answer is no
>
> Indeed, since there is no single line, l,
> such that every initial segment of s is contained
> in l, we do not even have to talk about complete s.-
In fact we can say that in a suitable list "every" initial segment of
s is contained in some line, since there is no s(n) = (s1, s2, ...,
sn) missing. But there is no sensible way of saying "all" initial
segment. (There is no last line in the Binary Tree.)
The same holds for lists of natural numbers and rational numbers.
Therefore there is no difference concerning "countability".
Thank you for this discussion it has also helped me to clear thoughts.
Regards, WM
|
|
