```Date: Feb 20, 2013 11:44 AM
Author: William Hughes
Subject: Re: Matheology § 222 Back to the roots

On Feb 20, 5:29 pm, WM <mueck...@rz.fh-augsburg.de> wrote:> On 20 Feb., 13:31, William Hughes <wpihug...@gmail.com> wrote:>> > On Feb 20, 12:37 pm, WM <mueck...@rz.fh-augsburg.de> wrote:>> > > On 20 Feb., 11:30, William Hughes <wpihug...@gmail.com> wrote:> > <snip>> > > > A  statement you can make is that there> > > > is no line of the list with the property> > > > that it is coFIS to d.> > > > (you do not need every line or every FIS to "actually> > > > exist" to make this statement)>> > > You need all FISs of d to make this statement.>> > No, for each line of L you only need some> > of the FISs.  For every line you need every> > not all.>> Then you have a statement for finitely many lines and none for> infinitely many lines.>>>>>>>>>>>> > <snip>>> > > > Let z be a potentially infinite sequence such that> > > > for some natural number m, the mth FIS of> > > > z contains a zero.>> > > > Are z and x coFIS?>> > > No.>> > Is the following statement true>> > For every natural number n we have>> >     the (n+1)st FIS of the nth line> >     of L contains a 0.>> Of course.Is the statementFor every natural number n we have    the nth line of L and x    are not coFIStrue?
```