Date: Feb 20, 2013 11:29 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 222 Back to the roots

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. Similarly we have "for every natural number there are
infinitely many FIS of infinitely many lines that do not contain a 0".

Again: Do not confuse every and all.
After "all natural numbers" no natural number is following.
After every natural number, there are infinitely many natural numbers
following.

Regards, WM