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