Date: Feb 20, 2013 3:39 PM
Subject: Re: Matheology � 222 Back to the roots
WM <firstname.lastname@example.org> 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.
When one has every, one has all.
> > <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.
When one has every one has all.
> After "all natural numbers" no natural number is following.
> After every natural number, there are infinitely many natural numbers
For all naturals numbers, n, there are infinitely many following n.
When something is true for every natural number it is true of all of
The set of every natural number contains all of them.
Lets see WM distinguish between "for all x, f(x)" and "for every x, f(x)