Date: Mar 10, 2013 7:08 AM
Subject: Re: Matheology § 222 Back to the roots
On 10 Mrz., 11:12, William Hughes <wpihug...@gmail.com> wrote:
> On Mar 10, 10:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 10 Mrz., 10:28, William Hughes <wpihug...@gmail.com> wrote:
> > > On Mar 10, 10:02 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > On 9 Mrz., 23:53, William Hughes <wpihug...@gmail.com> wrote:
> > > > > We will say x is coFIS to (y) iff
> > > > > i. We have (x) associated to x and
> > > > > (y) associated to y
> > > > > ii. For every n, (x) and (y) produce the same
> > > > > finite string.
> > > > "Every given n" is tantamount to "there is a last given n".
> > > I do not talk about "every given n" but about "every n"
> > > (this means from 1 to n for every n). Note that
> > > "there is a last n" but it is not a findable natural
> > > number.
> > Better say "a not fixable natural number".
> > > Note that you do not need the x_n to exist to say
> > > something about them.
> > That is true. For instance we can say that a natural number is either
> > even or odd.
> > > For example, you can say no
> > > x_n that will ever exist will be equal to 0.
> > Yes, that is another example.
> > > If you say x is coFIS to y you are saying something
> > > about x_n and y_n that may not exist at this time.
> > That is the hardest problem. There are some properties which can be
> > determined (like the examples above). There are other which cannot.
> > But in every case we know that there is a line of the list that is
> > identical with the FIS of d, both existing or not existing yet.
> However this is not a findable line.-
It is not a fixable line, say.
But here we have the case where we can know properties of not (yet)
existing entities. We know that every natural number (without 0) is
positive. In a similar way we know that for every line of the list
there is a FIS of the diagonal and vice versa. That's why I refuse to
accept that the diagonal is longer than every line.
We can also say, the diagonal is coFIS with the list. But that would
not change anything, because the list has always a finite maximal line
(which has *never* more elements than can be counted by a natural
number). You would like to state that the diagonal has more FISs than
any natural number can count. That is not possible. The resaon is that
the list has no actually infinite line which is correct even when
accepting actual infinity. And just this fact contradictis actual