Date: Mar 3, 2013 5:22 AM
Subject: Re: Matheology § 222 Back to the roots
On 3 Mrz., 00:08, William Hughes <wpihug...@gmail.com> wrote:
> On Mar 2, 11:10 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 2 Mrz., 19:24, William Hughes <wpihug...@gmail.com> wrote:
> > > On Mar 2, 6:27 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > We both agree that there is a natural number
> > > valued function of time, m(t), such that
> > > at any time t, m(t) is the index of an existing
> > > line which contains all existing FIS of d.
> > > We each believe that our m(t) is not constant.
> > > We also agree that there does not exist
> > > (in the sense of not able to find) a
> > > natural number n such that the
> > > nth line of L is coFIS with the
> > > diagonal.
> > > I find your characterization of this
> > > situation as "there is a natural
> > > number m such that the mth line
> > > of L is coFIS with the diagonal"
> > since there do not exist more than m FIS of the diagonal.
> > > to be silly.
> > Because you do not yet fully understand potential infinity: There do
> > not exist more than m FIS of the diagonal.
> Oh, I understand all right. It is just that I think
> calling m (which cannot be a findable
> natural number and behaves exactly like m(t))
> a natural number is silly.
> > Question: Do you find your characterization of the situation in
> > finished infinity not silly? Don't you see a mathematical
> > contradiction of the sentence: There are all FIS of d in the list but
> > not in one single line?
> Not at all. Clearly
> there are all FIS of d in one single line
> iff there is a last line.
> I do not consider the sentence
> "There is no last line"
> to be a contradiction.-
But "there are all FIS" is not a contradiction, if they must be in
more than one line of a list that contains only lines that contain
everything that is in all preceding lines?