In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 27 Nov., 19:21, William Hughes <wpihug...@gmail.com> wrote: > > On Nov 27, 2:11 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 27 Nov., 18:42, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Nov 27, 4:37 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > <snip> > > > > > > > According to set theory the limit set is empty. > > > > > > Indeed. > > > > > Nice to hear that. > > > > > > And the proof of this does not depend > > > > on "actual infinity". > > > > > Of course it does, because every digits moves to the right side. > > > > Stick with > > > > 1 > > 10 > > 100 > > .... > > > > The 1 moves to the left. > > > > We do not have to check the positions one by one. > > It is easy to show by induction (which requires only > > potential infinity) that for each n in |N > > the nth position is not occupied. > > Correct. Every position is unoccupied, but that does not mean that all > positions are unoccupied: It is as easy to show by induction that > there are never aleph_0 zeros right to the 1.
But there is no position to the left of the radix point which in the limit has not become zero. If there were, even someone as thick as WM could name it, but even he cannot. > > > This is equivalent to > > saying that the set of all possible occupied positions is empty.- > > Only if the mistake is made to consider infinity as finished, every as > synonymous to all, namely to consider all finite positions occupied by > zeros.
Name one which is not! > > But as I mentioned already, this example has nothing at all to do with > my proof.
Just as WM's proof, as usual, has nothing to do with reality. --