On 28 Nov., 00:03, William Hughes <wpihug...@gmail.com> wrote: > On Nov 27, 5:17 pm, WM <mueck...@rz.fh-augsburg.de> wrote:> On 27 Nov., 21:48, William Hughes <wpihug...@gmail.com> wrote: > > <snip> > > > > > but that does not mean that all > > > > positions are unoccupied: > > > > But this is not what I am saying. > > This is an important point. Clearly "each n in |N has > property P" does not mean "all n in |N have property P" > if we assume that "all n in |N" might not exist. > However from "each n in |N has property P" we can > conclude that no n in |N has property (not P).
But even this seemingly innocent conclusion can lead to contradictions. Compare the sequence: > > 01. > > 0.1 > > 010.1 > > 01.01 > > 0101.01 > > 010.101 > > 01010.101 > > 0101.0101 > > ... No term has more digits left than right. This includes, according to analysis (potential infinity = always finite although not limited), the limit. > > > > Compare > > > > Each natural number >1 is either prime or composite. > > > The set of natural numbers >1 that are neither prime > > > nor composite is empty. > > > Yes, but here we have a property that is a property of the set like: > > Every natural number is a natural number. > > Induction works for any property P. E.g Each natural number > is a natural number can be proved by induction. > > 1 is a natural number. > If n is a natural number n+1 is a natural number > Each n in |N is a natural number.
Induction proves also: Every set of natural numbers is finite. Why do you overlook this simple proof? > > One gets the same "contradictions" if one uses potential > infinity rather than actual infinity.
No. Only by actual infinity one is forced to accept that the limit of my sequence has no digits (neither 1 nor 0) left of the point and thus is less than 1.