On Nov 22, 5:22 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > On 22 Nov., 22:09, William Hughes <wpihug...@gmail.com> wrote: > > > > > > > > > > > On Nov 22, 4:54 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > On 22 Nov., 20:22, William Hughes <wpihug...@gmail.com> wrote: > > > > > Note that I was able to handle your > > > > "simple" case using induction. > > > > > I consider the following case easy. > > > > If you disagree maybe you can say > > > > why? > > > > > Consider the sequence of real numbers > > > > > 1.0 > > > > 10.0 > > > > 100.0 > > > > ... > > > > > The limit is oo (unbounded) > > > > > According to set theory, the number of 1's in the limit > > > > is 0. (The limit of the set of positions at which we > > > > have a 1 is the empty set). > > > > Why should the 1 disappear completely? > > > The limit of the set of positions at which we > > have a 1 is the empty set. If there is a 1 it has to > > be a one without a position. > > Iff infinity can be finished.
Nope. We can establish that each integer has the property that it is not the index of a position with a 1 without using completed infinity.
P(n): n is not the index of a position with a 1.
P(1) is true. If P(n) is true then P(n+1) is true.
