On 22 Nov., 22:38, William Hughes <wpihug...@gmail.com> wrote: > 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. > > For each natural number n, P(n) is true.-
Great! What about the digits in the sequence 1. 12. 123. ...
Of course, this sequence is a little bit more sophisticated than yours. In order to avoid complications with numbers > 9 consider the indices only. Here induction shows: n is not the index of a position with a finite index. According to set theory all digits vanish in the infinite. Set theory is really a theory of spirit. But analysis runs differently. And say, why is the anti-diagonal determined in such a boring way? We could make it vanish completely. Then Cantorian disciples would come to rest.
