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. > > > But let's assume it. > > > According to analysis the number of 1's in the limit is 1 and the > > number of zeros left to the point is infinite. > > Nonsense. There is no such thing as a number with an > infinite number of zeros left of the decimal point.
Infinity is certainly not a number with a finite number of zeros left of the decimal point. In analysis oo has more digits than any finite number. Do you know the definition: For every n in |N : n < oo. > > According to analysis the sequence grows without bound.-
Correct. For every finite number of digits you find another digit. That means there are more digits than any finite number of digits. But the first one, namely the 1, certainly never disappears. (And not even in set theory, because according to set theory there are infinitely many *finite* positions in the limit. Now you try to rob the 1 of its finite place? Do you think that the limit has more than the aleph_0 finite indexes?)