> > Those sets are completely identical. > > If all of the 1's in 0.111... are totally identical, then they must all > have the same lcation locations,
That is wrong. The following two sets are identical although they have not the same location:
{a} {a} Same holds for 0.11 0.11 > > But that is without relevance for the argument. (WH's arguing is > > simply a red herring.) > > > This single sequence of triangles: > > > 1 > > > 1 > > 23 > > > 1 > > 23 > > 456 > > > ... > > > either has a limit triangle, then every side has the same limit, > > namely aleph_0 elements as a maximum or aleph_0 elements as a > > supremum, or none of both. > > It does not have a limit triangle unless one can identify each of the > three necessary vertices of that limit triangle, just as one can > identify each of the three vertices of any of the finite triangles in > your sequence.
Then also the limit of {1,...n} does not exist. The limit has only one end. > > > > > That follows by symmetry - at least in my version of mathematics. > > In my version of mathematics one does not have a triangle without having > 3 non-colinear points as its vertices. > > As far as I can see, the limit of WM's sequence of "triangles" has at > most one identifiable vertex at its top and two identifiable sides, a > vertical left side and a diagonal right side, which in standard math > does not qualify it as anything more then a uni-angle, and certainly not > a tri-angle. > > Now if WM can locate his mythical other two vertices and the mythical > side connecting them, I will reconsider
Consider simply the limits of the three sides, each of which is for any finite step as well defined and located as the terms of the following sequence:
