In article <1a1352d6-772d-4e2d-8f9d-7e56635b6604@x7g2000yqn.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 16 Okt., 22:33, SPQR <S...@roman.gov> wrote: > > In article > > <6c294e30-0b1e-40d0-879a-70cc3a550...@u13g2000vbx.googlegroups.com>, > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 16 Okt., 01:49, Virgil <vir...@ligriv.com> wrote: > > > > > > Since a square must have 4 vertices as well as 4 sides, where does one > > > > find more that one vertex for the "infinite square"? > > > > > Since a completed linear set must have two ends, where does one find > > > the end of 1, 2, 3, ... ? > > > > > Hope, you see the similarity. > > > > But there is no requirement that those "ends" be members of the set. > > Open intervals are "complete" sets but do not contain their endpoints. > > The set of rationals, {1/n: n in N}, is quite complete without > > containing 0. > > And the right side of the triangle > > 0. > 0.1 > 0.11 > 0.111 > ... > > is quite complete without containing 1/9 - as complete as the bootom > side.
What "bootom side" when all finite lines are present?
When all finite lines are present, there is no "bootom side" and only one vertex, at the top, and the whole is no longer a triangle at all.
