On 17 Okt., 21:06, SPQR <S...@roman.gov> wrote: > In article > <1a1352d6-772d-4e2d-8f9d-7e56635b6...@x7g2000yqn.googlegroups.com>, > > > > > > WM <mueck...@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 naturals are in the left side, then they are also in the bottom. Look here:
1
2 12
3 2 123
4 3 2 1234
continue in infinity. Can there be a bottom side shorter than the left side?
