Timeline for Is $\mathbb{Z}^2$ endowed with the square of the strict order, a lattice-ordered group?

8 events

when toggle format	what		by	license	comment
Feb 8, 2020 at 18:02	history	edited	YCor	CC BY-SA 4.0	edited tags, added definition
Feb 8, 2020 at 17:57	history	edited	YCor		edited tags
Mar 25, 2012 at 0:48	vote	accept	Rajnish
Mar 25, 2012 at 0:47	vote	accept	Rajnish
Mar 25, 2012 at 0:47
Mar 13, 2012 at 15:25	comment	added	boumol		[Updated comment (after Aaron answer)] Let us call P to your proposal for positive cone. Since P is closed under addition in $\mathbb{Z}^2$ and it contains the neutral element, it is obvious that we have a partially ordered group under the following order definition: $x \leq y$ iff $x - y \in P$. Thus, it is enough to check that this order is a lattice, i.e., every two elements have an infimum and a supremum. I thought this could be straightforwardly checked, but as Aaron says in his answer this is not the case; e.g., there is no infimum of $(a,b)$ and $(c,d)$ when $a<c$ and $b>d$.
Mar 13, 2012 at 13:55	answer	added	Aaron Tikuisis		timeline score: 4
Mar 13, 2012 at 1:00	history	edited	David Roberts♦	CC BY-SA 3.0	Edited notation/cleaned LaTeX
Mar 13, 2012 at 0:56	history	asked	Rajnish	CC BY-SA 3.0