Timeline for Is $\mathbb{Z}^2$ endowed with the square of the strict order, a lattice-ordered group?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 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 16:51 | comment | added | boumol | @Aaron: As you say in the last remark your characterization is very useful from a conceptual point of view. Let me add some clarification. For arbitrary partial orders it is false that the existence of supremum of two elements also gives the existence of infimum of two elements. However, using the group structure it is possible to prove that this is the case (because the infimum of $x$ and $y$ can be defined as $- \sup \{-x,-y\}$). | |
| Mar 13, 2012 at 16:31 | comment | added | Aaron Tikuisis | @Andreas: You're right (as long as you replace $(2,1)$ and $(1,2)$ by $(3,2)$ and $(2,3)$), this is a more elementary proof. I would like to nonetheless point out that the characterization that I gave can make it easier (at least conceptually) to check if other examples are lattice-ordered. | |
| Mar 13, 2012 at 15:21 | comment | added | boumol | @Aaron: Thanks for the clarification why my previous suggestion is wrong. @Andreas: I am afraid that in the partially ordered group proposed in the question it not true that $(2,1)$ is greater or equal than $(0,1)$ (because the difference is $(2,0)$, and hence not in the proposed positive cone). | |
| Mar 13, 2012 at 14:28 | comment | added | Andreas Blass | Alternatively, you could just check directly that $(0,1)$ and $(1,0)$ have no least upper bound. They have upper bounds $(2,1)$ and $(1,2)$ but no upper bound below both of these. | |
| Mar 13, 2012 at 14:00 | comment | added | Aaron Tikuisis | I don't have sufficient reputation to leave comments, so this comment is out of place. For clarity, I would point out why $(a-1,d-1)$ is not the infimum of $(a,b)$ and $(c,d)$ when $a < c$ and $b > d$: we have, for example, $(a-1,d-2) \leq (a,b), (c,d)$ yet $(a-1,d-2) \not\leq (a-1,d-1)$. | |
| Mar 13, 2012 at 13:55 | history | answered | Aaron Tikuisis | CC BY-SA 3.0 |