Let $(A,+)$ be some abelian group that we totally order, the definition can be done in two ways which are equivalent.
Definition 1
Let $\preceq$ be a given endorelation on $A$ such that
- $\preceq$ is transitive
- $\preceq$ is reflexive
- $\preceq$ is antisymmetric
- For $x,y\in A$, either $x \preceq y$ or $y \preceq x$
- $\preceq$ is compatible with $+$, that is $x\preceq y \implies x+z\preceq y+z$
Definition 2 Let $P\subset A$ be given such that
- $P+P\subset A$
- $P\cap-P=\emptyset$
- $P \cup-P=A-0$
These two definitions can be shown to be equivalent which is not an issue to me, I was however thinking about an ordering that is dense. For the first definition we have it being simple to state, namely if $x\preceq y$ then there exist a $z-\{x,y\}$ such that $x\preceq z \preceq y$. I was thinking however how would one state it for the later one in a nice succint manner? I was thinking using cosets, that is for $x,y\in A$ with $y-x\in P$ we have that $$(P-x)\cap(y-P)\neq\emptyset$$ would suffice, as we can make $\preceq$ from the second by having $y-x\in P\implies x\preceq y$, which gives us from the definiton of dense in the first case that there exist a $z\in A$ such that $z-x\in P$ and $y-z\in P$. However I feel that this does not necciserily hold as after all, we are not garantueed that $z-x=y-z$ as that would imply $x=y$ which is not the case most of the time. So that the intersection is non-emtpy may not hold, what would be proper manner to define the property of being dense using the language of the second definition?
Edit: After some additional work I think I have found the answer, while cosets are not entirely correct, as $P$ is not a subgroup, the general idea is correct. My line of reasoning is this, we are given that $x\preceq z$ and $z\preceq y$ is the property we are looking for, with $z$ being neither $x$ nor $y$. That means we have $z-x\in P$ and $y-z\in P$ which we can rewrite to be $z\in x+P$ and $-z\in -y+p$ or $z\in y-P$. This gives us then that $z$ is in both those "cosets" and in turn we have $$y-P\cap x+P\neq\emptyset$$ Is this correct?
A very simple way to define density in terms of $P$ is that the ordering is dense iff $P+P=P$. First, if the ordering is dense and $x\succ0$, then there exists $y$ such that $0\prec y\prec x$ and then $x=y+(x-y)$ is a sum of two positive elements, so $P+P=P$. Conversely, if $P+P=P$ and $x\prec y$, then we can write $y-x=a+b$ where $a,b\in P$, and then $x\prec x+a\prec y$.