A graph $G$ with vertex set $V$ has $\dim(G) \leq d$ if and only if there exists a sequence $<_{1},<_{2}, \ldots , <_{d}$ of total orders on $V$ satisfying the following conditions:
- the intersection of $<_{1},<_{2}, \ldots , <_{d}$ is empty;
- for each edge $\{x,y\}$ and each vertex $z \notin \{x,y\}$ of $G$, there is at least one order $<_{i}$ in the sequence such that $x<_{i}z$ and $y<_{i}z$. A $d$-dimensional representation on a set $V$ is a sequence of $d$ total orders on $V$ that satisfies (1). The dual relation $<_k^*$ is defined as $ x <_k^* y \iff x <_k y \land y <_i x $ for $k\neq i$.
I have the following proposition:
For a $3$-dimensional representation $<_1,<_2,<_3$ on a set $V$, it holds: $$\{ (x,y) : \{x,y\}\in E(G) \land x <_k^*y\}=\{(x,y) : y = \min_{<_k} \{v \in V: x<_k^* v\}\}.$$
I do not see the equality right away, but I think that using (1) and (2) should help. Can anyone point me in the right direction to understand the equality?