Let $(P,\leq)$ be a finit poset, $w$ its width. Then, by Dilworth's theorem, I know that $P$ is a union of $w$ chains.
My question is: is there also a relation between the order dimension of a poset and its width?
Thanks very much for the help. Also if there are any results that some how disciplines the order dimension of a poset based on any order relation's properties it would be of great help.
Dilworth also proved that the order dimension of a (finite) poset is less than or equal to its width. This was in the same paper as the result you quoted:
(By the way, the same goes for infinite posets, provided the width is finite.)
Here is an outline of the proof. Suppose the (finite) poset $P$ has width $n.$ Then $P$ is the union of $n$ disjoint chains $C_1,\dots,C_n.$ (This is the tricky part but you already know it.) For each $i\in\{1,\dots,n\},$ extend the given partial order to a total order $\le_i$ so that,if $x,y$ are incomparable in $P$ and $y\in C_i,$ then $x\le_iy.$ (Exercise.) Then the given partial order is the intersection of the $n$ total orders $\le_1,\dots,\le_n.$