# Dimension of a partially ordered set

Let $P = (X,{\le})$ be a partially ordered set. A linear extension $L$ of $P$ is a total ordering $\le_1$ (see also Totally ordered set) on $X$ such that $\mathrm{id} : P \rightarrow L$ is order preserving, i.e. such that $x \le y$ in $P$ implies $x \le_1 y$ in $L$. The existence of linear extensions was proved by E. Szpilrain, [a4], in 1931.
The dimension of a poset $P$ is the least $d$ for which there exists a family of linear extensions $L_1=(X,{\le_1}),\ldots,L_d=(X,{\le_d})$ such that $$x \le y \Leftrightarrow x \le_1 y,\ldots,x \le_d y \ .$$
A chain in $P$ is a partially ordered subset of $P$ (a sub-poset) that is totally ordered. An anti-chain in $P$ is a set of elements $x_1,\ldots,x_n$ such that no $x_i, x_j$, $i \ne j$, are comparable. The height of $P$ is the maximal length of a chain. The width of $P$ is the maximal size of an anti-chain. The Dilworth decomposition theorem says that if $\mathrm{width}(P) = w$, then $P$ decomposes as the disjoint sum of $d$ chains (cf. Disjoint sum of partially ordered sets; cf. also (the editorial comments to) Partially ordered set). Using this, T. Hiraguchi [a2] proved $$\mathrm{dim}(P) \le \mathrm{width}(P) \ .$$ The concept of dimension was introduced by B. Dushnik and E.W. Miler [a1].