dimension of a poset

Let $P$ be a finite poset and $ℛ$ be the family of all realizers of $P$. The dimension of $P$, written $\dim (P)$, is the cardinality of a member $E\in ℛ$ with the smallest cardinality. In other words, the dimension $n$ of $P$ is the least number of linear extensions $L_1,\mathrm{\dots },L_n$ of $P$ such that $P=L_1\cap \mathrm{\cdots }\cap L_n$. ($E$ can be chosen to be $\{L_1,\mathrm{\dots },L_n\}$).

If $P$ is a chain, then $\dim (P)=1$. The converse is clearly true too. An example of a poset with dimension 2 is an antichain with at least $2$ elements. For if $P=\{a_1,\mathrm{\dots },a_m\}$ is an antichain, then one way to linearly extend $P$ is to simply put $a_i\le a_j$ iff $i\le j$. Called this extension $L_1$. Another way to order $P$ is to reverse $L_1$, by $a_i\le a_j$ iff $j\le i$. Call this $L_2$. Note that $L_1$ and $L_2$ are duals of each other. Let $L=L_1\cap L_2$. As both $L_1$ and $L_2$ are linear extensions of $P$, $P\subseteq L$. On the other hand, if $(a_i,a_j)\in L$, then $a_i\le a_j$ in both $L_1$ and $L_2$, so that $i\le j$ and $j\le i$, or $i=j$ and whence $a_i=a_j$, which implies $(a_i,a_j)=(a_i,a_i)\in P$. $L\subseteq P$ and thus $\dim (P)=2$.

Remark. Let $P$ be a finite poset. A theorem of Dushnik and Miller states that the smallest $n$ such that $P$ can be embedded in ${ℝ}^n$, considered as the $n$-fold product of posets, or chains of real numbers $ℝ$, is the dimension of $P$.