Disjoint union of posets

Remarks

If $P$ and $Q$ are not disjoint, first replace them by disjoint copies, such as $P \times \set{0}$ and $Q \times \set{1}$. The same construction works for any family of pairwise disjoint posets. Example: if $P$ is the chain $a < b$ and $Q$ is the chain $c < d$, then $P + Q$ is the poset $\mathbf{2} + \mathbf{2}$ with exactly the strict relations $a < b$ and $c < d$. Non-example: the poset with $a < b$, $c < d$ and also $a < d$ is not $P + Q$, since it relates an element of $P$ to an element of $Q$. An antichain with $n$ elements is the disjoint union of $n$ one-element posets. The width of $P + Q$ is the sum of the widths, and its height is the larger of the two heights.