Width and height of order extensions

Let $S(X)$ be the semilattice of posets on a set $X$. The height and width functions of $S(X)$ are monotone and antitone respectively. This allows us to place bounds on the height and width of extensions of posets.

Lemma. let $P,E$ be posets with $P \subseteq E$ then $width(E) \subseteq width(P)$.

Proof. let $A$ be an antichain in $E$. Then $\forall x,y \in A : x \not\subseteq y \wedge y \not\subseteq x$. Then let $A$ be the same set embedded in $P$. Suppose $A$ is not an antichain so there exists $x,y : x \subseteq y$ then because $P \subseteq E$ this means that $x \subseteq y$ in $E$ which contradicts the fact that $A$ is an antichain in $E$. So $A$ must be an antichain in $P$. $\square$

Lemma. let $P,E$ be posets with $P \subseteq E$ then $height(P) \subseteq height(E)$.

Proof. let $C$ be a chain in $P$ then $\forall x,y \in C : x \subseteq y \vee y \subseteq x$ in $P$. Then by extension $\forall x,y \in C : x \subseteq y \vee y \subseteq x$ with respect to $E$. It follows that $E$ preserves the chains of $P$. $\square$

These two lemmas immediately lead to the following theorem:

Theorem. let $S(X)$ be the semilattice of posets on a finite set $X$ then:

• $width : S(X) \to \mathbb{N}$ is antitone
• $height : S(X) \to \mathbb{N}$ is monotone

The width function forms a ranking on the semilattice of posets on a set, but not every width class is an antichain. The exceptions to this are the minimal width class consisting of the unique antichain on the set and the maximal width class consisting of all total orders on the set (which has cardinality $n!$ for a finite set $n$).

Total orders have the least width so they are the maximal elements of the semilattice of posets. That is why in dimension theory we study posets by the intersection of total orders. This theorem can now be used to establish bounds on the width of order extensions. For example, suppose we have a J-trivial semigroup extending some other semigroup $S$, then its width must be less then $S$.