Hasse Diagram of Dual Ordering

Theorem

Let $D$ be the Hasse diagram of an ordering $\RR$.

Then the Hasse diagram $D'$ of the dual ordering $\RR^{-1}$ of $\RR$ is obtained by turning $D$ upside down.

Proof

Let $\tuple {x, y}$ be an ordered pair in $\RR$ which is represented by a line in $D$ joining $x$ and $y$ such that $x$ is below $y$.

Then $x \mathrel \RR y$ such that:

$\not \exists z: x \mathrel \RR z \land \mathrel \RR y$

As $\RR^{-1}$ is the dual of $\RR$, we have that $y \mathrel {\RR^{-1} } x$ such that:

$\not \exists z: y \mathrel {\RR^{-1} } z \land \mathrel {\RR^{-1} }$

and so there will be a line in $D'$ joining $x$ and $y$ such that $x$ is above $y$.

The result follows.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.2$