Dual of Ordered Semigroup is Ordered Semigroup
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Theorem
Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.
Then its dual $\struct {S, \circ, \succcurlyeq}$ is also an ordered semigroup.
Proof
From Dual Ordering is Ordering, we have that $\struct {S, \succcurlyeq}$ is an ordered set.
We also note from the definition that $\struct {S, \circ}$ is a semigroup.
It remains to be demonstrated that $\succcurlyeq$ is compatible with $\circ$.
Recall that $\struct {S, \circ, \preccurlyeq}$ is an ordered semigroup.
Hence a fortiori $\preccurlyeq$ is compatible with $\circ$.
Let $x, y \in S$ be arbitrary such that $x \succcurlyeq y$.
We have:
\(\ds x\) | \(\succcurlyeq\) | \(\ds y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\preccurlyeq\) | \(\ds x\) | Definition of Dual Ordering | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {y \circ z}\) | \(\preccurlyeq\) | \(\ds \paren {x \circ z}\) | Definition of Relation Compatible with Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ z}\) | \(\succcurlyeq\) | \(\ds \paren {y \circ z}\) |
and similarly:
\(\ds x\) | \(\succcurlyeq\) | \(\ds y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\preccurlyeq\) | \(\ds x\) | Definition of Dual Ordering | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ y}\) | \(\preccurlyeq\) | \(\ds \paren {z \circ x}\) | Definition of Relation Compatible with Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ x}\) | \(\succcurlyeq\) | \(\ds \paren {z \circ y}\) |
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.3$