Power Set with Union and Superset Relation is Ordered Semigroup

Theorem

Let $S$ be a set and let $\powerset S$ be its power set.

Let $\struct {\powerset S, \cup, \supseteq}$ be the ordered structure formed from the set union operation and superset relation.

Then $\struct {\powerset S, \cup, \supseteq}$ is an ordered semigroup.

Proof

From Power Set with Union is Commutative Monoid, $\struct {\powerset S, \cup}$ is a fortiori a semigroup.

From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.

We have that $\supseteq$ is the dual to $\subseteq$.

Hence $\struct {\powerset S, \supseteq}$ is an ordered set.

It remains to be shown that $\supseteq$ is compatible with $\cup$.

Let $A, B \subseteq S$ be arbitrary such that $A \supseteq B$.

Thus:

\(\ds A\)

\(\supseteq\)

\(\ds B\)

\(\ds \leadsto \ \ \)

\(\ds \forall T \subseteq S: \, \)

\(\ds A \cup T\)

\(\supseteq\)

\(\ds B \cup T\)

Set Union Preserves Subsets: Corollary

\(\ds \leadsto \ \ \)

\(\ds \forall T \subseteq S: \, \)

\(\ds T \cup A\)

\(\supseteq\)

\(\ds T \cup B\)

Union is Commutative

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$