Power Set with Intersection and Subset Relation is Ordered Semigroup
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Theorem
Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \subseteq}$ be the ordered structure formed from the set intersection operation and subset relation.
Then $\struct {\powerset S, \cap, \subseteq}$ is an ordered semigroup.
Proof
From Power Set with Intersection is Commutative Monoid, $\struct {\powerset S, \cap}$ is a fortiori a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
It remains to be shown that $\subseteq$ is compatible with $\cap$.
Let $A, B \subseteq S$ be arbitrary such that $A \subseteq B$.
Thus:
\(\ds A\) | \(\subseteq\) | \(\ds B\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall T \subseteq S: \, \) | \(\ds A \cap T\) | \(\subseteq\) | \(\ds B \cap T\) | Set Intersection Preserves Subsets: Corollary | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall T \subseteq S: \, \) | \(\ds T \cap A\) | \(\subseteq\) | \(\ds T \cap B\) | Intersection 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 $15.8$