Intersection of Subsemigroups/General Result/Proof 1
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Theorem
Let $\struct {S, \circ}$ be a semigroup.
Let $\mathbb H$ be a set of subsemigroups of $\struct {S, \circ}$, where $\mathbb H \ne \O$.
Then the intersection $\bigcap \mathbb H$ of the members of $\mathbb H$ is the largest subsemigroup of $\struct {S, \circ}$ contained in each member of $\mathbb H$.
Proof
Let $T = \bigcap \mathbb S$.
Then:
\(\ds a, b\) | \(\in\) | \(\ds T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall K \in \mathbb S: \, \) | \(\ds a, b\) | \(\in\) | \(\ds K\) | Definition of Set Intersection | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall K \in \mathbb S: \, \) | \(\ds a \circ b\) | \(\in\) | \(\ds K\) | Subsemigroups are closed | |||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ b\) | \(\in\) | \(\ds T\) | Definition of Set Intersection |
So by Subsemigroup Closure Test, $\struct {T, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Now to show that $\struct {T, \circ}$ is the largest such subsemigroup.
Let $x, y \in T$.
Then $\forall K \subseteq T: x \circ y \in K \implies x \circ y \in T$.
Thus $\forall K \in \mathbb S: K \subseteq T$.
Thus $T$ is the largest subsemigroup of $S$ contained in each member of $\mathbb S$.
$\blacksquare$