Intersection of Subsemigroups/General Result

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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 1

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 $\mathbb S$.

$\blacksquare$


Proof 2

From Set of Subsemigroups forms Complete Lattice:

$\struct {\mathbb S, \subseteq}$ is a complete lattice.

where for every set $\mathbb H$ of subsemigroups of $S$:

the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.

Hence the result, by definition of infimum.

$\blacksquare$


Sources

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