Intersection of Subsemigroups

Theorem

Let $\struct {S, \circ}$ be a semigroup.

Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be subsemigroups of $\struct {S, \circ}$.

Then the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a subsemigroup of that $\struct {S, \circ}$.

General Result

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

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Suppose $\struct {S, \circ}$ is a semigroup where $S$ is the empty set.

Suppose $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ are subsemigroups of $\struct {S, \circ}$.

Then it follows that $T_1$ and $T_2$ are both empty.

Since $S$, $T_1$ and $T_2$ are empty, by Intersection with Empty Set, it follows that the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is equal to $\struct {S, \circ}$.

Hence, by Semigroup is Subsemigroup of Itself, it follows that the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a subsemigroup of that $\struct {S, \circ}$.

Now suppose $\struct {S, \circ}$ is a semigroup, where $S$ is non-empty.

Let $T = T_1 \cap T_2$ where $T_1, T_2$ are subsemigroups of $\struct {S, \circ}$. Then:

\(\ds \)

\(\)

\(\ds a, b \in T\)

\(\ds \)

\(\leadsto\)

\(\ds a, b \in T_1 \land a, b \in T_2\)

Definition of Set Intersection

\(\ds \)

\(\leadsto\)

\(\ds a \circ b \in T_1 \land a \circ b \in T_2\)

Subsemigroups are closed

\(\ds \)

\(\leadsto\)

\(\ds a \circ b \in T\)

Definition of Set Intersection

Thus $\struct {T, \circ}$ is closed, and is therefore a semigroup from the Subsemigroup Closure Test.

$\blacksquare$