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 $\struct {S, \circ}$.

Proof
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:

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