Intersection of Subsemigroups

Theorem
Let $\left({S, \circ}\right)$ be a semigroup.

Let $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$ be subsemigroups of $\left({S, \circ}\right)$.

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

If $\left({T, \circ}\right)$ is that intersection of $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$, it follows that $\left({T, \circ}\right)$ is also a subsemigroup of both $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$.

General Result
Let $\mathbb S$ be a set of subsemigroups of $\left({S, \circ}\right)$, where $\mathbb S \ne \varnothing$.

The intersection $\bigcap \mathbb S$ of the members of $\mathbb S$ is itself a subsemigroup of $\left({S, \circ}\right)$.

Also, $\bigcap \mathbb S$ is the largest subsemigroup of $\left({S, \circ}\right)$ contained in each member of $\mathbb S$.

Proof
Let $T = T_1 \cap T_2$ where $T_1, T_2$ are subsemigroups of $\left({S, \circ}\right)$. Then:

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

The other results follow from this and Intersection Subset.

Proof of General Result
Let $T = \bigcap \mathbb S$.

Let $T_k$ be any element of $\mathbb S$. Then:

So $\left({T, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$.


 * Now to show that $\left({T, \circ}\right)$ 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 any $K \in \mathbb S: K \subseteq T$ and thus $T$ is the largest subsemigroup of $S$ contained in each member of $\mathbb S$.