# 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 $\mathbb S \ne \varnothing$.

Then 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:

 $\displaystyle$  $\displaystyle a, b \in T$ $\displaystyle$ $\implies$ $\displaystyle a, b \in T_1 \land a, b \in T_2$ Definition of Set Intersection $\displaystyle$ $\implies$ $\displaystyle a \circ b \in T_1 \land a \circ b \in T_2$ Subsemigroups are closed $\displaystyle$ $\implies$ $\displaystyle a \circ b \in T$ Definition of Set Intersection

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

$\blacksquare$

The other results follow from this and Intersection is Subset.