Intersection of Subsemigroups

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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.