Intersection of Subsemigroups/General Result

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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 $T = \bigcap \mathbb S$.

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

\(\displaystyle \) \(\) \(\displaystyle a, b \in T\)
\(\displaystyle \) \(\implies\) \(\displaystyle \forall k: a, b \in T_k\) Definition of Set Intersection
\(\displaystyle \) \(\implies\) \(\displaystyle \forall k: a \circ b \in T_k\) Subsemigroups are closed
\(\displaystyle \) \(\implies\) \(\displaystyle a \circ b \in T\) Definition of Set Intersection


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 $\mathbb S$.

$\blacksquare$


Sources