Intersection of Submonoids with Monoid Identity is Submonoid
Theorem
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $I$ be an indexing set.
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of submonoids of $S$.
For each $S_\alpha \in \family {S_\alpha}_{\alpha \mathop \in I}$, let $e_S \in S_\alpha$.
Let $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ denote the intersection of $\family {S_\alpha}$
Then $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is a submonoid of $S$.
Proof
First we show that $\struct {\ds \bigcap_{\alpha \mathop \in I} S_\alpha, \circ}$ is a semigroup:
Semigroup Axiom $\text S 0$: Closure
\(\ds a, b\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} S_\alpha\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds a, b\) | \(\in\) | \(\ds S_\alpha\) | Definition of Intersection of Family | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds a \circ b\) | \(\in\) | \(\ds S_\alpha\) | Semigroup Axiom $\text S 0$: Closure for all $S_\alpha$ | |||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ b\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} S_\alpha\) |
$\Box$
Semigroup Axiom $\text S 1$: Associativity
From the above we have that $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is closed under $\circ$.
From Restriction of Associative Operation is Associative we have that $\circ$ is associative on $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$.
Hence we have that $\struct {\ds \bigcap_{\alpha \mathop \in I} S_\alpha, \circ}$ is a semigroup.
$\Box$
Identity Element
We are given that:
- $\forall \alpha \in I: e_S \in S_\alpha$
\(\ds \forall a \in S: \, \) | \(\ds a \circ e_S\) | \(=\) | \(\ds e_S\) | Definition of Identity Element | ||||||||||
\(\ds \forall \alpha \in I: \, \) | \(\ds e_S\) | \(\in\) | \(\ds S_\alpha\) | by hypothesis | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \forall a \in S_\alpha: \, \) | \(\ds a \circ e_S\) | \(=\) | \(\ds e_S\) | by hypothesis |
That is, for all $S_\alpha$, $e_S$ is the identity element of $S_\alpha$
Thus for all $S_\alpha$, $\struct {S_\alpha, \circ}$ is a monoid.
$\blacksquare$
Sources
- 1999: J.C. Rosales and P.A. García-Sánchez: Finitely Generated Commutative Monoids ... (previous) ... (next): Chapter $1$: Basic Definitions and Results