How to determine a set a semiring or a ring?

[Dhamnekar Win,odd]

Let E be a finite nonempty set and let $\Omega := E^{\mathbb{N}}$ be the set of all E-valued
sequences $\omega = (\omega_n)_{n\in \mathbb{N}}$. For any $\omega_1, \dots,\omega_n \in E$ Let

$[\omega_1, \dots,\omega_n]= \{\omega^, \in \Omega : \omega^,_i = \omega_i  \forall i =1,\dots,n \}$

be the set of all sequences whose first n values are $\omega_1,\dots, \omega_n$. Let $\mathcal{A}_0 =\{\emptyset\}$ for $n\in \mathbb{N}$ define

$\mathcal{A}_n :=\{[\omega_1,\dots,\omega_n] : \omega_1,\dots, \omega_n \in E\}$.

 

Hence  show that $\mathcal{A}= \bigcup_{n=0}^\infty \mathcal{A}_n$ is a semiring but is not a ring if (#E >1).  

My answer:

Let's consider an example where $E = \{0,1\}$ and $\Omega = E^{\mathbb{N}}$ is the set of all E-valued sequences. For any $\omega_1,\dots,\omega_n \in E$, we have

$[\omega_1,\dots,\omega_n] = \{\omega \in \Omega : \omega_i = \omega_i \forall i = 1,\dots,n\}$ which is the set of all sequences whose first $n$ values are $\omega_1,\dots,\omega_n$. Let $\mathcal{A}_0 = \{\emptyset\}$ and for $n \in \mathbb{N}$ define

$\mathcal{A}_n := \{[\omega_1,\dots,\omega_n] : \omega_1,\dots,\omega_n \in E\}.$

Hence $\mathcal{A} = \bigcup_{n=0}^\infty \mathcal{A}_n$ is a semiring but not a ring if # E > 1.

 

To see why $\mathcal{A}$ is a semiring, let's verify that it satisfies the three conditions for a semiring. First, it contains the empty set because $\mathcal{A}_0 = \{\emptyset\}$ and $\mathcal{A} = \bigcup_{n=0}^\infty \mathcal{A}_n$. Second, for any two sets $A,B \in \mathcal{A}$, their difference $B \setminus A$ is a finite union of mutually disjoint sets in $\mathcal{A}$. For example, let $A = [0]$ and $B = [1]$, then $B \setminus A = [1]$, which is in $\mathcal{A}$. Third, $\mathcal{A}$ is closed under intersection. For example, let $A = [0]$ and $B = [1]$, then $A \cap B = \emptyset$, which is in $\mathcal{A}$.

 

However, $\mathcal{A}$ is not a ring because it does not satisfy all three conditions for a ring. Specifically, it does not satisfy condition (ii) for a ring, which requires that $\mathcal{A}$ be closed under set difference. For example, let $A = [0,0]]$ and $B = [0,1]$, then $B \setminus A = [0,1] \setminus [0,0] = [0,1]$, which is not in $\mathcal{A}$.

 

I hope this example helps to illustrate why $\mathcal{A}$ is a semiring but not a ring if the cardinality of $E$ is greater than 1.  Is this answer correct?

Edited May 6, 2023 by Dhamnekar Win,odd
Latex errors rectifications