How to determine a set a semiring or a ring?

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

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

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

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

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

My answer:

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

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

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

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

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

However, [math]\mathcal{A}[/math] 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 [math]\mathcal{A}[/math] be closed under set difference. For example, let [math]A = [0,0]][/math] and [math]B = [0,1][/math], then [math]B \setminus A = [0,1] \setminus [0,0] = [0,1][/math], which is not in [math]\mathcal{A}[/math].

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

Edited May 6, 20232 yr by Dhamnekar Win,odd
Latex errors rectifications