How to determine a set is semiring or ring?

[Win_odd Dhamnekar]

Let E be a finite nonempty set and let [imath]\Omega := E^{\mathbb{N}}[/imath] be the set of all E-valued
sequences [imath] \omega = (\omega_n)_{n\in \mathbb{N}}[/imath]. For any [imath] \omega_1, \dots,\omega_n \in E [/imath] 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 [imath]\omega_1,\dots, \omega_n[/imath]. Let [imath]\mathcal{A}_0 =\{ \emptyset \}[/imath] for [imath]n \in \mathbb{N}[/imath] define [math]\mathcal{A}_n :=\{[\omega_1,\dots,\omega_n] : \omega_1,\dots, \omega_n \in E\}[/math]. Hence [imath]\mathcal{A}= \bigcup_{n=0}^\infty \mathcal{A}_n[/imath] is a semiring but is not a ring if (#E >1).

My attempt to answer this question:


A class of sets [imath]A \subset 2^\Omega[/imath] is called a ring if the following three conditions hold: [imath](i) \emptyset \in \mathcal{A}.[/imath] (ii) [imath]\mathcal{A}[/imath] is [imath]\setminus[/imath]-closed. (iii) [imath]\mathcal{A}[/imath] is [imath]\cup[/imath]-closed.

A class of sets [imath]\mathcal{A} \subset 2^\Omega[/imath] is called a semiring if (i) [imath]\emptyset \in \mathcal{A},[/imath] (ii) for any two sets [imath]A,B \in \mathcal{A}[/imath] the difference set [imath]B \setminus A[/imath] is a finite union of mutually disjoint sets in [imath]\mathcal{A}[/imath], (iii)[imath] \mathcal{A}[/imath] is [imath]\cap[/imath]-closed.

Now, How did the author say [imath]\mathcal{A}= \bigcup_{n=0}^\infty \mathcal{A}_n[/imath] is a semiring but not a ring if (#E >1)

 

[Win_odd Dhamnekar]

Let E be a finite nonempty set and let [imath]\Omega := E^{\mathbb{N}}[/imath] be the set of all E-valued
sequences [imath] \omega = (\omega_n)_{n\in \mathbb{N}}[/imath]. For any [imath] \omega_1, \dots,\omega_n \in E [/imath] 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 [imath]\omega_1,\dots, \omega_n[/imath]. Let [imath]\mathcal{A}_0 =\{ \emptyset \}[/imath] for [imath]n \in \mathbb{N}[/imath] define [math]\mathcal{A}_n :=\{[\omega_1,\dots,\omega_n] : \omega_1,\dots, \omega_n \in E\}[/math]. Hence [imath]\mathcal{A}= \bigcup_{n=0}^\infty \mathcal{A}_n[/imath] is a semiring but is not a ring if (#E >1).

My attempt to answer this question:


A class of sets [imath]A \subset 2^\Omega[/imath] is called a ring if the following three conditions hold: [imath](i) \emptyset \in \mathcal{A}.[/imath] (ii) [imath]\mathcal{A}[/imath] is [imath]\setminus[/imath]-closed. (iii) [imath]\mathcal{A}[/imath] is [imath]\cup[/imath]-closed.

A class of sets [imath]\mathcal{A} \subset 2^\Omega[/imath] is called a semiring if (i) [imath]\emptyset \in \mathcal{A},[/imath] (ii) for any two sets [imath]A,B \in \mathcal{A}[/imath] the difference set [imath]B \setminus A[/imath] is a finite union of mutually disjoint sets in [imath]\mathcal{A}[/imath], (iii)[imath] \mathcal{A}[/imath] is [imath]\cap[/imath]-closed.

Now, How did the author say [imath]\mathcal{A}= \bigcup_{n=0}^\infty \mathcal{A}_n[/imath] is a semiring but not a ring if (#E >1)

My answer:
Let's consider an example where [imath]E = \{0,1\}[/imath] and [imath]\Omega = E^{\mathbb{N}}[/imath] is the set of all [imath]E[/imath]-valued sequences. For any [imath]\omega_1,\dots,\omega_n \in E[/imath], 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 [imath]n[/imath] values are [imath]\omega_1,\dots,\omega_n[/imath]. Let [imath]\mathcal{A}_0 = \{\emptyset\}[/imath] and for [imath]n \in \mathbb{N}[/imath] define [math]\mathcal{A}_n := \{[\omega_1,\dots,\omega_n] : \omega_1,\dots,\omega_n \in E\}.[/math] Hence [imath]\mathcal{A} = \bigcup_{n=0}^\infty \mathcal{A}_n[/imath] is a semiring but not a ring if [imath]|E| > 1[/imath].

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

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

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