How to determine if a set is a semiring or a ring?

[WMDhamnekar]

Let E be a finite nonempty set and let $\Omega := E^{\mathbb{N}}$be the set of all E-valued
sequences $\omega = (\omega_n)_{n\in \mathbb{N}}F$or 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?

 

[bpet]

In order to determine if the set is a semiring you first need to specify exactly what are the addition and multiplication operations.

 

[andrewkirk]

In order to determine if the set is a semiring you first need to specify exactly what are the addition and multiplication operations.

It refers to a semi-ring of sets, defined here, not a semi-ring in the algebraic sense.

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 $\mathcal{A}$.

That is not correct. [0,1] is indeed an element of $\mathcal A$ so this example does not demonstrate non-closure. That example was never going to work because you are subtracting from a set another set that has no overlap with it, so the subtraction doesn't change the set. I don't want to give too much away so I'll keep the hint general. Why not instead try taking the union of two sets to give a set that can't be expressed as any [a, b, ..., c]? Since the whole structure is based on fixing the first n components of a sequence, try choosing two sets that are the same in their first component ($\omega_1$), but differ in their second ($\omega_2$). That should give you a set of all sequences with a fixed second component but arbitrary first component, and that might not correspond to any equivalence class in $\mathcal A$.