- #1

WMDhamnekar

MHB

- 376

- 28

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?