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

• I
• WMDhamnekar
In summary: If we only consider addition then it is not a semiring because the set contains the empty set. If we only consider multiplication then it is a semiring because it contains the set [0,1].
WMDhamnekar
MHB
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?

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

bpet said:
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.
WMDhamnekar said:
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##.

### Similar threads

• Set Theory, Logic, Probability, Statistics
Replies
6
Views
926
• Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
27
Views
3K
• Set Theory, Logic, Probability, Statistics
Replies
14
Views
1K
• Topology and Analysis
Replies
2
Views
323
• Set Theory, Logic, Probability, Statistics
Replies
15
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
255
• Topology and Analysis
Replies
4
Views
472
• Topology and Analysis
Replies
1
Views
382