# Ring of Sets vs Ring in Universal Algebra

I actually want to continue this post. I believe the naming convention in Mathematics is consistent, such that there are no clearly distinguish objects have the same name. However, I'm not sure how to relate the Ring of Sets and the Ring in Universal Algebra.

The definition of Ring in Universal Algebra is

1. $R$ is an abelian group under addition, meaning that:
• $(a + b) + c = a + (b + c)$ for all $a, b, c \in R$ ($+$ is associative).
• $a + b = b + a$ for all $a, b \in R$ ($+$ is commutative).
• There is an element $0 \in R$ such that $a + 0 = a$ for all $a \in R$ ($0$ is the additive identity).
• For each $a \in R$ there exists $−a \in R$ such that $a + (−a) = 0$ ($−a$ is the additive inverse of $a$).
2. $R$ is a monoid under multiplication, meaning that:
• $(a · b) · c = a · (b · c)$ for all $a, b, c \in R$ ($·$ is associative).
• There is an element $1 \in R$ such that $a · 1 = a$ and $1 · a = a$ for all $a \in R$ ($1$ is the multiplicative identity).
3. Multiplication is distributive with respect to addition:
• $a ⋅ (b + c) = (a · b) + (a · c)$ for all $a, b, c \in R$ (left distributivity).
• $(b + c) · a = (b · a) + (c · a)$ for all $a, b, c \in R$ (right distributivity).

Next, the definition of Ring of Sets is

1. $A,B\in\mathcal{R}$ implies $A\cap B\in \mathcal {R}$ and
2. $A,B\in \mathcal {R}$ implies $A\cup B\in \mathcal {R}$.

Then, I tried to relate between this two definition.

1. Let assume that $\cap$ is the additive operator. Then, $\mathcal {R}$ must contain the additive identity and inverse for all $A \in \mathcal {R}$.

• The additive identity is the whole element set $S$, such that for all $A \in \mathcal {R}, A \subseteq S$. It can be easily seen that $A \cap S = A$, for all $A \in \mathcal {R}$.
• Nevertheless, pick $A \in \mathcal{R}$, such that $A \subset S$. Then, $A$ doesn't have the inverse, i.e. no $B \in \mathcal{R}$, such that $A \cap B = S$.
2. Let assume that $\cup$ is the additive operator. The argument is similar to (1).

I might be wrong, but what I'm trying to say is the structure of Ring of Sets isn't consistent with the structure of general Ring. Kindly need your explanation. Thank you.

• In a ring of sets, the additive operation is symmetric difference and the multiplicative operation is intersection.
– bof
Commented Nov 5, 2016 at 4:41
• I believe the naming convention in Mathematics is consistent, such that there are no clearly distinguish objects have the same name. This belief is simply false, and you’ve found one of the counterexamples. Commented Nov 5, 2016 at 4:42
• Actually, "closed under union and intersection" is properly called a lattice of sets. An actual ring oif sets, what that silly Wikipedia article calls a ring of sets in the measure theory sense, is closed under intersection and subtraction (whence of course it's also closed under union). And, yes, the "measure theory" rings are rings in the sense of abstract algebra, with symmetric difference as addition.
– bof
Commented Nov 5, 2016 at 4:45
• As for your assumption of "consistent naming", that Wikipedia article you linked to states that there are two different notions called a "ring of sets". This is the first time I ever heard that some people call a lattice of sets a "ring of sets". Bad people.
– bof
Commented Nov 5, 2016 at 4:49
• Did you know that a quantum group is not a group? Commented Nov 5, 2016 at 4:56

Ring of Sets by corresponding union to addition and intersection to multiplication ($$\varnothing$$ to $$0$$ and $$E$$ (universe of set) to $$1$$ as well) satisfy most ring axioms. However the fundamental difference is that it doesn't have an additive inverse because $$X\cup Y=\varnothing$$ if and only if $$X=\varnothing$$ and $$Y=\varnothing$$.

Thus your definition of Ring of Sets is not a ring rigorously.

Another way to define Ring of Sets is known as Boolean Ring, i.e by corresponding symmetric difference of sets to addition and intersection to multiplication ($$\varnothing$$ to $$0$$ and $$E$$ (universe of set) to $$1$$ as well). It satisfies all ring axioms with an additive inverse.

• @user259242, it seems reasonable to define that a representation of a distributive lattice $L$ is a set $X$ together with a lattice homomorphism $L \rightarrow \mathcal{P}(X)$, by analogy with how a representation of a ring $R$ is an abelian group $X$ together with a ring homomorphism $R \rightarrow \mathrm{End}(X)$. Commented Nov 5, 2016 at 5:02