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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.

Eugene Zhang
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    In a ring of sets, the additive operation is symmetric difference and the multiplicative operation is intersection. – bof Nov 05 '16 at 04:41
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    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. – Brian M. Scott Nov 05 '16 at 04:42
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    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 Nov 05 '16 at 04:45
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    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 Nov 05 '16 at 04:49
  • Did you know that a quantum group is not a group? – user259242 Nov 05 '16 at 04:56
  • @BrianM.Scott, it would be so so nice if that belief was true, though. – goblin GONE Nov 05 '16 at 05:03
  • @user259242 Well a quantum group is a Hopf algebra, and Hopf monoids (of which a Hopf algebra is an example) are a generalization of group objects to monoidal categories. For example, a Hopf monoid in the category of sets is a group, and a Hopf monoid in the category of smooth manifolds is a Lie group. – ಠ_ಠ Nov 05 '16 at 21:30

2 Answers2

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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.

Please read this post for more detail.

Eugene Zhang
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A "ring of sets" should really be called a "distributive lattice of sets." Does that answer your question?

goblin GONE
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  • Just reading that wiki article I noticed some more confusing nomenclature, specifically in the section titled Representation theory, which, of course, has exactly nothing to do with representation theory. – user259242 Nov 05 '16 at 05:00
  • @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 $R \rightarrow \mathrm{End}(X)$. – goblin GONE Nov 05 '16 at 05:02
  • Entirely reasonable indeed. But to call the study of such a gadget 'representation theory' is another kettle of fish. – user259242 Nov 05 '16 at 05:03
  • @user259242, I guess I basically agree. – goblin GONE Nov 05 '16 at 05:04
  • Well I suppose mathematics is not that consistent then. Thanks! – Abe Vallerian Nov 05 '16 at 05:17
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    @AbeVallerian No, mathematics is perfectly consistent. Mathematical notation adopted by different mathematicians is not consistent. – Jacob Wakem Nov 05 '16 at 21:09