# inverse element in a field of sets

A field is an abelian group under addition and a group under multiplication. But for the power set of a set S, under union and intersection operations and with S and the empty set being their identities, the inverse of any set is not unique for either union or intersection. So is the power set a field of sets?

Must the inverse of a set for either union or intersection be restricted to the complement of the set?

Thanks!

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A field is an abelian group under addition, the nonzero elements are an abelian group under multiplication, and multiplication distributes over the sum. – Arturo Magidin Oct 12 '10 at 2:39
Thank you, Professor, for pointing out the errors! But what I would like to know is if a field of sets is a field under union and intersection, and also an algebra? If yes, why the inverse element seems not unique? Is a Boolean algebra a field and an algebra? – Tim Oct 12 '10 at 2:52
The problem is that you have several meanings of "field" and "algebra" floating around here. See my response: "field of sets" is not "field" in the sense of ring theory, unless it is the trivial one. "Algebra" for fields of sets refers to Boolean algebra structure, which is really a lattice structure, not a ring-theoretic one. – Arturo Magidin Oct 12 '10 at 3:00
I'd just like to point out that it's not the case that inverse elements for union or intersection aren't unique. Rather, they simply don't exist for all but the most trivial sets. For example, the identity element for union is $\emptyset$, and unless $A=\emptyset$, there is no $B$ such that $A\cup B=\emptyset$. The identity for intersection is $S$ and a similar argument holds. – Paul VanKoughnett Oct 12 '10 at 4:45

A "field of sets" is a collection $\mathcal{F}$ of subsets of a given set $X$ which is closed under binary unions, intersections, and under complementation. Alas, it is not an "field" in the sense of abstract algebra.

It is a "subalgebra" of the Boolean algebra of the power set of $X$. Alas, here as well, "Boolean algebra" and "subalgebra" do not have the meaning they have in abstract algebra/ring theory: a "Boolean algebra" is not really an algebra: rather, it is a special kind of lattice.

Namely, a Boolean algebra is a set $A$ together with two binary operations, $\wedge$ (meet) and $\vee$ (join), one unary operation ${}^c$ (complementation), and two nulary operations (distinguished elements) $0$ and $1$. We require $\vee$ and $\wedge$ to be associative, commutative, and to distribute over each other (this is already different from the case of "algebra" in the ring-theoretic sense, in which only multiplication distributes over sums, and not the other way around); we also have two rules of "absorption" that further describe how $\wedge$ and $\vee$ interact: $x\wedge(x\vee y) = x$ and $x\vee(x\wedge y) = x$. Finally, we require $x\vee x^{c}=1$ and $x\wedge x^{c}=0$.

The typical example of a Boolean algebra/Boolean lattice is the lattice of subsets of a given set $X$, with $\wedge$ corresponding to intersection, $\vee$ to union, ${}^c$ to complementation, $0$ to $\emptyset$, and $1$ to $X$. Another example is propositional calculus, with $\wedge$ corresponding to conjunction, $\vee$ to disjunction, ${}^{c}$ to negation, $1$ to the class of tautologies, and $0$ to the class of contradictions.

The reason for calling it algebra, in defiance of the meaning of "algebra" from ring theory, is historical: Boole talked about "algebra of thought" or "algebra of logic".

Now, any Boolean algebra gives rise to a ring. We define $+$ by "symmetric difference": $x+y = (x\wedge y^{c})\vee(x^{c}\wedge y)$. If you do this, then you get an abelian group, with identity element being $0$, and with every element being its own inverse (that is, every nontrivial element is of additive order $2$). We define $*$ by $x*y=x\wedge y$. The unity of this ring is the $1$ from the Boolean lattice. This ring is an algebra over the field of $2$ elements, since it has characteristic $2$ and a unity.

You may ask: will this ring be a field? The only time it is a field is if the original Boolean algebra was the trivial algebra, with $0$ and $1$ the only elements. For if $x$ is any other element, $0\neq x$, $x\neq 1$, then if you look at the order induced by the lattice structure ($x\preceq y$ if and only if $x\wedge y = x$) then you have $0\prec x \prec 1$, and $x\wedge y \preceq x$ for all $y$. Therefore, since $x$ is strictly smaller than $1$, then so will $x*y$ for any $y$, so $x$ cannot have a multiplicative inverse in the corresponding ring.

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More simply, a nontrivial Boolean ring isn't a field since every element is idempotent $x^2 = x \wedge x = x$, but a field has only trivial idempotents. – Bill Dubuque Oct 12 '10 at 4:04

One word, two meanings. A 'field of sets' is not a field in the abstract algebra sense.

EDIT: While a 'field of sets' is not a field, a Boolean Algebra (BA) of sets (of which a field of sets is a special case) is indeed an algebra. One can check that a BA is a ring, with symmetric difference $(A \cap B^c) \cup (A^c \cap B)$ as addition and intersection as multiplication. It is then trivially an algebra over the two-element field $\mathbb{F}_2$ by defining $1\cdot A=A$ and $0\cdot A=\emptyset$ and checking that $A+A=\emptyset$.

[Arturo's answer discussed this before me in a more general setting. The edit above is to set right my incorrect assertion in the comments that a BA is not an algebra.]

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If a field of sets not a field, what kind of algebraic structure is it? – Tim Oct 12 '10 at 1:44
As Wikipedia says, it's a "subalgebra of the power set Boolean algebra". Follow the links for each of the terms. en.wikipedia.org/wiki/Field_of_sets – Jyotirmoy Bhattacharya Oct 12 '10 at 1:49
Still, it is an algebra, and therefore a vector field. It must be an abelian group under addition (either union or intersection). So every set should have a unique inverse. If what I said can be true, the field of sets is a field under union and intersection. – Tim Oct 12 '10 at 1:56
It so happens that a Boolean algebra is not an algebra! – Jyotirmoy Bhattacharya Oct 12 '10 at 2:03
But it's not. As you yourself pointed out, there are inverses for neither intersection nor union. – Paul VanKoughnett Oct 12 '10 at 2:20