# Equivalence of Definitions of Algebra of Sets

## Theorem

The following definitions of the concept of Algebra of Sets are equivalent:

### Definition 1

Let $S$ be a set.

Let $\powerset S$ be the power set of $S$.

Let $\RR \subseteq \powerset S$ be a set of subsets of $S$.

$\RR$ is an algebra of sets over $S$ if and only if $\RR$ satisfies the algebra of sets axioms:

 $(\text {AS} 1)$ $:$ Unit: $\ds S \in \RR$ $(\text {AS} 2)$ $:$ Closure under Union: $\ds \forall A, B \in \RR:$ $\ds A \cup B \in \RR$ $(\text {AS} 3)$ $:$ Closure under Complement Relative to $S$: $\ds \forall A \in \RR:$ $\ds \relcomp S A \in \RR$

### Definition 2

An algebra of sets is a ring of sets with a unit.

## Proof

### Definition 1 implies Definition 2

Let $\RR$ be a system of sets such that $\forall A, B \in \RR$:

$(1): \quad A \cup B \in \RR$
$(2): \quad \relcomp X A \in \RR$

Let $A, B \in \RR$.

From the definition:

$\forall A \in \RR: A \subseteq X$.

Hence from Intersection with Subset is Subset:

$\forall A \in \RR: A \cap X = A$

Hence $X$ is the unit of $\RR$.

From Properties of Algebras of Sets, we have that $\RR$ is closed under set intersection.

From the definition of symmetric difference:

$A \symdif B = \paren {A \setminus B} \cup \paren {B \setminus A}$

Since both set union and set difference are closed operations, it follows that symmetric difference is also closed.

So by definition 1 of ring of sets, it follows that $\RR$ is indeed a ring of sets.

$\Box$

### Definition 2 implies Definition 1

Let $\RR$ be a ring of sets with a unit $X$.

By definition, $X \in \RR$.

From definition 2 of ring of sets, $\RR$ is:

$(1) \quad$ closed under set union
$(2) \quad$ closed under set difference.

From Unit of System of Sets is Unique, we have that:

$\forall A \in \RR: A \subseteq X$

from which we have that $X \setminus A = \relcomp X A$.

So $\RR$ is an algebra of sets by definition 1.

$\blacksquare$