# Equivalence of Definitions of Algebra of Sets

## Contents

## Theorem

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

### Definition 1

Let $X$ be a set.

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

Let $\mathcal R \subseteq \powerset X$ be a set of subsets of $X$.

Then $\mathcal R$ is an **algebra of sets over $X$** if and only if the following conditions hold:

\((AS \, 1)\) | $:$ | Unit: | \(\displaystyle X \in \mathcal R \) | |||||

\((AS \, 2)\) | $:$ | Closure under Union: | \(\displaystyle \forall A, B \in \mathcal R:\) | \(\displaystyle A \cup B \in \mathcal R \) | ||||

\((AS \, 3)\) | $:$ | Closure under Complement Relative to $X$: | \(\displaystyle \forall A \in \mathcal R:\) | \(\displaystyle \relcomp X A \in \mathcal R \) |

### Definition 2

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

## Proof

### Definition 1 implies Definition 2

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

- $(1): \quad A \cup B \in \mathcal R$
- $(2): \quad \complement_X \left({A}\right) \in \mathcal R$

Let $A, B \in \mathcal R$.

From the definition:

- $\forall A \in \mathcal R: A \subseteq X$.

Hence from Intersection with Subset is Subset:

- $\forall A \in \mathcal R: A \cap X = A$

Hence $X$ is the unit of $\mathcal R$.

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

From the definition of symmetric difference:

- $A * B = \left({A \setminus B}\right) \cup \left({B \setminus A}\right)$

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 $\mathcal R$ is indeed a ring of sets.

$\Box$

### Definition 2 implies Definition 1

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

By definition, $X \in \mathcal R$.

From definition 2 of ring of sets, $\mathcal R$ is:

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

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

- $\forall A \in \mathcal R: A \subseteq X$

from which we have that $X \setminus A = \complement_X \left({A}\right)$.

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

$\blacksquare$

## Sources

- 1970: Avner Friedman:
*Foundations of Modern Analysis*... (previous) ... (next): $\S 1.1$: Rings and Algebras