# Power Set with Union and Intersection forms Boolean Algebra

## Theorem

Let $S$ be a set, and let $\powerset S$ be its power set.

Denote with $\cup$, $\cap$ and $\complement$ the operations of union, intersection and complement on $\powerset S$, respectively.

Then $\struct {\powerset S, \cup, \cap, \complement}$ is a Boolean algebra.

## Proof

Taking the criteria for definition 1 of a Boolean algebra in turn:

### $(\text {BA} 0):$ Closure

$\powerset S$ is closed under both $\cup$ and $\cap$:

- Power Set is Closed under Intersection
- Power Set is Closed under Union
- Power Set is Closed under Complement

$\Box$

### $(\text {BA} 1):$ Commutativity

Both $\cup$ and $\cap$ are commutative from Intersection is Commutative and Union is Commutative.

$\Box$

### $(\text {BA} 2):$ Distributivity

Both $\cup$ and $\cap$ distribute over the other, from Union Distributes over Intersection and Intersection Distributes over Union.

$\Box$

### $(\text {BA} 3):$ Identity Elements

Both $\cup$ and $\cap$ have identities:

From Power Set with Intersection is Monoidâ€Ž, $S$ is the identity for $\cap$.

From Power Set with Union is Monoid, $\O$ is the identity for $\cup$.

$\Box$

### $(\text {BA} 4):$ Complements

From Union with Complement:

- $\forall A \in S: A \cup \map \complement A = S$

which is the identity for $\cap$.

From Intersection with Complement:

- $\forall A \in S: A \cap \map \complement A = \O$

which is the identity for $\cup$.

$\Box$

All the criteria for a Boolean algebra are therefore fulfilled.

$\blacksquare$

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): $\S 1.5$ - 1964: Steven A. Gaal:
*Point Set Topology*... (previous) ... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets - 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): $\S 2$ - 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): $\S 2$: Exercise $1$