# Power Set with Intersection is Commutative Monoid

## Contents

## Theorem

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

Then $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.

The only invertible element of this structure is $S$.

Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cap}$ cannot be a group.

## Proof

From Power Set is Closed under Intersection:

- $\forall A, B \in \powerset S: A \cap B \in \powerset S$

From Set System Closed under Intersection is Commutative Semigroup, $\struct {\powerset S, \cap}$ is a commutative semigroup.

From Identity of Power Set with Intersection, we have that $S$ acts as the identity.

It remains to be shown that only $S$ has an inverse:

For $T \subseteq S$ to have an inverse under $\cap$, we require $T^{-1} \cap T = S$.

From this it follows that $T = S = T^{-1}$.

The result follows by definition of commutative monoid.

$\blacksquare$

## Also see

## Sources

- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: I. Basic Concepts*... (previous) ... (next): $\text{I}.1$: Definition and examples of semigroups: Example $7$ - 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 4.4$. Gruppoids, semigroups and groups: Example $77$ - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $4.4$ - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{II}$: Exercise $\text{T}$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 29$. Semigroups: definition and examples: $(3)$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $5$: Semigroups: Exercise $2$