Power Set Closed under Set Difference

Theorem

Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Then:

$\forall A, B \in \mathcal P \left({S}\right): A \setminus B \in \mathcal P \left({S}\right)$

where $A \setminus B$ denotes the set difference of $A$ and $B$.

Proof

Let $A, B \in \mathcal P \left({S}\right)$.

Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.

We also have $A \setminus B \subseteq A$ from Set Difference Subset.

Thus by Subsets Transitive, $A \setminus B \subseteq S$.

Thus $A \setminus B \in \mathcal P \left({S}\right)$, and closure is proved.

$\blacksquare$

Also see

• Power Set Closed under Intersection
• Power Set Closed under Union

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 4.1$: Example $60$