Power Set is Closed under Set Difference

Theorem

Let $S$ be a set.

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


Then:

$\forall A, B \in \powerset S: A \setminus B \in \powerset S$

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


Proof

Let $A, B \in \powerset S$.

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 is Subset.

Thus by Subset Relation is Transitive, $A \setminus B \subseteq S$.

Thus $A \setminus B \in \powerset S$, and closure is proved.

$\blacksquare$


Also see


Sources

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