Power Set is Closed under Set Complement

Theorem

Let $S$ be a set.

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

Then:

$\forall A \in \powerset S: \relcomp S A \in \powerset S$

Proof

Let $A \in \powerset S$.

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

By definition of relative complement:

$\relcomp S A = \set {x \in S: x \notin A}$

Hence $\relcomp S A$ is a subset of $S$.

That is:

$\relcomp S A \in \powerset S$

and closure is proved.

$\blacksquare$

Also see

• Power Set is Closed under Union
• Power Set is Closed under Intersection
• Power Set is Closed under Set Difference

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): algebra of sets