Power Set is Closed under Complement

Theorem

Let $S$ be a set.

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

Then:

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

where $\relcomp S A$ denotes the complement of $A$ relative to $S$.

Proof

\(\ds A\)

\(\in\)

\(\ds \powerset S\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(\subseteq\)

\(\ds S\)

Definition of Power Set

\(\ds \leadsto \ \ \)

\(\ds S \setminus A\)

\(\subseteq\)

\(\ds S\)

Set Difference is Subset

\(\ds \leadsto \ \ \)

\(\ds \relcomp S A\)

\(\subseteq\)

\(\ds S\)

Definition of Relative Complement

\(\ds \leadsto \ \ \)

\(\ds \relcomp S A\)

\(\in\)

\(\ds \powerset S\)

Definition of Power Set

$\blacksquare$

Also see

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