Complement of Relative Complement is Union with Complement

Theorem

Let $A, B, C$ be sets such that $A \subseteq B \subseteq C$.

Then:

$\relcomp C {\relcomp B A} = A \cup \relcomp C B$

Proof

\(\ds \relcomp C {\relcomp B A}\)

\(=\)

\(\ds C \setminus \paren {B \setminus A}\)

Definition of Relative Complement

\(\ds \)

\(=\)

\(\ds \paren {C \setminus B} \cup \paren {C \cap A}\)

Set Difference with Set Difference is Union of Set Difference with Intersection

\(\ds \)

\(=\)

\(\ds \paren {C \setminus B} \cup A\)

Intersection with Subset is Subset

\(\ds \)

\(=\)

\(\ds A \cup \paren {C \setminus B}\)

Union is Commutative

\(\ds \)

\(=\)

\(\ds A \cup \relcomp C B\)

Definition of Relative Complement

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $2 \ \text{(b)}$