(A cap C) cup (B cap Complement C) = Empty iff B subset C subset Complement A

Theorem

Let $A$, $B$ and $C$ be subsets of a universe $\Bbb U$.

Then:

$\paren {A \cap C} \cup \paren {B \cap \map \complement C} = \O \iff B \subseteq C \subseteq \map \complement A$

where $\map \complement C$ denotes the complement of $C$ in $\Bbb U$.

Proof

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

\(=\)

\(\ds \O\)

\(\ds \leadstoandfrom \ \ \)

\(\ds A \cap C\)

\(=\)

\(\ds \O\)

Union is Empty iff Sets are Empty

\(\, \ds \land \, \)

\(\ds B \cap \map \complement C\)

\(=\)

\(\ds \O\)

\(\ds \leadstoandfrom \ \ \)

\(\ds C\)

\(\subseteq\)

\(\ds \map \complement A\)

Empty Intersection iff Subset of Complement

\(\, \ds \land \, \)

\(\ds B\)

\(\subseteq\)

\(\ds C\)

Intersection with Complement is Empty iff Subset

\(\ds \leadstoandfrom \ \ \)

\(\ds B\)

\(\subseteq\)

\(\ds C \subseteq \map \complement A\)

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(e)}$