Set Difference Equals First Set iff Empty Intersection

Theorem

$S \setminus T = S \iff S \cap T = \O$

Proof

Assume $S, T \subseteq \Bbb U$ where $\Bbb U$ is a universal set.

\(\ds S \setminus T\)

\(=\)

\(\ds S\)

\(\ds \leadstoandfrom \ \ \)

\(\ds S \cap \map \complement T\)

\(=\)

\(\ds S\)

Set Difference as Intersection with Complement

\(\ds \leadstoandfrom \ \ \)

\(\ds S\)

\(\subseteq\)

\(\ds \map \complement T\)

Intersection with Subset is Subset‎

\(\ds \leadstoandfrom \ \ \)

\(\ds S \cap \map \complement {\map \complement T}\)

\(=\)

\(\ds \O\)

Intersection with Complement is Empty iff Subset

\(\ds \leadstoandfrom \ \ \)

\(\ds S \cap T\)

\(=\)

\(\ds \O\)

Complement of Complement

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \ \text{(b)}$
• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.5$: Complementation