Set Difference with Disjoint Set

Theorem

Let $S, T$ be sets.

Then:

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

where:

$S \cap T$ denotes set intersection

$\O$ denotes the empty set

$S \setminus T$ denotes set difference.

Proof

\(\ds S \cap T\)

\(=\)

\(\ds \O\)

\(\ds \leadstoandfrom \ \ \)

\(\ds S\)

\(\subseteq\)

\(\ds \map \complement T\)

Intersection with Complement is Empty iff Subset

\(\ds \leadstoandfrom \ \ \)

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

\(=\)

\(\ds S\)

Intersection with Subset is Subset‎‎

\(\ds \leadstoandfrom \ \ \)

\(\ds S \setminus T\)

\(=\)

\(\ds S\)

Set Difference as Intersection with Complement

$\blacksquare$

Sources

• 1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
• 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): Appendix $\text{A}.2$: Theorem $\text{A}.11$