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 $\text{A}.11$