Set Difference Equals First Set iff Empty Intersection

Theorem

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

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

$\ds S \setminus T$

$=$

$\ds S$

$\ds \leadstoandfrom \ \ $

$\ds S \cap \map \complement T$

$=$

$\ds S$

$\ds \leadstoandfrom \ \ $

$\ds S$

$\subseteq$

$\ds \map \complement T$

$\ds \leadstoandfrom \ \ $

$\ds S \cap \map \complement {\map \complement T}$

$=$

$\ds \O$

$\ds \leadstoandfrom \ \ $

$\ds S \cap T$

$=$

$\ds \O$

$\blacksquare$

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