Set Difference Equals First Set iff Empty Intersection
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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