Set Difference as Intersection with Relative Complement
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Theorem
Let $A, B \subseteq S$.
Then the set difference between $A$ and $B$ can be expressed as the intersection with the relative complement with respect to $S$:
- $A \setminus B = A \cap \relcomp S B$
Proof
| \(\ds A \setminus B\) | \(=\) | \(\ds \set {x: x \in A \land x \notin B}\) | Definition of Set Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {x: \paren {x \in A \land x \in X} \land x \notin B}\) | Definition of Subset, Modus Ponens and Rule of Conjunction | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {x: x \in A \land \paren {x \in X \land x \notin B} }\) | Conjunction is Associative | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {x: x \in A \land x \in \relcomp S B}\) | Definition of Relative Complement | |||||||||||
| \(\ds \) | \(=\) | \(\ds A \cap \relcomp S B\) | Definition of Set Intersection |
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 5$: Complements and Powers
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $5$
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $1$: Events and probabilities: $1.2$: Outcomes and events: Exercise $2$