Set Difference as Intersection with Complement
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Theorem
Set difference can be expressed as the intersection with the set complement:
- $A \setminus B = A \cap \map \complement B$
Proof
This follows directly from Set Difference as Intersection with Relative Complement:
- $A \setminus B = A \cap \relcomp S B$
Let $S = \Bbb U$.
Since $A, B \subseteq \Bbb U$ by definition of the universe, the result follows.
$\blacksquare$
Sources
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Subsets and Complements; Union and Intersection
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{B x}$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $1 \ \text{(i)}$