Set Difference as Intersection with Complement

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 universal set, 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 $1 \ \text{(i)}$
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