Symmetric Difference with Universal Set
Theorem
- $\mathbb U \symdif S = \map \complement S$
where:
- $\mathbb U$ denotes the universal set
- $\symdif$ denotes symmetric difference.
Proof
| \(\ds \mathbb U \symdif S\) | \(=\) | \(\ds \mathbb U \cup S \setminus \mathbb U \cap S\) | Definition 2 of Symmetric Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds \mathbb U \cup S \setminus S\) | Intersection with Universal Set | |||||||||||
| \(\ds \) | \(=\) | \(\ds \mathbb U \setminus S\) | Union with Universal Set | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \complement S\) | Definition of Set Complement |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $9 \ \text{(i)}$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): symmetric difference: $\text {(i)}$