Symmetric Difference is Commutative
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Theorem
Symmetric difference is commutative:
- $S \symdif T = T \symdif S$
Proof
\(\ds S \symdif T\) | \(=\) | \(\ds \paren {S \setminus T} \cup \paren {T \setminus S}\) | Definition of Symmetric Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {T \setminus S} \cup \paren {S \setminus T}\) | Union is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds T \symdif S\) | Definition of Symmetric Difference |
$\blacksquare$
Also see
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 5$: Complements and Powers
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $1$. Sets: Exercise $7 \ \text{(i)}$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $14$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): symmetric difference: $\text {(iii)}$