Union of Symmetric Differences
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Theorem
Let $R, S, T$ be sets.
Then:
- $\paren {R \symdif S} \cup \paren {S \symdif T} = \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.
Proof
From the definition of symmetric difference, we have:
- $R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Thus, expanding:
\(\ds \paren {R \symdif S} \cup \paren {S \symdif T}\) | \(=\) | \(\ds \paren {R \setminus S} \cup \paren {S \setminus R} \cup \paren {S \setminus T} \cup \paren {T \setminus S}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {R \setminus S} \cup \paren {T \setminus S} } \cup \paren {\paren {S \setminus R} \cup \paren {S \setminus T} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {R \cup T} \setminus S} \cup \paren {\paren {S \setminus R} \cup \paren {S \setminus T} }\) | Set Difference is Right Distributive over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {R \cup T} \setminus S} \cup \paren {S \setminus \paren {R \cap T} }\) | De Morgan's Laws: Difference with Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {R \cup S \cup T} \setminus S} \cup \paren {S \setminus \paren {R \cap T} }\) | Set Difference with Union is Set Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}\) | De Morgan's Laws for Difference with Intersection: Corollary |
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next) $\S 2$: Problem $2$