Symmetric Difference of Unions is Subset of Union of Symmetric Differences
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Theorem
Let $I$ be an indexing set. Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
- $\ds \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}$
where $S \symdif T$ is the symmetric difference between $S$ and $T$.
Proof
From Difference of Unions is Subset of Union of Differences, we have:
- $\ds \bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
- $\ds \bigcup_{\alpha \mathop \in I} T_i \setminus \bigcup_{\alpha \mathop \in I} S_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}$
where $\setminus$ denotes set difference.
Thus we have:
\(\ds \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha\) | \(=\) | \(\ds \paren {\bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} T_\alpha \setminus \bigcup_{\alpha \mathop \in I} S_\alpha}\) | Definition of Symmetric Difference | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha} \cup \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}\) | Difference of Unions is Subset of Union of Differences | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{\alpha \mathop \in I} \paren {\paren {S_\alpha \setminus T_\alpha} \cup \paren {T_\alpha \setminus S_\alpha} }\) | Union is Associative and Union is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}\) | Definition of Symmetric Difference |
$\blacksquare$