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

$\blacksquare$