Symmetric Difference of Unions is Subset of Union of Symmetric Differences

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: