Difference of Unions is Subset of Union of Differences

Theorem
Let $I$ be an indexing set.

Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.

Then:
 * $\displaystyle \left({\bigcup_{\alpha \in I} S_\alpha}\right) \setminus \left({\bigcup_{\alpha \in I} T_\alpha}\right) \subseteq \bigcup_{\alpha \in I} \left({S_\alpha \setminus T_\alpha}\right)$

where $S_\alpha \setminus T_\alpha$ denotes set difference.

Proof
Let $\displaystyle x \in \left({\bigcup_{\alpha \in I} S_\alpha}\right) \setminus \left({\bigcup_{\alpha \in I} T_\alpha}\right)$.

Then by definition of set difference:

By definition of set union, it follows that:

and so:
 * $\exists \beta \in I: x \in S_\beta \setminus T_\beta$

Hence:
 * $\displaystyle x \in \bigcup_{\alpha \in I} \left({S_\alpha \setminus T_\alpha}\right)$

by defintion of set union.

The result follows by definition of subset.