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:
 * $\ds \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha} \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$

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

Proof
Let $\ds x \in \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha}$.

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:
 * $\ds x \in \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$

by definition of set union.

The result follows by definition of subset.