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

where $$S_\alpha \setminus T_\alpha$$ denotes Set Difference.

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

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

by defintion of set union.

The result follows by definition of subset.