# 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:

$\displaystyle \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha} \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$

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

## Proof

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

Then by definition of set difference:

 $\displaystyle x$ $\in$ $\displaystyle \bigcup_{\alpha \mathop \in I} S_\alpha$ $\displaystyle x$ $\notin$ $\displaystyle \bigcup_{\alpha \mathop \in I} T_\alpha$

By definition of set union, it follows that:

 $\displaystyle \exists \beta \in I: x$ $\in$ $\displaystyle S_\beta$ $\displaystyle \neg \exists \beta \in I: x$ $\in$ $\displaystyle T_\beta$ $\displaystyle \leadsto \ \$ $\displaystyle \forall \beta \in I: x$ $\notin$ $\displaystyle T_\beta$ De Morgan's Laws (Predicate Logic)

and so:

$\exists \beta \in I: x \in S_\beta \setminus T_\beta$

Hence:

$\displaystyle x \in \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$

by defintion of set union.

The result follows by definition of subset.

$\blacksquare$