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:

\(\ds x\)

\(\in\)

\(\ds \bigcup_{\alpha \mathop \in I} S_\alpha\)

\(\ds x\)

\(\notin\)

\(\ds \bigcup_{\alpha \mathop \in I} T_\alpha\)

By definition of set union, it follows that:

\(\ds \exists \beta \in I: \, \)

\(\ds x\)

\(\in\)

\(\ds S_\beta\)

\(\ds \neg \exists \beta \in I: \, \)

\(\ds x\)

\(\in\)

\(\ds T_\beta\)

\(\ds \leadsto \ \ \)

\(\ds \forall \beta \in I: \, \)

\(\ds x\)

\(\notin\)

\(\ds T_\beta\)

De Morgan's Laws (Predicate Logic)

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.

$\blacksquare$

Sources

• 1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 1$: Sets and Functions: Problem $5$