Prove relationships between complements, unions and intersections for classes of sets

Prove that for a class of sets $\mathcal{F}$ we have,

$B \smallsetminus \bigcup_{A \in \mathcal{F}} A = \bigcap_{A \in \mathcal{F}} (B \smallsetminus A) \qquad \text{and} \qquad B \smallsetminus \bigcap_{A \in \mathcal{F}} A = \bigcup_{A \in \mathcal{F}} (B \smallsetminus A).$

Proof. Let $x$ be an arbitrary element of $B \smallsetminus \bigcup_{A \in \mathcal{F}} A$ . This means that $x \in B$ and $x \notin \bigcup_{A \in \mathcal{F}} A$ , which means that $x$ is not in $A$ for any $A$ in the class $\mathcal{F}$ . Hence, for every $A \in \mathcal{F}$ we have $x \in (B \smallsetminus A)$ (since $x$ is in $B$ no matter what, and $x$ is not in $A$ for any $A$ that we choose, so it must be in $B \smallsetminus A$ ). But, $x \in (B \smallsetminus A)$ for all $A \in \mathcal{F}$ means that $x$ is in the intersection $\bigcap_{A \in \mathcal{F}} (B \smallsetminus A)$ ; and hence, $B \smallsetminus \bigcup_{A \in \mathcal{F}} A \subseteq \bigcap_{A \in \mathcal{F}} (B \smallsetminus A)$ .
For the reverse inclusion, let $x$ be any element in $\bigcap_{A \in \mathcal{F}} (B \smallsetminus A)$ . This means that $x \in (B \smallsetminus A)$ for every $A \in \mathcal{F}$ , i.e., $x \in B$ and $x \notin A$ for every $A \in \mathcal{F}$ . Since $x \notin A$ for every $A$ , we then have $x \notin \bigcup_{A \in \mathcal{F}}$ . Hence, $x \in B \smallsetminus \bigcup_{A \in \mathcal{F}}$ . Therefore, $\bigcap_{A \in \mathcal{F}} (B \smallsetminus A) \subseteq B \smallsetminus \bigcup_{A \in \mathcal{F}} A$ .
Hence, $B \smallsetminus \bigcup_{A \in \mathcal{F}} A = \bigcap_{A \in \mathcal{F}} (B \smallsetminus A)$ . ∎

Proof. Let $x$ be any element of $B \smallsetminus \bigcap_{A \in \mathcal{F}} A$ . This means that $x \in B$ and $x \notin \bigcap_{A \in \mathcal{F}} A$ . Further, $x$ not in the intersection of the sets $A \in \mathcal{F}$ means that there is at least one $A \in \mathcal{F}$ , say $A'$ , such that $x \notin A'$ . Since $x \notin A'$ and $x \in B$ , we know $x \in (B \smallsetminus A')$ . Then we can conclude $x \in \bigcup_{A \in \mathcal{F}} (B \smallsetminus A)$ . Thus, $B \smallsetminus \bigcap_{A \in \mathcal{F}} \subseteq \bigcup_{A \in \mathcal{F}} (B \smallsetminus A)$ .
For the reverse inclusion, we let $x$ be any element in $\bigcup_{A \in \mathcal{F}} (B \smallsetminus A)$ . This means there is at least one $A \in \mathcal{F}$ , say $A'$ , such that $x \in (B \smallsetminus A')$ , which means $x \in B$ and $x \notin A'$ . Since $x \notin A'$ , we know $x \notin \bigcap_{A \in \mathcal{F}} A$ ; therefore, $x \in B \smallsetminus \bigcap_{A \in \mathcal{F}} A$ . Hence, $\bigcup_{A \in \mathcal{F}} (B \smallsetminus A) \subseteq B \smallsetminus \bigcap_{A \in \mathcal{F}} A$ .
Therefore, $B \smallsetminus \bigcap_{A \in \mathcal{F}} A = \bigcup_{A \in \mathcal{F}} (B \smallsetminus A).$ ∎