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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).

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