What does $\cap_{A \in \mathcal{F}} (B \cup A)$ mean? This is from Velleman page 143, problem 16b.
$\mathcal{F}$ is a family of sets.
B is a set.
The statement is $B \cup (\cap \mathcal{F} ) = \cap_{A \in \mathcal{F}} (B \cup A)$.
I don't understand what the right hand side means.
As far as I understand a leading $\cup$, as in $\cap \mathcal{G}$, is an operator that can only apply to a family of sets. However, $(B \cup A)$ is not a family; it's just a set. So what is that leading $\cap_{A \in \mathcal{F}}$ supposed to mean?
 A: The expression
$$\bigcap_{A \in \mathcal{F}} B \cup A$$
means
$$\bigcap \{ B \cup A : A \in \mathcal{F} \}$$
where the last union is interpreted as an operator on a collection of sets. More generally, if $f$ is some operator that we can apply on sets from a collection $\mathcal{F}$ to get other sets, then
$$\bigcap_{A \in \mathcal{F}} f(A) = \bigcap \{ f(A) : A \in \mathcal{F} \}$$
A: "$B \cup A$" (in the this context) is a family of sets, there is one for each $A \in \mathcal F$.
This statement is about the distributive property of union and intersection. For a concrete example, suppose $\mathcal F = \{A_1,A_2\}$:
$$
B \cup \left(A_1 \cap A_2\right) = \left(B \cup A_1\right) \cap \left( B \cup A_2 \right).
$$
A: It is just this.
$$B \cup (\cap \mathcal{F} ) = B\cup\bigcap_{A \in \mathcal{F}} A = \bigcap_{A \in \mathcal{F}} (B\cup A)$$
It means that if you take the union of $B$ with the intersection of all elements of the family $\mathcal{F}$, this is the same as take the intersection of all elements in the form $B\cup A$ as $A$ ranges over all the family $\mathcal{F}$.
A: When I did the exercise, I interpreted $x \in \cap_{A \in \mathcal{F}} (B \cup A)$ as $\forall A \in \mathcal{F} (x \in (B \cup A))$, which did its job, since dealing with elementhood is the main task in this exercise.
