Question about distributive laws for indexed families of sets Let $X=\{ A_i \}_{i \in I} \cup \{ B_j \}_{j \in J} \cup \{ C_k \}_{k \in K}$
Is it the case that $\bigcap X = \bigcap \{ A_i \}_{i \in I} \cup 
\bigcap \{ B_j \}_{j \in J} \cup \bigcap \{ C_k \}_{k \in K}$ ?
This is a rather general question, so it might be a duplicate. If this is the case, I apologize; I didn't find the answer on a different question.
Thanks.
 A: Nope, here's a counter-example: Take $I=J=K=\{1\}$ and let $A_1=\{\emptyset\},B_1=\{\emptyset\},C_1=\{\{\emptyset\}\}$.  Then $X=\{\emptyset,\{\emptyset\}\}$ and $\bigcap{X}=\emptyset$.  However, $\bigcap\{\emptyset\}=\emptyset$ and $\bigcap\{\{\emptyset\}\}=\{\emptyset\}$, and $\emptyset \cup \{\emptyset\}=\{\emptyset\}\neq \emptyset$.
What is true is that
$$\bigcap{X} = \bigcap{\{A_i\}_{i\in I}}\cap \bigcap{\{B_j\}_{j\in J}} \cap \bigcap{\{C_k\}_{k\in K}}$$
when each of the collections $\{A_i\}_{i\in I}, \{B_j\}_{j\in J},\{C_k\}_{k\in K}$ are non-empty, which can be proven by supposing $x\in \bigcap{X}$, which means that $x\in A_i,B_j,C_k$ for every $i,j,k$.  Hence, $x\in \bigcap{\{A_i\}_{i\in I}}\cap \bigcap{\{B_j\}_{j\in J}}\cap \bigcap{\{C_k\}_{k\in K}}$, as desired.  Likewise, if $x\in \bigcap{\{A_i\}_{i\in I}}\cap \bigcap{\{B_j\}_{j\in J}}\cap \bigcap{\{C_k\}_{k\in K}}$, then $x$ is in every $A_i,B_j,C_k$ for any $i,j,k$ and so is in $\bigcap{X}$. Thus, we get equality of the two sets.
Perhaps what you're thinking when you think of this as a potential application of distributivity is that $\bigcap$ is going to distribute over a union.  As is shown above, this doesn't work.  Instead, $A \cap \cdot$ will distribute over a union, so we have $A \cap (B\cup C) = (A\cap B)\cup (A\cap C)$, but not $\bigcap{(A\cup B)} = \bigcap{A} \cup \bigcap{B}$.
