Does Indexed Union distribute over Indexed Intersection? (same index) I am wondering whether this is true: 
$$
\big(\bigcup_i A_i\big)\ \cap\ \big(\bigcup_i B_i\big) = \bigcup_i \big( A_i \cap B_i \big)
$$
I have found proof that union distributes over intersection of not indexed sets and some proof where only one of the two sets is indexed, but I cannot prove myself whether this holds or not. 
Any ideas? 
 A: It doesn't hold, not even for finite index sets.
For example
$$(\{1\} \cup \{2\}) \cap (\{2\} \cup \{1\}) = \{1,2\}$$
but
$$(\{1\} \cap \{2\}) \cup (\{2\} \cap \{1\}) = \emptyset$$
What is true is this:
$$\left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) = \bigcup_{i\in I, j \in J} A_i \cap B_j$$
A: A proof of the statement that mechanodroid has indicated to be true:
Let $X$ be a nonempty set and let $\{A_i : i \in I \}, \{ B_j : j \in J\} \subset P(X)$ (to avoid trivialities, we also assume that there is a nonempty set in each of these indexed families of sets). We will show that
\begin{equation} \left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) = \bigcup_{i\in I,\, j \in J} A_i \cap B_j . \end{equation}
Suppose $x \in \left(\bigcup_{i\in I} A_i \right) \cap \left(\bigcup_{j\in J} B_j\right)$. So $x \in \bigcup_{i\in I} A_i$ and $x \in \bigcup_{j\in J} B_j$ (definition of set intersection). By the definition of set union for an indexed family of sets, there exists $k \in I$ and there exists $l \in J$ such that $x \in A_k$ and $x \in B_l$. Hence $x \in A_k \cap B_l$. Thus, $\left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) \subseteq \bigcup_{i\in I,\, j \in J} A_i \cap B_j $, since clearly $x \in \bigcup_{i\in I,\, j \in J} A_i \cap B_j \left( \supseteq A_k \cap B_l \right)$.
Suppose $x \in \bigcup_{i\in I,\, j \in J} A_i \cap B_j $. Then there exists $(k,l) \in I \times J$ such that $x \in A_k \cap B_l$. So there exists $k \in I$ and $l \in J$ (definition of the Cartesian product) such that $x \in A_k$ and $x \in B_l$. Thus we have that $x \in \bigcup_{i\in I} A_i$ and $x \in \bigcup_{j\in J} B_j$, and so $x \in \left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right)$. Therefore, $\left(\bigcup_{i\in I} A_i\right) \cap \left(\bigcup_{j\in J} B_j\right) = \bigcup_{i\in I,\, j \in J} A_i \cap B_j $.

As a counterexample (to the claim in aky's question) with a countably infinite indexing set ($\mathbb{N}$) and sets that contain more than one point (for fun, since mechanodroid's answer is already sufficient, as well as a stronger counterexample): 
For $n=1,2, \ldots$ we define the following subsets of $\mathbb{R}$
\begin{align} & A_n:=\Big[-1, \frac{1}{n}-\frac{1}{2} \Big] \text{ , and} \\&
B_n:=\Big[\frac{1}{n} , \frac{3}{2} \Big] .
\end{align}
So 
\begin{align}\left(\bigcup_{n\in \mathbb{N}} A_n\right) \bigcap \left(\bigcup_{n\in \mathbb{N}} B_n\right) & = [-1,1/2 ] \cap (0, 3/2] \\& = (0, 1/2],
\end{align}
BUT
\begin{align}\bigcup_{n\in \mathbb{N}} \left( A_n \cap B_n \right) & = \bigcup_{n\in \mathbb{N}} \emptyset \\& = \emptyset.
\end{align}
