Is it true that finite intersection distributes over arbitrary unions? I have come across the problem of showing that $$\bigcap_{i=1}^n \Big ( \bigcup_{\alpha\in A} X_\alpha^{(i)}\Big) = \bigcup_{\alpha\in A} \Big ( \bigcap_{i=1}^n X_\alpha^{(i)} \Big)$$ for some family of sets $\{ A_\alpha^{(i)} \}$. Is this statement even true for arbitrary sets? If not, what conditions do I need to assert this. 
 A: It’s false in general. Let $n=2$ and $A=\{0,1\}$. Let $X_0^{(1)}=X_1^{(2)}=\{0\}$ and $X_0^{(2)}=X_1^{(1)}=\{1\}$. Then
$$\bigcap_{i=1}^2\bigcup_{\alpha\in A}X_\alpha^{(i)}=\left(X_0^{(1)}\cup X_1^{(1)}\right)\cap\left(X_0^{(2)}\cup X_1^{(2)}\right)=\{0,1\}\cap\{0,1\}=\{0,1\}\;,$$
but
$$\bigcup_{\alpha\in A}\bigcap_{i=1}^2X_\alpha^{(i)}=\left(X_0^{(1)}\cap X_0^{(2)}\right)\cup\left(X_1^{(1)}\cap X_1^{(2)}\right)=\varnothing\cup\varnothing=\varnothing\;.$$
This is such a simple example that I suspect that you’ll need very stringent conditions indeed to make it hold.
A: Actually it can be proved that
$$
\bigcap_{i=1}^n \Big( \bigcup_{\alpha\in A} X_\alpha^{(i)}\Big) = \bigcup_{\overrightarrow{\alpha}\in \overrightarrow{A}} \Big ( \bigcap_{\beta\in A_{\overrightarrow{\alpha}}} X_\beta^{(i)} \Big)
$$
where $\overrightarrow{A}=\underbrace{A\times \cdots \times A}_n$ and $A_{\overrightarrow{\alpha}}$ is the subset of $A$ with $n$ members (members may not be distinct). For example, if $A=\{1,2,3\}$, $A_{\alpha}$ can be $\{\underbrace{1,1,\cdots,1}_n\}$, $\{\underbrace{1,1,\cdots,1}_{n-1},2\}$, $\{1,\underbrace{2,\cdots,2}_{n-1}\}$, etc.
