How to prove following equality if $I = \left\{\alpha \right\}$ and $J = \left\{\beta\right\}$ sets of indecies Let $I = \left\{\alpha \right\}$ and $J = \left\{\beta\right\}$ arbitrary sets of indices
$$(\bigcup_{\alpha \in I}{A_{\alpha}}) \bigcap{(\bigcup_{\beta\in J}{B_{\beta}})} = \bigcup_{\alpha, \beta \in I\times J}{(A_{\alpha}\bigcap{B_{\beta}})}$$
Can someone give me a hint or show how to prove above equality?

The most usual way of showing equality between sets is two show two inclusions:
Let $x \in \left( \bigcup_{\alpha \in I} A_\alpha \right) \cap \left( \bigcup_{\beta \in J} B_\beta \right)$
Then $x \in \bigcup_{\alpha \in I} A_\alpha$ so there is some $\alpha_x$ such that $x \in A_{\alpha_x}$, and similarly there is a $\beta_x \in J$ such that $x \in B_{\beta_x}$. This means that $(\alpha_x,\beta_x) \in I \times J$ and $x \in A_{\alpha_x} \cap B_{\beta_x}$, which makes $x$ a member of the right hand union
$\bigcap_{(\alpha,\beta) \in I \times J} (A_{\alpha} \cap B_{\beta})$.
Now show the reverse inclusion yourself, it's quite similar.
