Proving the properties of big union of unions for indexed sets Let $I$ be an index set, and for each $i \in I$, let $J_{i}$, be another index set. For each $i \in I$ and $j \in J_{i}$, let $U_{j}$ be a set. Set X = $\bigcup\limits_{i\in I}J_{i}$.
Prove that:
$$\bigcup\limits_{i\in I} (\bigcup\limits_{j\in J_{i}}U_{j}) = \bigcup\limits_{x\in X}U_{X}$$
I understand that informally, I have to prove that "a big union of big unions is a big union". How would I write a proof for this example? Thanks.
 A: Just to give you an idea, I'll prove that $\bigcup\limits_{i\in I} (\bigcup\limits_{j\in J_i}U_{j}) \subseteq \bigcup\limits_{x\in X}U_{x}$. Then you have to prove that $\bigcup\limits_{x\in X}U_{x} \subseteq \bigcup\limits_{i\in I} (\bigcup\limits_{j\in J_i}U_{j})$. I made some corrections. Please beware of the typos.  
Proof. Suppose $a \in \bigcup\limits_{i\in I} (\bigcup\limits_{j\in J_i}U_{j})$. Then we can choose some $i_0 \in I$ such that $a \in \bigcup\limits_{j\in J_{i_0}}U_{j}$. Since $a \in \bigcup\limits_{j\in J_{i_0}}U_{j}$, we can choose some $j_0 \in J_{i_0}$ such that $a \in U_{j_{0}}$. But then since $i_0 \in I$ and $j_0 \in J_{i_0}$, $j_0 \in \bigcup\limits_{i\in I}J_{i} = X$. Since $j_0 \in X$ and $a \in U_{j_{0}}$, we can conclude that $a \in \bigcup\limits_{x\in X}U_{x}$. But $a$ was an arbitrary element of $\bigcup\limits_{i\in I} (\bigcup\limits_{j\in J_i}U_{j})$, so this shows that $\bigcup\limits_{i\in I} (\bigcup\limits_{j\in J_i}U_{j}) \subseteq \bigcup\limits_{x\in X}U_{x}$, as required.
