Reformulation of intersection of two unions

Question

I want to prove a relation: $\left(\cup_{i \in I}A_i\right) \cap \left(\cup_{j \in J}B_j \right)= \cup_{i \in I, j\in J}\left(A_i \cap B_j\right)$. Would someone check if the following my attempt is correct or not? \begin{align*} x\in\left(\cup_{i \in I}A_i\right) \cap \left(\cup_{j \in J}B_j \right) &\Leftrightarrow \left(\exists i\in I \right)\left[x \in A_i\cap\left(\bigcup_{j\in J}B_j\right)\right]\\ &\Leftrightarrow \left(\exists i\in I \right)\left[\left(\exists j \in J\right)\left[x \in A_i \cap B_j\right]\right)\\ &\Leftrightarrow \left(\exists i\in I, j\in J\right)\left[x\in A_i \cap B_j\right]\\ &\Leftrightarrow x \in \bigcup_{i\in I, j\in J}\left(A_i \cap B_j\right) \end{align*}

Answer 1

It is correct, yes. I personally would use the notation $\bigcup_{(i,j) \in I \times J} (A_i \cap B_j)$ to indicate that $i \in I$ and $j \in J$ can be chosen independently of each other and that we take the union over all such combinations. But this is what conventionally $\bigcup_{i \in I, j \in J} (A_i \cap B_j)$ is also taken to mean. As long as we're clear on that.

On another note, you don't need to go all symbolic and bi-implicational. Showing two inclusions in words and fewer symbols is often more readable.