Proof of $(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)=\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)$ Here is how I think I prooved $(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)=\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)$
If $\omega\in(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)$ suppose for contradiction that it $\exists(i,j)\in I\times J$ such that $\omega\notin A_i\cup B_j$.
Then $\omega\notin A_i$ and $\omega\notin B_j\implies \omega\notin\bigcap A_i$ and $\omega\notin \bigcap B_j\implies \omega \notin (\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)$ 
Now the other way:
If $\omega\in\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)$ suppose for contradiction that $\omega\notin\cap A_i$ and $\omega\notin\cap B_j\implies\exists(i,j)\in I\times J$ such that $\omega\notin A_i$ and $\omega\notin B_j\implies \omega\notin A_i\cup B_j\implies \omega\notin\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)$ 
Is everything ok in this proof?
Thanks
 A: You don't need contradiction (but your proof is good nonetheless).
Suppose
$$
\omega\in\Bigl(\bigcap_{i\in I}A_i\Bigr)\cup\Bigl(\bigcap_{j\in J}B_j\Bigr)
$$
Then one of the following two cases holds


*

*$\omega\in\Bigl(\bigcap_{i\in I}A_i\Bigr)$

*$\omega\in\Bigl(\bigcap_{j\in J}B_j\Bigr)$
Let $i_0\in I$ and $j_0\in J$ be given. In case 1, $\omega\in A_{i_0}$; in case 2, $\omega\in B_{j_0}$. Therefore, in either case $\omega\in A_{i_0}\cup B_{j_0}$. Since $i_0$ and $j_0$ are arbitrary, we conclude that
$$
\omega\in\bigcap_{\substack{i\in I\\j\in J}}(A_i\cup B_j)
$$
Conversely, suppose
$$
\omega\in\bigcap_{\substack{i\in I\\j\in J}}(A_i\cup B_j)
$$
This means that for every $i\in I$ and $j\in J$, $\omega\in A_i\cup B_j$. If $\omega\notin\bigcap_{i\in I}A_i$, then there is $i_0\in I$ with $\omega\notin A_{i_0}$. Since, for every $j\in J$, $\omega\in A_{i_0}\cup B_j$, we conclude that, for every $j\in J$, $\omega\in B_j$; therefore $\omega\in\bigcap_{j\in J}B_j$.
A: The proof looks OK, I would just add another word or two to explain how you know this:

suppose for contradiction that $\omega\notin\cap A_i$ and $\omega\notin\cap B 
_j\implies\exists(i,j)\in I\times J$ such that $\omega\notin A_i$ and $\omega\notin B_j$

Instead of writing it like that, I would first explain that from $\omega\notin \bigcap A_i$, we know there exists some $i\in I$ such that $\omega\notin A_i$. Then similarly how we get $j$, and then conclude what is true for $(i,j)$.
The way it is written now skips this step.
A: Proving equality $(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)=\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)$  means proving the two inclusions
$$
(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)
\subseteq 
\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)
\qquad 
\mbox{ and }
\qquad 
(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)
\supseteq 
\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j).
$$
And proving these two inclusions means (respectively) proving that
$$
\omega\in 
(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)
\implies \omega \in 
\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)
\\
\mbox{ and }\\
\\ 
\omega \in
\bigcap\limits_{(i,j)\in I\times J}(A_i\cup B_j)
\implies 
\omega \in
(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)
$$
We provide the first implication. Just note that each of the statements below implies the following.


*

*$\omega\in(\bigcap\limits_{i\in I} A_i)\bigcup(\bigcap\limits_{j\in J} B_j)$

*$\omega\in\bigcap\limits_{i\in I} A_i$ or $\omega \in\bigcap\limits_{j\in J} B_j$

*For all $i\in I$ we have $\omega\in A_i$ or for all $j\in J$ we have $j\in B_j$

*For all $(i,j)\in I\times J$ we have $\omega\in A_i\cup B_j$

*$\omega \in \bigcap_{(i, j)\in I\times J} A_i\cup B_j$
In a similar way (with proper use of the connectives "for all" and "there is" ) you prove the second implication.
