# 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

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

1. $$\omega\in\Bigl(\bigcap_{i\in I}A_i\Bigr)$$
2. $$\omega\in\Bigl(\bigcap_{j\in J}B_j\Bigr)$$

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 $$\omega\in\bigcap_{j\in J}B_j$$.

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.

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.