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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

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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

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

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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.

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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.

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