Let $\{ C_{i,j} : i\in I \text{ and } j\in J \}$ be a family of sets. Assume the set $I$ (of indices $i$) is arbitrary, non-empty, and non-enumerable. Similarly, assume the set $J$ (of indices $j$) is arbitrary, non-empty, and non-enumerable. Prove or disprove the equality $$\bigcup_{i\in I} \bigcap_{j\in J} C_{i,j} = \bigcap_{j\in J} \bigcup_{i\in I} C_{i,j}.$$
Check my proof of this equality. Any errors?
To prove that the equality is true, we need to show two inclusions. First, we will show that
$$\bigcup_{i\in I} \bigcap_{j\in J} C_{i,j} \subseteq \bigcap_{j\in J} \bigcup_{i\in I} C_{i,j}.$$
Suppose $x$ belongs to $\bigcup_{i\in I} \bigcap_{j\in J} C_{i,j}$. This means that there exists an index $i_0$ in $I$ such that $x \in \bigcap_{j\in J} C_{i_0,j}$. For all $j \in J$, we have $x \in C_{i_0,j}$, therefore $x \in \bigcup_{i\in I} C_{i,j}$ for all $j \in J$. This implies that $x$ belongs to $\bigcap_{j\in J} \bigcup_{i\in I} C_{i,j}$, as desired.
Now, let's show that
$$\bigcap_{j\in J} \bigcup_{i\in I} C_{i,j} \subseteq \bigcup_{i\in I} \bigcap_{j\in J} C_{i,j}.$$
Suppose $x$ belongs to $\bigcap_{j\in J} \bigcup_{i\in I} C_{i,j}$. This means that for all $j \in J$, we have $x \in \bigcup_{i\in I} C_{i,j}$. Therefore, for each $j \in J$, there exists an index $i_j$ in $I$ such that $x \in C_{i_j,j}$. If for each index $j$ we choose a unique index $i_j$, we can define a function $\varphi: J\to I$ by setting $\varphi(j)=i_{j}$. Then, for all $j \in J$, we have $x \in C_{i_j,j}$, which implies that $x$ belongs to $\bigcap_{j\in J} C_{i_j,j}$. Therefore, $x$ belongs to $\bigcup_{i\in \varphi(J)} \bigcap_{j\in J} C_{i,j}$. On the other hand, since $\varphi(J)\subset I$, we have $$\bigcup_{i\in \varphi(J)} \bigcap_{j\in J} C_{i,j} \subseteq \bigcup_{i\in I} \bigcap_{j\in J} C_{i,j}$$ With this, we conclude that $$\bigcup_{i\in I} \bigcap_{j\in J} C_{i,j} = \bigcap_{j\in J} \bigcup_{i\in I} C_{i,j}$$ is true.
The inclusion $$\bigcup_{i\in I} \bigcap_{j\in J} C_{i,j} \subseteq \bigcap_{j\in J} \bigcup_{i\in I} C_{i,j}$$ is valid and your proof is correct. However, the reverse inclusion $$\bigcap_{j\in J} \bigcup_{i\in I} C_{i,j} \subseteq \bigcup_{i\in I} \bigcap_{j\in J} C_{i,j}.$$ is false. Can you find a counterexample? (Hint: "If for each index $j$ we choose a unique index $i_j$")