Contention of arbitrary families of sets

Question

Apparently for $\{ A_{ij} \}_{(i,j)\in I \times J}$, $$\bigcup\limits_{j \in J} \left(\bigcap\limits_{i \in I} A_{ij}\right) \subseteq \bigcap\limits_{i \in I} \left(\bigcup\limits_{j \in J} A_{ij}\right).$$ I'm having a hard time proving that contention. Here's what I've got so far: "Let $x \in \left( \bigcup _{i \in I} \left( \bigcup _{i \in J} A_{ij} \right) \right)$. That means there exists some $j \in J$ such that, for all $i \in I$ $x$ is in $A_{ij}$." I'm not sure if I can therefore argue that for every $i \in I$ there is a $j \in J$ such that $x \in A_{ij}$, especially because the other contention doesn't always hold according to my textbook. I.e., $$\bigcup\limits_{j \in J} \left(\bigcap\limits_{i \in I} A_{ij}\right) \nsupseteq \bigcap\limits_{i \in I} \left(\bigcup\limits_{j \in J} A_{ij}\right).$$ I suppose this is a matter of quantifiers, but I just don't know what I'm missing.

Answer 1

You’ve started off fine. Fix $j_0\in J$ such that $x\in A_{ij_0}$ for each $i\in I$. Then for each $i\in I$ we have

$$x\in A_{ij_0}\subseteq\bigcup_{j\in J}A_{ij}\;,\tag{1}$$

and since $(1)$ is true for every $i\in I$, we must have

$$x\in\bigcap_{i\in I}\bigcup_{j\in J}A_{ij}\;.$$

Finally, $x$ was an arbitrary element of $\bigcup_{j\in J}\bigcap_{i\in I}A_{ij}$, so

$$\bigcup_{j\in J}\bigcap_{i\in I}A_{ij}\subseteq\bigcap_{i\in I}\bigcup_{j\in J}A_{ij}\;.$$