If $A_{nm}$ is a subset of $A$ for $n = 1,2, \dots$ and $m = 1,2, \dots$ is it necessarily true that
$$\displaystyle\bigcup_{n=1}^\infty \Big[ \displaystyle\bigcap_{m=1}^\infty A_{nm} \Big] = \displaystyle\bigcap_{m=1}^\infty \Big[ \displaystyle\bigcup_{n=1}^\infty A_{nm} \Big] \, \, ? $$
I claim that this is the case. So I attempt to prove it by showing double inclusion which implies equality.
Proof.
Let $x \in \displaystyle\bigcup_{n=1}^\infty \Big[ \displaystyle\bigcap_{m=1}^\infty A_{nm} \Big]$. By definition of union, there exists $\displaystyle\bigcap_{m=1}^\infty A_{n_0 m}$ so that $x \in \displaystyle\bigcap_{m=1}^\infty A_{n_0 m}$. By definition of intersection, we have
$$\forall \, m \in \mathbb{N}, x \in A_{n_0 m}$$
Now notice that for any $m \in \mathbb{N}$, we also have $x \in A_{n_0 m} \subset \displaystyle\bigcup_{n=1}^\infty A_{nm}$.
Now from here, I want to somehow get an intersection on the outside of that union to show inclusion the first direction, but I'm not sure how to do so. Maybe union and intersection aren't interchangeable. Maybe there's a counterexample to this claim. Any advice?
Best Answer
This is false. For every $n$, let $A_{n1},A_{n2},\dots$ partition the entire space $X$ (so are disjoint). Then $$\bigcup_{n=1}^\infty \Big[ \displaystyle\bigcap_{m=1}^\infty A_{nm} \Big] = \bigcup_{n=1} \emptyset=\emptyset$$
and
$$\bigcap_{m=1}^\infty \Big[ \displaystyle\bigcup_{n=1}^\infty A_{nm} \Big]=\bigcap_{m=1}^\infty X=X$$