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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?

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    $\begingroup$ Suppose $A_{m,n}=\{1\}$ if $m=n$ and $A_{m,n}=\varnothing$ if $m\ne n$. What do the two sides of your identity turn out to be in that case? $\endgroup$
    – bof
    Jan 12, 2022 at 2:22
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    $\begingroup$ Hint: even with the $\infty$s replaced by $2$s, you can find a counterexample. $\endgroup$ Jan 12, 2022 at 2:23

1 Answer 1

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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

$$\bigcap_{m=1}^\infty \Big[ \displaystyle\bigcup_{n=1}^\infty A_{nm} \Big]=\bigcap_{m=1}^\infty X=X$$

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  • $\begingroup$ As noted in Greg Martin's comment, this can be done with $m$ and $n$ just ranging from 1 to 2. Just partition into two pieces for each $n$. $\endgroup$
    – ndhanson3
    Jan 12, 2022 at 2:25

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