Showing a counterexample of $\cap\left(\cup A_{ij}\right)\subset \cup\left(\cap A_{ij}\right)$ 
Prove, or disprove with a counterexample, the equality:
  $$
\bigcup^{\infty}_{j = 1}\left(\bigcap^{\infty}_{i = 1}A_{ij}\right) =\bigcap^{\infty}_{i = 1}\left(\bigcup^{\infty}_{j = 1}A_{ij}\right)
$$

Let's call the left set $A$ and $B$ the right one. I gave a proof for $A\subset B$ and (if it's correct) it sugests that $B\subset A$ is not always valid. It's as follows:
Let $x\in A$, then there exists $j$ such that $x\in\cap_{i=1}^{\infty} A_{ij}$ $\Rightarrow$ $\exists j\in \mathbb{N}:\forall i\in\mathbb{N}, x\in A_{ij} \Longrightarrow \forall i\in\mathbb{N},\exists j\in\mathbb{N}: x\in A_{ij} \Rightarrow \forall i\in\mathbb{N}, x\in \cup_{j=1}^{\infty}A_{ij} \Rightarrow x\in B$
I think the crucial step to why we don't always have $B\subset A$ is the implication marked above with a longer arrow. However I'm stuck in trying to get a counterexample. Any help would be appreciated.
 A: Hint: Take two disjoint non-empty sets: $E$ and $F$ and set 
 $$ A_{11}=A_{22}=E , \; \; A_{12}=A_{21}= F$$
and calculate the two sets in question.
A: On the one hand, $
x\in\bigcup^{\infty}_{j = 1}\left(\bigcap^{\infty}_{i = 1}A_{ij}\right)=:A
$
means

there exists a (positive integer) $j$ such that $x\in A_{ij}$ for all (positive integers) $i$.$\tag{1}$

If you arrange the sets $A_{ij}$ as an "infinite matrix":
$$
\begin{pmatrix}
A_{11} &   A_{12} &\cdots\\
A_{21} &   A_{22} &\cdots\\
\vdots & \vdots  &\ddots\\
\end{pmatrix}
$$
then (1) means $x$ belongs to all the sets in some column. 
On the other hand, $x\in\bigcap^{\infty}_{i = 1}\left(\bigcup^{\infty}_{j = 1}A_{ij}\right)=:B$ means

for all $i$, there exists $j$ such that $x\in A_{ij}$.$\tag{2}$

By the language of matrices, (2) means that for each row of the matrix, $x$ belongs to some set, but those sets are not necessarily in the same column. 
Now to construct an example such that $x\in B$ but $x\not\in A$, all you need to do is "assigning" $x$ to some set on each row of the matrix, such that these sets are not in the same column. 
For instance, 


*

*make $x$ belong to $A_{11}$ but not belong to other sets in the first row;

*make $x\in A_{i2}$ for all $i>1$;

*make $x$ not belong to any sets not mentioned above. 

A: Your proof of $A\subseteq B$ is correct.
Suppose that $A_{ii}=\{x\}$ for every $i\in\mathbb N$ and $A_{ij}=\varnothing$ if $i\neq j$.
Then $A=\varnothing$ as a union of empty sets and $B=\{x\}$ as an intersection of sets $\{x\}$.
