How to interpret big $\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}$ and $\bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$ Could someone please explain to me how to interpret the notation $$\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}$$ and   $$\bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$$ ?
I've encountered the following problem: 
Given two sets $I$ and $J$. For each $i\in I$ and $j\in J$, $M_{i,j}$ is a set. Prove that:
$$\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}\subseteq \bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$$

Analogous to the capital sigma notation, I iterated first over the second symbol and then over the first. I've tried both with partially overlapping $I$ and $J$ and with disjoint $I$ and $J$, however in both cases I got 
$$\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j} = \bigcap_{j\in J}\bigcup_{i\in I}M_{i,j} = I$$
which wasn't what I was supposed to prove. Therefore, I am doing something wrong, but I can't figure out what exactly.
 A: Suppose that $x\in\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}$; then there is at least one $i_0\in I$ such that $x\in\bigcap_{j\in J}M_{i_0,j}$. Conversely, if there is such an $i_0\in I$, then $x\in\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}$. Thus, $x\in\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}$ if and only if 
$$\exists i\in I\,\forall j\in J\,(x\in M_{i,j})\;.\tag{1}$$
Now suppose that $x\in\bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$; then for each $j\in J$ there is at least one $i(j)\in I$ such that $x\in M_{i(j),j}$, and again the converse holds as well. Thus, $x\in\bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$ if and only if 
$$\forall j\in J\,\exists i\in I\,(x\in M_{i,j})\;.\tag{2}$$
On purely logical grounds statement $(1)$ is much stronger than statement $(2)$: $(1)$ implies $(2)$, but not conversely. (As a more familiar example compare $\forall m\in\Bbb Z\,\exists n\in\Bbb Z\,(m+n=0)$, which is true, with the obviously false statement $\exists m\in\Bbb Z\,\forall n\in\Bbb Z\,(m+n=0)$.)
It may be helpful to look at the following schematic representation of what’s going on here; it’s for the particular case $I=J=\Bbb N$, but it gives the right idea for the general case. We can arrange the sets $M_{i,j}$ in an $I\times J$ matrix:
$$\begin{array}{c|cccccc|c}
&0&1&2&3&4&\ldots\\ \hline
0&M_{0,0}&M_{0,1}&M_{0,2}&M_{0,3}&M_{0,4}&\ldots&\bigcap_{j\in J}M_{0,j}\\
1&M_{1,0}&M_{1,1}&M_{1,2}&M_{1,3}&M_{1,4}&\ldots&\bigcap_{j\in J}M_{1,j}\\
2&M_{2,0}&M_{2,1}&M_{2,2}&M_{2,3}&M_{2,4}&\ldots&\bigcap_{j\in J}M_{2,j}\\
3&M_{3,0}&M_{3,1}&M_{3,2}&M_{3,3}&M_{3,4}&\ldots&\bigcap_{j\in J}M_{3,j}\\
4&M_{4,0}&M_{4,1}&M_{4,2}&M_{4,3}&M_{4,4}&\ldots&\bigcap_{j\in J}M_{4,j}\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ \hline
&\bigcup_{i\in I}M_{i,0}&\bigcup_{i\in I}M_{i,1}&\bigcup_{i\in I}M_{i,2}&\bigcup_{i\in I}M_{i,3}&\bigcup_{i\in I}M_{i,4}&\ldots
\end{array}$$
Then $x\in\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}$ if and only if $x$ belongs to every set in some row of the array; in the first paragraph I used $i_0$ to designate that row.
In order for $x$ to be in $\bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$, however, $x$ doesn’t have to be in every set in some row: all that’s necessary is that it be in some member of each column. If $I=J=\Bbb N$, for instance, $x$ might belong to the sets $M_{2j,j}$ and no others. Then $x\in\bigcup_{i\in I}M_{i,j}$ for each $j\in J$, so $x\in\bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$, but there is no row such that $x$ is in every set in that row.
A: Very often it's good to to a trivial example, e.g. Let $I = J = \Bbb N$ and  $$M_{ij} = \begin{cases} \Bbb \emptyset, & j=i \\ \Bbb N, & \text{otherwise} \end{cases}$$
then 
$$ \bigcup_{i\in I}\bigcap_{j\in J}M_{i,j} = \emptyset \subseteq \Bbb N =  \bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$$
Don't know exactly where you made an error in your proof bis but the proof is very simple:
1.) Obviously it holds: $$\bigcap_{j\in J} M_{i,j} \subseteq M_{i,j}$$
2.) That's why 
$$\bigcup_{i\in I}\left(\bigcap_{j\in J} M_{i,j}\right) \subseteq \bigcup_{i\in I} M_{i,j}$$
3.) And now the left hand side does not depend on j anymore, but the right hand side does. So taking the intersection over J leads to
$$\bigcup_{i\in I}\bigcap_{j\in J}M_{i,j}\subseteq \bigcap_{j\in J}\bigcup_{i\in I}M_{i,j}$$
