Indexed family by two indexes Show an example of indexed family $A (i, j)$ by two indexes $i,j$, such that each set will be different: every combination of (union over $i$/intersection over $i$), (union over $j$/intersection over $j$). E.g.: $\bigcup_i\bigcup_j A(i, j)$, $\bigcup_i\bigcap_j A(i, j)$, etc.
Thanks for any advice! 
 A: An index set can be of any size, so let’s try for a simple example with a $2$-element index set $I=\{0,1\}$. Then
$$\begin{align*}
\bigcup_{i\in I}\bigcup_{j\in I}A(i,j)&=\big(A(0,0)\cup A(0,1)\big)\cup\big(A(1,0)\cup A(1,1)\big)\;,\\
\bigcup_{i\in I}\bigcap_{j\in I}A(i,j)&=\big(A(0,0)\cap A(0,1)\big)\cup\big(A(1,0)\cap A(1,1)\big)\;,\\
\bigcap_{i\in I}\bigcup_{j\in I}A(i,j)&=\big(A(0,0)\cup A(0,1)\big)\cap\big(A(1,0)\cup A(1,1)\big)\;,\text{ and}\\
\bigcap_{i\in I}\bigcap_{j\in I}A(i,j)&=\big(A(0,0)\cap A(0,1)\big)\cap\big(A(1,0)\cap A(1,1)\big)\;,
\end{align*}\tag{1}$$
and we’d like to choose $A(0,0),A(0,1),A(1,0)$, and $A(1,1)$ so that these four sets are all different. Still trying to keep things simple, let’s see if we can manage this with sets such that $A(0,0),A(0,1)$, and $A(1,0)$ are all subsets of $A(1,1)$. If we make that assumption, we can simplify $(1)$ quite a bit:
$$\begin{align*}
\bigcup_{i\in I}\bigcup_{j\in I}A(i,j)&=A(1,1)\;,\\
\bigcup_{i\in I}\bigcap_{j\in I}A(i,j)&=\big(A(0,0)\cap A(0,1)\big)\cup A(1,0)\;,\\
\bigcap_{i\in I}\bigcup_{j\in I}A(i,j)&=A(0,0)\cup A(0,1)\;,\text{ and}\\
\bigcap_{i\in I}\bigcap_{j\in I}A(i,j)&=\big(A(0,0)\cap A(0,1)\big)\cap A(1,0)\;.
\end{align*}\tag{2}$$
(You should try to justify this.) Clearly we need to have
$$\big(A(0,0)\cap A(0,1)\big)\cup A(1,0)\ne \big(A(0,0)\cap A(0,1)\big)\cap A(1,0)\;;$$
this will happen automatically if choose the sets so that $A(0,0)\cap A(0,1)\subsetneqq A(1,0)$, because then
$$\big(A(0,0)\cap A(0,1)\big)\cup A(1,0)=A(1,0)\;,$$
while
$$\big(A(0,0)\cap A(0,1)\big)\cap A(1,0)=A(0,0)\cap A(0,1)\;.$$
If we do this, $(2)$ further simplifies to
$$\begin{align*}
\bigcup_{i\in I}\bigcup_{j\in I}A(i,j)&=A(1,1)\;,\\
\bigcup_{i\in I}\bigcap_{j\in I}A(i,j)&=A(1,0)\;,\\
\bigcap_{i\in I}\bigcup_{j\in I}A(i,j)&=A(0,0)\cup A(0,1)\;,\text{ and}\\
\bigcap_{i\in I}\bigcap_{j\in I}A(i,j)&=A(0,0)\cap A(0,1)\;.
\end{align*}\tag{3}$$
We’re already assuming that $A(1,0)\subseteq A(1,1)$, so to keep these four sets distinct we need to require that $A(1,0)\subsetneqq A(1,1)$. We’ve also decided to arrange that $A(0,0)\cap A(0,1)\subsetneqq A(1,0)$. One simple way to do this is to make sure that $A(0,0)\cup A(0,1)\subsetneqq A(1,0)$; that ensures that all of the sets in $(3)$ are distinct except possibly the last two, and having $A(0,0)\subsetneqq A(0,1)$ takes care of that.
If you go back through the simplifying assumptions that I’ve made, you’ll see that they amount to requiring that
$$A(0,0)\subsetneqq A(0,1)\subsetneqq A(1,0)\subsetneqq A(1,1)\;:$$
any four sets satisfying this condition will do the trick. For instance, we could take $A(0,0)=\varnothing$, $A(0,1)=\Bbb Z$, $A(1,0)=\Bbb Q$, and $A(1,1)=\Bbb R$. Or we could be much more modest and take $A(0,0)=\varnothing$, $A(0,1)=\{0\}$, $A(1,0)=\{0,1\}$, and $A(1,1)=\{0,1,2\}$.
If you want an example with an infinite index set, say $\Bbb N$, just take either of the examples above, and for integers $i,j>1$ let $A(0,j)=A(0,1)$, $A(i,0)=A(1,0)$ and $A(i,j)=A(1,1)$.
