Give a counterexample to $\bigcup\limits_{t \in T} (A_t \cap B_t)= \bigcup\limits_{t \in T} A_t \cap \bigcup\limits_{t \in T} B_t$ Let $\{A_t\}_{t \in T}$ and  $\{B_t\}_{t \in T}$  be  two non-empty indexed families of set. 
Find a counterexample to
$$\bigcup\limits_{t \in T} (A_t \cap B_t)= \bigcup\limits_{t \in T} A_t \cap  \bigcup\limits_{t \in T} B_t.$$
It is clear that $T$  must be an infinite set but I have no idea about the families $\{A_t\}_{t \in T}$ and  $\{B_t\}_{t \in T}.$
 A: Hint: $T$ doesn't need to be infinite, for $T = \{1, 2\}$ you can easily find an example where the left-hand side is empty, while the right-hand side is not.
A: Let's analyze the two unions.


*

*$x\in\bigcup_{t\in T}(A_t\cap B_t)$ if there is some $t$ such that $x\in A_t$ and $x\in B_t$.

*$x\in\bigcup_{t\in T}A_t\cap\bigcup_{t\in T}B_t$ if there is some $t$ such that $x\in A_t$ and some $t'$ such that $x\in B_{t'}$.


One inclusion is trivial. Condition (1) is exactly when we can take $t=t'$ for (2). So if we want a counterexample, we need to find an example where the second condition holds, but the first does not.
For example, if every $A_t=B_{t'}$ for some $t\neq t'$, but $A_t\cap B_t=\varnothing$ for all $t\in T$. Then one side is empty, while the other is the entire union.
To find an example like that you can do with $T=\{1,2\}$. I'll leave you to it now.
A: To make it more concrete, perhaps, let $T=\{\alpha,\beta\}$, where $\alpha\neq\beta$. Now, let $A_\alpha=\{\alpha\}$, $A_\beta=\{\beta\}$, $B_\alpha=\{\beta\}$, and $B_\beta=\{\alpha\}$. Then we have that 
$$
\bigcup_{t\in T}(A_t\cap B_t)=(A_\alpha\cap B_\alpha)\cup(A_\beta\cap B_\beta)=\varnothing\cup\varnothing=\varnothing,
$$
but we also have that
$$
\bigcup_{t\in T}A_t\cap \bigcup_{t\in T}B_t=(A_\alpha\cup A_\beta)\cap(B_\alpha\cup B_\beta)=\{\alpha,\beta\}\cap\{\beta,\alpha\}=\{\alpha,\beta\}\neq\varnothing,
$$
and this serves as a counterexample. 
