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  $\{B_t\}_{t \in T}.$

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. 
