
Family of subset problem
Let * be a binary operation on a nonempty set X and let Y be a nonempty subset of X and $\displaystyle \{ Z_i \} _{i \in I} $ be a family of nonempty subsets of X. Prove that:
$\displaystyle Y* \bigcup _{i \in I} Z_i = \bigcup _{i \in I } (Y*Z_i ) $

$\displaystyle \left( {s,t} \right) \in Y*\left( {\bigcup\limits_{j \in I} {Z_j } } \right) \Rightarrow \quad s \in Y \wedge t \in \left( {\bigcup\limits_{j \in I} {Z_j } } \right)$
$\displaystyle t \in \left( {\bigcup\limits_{j \in I} {Z_j } } \right) \Rightarrow \quad \left( {\exists k \in I} \right)\left[ {t \in Z_k } \right]$
$\displaystyle \Rightarrow \quad \left( {s,t} \right) \in \left( {Y*Z_k } \right) \Rightarrow \quad \left( {s,t} \right) \in \left( {\bigcup\limits_{j \in I} {Y*Z_j } } \right)$
That is one direction. Can you use it to do the other direction?
I hope that I understand the notation you are using.
If this is incorrect, please tell us what is correct.

For the opposite direction, I have:
Let $\displaystyle (b,a) \in \bigcup _{j \in J } (B * A_j ) $
Then there exist an element $\displaystyle i \in J $ such that $\displaystyle (b,a) \in B*A_i $
So then $\displaystyle b \in B \ , \ a \in A_i \subseteq ( \bigcup _{j \in J} A_j ) $
Implies that $\displaystyle (b,a) \in B * \bigcup _{j \in J } A_j $
Thus complete the proof. I hope this is right.
There is a second part of the problem:
Prove that at least one of $\displaystyle B * \bigcap _{j \in J} A_j \subseteq \bigcap (B * A_j) $ and $\displaystyle B * \bigcap _{j \in J} A_j \supseteq \bigcap (B * A_j) $ holds, but they don't have to equal.
Now, so that means I would have two cases. But what gives me trouble is that what two different cases would give me those two different result? Would it be that $\displaystyle B \subseteq \bigcup _{j \in J} A_j $ and $\displaystyle B \supseteq \bigcup _{j \in J} A_j $?