# Family of subset problem

• August 26th 2008, 01:59 PM
Family of subset problem
Let * be a binary operation on a nonempty set X and let Y be a nonempty subset of X and $\{ Z_i \} _{i \in I}$ be a family of nonempty subsets of X. Prove that:

$Y* \bigcup _{i \in I} Z_i = \bigcup _{i \in I } (Y*Z_i )$
• August 26th 2008, 02:23 PM
Plato
$\left( {s,t} \right) \in Y*\left( {\bigcup\limits_{j \in I} {Z_j } } \right) \Rightarrow \quad s \in Y \wedge t \in \left( {\bigcup\limits_{j \in I} {Z_j } } \right)$
$t \in \left( {\bigcup\limits_{j \in I} {Z_j } } \right) \Rightarrow \quad \left( {\exists k \in I} \right)\left[ {t \in Z_k } \right]$
$\Rightarrow \quad \left( {s,t} \right) \in \left( {Y*Z_k } \right) \Rightarrow \quad \left( {s,t} \right) \in \left( {\bigcup\limits_{j \in I} {Y*Z_j } } \right)$

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.
• September 7th 2008, 02:43 PM
For the opposite direction, I have:

Let $(b,a) \in \bigcup _{j \in J } (B * A_j )$

Then there exist an element $i \in J$ such that $(b,a) \in B*A_i$

So then $b \in B \ , \ a \in A_i \subseteq ( \bigcup _{j \in J} A_j )$
Implies that $(b,a) \in B * \bigcup _{j \in J } A_j$

Thus complete the proof. I hope this is right.

There is a second part of the problem:

Prove that at least one of $B * \bigcap _{j \in J} A_j \subseteq \bigcap (B * A_j)$ and $B * \bigcap _{j \in J} A_j \supseteq \bigcap (B * A_j)$ 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 $B \subseteq \bigcup _{j \in J} A_j$ and $B \supseteq \bigcup _{j \in J} A_j$?