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Math Help - Family of subset problem

  1. #1
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    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 )
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  2. #2
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    \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.
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  3. #3
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    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 ?
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