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Math Help - Equality of two sets (using Boolean algebra)

  1. #1
    Member Pranas's Avatar
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    Equality of two sets (using Boolean algebra)

    Hello. I recently started my course, so this is pretty trivial.

    Prove:
    \displaystyle \[f\left( {\bigcup\limits_{i \in \ell } {{A_i}} } \right) = \bigcup\limits_{i \in \ell } {f\left( {{A_i}} \right)} \]


    Well, analogous examples demonstrated in my university begin with

    \displaystyle \[\forall y \in f\left( {\bigcup\limits_{i \in \ell } {{A_i}} } \right) \Rightarrow \left( {\exists x:f\left( x \right) = y} \right) \wedge \left( {x \in \bigcup\limits_{i \in \ell } {{A_i}} } \right)\]

    The problem is that I am not sure how to represent this highly abstract union of sets since
    \displaystyle \[x \in \bigcup\limits_{i \in \ell } {{A_i}}  \Rightarrow x \in {A_i} \wedge i \in \ell \]
    I guess wouldn't make enough sense (I could tell the same about intersection of those very same sets).

    So what is missing on my mind?
    Thanks for help.
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  2. #2
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    Quote Originally Posted by Pranas View Post
    The problem is that I am not sure how to represent this highly abstract union of sets since \displaystyle \[x \in \bigcup\limits_{i \in \ell } {{A_i}}  \Rightarrow x \in {A_i} \wedge i \in \ell \]
    I guess wouldn't make enough sense (I could tell the same about intersection of those very same sets
    This is one standard way: \left( {\exists j \in \ell } \right)\left[ {x \in A_j } \right]
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by Plato View Post
    This is one standard way: \left( {\exists j \in \ell } \right)\left[ {x \in A_j } \right]
    I believe it's all I needed. Thanks again

    And an intersection would be \left( {\forall j \in \ell } \right)\left[ {x \in A_j } \right] right?
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