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Thread: Inverse of intersections

  1. #1
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    Inverse of intersections

    Prove that $\displaystyle f^{-1} ( \bigcap _ { \alpha \in A } ) = \bigcap _{ \alpha \in A } f^ {-1} (E_ \alpha ) $

    My proof.

    Pick $\displaystyle x \in f^{-1} ( \bigcap _ { \alpha \in A } E_ \alpha) $

    implies that $\displaystyle \exists y \in \bigcap _{ \alpha \in A } E_ \alpha $ such that $\displaystyle f(y)=x $

    Note that $\displaystyle y \in E_ \alpha \ \ \ \ \ \forall \alpha \in A $

    implies that $\displaystyle x \in f^{-1} (E_ \alpha ) \ \ \ \ \ \forall \alpha \in A $

    Therefore we have $\displaystyle x \in \bigcap _ { \alpha \in A } f^{-1} (E_ \alpha ) $

    On the other hand, pick $\displaystyle x \in \bigcap _ { \alpha \in A } f^ {-1} (E _ \alpha ) $

    implies that $\displaystyle x \in f^{-1} (E _ \alpha ) \ \ \ \ \ \forall \alpha \in A $

    so $\displaystyle \exists y \in E_ \alpha $ such that $\displaystyle f(y) = x $

    Note that we can find such y for every $\displaystyle \alpha \in A $, therefore, $\displaystyle x \in f^{-1} ( \bigcap _ { \alpha \in A } E_ \alpha ) $

    I have doubts about the second last line of my proof, is it reasonable for me to state that? Thanks!
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  2. #2
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    $\displaystyle \begin{gathered}
    \hfill \\
    x \in f^{ - 1} \left( {\bigcap\limits_n {E_n } } \right)\, \Rightarrow \,f(x) \in \bigcap\limits_n {E_n } \, \Rightarrow \,\left( {\forall n} \right)\left[ {f(x) \in E_n } \right]\, \Rightarrow \,\left( {\forall n} \right)\left[ {x \in f^{ - 1} \left( {E_n } \right)} \right] \hfill \\
    \end{gathered}$
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