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Thread: Inverse functions of unions and intersections

  1. #1
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    Inverse functions of unions and intersections

    Here is the problem:

    show that if f:A $\displaystyle \rightarrow $ B and G, H are substes of B, then

    $\displaystyle f^{-1}$(G$\displaystyle \cup$H) = $\displaystyle f^{-1}$ (G)$\displaystyle \cup$ $\displaystyle f^{-1}$(H)

    and

    $\displaystyle f^{-1}$(G$\displaystyle \cap$H) = $\displaystyle f^{-1}$ (G)$\displaystyle \cap$ $\displaystyle f^{-1}$(H)






    So I understand the main principle behind an inverse, such that if f(x)$\displaystyle \in$H, then x$\displaystyle \in f^{-1}$(H). I also understand the proof for inverse composite functions. But this is confusing. How do I prove this?? Thank you
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by osudude View Post
    Here is the problem:

    show that if f:A $\displaystyle \rightarrow $ B and G, H are substes of B, then

    $\displaystyle f^{-1}$(G$\displaystyle \cup$H) = $\displaystyle f^{-1}$ (G)$\displaystyle \cup$ $\displaystyle f^{-1}$(H)

    and

    $\displaystyle f^{-1}$(G$\displaystyle \cap$H) = $\displaystyle f^{-1}$ (G)$\displaystyle \cap$ $\displaystyle f^{-1}$(H)



    So I understand the main principle behind an inverse, such that if f(x)$\displaystyle \in$H, then x$\displaystyle \in f^{-1}$(H). I also understand the proof for inverse composite functions. But this is confusing. How do I prove this?? Thank you
    If $\displaystyle G\cup H=\varnothing$ then $\displaystyle G=H=\varnothing$ and the conclusion readily follows. So, WLOG assume that $\displaystyle G,H\ne\varnothing$. Let $\displaystyle x\in f^{-1}\left(G\cup H\right)$ then $\displaystyle f(x)\in \left(G\cup H\right)$ so that $\displaystyle f(x)\in G\text{ or }f(x)\in H$ so $\displaystyle x\in f^{-1}\left(G\right)\text{ or }f^{-1}(x)\in H\Longleftrightarrow x\in\left( f^{-1}\left(G\right)\cup f^{-1}\left(H\right)\right)$. The opposite inclusion is similar.

    If $\displaystyle G\cap H=\varnothing$ the conclusion is clear. So, suppose WLOG that $\displaystyle G\cap H\ne\varnothing$. Let $\displaystyle x\in f^{-1}\left(G\cap H\right)$ then $\displaystyle f(x)\in \left(G\cap H\right)\Longleftrightarrow f(x)\in G\text{ and }f(x)\in H$. Clearly then we see that $\displaystyle x\in f^{-1}\left(G\right)\text{ and }x\in f^{-1}\left(H\right)\Longleftrightarrow x\in\left(f^{-1}\left(G\right)\cap f^{-1}\left(H\right)\right)$. Once again, the opposite inclusion is similar.
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  3. #3
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    haha seems so easy now!!! thank you soo much!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by osudude View Post
    haha seems so easy now!!! thank you soo much!
    I made a slight "typo". Find it for your own good.
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  5. #5
    Senior Member Shanks's Avatar
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    the $\displaystyle f^{-1}$ preserve the union, intersection, and complement of sets.
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