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Math Help - Proof of f(S U T) = f(S) U f(T) ?

  1. #1
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    Smile Proof of f(S U T) = f(S) U f(T) ?

    I have to prove that for my discrete class, and I know it's true, but I'm stuck. I stopped to see my professor but I am still confused. Any help is greatly appreciated.

    I started the proof going left to right:
    pf: Let x f(S U T). Then f(x) S U T and there exists a z S U T such that f(z) = x.
    Since z S U T then z S or z T.
    Assume z S. Since f(z) = x, then x f(S) and x f(S) U f(T).
    Assume z T. Since f(z) = x, then x f(T) and x f(S) U f(T).
    Therefore, since x f(S) U f(T), and f(S U T) is a subset of f(S) U f(T).

    I know I need to go back the other way, and as long as the part shown is correct then the other way is too. Any criticism is welcome, or tell me where I went wrong, etc.

    Thanks again in advance for any help!!
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  2. #2
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    Quote Originally Posted by mathgirl13 View Post
    I have to prove that for my discrete class, and I know it's true, but I'm stuck. I stopped to see my professor but I am still confused. Any help is greatly appreciated.

    I started the proof going left to right:
    pf: Let x f(S U T). Then f(x) S U T and there exists a z S U T such that f(z) = x

    \color{red}\mbox{This doesn't make sense}:

    \color{red}x \in f(S \cup T) \Longrightarrow \exists y \in S \cup T\,\,s.t.\,\, x = f(y).\,If \,\,y \in S \Longrightarrow \color{red}x \in f(S)\,,\,\,if\,\,y \in T \Longrightarrow x \in f(t) \color{red}\,,\,so\,\,anyway\,\,x \in f(s) \cup f(T)


    Since z S U T then z S or z T.
    Assume z S. Since f(z) = x, then x f(S) and x f(S) U f(T).
    Assume z T. Since f(z) = x, then x f(T) and x f(S) U f(T).
    Therefore, since x f(S) U f(T), and f(S U T) is a subset of f(S) U f(T).

    I know I need to go back the other way, and as long as the part shown is correct then the other way is too. Any criticism is welcome, or tell me where I went wrong, etc.

    Thanks again in advance for any help!!
    The other direction: x \in f(S) \cup f(T) \Longrightarrow x \in f(S) \vee x \in F(T) \Longrightarrow  x=f(y)\,,\,\,y \in S\,\,\vee \,x=f(y)\,,\,\,y \in T\,\Longrightarrow \,anyway\,,\,\,y \in S \cup T\,\,\,and\,\,\,x \in f(S \cup T)

    Tonio
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  3. #3
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    Thank you!! I didn't think it made sense either? but I thought that was what my professor said so I put it, thank you for correcting it and your help!!
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