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  1. MHB Master

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    #1
    Let $f:S\to T$ be a function.
    Prove that the following statements are equivalent.


    a
    $f$ is one-to-one on $S$.




    b
    $f(A\cap B) = f(A)\cap f(B)$ for all subsets $A,B$ of $S$.




    c
    $f^{-1}[f(A)] = A$ for every subset $A$ of $S$.




    d
    For all disjoint subsets $A$ and $B$ of $S$, the images $f(A)$ and $f(B)$ are disjoint.

    Having a tough time with this one.

  2. MHB Craftsman
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    #2
    Quote Originally Posted by dwsmith View Post
    Let $f:S\to T$ be a function.
    Prove that the following statements are equivalent.
    a
    $f$ is one-to-one on $S$.

    b
    $f(A\cap B) = f(A)\cap f(B)$ for all subsets $A,B$ of $S$.
    For any function $f$ we have $f(A\cap B)\subseteq f(A)\cap f(B)$.
    So suppose that $f$ is one-to-one and $t\in f(A)\cap f(B)$.
    $\left( {\exists {a_t} \in A} \right)\left[ {f({a_t}) = t} \right]~\&~\left( {\exists {b_t} \in B} \right)\left[ {f({b_t}) = t} \right]$.
    Use one-to-one to prove $t\in f(A\cap B)$.

    Now start with $f(A\cap B) = f(A)\cap f(B)$ and continue.
    Last edited by Jameson; September 10th, 2012 at 15:34. Reason: fixed small Latex typo

  3. MHB Master

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    #3 Thread Author
    Quote Originally Posted by Plato View Post
    For any function $f$ we have $f(A\cap B)\subseteq f(A)\cap f(B)$.
    So suppose that $f$ is one-to-one and $t\in f(A)\cap f(B)$.
    $\left( {\exists {a_t} \in A} \right)\left[ {f({a_t}) = t} \right]~\&~\left( {\exists {b_t} \in B} \right)\left[ {f({b_t}) = t} \right]$.
    Use one-to-one to prove $t\in f(A\cap B)$.

    Now start with $f(A\cap B) = f(A)\cap f(B)$ and continue.

    How can $f(A\cap B) = f(A)\cap f(B)$ be used to show c is true? I don't get it.

  4. MHB Craftsman
    MHB Math Helper

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    #4
    Quote Originally Posted by dwsmith View Post
    How can $f(A\cap B) = f(A)\cap f(B)$ be used to show c is true? I don't get it.
    I would show that $a) \Leftrightarrow c)$.

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    #5
    Note that $f^{-1}(f(A))\supseteq A$, so one only has to show $f^{-1}(f(A))\subseteq A$. Suppose $f^{-1}(f(A))=A\cup B$ where $A\cap B=\emptyset$. Using (b) show that $f(B)=\emptyset$ and therefore $B=\emptyset$.

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