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Thread: Equality of sets

  1. #1
    Senior Member oldguynewstudent's Avatar
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    Equality of sets

    I have a final tonight and I forgot how to do this:

    Let f: X $\displaystyle \longrightarrow$ Y be a function. Show that

    (a) f(S $\displaystyle \cap$ T)) $\displaystyle \subset$ f(S) $\displaystyle \cap$ f(T)

    (b) If f is 1-1, then show that the above inclusion is actually an equality of sets.

    (a) f(S $\displaystyle \cap$ T) $\displaystyle \equiv$ { y | y = f(x), x $\displaystyle \epsilon$ S and x $\displaystyle \epsilon$ T} $\displaystyle \equiv$ {y | y = f(x), x $\displaystyle \epsilon$ S} and {y | y = f(x), x $\displaystyle \epsilon$ T} $\displaystyle \subseteq$ f(S) $\displaystyle \cap$ f(T)

    (b) Is the following correct?

    If f 1-1 then f($\displaystyle x_1$) = f($\displaystyle x_2$) $\displaystyle \longrightarrow$ $\displaystyle x_1$ = $\displaystyle x_2$

    We have already shown that f(S $\displaystyle \cap$ T)) $\displaystyle \subset$ f(S) $\displaystyle \cap$ f(T). It is left to show that f(S) $\displaystyle \cap$ f(T) $\displaystyle \subset$ f(S $\displaystyle \cap$ T)).

    If f(x) $\displaystyle \epsilon$ f(S) $\displaystyle \cap$ f(T), then f(x) $\displaystyle \epsilon$ f(S) and f(x) $\displaystyle \epsilon$ f(T). Because f is 1-1, x $\displaystyle \epsilon$ S and x $\displaystyle \epsilon$ T. This means x $\displaystyle \epsilon$ S $\displaystyle \cap$ T. Again because f is 1-1, f(x) $\displaystyle \epsilon$ f(S $\displaystyle \cap$ T). This proves each is a subset of the other so they are equivalent.
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  2. #2
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    Quote Originally Posted by oldguynewstudent View Post
    We have already shown that f(S $\displaystyle \cap$ T)) $\displaystyle \subset$ f(S) $\displaystyle \cap$ f(T). It is left to show that f(S) $\displaystyle \cap$ f(T) $\displaystyle \subset$ f(S $\displaystyle \cap$ T)).

    If f(x) $\displaystyle \epsilon$ f(S) $\displaystyle \cap$ f(T), then f(x) $\displaystyle \epsilon$ f(S) and f(x) $\displaystyle \epsilon$ f(T). Because f is 1-1, x $\displaystyle \epsilon$ S and x $\displaystyle \epsilon$ T. This means x $\displaystyle \epsilon$ S $\displaystyle \cap$ T. Again because f is 1-1, f(x) $\displaystyle \epsilon$ f(S $\displaystyle \cap$ T). This proves each is a subset of the other so they are equivalent.
    The part in red has a logical error but the idea is correct.
    $\displaystyle \begin{gathered}
    z \in f(S) \cap f(T) \hfill \\
    z \in f(S)\;\& \;z \in f(T) \hfill \\
    \left( {\exists s \in S} \right)\left[ {f(s) = z} \right]\;\& \;\left( {\exists t \in T} \right)\left[ {f(t) = z} \right] \hfill \\
    z = f(s) = f(t)\; \Rightarrow \;s = t \hfill \\
    \; \Rightarrow \;z \in f(S \cap T) \hfill \\
    \end{gathered} $
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