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Math Help - Equality of sets

  1. #1
    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 \longrightarrow Y be a function. Show that

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

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

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

    (b) Is the following correct?

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

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

    If f(x) \epsilon f(S) \cap f(T), then f(x) \epsilon f(S) and f(x) \epsilon f(T). Because f is 1-1, x \epsilon S and x \epsilon T. This means x \epsilon S \cap T. Again because f is 1-1, f(x) \epsilon f(S \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 \cap T)) \subset f(S) \cap f(T). It is left to show that f(S) \cap f(T) \subset f(S \cap T)).

    If f(x) \epsilon f(S) \cap f(T), then f(x) \epsilon f(S) and f(x) \epsilon f(T). Because f is 1-1, x \epsilon S and x \epsilon T. This means x \epsilon S \cap T. Again because f is 1-1, f(x) \epsilon f(S \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.
    \begin{gathered}<br />
  z \in f(S) \cap f(T) \hfill \\<br />
  z \in f(S)\;\& \;z \in f(T) \hfill \\<br />
  \left( {\exists s \in S} \right)\left[ {f(s) = z} \right]\;\& \;\left( {\exists t \in T} \right)\left[ {f(t) = z} \right] \hfill \\<br />
  z = f(s) = f(t)\; \Rightarrow \;s = t \hfill \\<br />
  \; \Rightarrow \;z \in f(S \cap T) \hfill \\ <br />
\end{gathered}
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