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Thread: A problem with sets

  1. #1
    Member Mollier's Avatar
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    A problem with sets

    Hi.

    problem:

    Show that if $\displaystyle f:A\rightarrow B$ and E,F are subsets of A, then
    $\displaystyle f(E\cup F)=f(E)\cup f(F)$ and $\displaystyle f(E\cap F)\subseteq f(E) \cap f(F)$.

    ---------------------------------------------------------------------------------------------------------------

    attempt:


    First I try to show that $\displaystyle f(E\cup F)\subset f(E) \cup f(F)$ and then that $\displaystyle f(E) \cup f(F) \subset f(E\cup F)$.

    $\displaystyle y\in f(E\cup F) \Rightarrow f^{-1}(y)\in E\cup F \Rightarrow f^{-1}(y)\in E \; or \; f^{-1}(y)\in F$.
    $\displaystyle f(f^{-1}(y))=y\in f(E\cup F)$ and so $\displaystyle f(E\cup F)\subset f(E) \cup f(F)$.

    $\displaystyle y\in f(E)\cup f(F) \Rightarrow f^{-1}(y)\in E or f^{-1}(y)\in F \Rightarrow f^{-1}(y)\in E\cup F$.
    $\displaystyle f(f^{-1}(y))=y \in f(E\cup F)$ and so $\displaystyle f(E) \cup f(F) \subset f(E\cup F)$.

    A friend of mine pointed out that I could show this in the following manner:

    $\displaystyle
    \begin{aligned}
    f(E\cup F) =&\; \{f(x): x\in E \; or \; x\in F\}\\
    =&\; \{f(x): x\in E\} \cup \{f(x): x\in F\}\\
    =&\; f(E) \cup f(F)
    \end{aligned}
    $


    $\displaystyle y\in f(E\cap F) \Rightarrow f^{-1}(y)\in E\cap F \Rightarrow f^{-1}(y)\in E \; and \; f^{-1}(y)\in F$.
    Then $\displaystyle f(f^{-1}(y))\in f(E)\cap f(F) \Rightarrow f(E\cap F)\subset f(E) \cap f(F).$

    I do not know how to continue. If $\displaystyle y\in f(E) \cap f(F)$, what then? Should I be doing all of this differently?

    Thank you.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Mollier View Post
    Hi.

    problem:

    Show that if $\displaystyle f:A\rightarrow B$ and E,F are subsets of A, then
    $\displaystyle f(E\cup F)=f(E)\cup f(F)$ and $\displaystyle f(E\cap F)\subseteq f(E) \cap f(F)$.

    ---------------------------------------------------------------------------------------------------------------

    attempt:


    First I try to show that $\displaystyle f(E\cup F)\subset f(E) \cup f(F)$ and then that $\displaystyle f(E) \cup f(F) \subset f(E\cup F)$.

    $\displaystyle y\in f(E\cup F) \Rightarrow f^{-1}(y)\in E\cup F \Rightarrow f^{-1}(y)\in E \; or \; f^{-1}(y)\in F$.
    $\displaystyle f(f^{-1}(y))=y\in f(E\cup F)$ and so $\displaystyle f(E\cup F)\subset f(E) \cup f(F)$.
    I don't see anywhere in the hypotheses that says f is invertible so you should not be writing "$\displaystyle f^{-1}(y)$". Write, rather, "if $\displaystyle y\in f(E\cup F)$ then there exist x in $\displaystyle E\cup F$ such that f(x)= y. Since $\displaystyle x\in E\cup F$, either $\displaystyle x\in E$ or $\displaystyle x\in F$.

    Case 1: if $\displaystyle x\in E$ then $\displaystyle y\in f(E)$.

    Case 2: if $\displaystyle x\in F$ then $\displaystyle y\in f(F)$.

    $\displaystyle y\in f(E)\cup f(F) \Rightarrow f^{-1}(y)\in E or f^{-1}(y)\in F \Rightarrow f^{-1}(y)\in E\cup F$.
    $\displaystyle f(f^{-1}(y))=y \in f(E\cup F)$ and so $\displaystyle f(E) \cup f(F) \subset f(E\cup F)$.

    A friend of mine pointed out that I could show this in the following manner:

    $\displaystyle
    \begin{aligned}
    f(E\cup F) =&\; \{f(x): x\in E \; or \; x\in F\}\\
    =&\; \{f(x): x\in E\} \cup \{f(x): x\in F\}\\
    =&\; f(E) \cup f(F)
    \end{aligned}
    $
    So your friend was essentially telling you the same thing I just did!


    $\displaystyle y\in f(E\cap F) \Rightarrow f^{-1}(y)\in E\cap F \Rightarrow f^{-1}(y)\in E \; and \; f^{-1}(y)\in F$.
    Then $\displaystyle f(f^{-1}(y))\in f(E)\cap f(F) \Rightarrow f(E\cap F)\subset f(E) \cap f(F).$

    I do not know how to continue. If $\displaystyle y\in f(E) \cap f(F)$, what then? Should I be doing all of this differently?

    Thank you.
    If $\displaystyle y\in f(E\cap F)$ then there exist x in $\displaystyle E\cap F$ such that f(x)= E. Since $\displaystyle x\in E\cap F$, then $\displaystyle x\in E$ so $\displaystyle y= f(x)\in f(E)$ and $\displaystyle x\in F$ so $\displaystyle y= f(x)\in f(F)$. Therefore $\displaystyle y\in f(E)\cap f(F)$.
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  3. #3
    Member Mollier's Avatar
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    That was a crystal clear explanation, thank you very much!
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