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Thread: Inverse of sets

  1. #1
    Mel
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    Inverse of sets

    I have attached a question that I am having trouble with. Thanks to anyone who can help.DOC1.DOC
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  2. #2
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    Sets and Inverse functions

    Hello Mel

    Here is the problem.

    Suppose $\displaystyle X$ and $\displaystyle Y$ are nonempty sets and $\displaystyle f: X \rightarrow Y$ is a function. Let $\displaystyle A$ and $\displaystyle B$ be subsets of $\displaystyle Y$. Prove that $\displaystyle f^{-1} (A \cup B) = f^{-1}(A) \cup f^{-1}(B)$

    We do this by showing that $\displaystyle f^{-1} (A \cup B) \subseteq f^{-1}(A) \cup f^{-1}(B)$ and $\displaystyle f^{-1}(A) \cup f^{-1}(B) \subseteq f^{-1} (A \cup B)$


    So, for the first part:

    Suppose $\displaystyle x \in f^{-1}(A \cup B)$. Then for some $\displaystyle y \in Y$, $\displaystyle f(x) = y$, and $\displaystyle y \in A \cup B$.

    $\displaystyle \Rightarrow y \in A$ or $\displaystyle y \in B$ [Note: throughout this proof 'or' means 'inclusive or'. So $\displaystyle y \in A$ or $\displaystyle y \in B$ means $\displaystyle y \in A$ or $\displaystyle y \in B$ or both.

    $\displaystyle \Rightarrow f(x) \in A$ or $\displaystyle f(x) \in B$

    $\displaystyle \Rightarrow x \in f^{-1}(A)$ or $\displaystyle x \in f^{-1}(B)$

    $\displaystyle \Rightarrow x \in f^{-1}(A) \cup f^{-1}(B)$

    So $\displaystyle x \in f^{-1}(A \cup B) \Rightarrow x \in f^{-1}(A) \cup f^{-1}(B)$

    $\displaystyle \Rightarrow f^{-1}(A \cup B) \subseteq f^{-1}(A) \cup f^{-1}(B)$ (1)


    For the second part:

    Suppose $\displaystyle x \in f^{-1}(A) \cup f^{-1}(B)$

    Then $\displaystyle x \in f^{-1}(A)$ or $\displaystyle x \in f^{-1}(B)$

    Then for some $\displaystyle y \in Y$, $\displaystyle f(x) = y$ and $\displaystyle y \in A$ or $\displaystyle y \in B$

    $\displaystyle \Rightarrow y \in A \cup B $

    $\displaystyle \Rightarrow x \in f^{-1}(A \cup B)$

    So $\displaystyle x \in f^{-1}(A) \cup f^{-1}(B) \Rightarrow x \in f^{-1}(A \cup B)$

    $\displaystyle \Rightarrow f^{-1}(A) \cup f^{-1}(B) \subseteq f^{-1} (A \cup B)$ (2)


    So, from (1) and (2): $\displaystyle f^{-1} (A \cup B) = f^{-1}(A) \cup f^{-1}(B)$


    Grandad
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