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Thread: Graphs

  1. #1
    Senior Member tukeywilliams's Avatar
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    Graphs

    Given $\displaystyle f: X \to Y $ is a function with graph $\displaystyle G_f \subseteq X \times Y $ prove that $\displaystyle f $ is surjective $\displaystyle \Leftrightarrow \forall y \in Y, (X \times \{y \} \cap G_f) \neq \emptyset $.

    Proof: '$\displaystyle \Rightarrow $' direction: If $\displaystyle f $ is surjective, then $\displaystyle \forall y \in Y, \exists x \in X, y = f(x) $. Then $\displaystyle X \times \{y \} \cap G_f \Rightarrow x \in X $ and $\displaystyle y \in \{y \} $ or $\displaystyle y = y $. Then $\displaystyle X \times \{y \} \cap G_f \neq \emptyset $ because by definition $\displaystyle f $ is surjective.

    '$\displaystyle \Leftarrow $' direction: If $\displaystyle \forall y \in Y, (X \times \{y \} \cap G_f) \neq \emptyset $ then $\displaystyle x \in X $ and $\displaystyle y = y $ so that $\displaystyle X \times \{y \} \cap G_f \neq \emptyset$. Then $\displaystyle \forall y \in Y, \exists x \in X, y = f(x) $ so that $\displaystyle f $ is surjective. $\displaystyle \square $

    Is this proof correct?

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  2. #2
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    Quote Originally Posted by tukeywilliams View Post
    Is this proof correct?
    Yes it is correct and well done.
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