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Thread: Proofs: Domain of composite functions

  1. #1
    DHC
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    Proofs: Domain of composite functions

    Here is the question I am trying to solve:

    Let R be a relation from A to B and S be a relation from B to C

    Prove that: Dom($\displaystyle S \circ R$) $\displaystyle \subseteq $ Dom(R)
    I imagine that I would begin by trying to prove that:

    $\displaystyle \varphi \in$ Dom ($\displaystyle S \circ R$) $\displaystyle \Rightarrow \varphi \in$ Dom(R)

    however, I have no idea how to go about doing this. no doubt this stems from my inability to define the domain of a composite function


    any help would be much appreciated!
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  2. #2
    MHF Contributor

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    Quote Originally Posted by DHC View Post
    Let R be a relation from A to B and S be a relation from B to C, Prove that: Dom(SoR ) Dom(R)
    Here is the definition of the composition.
    $\displaystyle \left( {a,b} \right) \in S \circ R \Rightarrow \left( {\exists c \in B} \right)\left[ {\left( {a,c} \right) \in R \wedge \left( {c,b} \right) \in S} \right]$.
    From which the proof follows rather easily.
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  3. #3
    DHC
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    Quote Originally Posted by Plato View Post
    Here is the definition of the composition.
    $\displaystyle \left( {a,b} \right) \in S \circ R \Rightarrow \left( {\exists c \in B} \right)\left[ {\left( {a,c} \right) \in R \wedge \left( {c,b} \right) \in S} \right]$.
    From which the proof follows rather easily.
    how does this sound, then?


    $\displaystyle S \circ R = ${$\displaystyle (a,c): (\exists b \in B) (a,b) \in R \wedge (b,c) \in S$} $\displaystyle \Rightarrow Dom(S \circ R)=${$\displaystyle x \in Dom(R): R(x) \in Dom(S)$}


    hence for x to be in $\displaystyle Dom(S \circ R)$, x must be in Dom(R)
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