Proofs: Domain of composite functions

**DHC** · Apr 20th 2009, 02:24 PM

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( $S \circ R$ ) $\subseteq$ Dom(R)

I imagine that I would begin by trying to prove that:

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

**Plato** · Apr 20th 2009, 02:40 PM

Originally Posted by DHC

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.
$\left( {a,b} \right) \in S \circ R \Rightarrow \left( {\exists c \in B} \right)\left[ {\left( {a,c} \right) \in R \wedge \left( {c,b} \right) \in S} \right]$ .
From which the proof follows rather easily.

**DHC** · Apr 20th 2009, 02:49 PM

Originally Posted by Plato

Here is the definition of the composition.
$\left( {a,b} \right) \in S \circ R \Rightarrow \left( {\exists c \in B} \right)\left[ {\left( {a,c} \right) \in R \wedge \left( {c,b} \right) \in S} \right]$ .
From which the proof follows rather easily.

how does this sound, then?

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

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

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