Question

Proof problem. If anyone here can help, that would be great.

Answer 1

Suppose R is a relation from A to B and S a relation from B to C:

Prove that Dom(S o R) = Dom(R) if Ran(R) is a subset of Dom (S)

Answer 2

what is Dom?

Answer 3

domain
ran is range

Answer 4

domain of the composite function?

Answer 5

yes

Answer 6

What is your definition of composite function. Because this is true by definition...

Answer 7

a composite function \(f\circ g\) must have \(Ran(g)\) as a subset of \(Dom(f)\) for it to be called a function.

I guess you could go by contradiction.

Suppose that \(\exists x\in Ran(R)\) s.t. \(x\notin Dom(S), then \(\exists a\in A\) s.t. \(R(a) = x\) but since \(x\notin B\) we have that S(R(a)) is not defined and thus \(R\circ S\) is not a function.

Answer 8

that should say

\(x\notin Dom(S), \text{ then } \exists a\in A\)

Answer 9

the way my book defines it is this:

\[S o R=(a,c) \in A \times C | \exists b \in B((a,b) \in R and (b,c) \in S))\]

Answer 10

see if my proof makes since

Answer 11

okay, quick question though: why does Ran(r) have to be a subset of Som(s)

Answer 12

is it because the range can never occupy a greater space than its domain?

Answer 13

your proof makes sense but she wants to prove it directly

Answer 14

no its because a composition is two steps

step one takes an elements from set A and sends it to set B, the next step takes that element and sends it to set C

Now if the first step did not send the element to something that the second step is defined for, then we cant evaluate it.

Answer 15

but can you have that the dom(r) is a subset of the ran(s)

Answer 16

PF: \(S\circ R\) is a function composed with another function we can write is as such\(S(R(a,b))\). Since this is defined, it must be that \(R(a,b)\in Dom(S)\). This is true for all \((a,b)\in Dom(R)\)

Answer 17

okay thanks