Let R be a relation from the set A to the set B, then: Prove that Ran (R)=Dom (R-1 ).

Question

there is no other detail

Answer 1

Let's recall some definitions.

1) Let A and B be sets. A relation from A to B is a subset R ⊆ A x B of the cartesian product. That is to say, R consists of ordered pairs (a, b) where a ∈ A and b ∈ B.

2) Let R be a relation from A to B. The domain of R is the subset dom(R) ⊆ A consisting of all those a ∈ A for which there exists some b ∈ B with (a, b) ∈ R.

3) Let R be a relation from A to B. The range of R is the subset ran(R) ⊆ B consisting of all those b ∈ B for which there exists some a ∈ A with (a, b) ∈ R.

4) Let R be a relation from A to B. The inverse of R is the relation R^-1 from B to A consisting of all those pairs (b, a) for which (a, b) ∈ R.

Now, to answer your question, we need to prove that ran(R) ⊆ dom(R^-1) and dom(R^-1) ⊆ ran(R). To show the first inclusion, let b ∈ ran(R). By definition of the range, there exists some a ∈ A such that (a, b) ∈ R. Thus the pair (b, a) ∈ R^-1, and so a is a witness to the fact that b ∈ dom(R^-1). To show the second inclusion, we work backwards: let b ∈ dom(R^-1). By definition of the domain, there is some a ∈ A for which (b, a) ∈ R^-1. By definition of the inverse relation, this means that (a, b) ∈ R, and so b ∈ ran(R).

Since we have shown two mutual inclusions, we must conclude that the two sets are equal.