Let $R$ be a relation on $A$ and $S$ its transitive closure. Prove $\text{Dom}(S) = \text{Dom}(R)$ and $\text{Ran}(S) = \text{Ran}(R)$

Question

I have the following exercise from the book "How to prove it", chapter 4.5, p.211, problem 9.

Suppose $R$ is a relation on $A$, and let $S$ be the transitive closure of $R$. Prove that $\operatorname{Dom}(S) = \operatorname{Dom}(R)$ and $\operatorname{Ran}(S) = \operatorname{Ran}(R)$

Since I had no idea how to approach it, I looked at the hint in the appendix where it says:

Hint: Let $T = \lbrace(x,y) \in S \mid x \in \operatorname{Dom}(R) \text{ and } y \in \operatorname{Ran}(R)\rbrace$. Prove that $R \subseteq T$ and $T$ is transitive.

I really don't understand why this should then be the proof. Can somebody explain the rational behind this to me?

Answer 1

The idea is that the transitive closure of $R$ is the smallest transitive relation containing $R$. The hint gives a transitive relation that contains $R$ and has the same domain and range as $R$. So we have $R\subseteq S\subseteq T$.

Consequently, we have $\text{dom}(R)\subseteq\text{dom}(S)\subseteq\text{dom}(T) $, and similarly for the ranges. But the domains of $R$ and $T$ are equal, and so the domain of $S$ is the same also. Similarly for ranges.