Proof that $S$ is a partial order when it is the reflexive closure of a strict partial order

Question

Suppose $R$ is a strict partial order on $A$. Let $S$ be the reflexive closure of $R$. Show that $S$ is a partial order on A. ("How to Prove it", Chapter 4.5 exercise 4.a)

To prove that $S$ is a partial order on A, I must show the following three properties :

1. $S$ is reflexive
2. $S$ is transitive
3. $S$ is antisymmetric

Since reflexivity is implied in the definition of a reflexive closure, no need to prove anything. To prove property 2 and 3, I assumed that $S = R \cup i_A$, and then I applied proof by cases.

Is the assumption that $S = R \cup i_A$ legimiate in the proof or can it be proven without it?

Answer 1

Yes, is legitimate and you can quote Theorem 4.5.2 from the text, which says that the reflexive closure is $R\cup i_A$. No need to reprove it again.