Closure of symmetric relation

Question

I want to show that the symmetric closure on a relation $R$ is the smallest relation on $R$. I wrote the symmetric closure as $R \cup \{(b,a) | (a,b) \in R\}$ Now to prove that the above relation is indeed the smallest relation on $R$, I chose a symmetric relation $S$ that extends $R$ and then wanted to show that $R \cup \{(b,a) | (a,b) \in R\}$ is a subset of $S$, but I can't figure how.

Answer 1

If $S$ contains $R$ and is symmetric, then for all $(a,b) \in R$ we have that $(b, a) \in S$ by definition of symmetry. Thus, $R \cup \{(b,a): (a,b) \in R\} \subset S$.

Conversely, any symmetric relation containing $R$ also contains $\{(b,a): (a,b) \in R\}.$ Thus, we indeed have that $R \cup \{(b,a): (a,b) \in R\} \subset S$ is the smallest symmetric relation containing $R$.