Prove that $x_\text{sym}(R)=R \cup R^{T}$

Question

Prove that $x_\text{sym}(R)=R \cup R^{T}$, where $R$ is a relation.

So I'm supposed to show that the union of symmetric relations is symmetric.

I would start by setting $(a,b) \in x_{sym}(R)$ which means that

$(a,b) \in R \cup R^T \Leftrightarrow$

$\Leftrightarrow (a,b) \in R \text{ }\vee (a,b) \in R^T \Leftrightarrow$

$\Leftrightarrow (b,a) \in R \text{ }\vee (b,a) \in R^T \Leftrightarrow$

$\Leftrightarrow (b,a) \in R \cup R^T \Leftrightarrow$

$\Leftrightarrow (b,a) \in x_{sym}(R)$

I just used some known definitions (especially that of the union) but maybe I did a mistake because I wasn't sure about the tranposition and just considered it as another normal symmetric relation.

Did I do it correctly and is there a better way of proving this?

Answer 1

What bazaar, misleading notation. Much prefered is simply sym(R).

All you did was to show that sym(R) is symmetric.
To show that sym(R) is the symmetric closure of R,
you have to also show that sym(R) is the smallest symmetric relation containg R.

All of that did not touch the problem you are supposed
to prove, that a union of symmetric relations is a symmetric relation. That is an easy problem which does not need sym(R).

Just as easy an exercise is to show an intersection of symmetric relations is a symmetric relation.