Closure of symmetry

Let \(R\) and \(S\) be symmetric relations on the same set. Which of the following relations are also symmetric?

Variant
\(R\cup S\)
Solution
If \(x(R\cup S)y\), then either \(xRy\) or \(xSy\), and so either \(yRx\) or \(ySx\) and so \(y(R\cup S) x\).
Answer
The relation \(R\cup S\) is symmetric.
Variant
\(R\cap S\)
Solution
If \(x(R\cap S)y\), then \(xRy\) and also \(xSy\), and so \(yRx\) and also \(ySx\), and so \(y(R\cap S) x\).
Answer
The relation \(R\cap S\) is symmetric.
Variant
\(R\setminus S\)
Solution
If \(x (R\setminus S) y\), then \((x,y)\in R\) and \((x,y) \notin S\). Therefore \((y,x)\in R\) and \((y,x) \notin S\), and so \(y (R\setminus S) x\)
Answer
The relation \(R\setminus S\) is symmetric.
Variant
\(R\mathbin{\Delta}S\)
Solution
If \(x(R\mathbin{\Delta}S) y\), then one of these possibilities must hold: \((x,y)\in R\) and \((x,y) \notin S\), or else \((x,y)\notin R\) and \((x,y) \in S\). Then either \((y,x)\in R\) and \((y,x) \notin S\), or else \((y,x)\notin R\) and \((y,x) \in S\), and therefore \(y(R\mathbin{\Delta}S) x\).
Answer
The relation \(R\mathbin{\Delta}S\) is symmetric.
Variant
\(R\circ S\)
Solution
On the set \(X=\{a,b,c\}\) consider the symmetric relations \(R=\{(a,b),(b,a)\}\) and \(S=\{(b,c),(c,b)\}\). The composition \(R\circ S\) contains only the pair \((a,c)\).
Answer
The relation \(R\circ S\) need not be symmetric.
Variant
\(R^{-1}\)
Solution
If \(x R^{-1} y\), then \(y R x\), so by symmetry \(x R y\) and so \(y R^{-1} x\).
Answer
The relation \(R^{-1}\) is symmetric.