Closure of reflexivity

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

Variant
\(R\cup S\)
Solution
The relation \(R\cup S\) contains all pairs \((x,x)\) e.g. from the relation \(R\), so it is reflexive.
Answer
The relation \(R\cup S\) is reflexive.
Variant
\(R\cap S\)
Solution
Every pair \((x,x)\) is in the relation \(R\) and also in the relation \(S\), so it is also in their intersection.
Answer
The relation \(R\cap S\) is reflexive.
Variant
\(R\setminus S\)
Solution
Every pair \((x,x)\) is in both relations \(R\) and \(S\), so the difference \(R\setminus S\) will not contain any such pair.
Variant
\(R\mathbin{\Delta}S\)
Solution
Because \(R\) and \(S\) both contain the pair \((x,x)\), the symmetric difference of relations \(R\) and \(S\) will not contain any such pair.
Answer
The relation \(R\mathbin{\Delta}S\) is not reflexive.
Variant
\(R\circ S\)
Solution
Composing \(xRx\) with \(xSx\) yields \(x(R\circ S)x\).
Answer
The relation \(R\circ S\) is reflexive.
Variant
\(R^{-1}\)
Solution
From \(xRx\) we immediately obtain \(xR^{-1}x\).
Answer
The relation \(R^{-1}\) is reflexive.