Question

Let $R$ and $S$ be any two equivalence relations on a non-empty set $A$. Which one of the following statements is TRUE ?

• A. $R\cap S,R\cup S$ are both equivalence relations
• B. $R\cup S$ is an equivalence relation
• C. $R\cap S$ is an equivalence relation
• D. Neither $R\cup S$ nor $R\cap S$ is an equivalence relation

Answer 1

The correct option is C $R\cap S$ is an equivalence relation

$R\cap S$ is an equivalence relation as can be seen from proof given below.
Let $\forall xϵA(x,x)ϵR$ and $(x,x)ϵS$ (since $R$ and $S$ are reflexive)
$\therefore (x,x)ϵR\cap S$ also $\therefore R\cap S$ is reflexive
Now, $(x,y)ϵR\cap S$
$\Rightarrow (x,y)ϵR$ and $(x,y)ϵS$
$\Rightarrow (y,x)ϵR$ and $(y,x)ϵS$
(Since $R$ and $S$ are symmetric)
$\Rightarrow (y,x)ϵR\cap S$
$\therefore (x,y)ϵR\cap S$
$\Rightarrow (y,x)ϵR\cap S$
$R\cap S$ is therefore symmetric
Now consider
$(x,y)$ and $(y,z)ϵR\cap S$
$\Rightarrow (x,y)$ and $(y,z)ϵR$
and $(x,y)$ and $(y,z)ϵS$
$\Rightarrow (x,z)ϵR$ and $(x,z)ϵS$
(Since $R$ and $S$ are transitive)
$\Rightarrow (x,z)ϵR\cap S$
$\therefore R\cap S$ is transitive also. Since $R\cap S$ is reflexive, symmetric and transitive.
$\therefore R\cap S$ is equivalence relation.
Note: A similar argument cannot be made from $R\cup S$