Question

Let $R$ and $S$ be two equivalence relations in a set $A$. Then

• A. $R\cup S$ is an equivalence relation in $A$
• B. $R\cap S$ is an equivalence relation in $A$
• C. $R-S$ is an equivalence relation in $A$
• D. None of these

Answer 1

Answer: (B)

$R\cap S$ is an equivalence relation in $A$

Given $R$, and $S$ are relations on set $A$
$\therefore R\subseteq A\times A$ and $S\subseteq A\times A\Rightarrow R\cap S\subseteq A\times A$
$\Rightarrow R\cap S$ is also a relation on $A$
Reflexivity: Let $a$ be an arbitary element of $A$.
Then $a\in A\Rightarrow (a,a)\in R$ and $(a,a)\in S$ .... [$\because R$ and $S$ are reflexive]
$\Rightarrow (a,a)\in R\cap S$
Thus, $(a,a)\in R\cap S$ for all $a\in A$
So, $R\cap S$ is a reflexive relation on $A$
Symmetry: Let $a,b\in A$ such that $(a,b)\in R\cap S$
Then, $(a,b)\in R$ and $(a,b)\in S$
[$\because R$ and $S$ are symmetric]

$\Rightarrow (b,a)\in R$ and $(b,a)\in S$
$\Rightarrow (b,a)\in R\cap S$
Thus, $(a,b)\in R\cap S\Rightarrow (b,a)\in R\cap S$ for all $(a,b)\in R\cap S$.
So, $R\cap S$ is symmetric on $A$
Transitivity:: Let $a,b,c\in A$ such that $(a,b)\in R\cap S$ and $(b,c)\in R\cap S$
$\Rightarrow$ $(a,b)\in R$ and $(a,b)\in S$
and $(b,c)\in R$ and $(b,c)\in S$
$\Rightarrow \left\{\right(a,b)\in R,(b,c)\in R\}$
and $\left\{\right(a,b)\in S,(b,c)\in S\}$
$\Rightarrow (a,c)\in R$ and $(a,c)\in S$
[$\because R$ and $S$ transitive, So $(a,b)\in R$ and $(b,c)\in R\Rightarrow (a,c)\in R$; $(a,b)\in S$ and $(b,c)\in S\Rightarrow (a,c)\in S$]
$\Rightarrow (a,c)\in R\cap S$
Thus, $(a,b)\in R\cap S$ and $(b,c)\in R\cap S$
$\Rightarrow (a,c)\in R\cap S$. So $R\cap S$ is transitive on $A$
Hence, $R\cap S$ is an equivalence relation on $A$