Let $S=\{1,2,3, \ldots, 10\}$. Suppose $M$ is the set of all the subsets of $S$, then the relation

$\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} \neq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}\}$ is :

<p>Let $S=\{1,2,3, \ldots, 10\}$</p> <p>$R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}$</p> <p>For Reflexive,</p> <p>$M$ is subset of '$S$'</p> <p>So $\phi \in \mathrm{M}$</p> <p>for $\phi \cap \phi=\phi$</p> <p>$\Rightarrow$ but relation is $\mathrm{A} \cap \mathrm{B} \neq \phi$</p> <p>So it is not reflexive.</p> <p>For symmetric,</p> <p>$$\begin{array}{ll} \text { ARB } & \mathrm{A} \cap \mathrm{B} \neq \phi, \\ \Rightarrow \mathrm{BRA} & \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi, \end{array}$$</p> <p>So it is symmetric.</p> <p>For transitive,</p> <p>$$\begin{aligned} \text { If } A & =\{(1,2),(2,3)\} \\ B & =\{(2,3),(3,4)\} \\ C & =\{(3,4),(5,6)\} \end{aligned}$$</p> <p>$\mathrm{ARB}$ & $\mathrm{BRC}$ but $\mathrm{A}$ does not relate to $\mathrm{C}$ So it not transitive</p>