Let S = {1, 2, 3, ldots, 10}. Suppose M is th | vedclass

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(D) Let $S = \{1, 2, 3, \ldots, 10\}$.The relation is defined as $R = \{(A, B) : A \cap B 
eq \phi; A, B \in M\}$.$1$. Reflexivity: $A$ relation $R$

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symmetric only

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(D) Let $S = \{1, 2, 3, \ldots, 10\}$.The relation is defined as $R = \{(A, B) : A \cap B 
eq \phi; A, B \in M\}$.$1$. Reflexivity: $A$ relation $R$ is reflexive if $(A, A) \in R$ for all $A \in M$. This requires $A \cap A 
eq \phi$,which means $A 
eq \phi$. Since the empty set $\phi$ is a subset of $S$ and $\phi \cap \phi = \phi$,the condition $A \cap A 
eq \phi$ is not satisfied for $A = \phi$. Thus,$R$ is not reflexive.$2$. Symmetry: $A$ relation $R$ is symmetric if $(A, B) \in R \implies (B, A) \in R$. If $A \cap B 
eq \phi$,then $B \cap A 
eq \phi$ because intersection is commutative. Thus,$R$ is symmetric.$3$. Transitivity: $A$ relation $R$ is transitive if $(A, B) \in R$ and $(B, C) \in R \implies (A, C) \in R$. Let $S = \{1, 2, 3\}$. Let $A = \{1, 2\}$,$B = \{2, 3\}$,and $C = \{3\}$. Here,$A \cap B = \{2\} 
eq \phi$ and $B \cap C = \{3\} 
eq \phi$. However,$A \cap C = \phi$. Since $A \cap C = \phi$,the condition $(A, C) \in R$ is not satisfied. Thus,$R$ is not transitive.Therefore,the relation is symmetric only.

Let $S = \{1, 2, 3, \ldots, 10\}$. Suppose $M$ is the set of all subsets of $S$. Then the relation $R = \{(A, B) : A \cap B \neq \phi; A, B \in M\}$ is :

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