Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $X$ is reflexive if $(x, x) \in R$ for all $x \in X$.
• Symmetric Relation: A relation $R$ on a set $X$ is symmetric if for all $x, y \in X$, whenever $(x, y) \in R$, then $(y, x) \in R$.
• Transitive Relation: A relation $R$ on a set $X$ is transitive if for all $x, y, z \in X$, whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.
• Power Set: The power set of a set $S$, denoted by $\mathcal{P}(S)$ or $M$ in this case, is the set of all subsets of $S$.
• Set Intersection: The intersection of two sets $A$ and $B$, denoted by $A \cap B$, is the set of elements common to both $A$ and $B$.

Step-by-Step Solution

We are given the set $S = \{1, 2, 3, \ldots, 10\}$. The set $M$ is the power set of $S$, i.e., $M = \mathcal{P}(S)$. The relation $R$ is defined on $M$ as $R = \{(A, B) : A \cap B \neq \phi; A, B \in M\}$. We need to check if $R$ is reflexive, symmetric, and transitive.

Step 1: Checking for Reflexivity

To check if $R$ is reflexive, we need to determine if $(A, A) \in R$ for every set $A \in M$. According to the definition of $R$, $(A, A) \in R$ if and only if $A \cap A \neq \phi$. We know that for any set $A$, $A \cap A = A$. Therefore, $(A, A) \in R$ if and only if $A \neq \phi$. The set $M$ includes all subsets of $S$, which means it includes the empty set, $\phi$. Let's consider $A = \phi$. Since $\phi \subseteq S$, $\phi \in M$. For $(\phi, \phi)$ to be in $R$, we must have $\phi \cap \phi \neq \phi$. However, $\phi \cap \phi = \phi$. Thus, the condition $\phi \cap \phi \neq \phi$ is false. So, $(\phi, \phi) \notin R$. Since there exists at least one element in $M$ (the empty set) that is not related to itself, the relation $R$ is not reflexive.

Step 2: Checking for Symmetry

To check if $R$ is symmetric, we need to determine if for any $A, B \in M$, whenever $(A, B) \in R$, then $(B, A) \in R$. Assume $(A, B) \in R$. By the definition of $R$, this means $A \cap B \neq \phi$. Now, we need to check if $(B, A) \in R$. This would require $B \cap A \neq \phi$. We know that the intersection of sets is commutative, i.e., $A \cap B = B \cap A$. Therefore, if $A \cap B \neq \phi$, it directly implies that $B \cap A \neq \phi$. Thus, if $(A, B) \in R$, then $(B, A) \in R$ is always true. The relation $R$ is symmetric.

Step 3: Checking for Transitivity

To check if $R$ is transitive, we need to determine if for any $A, B, C \in M$, whenever $(A, B) \in R$ and $(B, C) \in R$, then $(A, C) \in R$. Assume $(A, B) \in R$, which means $A \cap B \neq \phi$. Assume $(B, C) \in R$, which means $B \cap C \neq \phi$. We need to determine if these conditions necessarily imply $A \cap C \neq \phi$. Let's construct a counterexample. Consider the following subsets of $S$: Let $A = \{1, 2\}$. Since $\{1, 2\} \subseteq S$, $A \in M$. Let $B = \{2, 3\}$. Since $\{2, 3\} \subseteq S$, $B \in M$. Let $C = \{3, 4\}$. Since $\{3, 4\} \subseteq S$, $C \in M$.

Check if $(A, B) \in R$: $A \cap B = \{1, 2\} \cap \{2, 3\} = \{2\}$. Since $\{2\} \neq \phi$, $(A, B) \in R$.

Check if $(B, C) \in R$: $B \cap C = \{2, 3\} \cap \{3, 4\} = \{3\}$. Since $\{3\} \neq \phi$, $(B, C) \in R$.

Now, check if $(A, C) \in R$: $A \cap C = \{1, 2\} \cap \{3, 4\} = \phi$. Since $A \cap C = \phi$, $(A, C) \notin R$.

We have found sets $A, B, C \in M$ such that $(A, B) \in R$ and $(B, C) \in R$, but $(A, C) \notin R$. Therefore, the relation $R$ is not transitive.

Common Mistakes & Tips

• When checking reflexivity, always remember to consider the empty set if it is a possible element of the domain.
• For symmetry, always rely on the commutative property of the set operation involved ($A \cap B = B \cap A$).
• To disprove transitivity, you need to find just one counterexample where the premise holds but the conclusion fails. Carefully choose your sets to make the intersections non-empty but the final intersection empty.

Summary

The relation $R$ is defined on the power set $M$ of $S=\{1, 2, \ldots, 10\}$, where $(A, B) \in R$ if $A \cap B \neq \phi$. We found that the relation is not reflexive because the empty set is not related to itself. The relation is symmetric because the intersection operation is commutative ($A \cap B = B \cap A$). Finally, the relation is not transitive, as demonstrated by a counterexample where $A \cap B \neq \phi$ and $B \cap C \neq \phi$, but $A \cap C = \phi$. Therefore, the relation $R$ is symmetric only.

The final answer is \boxed{A}.