To ascertain the properties of the relation \( R = \{ (A, B): A \cap B eq \emptyset \} \), we must evaluate its reflexivity, symmetry, and transitivity.
- Reflexivity: A relation \( R \) on a set \( M \) is reflexive if for all \( x \in M \), \( (x, x) \in R \). In this context, for a relation \( R \) on the power set of \( S \), every subset \( A \) of \( S \) must satisfy \( A \cap A eq \emptyset \). Since \( A \cap A = A \), this condition holds unless \( A \) is the empty set. As \( \emptyset \) is a subset of \( S \) and \( \emptyset \cap \emptyset = \emptyset \), the relation \( R \) is not reflexive because it fails for the empty set.
- Symmetry: A relation \( R \) is symmetric if for all \( A, B \in S \), if \( (A, B) \in R \), then \( (B, A) \in R \). If \( A \cap B eq \emptyset \), then it is also true that \( B \cap A eq \emptyset \). Therefore, the relation \( R \) is symmetric.
- Transitivity: A relation \( R \) is transitive if for all \( A, B, C \in S \), if \( (A, B) \in R \) and \( (B, C) \in R \), then \( (A, C) \in R \). It is possible to find subsets \( A \), \( B \), and \( C \) such that \( A \cap B eq \emptyset \) and \( B \cap C eq \emptyset \), but \( A \cap C = \emptyset \). For example, let \( S = \{1, 2, 3\} \), \( A = \{1\} \), \( B = \{1, 2\} \), and \( C = \{2\} \). Then \( A \cap B = \{1\} eq \emptyset \) and \( B \cap C = \{2\} eq \emptyset \), but \( A \cap C = \emptyset \). Thus, the relation \( R \) is not transitive.
Based on these observations, the relation is symmetric only.
The correct conclusion is:
symmetric only