Question:

Let $R$ and $S$ be two equivalence relations on a non-void set $A$. Then

Updated On: Apr 15, 2025
  • R ∪ S is equivalence relation
  • R ∩ S is equivalence relation
  • R ∩ S is not equivalence relation
  • R ∪ S is not equivalence relation
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The Correct Option is B, D

Solution and Explanation

We are given two equivalence relations \( R \) and \( S \) on a non-void set \( A \), and we are asked to analyze the properties of the union and intersection of these relations.

Step 1: Properties of Equivalence Relations
An equivalence relation on a set must satisfy three properties: 1. **Reflexivity**: \( a \sim a \) for all \( a \in A \). 2. **Symmetry**: If \( a \sim b \), then \( b \sim a \). 3. **Transitivity**: If \( a \sim b \) and \( b \sim c \), then \( a \sim c \). 

Step 2: Intersection of Equivalence Relations
The intersection of two equivalence relations \( R \cap S \) is an equivalence relation. To see this: - Reflexivity: Since both \( R \) and \( S \) are reflexive, \( a \sim a \) for all \( a \in A \), and thus \( a \sim a \) holds in \( R \cap S \). - Symmetry: If \( a \sim b \) in \( R \cap S \), then \( a \sim b \) in both \( R \) and \( S \), so \( b \sim a \) in both relations. Hence, \( a \sim b \) implies \( b \sim a \) in \( R \cap S \). - Transitivity: If \( a \sim b \) and \( b \sim c \) in \( R \cap S \), then \( a \sim b \) and \( b \sim c \) in both \( R \) and \( S \). Since both \( R \) and \( S \) are transitive, \( a \sim c \) in both \( R \) and \( S \), so \( a \sim c \) in \( R \cap S \). Thus, \( R \cap S \) is an equivalence relation. 

Step 3: Union of Equivalence Relations
The union of two equivalence relations \( R \cup S \) is **not** necessarily an equivalence relation. To see this, consider the transitivity property: - While both \( R \) and \( S \) are transitive, the union \( R \cup S \) may not be transitive. For example, if \( a \sim b \) in \( R \) and \( b \sim c \) in \( S \), but there is no direct relation between \( a \) and \( c \) in \( R \cup S \), transitivity will fail. Thus, \( R \cup S \) is not an equivalence relation.

Answer:

\[ \boxed{R \cap S \text{ is an equivalence relation, and } R \cup S \text{ is not an equivalence relation.}} \]

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