Question

If \( R_1 \) and \( R_2 \) be two equivalence relations on a set A, then prove that \( R_1 \cap R_2 \) be also an equivalence relation.

Hint:

When proving properties for intersections (or unions) of sets, always break down the condition into its components for each set. Use the given properties of the individual sets and then combine the results using the definition of the intersection.

Answer 1

Step 1: Understanding the Concept:
To prove that \( R_1 \cap R_2 \) is an equivalence relation, we must show that it satisfies the three properties of an equivalence relation: reflexivity, symmetry, and transitivity. We are given that \( R_1 \) and \( R_2 \) are already equivalence relations, so they satisfy these properties individually.
Step 2: Key Definitions:
Let R be a relation on a set A.
Reflexive: For all \( a \in A \), \( (a, a) \in R \).
Symmetric: If \( (a, b) \in R \), then \( (b, a) \in R \).
Transitive: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).
Intersection: \( (a, b) \in R_1 \cap R_2 \) if and only if \( (a, b) \in R_1 \) and \( (a, b) \in R_2 \).
Step 3: Detailed Explanation:
Let \( R = R_1 \cap R_2 \). We will prove that R is reflexive, symmetric, and transitive.
(i) Reflexivity:
Let \( a \) be an arbitrary element of A.
Since \( R_1 \) is an equivalence relation, it is reflexive. Therefore, \( (a, a) \in R_1 \).
Since \( R_2 \) is an equivalence relation, it is reflexive. Therefore, \( (a, a) \in R_2 \).
Since \( (a, a) \in R_1 \) and \( (a, a) \in R_2 \), by the definition of intersection, \( (a, a) \in R_1 \cap R_2 \).
Thus, \( R_1 \cap R_2 \) is reflexive.
(ii) Symmetry:
Let \( (a, b) \in R_1 \cap R_2 \).
This implies \( (a, b) \in R_1 \) and \( (a, b) \in R_2 \).
Since \( R_1 \) is symmetric, if \( (a, b) \in R_1 \), then \( (b, a) \in R_1 \).
Since \( R_2 \) is symmetric, if \( (a, b) \in R_2 \), then \( (b, a) \in R_2 \).
Since \( (b, a) \in R_1 \) and \( (b, a) \in R_2 \), by the definition of intersection, \( (b, a) \in R_1 \cap R_2 \).
Thus, \( R_1 \cap R_2 \) is symmetric.
(iii) Transitivity:
Let \( (a, b) \in R_1 \cap R_2 \) and \( (b, c) \in R_1 \cap R_2 \).
From \( (a, b) \in R_1 \cap R_2 \), we have \( (a, b) \in R_1 \) and \( (a, b) \in R_2 \).
From \( (b, c) \in R_1 \cap R_2 \), we have \( (b, c) \in R_1 \) and \( (b, c) \in R_2 \).
Since \( R_1 \) is transitive, \( (a, b) \in R_1 \) and \( (b, c) \in R_1 \) implies \( (a, c) \in R_1 \).
Since \( R_2 \) is transitive, \( (a, b) \in R_2 \) and \( (b, c) \in R_2 \) implies \( (a, c) \in R_2 \).
Since \( (a, c) \in R_1 \) and \( (a, c) \in R_2 \), by the definition of intersection, \( (a, c) \in R_1 \cap R_2 \).
Thus, \( R_1 \cap R_2 \) is transitive.
Step 4: Final Answer:
Since \( R_1 \cap R_2 \) is reflexive, symmetric, and transitive, it is an equivalence relation. Hence proved.