If R and S are two equivalence relations on a set A, then

The correct answer is Option (3) → \((R \cap S)^{-1}\) is also an equivalence relation.

First, \(R \cap S\) is an equivalence relation:

    Reflexive: every (a, a) is in R and in S, so it is in \(R \cap S\).
    Symmetric: if (a, b) is in R and in S, so is (b, a).
    Transitive: if (a, b), (b, c) are in R and in S, so is (a, c).

For a symmetric relation T, (a, b) is in T exactly when (b, a) is in T, so \(T^{-1} = T\). Hence \((R \cap S)^{-1} = R \cap S\), which is an equivalence relation.

Why the other options are incorrect

• Option (1) \(R \cup S\) is also an equivalence relation. — Incorrect. The union need not be transitive. On A = {1, 2, 3}, let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} and S = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}. Then \(R \cup S\) has (1, 2) and (2, 3) but not (1, 3).

• Option (2) \((R \cup S)^{-1}\) is also an equivalence relation. — Incorrect. \(R \cup S\) is symmetric, so \((R \cup S)^{-1} = R \cup S\), which fails in the same example.

• Option (4) \(R \cap S\) is not an equivalence relation. — Incorrect. \(R \cap S\) is always an equivalence relation, since R and S are both reflexive, symmetric and transitive.