Let R₁ and R₂ be two equivalence relations on a set. Consider the following…

Concept

An equivalence relation is reflexive, symmetric, and transitive. An operation preserves equivalence only when every relation produced by that operation retains all three properties.

Intersection keeps only pairs shared by both input relations, so each required pair is protected by both relations. Union may combine pairs contributed by different relations, and that mixed chain need not include its composite pair.

Application

1. For R₁ ∩ R₂, every element x has (x, x) in both R₁ and R₂, so reflexivity is preserved. If (x, y) lies in both relations, then (y, x) lies in both, so symmetry is preserved.

2. If (x, y) and (y, z) lie in R₁ ∩ R₂, both pairs lie in each transitive relation. Therefore (x, z) lies in each relation and hence in the intersection. Thus assertion (ii) holds for every pair of equivalence relations.

3. For a counterexample to assertion (i), take the set {1, 2, 3}. Let R₁ be induced by the partition {{1, 2}, {3}} and R₂ by {{1}, {2, 3}}. Their union contains (1, 2) and (2, 3), but neither relation contains (1, 3), so the union is not transitive.

Contrast

• “Both assertions are true” overlooks the mixed-chain transitivity failure possible in a union.

• “Assertion (i) is true, but assertion (ii) is false” reverses the preservation behavior of the two operations.

• “Neither (i) nor (ii) is true” overlooks that intersection preserves reflexivity, symmetry, and transitivity.

Result

Assertion (ii) is true, but assertion (i) is false in general.