Question

Let a set A = A₁ ⋃ A₂ ⋃ …⋃ A_k, where A_i ⋂ A_j = Φ for i ≠ j, 1 ≤ i, j ≤ k. Define the relation R from A to A by R = {(x, y) : y ∈ A_i if and only if x ∈ A_i, 1 ≤ i ≤ k}. Then, R is

• A. reflexive, symmetric but not transitive
• B. reflexive, transitive but not symmetric
• C. reflexive but not symmetric and transitive
• D. an equivalence relation

Answer 1

The Correct Option is D

To determine the nature of the relation R defined from the set A to A, let's analyze its properties based on the given definition:

The set A is defined as the union of disjoint subsets A_1, A_2, \ldots, A_k. The relation R is specified as follows:

R = \{(x, y) : y \in A_i \text{ if and only if } x \in A_i, 1 \leq i \leq k\}.

1. To check if R is reflexive:
   • For R to be reflexive, each element x \in A must be related to itself. Since y \in A_i if and only if x \in A_i, for any x \in A_i, we have (x, x) \in R. Hence, R is reflexive.
2. To check if R is symmetric:
   • For R to be symmetric, if (x, y) \in R, then (y, x) must also be in R. Given (x, y) \in R implies both x and y belong to the same subset A_i, it follows that (y, x) \in R. Thus, R is symmetric.
3. To check if R is transitive:
   • For R to be transitive, if (x, y) \in R and (y, z) \in R, then (x, z) \in R must hold. Since x, y, z are all in the same subset A_i, (x, z) \in R follows. Hence, R is transitive.

Since R is reflexive, symmetric, and transitive, it satisfies the conditions of an equivalence relation. Therefore, the correct answer is that R is an equivalence relation.