Question

Let X be a family of sets and R be a relation on X defined by ‘A is disjoint from B’. Then R is
A. Reflexive
B. Symmetric
C. Anti-symmetric
D. Transitive

Answer 1

Hint: Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive.

Complete step-by-step answer:
R is a relation on X defined by ‘A is disjoint from B’. This means that the subsets A and B in X have no elements in common or $A \cap B = \emptyset$.

For R to be reflexive, aRa should exist., where A is a non-empty subset of X. This means that-
$A \cap A = \emptyset$
But this is not possible as every non-empty set has all the elements common with itself. Hence, R is not reflexive.

For R to be symmetric, if aRb exists, then bRa should also exist. So we can proceed as-
Let $A \cap B = \emptyset$ where A and B are non-empty subsets of X.
Then clearly, $B \cap A = \emptyset$. This implies that bRa also exists.
Hence, R is symmetric.

For R to be transitive, if aRb and bRc exist, then aRc should also exist. But this is not necessarily true. We can show this by an example-
Let A = {1, 2, 3}, B = {4, 5, 6} and C = {1, 7, 8}
It is clear that A and B have no elements in common, and B and C have no elements in common as well. This implies that $A \cap B = \emptyset {\text{ and}} \;B \cap C = \emptyset$. But , A and C have one element in common, hence the relation cannot be transitive.
The given relation R is symmetric. The correct option is B.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified. Also if it is not possible to prove that relation is symmetric, reflexive or transitive, then use a suitable example to show that it is not.