A relation \(R\) is said to be circular if \(aRb\) and bRc together imply…

Definition: A relation S is circular if whenever aSb and bSc hold, then cSa holds (for all a,b,c).

Claim: If S is reflexive and circular, then S is an equivalence relation.

• Symmetric: Take any a and b with aSb. Because S is reflexive, bSb holds. Then aSb and bSb together trigger circularity, which yields bSa. Hence S is symmetric.

• Transitive: Take any a,b,c with aSb and bSc. Circularity gives cSa. Using the symmetry already proved, from cSa we get aSc. Therefore S is transitive.

Since S is reflexive by hypothesis and we have shown S is symmetric and transitive, S is an equivalence relation.

Remarks on the other possible statements:

• Reflexive and symmetric does not imply equivalence: Example: on {1,2,3} include all reflexive pairs and the symmetric pairs 1~2 and 2~3 (and their symmetric counterparts) but do not include 1~3. This relation is reflexive and symmetric but fails transitivity.

• Circular and symmetric does not imply equivalence: The empty relation on a nonempty set is symmetric and vacuously circular but not reflexive, so it is not an equivalence relation.

• Transitive and circular does not imply equivalence: Again, the empty relation on a nonempty set is transitive and vacuously circular but not reflexive, so transitivity together with circularity alone does not guarantee an equivalence relation.