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If R and S are two equivalence relations on a set A, then which of the following is always an equivalence relation?

RankGuru · Accepted Answer

1. **Reflexive Check**: If $xRx$ and $xSx$, then $(x, x)$ must be in both $R$ and $S$. Thus $(x, x) \in R \cap S$. (Reflexive)
2. **Symmetric Check**: If $(x, y) \in R \cap S$, then $(x, y) \in R$ and $(x, y) \in S$.
   - By symmetry of $R, S$, both $(y, x) \in R$ and $(y, x) \in S$.
   - Thus $(y, x) \in R \cap S$. (Symmetric)
3. **Transitive Check**: If $(x, y) \in R \cap S$ and $(y, z) \in R \cap S$, then the pairs are in both $R$ and $S$.
   - By transitivity of $R, S$, $(x, z) \in R$ and $(x, z) \in S$.
   - Thus $(x, z) \in R \cap S$. (Transitive)
4. **Counter-example for Union**: If $R = \{(1,1), (2,2), (1,2), (2,1)\}$ and $S = \{(2,2), (3,3), (2,3), (3,2)\}$, then $R \cup S$ contains $(1,2)$ and $(2,3)$ but not $(1,3)$. Violates transitivity.

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