Let rho_{1} and rho_{2} be two equivalence re | vedclass

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(B) An equivalence relation must be reflexive, symmetric, and transitive. $1$. Intersection: If $\rho_{1}$ and $\rho_{2}$ are equivalence relations, t

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$\rho_{1} \cap \rho_{2}$ is an equivalence relation but $\rho_{1} \cup \rho_{2}$ is not necessarily so.

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(B) An equivalence relation must be reflexive, symmetric, and transitive. $1$. Intersection: If $\rho_{1}$ and $\rho_{2}$ are equivalence relations, then $\rho_{1} \cap \rho_{2}$ is always reflexive, symmetric, and transitive. Thus, $\rho_{1} \cap \rho_{2}$ is an equivalence relation. $2$. Union: The union $\rho_{1} \cup \rho_{2}$ is always reflexive and symmetric, but it is not necessarily transitive. For example, let $S = \{1, 2, 3\}$. Let $\rho_{1} = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ and $\rho_{2} = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}$. Then $\rho_{1} \cup \rho_{2}$ contains $(1,2)$ and $(2,3)$, but it does not contain $(1,3)$, so it is not transitive. Therefore, $\rho_{1} \cap \rho_{2}$ is an equivalence relation, but $\rho_{1} \cup \rho_{2}$ is not necessarily so.

Let $\rho_{1}$ and $\rho_{2}$ be two equivalence relations defined on a non-void set $S$. Then

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