If $R_{1}$ and $R_{2}$ are equivalence relations in a set $A$,show that $R_{1} \cap R_{2}$ is also an equivalence relation.
(N/A) Since $R_{1}$ and $R_{2}$ are equivalence relations,$(a, a) \in R_{1}$ and $(a, a) \in R_{2}$ for all $a \in A$.
This implies that $(a, a) \in R_{1} \cap R_{2}$ for all $a \in A$,which shows that $R_{1} \cap R_{2}$ is reflexive.
Further,if $(a, b) \in R_{1} \cap R_{2}$,then $(a, b) \in R_{1}$ and $(a, b) \in R_{2}$.
Since $R_{1}$ and $R_{2}$ are symmetric,$(b, a) \in R_{1}$ and $(b, a) \in R_{2}$,which implies $(b, a) \in R_{1} \cap R_{2}$. Thus,$R_{1} \cap R_{2}$ is symmetric.
Finally,if $(a, b) \in R_{1} \cap R_{2}$ and $(b, c) \in R_{1} \cap R_{2}$,then $(a, b) \in R_{1}, (b, c) \in R_{1}$ and $(a, b) \in R_{2}, (b, c) \in R_{2}$.
Since $R_{1}$ and $R_{2}$ are transitive,$(a, c) \in R_{1}$ and $(a, c) \in R_{2}$,which implies $(a, c) \in R_{1} \cap R_{2}$.
This shows that $R_{1} \cap R_{2}$ is transitive.
Since $R_{1} \cap R_{2}$ is reflexive,symmetric,and transitive,it is an equivalence relation.
This implies that $(a, a) \in R_{1} \cap R_{2}$ for all $a \in A$,which shows that $R_{1} \cap R_{2}$ is reflexive.
Further,if $(a, b) \in R_{1} \cap R_{2}$,then $(a, b) \in R_{1}$ and $(a, b) \in R_{2}$.
Since $R_{1}$ and $R_{2}$ are symmetric,$(b, a) \in R_{1}$ and $(b, a) \in R_{2}$,which implies $(b, a) \in R_{1} \cap R_{2}$. Thus,$R_{1} \cap R_{2}$ is symmetric.
Finally,if $(a, b) \in R_{1} \cap R_{2}$ and $(b, c) \in R_{1} \cap R_{2}$,then $(a, b) \in R_{1}, (b, c) \in R_{1}$ and $(a, b) \in R_{2}, (b, c) \in R_{2}$.
Since $R_{1}$ and $R_{2}$ are transitive,$(a, c) \in R_{1}$ and $(a, c) \in R_{2}$,which implies $(a, c) \in R_{1} \cap R_{2}$.
This shows that $R_{1} \cap R_{2}$ is transitive.
Since $R_{1} \cap R_{2}$ is reflexive,symmetric,and transitive,it is an equivalence relation.
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