Equivalence relations proof

Question

I need to prove that if $R_1$ and $R_2$ are equivalence relations on the set $A$, then $R_1\cap R_2$ is an equivalence relation. Problem is I dont know how. Please help!

Answer 1

Since both are reflexive, $(x,x) \in R_1$ and $(x,x) \in R_2$. $(x,x) \in R_1 \cap R_2$.

Suppose $(x,y) \in R_1 \cap R_2$. $(x,y) \in R_1$ and $(x,y) \in R_2$. Since both are symmetric $(y,x) \in R_1$ and $(y,x) \in R_2$. Hence $(y,x) \in R_1 \cap R_2$.

Suppose $(x,y), (y,x) \in R_1 \cap R_2$. Hence $(x,y), (y,z) \in R_1$ and $(x,y), (y,z) \in R_2$. Since both are transitive, $(x,z) \in R_1$ and $(x,z) \in R_2)$. Hence $(x,z) \in R_1 \cap R_2$.

It has been shown that $R_1 \cap R_2$ is an equivalence relation.