Intersection of two sets which are equivalence on set A is always equivalence?

Question

If $R_{1}$ and $R_{2}$ are equivalence relations on set A ,then$ R_{1}\bigcap R_{2}$ must be equivalence relation.

firstly, I am not understanding the function of R,I think that, this is only a relation (according to me) how can it be an equivalence it's not a set.I have been trying to solve this problem for 3 hrs now I lift up my hands.

Answer 1

firstly, I am not understanding the function of R,I think that, this is only a relation (according to me) how can it be an equivalence it's not a set.

A relation $R$ is a set of pairs, $R\subseteq A\times A$. $aRb$ is just a short form to write $(a,b)\in R$.

Now if we define $R=R_1\cap R_2$, we have $$\begin{align} aRb &\iff (a,b)\in R &&\text{(expansion of short form)}\\ &\iff (a,b)\in R_1\cap R_2 &&\text{(definition of R)}\\ &\iff (a,b)\in R_1\land (a,b)\in R_2 &&\text{(definition of intersection)}\\ &\iff aR_1b\land aR_2b &&\text{(rewrite back into short form)} \end{align}$$

So your task is to show that $R$ is an equivalence relation, that is, is reflexive, symmetric and transitive.