Proving that union of two equivalence relations may not be equivalence but intersection is always

Question

How do we show that if $R$ and $S$ are two equivalence relations, then $R\cup S$ may/may not be a equivalence relation but $R\cap S$ is always a equivalence relation rigorously.

i can show it by example as take $A$ set to be $\{1,2,3\}$, then $R=\{ (1,1),(2,2),(3,3),(1,2),(2,1)\}$ and $S = \{(1,1),(2,2),(3,3),(1,3),(3,1)\}$ are both equivalence but $R\cup S$ is not since it needs to have $(2,3)$ and $(3,2)$ in it too but intersection contains the identity one which is equivalence relation.

Answer 1

The counterexample is sufficient to show that such a union is not always an equivalence relation. So you have done that.

However, a single example is insufficient to show that such intersections will always be an equivalence relation. Rather, you need to prove that there can be no counterexamples.

You may do this by proving that $R\cap S$ will be reflexive, symmetric, and transitive, whenever both $R$ and $S$ are. To begin, here's a proof for the inheritance of reflexivity.

For any $\def\<{\langle}\def\>{\rangle} x\in A$, it is that $\<x,x\>\in R$ and $\<x,x\>\in S$, because $R$ and $S$ are reflexive, and therefore $\<x,x\>\in R\cap S$ (by definition of 'intersection'), meaning the intersection is reflexive too.