Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 4 / 2D / i / Solution

The intersection is always an equivalence relation. Both component relations contain $(x,x)$, so $V$ is a reflexive relation. If $xVy$, both relations relate $x$ to $y$, and symmetry in each gives $yVx$. If $xVy$ and $yVz$, transitivity in each component gives both $xRz$ and $xSz$, hence $xVz$. This is the intersection of equivalence relations; the same argument works for any family of such relations.