If $f: R \to R$ be defined by $f(x) = \left\{ \begin{array} {1 1} 1 & \quad\text{ if x is rational } \\ 0 ,& \quad \text{if x is irrational }\\ \end{array} \right.$ then f is continous at

Question

$\begin{array}{1 1} \text{all rational points} \\ \text{all irrational points} \\ \text{all real points} \\ \text{none of these} \end{array}$