# a function that continuous at every irrational but discontinuous at rational

Does exist a function $f$ that discontinuous at rational and continuous at every irrational but the restriction $f$ to the set of all irrational numbers is not constant and $f(q_n)$ is convergent where $\{q_n\}$ is a sequence of rationals.

Thomae function is one of an example of function that satisfies above condition if the restriction $f$ to the set of all irrational numbers is constant.

• Do you mean that $f(q_n)$ is always convergent for any arbitrary sequence $\{q_n\}$ of rational number? May 18, 2016 at 10:14
• i don't think "convergence" suit this question May 18, 2016 at 11:15

Denote by $T(x)$ Thomae's function. Then $$f(x)=T(x)+x$$ is a function satisfying your condition.