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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.

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

1 Answer 1

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

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