The famous Thomae's function provides an example which is continuous only on irrationals. Similarly, can we have a function which is continuous only on rationals and nowhere else?
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2$f\colon \mathbb{Q} \rightarrow \mathbb{Q}$ with $f(x)=0$ is continuous on the rationals and there is nowhere else. – Henry Oct 09 '15 at 07:06
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Thanks for pointing it out. – Debdas Ghosh Oct 10 '15 at 07:52
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No.
The Wikipedia article on Thomae's function gives an elegant proof of why this is impossible. It includes links to terms it uses. I'll present what it says here but I suggest you view it on Wikipedia.
A natural follow-up question one might ask is if there is a function which is continuous on the rational numbers and discontinuous on the irrational numbers.
This turns out to be impossible; the set of discontinuities of any function must be an Fσ set. If such a function existed, then the irrationals would be an Fσ set and hence, as they don't contain an interval, would also be a meager set.
It would follow that the real numbers, being a union of the irrationals and the rationals (which is evidently meager), would also be a meager set. This would contradict the Baire category theorem.
