There exists a function that is continuous on the irrationals and discontinuous on the rationals, for example, the function where is the denominator of when written in lowest terms.

Does there exist a function that is continuous on the rationals and discontinuous on the irrationals? This would seem harder- as the rationals are a much smaller set than the irrationals.

In fact, we show that no function can be continuous only on a countable dense set of , such as . The main idea is that if the set of continuity points were countable, then we could choose a nested sequence of intervals around points of where the variation in goes to 0, that eventually avoids all points of . But the common point of the intervals would be a continuity point, contradiction.

Suppose by way of contradiction that is a function with a countable set of continuity points . Let be any positive sequence converging to 0. By continuity at , we can choose a nondegenerate closed interval around such that for all . (By definition we can choose an open interval, but every open interval contains a closed one.) Now define nested intervals inductively as follows: once has been defined, if let , and if let be an interval around that is contained in , does not contain , and such that whenever . Note that each avoids ; therefore contains no continuity point. However, as a series of nested intervals it contains a point .

We can continue our construction forever- we cannot have since the set of continuity points is dense. We show is a continuity point. Given we can choose so that and . Now so for any , imply that

Since there is an interval around where differs from by at most , is a continuity point not in our list , contradiction.

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The subsets of that can form the set of discontinuity points of a set are exactly sets, i.e. sets that are a countable union of closed sets.

[Counterexamples in Analysis, p. 31]

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holdenleeon April 26, 2010at 8:04 pm

[…] Can a function be continuous only on rationals? […]

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Can a function be continuous only on rationals? | Anonymous Mathon May 17, 2013at 9:11 am

Thanks for the helpful explanation. I think the $\latex \bigcup_{n}I_n$ should be an intersection, but we can still understand.

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Dennison June 11, 2015at 2:14 pm

Thanks – fixed.

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holdenleeon July 10, 2015at 11:43 am