Posted by: holdenlee | April 26, 2010

Can a function be continuous only on rationals?

There exists a function that is continuous on the irrationals and discontinuous on the rationals, for example, the function f(x)=\begin{cases} d(x) & x\in \mathbb{Q}\\ 0 & x\not\in\mathbb{Q}\end{cases} where d(x) is the denominator of x 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 \mathbb{R}, such as \mathbb{Q}. The main idea is that if the set S of continuity points were countable, then we could choose a nested sequence of intervals around points of S where the variation in f goes to 0, that eventually avoids all points of S. But the common point of the intervals would be a continuity point, contradiction.

Suppose by way of contradiction that f is a function with a countable set of continuity points x_1,x_2,\ldots . Let \epsilon_i be any positive sequence converging to 0. By continuity at x_1, we can choose a nondegenerate closed interval I_1 around x_1 such that |f(x)-f(x_1)|<\epsilon_1 for all x\in I_1. (By definition we can choose an open interval, but every open interval contains a closed one.) Now define nested intervals I_n inductively as follows: once I_n has been defined, if x_{n+1}\not\in I_n let I_{n+1}=I_n, and if x_{n+1}\in I_n let I_{n+1} be an interval around x_{n+1} that is contained in I_n, does not contain x_n, and such that |f(x)-f(x_{n+1})|<\epsilon_{n+1} whenever x\in I_{n+1}. Note that each I_{n+1} avoids x_n; therefore \bigcap_n I_n contains no continuity point. However, as a series of nested intervals it contains a point y.

We can continue our construction forever- we cannot have I_n=I_{n+1}=\cdots since the set of continuity points is dense. We show y is a continuity point. Given \epsilon>0 we can choose n so that I_n\neq I_{n-1} and \epsilon_n<\frac{\epsilon}{2}. Now y\in I_n so for any z\in I_n, y,z\in I_n imply that

\displaystyle |f(y)-f(z)|\leq |f(y)-f(x_n)|+|f(x_n)-f(z)|<\epsilon_n+\epsilon_n<\epsilon.

Since there is an interval around y where f(z) differs from f(y) by at most \epsilon, y is a continuity point not in our list x_1, x_2,\ldots, contradiction.



  1. The subsets of \mathbb{R} that can form the set of discontinuity points of a set f are exactly F_{\sigma} sets, i.e. sets that are a countable union of closed sets.
    [Counterexamples in Analysis, p. 31]

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

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

    • Thanks – fixed.

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