Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We know that there exist real functions which are continuous at each irrational and dis- continuous at each rational number.

But does there exist a function $f: \mathbb{R} \to \mathbb{R}$ that is differentiable at every irrational and discontinuous at every rational?

share|improve this question
In case anyone isn't familiar with the functions mentioned in the first sentence –  Casebash Aug 11 '10 at 7:48
@Casebash: Here is one. Define $f$ in the following manner. $f(p/q)=\frac{1}{q}$, if $p/q$ is in its lowest terms with $q>0$ and $f(x)=0$ for $x \in \mathbb{R}\setminus \mathbb{Q}$. You can see this in "Methods of Real Analysis" book by R.R.Goldberg –  anonymous Aug 11 '10 at 8:14
The title could be more specific: at least the word "differentiable" should be in there somewhere. –  Pete L. Clark Aug 11 '10 at 14:43
Hi- Can i interchange the question and post it. New question would be does there exists functions $f:\mathbb{R} \to \mathbb{R}$ which is differentiable at each rational and discontinuous at every irrational –  anonymous Aug 12 '10 at 12:47
There are no functions continuous on precisely the rationals, since the irrationals are not an $F_\sigma$ set. –  user641 Aug 12 '10 at 21:56
add comment

2 Answers

up vote 3 down vote accepted

I think that there can be no such function.

Suppose that $f$ is discontinuous at some point $q$. There must be constants $A,B$ so that $f(s) < A < B < f(t)$ holds for points $s,t$ arbitrarily near to $q$.

Suppose that $x_0<q$ satisfies $q-x_0 < B-A$. Choose sequences $s_n, t_n > x_0$ converging to $q$ and satisfying $f(s_n) < A < B < f(t_n)$. Then we have

$$\frac{f(t_n) - f(x_0)}{t_n - x_0} - \frac{f(s_n) - f(x_0)}{s_n - x_0} > \frac{B - f(x_0)}{t_n - x_0} - \frac{A - f(x_0)}{s_n - x_0} \to \frac{B-A}{q - x_0} > 1$$

and may conclude that there exist $s,t$ as close to $q$ as we like satisfying

$$\frac{f(t) - f(x_0)}{t - x_0} - \frac{f(s) - f(x_0)}{s - x_0} > 1$$

Arguing similarly when $x_0 > q$ and trivially (since the Newton quotients will be unbounded as a result of the discontinuity) when $x_0 = q$ we get the same conclusion with only $|x_0 - q| < B-A$.

For $n$ a positive integer let $X_n$ denote all of the points $x_0$ in $\mathbf{R}$ for which there exist $s,t$ at distance less than $1/n$ from $x_0$ satisfying the preceding inequality. Our argument implies that, for every $n$, $X_n$ is a neighbourhood of every point $q$ at which $f$ is discontinuous (consider the points $x_0$ whose distance from $q$ is less than $1/n$ and $B-A = B_q - A_q$). If it happens that the points of discontinuity are dense (as in the case of $\mathbf{Q}$) then this implies that every $X_n$ contains an open dense set and thus $\bigcap_n X_n$ is second category in $\mathbf{R}$.

The point is of course that no point at which $f$ is differentiable can be in all the $X_n$ so if $f$ is discontinuous on the rationals then $f$ is differentiable on at most a set of 1st category in $\mathbf{R}$ (which the irrationals are not).

share|improve this answer
add comment

It came up in an answer that has been deleted that a solution to this problem can be found as "solution 2" in the following file: http://www.isibang.ac.in/~statmath/problems/soljan09.pdf. Another reference is "A theorem concerning functions discontinuous on a dense set" by Fort.

share|improve this answer
Yes; it seems Chandru1 has left the building. –  Arturo Magidin Apr 2 '11 at 5:04
The argument in Fort is the same. I think I'll just delete mine. –  Mike Apr 2 '11 at 5:19
Actually I don't know how to do that... –  Mike Apr 2 '11 at 5:22
@Mike: I don't know why you want to delete it, but if you do, there should be a link at the bottom left of your answer (when you're logged in) that says "delete". (The full list is "list edit delete flag".) –  Jonas Meyer Apr 2 '11 at 5:27
@Mike You should continue to drink this brand of tea, it has some beneficial effects. :-) –  Did Apr 2 '11 at 10:17
show 3 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.