Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove or disprove: if the set of continuous points of function $f\colon \mathbb{R}\to\mathbb{R}$ is dense everywhere, then

  1. the set of continuous points of $f$ is uncountable;

  2. the set of discontinuous points of $f$ is countable.

share|cite|improve this question
@today: What have you tried? – user9413 May 24 '11 at 15:01
today-: Have you seen that the set of points of discontinuity of a function must be an $F_\sigma$? This could help you answer 1. The converse could help with 2, but it can be answered more directly. Do you have any thoughts to add, such as where you got stuck? – Jonas Meyer May 24 '11 at 15:03
@today: Show that the set of points of continuity is $G_{\delta}$ – user9413 May 24 '11 at 15:07
today-: Welcome to the website. Around here, people are usually happier to invest time to help if you give more than just a problem statement. You could ask particular questions, and indicating difficulties you've had (or any thoughts at all) would make it easier for people to direct their answers to be more helpful. You can edit your question by clicking "edit" in the lower left. – Jonas Meyer May 24 '11 at 15:15
@today: That's good to hear! If you're willing to elaborate here on how your question has been answered, it could still be of benefit. For one thing, users here could help you clear up any possible remaining doubts. For another, it would benefit future readers interested in the question. – Jonas Meyer May 24 '11 at 15:26

The following observations give an outline of one way to answer 1.

  • The set of discontinuities is an $F_\sigma$.
  • An $F_\sigma$ with dense complement is a countable union of nowhere dense sets.
  • A countable set is a countable union of nowhere dense sets.
  • $\mathbb{R}$ is not a countable union of nowhere dense sets.

The following observations give an outline of one way to answer 2.

  • Every $F_\sigma$ is the set of discontinuities of some function.
  • There exist uncountable $F_\sigma$ sets with dense complement.

But you don't need such generalities for 2. You may want to look at your favorite example of a closed uncountable set with dense complement to get a more direct solution. (I'm assuming your favorite is the same as mine.)

share|cite|improve this answer
Do you have an idea how many of your answers on this site involve this favorite example of yours? 20, 50 or even 100? :) (no criticism of course, it only shows that it is a very good example to have in one's bag of tricks) – t.b. May 24 '11 at 16:49

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.