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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.

  • 0
    @Thomas: True, but of course$2$isn't :).2011-05-24

1 Answers 1

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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.)

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    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)2011-05-24