1
$\begingroup$

Can a real value function, defined for every real number, have finite (or countable) points of continuity ?

As for the not countable case, the answer is trivial: any polynomial has not countable points of continuity.

We can see the opposite problem: points of discontinuity.

Dirichlet function has not countable points of discontinuity. Functions with finite or countable points of discontinuity are trivial examples.

$\endgroup$
  • $\begingroup$ yes. Try $f(x)=xg(x)$ where $g(x) \in [0,1]$ and is discontinuous everywhere. $\endgroup$ – Michael May 26 '15 at 5:19
  • $\begingroup$ @Michael isn't that function continuous at every point in the interval (of uncountably many points) $(0,1)$? $\endgroup$ – ASKASK May 26 '15 at 5:22
  • $\begingroup$ Oh my bad, I misread it as "$g(x) \in [0,1]$ and is discontinuous everywhere else" (which to me implied that it was continuous in $[0,1]$ $\endgroup$ – ASKASK May 26 '15 at 5:24
  • $\begingroup$ I still don't see how that is continuous anywhere though $\endgroup$ – ASKASK May 26 '15 at 5:25
  • $\begingroup$ I meant that $0 \leq g(x) \leq 1$ for all $x \in \mathbb{R}$, and is discontinuous at all points $x \in \mathbb{R}$. You can come up with an explicit $g(x)$ of this form. $\endgroup$ – Michael May 26 '15 at 5:25
1
$\begingroup$

Here's a concrete example: let $g(x)$ be the characteristic function of the rationals, and let $f(x)=xg(x)$. Then basically the graph of $f$ looks like a big dotted "V" (the part corresponding to the rationals) with a dotted line running underneath it (the part corresponding to the irrationals), and these meet up at the origin. Motivated by this, it's easy to check that $f$ is continuous at exactly the origin.

Similar arguments give you functions whose points of continuity are an arbitrary finite set.


What if we keep going? Well, by copying the above $f$ - restricted to $[-1, 1]$ - over and over, we get a function which is continuous at countably many points. But what sort of restrictions are there? Let $C(h)$ be the set of points at which $h$ is continuous, and let $\mathcal{C}=\{C(h): h:\mathbb{R}\rightarrow\mathbb{R}\}$ be the set of all possible sets of continuity. Is every countable set in $\mathcal{C}$? This is a fun exercise . . .

$\endgroup$
  • $\begingroup$ And a mild modification gives a function whose points of discontinuity are the natural numbers. $\endgroup$ – André Nicolas May 26 '15 at 5:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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