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.

  • $\begingroup$ 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

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 . . .

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

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