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

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.

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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 . . .