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Someone told me this problem long ago and I have made no progress. Let $f:U\to\mathbb{C}$, with $U\subseteq\mathbb{C}$ open, be a continuous function such that $f'(z):=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ is defined at every point in $U\setminus A$, where $A$ is a countable subset of $U$. Is $f$ holomorphic?

Due to Riemann's Theorem we can suppose that $A$ has no isolated points. That implies that $A$ is homeomorphic to $\mathbb{Q}$, by Sierpinski´s theorem (concretely, the compact case has been solved as mentioned by Martin R in the comments). I usually imagine the case when $A$ is dense in $U$, that will probably be the toughest one.

Saúl RM
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    Are you assuming that $f$ is continuous? – José Carlos Santos Jan 21 '22 at 18:54
  • Oops, that was important. I´ll add it. Apologies to the person that already answered. – Saúl RM Jan 21 '22 at 19:07
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    See https://math.stackexchange.com/q/432298/42969: Any countable compact set is “removable”. – Martin R Jan 21 '22 at 19:25
  • If I´m not mistaken, that only talks about countable compact sets. That case is not very hard because for every countable compact sets, you can remove isolated points one by one until you are left with the empty set (the points you remove can be numbered by ordinals, not necessarily the naturals). This is because any countable set without isolated points will be homeomorphic to $\mathbb{Q}$ (Sierpinski´s theorem) – Saúl RM Jan 21 '22 at 19:27
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    Yes, and I did not claim that it is the identical question. – Martin R Jan 21 '22 at 19:30
  • Btw, if you already have the answer for a countable compact set then you might want to mention that in your question. – Martin R Jan 21 '22 at 19:32
  • I´ll do that. I actually hadn´t thought of the problem for compacts until I saw your comment, thanks – Saúl RM Jan 21 '22 at 19:33
  • The answer is positive if, additionally, you assume that $f$ is absolutely continuous on almost every coordinate line (in this case, it is even enough to assume that $A$ has zero Lebesgue measure). Not sure if continuity alone will suffice. – Moishe Kohan Jan 21 '22 at 22:57
  • Interesting, how could you prove it using absolute continuity in the coordinate lines? – Saúl RM Jan 21 '22 at 23:10
  • @SaúlRodríguezMartín It is a standard result about quasiconformal maps, I think due to Weyl. I will dig up a reference. – Moishe Kohan Jan 22 '22 at 01:47
  • Noting that at any point of differentiability, the partial derivatives exist and satisfy Cauchy Riemann, the result follows from the Looman Menchoff theorem; that is a fairly subtle measure theoretic result though, so there may be an easier direct proof here as the hypothesis is stronger – Conrad Jan 22 '22 at 09:33

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As Conrad noted in a comment, the result is true and it is a direct consequence of the Looman Menchoff theorem.

Saúl RM
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