I've been trying to use the enumeration of the set of rationals and the fact that $|x-a|$ is non differentiable at a. So let $r_1,r_2,r_3,...$ be an enumeration then $|x-r_1| + |x-r_2| + ...$ must satisfy. Am I right? Also is it true that the set of differentiable points is open?
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1Doesn't that series you wrote diverge? How about taking a strictly increasing function $g:\mathbb R\to\mathbb R$ which is discontinuous precisely at the rational numbers and then defining $f(x)=\int_0^xg(t)dt$? Does that work? The function $g(x)$, as you probably know, can be defined as $\sum_{r_n\lt x}2^{-n}$. – bof Nov 16 '19 at 10:36