There is no differentiable function $f:(0,1)\to \mathbb{R}$ whose derivative is continuous on $(0,1)\cap \mathbb{Q}$

Asked Dec 05 '17 at 21:51

Active Dec 05 '17 at 23:44

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Theorem: There is no differentiable function $f:(0,1)\to \mathbb{R}$ whose derivative is continuous EXACTLY on $E=(0,1)\cap \mathbb{Q}$.

Proof: Since $E$ is a meagre set, it is not $G_\delta$, thus there cannot exist a set $D(f)$ of continuity of $f$ such that $D(f)\subset E$, since $D(f)$ must be a $G_\delta$ dense set in $E$.

Please let me know if my idea is correct. I think that I may not yet have grasped the related concepts very well.

edited Dec 05 '17 at 23:44

asked Dec 05 '17 at 21:51

sequence

9,638

Is this some non-standard definition of continuity? The usual epsilon-delta one works fine for intervals in $\mathbb{Q}$. – Connor Harris Dec 05 '17 at 21:56
what do you call continuity on a disconnected set ? because $f=1$ and $f'=0$ seem pretty continuous to me. – zwim Dec 05 '17 at 21:56
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I suspect the OP means "whose derivative is continuous exactly on $E$", i.e. for which $E$ is the set of points at which $f'$ is continuous. – Alex Kruckman Dec 05 '17 at 21:58
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Please restate the problem. There are things you have misstated. – zhw. Dec 05 '17 at 22:36
Related: https://math.stackexchange.com/a/112133/44121 – Jack D'Aurizio Dec 05 '17 at 22:44
I apologze. @AlexKruckman is right. Edited. – sequence Dec 05 '17 at 23:45

There is no differentiable function $f:(0,1)\to \mathbb{R}$ whose derivative is continuous on $(0,1)\cap \mathbb{Q}$

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