I would like to know if there exists a differentiable vector-valued function $f:[a,b]\rightarrow \Bbb R^2$ such that the image $f'([a,b])$ of its derivative is disconnected.

My Attempt

First of all I asked for $f$ to be vector-valued because, if it were real-valued, then Darboux's Theorem would imply that $f'([a,b])$ be connected. For a similar reason, if we let $x(t)$ and $y(t)$ denote the component functions of $f$, then their derivatives can not both be continuous, for otherwise $f'$ would also be continuous hence preserving connectedness. Therefore, WLOG, $x'$ has to be discontinuous; however, Darboux's Theorem imply that it can not have discontinuities of the I kind or discontinuities of the II kind with infinite directional limits on $]a, b[$. Now, all the functions I came up for $x$ are not enough to construct the desired $f$ and I am starting to believe that such a function doesn't actually exist but I am not able to prove it.

Any help or hint, as always, is highly appreciated!

  • $\begingroup$ @Aname I do not understand the point you are trying to make $\endgroup$ Commented Apr 21, 2022 at 10:17

1 Answer 1


tl; dr: Yes, the image can be disconnected. (!)

Consider the curve \begin{align*} f(t) &= (t^{2}\sin(1/t), t^{2} \cos(1/t)) \\ &= t^{2}(\sin(1/t), \cos(1/t)), \end{align*} extended to be $(0,0)$ at $0$. The standard calculation shows \begin{align*} f'(t) &= (2t\sin(1/t) - \cos(1/t), 2t\cos(1/t) + \sin(1/t) \\ &= 2t(\sin(1/t), \cos(1/t)) + (-\cos(1/t), \sin(1/t)) \end{align*} for $t \neq 0$ and $f'(0) = (0, 0)$.

Since $|f'(t)| \geq |1 - 2t|$ by the reverse triangle inequality, in a sufficiently small deleted neighborhood of $t = 0$ the derivative is bounded away from $(0, 0)$. The image of the derivative is therefore the union of the origin and a (very fast spiral) curve lying in an annulus about the unit circle.

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    $\begingroup$ Thank you very much for the answer! I did think about using the unit circle parametrization by sine and cosine and the function $t^2\sin(1/t)$ but it didn’t came to my mind to combine these two ideas! $\endgroup$ Commented Apr 21, 2022 at 14:54

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