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Function $f(x)$ maps $(-a,a)$ to $\mathbb{R}^2$ and $f \in C^1$ (continously differentiable). Is it possible that image of every open interval $(-b,b)$ (for $b<a$ of course) contains neighborhood of $f(0)$?

I've tried to figure out this problem, but I have no idea how to do this. I know that inverse function theorem or implicit function theorem can be helpful, anyway I don't see any way in which I could apply these theorems in this problem.

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  • $\begingroup$ Let $r$ be the slope of the tangent at $x=0$. Take $\epsilon>0$ so small that the angle between the slopes $r\pm \epsilon$ is less than $\pi/2$. Then, by definition of differentiable there is $c>0$, such that for all $0\leq b<c$ the curve on $(-b,b)$ is between the lines passing through $f(0)$ with slopes $r\pm\epsilon$. The cone covered by these two lines doesn't cover any disc with center $f(0)$. $\endgroup$ – conditionalMethod Nov 8 '19 at 7:39
  • $\begingroup$ yes it is possible : take $f$ any constant function. $\endgroup$ – Olivier Roche Nov 8 '19 at 13:16
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    $\begingroup$ @Olivier a constant function from $\Bbb R $ to $\Bbb R ^2$ have a nowhere density image (a singleton), so certainly it cannot contain any open set $\endgroup$ – Masacroso Nov 8 '19 at 13:33
  • $\begingroup$ @Masacroso Indeed, thanks $\endgroup$ – Olivier Roche Nov 8 '19 at 14:48
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The answer is no. One way to show it is showing that a compact rectifiable curve in $\Bbb R ^2$ cannot contain an open disc.

Now note that if $f|_{[-b,b]}$ is continuously differentiable then necessarily it image is a rectifiable compact curve in $\Bbb R ^2$.

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