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Consider a smooth curve $\gamma: [a,b] \to \mathbb S^2$ on the two-sphere in $\mathbb R^3$. Consider then the set defined by \begin{align*} \{ \alpha \gamma(t) \; : \; \alpha \in \mathbb R, \; t \in [a,b] \}. \end{align*} Question: Is there a curve such that the above set contains an $\epsilon$-ball? I came up with this question in my research and I have to admit that my visual thinking is letting me down.

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    $\begingroup$ The projection $\pi:{\mathbb R}^3\setminus\{0\}\to{\mathbb S}^2$ consisting in quotiening by the group ${\mathbb R}^{>0}$ is open (one can take $\pi$ for defining on the $2$-sphere the quotient structure of a smooth manifold). So, you are actually asking if a smooth curve can be open on a $2$-manifold. Of course, this is impossible. (For instance, because the image has measure $0$.) And still impossible after your edit for the same reason. $\endgroup$ Jan 14, 2014 at 14:42
  • $\begingroup$ Seconding Sasha's comment, your question admits a positive answer with Peano-like curves. $\endgroup$ Jan 14, 2014 at 15:18
  • $\begingroup$ Summarizing the two above comments, the answer to your question is no, but turns to yes if you lower the regularity of the curve from smooth to continuous. It seems that Sasha Anan'in should write his comment as an answer so that it can be accepted, or we should close the question, which seems borderline. $\endgroup$ Jan 14, 2014 at 21:20

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I have an idea, but it needs to be made rigorous. It might be helpful to consider the smooth map $f:[a,b]\times\mathbb{R}\rightarrow\mathbb{R}^3$, $$f(t,\alpha)=\alpha\gamma(t).$$ This is a map from a $2$-manifold with boundary to a $3$-manifold. So, its differential has rank at most $2$ at a given point, while the tangent space to a point in an open subset of $\mathbb{R}^3$ has dimension $3$. So, it would seem that the answer is no, but I am far from certain.

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  • $\begingroup$ Thanks for your effort! I did make a stupid mistake in the problem formulation. I wanted to ask if there exists a curve such that the set contains an epsilon-Ball. This is enough already. The set does not need to be open. $\endgroup$
    – Klein31
    Jan 14, 2014 at 14:40

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