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$\mathbb R^n\supset[a,b]$ is domain of definition $f:[a,b]\to\mathbb R^n:t\mapsto(f^1(t),f^2(t),\ldots,f^n(t))$ where $f^i\in C^\infty$

I need to prove that image of $f$, that means $f[a,b]$, doesn't contain any open ball in it. Intuitively, if image contains an open ball, then $f$ has a lot of singular points in which it has no derivative. But I have difficulties with the proof of the statement. Can you help me?

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    There are many ways that one could do this. I am not entirely sure what you're allowed to use. For example, you could try the following. Say $f([a,b])$ contains an open ball. Then there exists $(\alpha,\beta)\subset [a,b]$ such that $f(\alpha, \beta)$ contains an open ball and $f'|_{(\alpha,\beta)}\neq 0$ (why?). Then $(a,b)$ is homeomorphic to an open ball in $\mathbb{R}^n$ (with $n>1$) (why?). This cannot be (why? - you don't have to use full force of the Invariance of Domain theorem; in this case there are *simple* topological reasons).2012-06-24
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    The quickest way is to show that the image has measure zero, using the Lipschitz property of $f$.2012-06-24
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    a . . . z, a...z The first of these is what you'll get from $\TeX$ code that says "a\ldots z", and the second from "a...z", at least if you're using $\TeX$ or $\LateX$ in the normal way as opposed to the way it's used on web sites. The first alternative is generally considered correct and the second incorrect; that's why \dots and \ldots work the way they do.2012-06-24
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    Doesn't $f$ have the same image as the path $c:[0,1] \to \mathbb{R}^n, c(s)=f(a+s(b-a))$?2012-06-24

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