How can I prove that there is no $C^1$ - class curve $\gamma : [0,1] \rightarrow \mathbb{R^2}$ that can fill the unit disk?
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2 Answers
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Well, note that $\gamma$ being $C^1$ implies that $\gamma$ is rectifiable. Hence, the Hausdorff dimension of the image of $\gamma$ is $1$ (unless, of course, $\gamma$ is constant). However, the Hausdorff dimension of the unit disk is $2$, by the existence of the Lesbegue measure.
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1$\begingroup$ Note that all we need for this argument is that $|\gamma|$ has bounded variation. $\endgroup$ Feb 4, 2020 at 12:28
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$\begingroup$ Thank you. I'm not familliar with Hausdorff dimension. Is there an easier way to prove this? $\endgroup$ Feb 4, 2020 at 23:08
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$\begingroup$ I don't think that there is a good proof of this fact that doesn't essentially state: "The image of $\gamma$ has a curve-length, but the curve-length of the unit disk is infinite." $\endgroup$ Feb 5, 2020 at 6:47
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The image of a $C^1$ map $\Bbb R \to \Bbb R^2$ has measure zero. This is Sard's theorem.
