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Let $R$ be a simply-connected compact region in the plane and assume that there is a closed curve $\gamma\colon [0,1]\longrightarrow\mathbb R^2$ such that $\partial R=\gamma([0,1])$. Can $\gamma$ then also be chosen to be weakly simple? Here, weakly simple means that it can be made simple by an infinitesimal perturbation or, more formally, that there exists a sequence of simple closed curves $\gamma_1,\gamma_2,\ldots$ that converge to $\gamma$ in Fréchet distance. For the definition of Fréchet distance, see: https://en.wikipedia.org/wiki/Fr%C3%A9chet_distance

Mikkel Abrahamsen
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This follows from the Caratheodory-Torhorst extension theorem, see my answer here. To get the desired conclusion take a Riemann mapping $f: \{z: |z|<1\} \to R$ and consider the sequence of its restrictions to the circles $\{z: |z|= r_n=1- \frac{1}{n}\}$ precomposed with dilations $z\mapsto r_n z$. Because $f$ extends continuously to the closed disk, these compositions $g_n: C\to {\mathbb C}, C=\{z: |z|=1\}$, will converge to a loop $h: C\to \partial R$ (parameterizing $\partial R$) uniformly. Hence, the simple loops $C_n=g_n(C)$ converge to $\partial R$ in the Frechet topology.

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Moishe Kohan
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  • Thanks for the answer. Does this argument also work if $R$ has no interior points, i.e., if $R=\partial R$? Or if the set of interior points is disconnected? – Mikkel Abrahamsen Jan 24 '17 at 08:59
  • @MikkelAbrahamsen: For this to work you need the interior to $R$ to be dense in $R$ and connected. Otherwise, the claim is false. – Moishe Kohan Jan 24 '17 at 15:04
  • Can you give me an example where it's false? I imagined that it might always be possible to choose a sequence of simple curves $\gamma_1,\ldots$ in the complement of $R$ that converges to $\partial R$ in Fréchet distance. – Mikkel Abrahamsen Jan 25 '17 at 08:49
  • If there is an interior point $r$ in $R$, we can maybe consider the inverse $R'$ of $R$ with respect to a circle centered at $r$ and containing $R$ in its interior. Then the inverse of the complement of $R$ is an open region bounded by $\partial R'$, where we can use your argument? – Mikkel Abrahamsen Jan 25 '17 at 09:11