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Let $\Sigma$ be a sphere topologically embedded into $\mathbb{R}^3$.

Is it always possible to find a disc $\Delta\subset\Sigma$ which is bounded by a plane curve?

It is easy to find an open disc which boundary lies in a plane, but the boundary might be crazy; for example it might be Polish circle shown on the diagram.

enter image description here

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    $\begingroup$ The title and the question don't quite match... Did you mean to say, "by a closed curve in the plane?" $\endgroup$ – Joseph O'Rourke Dec 16 '16 at 20:02
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    $\begingroup$ probably rather "a closed plane curve" (i.e. contained in some plane, not a given plane) $\endgroup$ – YCor Dec 16 '16 at 20:10
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    $\begingroup$ The main body of the question still doesn't mention that the curve is supposed to be planar! Also, do you have an example of the situation in the last sentence, where the boundary doesn't contain simple curve? $\endgroup$ – Jim Conant Dec 16 '16 at 21:14
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    $\begingroup$ @JimConant take any open disk $D$ of the plane with some weird boundary: since it is closed, there is a smooth function whose zero set is exactly $\partial D$. Then the graph of $f$ is a smooth surface with an open disk $D'$ whose boundary is $\partial D'=\partial D$, the intersection with the plane. $\endgroup$ – Pietro Majer Dec 16 '16 at 23:09
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    $\begingroup$ @PietroMajer Consider a boundary which is a circle union a line segment. You can certainly make this the cross section of a map of a sphere, but there is pinching, so it is not embedded. $\endgroup$ – Jim Conant Dec 16 '16 at 23:40

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