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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, 2016 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, 2016 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, 2016 at 21:14
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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, 2016 at 23:40
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    $\begingroup$ Every simple closed curve on the sphere bounds two discs on it, therefore your question can be phrased equivalently: "Does every sphere topologically embedded in $\mathbb{R}^3$ contain a planar simple closed curve?" How about a planar arc? $\endgroup$
    – Wlodek Kuperberg
    Feb 1, 2019 at 19:35

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As requested, here is a related result:

Theorem. There exists a continuous (also also open) real-valued function $R^2\to R$ whose level sets are all homeomorphic to the open pseudo-arc. In particular, level sets contain no nondegenerate arcs.

Similarly, one can construct a continuous monotone map $S^2\to [0,1]$ such that all level sets are homeomorphic to the pseudo-circle, i.e. the unique (up to homeomorphism) hereditarily indecomposable circularly chainable continuum. Again, pseudo-circle contains no nondegenerate arcs.

The story of this theorem is rather interesting. Quoting from

Prajs, Janusz R., A continuous circle of pseudo-arcs filling up the annulus, Trans. Am. Math. Soc. 352, No. 4, 1743-1757 (2000). ZBL0936.54019.

where a proof of this theorem is given:

Among the results obtained by Knaster during World War II one can find the following announcement, originally presented in Kiev in 1940:

There exists a real-valued, monotone mapping from the plane that is not constant on any arc. In the construction Knaster’s hereditarily indecomposable continua were exploited. Actually, Knaster’s result can be reformulated in the following stronger version: There exists a real valued, monotone mapping from the plane such that all pointinverses are hereditarily indecomposable.

Unfortunately, Knaster’s notes concerning this result were burned during the war, Knaster had never written down the result again, and even his closest exstudents do not know his original idea of construction...

A result similar to that announced by Knaster (with higher dimensional analogues) was proved by Brown in 1958:

M. Brown, Continuous collections of higher dimensional continua, Ph. D. Thesis, University of Wisconsin, 1958.

Of course, Brown did not publish his proof either, but Prajs finally did, see above.

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