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Let $A$ and $B$ be non-empty open and disjoint sets on an $S^2$. Does there exist (at least) one closed $C^0$ curve in $S^2\setminus(A\cup B)$.

If it is not true, what other condition has to be required to make it true.

Comment: It seems clear that any continuous path $\alpha(\tau)$ defined for $\tau\in[0,1]$ with $\alpha(0)\in A$ and $\alpha(1)\in B$ has to intersect $S^2\setminus(A\cup B)$ in at least one point.

Interpreting the $S^2$ as the Riemannsphere this can also be interpreted as a question about open sets on the plane.

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    $\begingroup$ Since $S^2$ is connected, $S^2\setminus (A\cup B)$ is non-empty. So a constant curve will do. $\endgroup$ Oct 10 '17 at 16:23
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    $\begingroup$ @JohnGowers I guess that the OP means "$C^0$ Jordan curve. $\endgroup$ Oct 10 '17 at 16:27
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    $\begingroup$ I would indeed be interested in Jordan curves. $\endgroup$ Oct 10 '17 at 17:03
  • $\begingroup$ One should also require the closures of A and B be disjoint. $\endgroup$ Oct 10 '17 at 17:06
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    $\begingroup$ @JosephVanName no this is not a necessary requirement. take any circle on $S^2$ and let one side be $A$ and the other side $B$ then the closure of $A$ and $B$ intersect exaclty in the complement of their union $\endgroup$ Oct 10 '17 at 17:12
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Take your favorite connected but not locally connected compact $K$ in the real plane, assuming $\mathbb R^2 \setminus K$ to be connected. There exists two distinct points $p,q\in K$ that cannot be joined by a curve ($C^0$ map) within $K$, but are nonetheless accessible from the exterior. Embed $K$ in $\mathbb S^2\simeq \mathbb R^2\cup \{\infty\}$ and join $\infty$ to $p$ and $q$ with two simple, continuous paths that do not cross one another nor $K$. The complement of this compact set $C$ has two open connected components $A$ and $B$, but $C$ does not contain the image of any Jordan curve.

Conversely, if no such curve exists in $C:=\mathbb S^2\setminus(A\cup B)$ for general $A$ and $B$ then $C$ has to be not locally connected.

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  • $\begingroup$ What is the set $C$ here exactly? the two path from $\infty$ to $p$ and $q$ together with K? So the condition for a $C^0$ curve to exist would be that the complement of $A\cup B$ is locally connected? Is it possible to put requirements on the open sets $A$ and $B$ such as to guarantee the complement to be locally connected? $\endgroup$ Oct 11 '17 at 9:36
  • $\begingroup$ Yes, $C$ is the union of the paths and $K$. If the boundary of $A$ and $B$ are the image of a $C^0$ path then (almost tautologically) the complement is locally connected. This is the case if $A$ and $B$ are the inverse image of "reasonnable" functions, for instance, but I don't know any more general formulation. $\endgroup$ Oct 11 '17 at 11:20
  • $\begingroup$ So you re saying that if $A$ and $B$ are locally connected then there exists a $C^0$ curve in the complement? $\endgroup$ Oct 11 '17 at 12:46
  • $\begingroup$ Not $A$ and $B$, open sets are always locally connected, but their adherences. $\endgroup$ Oct 11 '17 at 15:36
  • $\begingroup$ I don't quite understand your last two points. You re saying if $A$ and $B$ are the inverse of "reasonable" functions the boundary is a $C^0$ curve? Does it help if i have a way to construct a continuous (probably smooth) function on $A$ that goes to zero towards the boundary of $A$? $\endgroup$ Oct 12 '17 at 14:53
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Take a copy $W$ of the Warsaw circle in the sphere $S^2$. Its complement $S^2\setminus W$ decomposes into two open disjoint sets $A,B$ such that $S^2\setminus (A\cup B)=W$ does not contain any closed Jordan curve.

Instead of the Warsaw circle we can also take any circle-like continuum $K$ in $S^2$ which does not contain a non-trivial continuous path. In this case the complement $S^2\setminus (A\cup B)$ will not contain non-trivial curves.

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