Suppose i have two non-empty disjoint open sets on $S^2$ does there always exist a $C^0$ curve in the complement of the union of the two sets

Question

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.

Answer 1

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.

Answer 2

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.