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.

  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ @JohnGowers I guess that the OP means "$C^0$ Jordan curve. $\endgroup$ Oct 10 '17 at 16:27
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ @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

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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.