# Do simply connected open sets in $\Bbb R^2$ always have continuous boundaries?

A Jordan curve is a continuous closed curve in $\Bbb R^2$ which is simple, i.e. has no self-intersections. The Jordan curve theorem states that the complement of any Jordan curve has two connected components, an interior and an exterior.

Let us define an unbounded curve to be a continuous map $f: \Bbb R\to\Bbb R^2$ such that the limit of $|f(t)|$ as $t$ goes to plus or minus infinity is infinity. Then as discussed in the comments to my question here, the complement of an unbounded simple curve has two connected components:
Does the Jordan curve theorem apply to non-closed curves?

My question is, is every simply connected open set in $\Bbb R^2$ a connected component of the complement of either a Jordan curve or an unbounded simple curve? To put it another way, is the boundary of a simply connected open set always a continuous curve, or do there exist sets with weirder boundaries than that?

If they do always have continuous boundaries, can this be generalized to higher dimensions?

Any help would be greatly appreciated.

#### Solutions Collecting From Web of "Do simply connected open sets in $\Bbb R^2$ always have continuous boundaries?"

Even ignoring the trivial counterexample $\mathbb{R}^2$, we can generate a number of counterexamples.

Take $\mathbb{R}^2$ and delete the entire $x$ axis aside from a small interval around the origin. This will be simply connected with disconnected boundary.

Demanding that our set be bounded, we intersect our previous counterexample with the unit open ball centered at the origin. This will again be open, simply connected, and will have the unfortunate fact that it’s boundary is not a Jordan curve.

It possible that if you require that your open set be a bounded regular open set i.e. that $A = Int(Cl(A))$ that you will get an affirmative answer.

Consider the open box $B = (-1,1) \times (-2,2)$. Let $C$ be the closed topologist’s sine curve, consisting of the points $(x, \sin(1/x))$ for $0 < x \le 1$, and the segment $\{0\} \times [-1,1]$. $C$ is compact, so $U := B \setminus C$ is open, and also bounded. $U$ is also simply connected, and its boundary $\partial U$ consists of the rectangle $\partial B$ together with $C$. But $C$ is not path connected, and neither is $\partial U = \partial B \cup C$, so $\partial U$ is not the continuous image of the path connected space $[0,1]$. So it is not even a continuous curve, let alone a Jordan curve.