4

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.

Thank You in Advance.

Community
  • 1
Keshav Srinivasan
  • 10,204
  • 2
    The complement of the Mandelbrot set (in the sphere $\widehat{\mathbb{C}}$) is simply connceted. Its boundary is pretty wild. – Daniel Fischer Aug 09 '13 at 20:12
  • @DanielFischer I'm talking about sets in R^2, not the Riemann sphere. – Keshav Srinivasan Aug 09 '13 at 20:40
  • 1
    Well, consider the inversion $z \mapsto \frac1z$. That maps the complement to a subset of $\mathbb{C}$, which can be canonically identified with $\mathbb{R}^2$. – Daniel Fischer Aug 09 '13 at 20:41
  • Donald Sarason from Berkeley has a picture of a drawing he uses in his complex analysis class, of a simply connected open set that looks like a dragon with infinitely many legs, the eye is a spiral. It is as far from a "continuous boundary" as you could imagine. (I have a copy of the picture, but am currently without access to a scanner. I'll try to remember in a few days and post a copy.) – Andrés E. Caicedo Aug 09 '13 at 23:40

2 Answers2

4

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.

Devin Murray
  • 1,028
3

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.

See this related answer.

Community
  • 1
Nate Eldredge
  • 97,710