Space-Filling Jordan Curve

Question

My question is about a simple closed curve that is also a space-filling curve. The figure shows 6 iterations of the formation of a Hilbert curve (limit), whose trace is a solid square. I think we may, at each iteration, connect the endpoints of the curve in order to obtain a Jordan curve (simple closed curve), preserving the limit of this sequence of curves (a solid square). So, at the limit, we will have a space-filling, simple closed curve. By the Jordan curve theorem, every simple closed curve "divides the plane into an "interior" region bounded by the curve and an "exterior" region containing all of the nearby and far away exterior points, so that every continuous path connecting a point of one region to a point of the other intersects with that loop somewhere" (Wikipedia).

Question: Does there really exist a space-filling, simple closed curve? What is the interior region of a space-filling, simple closed curve? The empty set?

Answer 1

In Osgood's paper "A Jordan Curve of Positive Area" you have the PDF here he provides a construction for a space-filling curve $[0,1]\hookrightarrow [0,1]^2$ but it is not a closed curve: it is, using nomenclature of the Jordan Curve Theorem, a Jordan Arc. Still, at the end of the paper, he provides the construction of a closed jordan curve. Hope it satisfies your curiosity!

Answer 2

The Moore curve is closed in the limit. Start and end are neighboring in each iteration, and the distance between them decreases approximately as $2^{-n}$.

Answer 3

What is happening in the pictures that you drew is that you have a sequence of parameterized Jordan curves, i.e. continuous injective maps $f_n: S^1\to R^2$, and this sequence converges uniformly to a continuous map. However, a uniform limit of injective continuous maps need not be injective. The correct statement is:

Proposition. Suppose that $f: [0,1]\to R^2$ is a continuous injective map. Then the image of $f$ has empty interior. In particular, there cannot be a simple space filling curve or arc.

Proof. First of all, since $[0,1]$ is compact and $R^2$ is Hausdorff, $f: [0,1]\to E=f([0,1])$ is a homeomorphism.

Suppose that there exists an interior point $p\in E$. The set $E$ is path-connected. For every path-connected subset $A\subset R^2$ and every interior point $a\in A$, the complement $A\setminus \{a\}$ is path-connected. But $x=f^{-1}(p)$ disconnects $[0,1]$ unless $x$ is one of the end-points of $[0,1]$. Since the interior of $E$ contains infinitely many points, we can assume that $p$ is chosen so that $x\notin \{0,1\}$. Thus, $f^{-1}(E\setminus \{p\})$ is disconnected. It follows that $f: [0,1]\to E$ cannot be a homeomorphism. A contradiction. qed