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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?

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    $\begingroup$ I don't think the curve that you get in the limit is still simple. It seems to me that some, if not all, points of the square are (in the limit) covered several times. $\endgroup$ Jun 14, 2019 at 20:21
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    $\begingroup$ From Wikipedia: There exist non-self-intersecting curves of nonzero area, the Osgood curves, but they are not space-filling. $\endgroup$ Jun 14, 2019 at 20:23
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    $\begingroup$ Of course its not simple: a simple curve is the image of a closed interval (or circle, if it's a closed curve) under a continuous one-to-one map. A continuous one-to-one function from a compact set to Hausdorff space is a homeomorphism. A (filled) square is not homeomorphic to an interval or circle. $\endgroup$ Jun 14, 2019 at 20:25

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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

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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!

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  • $\begingroup$ This answer is wrong, Osgood does none of these. In fact, if $f: [0,1]\to R^2$ is a continuous injective map, then its image has empty interior. What Osgood constructs are Jordan curves of positive area. $\endgroup$ Jun 28, 2022 at 12:32
  • $\begingroup$ @MoisheKohan I am not sure how you have read Osgood's paper, but he does actually do "all of these". He does use, as you have said, the term Jordan curva: I find it missleading because, usually, Jordan curve refers to closed curves; that's why I use the term Jordan arc. The map is indeed injective, because every two different values $t_1,t_2\in [0,1]$ are mapped to different points in the square $[0,1]^2$. I would suggest you read Sagan's "Space-filling curves$ as well as this paper sciencedirect.com/science/article/pii/… Hope you finally understand it $\endgroup$ Jun 29, 2022 at 13:45
  • $\begingroup$ It is not the matter of the end-points. Osgood constructs Jordan curves of positive area, now called Osgood curves. He does not construct bijective continuous maps $[0,1]\to [0,1]^2$ because such maps do not exist. $\endgroup$ Jun 29, 2022 at 15:26
  • $\begingroup$ I haven't said, at no point, that such curves are bijective. I have stated that they are INJECTIVE, becase they are. As a matter of fact, "osgood curves" is a term used for INJECTIVE curves, while the used term in general is "peano curve". Osgood's paper was interesting because he showed, for the first time, that there could be an injective space filling curve. I insist: you should read the paper cited above. $\endgroup$ Jun 29, 2022 at 18:08
  • $\begingroup$ Space filling curves are required to be onto a square or a subset with nonempty interior. What is your definition of space-filling? $\endgroup$ Jun 29, 2022 at 18:37
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The Moore curve is closed in the limit. Start and end are neighboring in each iteration, and the distance between them decreases approximately as $2^{-n}$.

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  • $\begingroup$ Moore curve is not simple (and cannot be simple). $\endgroup$ Jun 28, 2022 at 12:33

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