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Is there a curve $\gamma : \mathbb{R} \to \mathbb{R}^2$ injective and $\mathcal{C}^1$ whose range is dense in $\mathbb{R}^2$?

PatrickR
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Seirios
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    Probably useful: the image of a smooth curve defined in a compact interval is nowhere dense. – Giuseppe Negro Mar 05 '13 at 17:44
  • @GiuseppeNegro Now apply Baire Category. – JSchlather Mar 05 '13 at 18:03
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    @JSchlather: This shows the image of $\gamma$ must be meager, but that doesn't mean it can't be dense. – Nate Eldredge Mar 05 '13 at 18:05
  • @NateEldredge Ah, yes I'm mixing up parts. My mistake. – JSchlather Mar 05 '13 at 18:07
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    Here's a conjecture: if $E \subset \mathbb{R}^2$ is homeomorphic to $[0,1]$, then $\mathbb{R}^2 \setminus E$ is connected. Being connected and open, it is path connected. If this holds, it should be possible to use induction to construct an injective curve passing through every point in a countable dense subset. – Nate Eldredge Mar 05 '13 at 18:10
  • Yeah, it seems like that should be the case. Is that really only conjectured? It seems like, even if you couldn't prove it, you could construct the curve inductively so that you knew it was true for $E_n$ the image of $[0,n]$. @NateEldredge – Thomas Andrews Mar 05 '13 at 18:13
  • @ThomasAndrews: by "conjecture" I mean "seems like it should be true, but I couldn't think of a proof within 5 minutes". Someone must know whether it's true or not. – Nate Eldredge Mar 05 '13 at 18:15
  • @ThomasAndrews The issue is how to ensure the curve is $C^1$? – JSchlather Mar 05 '13 at 18:15
  • @JSchlather: I don't think that's a serious problem. In any open path-connected set, you can join any two points by a $C^\infty$ curve (take a continuous curve and mollify it). Then you just have to join the endpoint of your existing curve with the next target point, and tweak it a little to match the derivatives. – Nate Eldredge Mar 05 '13 at 18:17
  • Yeah, $C^1$ is relatively easy, it would seem to me. – Thomas Andrews Mar 05 '13 at 18:24
  • Ah, misread "Here's a conjecture..." as "There's a conjecture..." @NateEldredge :) – Thomas Andrews Mar 05 '13 at 18:25
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    @NateEldredge: In fact, without loss of generality you can take $E=[0,1] \times {0}$, so your conjecture is true: http://math.stackexchange.com/questions/287700/extending-a-homeomorphism-between-two-curves – Seirios Mar 05 '13 at 19:32
  • The number of connected components of the complement of a compact $K$ in $\mathbb R^n$ is a topological invariant of the compact. It's the generalized Jordan's theorem. – Pietro Majer Jan 03 '25 at 16:17

1 Answers1

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Given a countable subset $E=\{x_n\mid n\in\mathbb N_0\}\subset \mathbb R^2$, there is an injective $C^\infty$ curve $\gamma\colon [0,\infty)\to\mathbb R^2$ passing through all points of $E$. To show this, we define $\gamma_n\colon [0,n]\to\mathbb R^2$ with

  1. $\gamma_n\in C^\infty([0,n],\mathbb R^2)$
  2. $\gamma_n$ is injective
  3. $\{x_0,\ldots,x_n\}\subset \gamma_n([0,n])$
  4. There exist $ t_n\in[0,n)$, $r_n,s_n\in\mathbb R^2$ such that $\gamma_n(t)=t\cdot r_n+s_n$ for $t_n\le t\le n$

As a start, let $\gamma_1$ be the straight line segment from $x_0$ to $x_1$.

Now assume we are given $\gamma_{n-1}$ that fulfills conditions 1.-4. (with $n$ replaced by $n-1$). Let $S$ be the set of points $x$ such that there exists a curve $\gamma_n$ with $\gamma_n(n)=x$ and $\gamma_n|_{[0,n-1]}=\gamma_{n-1}$ and fulfiling conditions 1., 2. and 4. (but not necessarily 3.).

Then $S$ contains $B(\gamma_{n-1}(n-1),\delta)\setminus\gamma_{n-1}([t_{n-1},n-1])$ where $\delta$ is the (positive) distance between $\gamma_{n-1}(n-1)$ and the compact set $\gamma_{n-1}([0,t_{n-1}])$. This is so because we can add a tiny linear segment to the desired endpoint and make a smooth connection between two line segments. By a similar construction, $S$ is open. Again by asimilar construction, $\mathbb R^2\setminus(S\cup\gamma_{n-1}([0,n-1]))$ is open. Since $\mathbb R^2\setminus\gamma_{n-1}([0,n-1])$ is connected, we conclude $S=\mathbb R^2\setminus\gamma_{n-1}([0,n-1])$. Especially, we can either select $x=x_{n}$ and thus obtain $\gamma_n$ with conditions 1.-4. fulfilled and $\gamma_n|_{[0,n-1]}=\gamma_{n-1}$. Or we already have $x_n\in\gamma_{n-1}([0,n-1])$ "by accident" and can choose $x$ arbitrary and reach the same goal.

The curve $\gamma\colon [0,\infty)\to\mathbb R^2$ with $\gamma(t)=\gamma_n(t)$ for some $n>t$ then fulfills

  1. $\gamma\in C^\infty([0,\infty),\mathbb R^2)$
  2. $\gamma$ is injective
  3. $E\subset\gamma([0,\infty))$

If $E$ is a dense set, we conclude that $\gamma([0,\infty))$ is dense.

Seirios
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Hagen von Eitzen
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