Is there an injective $C^1$ curve dense in the plane?

Question

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

Answer 1

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.