I was reviewing my notes from a Differential Geometry course I took last year, and I started thinking of the following problem$\color{blue}{^1}$.
Let $n \ge 2$. Does there exist a smooth curve $\alpha: \Bbb R\to \Bbb R^n$ satisfying $\overline{\alpha(\Bbb R)} = \Bbb R^n$, i.e., the trajectory of the curve is dense in $\Bbb R^n$?
It is not immediately clear that such a curve exists in the first place, but an explicit construction shall take that problem off our hands.
Some Thoughts: Since $\Bbb R^n$ is separable, let us choose a countable dense subset $S:= \{x_k\}_{k\in \Bbb Z}$ of $\Bbb R^n$. For every $k\in \Bbb Z$, define $\alpha$ on the closed interval $[k,k+1]$ as the segment joining $x_k$ to $x_{k+1}$. Then, the trajectory of $\alpha$ is dense in $\Bbb R^n$, but the curve obtained is not necessarily smooth. Perhaps there is some way to modify the $\alpha$ so obtained, to "smoothen" it around the corners?
Is the problem easier to solve when $n = 2$? Maybe we can take $n = 2$ as our guiding light to generalize to higher dimensions.
I'd appreciate any help! Thanks a lot.
P.S.
By a "curve", I really mean a regular curve, i.e., $\alpha'(t) \ne 0$ for all $t\in \Bbb R$, but please feel free to drop this assumption if that is really needed. Furthermore, it would be good if we can prevent the curve from intersecting itself.
Smooth means $\mathcal C^\infty$.
$\color{blue}{1.}$ The source of this problem is my imagination, so I cannot comment on the difficulty and/or solvability of the same. It doesn't seem out of the scope of a first course in differential geometry, though.