How big can the Hausdorff dimension of the closure of a smooth curve be?

Question

Consider curves in $\mathbb{R}^n$. Smooth curves have Hausdorff dimension $1$.

The closure of a smooth curve can have Hausdorff dimension $> 1$. (For example, a curve dense in a torus.)

How big can the Hausdorff dimension of the closure of a smooth curve be?

(Can it be $> 2$? For instance, can there be a smooth curve dense in a solid torus?)

Answer 1

We can extend the example you mention, sometimes called an irrational winding of a torus, to higher-dimensional tori $\Bbb T_n$:

Pick a constants $\alpha_1, \ldots, \alpha_n \in \Bbb R$ such that $1, \alpha_1, \ldots, \alpha_n$ are linearly independent over $\Bbb Q$, and denote $$\alpha := (\alpha_1, \ldots, \alpha_n) \in \Bbb R^n .$$ Then, the image of the curve $$\gamma : \Bbb R \to \Bbb T_n \cong \Bbb R^n / \Bbb Z^n, \qquad\gamma(t) := t \alpha \pmod {\Bbb Z^n}$$ is dense in $\Bbb T_n$.

(For $n = 2$ the linear independence condition is just that the ratio $\frac{\alpha_2}{\alpha_1}$ is irrational.)

Answer 2

Consider all points in $[0,1]^n$ with rational coordinates. We can enumerate them with natural numbers, so they produce a sequence $P_i$. We can also enumerate them in such a way so $|P_{i+1}-P_i|>\epsilon$ for every $i$ and fixed $\epsilon$. Now you can assign a spline that goes through all points in a succession. By construction, the closure of the curve will include $[0,1]^n$ and thus will have the dimension $n$