Hausdorff dimension of a dense orbit in the Lorenz attractor

If I am not mistaken, then:

• the Lorenz attractor $$\mathcal{A}$$ has Hausdorff dimension $$\dim_H(\mathcal{A}) > 2$$, and
• the Lorenz attractor $$\mathcal{A}$$ contains a dense orbit $$\mathcal{O}$$, i.e. such that $$\bar{\mathcal{O}} = \mathcal{A}$$.

But:

• the Hausdorff dimension does not satisfy $$\dim_H(X) = \dim_H(\bar{X})$$ in general,
• for example $$\dim_H([0,1] \cap \mathbb{Q}) = 0$$ while $$\dim_H([0,1]) = 1$$.

What is the Hausdorff dimension of a dense orbit in the Lorenz attractor? Is it $$\dim_H(\mathcal{A})$$, or smaller?

• Any single orbit of the Lorenz system is a smooth curve and, therefore, has Hausdorff dimension $1$. I think this is crystal clear for a curve defined over $[0,N]$ for any $N\in\mathbb N$. To extend the result a curve defined over $[0,\infty)$, simply take a countable union of intervals as $N\to\infty$. – Mark McClure Aug 9 '19 at 11:12
• @MarkMcClure Thanks, I just realized the same. I would accept your answer, but it's written in a comment. – Ricardo Buring Aug 9 '19 at 11:45
• – Ricardo Buring Aug 14 '19 at 14:18