Let $A, B \subset \mathbb R$ be IID random closed subset. Suppose that there exists $d \in (1/2, 1]$ such that the Hausdorff dimension of $A$ is equal to $d$ almost surely. Is it true that $\mathbf P\lbrack A \cap B \neq \emptyset\rbrack > 0$?
(Note that zero sets of two independent Brownian motions almost surely do not intersect. Thus we have to require at least $d > 1/2$.)
No. They may be almost surely disjoint.
If $y\in\{0,1\}^{\mathbb N}$, let $S_y$ denote the (closure of the) set of $x$'s in $[0,1]$ such that the $2^n$th binary digit of $x$ is $y_n$; and the $(2^n+1)$st and $(2^n-1)st$ digits of $x$ are 0. [These additional conditions are just to prevent issues with multiple binary expansions].
If $y$ is selected uniformly from a coin-tossing measure, then $S_y$ and $S_{y'}$ are almost surely disjoint. And the Hausdorff dimension of each $S_y$ is 1: The box dimension is obviously 1: the number of balls of diameter $2^{-N}$ to cover $S_y$ is $2^{N-3\log_2 N}$; obtaining the Hausdorff dimension requires Frostman's lemma.
