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For a nonempty bounded set, let $N_{\delta }(E)$ be the least number of radius-$\delta$ balls covering $E$. The lower and upper Minkowski dimensions are respectivelyUsing comparable cubes or ball diameters gives the same limits. For an unbounded set one must specify a local or bounded-piece convention; these are the usual bounded-set definitions used in the Kakeya set problem.

For the digit set, fixing the first $k$ digits gives exactly $3^k$ cylinders in intervals of length $9^{-k}$. They cover the set with at most $3^k$ balls of comparable radius. Their base-nine prefix integers are separated by at least two, so these containing intervals have gaps at least $9^{-k}$. Choosing one point in each cylinder shows that a radius-$9^{-k}$ ball meets at most a bounded number of chosen points. Hence $N_{9^{-k}}(Q)\asymp 3^k$. For $9^{-(k+1)}<\delta \le 9^{-k}$, monotonicity sandwiches the covering number between constant multiples of $3^k$ and $3^{k+1}$. This proves the box-counting dimension of separated digit sets formula here:A planar Besicovitch set is a bounded set of zero Lebesgue measure containing a unit line segment in every unoriented direction. The following proof in fact applies to every bounded planar Kakeya set, whether its area is zero or positive.

Choose $M\asymp {\delta }^{-1}$ directions equally spaced in an angular interval of length, say, $\pi /2$, and one segment of the set in each direction. Let $T_j$ be its rectangle of length one and width comparable to $\delta$, lying in the $C\delta$ neighborhood $E_{C\delta }$. Put $F={\sum }_j{1}_{T_j}$. Then $\int F\asymp M\delta \asymp 1$. Two such Kakeya tubes at angle $\alpha$ have overlap at most $C{\delta }^2/\alpha$, since the intersection of their supporting strips is a parallelogram of that area. The trivial bound $C\delta$ applies for nearly parallel tubes. Thus, uniformly over their translations,Sum these intersections. Each row contributes at most $C\delta \log (2/\delta )$, so $\int F^2\lesssim \log (2/\delta )$. The Cauchy-Schwarz inequality on the union now gives the planar Kakeya neighborhood lower boundA cover of $E$ by $N_{\delta }(E)$ radius-$\delta$ balls covers $E_{C\delta }$ by their fixed-factor enlargements. Hence $N_{\delta }(E)\gtrsim {\delta }^{-2}/\log (2/\delta )$. Boundedness gives the matching upper exponent $N_{\delta }(E){\lesssim }_E{\delta }^{-2}$. Taking the lower and upper limits provesThe logarithmic neighborhood loss is compatible with zero area; it still forces full dimension.