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Call a Borel set $A \subseteq [0,1]$ good if $$0 < \dim(A) \leq \overline{\dim_\text{M}}(A) < 2 \dim(A),$$ where $\dim(A)$ is the Hausdorff dimension of $A$ and $\overline{\dim_\text{M}}(A)$ is the upper Minkowski dimension of $A$. (Note that the second inequality in the string of three inequalities holds automatically.)

Does a compact set $X \subseteq [0,1]$ of positive Hausdorff dimension necessarily contain a subset which is a good set?

A positive answer leads to some more questions: Is there a subset of $X$ with the same Hausdorff dimension as $X$ which is also good? To what extent can the $2$ in the definition of good be reduced while maintaining a positive answer? Is it possible to find a subset for which the Hausdorff and upper Minkowski dimensions agree?

Otherwise, I'm hoping for a counterexample: a set with positive Hausdorff dimension with the property that all positive-dimension subsets have "large" upper Minkowski dimension. Given such an example, would it further be possible to have an $\epsilon > 0$ such that all positive-dimension subsets have upper Minkowski dimension greater than $\epsilon$?

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A counterexample:

Let $E = \{0,1\}$ and let $\pi:E^{\mathbb N}\to [0,1]$ be the coding map for the Cantor set. Now fix $N\in\mathbb N$, let $$ A = \{n\in\mathbb N : \lfloor \log_N(n)\rfloor \text{ is even}\}, $$ let $S = \{\omega\in E^{\mathbb N} : \omega_n = 0 \; \forall n\in A\}$, and let $X = \pi(S)$. A standard computation (e.g. mass distribution principle on the Bernoulli measure on $X$) shows that $$ \dim(X) = \log_3(2)\sum_{n\text{ odd}} \left[\frac{1}{N^n} - \frac{1}{N^{n + 1}}\right] = \frac{\log_3(2)}{N + 1} > 0. $$ Now let $A$ be a subset of $X$, and fix $s < \dim(A)$. By Frostman's lemma, there exists a probability measure $\mu$ supported on $A$ such that for all $B(x,r)$, $\mu(B(x,r)) \leq C r^s$, where $C > 0$ is fixed. Now fix an even integer $n$ and let $$ r_n = 3^{-N^n}, \;\; r_{n + 1} = 3^{-N^{n + 1}}. $$ Then for all $x\in X$, $B(x,r_n)\cap X = B(x,r_{n + 1})\cap X$ (ignoring endpoints), so $$ \mu(B(x,r_n)) \leq C r_{n + 1}^s. $$ So if $B\subseteq A$ is an $r_n$-net, then $$ \#(B) \geq \frac{1}{C r_{n + 1}^s} = \frac{1}{C} r_n^{-Ns}. $$ Thus $\overline{\dim_M}(A) \geq Ns$; since $s$ was arbitrary, $$ \overline{\dim_M}(A) \geq N \dim(A). $$

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