The Minkowski dimension of a set $S$ is defined as: dim$_S = \lim_{\epsilon\to 0}\frac{\text{log}(N_S(\epsilon))}{\text{log}(1/\epsilon)}$. Where $N_S(\epsilon)$ is the number of boxes of side length $\epsilon$ required to cover the set $S$.
Let $U, V \subset \mathbb{R}$. Prove or disprove using a counterexample that dim$_{(U \cup V)}$ = max$\{\text{dim}_U, \text{dim}_V\}$.
My instinct is that this is true, but I'm unsure how you would prove it. Thoughts:
$\exists \epsilon^* > 0$ such that $\forall \epsilon < \epsilon*$ \begin{equation} \left|\frac{\text{log}(N_U(\epsilon))}{\text{log}(1/\epsilon)} - \text{dim}_U\right| < \epsilon \text{ and } \left|\frac{\text{log}(N_V(\epsilon))}{\text{log}(1/\epsilon)} - \text{dim}_V\right| < \epsilon \end{equation} Then we want $|\frac{\text{log}(N_U(\epsilon)+N_V(\epsilon))}{\text{log}(1/\epsilon)} - \text{max}\{\text{dim}_U,\text{dim}_V\}| < \epsilon*\text{constant}$. Any thoughts on how to proceed?
This is true. One direction follows from monotonicity: since $A\cup B$ contains $A$, we have $\dim (A\cup B)\ge \dim A$, and similarly $\dim (A\cup B)\ge \dim B$.
For the other, notice that $$N_{A\cup B}(\epsilon)\le N_A(\epsilon)+N_B(\epsilon)\le 2\max(N_A(\epsilon), N_B(\epsilon))$$ hence $$\frac{\log(N_{A\cup B}(\epsilon))}{\log(1/\epsilon)} \le \max\left(\frac{\log(N_{A}(\epsilon))}{\log(1/\epsilon)} , \frac{\log(N_{B}(\epsilon))}{\log(1/\epsilon)} \right) + \frac{\log 2}{\log(1/\epsilon)} $$ Let $\epsilon\to 0$ here.