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Exercise: Show that, for bounded sets $A$ and $B$, $$\overline\dim_{\text{box}}(A\cup B)=\max\{\overline\dim_{\text{box}}(A),\overline\dim_{\text{box}}(B)\},$$ where $\overline\dim_{\text{box}}(A)$ is the upper box dimension of the set $A$.

So far: I have that $A\subset A\cup B$ and $B\subset A\cup B$, so clearly $\max\{\overline\dim_{\text{box}}(A),\overline\dim_{\text{box}}(B)\}\leqslant\overline\dim_{\text{box}}(A\cup B)$.

For the other direction: I have that $N_{\delta}(A)+N_{\delta}(B)\leqslant2\max\{N_{\delta}(A),N_{\delta}(B)\}$ where $N_{\delta}(A)$ is the smallest number of closed balls which covers $A$. Clearly $\overline\dim_{\text{box}}(A\cup B)\leqslant N_{\delta}(A)+N_{\delta}(B)$, but I don't know where to go from there.

Can someone tell me if what I have done is correct, whether or not there is an easier way, and how to finish it off please?

user458984
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  • you've almost done: you have $max\left { N_\delta (U), N_\delta (V) \right }\leqslant N_\delta (U\cup V)\leqslant 2\cdot max\left { N_\delta (U), N_\delta (V) \right }$. Any troubles with $2$ as a multiplier? – serg_1 Jun 29 '17 at 23:53
  • Sorry, I do not know what your question means. Trouble with $2$? – user458984 Jul 02 '17 at 08:58
  • @serg_1, is it anything to do with the fact that 'taking' the dimension of either side removes the $2$? – user458984 Jul 02 '17 at 17:30

1 Answers1

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I think I have finally found the answer!

Let $N_{\delta}(S)$ the smallest number of closed balls of radius $\delta$ which cover a set $S$. Clearly $E,F\subset E\cup F$, so $\max\{\overline{\text{dim}}_{box}(E),\overline{\text{dim}}_{box}(F)\}\leq\overline{\text{dim}}_{box}(E\cup F)$.

Now, $N_{\delta}(E\cup F)\leq N_{\delta}(E)+N_{\delta}(F)\leq2\max\{N_{\delta}(E),N_{\delta}(F)\}$ and so $$\frac{\log(N_{\delta}(E\cup F))}{-\log(\delta)}\leq\max\left\{\frac{\log(N_{\delta}(E))}{-\log(\delta)},\frac{\log(N_{\delta}(F))}{-\log(\delta)}\right\}+\frac{\log(2)}{-\log(\delta)}.$$ Letting $\delta\to0$ gives that $\overline{\text{dim}}_{box}(E\cup F)\leq\max\{\overline{\text{dim}}_{box}(E),\overline{\text{dim}}_{box}(F)\}$. Thus $\overline{\text{dim}}_{box}(E\cup F)=\max\{\overline{\text{dim}}_{box}(E),\overline{\text{dim}}_{box}(F)\}$.

user458984
  • 325