I have been asked to prove that for a bounded set $F\subset\mathbb{R}^n$,
$\dim_H F\le \underline{\dim}_B F \le \overline{\dim}_B F$
where $\dim_H F$ is the Hausdorff dimension, $\underline{\dim}_B F$ is the lower box counting dimension and $\overline{\dim}_B F$ is the upper box counting (Minkowski) dimension of $F$ (and I have done that).
I understand that $F$ is needed to be bounded, since otherwise we cannot really count the box-covering for $F$.
Now here I found a definition for unbounded sets $F \subset \mathbb{R}^n$:
$\underline{\dim}_B F=\sup_{E \subset F, \rm ~bounded} \underline{\dim}_B E$
$\overline{\dim}_B F=\sup_{E \subset F,\rm ~bounded} \overline{\dim}_B E$
But I failed to pove that $\dim_H F\le \underline{\dim}_B F \le \overline{\dim}_B F$ for any unbounded set $F \subset \mathbb{R}^n$.
Here's where I'm stuck:
$\underline{\dim}_B F = \sup_{E \subset F, \rm ~bounded} \underline{\dim}_B E \ge \sup_{E \subset F, \rm ~ bounded} {\dim}_H E \le \sup_{E \subset F} \dim_H E = \dim_H F$
Any help is appreciated!
Best Answer
As said in a comment, the proof of $\underline{\dim}_B F \ge \dim_H F$ for unbounded sets can proceed as $$\underline{\dim}_B F = \sup_{E \subset F, \rm ~bounded} \underline{\dim}_B E \ge \sup_{E \subset F, \rm ~ bounded} {\dim}_H E = \dim_H F$$ where the last step is based on the fact that the Hausdorff dimension is countably stable: $$ F = \bigcup_{n=1}^\infty (F\cap B(0,n)) \implies \dim_H F = \sup \dim_H(F\cap B(0,n)) $$
The inequality $\underline{\dim}_B F \le \overline{\dim}_B F$ for unbounded sets follows directly from the definition, since both sides are suprema taken over bounded sets, and the upper dimension is known to be larger for those.