Let $(X,d)$ be a metric space. Let $E\in X$ be a totally bounded set. For each $\delta > 0$, let $N(E,\delta)$ be the minimal number of balls of radius $\delta$ needed to cover $E$ (note that the center of these balls are in $X$ and are not required to be in $E$).
The function $\delta$ →$N(E,\delta)$ is called the metric entropy function of $E$. And we care about the behavior of $N(E,\delta)$ as $\delta$ → 0. Define
$dim(E)$ = lim sup ($\delta$ → 0) log $N(E,\delta)$ / log(1/$\delta$)
$\underline{dim(E)}$ = lim inf ($\delta$ → 0) log $N(E,\delta)$ / log(1/$\delta$)
The terms $dim(E)$ and $\underline{dim(E)}$ are called the upper and lower metric dimension of $E$. In case $dim(E)$ = $\underline{dim(E)}$ = $\alpha$, we call $\alpha$ the metric dimension of $E$.
$(b)$ Let $\alpha$ > $0$ be given. Assume that for each $\epsilon$ > $0$, there exist $\delta_0$ = $\delta_0(\epsilon)$ and $C = C(\epsilon) \geq 1$ such that
$\delta^{-\alpha+\epsilon}/C \leq N(E,\delta) \leq C\delta^{-\alpha-\epsilon}$ for all $\delta\in (0,\delta_0)$.
Show that $E$ has the metric dimension $\alpha$.
Sorry for the long problem...I attempted to start $(b)$ by using the given inequality and manipulating it until I had log $N(E,\delta)$ / log(1/$\delta$). However, I ended with something like:
$(-\alpha+\epsilon)(($log$\delta) - log($C$)$) / log(1/$\delta$) $\leq$ log $N(E,\delta)$ / log(1/$\delta$) $\leq$ $(-\alpha-\epsilon)($(log($\delta$) + log(C)) / log(1/$\delta$).
I am now unsure what to do. Is there some definition I can pull out of this that will help me out or is there some algebraic manipulation I need to do? Any help is very appreciated.
edit: I've cleared up my mistake. I now have:
$\alpha-\epsilon \leq lim$ (log $N(E,\delta)$ / log$(1/\delta)$) $\leq \alpha+\epsilon$.
However, I feel as though I may have messed something up, as we're looking for the limsup and liminf to be equal, so I should have taken the sup and inf of the middle before the limit. I can see that $\forall \epsilon > 0$ my $dim(E)$ terms are squeezed in between $\alpha$ +/- $\epsilon$ and thus equal, but I'm not sure how to move on with a technical proof of this. Any help is, again, very appreciated!