We define $H_{\delta}^{d}(S):= \inf\Big\{\sum_{i=1}^{\infty}diam(U_{i})^{d}: S \subset \bigcup_{i=1}^{\infty}U_{i}, diam(U_{i}) < \delta\Big\}$
where $diam(U):= \sup\{d(x,y): x,y \in U\}$.
$$H^{d}(S):= \sup_{\delta >0}H_{\delta}^{d}(S)$$
Show that $H^{d}(S) = \lim_{\delta \rightarrow 0^{+}}(S)$.
Attempt:
As $\delta \rightarrow 0$ the infimum is taken over a decreasing set so is getting larger. I then need to take the sup of this, but am not sure how the result then follows?
For $t > 0$, set $F(t) = H_{t}^{d}(S)$. Note that $s > t$ implies $F(s) \leq F(t)$. Set $L = \sup_{t > 0}F(t)$. Let $L' < L$ be arbitrary. By definition of $\sup$, there is $s > 0$ such that $L' < F(s) \leq L$. Thus for all $t < s$, $L' < F(s) \leq F(t) \leq L$. Since $L'$ was arbitrary, this means $\lim_{t \searrow 0}F(t) = L$.