Let $F\subseteq \mathbb{R}^n$ be non-empty and bounded.
We say that a countable family of set $\{$ $U_i$ $\}$ is a $\delta-cover$ for $F$ if $F\subseteq \bigcup_{I\in I}U_i$. Let $N_{\delta}(F)$ $=$ $min \{$ # of sets in any $\delta-cover$ of $F$ $\}$. Where $U_i$ has diameter at most $\delta$ (And greater than 0)
Then consider the unit interval $F=[0,1]$. For $0<\delta<1$, we get:
$\frac{1}{\delta} \leq N_{\delta}([0,1])<1+\frac{1}{\delta}$
My questions are:
(1) How do you prove the above inequalities?
(2) Why does $N_\delta(F)$ exist?
(1) Let $N=N_{\delta}([0, 1])$. If $N < \frac{1}{\delta}$ then there must be a $\delta -cover$ $\{U_i\}_{i=1}^N$ of $[0, 1]$. But since $[0, 1] \subset \cup U_i$, we have $1 \le \sum \delta = N\delta < 1$, absurd.
(2) Note that, since $F$ is bounded, the closure $\overline{F}$ of $F$ is bounded, so it is also compact. Now, the collection $ \{ B(x; \frac{\delta}{2}) \}_{x \in \overline{F} }$ is a $\delta -cover$ of $\overline{F}$, so it must have a finite subcover. Therefore $N_{\delta}(F)$ exists.