Let $(X, d)$ be a metric space and $A \subseteq X$ a totally bounded subset. We call $\Gamma = \{U_i\}_{i=1}^n$ an $\varepsilon$-covering of $A$, if $A \subseteq \bigcup_{i=1}^n U_i$ and $diam(U_i) \leq 2\varepsilon$ for all $i = 1,\dots,n$. For such a set we may define the quantity $\mathcal N_\varepsilon(A)$, which is the smallest possible size of an $\varepsilon$-covering. My question can now be phrased as follows:
Suppose $A \subseteq X$ is an infinite totally bounded set. Does there exist a finite subset $B \subseteq A$ such that $\mathcal N_\epsilon(A) = \mathcal N_\epsilon(B)$
If $\mathcal N_\epsilon(A) = 1$, this is trivial. It is also is easy to prove that this is true when $\mathcal N_\epsilon(A) = 2$. Simply note that $A$ itself can't be an $\varepsilon$-covering of $A$ since $\mathcal N_\epsilon(A) > 1$. Therefore $diam(A) > 2\varepsilon$, which implies that there exist two elements $x$ and $y$ in $A$ with $d(x, y) > 2\varepsilon$. Finally let $B = \{x, y\}$. Clearly $\mathcal N_\epsilon(B) = 2$. Unfortunately I couldn't succeed in finding a similar proof for higher values of $\mathcal N_\epsilon(A)$. Is this even true?
