Let $(E, d)$ be a metric space. For $\varepsilon>0$, we define two notions of $\varepsilon$-covering number as follows, i.e.,


* $N\_\varepsilon^o (E)$ is the smallest number of *open* balls whose radii are $\varepsilon$ that cover $E$.
* $N\_\varepsilon^c (E)$ is the smallest number of *closed* balls whose radii are $\varepsilon$ that cover $E$.


Then $N\_\varepsilon^o (E)$ is not necessarily equal to $N\_\varepsilon^c (E)$, for example, take $E = \{0, 1\}$ and $\varepsilon=1$. However, if $D$ is a dense subset of $E$, then $N\_\varepsilon^c (E) = N\_\varepsilon^c (D)$.


Let $(E, d)$ and $(E', d')$ be metric spaces. The spaces $E$ and $E'$ are said to be isometric (denoted by $E \cong E'$) if there is a bijective isometry between them.


I would like to ask if any of below statements is true, i.e.,



> 
> 1. If $N\_\varepsilon^o (E) = N\_\varepsilon^o (E')$ for all $\varepsilon>0$, then $E \cong E'$.
> 2. If $N\_\varepsilon^o (E) = N\_\varepsilon^o (E')$ and $N\_\varepsilon^c (E) = N\_\varepsilon^c (E')$ for all $\varepsilon>0$, then $E \cong E'$.
> 
> 
> 


Thank you so much for your elaboration!



Are two metric spaces isometric if they have the same $\varepsilon$-covering numbers for all $\varepsilon>0$?
No. For $0 < \delta \leq 2$ let $E\_\delta$ be the metric space
consisting of three points $A,B,C$ with
$d(A,B) = d(A,C) = 1$ and $d(B,C) = \delta$.
I claim that the $E\_\delta$ for $1 \leq \delta \leq 2$ all have
the same covering numbers for all radii $\varepsilon$.
Indeed in every such $E\_\delta$
all open $\varepsilon$-balls with $\varepsilon\leq 1$
and all closed $\varepsilon$-balls with $\varepsilon < 1$
contain only their centers, so it takes $3$ such balls to cover the space;
but $E\_\delta$ is covered by
a single open $\varepsilon$-ball centered at $A$
once $\varepsilon > 1$,
and also by a single closed $\varepsilon$-ball centered at $A$
once $\varepsilon \geq 1$. Clearly no two $E\_\delta$ with distinct $\delta$ are isometric.


