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.,

- If $N_\varepsilon^o (E) = N_\varepsilon^o (E')$ for all $\varepsilon>0$, then $E \cong E'$.
- 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!