Let $(X, d)$ be a compact metric space.


* We say that $\{x\_1, \cdots, x\_n\} \subseteq X$ is an $\varepsilon$-**covering** of $X$ if for any $x \in X$, there exists $i \in \{1, \ldots, n\}$ such that $d(x, x\_i) \leq \varepsilon$. Let
$$
\operatorname{Cov} (X, \varepsilon) := \min \{n: \exists \varepsilon \text {-covering of } X \text { with size } n\}
$$
be the $\varepsilon$-covering number of $X$.
* We say that $\{x\_1, \cdots, x\_n\} \subseteq X$ is an $\varepsilon$-**packing** of $X$ if $d(x\_i, x\_j)>\varepsilon$ for all distinct $i, j$. Let
$$
\operatorname{Pack} (X, \varepsilon) := \max \{n: \exists \varepsilon \text {-packing of } A \text { with size } n\}
$$
be the packing number of $A$.


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


@Noam gave below example in his [answer](https://mathoverflow.net/questions/442063/are-two-metric-spaces-isometric-if-they-have-the-same-varepsilon-covering-num):



> 
> For $\delta \in (0, 2]$, let $E\_\delta$ be the metric space consisting of three points $A,B,C$ with $d(A,B) = d(A,C) = 1$ and $d(B,C) = \delta$. Then for all $\delta, \delta' \in [1, 2]$,
> $$
> \operatorname{Cov} (E\_\delta, \varepsilon) = \operatorname{Cov} (E\_{\delta'}, \varepsilon) =
> \begin{cases}
>  3 & \text{if} \quad \varepsilon < 1, \\
>  1 & \text{if} \quad \varepsilon \ge 1.
> \end{cases}
> $$
> 
> 
> 


This example shows that $\operatorname{Cov} (X, \varepsilon) = \operatorname{Cov} (X', \varepsilon)$ for all $\varepsilon>0$ does not necessarily imply $X \cong X'$. Back to @Noam example, it's clear that
$$
\operatorname{Pack} (E\_1, 1) = 0 \neq 2= \operatorname{Pack} (E\_{2}, 1).
$$


I would like to ask if below statement is true, i.e.,



> 
> If $\operatorname{Cov} (X, \varepsilon) = \operatorname{Cov} (X', \varepsilon)$ and $\operatorname{Pack} (X, \varepsilon) = \operatorname{Pack} (X', \varepsilon)$ for all $\varepsilon>0$, then $X \cong X'$.
> 
> 
> 


Thank you so much for your elaboration!



Are two metric spaces isometric if they have the same $\varepsilon$-covering and $\varepsilon$-packing numbers for all $\varepsilon>0$?
Certainly no. Consider metric spaces on $n$ points and all distance 1 and 2. There are $2^{n^2/2+o(n^2)}$ such spaces. But only polynomially many different covering and packing functions.


