# Are two metric spaces isometric if they have the same $\varepsilon$-covering and $\varepsilon$-packing numbers for all $\varepsilon>0$?

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:

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!

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.