2
$\begingroup$

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!

$\endgroup$

1 Answer 1

10
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.