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!

  • 3
    $\begingroup$ BTW the assertion about dense subsets is not quite correct because $D$ may be missing the center of a necessary ball. For example, let $E$ be the interval $[-1,1]$, and $D$ be the dense subset $[-1,0) \cup (0,1]$. Then $N_1^c(E) = 1$ but $N_1^c(D) = 2$. $\endgroup$ Commented Mar 4, 2023 at 4:39
  • 1
    $\begingroup$ Stupid answer: there are only continuum many possible functions $\varepsilon\mapsto (N^o(E),N^c(E))$. There are more than continuum many metric spaces up to isometry. So the assignment can't be injective. So the question is more interesting by restricting to compact metric spaces. (Anyway Noam Elkies's answer addresses this more interesting setting, and even finite metrics paces.) $\endgroup$
    – YCor
    Commented Mar 4, 2023 at 10:12

1 Answer 1


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.

  • $\begingroup$ Thank you so much for your help! $\endgroup$
    – Akira
    Commented Mar 4, 2023 at 4:44

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