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!
Best Answer
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.